Q
stringlengths 18
13.7k
| A
stringlengths 1
16.1k
| meta
dict |
|---|---|---|
If the faster I go, the slower time becomes for me, then does that mean the slower I go the faster time becomes? Disclaimer - I’m not a trained physicist. I just have an interest in the subject so this question is probably going to come across really confusing but I’m trying my hardest to explain.
So firstly, this is my assumption of how time dilation works. So say, you travel in a spaceship at almost the speed of light for one day. You might only spend 1 day at that speed, but because time passes slower for you than people on Earth, so then that 1 day is actually 10 years in Earth time and when you might come back, everyone is 10 years older while you’ve only aged a day.
So given this, how would I achieve the opposite? How could I slow myself down in comparison to everyone else on Earth that me spending 10 years at that speed is actually only 1 day for everyone else?
I ask because there really isn’t an absolute velocity that everyone is travelling at right? We are stationary right now but the Earth is moving around the Sun which is moving around the Milky Way and who knows where that’s orbiting. How can I go slower than how fast I’m travelling currently?
| No. If you are at rest with respect to the earth, you experience time the same way people on earth do.
If you are at motion with respect to the earth, any kind of motion, you experience the time dilation the way you sketched.
The relative motion between you and the earth cannot be "slower" than zero, which is the case when you are at rest with respect to it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/318458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Ray Optics: Reflection and Refraction Suppose you have extremely sensitive photographic film and you can expose it in an optical system for a very long time. Will you be able to record a virtual image on film?
| No. Remember that images look like objects because light comes from images in the same way that it comes from objects.
Consider a point source of light. (More complicated objects can be treated a set of point sources.) If a lens, mirror etc. produces a situation where rays of light pass through a point and then diverge in the same way that they diverge from the source then we say a real image is formed at that point. You can take some photographic film, put it there and one point on the film will be exposed - you have photographed the image.
A simple scenario that produces a virtual image is plane mirror. Rays of light diverge from the source and reflect off of the plane mirror is such a way that they diverge as if the were all coming from a point behind the mirror. We say this is where the virtual image is located. The conventional way to show this on a ray diagram is to extend the reflected rays behind with mirror as shown with a dotted line.
Unlike with a real image, if you were to take some photographic film and put it behind the mirror at the point where the rays of light seem to diverge from the film would not be exposed no matter how long you wait because there is actually no light there. It's probably in the middle of a brick wall or a dark medicine cabinet.
You can photograph a virtual image just by pointing your camera at the mirror and focussing at the right distance behind the mirror. Now, when you photograph a virtual image you are actually forming a real image on the film. The same could be said if you simply see a virtual image - your eye produces a real image of the virtual image on your retina.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/318580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why doesn't saturation current in the photoelectric effect depend on the frequency of light absorbed by the metal emitter? If current $I$ is given by $I = nAev$, where $n$ is the number of electrons per unit volume, $A$ is the area, $e$ is the charge of an electron and $v$ is the velocity of the electron, it must mean that the current increases with increase in velocity of the electron which increases with the frequency of light incident on the metal emitter. Why doesn't then saturation current increase with increase in frequency?
| Increasing intensity means increase in rays of light, so it will hit more numbers of electrons but doesn't give enough speed to the electrons.
Increasing frequency means increase in energy of photon. So electrons will go faster, and more regarding potential will be needed i.e. stopping potential increased.
Electric current is flow of number of charges per unit time, not speed of charges per unit time.
Hence, intensity increase number of electrons, frequency increase speed of electrons. So that saturation current don't depend on frequency of light.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/318668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Why does energy always wants to propagate? This might sound easy but its really a extremely fundamental question. The answer of this question has diverse implications. From thermodynamics to dark energy
The question is Why does energy always wants to propagate? In other words why does energy density always decrease per unit volume per unit time????
| It is called the Second Law of Thermodynamics : The entropy of any isolated system always increases.
To understand that fully, you need to understand the definition of Entropy, which is not trivial.
Simplified, Entropy is a measure of how unorganized / random a system is.
Ordered systems (like one place hot, another one cold) have a low Entropy ('unorganization'); left alone, the entropy will increase, resulting in higher 'unorganization' (all place about equally warm).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/318854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Are position and momentum spaces of a particle in classical mechanics related by Fourier transform? Please don't be so harsh on me and correct me if I'm wrong, since physics is not my major and I am not native English speaker.
The article position and momentum space (https://en.wikipedia.org/wiki/Position_and_momentum_space) on Wikipedia stated that in quantum mechanics, the position and momentum spaces are related by Fourier transform since they are Pontryagin dual.
I am not clear if this conjugation applies in quantum mechanics only or if it applies for position and momentum spaces of a particle in general, including classical, Lagrangian and Hamiltonian mechanics also? If yes, how is it formulated?
| In classical mechanics, position and momentum are independent variables on phase space and are not related by Fourier transformation at all.
Their Fourier relation in quantum mechanics arises from the Stone-von Neumann theorem, saying that all unitary representations of the canonical commutation relations $[x,p] = \mathrm{i}\hbar$ (which are, apart from the appearance of the $\hbar$, essentially classical, since $\{x,p\} = 1$ for the Poisson bracket on phase space) are unitarily equivalent to representing $x$ as the multiplication and $p$ as the differentiation operator on $L^2(\mathbb{R})$. Now, the Fourier transformation famously interchanges multiplication and differentiation and therefore relates position and momentum representations in quantum mechanics.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/318962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The Science of Tearing Paper-bag Handles My mother came back from a market which bags the products in paper-bags with handles, and asked me to move the bags from the trunk of the car to the house. Being the lazy human I am, I hung a few bags on each arm so I could cut the number of trips back and forth. As I was walking to the front door, the handles of a bag tore, the bag plummeting to the concrete ground. A glass jar of peppers had been smashed in to a zillion little pieces. As you might expect, my mother was furious. "You're so lazy! If you hadn't hung so many on your arm, the peppers and their jar would still be intact!"
I disagree, here's why...
Scenario Lazy
Scenario Peppers
Conclusion
Note that Bag A will have $N_A$ on it regardless of Bag B's existence. Sure, my arm had $N_A + N_B$ ($> N_A$), but it wasn't the thing that broke. So, I conclude, that the tearing of the bag was inevitable, and that the peppers' fates were written by someone other than me (e.g.: manufacturer didn't put enough glue to handle expected weight, cashier put more weight than permitted, etc).
Is my reasoning correct? Or am I missing something that proves that I'm guilty?
| Cort Ammon is correct. It might be added that when a human carries a load on a weak handle he minimizes by subtle countermovements the inertial forces which are exerted on the handle due to the walking movements in addition to the weight of the load. This is obviously more difficult to do for more than one load.
Thus you mother is right in scolding you!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/319040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "50",
"answer_count": 5,
"answer_id": 2
} |
Electric field from a varying magnetic field and a solid conductive disk I am given this question:
On the second to last line, it asked for an electric field. However, I'm wondering where does the electric field come from?
Can somebody give me an idea? I know that a varying magnetic field produces an electric field, but the way the question is worded doesn't suggest it's asking for the electric field produced by the varying magnetic field (since it seems to suggest that the magnitude of electric field varies with "distance r
| The electric field $\vec E$ appears in Faraday's law
$\displaystyle \oint_{\rm loop}\vec E \cdot d\vec l= - \dfrac {d\Phi_{\rm B,surface}}{dt}$
Note that
$\displaystyle \oint_{\rm loop}\vec E \cdot d\vec l$ is the induced emf.
So choose a loop which is a circle centred at the centre of the disc.
The bit about the magnetic flux due to the induced current being negligible is so that you do not need to worry about the self inductance of the disc because the induced current will oppose the change producing it - Lenz.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/319129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is $\delta Q – \delta W$ a state function? I know that internal energy, $Q+W$ is a state function.
But
$$dU=\delta Q-\delta W,$$
is the change in internal energy, where $dU$ is change in internal energy, $\delta Q$ is the heat supplied to the system and $\delta W$ is the work done by the system. Is this a state function too?
| The quantity $dU=\delta Q-\delta W$ is not a state function simply because it is not a function, it is a differential of a function. In that case a differential of the state function $U$.
The fact that there is a function $U$ such that the differential $\delta Q-\delta W$ is the differential of $U$ means that the differential is exact or $U$ is a state function, i.e., it has a definite value for every point in the space of configuration. In another words, the change in energy depends only on the initial and final states.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/319236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Given torque, what is the direction of motion? Torque is the cross product $\vec \tau = \vec r \times \vec F$, which means it is perpendicular to both $\vec r$ and $\vec F$.
Consider some essentially two-dimension problem, like a horizontal iron bar with one end fixed, affected by gravity. The direction of the torque is perpendicular to the bar and gravity.
I also see a vector formula like $\vec \tau = I \vec \alpha $. Since the moment of inertia $I$ is a positive scalar, it does not change the direction of vectors. Hence, this kind of formula implies that the angular acceleration is perpendicular to the force causing it.
In our example, the non-fixed end of the iron bar would start moving down, but this acceleration is perpendicular to torque. This implies that it is perpendicular to $\vec \alpha$, above.
This leaves me quite confused; given torque, how can I determine how an object starts moving? There should be a cross product involved, somewhere; otherwise, the perpendicularity do not work out correctly, I think.
| Suppose the force $\vec F$ and displacement $\vec r$ are in the xy plane.
Then you would expect the the angular acceleration would be anticlockwise looking down from the top just from the direction of the force.
The torque $\vec \tau = \vec r \times \vec F$ is in the $\hat z$ direction but so is the the direction of the angular acceleration $\vec\alpha$ if you use the right hand grip rule.
The linear acceleration is in the same direction as the five which is in the xy-plane and hence perpendicular to $\hat z$ and the angular acceleration $\vec \alpha = \alpha \hat z$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/319456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Why do air bubbles rise to surface of water? I was asked this question by my son and I want to give him the correct answer. So keep it simple please.
Here is what he asked. Why do air bubbles rise when they are released underwater?
| The answer you seek depends on the age of your son.
You could use Archimedes principle which states the upward force on a body in a fluid is equal to the weight of the fluid displaced by the body.
So the two parameters are the weight of the body and the weight of the fluid it displaces.
If the weight of the body (rock) is greater than the weight of the fluid it displaces (air) there is a net downward force on the body and it falls.
If the weight of the body (air bubble) is less than the weight of the fluid it displaces (water) there is a net upward force on the body and it rises.
You can in turn categorise the three conditions in terms of densities and state that an object with a higher (average) density than the density of the fluid will sink, an object with a lower (average) density than the density of the fluid will rise.
If the weight of the body (ship) is equal to the weight of the fluid it displaces (water) there is a no net force on the body and it floats.
If you son is happy with falling objects in air then you can tell your son that what he is seeing is water falling in air.
The falling of an object in air can be thought of the potential energy (energy of a body due to its position) of the air and the object as being reduced.
When the object goes down the air it displaces goes up.
The object falling loses more potential energy that the air which it displaces gains in rising.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/319834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 2
} |
Actual classical time evolution in a closed Cavity I'm stuck with the question how the actual time evolution of a given classical electric field in a closed cavity will look like.
The initial conditions are a given function $u(x)$ in a one dimensional region between the points $0$ and $L$. At those points there are mirrors, that are said to reflect 100 % of the intensity. The function is to represent an electrical field.
I now want to know what equations will govern the time evolution. Surely, inside the cavity, the time evolution is given by a one dimensional wave equation $\partial_x^2 u - \partial_t^2 \frac{1}{c^2} u = 0$.
But how to account for the mirrors?
I'm stuck with this question because usualy reflection at a dielectric mirror is modeled by two regions with different dielectric constant. There, you solve the wave equations for both regions, apply proper boundary conditions for the boundary layer, and are finished. But I can't model a 100 % reflecting mirror as a dielectric mirror, so I'm clueless on this.
| It's a perfect electrical conductor (PEC) at the endpoints. Assuming $u(x)$ represents a component of the electric field perpendicular to the $x$-axis, the boundary conditions are $u(0)=u(L)=0$, which implies that the tangential component of the electric field is zero.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/319993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
what is meant by solar spectrum? How can we get absorption lines in solar spectrum instead of emissions lines at such high temperature?
Is solar spectrum contains emission lines?
| The solar spectrum itself originates as a close-to-perfect blackbody emission for an object at about 5800K. That is what produces the nice curve you see. The absorption lines in the spectrum are not always something that comes from the Sun itself but, in most common situations, the absorption lines you will hear referred to are from the atmosphere of Earth.
The multitudinous gases in Earth's atmosphere each have their own spectra of wavelengths that they will either absorb to become excited or emit when they are excited. As the light from the Sun passes through the atmosphere, it encounters these gases and some of the light is absorbed. Gases like Nitrogen, which makes up a majority of the atmosphere, will contribute more to the absorption spectrum; it has a fairly broad spectrum itself, so it tends to decrease the relative intensity across most wavelengths. More active gases like oxygen, water, and ozone have a more noticeable impact on the absorption spectrum; you'll see deeper absorption bands at some of their more dominant wavelengths.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/320126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is the existence of a photon relative? If an observer passes an electron, in such a way that the observer is accelerating, the observer would see photons because accelerating charges induce electromagnetic waves.
But from point of view of the electron or an inertial observer there is no magnetic field nor an acceleration which could 'produce' an electromagnetic wave.
So for the first observer there exists a photon but not for the second observer. How is this possible, are photons relative?
| To start with, accelerated frames are not inertial frames.
Secondly electrons and photons are elementary particles and are described in a quantum mechanical framework, where everything is particles and interactions of particles.
The blanket term "observer" has to be defined in terms of interactions, in order to be able to write down the mathematics of the system.
For an electron to be accelerated , it has to interact with a field and lose energy emitting a photon, as an example, in this diagram it is the field of another electron with which the upper electron interacts and radiates a photon:
But from point of view of the electron
Here the lower electron may be considered to start in its rest frame.
The whole diagram is Lorenz invariant, and the only thing that will be changing if one goes to the center of mass of the other electron , is how the energy is supplied to the photons.
In general photons do not disappear in Lorenz transformations, they may increase or decrease in frequency, decrease to the point of huge wavelengths which means that the light beam emergent from zillions such photons may have a wavelength the size of the universe and in the limit represent a static field.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/320629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Why is the mol a fundamental physical quantity? I am starting to study physics in detail and as I read about physical quantities, I was puzzled why mol (amount of substance) is taken as a physical quantity.
A physical quantity is any quantity which we can measure and has a unit associated with it. But a mol represents the amount of substance by telling the number of particles (atoms, molecules, ions, etc.) present. So it is a pure number and numbers are dimensionless. So mol should not be considered a physical quantity.
Also, fundamental physical quantities should be independent of each other. I am wondering whether mass and mol are independent. This is so as they surely affect each other as we can evidently see while calculating the number of moles and using the mass of that sample for calculation.
So how is the mol a fundamental physical quantity and independent of mass?
| The mole definitely isn't a fundamental physical quantity. It's just a shorthand for Avogadro's number, to make really big numbers more tractable. It's purely there for convenience, there's nothing fundamentally physically significant about it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/320784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 4,
"answer_id": 1
} |
Adiabatic free expansion of real (Van der Waal's model) gas below/at/above inversion temperature For an adiabatic free expansion, $W = 0$ and $Q=0$. Therefore, by the first law of thermodynamics, $\Delta U = Q -W = 0$.
For a Van der Waals model of a real gas, $\Delta U = n C_v \Delta T - a\,n^2 \left(\frac{1}{V_2}-\frac{1}{V_1} \right)$. This means that $n C_v \Delta T - a\,n^2 \left(\frac{1}{V_2} -\frac{1}{V_1} \right) = 0$, or $n C_v \Delta T = a\,n^2 \left(\frac{1}{V_2} -\frac{1}{V_1} \right)$. Now, since the gas is expanding, $V_2 > V_1$. So, $a \, n^2 \left(\frac{1}{V_2}-\frac{1}{V_1} \right) < 0$. Therefore, $n C_v \Delta T<0$. This means that the temperature of the gas decreases in an adiabatic free expansion.
Will the temperature always decrease? I mean, what if the gas is above its inversion temperature or at it? Does inversion temperature play no role in adiabatic free expansion?
| this is more a comment on your answer: in any free expansion the pressures in each side are not kept constant: the pressure decreases in one side and increases in the other side. The correct statement is: "It is only when the gas flows between two regions of different pressures (each kept constant) that the total entalpy $H_1 +H_2$ is conserved. For a real gas, it follows that the sign of $\Delta T$ depends on T".
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/321055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Why most of the minority carriers make it across the base as the base width is small? I am studying about BJT from an online lecture note. I made a question in the image below. The reference in the image is from the two lectures, lecture 18 and lecture 19.
Could anyone help me with the question in the image?
| When the hole concentration at the $x=0$ boundary is held constant at $n_0$ and holes are allowed to diffuse in the $0 < x < W$ region, they each travel a certain distance before being annihilated through recombination. The quantity $L_B$ in your notation represents the hole diffusion length, or the average distance traveled by diffusing holes during their average lifetime $\tau$.
When the diffusion length is large compared to the base width, $L_B >> W$, it means that most holes travel distances larger than $W$ before recombination, or else, that very few holes recombine while traveling a distance $W$. So the average concentration of holes surviving at the distal boundary of the base is comparable to their original concentration at the opposite boundary. And the lower the width $W$, the less the fraction of holes annihilated within the base.
If you prefer a quantitive treatment, the hole concentration at time $t$ and distance $x$ from the boundary is given by Ficks's diffusion laws as
$$
n(x, t) = n_0 \,\text{erfc}\Big(\frac{x}{L_B}\sqrt{\frac{\tau}{t}}\Big)
$$
where erfc(x) is known as the complementary error function. Do not worry about its exact expression. What is important is that for distances $x$ much smaller than the diffussion length $L_B$, $x/L_B << 1$, its Taylor expansion is just $\text{erfc}(x) = 1 - 2 x/\sqrt{\pi}$. Which means that for small base widths, such that $x/L_B < W/ L_B << 1$, and for time scales on the order of the hole lifetime $t \sim \tau$, the hole concentration varies linearly with the distance $x$,
$$
n(x) \approx n_0 \Big(1 - 2 \frac{x}{L_B}\Big)
$$
So at the other boundary of the base, for $x = W$, the concentration reads $n(W) \approx n_0 \Big(1 - 2 \frac{W}{L_B}\Big)$, which shows that the lower the width, the higher the surviving concentration.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/321379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the change in net internal energy? Suppose a person is hanging on a tree and he falls down to the ground. I consider the person, and the tree as a system and the Earth as surrounding. Applying the energy principle,
$$\Delta K_{trans} + \Delta K_{rel} + \Delta E_{int} = W_{net, ext}$$
What is the work done by external force? Should it be $Mgh$, where $h$ is the height the person is above the ground?
| If you are considering the person (and the tree) as the system then as the person falls the external force acting on the person is the gravitational attraction of the Earth on the person $mg$.
If the person (centre of mass) falls a distance $h$ then the work done by the external force on the person is $mgh$ and so the answer to you question is "Yes".
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/321602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Do selection rules apply only for radiative transitions or any other type of transition as well? Selection rules mentioned in all books on atomic physics state that for example orbital quantum number must change by 1 when electron does transition between energy states of an atom. Does this apply only for radiative transitions (that is transitions that involve photon emission/absorption) or does it apply generally for any other type of transitions (eg. collisions)?
What I am basically trying to clear up in my head is: when I see "selection rules" in a textbook, should I immediately associate that with radiative transitions only or should I also think about other types of transitions?
| As far as I can tell, "selection rules" is terminology applicable only to radiative transitions. The rules originate in the multiple expansion of the potential.
Dipole transitions change angular momentum by $0$ or $\hbar$ because the dipole moment is the $L=1$ multipole. Quadrupole transitions change angular momentum by up to $2\hbar$ because the quadrupole moment is the $L=2$ multipole.
In atoms dipole transitions are dominant but other transitions are also possible. In nuclear physics many transitions are quadripolar in nature, especially in rotational bands. Thus it is not so much that "angular momentum much change by $1$" in atomic transitions but rather that the quadrupole transitions are much much less intense than dipole transitions in atoms. Higher multipole beyond $L=1$ will be seen in atoms where some dipole transitions are forbidden.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/321762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Does Earth experience any significant, measurable time dilation at perihelion? Is there any measurable time dilation when Earth reaches perihelion? Can we measure such a phenomena relative to the motion of the outer planets?
| We can calculate the time dilation for an object moving in the Sun's gravitational field using the Schwarzschild metric. Strictly speaking this is an approximation since the Sun is rotating and not spherical, but it will give us a pretty good answer.
The Schwarzschild metric is (writing it in terms of the proper time):
$$ c^2d\tau^2 = \left( 1 - \frac{r_s}{r}\right) c^2dt^2 - \frac{dr^2}{1 - r_s/r} - r^2d\theta^2 - r^2\sin^2\theta d\phi^2 \tag{1} $$
where $r_s$ is the Schwarzschild radius of the Sun:
$$ r_s = \frac{2GM_{Sun}}{c^2} $$
At the perihelion and aphelion the motion is tangential so the radial velocity is zero and therefore $dr=0$. Also we will arrange our coordinates so that all motion is in the equatorial plane so $\theta=\pi/2$ and $d\theta=0$. Substituting these values into equation (1) we find the metric simplifies considerably to:
$$ c^2d\tau^2 = \left( 1 - \frac{r_s}{r}\right) c^2dt^2 - r^2 d\phi^2 \tag{2} $$
If the tangential velocity is $v$ then the angle $d\phi$ moved in a time $dt$ is just:
$$ d\phi = \omega dt = \frac{v}{r}dt $$
and we substitute this into equation (2) to get:
$$ c^2d\tau^2 = \left( 1 - \frac{r_s}{r}\right) c^2dt^2 - r^2 \left(\frac{v}{r}\right)^2dt^2 $$
which we rearrange to give us the equation for the time dilation:
$$ \frac{d\tau}{dt} = \sqrt{1 - \frac{r_s}{r} - \frac{v^2}{c^2}} \tag{3} $$
According to NASA's fact sheet the values of $v$ and $r$ at perihelion and aphelion are:
$$\begin{align}
r_p &= 1.4709 \times 10^{11} \,m \\
v_p &= 30290 \,m/s \\
r_a &= 1.5210 \times 10^{11} \,m\\
v_a &= 29190 \,m/s
\end{align}$$
And the Schwarzschild radius of the Sun is $r_s \approx 2953$ m. Putting these figures into our equation (3) gives us:
$$\begin{align}
d\tau/dt \,\text{perihelion} &= 0.99999998486 \\
d\tau/dt \,\text{aphelion} &= 0.99999998555
\end{align}$$
we can make these numbers a bit more digestible by expressing them as time lost per day e.g. how many seconds a day do clocks on the Earth run slower as a result of the time dilation. If we do this we find:
$$\begin{align}
\text{perihelion loss} &= 1.308 \,\text{ms/day} \\
\text{aphelion loss} &= 1.248 \,\text{ms/day}
\end{align}$$
And the difference between the two is about $60\mu$s per day. So clocks run about $60\mu$s per day more slowly at perihelion than they do at aphelion.
This is easily measurable in principle since atomic clocks have the accuracy to measure shifts this small. However there are practical difficulties. The time dilation is measured relative to a stationary observer outside the Sun's gravitational influence, and we can't easily put an atomic clock somewhere outside the orbit of Pluto to make the comparison. We could put a satellite in an exactly circular orbit at the average orbital radius of the Earth, and in that case our clocks would run about about $30\mu$s per day faster than the satellite clock at aphelion and the same amount slower at perihelion.
A quick footnote:
Count Iblis points out that pulsars make an excellent clock outside the gravitational influence of the Sun, and we can measure pulsar frequencies with enough accuracy to detect the $60\mu$s per day change between perihelion and aphelion. If anyone has a reference for this feel free to edit it into this answer.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/321910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Why does the proton have a parity $P=1$ and an anti-symmetric wave function? The overall parity of a proton is 1 because the parity of a quark is 1. How does this go together with the proton's wave function being anti-symmetric?
Is the reason for the proton's wave function's anti-symmetry the fact that in $SU(3)_C$ you consider the $u,d,s$ quark flavors to be identical for the strong interaction?
| The spatial parity of the proton (and the quark) is +1 by convention. Fermion and antifermion have opposite spatial parity. So the antiproton has spatial parity -1, by the same convention. But it could equally well be the other way about, provided you're consistent.
The proton wave function is antisymmetric under permutation of labels, as @CosmasZachos says. So is the antiproton wave function. That's not a convention, that's the spin statistics theorem.
There is no direct link (or conflict) between spatial parity and permutation symmetry here.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/322042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Is the speed of light the only speed that exists? Well, it seems to me that if I move faster in space I move slower in the dimension of time which is orthogonal to the dimension of space.
All speeds are then equal. Is this statement correct?
| The norm of the four velocity is always $c$, but of course not all four velocities are equal because they can point in different directions and vectors that point in different directions are not identical.
But to claim the word speed means the norm of the four-velocity seems unjustified. By speed we normally mean (the magnitude of) the coordinate velocity, and the coordinate velocity is the derivative of position in our coordinate system with time in our coordinate system. This can of course have any magnitude from zero up to the speed of light.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/322145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
Is the time period of oscillations in the underground tunnel constant for all lengths? I know that we can very easily derive an expression for the time period of oscillations in an underground tunnel due to gravity. It does not depend on length or inclination of the tunnel, and is 5050 seconds.
But if that's true it should be true even for tunnels that are very small in length. What I can not understand is why we don't just start falling and oscillating.
| Yes, it is the same for all lengths of tunnel. This is really no more surprising than that the period of a mass on an ideal spring is independent of its amplitude. A larger amplitude causes the mass to move more rapidly, covering a greater distance in the same time.
The formula derived by Naveen Balaji in Period of oscillation through a hole in the earth is independent of amplitude :
$$T=2\pi\sqrt{\frac{R^3}{GM}}$$
Naveen also shows in his derivation that the period is independent of the length of the tunnel through the Earth.
It is not clear what you are getting at when you ask "Why don't we just start falling and oscillating?" An obvious answer is : Because there are no tunnels available to fall through.
You could dig a very shallow tunnel close to the surface of the Earth, perhaps only a few km long. Then you would oscillate through that tunnel with a period of 42 minutes, the same as when it passes through the centre of the Earth, provided that the friction from the sides of the tunnel is negligible. You would still be attracted toward the centre of the Earth; this force causes your acceleration but it also leads to friction. At such a shallow angle of descent the normal reaction from the sides of the tunnel (hence also friction) is much greater than the accelerating force.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/322253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Shooting a bullet into an MRI What happens, if one shoots a bullet into an MRI system? Is there any chance to fully stop the bullet?
Boundary condifitions:
Consider an MRI machine with a reasonalby strong magnetic field (~3T) and tunable gradient coils. Furthermore, suppose we have a standard gun and bullet to shoot into the field of the machine.
Can we find a trajectory, that stops the bullet from leaving the MRI system?
For example, stopping it, reversing it, leading it on a circular path inside the machine etc. etc.?
Update: The stationary field inside the MRI machine seems to be a bad candidate, as it follows the centerline in a parallel fashion. So, is there any chance of eddy currents inside the bullet that help slowing it down? I found the x-y-gradient fields in a typical MRI system, the peak gradient is about 35-45 mT/m.The slew rate, which is defined as peak gradient over rise time can reach up to 200 T/m/s.
| New Version (as I had my doubts on the orders of magnetude...and it turned out that I made a mistake)
To simplify things I assume:
*
*velocity $v=1000\; \mathrm m /\mathrm s$
*perfectly conducting bullet
*density of $\rho=10000\; \mathrm{kg}/\mathrm{m}^3$
*A $B_\mathrm{max}=5\;\mathrm T$ and constant gradient of $B'=5\;\mathrm{T}/\mathrm{m}$
If the bullet is a considered a single winding solenoid field inside is
$$B=\mu_0 \lambda(x),$$
where $\lambda$ is the surface charge density. The dipole moment is
$$P=\mu_0 A I=\mu_0 V \lambda,$$
with current $I=l\lambda$,length $l$, area $A$, and Volume $V$. In this geometry the force on the dipole is
$$F(x)=\frac{P}{\mu_0} \frac{\mathrm d B}{\mathrm d x}=\frac{\mu_0 V \lambda}{\mu_0} B'=\frac{V B(x)}{\mu_0} B'$$
and as $B(x)=xB'$,i.e. constant gradient
$$F(x)=V\frac{(B')^2}{\mu_0}x$$
the integrated energy then is
$$E(x)=V\frac{(B')^2}{\mu_0}\frac{x^2}{2}$$
and we should have
$$V\frac{(B')^2}{\mu_0}\frac{x^2}{2}=\frac 1 2 m v^2 = \frac{\rho V}{2}v^2$$
hence
$$\frac{(B')^2}{\mu_0}x^2=\rho v^2$$
i.e.
$$x=\frac{v}{B'}\sqrt{\mu_0 \rho},$$
which in our case is $22\;\mathrm m$ and not one.
Again looking at energy we could try to figure out the speed reduction
$$v=\sqrt{v_0^2-\frac{(B')^2}{\rho \mu_0}x^2}$$
which is insignificant. Moreover I guess that there are no reasonable $B'$ and $\rho$ to make it work. But you could probably stop a $150\;\mathrm m/\mathrm s$ aluminum arrow shot by an archer.
I hope that I have my math correct now.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/322393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Resolution of acceleration due to gravity along initial direction of projectile While solving problems regarding projectiles cant we resolve the acceleration due to gravity along (or against) the direction of initial velocity and use this in the equations of motion to derive the answer?
| You could do that, but you would need to write equations of motion using coordinates which are not horizontal and vertical coordinates. You would have a zero initial velocity in the coordinate perpendicular to the initial velocity, but you would have accelerations in both coordinates. The coordinate directions would be tilted compared to gravity.
The final answers would be in that tilted coordinate system. You could then rotate them back to a horizontal system. But that's adding a lot of extra work when it's not necessary.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/322676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Sound wave equation in 3D closed Box We have the sound wave equation
$\Delta p - \frac{1}{v^2} \frac{d^2}{dt^2} p = 0$
in a closed Box. So we got Dirichlet boundary conditions and I can combine the solution for the 1D case to a 3D solution. In my quantum physics course we multiplied the 1D solution but in this lecture the professor actually added the solution.
$ p(x, y, z, t) = (\cos(\frac{n\pi}{L}x)+ \cos(\frac{k\pi}{L}y) +\cos(\frac{l\pi}{L}z)) \cdot \exp(i\omega t)$
My exam is tomorrow and I am confused why this is the solution.
| Either you are dismembering what was taught or your professor. I don't want to speculate who made the mistake, but the solution by your quantum physics professor is correct. The separation of variables method assumes that the solution is of the form:
$$ p(x,y,z,t) = X(x).Y(y).Z(z)e^{i\omega t}$$
So all your cosines need to be multiplied. I hope this helps. Good luck!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/323079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What are the differences between light from the Sun and light from a laser? The sun and the laser both give off photons, but there is clearly going to be a difference(s) in the light I receive from both sources. What are those differences (energy, intensity, orientation of photons, phases etc... literally any difference you can think of)? Imagine that I put a universal measuring device (something that can detect anything imaginable) in front of light from sun and also one in front of the laser. What are the differences?
The only one I know of is that the energy and therefore the frequency of the photons coming from the sun will all be higher than the ones from the laser (standard laser, nothing too mega and nothing out of sci-fi).
| No difference between "single" photons, however for a laser you can think of the photons as in step with each other (technical term would be "in phase"). Light from the laser is generated in a special ordered way of (more or less) one wavelength. The light from the sun is generated in a higgledy-piggledy way consisting of a whole spectrum of frequencies.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/323504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Noether's Theorem: form of infinitesimal transformation Noether's theorem states that if the functional $J$ is an extremal and invariant under infinitesimal transformation,
$$ t' = t+ \epsilon \tau + ...,\tag{1}$$
$$ q^{\mu'} = q^{\mu} + \epsilon \zeta^{\mu} +... .\tag{2}$$
Then the following conservation law holds:
$$p_\mu \zeta^{\mu} -H \tau - F = const.\tag{3}$$
What I am curious about is the forms of $\tau$ and $\zeta$. We can find strange conservation laws for any system so long as we find the right $\tau$ and $\zeta$. For instance, a free particle will admit to energy conservation when $ \tau = 1$ and $\zeta =0$. Also, a strange conservation for a damped oscillator can be found if $\tau =1$ and $\zeta = \frac{-bx}{2m}$.
My question is: What does the form taken by $\tau$ and $\zeta$ tell us about the system (or the laws of nature)? It seems to truly matter. Gallilean transformations ($\tau =0$, $ \zeta =t$) gives us regular old momentum conservation... Lorentz transformations do the same for relativity... But what do $\tau$ and $\zeta$ mean? What do they say? Why are they what they are?
| As a partial answer let us mention that (i) in a Hamiltonian formulation and (ii) for purely vertical infinitesimal quasi-symmetry transformations
$$\delta z^I~=~\epsilon \zeta^Iz^I,\tag{A}$$
meaning that OP's $\tau=0$ is assumed zero,
$$ \delta t~=~\epsilon\tau~=~0, \tag{B}$$
then the vertical generator
$$\zeta^I~=~\{z^I,Q\}\tag{C}$$
happens to be the Poisson bracket between the corresponding phase space coordinate $z^I$ and the conserved Noether charge $Q$. See e.g. this Phys.SE post for details.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/323923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Motion of bodies connected by springs Two blocks $A$ and $B$ of masses $m$ are connected by a spring of length $L$ and spring constant $k$. They rest on the frictionless floor. Another body of mass m moving with velocity $v$ collides elastically with $A$. The spring compresses and at maximum compression velocity of both $A$ and $B$ are $v/2$ each. Why did these bodies get equal velocities?
| Welcome to StackExchange!
To answer your question, apply conservation of momentum. I will assume the collision gave the block A some velocity v. Now the initial momentum of the system would be p=mv.
Suppose at some time, lets say the block A now has a speed u. By momentum conservation, you would get speed of block B to me v-u.
At maximum compression of spring, we will have spring to hold maximum potential energy. This means the kinetic energy of block system will be minimum. Which means
Kinetic Energy=0.5*m*(u)^2+0.5*m*(v-u)^2.
If you use derivative methods to minimize the kinetic energy, you will get u=v/2.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/324039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Books on undergraduate experimental physics? In undergraduate exams, problems sets etc, their are often questions that take the form:
Describe an experiment in which you can measure $x$, $y$ and $z$.
Does anyone known of any resources, covering a wide range of topics, where one can look up e.g. 'Hall coefficient' and find experimental details of how this could be measured (at an undergrad level)?
| A classic text recommended in many advanced undergraduate physics lab courses is Experiments in Modern Physics by Melissinos. It covers everything from alpha particle ionization to the Zeeman effect, including the Hall coefficient, and is most suitable for upper-level labs after first year. The 2003 Second Edition by Adrian C. Melissinos and Jim Napolitano is widely available new in both electronic and paper versions, e.g. from Amazon. If you search around you will also find the original 1966 version which may be better for some topics. For example, if you are interested in the Hall effect in semiconductors, you would want to look the 1966 edition, since 2003 edition Hall effect experiment is on bismuth, a semimetal.
Another well known text is The Art of Experimental Physics (1991) by Daryl W. Preston and Eric R. Dietz, also aimed at upper-level labs. This book has more background and less specific experimental detail, but this may be preferable if you only want a sketch of how to measure something, as would typically be expected for the answer to the exam question you give.
There are actually not a lot of books on how to do specific undergraduate experiments, but there is much information available from individual university lab courses. (Melissino's original book was based on material for an experimental physics lab course given to Junior/Senior students at the University of Rochester.) Students working on an experiment often find that looking at descriptions from other universities can be helpful. For example, you can find descriptions of how to make Hall effect measurements (and many others) by looking at lists of experiments from UC Berkeley, Colorado, Toronto (advanced, intermediate), and many other universities. Simply searching for Hall coefficient undergraduate lab will turn up many other sets of instructions. The various professors who author the experiment instructions have different viewpoints and backgrounds, so materials from different universities often complement each other.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/324115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Solar neutrino momentum flux through Earth According to wikipedia, the sun emits enough neutrinos that the number passing through a square meter of area oriented perpendicular to the sun at Earth distance is around $6.5 \times 10^{14}$ per second.
What is the momentum flux of these neutrinos? If you counted up the momentum of all the solar neutrinos passing through that square meter of area at $1 \cdot au$ from the sun, what's the order of magnitude? For example, if you made an impossible sci-fi material that was opaque to neutrinos, could the "lift" generated by the neutrino "wind" through a $1\cdot m^2$ "sail" overcome Earth gravity?
| There are two approaches to the calculation.
One is to know that the vast majority of solar neutrinos come from a p+p reaction that yields neutrinos with a maximum of 0.42 MeV. The spectrum yields an average neutrino energy of around 0.3 MeV.
As the neutrinos are highly relativistic, then $p = E/c$. The momentum flux is therefore $1.04\times 10^{-7}$ N/m$^2$.
The other approach is to know that 2.3% of the Sun's luminosity emerges in the form of neutrinos. So the neutrino "solar constant" at the Earth is $0.023 \times 1.37\times 10^3$ W/m$^2$. To get the momentum flux, one divides by $c$ to get $1.05\times 10^{-7}$ N/m$^2$.
Can this overcome gravity? Your perfectly absorbing neutrino sail (which we assume is transparent to electromagnetic radiation, which has 1/0.023 times more momentum flux?) would have to have a mass of less than $10^{-8}$ kg/m$^2$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/325339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Is the Charge on proton is $3.2\times10^{-19}$ greater than that of charge on electron? As the charge on electron is $$e^-=-1.6\times10^{-19}C$$ and charge on proton is $$p^+=+1.6\times10^{-19}C$$
Does this mean that the charge on electron is $3.2\times10^{-19}C$ less than that of charge on proton?
| The sign has nothing to do with the charge itself. It is just a convention. Just because we write an electron's charge as $-1.6\times10^{-19}$, it does not make the electron negative or its charge to be less than that of a proton.
We could have called charge on a proton to be $-1.6\times 10^{-19} C$ and the charge on an electron to be $1.6\times 10^{-19} C$. It does not make any difference in physics. Everything will continue to work the way it should.
Saying that a proton has $3.2\times 10^{-19}C$ of charge is not wrong. However, it is more useful to refer to the magnitudes while comparing charges. I would personally prefer to say negative $5\mu C$ is greater than positive $3\mu C$ as it conveys something useful.
However, when you are using those values in mathematical equations, you need to take their sign into account. In this case, $(p^+ - e^-)$ does give you $3.2\times10^{-19}C$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/325447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Is there a Birkhoff-like theorem for stationary axisymmetric metrics? I know about the theorem by Robinson and Carter about the uniqueness of the Kerr metric in the case of stationary axisymmetric (SA) black holes. Are there any uniqueness theorems like Birkhoff's theorem for stationary axisymmetric metrics?
| There is nothing as strong as the Birkhoff theorem in the case of stationarity and axisymmetry. Note that the Birkhoff theorem in its strongest form can be stated as:
"If even a piece of the space-time is spherically symmetric and a vacuum, then it is a piece of the Schwarzschild space-time."
There are various theorems about axially symmetric and stationary space-times which say that we reduce to Kerr under various conditions such as e.g. regularity outside of and on the horizon, connectedness of the horizon, asymptotic flatness, and global vacuum. I.e., with some physicist-faith leeway, we can construct arguments that as a reasonable, globally strictly vacuum and asymptotically flat space-time, the Kerr space-time is unique.
However, we do not know any reasonable matter solution which would match the Kerr space-time as an "exterior" solution. Geroch has even conjectured that there is no "interior" solution to the Kerr metric, i.e. a non-black-hole star which would reduce to Kerr outside its surface. (I personally believe that Geroch's conjecture is true.)
In practice, when we construct solutions of neutron stars, we find that they always differ from the Kerr case in the quadrupole and higher mass-multipole momenta of the space-time and we have to match them to approximately constructed non-Kerr metrics. I.e., when we are not globally vacuous, the uniqueness of Kerr is broken for sure.
There is even a rather well-known class of solutions derived by Manko & Novikov in 1992 which allow to set all of the infinite asymptotic values of mass-multipole momenta to arbitrary values. This, however, comes at the cost of weird singularities at the horizon and/or singular matter sources outside it. If you want a simpler playground for gaining some intuition on this, you can check out the axisymmetric and static Weyl metrics where you can plug in any Newtonian (axisymmetric and stationary) gravitational potential to generate a new space-time deviating from Kerr in its vacuum regions.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/325611",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Ratio of Forces Problem. An object is dropped from a building. After travelling $\text{10 m}$ freely, a force $F_1$ is applied on the object for $1$ second and the object comes to rest. If one would apply another force $F_2$ for $2$ seconds instead of $F_1,$ the body would come to rest. Find $F_1:F_2.$
Progress. We know that $v^2=u^2+2as.$
After travelling $10$ meters, the velocity of the object would be $\sqrt{0^2+2\times 9.8\times 10}=14\text{ m/s.}$
Now, moment of force = change of momentum.
This gives $F_1\times 1=F_2\times 2=\text{Mass of the object}\times 14\implies F_1:F_2=2:1.$
But the given answer says $F_1:F_2=17:12.$
Please point out my error.
Edit. Mr. Basu turned out to be right!
Actually, moment of net force = change in momentum.
This means $(F_1-mg)\times 1=(F_2-mg)\times 2=14m.$
$\implies F_1=m(g+14)~;~F_2=m(g+7).$
$\implies F_1:F_2=(g+14):(g+7)=17:12.$
| I think that this question is not homework question as @Eeshan has posted his procedure for solving the question. So this question deserves an explanation.
My answer:
The velocity of the particle after the fall of $10m$ is $10\sqrt{2}\frac{m}{s}$.
Let Force $F_1$ and $F_2$ produce an acceleration of $a_1$ and $a_2$ respectively.
Using the formula, $v=u+at$, and the fact that final velocity '$v$' is $0$ for both the cases, we get,
$at=-u$
For $F_1$,
$(a_1-10)1=10\sqrt{2}$
Similarly for $F_2$,
$(a_2-10)2=10\sqrt{2}$
On dividing the values of $a_1$ and $a_2$ , you will get,
$\frac{F_1}{F_2}=\frac{ma_1}{ma_2}=\frac{\sqrt{2}}{1}$
But, $\frac{\sqrt{2}}{1}=\frac{17}{12}\approx1.41$,
$\frac{F_1}{F_2}=\frac{17}{12}$.
So, you missed to include the effect of gravitation while solving your question.
And also, your edit is correct as $\frac{24}{17}\approx1.41$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/325721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Van der Waals forces & Partial charges We're told that charge is quantized. That the charge any body carries has to be an integral multiple of the fundamental unit of charge: $e=1.602×10^{-19}C$ such that $q=ne$ where $n$ is an integer.
But my book says (while explaining van der Waals forces) that the partial charges on a molecule ($\delta$) are always less than the fundamental unit of charge $e$.
How can this be? Isn't this against the quantization condition for charges?
Edit
| Imagine that you live in a world where the fundamental particles are cones with mass 1. But at some moment you want to know the weight of the halfs (the tip and the basis). Nobody forbids to calculate this masses. And this could be useful to calculate vibrations and rotations of such fundamental particles.
So it makes sense to calculate the charge distribution in molecules and it's not surprising that the calculations of this distribution are not integers of the fundamental electric charge q.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/325795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is there any direction of pressure i.e. according to force applied or something..? I was reading about pressure in general and came to know that it is a scalar and being scalar it has no direction(by definition of scalars).
But I think that there is some direction as when we push a wall we apply pressure on it in a particular direction,while cutting an object with a knife etc.
I don't know if its correct.Please help.
Any help and edits are appreciated
| Pressure (in the sense of the hydrostatic pressure $P$, i.e., the normal stress on all sides) is a scalar. Pressure (in the sense of the stress $\sigma$ in the context of materials science) is not a scalar but is instead a 3×3 matrix.
The matrix $\sigma=\sigma_{ij}$ arises because of the variety in possible directions, as you intuit: there are three orthogonal directions (e.g., $1$, $2$, and $3$, corresponding to the $x$, $y$, and $z$ directions, respectively) that the surface might face, and there are three independent orthogonal directions along which a force might be applied to that surface.
Thus, the stress matrix (aka the Cauchy stress tensor), as a 3-D generalization of "pressure", is $$\sigma_{ij}=\begin{pmatrix}\sigma_{11} & \sigma_{12} & \sigma_{13}\\&\sigma_{22}&\sigma_{23}\\&&\sigma_{33}\\\end{pmatrix}$$ where the first index is the direction of a vector normal to the surface and the second index is the direction of the force vector.
The matrix is diagonally symmetric because static equilibrium tells us that certain elements must be identical; for example, the magnitude of the force in the $1$ direction on the surface pointing in the $2$ direction must be equal to the magnitude of the force in the $-1$ direction on the surface pointing in the $-2$ direction, or the object would start accelerating along the $2$-axis. Furthermore, it must also be equal to the magnitude of the force in the $2$ direction on the surface pointing in the $1$ direction, or the objective would start rotating around the $3$ axis. Consequently, there are six independent stress values.
Finally, the scalar pressure $P$ corresponds to a stress matrix of $$\sigma_{ij}=\begin{pmatrix}-P & 0 & 0\\0&-P&0\\0&0&-P\\\end{pmatrix}$$ because hydrostatic pressure is compressive.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/325980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
What is the complete proof that the speed of light in vacuum is constant in relativistic mechanics? I study maths in uni and we have a course about relativity.
In the main principles I've read that the speed of light is invariant since we can calculate it from the Maxwell equations.
My problem is that the Maxwell equations I know are not relativistic.
What is the clear way to formulate the Maxwell equations with respect to the relativistic spacetime? Using that formulation do we get the same value for $c$? How do we do that?
Edit:
It is clear now what was my problem after the answers.
The wrong concept I had was that:
From the classical maxwell equations we can calculate the speed of light, and with that information we can build up the relativistic spacetime where the maxwell equations might look different. And it was weird for me.
From the answers it became clear that the invariant speed of light is an observation not a result.
Ps: I find it interesting that my maths like approach did not consider the possibility of something just being an observation, not a result.
| Maxwell equations are invariant under Lorentz transformations, which is the same as saying that they follow special relativity.
You can try to convince yourself by transforming to another inertial reference frame $S'$ and deriving the wave-equation of the fields, you should find that $c' = c$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/326114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 7,
"answer_id": 0
} |
How does centripetal force changes the direction of velocity? Isn't it a violation to Newton's second law of motion? Newton's second law is a vector law.
When when we resolve it in component form along the x, y and z axes we can conclude that force changes only the component of velocity along it, for example if the only force is along x axis then only the velocity along x changes but not other two. So why, in uniform circular motion, does the centripetal force changes the direction of velocity even though it is perpendicular to velocity?
My reasoning is that at any instant say at $t = 0$ the force is along radius and perpendicular to velocity, at $t = dt$ the velocity perpendicular to force is unchanged both in magnitude and direction but a new velocity is gained $dv$ in $dt$ time which is along the radius and now the resultant velocity has the same magnitude as before approximately but a different direction.
Is my reasoning correct or is there some a other explanation?
| One can always express the position of some object with respect to some origin as a vector in as $\mathbf r = r \hat r$ where $\mathbf r$ is the object's position vector, $r$ is the magnitude of that position vector, and $\hat r$ is the unit vector parallel to $\mathbf r$. Differentiating with respect to time yields $\mathbf v \equiv \dot{\mathbf r} = \dot r \hat r + r \dot{\hat r}$.
If the radial distance is constant, $\dot r$ is identically zero. But what about $\dot{\hat r}$? This is a unit vector, which is a special case of a constant length vector. Consider a vector $\mathbf x$ whose length is constant with respect to time: $||\mathbf x||^2 = \mathbf x \cdot \mathbf x= \text{const}$. Differentiating with respect to time yields $\mathbf x \cdot \dot{\mathbf x} = 0$. In other words, the time derivative of a constant length vector is either zero or is normal to that constant length vector. A unit vector is obviously a special case of a constant length vector.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/326214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 1
} |
Is light deflected by external electric and magnetic field? I recently read about the Maxwell's laws of Electromagnetic Waves and I found that Light is made up of both Electric and magnetic fields.
So now if i pass the light through a capacitor such that the plates are parallel to the light will the light be deflected?
If it is deflected then what about the particle nature of light in which the photons are neutral without any charge(as far as i know charge do not exist without mass).
If not why are the electric and magnetic fields not affecting the light in the wave nature if it is solely due to the wave nature why are electrons being deflected in the external fields.
I hope someone give me a clear idea of what is wrong with my idea.
| All electromagnetic waves have the electric and magnetic fields vibrating perpendicular to each other. But this does not mean that they are affected by electric and magnetic fields.
The quantum theory shows that light is made up of photons which do not carry charge, during to which it is unaffected by electric or magnetic fields.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/326324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
} |
Absorbtion of photon Electrons in an atom absorbs an incoming Photon to gain energy and jump to a higher energy level (shell or different orbital) considering the fact that the photon doesn't have enough energy to make the electron jump to any higher energy level would the electron still absorb the photon?
| First of all , the electron and the atom are in one quantum mechanical state. The whole atom will be aborbing the photon if there exists a difference between energy levels that, within widths, equal to the energy of the photon. If there exists, then a photon hitting an atom will turn it into an excited state, with the electron in a higher energy orbital and no photon.
If there is not fitting band energy difference, the photon will scatter off elastically on the residual electric fields of the atoms and mollecules.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/326751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why a circular camera lens captures rectangular pictures? I want to know how a circular camera lens captures a rectangular picture.
Moreover what will happen if the lens was rectangular.
I have tried a lot to think about it but i could not figure out anything.
| In short, cameras produce rectangular pictures because the sensor is rectangular; the illuminated area is circular and some light is wasted off the edges of the sensor. In film cameras the sensor is the film, which is stored in a roll as tape, and making the frames bigger so they catch all the light would waste film. In digital cameras you could make a round sensor if you so desired, but the language of rectangular images is so deeply entrenched that no one does this.
Image source
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/326938",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Geocentric Ecliptic to Perifocal Coordinate Transformation I'm self-studying elementary orbital mechanics and am experimenting with transforming satellite position and velocity vectors between different coordinate systems. I know there are coordinate transformation matrices between Geocentric Ecliptic and Geocentric Equatorial coordinates (simple rotation around one axis), and between Geocentric Equatorial and Perifocal (P, Q, W) coordinates (utilizing inclination, longitude of the ascending node, and argument of periapsis data). But I can't seem to develop a single coordinate transformation matrix from Geocentric Ecliptic directly to Perifocal. Is there an Euler angle rotation sequence that accomplishes this?
| After thinking about this problem further I finally figured out that making the coordinate transformation from ecliptic to perifocal is nearly the same as from geocentric to perifocal. The only difference is that you use an inclination value 23.4 degrees less, which is the angle between the ecliptic plane and the equatorial plane. Not sure why I didn't see in the first place.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/327089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to deduce the charge of an electron from the charge of an oil drop in the Millikan experiment? In the Millikan experiment we measure the charge of one oil drop. But how can I measure charge of one electron when I’m not sure how many electrons are contained within one oil drop?
| Strong peaks on a graph are expected if charge is packaged in discrete
units, but not if charge is infinitely divisible.
One constructs a histogram of the charge measured, and accumulates a few thousand
data points, with very fine drops in a relatively high field. The intent
was to only observe oil drops with small charge, such as would be held
stationary in gravity with a high E field if the droplet was of low mass.
So, if the entire range of charges measured is 1 to 20 electrons, and the
resolution of the experimental charge/mass and radius-of-droplet measurements
is sufficient, a few thousand measurements in a histogram would show
strong peaks. The distance between the peaks is the increment of
charge corresponding to addition of one electron to a droplet.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/327204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do you measure entanglement for type-II downconversion? I am going to do an experiment for a class, which will be using type-II downconversion to entangle pairs of photons pumped from a laser diode. We're planning on using a camera to take pics of the rings that'd be formed, but we need to also measure/quantify our results somehow. The only possibility I've been able to find from research online is measuring coincidence counts for the photon pairs, but I don't see how that really provides any interesting/useful information, and I'd have to find someone to loan us their single photon detectors or whatever equipment would need to be used.
So my question is, how can the entanglement be measured? Can the quality of entanglement be quantified, and how? Or is there just anything measureable/testable that wouldn't be too hard or too inane?
Please help me, really any info at all helps! Thank you!!
| So from what I know, proving entanglement in formal terms can be tricky since you need to do a succession of experiments to show that your system breaks Bell's inequalities. I never did it myself but it is documented in literature and even in Wikipedia.
In your case, if you only want to give a flavour of entanglement, you could simply show that detecting a horizontally polarised photon on one detector is always coincidental with detecting a vertically polarised photon on the other detector. At least to show the polarisation entanglement in type II downconversion it is the simplest way I can think of – but I would be happy to be proven wrong!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/327561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Can force change the mass of an object? I have a difficulty in understanding this 'mass never changes'. Say, i punch/hit a ball until it deflates, or bursting a balloon with a needle - doesn't the mass of the ball/balloon change? the air already diffuses out from the ball.
| Let's look at your system: a needle, a balloon skin, and a gas inside the skin. There is some total mass in this system.
Force of the needle pressed against the skin causes a stress which exceeds the strength of the skin material, making a hole. The air escapes through the hole after the force is gone, but the mass of the air doesn't change. The mass of the needle and the mass of the balloon skin don't change.
Force didn't make the mass of the system change. The mass of the system is simply redistributed. This would be the same as pulling apart a lump of clay and making two smaller lumps: the total mass is the same, but has been redistributed.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/327659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Is really the electric field felt in every thing in space? I was wondering if the action/force of the electric field is really felt everywhere. I know it does reduce as you get further, but my thoughts concerned more about materials.
So, I know that the electric field is strictly related to Coulomb's Law, and Coulomb's Law is different between materials. So my questions are,
*
*Does a net elctric field (generated by two oppositely charged object)
is felt in every material in space? (In the picture, bottles,wood,copper,air,etc..)
*If it does, does it change its intensity based on materials? How?
I made a picture to try to help you understand my strange question, I drew the electric field reversed, because I like to imagine electrons moving.
If there are problem with the question, please let me know in a comment. I will try immediately to fix them, editing the question
| Look up something called dielectric constant. It's a measure of how the electric field due to a charge is affected when the charge is placed in another medium. The reference, of course, is made with respect to vacuum.
Vacuum has a dielectric constant $K$=1. While other material media have $K>1$.
Basically the term $K$ occurs in the denominator of the expression for electric field or force and is useful in determining the relative electric permittivity of a medium. The relative electric permittivity $\epsilon_r$, electric permittivity of vacuum $\epsilon_0$ and the dielectric constant $K$ are related by the expression
$\epsilon_r=K\epsilon_0$.
The relative permittivity is a measure of how the electric field/force is affected by the presence of a medium other than vacuum.
For vacuum, any expression for the electric field will be something like $\frac{1}{4π\epsilon_0} \frac{q}{r²}$
For a medium other than vacuum, the expression becomes $\frac{1}{4π\epsilon_r} \frac{q}{r²}$ or
$\frac{1}{4πK\epsilon_0} \frac{q}{r²}$
An example will help. When we say that $K$ is 80 for water, we mean that the electric field or force would be $\frac{1}{80}$ of its value in vacuum.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/327808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Potential energy for a crystal lattice I would like some help understanding the following passage:
Consider a crystal lattice such that its unit cell has 27 ions arranged such that there are alternative positive and negative ions for same magnitude.
Then the electrical potential of the crystal lattice is just the potential of arrangement of any one of the ion taken with all the ions in the crystal times times the total number of ions halved.
i.e,
$$U = {1\over2}nN_0\sum^{nN_0}_{k = 2} {q_1 q_k \over r_{1k}}, $$
where $n$ is the number of moles of the solid.
I understand what author is saying but not why he is saying so. I don't understand why taking total potential of one ion multiplied by total number of ions gives the total potential. For instance take a ion at the bottom right corner of the crystal, clearly the potential of this ion wrt to a ion in middle and an ion at top left corner of lattice will differ. What is the author's reasoning here?
| You are correct in asserting that the equation gives the potential energy for a single ion and then multiplies it by the number of ions to obtain the total potential energy.
The reason the effects of the boundary can be ignored is based on two factors:
*
*Nearby ions have more influence on the potential of a given ion than far away ions. (the influence drops of with 1/r)
*The majority of the ions is not at the border of the crystal. If there are several moles of atoms, only a very small percentage is at the border.
Number 1 shows that for the ions that are not close to the boundary, the boundary is unimportant. Number 2 shows that this holds for almost all ions.
This approximation is of course only valid for huge numbers of ions
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/328365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why are lower energy systems stable? Apart form the practical evidence that the systems that exist in nature try to attain lowest energy possible and hence, maximum stability, and atoms forms bonds to attain low potential energy but do we have any reason why is that ? Why low energy systems are stable?
What happens when a system loses its energy and why does losing energy (generally) signify stability?
This makes me wonder, what exactly is stability?
| Atoms and molecules are coupled with the electromagnetic field. The electrons of the atom and molecule are accelerating and then it will emit electromagnetic radiation, making the system losing energy.
Classical mechanics will predict a total collapse of the atom. Turns out that Quantum Mechanics saves the day, introducing the concept of discrete spectrum and ground state, and the ground state will be a stable state for the system.
Quantum mechanics can describe the process of radiation of atoms and molecules, see this for more details of how this works.
Now, in general, systems are stable when they minimize the potential energy because there is always a force pushing the system towards the direction that minimizes the potential energy, i.e. the force $f_k$ applied on a generalized coordinate $q_k$ is equal
$$
f_k=-\frac{\partial U}{\partial q_k}
$$
if the system is conservative. If the system is perturbed at a local minimum of $U$, this force will always try to bring the system back to the minimum, and then will be stable if the perturbations are sufficient small.
Atoms and molecules are stable under sufficient small perturbation for other reasons. They are at the ground state, and any perturbation of the system will be reverted into electromagnetic radiation and sent to infinity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/328487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Degree of freedom of a BEC I would like to know how many degrees of freedom an atomic gas has. What about the Bose-Einstein condensation from that gas?
| It all depends on the type of gas and on its temperature.
If the gas is made of single atoms like $^{87}Rb$ or $^{39}K$ (typical atoms that are Bose-Condensed, since you are asking), then it as $3$ degrees of freedom - motion along $x, y,$ and $z$.
If the constituents are diatomic molecules, like $H_2$, then the two atoms could also rotate about the centre of mass, which introduces $2$ new degrees of freedom - clockwise, counterclockwise.
They can also vibrate, as if they were connected by a spring - this introduces $2$ new modes, for in-phase and out-of-phase vibration.
Evidence for the $3$ translational, $2$ rotational and $2$ vibrational degrees of freedom is in the plot of heat capacities (below), where it can also be seen that, when present, the latter two modes only get unlocked (unfrozen) at higher temperatures. This is because the energy associated with that motion is on the order of several $\gg k_B T$:
So even it you made a BEC of diatomic molecules, you will not see the rotational and vibrational modes, as you are too cold. This is why a very active field is the one of cold molecules, where you bind two (cold) atoms together with, e.g., a Feshbach resonance, so as for the energy of the mode to be much lower and hence accessible.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/328613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why solid angle of a closed surface from exterior is zero? Actually my question comes from Gauss law which says charges outside a closed surface, doesn't effect on the whole electric flux of the surface.
The reason of that is that the solid angle of a closed surface while measuring it from outside of the surface, is zero. ( on the other hand, the solid angle of a closed surface from inside the closed surface is $4\pi$)
My question is why solid angle of a closed surface from exterior is zero?
Please don't use Stokes' theorem!
| As viewed from a point outside the surface, the closed surface has a far side and a near side. For example, for a cube there is a front face and a back face.
If there were a point charge situated where the viewer is, the flux from this charge goes into the enclosed volume across one surface and out of the enclosed volume across the other face. The total flux into or out of the volume is zero. Likewise when measuring solid angle the angle subtended by one of these surfaces is taken as being +ve and that subtended by the other is -ve. The front and back faces subtend the same angle, but these angles add up to zero.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/328879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why does gravity act at the center of mass? Sorry if this is a trivial question.
Why does gravity act at the center of mass?
If we have a solid $E$, shouldn't gravity act on all the points $(x,y,z)$ in $E$? Why then when we do problems we only only consider the weight force from the center of mass?
| Suppose I have a collection of n vectors $x_i\quad \forall i\in(1,n), i\in \mathbb{Z}$ such that the corresponding masses at each $x_i$ is $m_i$. This is your body $E$ and if the total mass of your body is $M$, then $$M=\sum_{i=1}^{n}m_i$$
In that case, if $E$ is subjected to a uniform acceleration field $\vec{g}$, as specified in the answer above, then the net force acting on the body is
$$F=\sum_{i=1}^{n}m_i \ddot{x}_i$$
But, the force on the entire body would be $F=Mg$. Let there be a point $X$ on the body such that I can say that $\ddot{X}=g$, Then I can write $F= M\ddot{X}=\sum_{i=1}^{n}m_i \ddot{x}_i$.
From this you can interpret that
$$\ddot{X}=\frac{\sum_{i=1}^{n}m_i \ddot{x}_i}{\sum_{i=1}^{n}m_i}$$
And the centre of mass is defined as
$$\begin{equation}\label{com}
x_{com}=\frac{\sum_{i=1}^{n}m_ix_i}{\sum_{i=1}^{n}m_i}
\end{equation}$$
Since the body $E$ has constant mass, you can get the definition of center of mass above by simple integration.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/329044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 7,
"answer_id": 1
} |
Explicit calculation of the two-point function by path integrals I need help carrying out the following calculation:
We have the generating functional of free theory:
$$Z[f] = \exp\left(\frac{i}{2} \int d^4xd^4y f(x)f(y)\Delta(x-y)\right) $$
where $f$ is an external field and $\Delta(x-y)$ the Green function of the Klein-Gordon equation.
The two-point function $\langle{0}| T\phi(x) \phi(y) |0\rangle$ is then calculated by the rule
$$\frac{1}{i^2} \frac{\delta Z[f]}{\delta f (x_1) \delta f (x_2)}\bigg|_{f=0}$$
The result should of course be $i\Delta(x-y)$.
Can someone show me the explicit steps? I just obtain gibberish.
| Omitting space-time indices (cf. DeWitt notation):
$$
Z[f]=\mathrm e^{\frac12f\cdot\Delta\cdot f}
$$
Therefore,
$$
Z'[f]=\Delta\cdot f\ \mathrm e^{\frac12f\cdot\Delta\cdot f}
$$
and
$$
Z''[f]=\Delta\ \mathrm e^{\frac12f\cdot\Delta\cdot f}+(\Delta\cdot f)^2\ \mathrm e^{\frac12f\cdot\Delta\cdot f}
$$
Therefore, setting $f=0$, we get
$$
Z''[0]=\Delta
$$
as required.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/329123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Why is dipole the simplest source in electrodynamics? I see this sort of statement in many materials, for example this:
The smallest radiating unit is a dipole, an electromagnetic point source.
and this:
The simplest infinitesimal radiating element, called a Hertzian dipole…
However, none of them includes a clear explanation. Why a dipole, which is consist of 2 charges? Isn't a single non-uniformly moving charge enough?
| The smallest radiating unit is an accelerating dipole moment. That can of course be produced by an accelerated single charge, which can be made equivalent to an oscillating dipole.
$$ \ddot{p} = q\ddot{r},$$
where $r$ is a displacement of the charge around some fiducial point.
You don't get a radiation field unless the charged particle is accelerating and because of this, the radiation "source" has to have a finite size. For a sinusoidal oscillation of acceleration amplitude ${a_0}$, where $\ddot{r}= a_0\sin \omega t$,then that size is $a_0/\omega^2$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/329292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 3,
"answer_id": 0
} |
What does a volume integral of $\textbf{J}$ mean? Assume stationary currents in vacuum, $\text{curl}\textbf{B} = \mu_0\textbf{J}$. With $\text{curl}\textbf{A}=\textbf{B}$ and $\text{div}\textbf{A} =0$ the vector potential $\textbf{A}$ can be written explicitly as a spatial integral over all space of the current density $\textbf{J}$, (see for example Jackson, section 5.4): $$\textbf{A(x)} = \frac{\mu_0}{4\pi}\int \textbf{J(x')}\frac{d^3x'}{\textbf{|x-x'|}} \tag{1}$$
So far so good, but what is puzzling about this integral is that the current density $\textbf{J}$ is associated with a 2-form and we are integrating it with differential 3D volume elements not 2D surface elements as I would have naively expected it.
Compare (1) with a similar formula for the scalar potential $\phi$ where $\textbf{E}=-\text{grad} \phi $, and now the charge density $\rho$ is given:
$$\phi{(\textbf{x})} = \frac{1}{4\pi\epsilon_0 }\int \rho{(\textbf{x}')}\frac{d^3x'}{\textbf{|x-x'|}} \tag{2}$$
Charge density $\rho$ is a spatial density and as such is associated with a 3-form, and as I would expect it is integrated with a 3D differential volume element.
My question is what does this all mean, what happens when something that is apparently a natural 2-form is integrated if it was something else?
| It's best to think relativisticly. We are solving the Maxwell equation $d\star (d A)=\star J$ where
$$
A=-\phi dt+ A_x dx +A_ydy+A_zdz\\
J= -\rho dt + j_x dx+j_ydy+j_z dz
$$
are 1-forms in four dimensions and $\star$ is the Hodge dual. So $\star J$ is a three form.
The $j_x$ appearsin $\star J$ with $dy\wedge dz \wedge dt$ so it's only a 2-form if you ignore the $dt$.
$\rho$, in the other hand, is the coefficient $dx\wedge dy\wedge dz$ and so in $\star J$ it is a three-form.
For time-dependent sources we integrate over four dimensions to get $A$ --- so the Green function must be a form which contains a delta function that only keeps contributions from the retarded time. For static current/charge distributions. The $t$ integral can be done leaving the 1-form answers you cite.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/329683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Crash simulation on Mythbusters I remember an episode of mythbusters where they were busting myths to do with a head on collision between two cars.
They said that instead of crashing two cars into each other at 50mph they would crash a car into a stationary object at 100mph because the energy involved in the crash would be the same.
Later on they corrected themselves to say that the energy is not the same. But I can't figure out why this would be the case?
Can someone explain if these two scenarios are the same or not. And why?
| Imagine a thought experiment in which the two cars and their contents are identical mirror images of each other.
In this perfectly symmetrical universe, after a 50 mph crash, the cars break up symmetrically: because the cars and contents are identical in every way, every fragment is emitted at exactly the same time from the same position and at the same speed, except for reflection.
Not only does every particle of one car collides with its corresponding particle from the other car, every collision happens in the plane of symmetry, with each pair of particles either stopping dead or bouncing away from each other.
But this is identical to the result of one car crashing at 50 mph into an immovable infinitely large wall where the plane of symmetry was.
Therefore, two identical cars at 50 mph each crashing is, in principle, the same as one car crashing into a wall at 50 mph — and different from a car crashing into the wall at 100 mph!
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/329831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How many photons enter our eyes per second when looking at the blue sky on a sunny day? How many photons enter our eyes per second when looking at the blue sky on a sunny day? Say the sun is directly over head and you are looking at the blue sky on the horizon. Say that the pupil is 2mm in diameter. I'm looking for an order of magnitude calculation here.
Update: the light hitting your eyes would only include the blue light scattered by the atmosphere. Not the directional light that is hitting the top of your head.
| The surface brightness of the Sun is -10.6 mag per square arcsecond.
The full moon on the other hand is about 14.5 (astronomical) magnitudes fainter than the Sun, has a similar apparent angular size and is just visible in a bright daytime sky.
The flux from the daylight sky incident upon the eye is therefore around $10^{14.5/2.5}$ times less than the solar constant. i.e. About $2\times 10^{-3}$ W/m$^2$.
The pupils of the eye might have a 2mm diameter in bright light, so receive around $6.2\times 10^{-9}$ W.
Let's assume that the average blue sky photon is at 400 nm with an energy of 3.1 eV, then you receive about $10^{10}$ per second (in each eye).
Ah, but this would be correct for a small patch of blue sky with the same angular extent as the full moon (about $7\times 10^{-5}$ steradians). The eye actually collects light from $\sim \pi$ steradians, but then the projected area of the pupil is reduced by a small factor (I think 0.75) because of the $\cos \theta$ term.
So the final result is $3\times 10^{14}$ photons per eye.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/329971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Example of Poincare recurrence theorem? Is it possible to explain Milankovitch cycles (or some other arbitrary planetary configuration that recurs to some approximation) in terms of the Poincare recurrence theorem?
More generally, is there a good physical example of the Poincare recurrence theorem?
| We just had a pedagogical paper on Poincare recurrence:
https://arxiv.org/abs/1705.01444
Yes, for the planetary configuration problem, some of the recurrences can be predicted accurately. It reduces to a classic problem in number theory, namely, the simultaneous Diophantine approximation problem for real numbers. Mathematicians have done a lot on this problem and in particular, a famous algorithm (the LLL algorithm) exists.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/330139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
In flat spacetime, what is the mixed (invariant?) form of the metric tensor? In flat space, the metric tensor is (in one of the two conventions)
$$\eta^\mu{} ^\nu = \begin{bmatrix}1&0&0&0\\0&-1&0&0\\0&0&-1&0\\0&0&0&-1\end{bmatrix} = \eta_\mu{}_\nu$$
What is $\eta^\mu{}_\nu$ or $\eta_\mu{}^\nu$? I read here it was the same! But if we use the metric tensor to convert covariant components and contravariant components, it seems like the answer should be the identity matrix.
If we use the definition of the metric as the cofficients of an infitesimal arc length element $ds^2$, I think this also shows that the answer should be the identity matrix.
Can someone explain (in a simple way for someone who doesn't know much about differential geometry) how to obtain the correct answer, whatever it is? Preferably with an answer that doesn't just rely on the memorized rules of einstein index notation.
| For any metric (either on flat or curved spacetime), $g^\mu_{\ \ \nu} = g^{\ \ \nu}_\mu = \delta^\mu_\nu$. See here for the explanation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/330313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Thin Film Interference, thinning each layer one by one I have three layers of different complex refraction index.
I am shining with a laser of wavelength lambda onto the layers.
Through etching, each layer is etched one by one, and hence one by one, the thickness is reduced of each layer until zero.
Experimentally, I am getting this intensity plot (the first and last layer have the same refraction index (blue line):
How can I obtain this with an analytic solution ?
(or iterative process)
I am mostly interested in normal incidence !
The plot is just there for reference. It only shows the shape of the final solution. The y-Axis represents the light intensity. The x-Axis the time.
Blue: Line -> Light intensity
Yellow Line: It's second derivative
| This isn't directly an answer since the method of doing the calculation is rather long and tedious, if basically straightforward, but I can tell you where to find the answer because I did precisely this as part of my PhD. The method is described in Optical Properties of Thin Solid Films by O. S. Heavens, Butterworths Scientific Publications, London 1955. It is on Google Books, though sadly not a scan, and I see there is a 1991 edition so at least you won't be trying to find the 1955 edition that I had to use.
Basically at each interface you calculate a relationship between the forward going and reflected waves either side of the layer, then start at the bottom layer where you know the forward going wave is zero and work your way back up.
It's not too bad for normally incident light but becomes very messy indeed for light incident at any significant angle to the normal because then you need to account separately for the two polarisations of the light.
I used it for films that were reacting so their thickness was changing because a layer of reaction product was growing, but the method will work just as well for films that are being etched.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/330461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
When can I pull out the identity operator? I have a problem that requires figuring out the sum:
$\sum_{n>1} |\langle n10| \hat{z} |100 \rangle|^2 = \sum_{n>1} \langle 100|\hat{z}|n10 \rangle \langle n10|\hat{z}|100 \rangle$
where $|nlm\rangle$ are Hydrogen wave functions, and $\hat{z} = r\cos\theta$. For n=1, this term is zero; so then this sum is
$\sum_n ...|n10\rangle\langle n10| ...$
Is this the identity operator? Can I simply pull that out of this equation and say this is equal to $\langle 100|\hat{z}^2|100 \rangle$?
I've checked these two values (my assumption which is just the one term, and the explicit sum to infinity) in Mathematica, and it seems like no. The expectation value of $\hat{z}^2$ with $|100\rangle$ is just $a_0^2$, whereas the sum (over 100 terms) seems to converge to about 0.717$a_0^2$. So it could be that a) I'm doing this right, and my Mathematica code is wrong, or b) my suspicions are confirmed that no, this is not the correct way to do this.
If this is the case, I'd also love some help actually computing this sum without doing hundreds of integrals.
Thanks!
RS
| I've figured it out. In this case, this is the identity operator, in the sense that we could change this to a sum over all the hydrogen wavefunctions since this integral over all of these other terms is zero.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/330580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to tell whether a Feynman diagram is $t$-channel or $s$-channel by looking? By looking at a diagram, how does one tell whether it represents a $s$-channel process or a $t$-channel process i.e., without finding the amplitude? I'm familiar with Mandelstam variables but I've trouble understanding what a $s$-channel or $t$-channel process would mean and how to tell the difference.
| I think the easiest way is to check the agreement beetween fermions/antifermions and the directions of their arrows.
Take this for example:
You can label all the fermions $e^-$ and it's t-channel. If you label the top left and bottom right $e^-$ and the other two $e^+$, it's s-channel.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/330717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Spring-Mass-Pendulum "via Newton's Laws" Good Night everyone:
I have one problem here that I KNOW how to solve using Lagragian Dynamics. But, I really want to know how to solve using Vector decomposition, Newton's Laws, first-year physics and so on.....
I really apreciate tips and hints, both mathematical and physical.
I DON'T WANT A SOLUTION OR STEP-BY-STEP.
Thank you.
(*) The "motivation" for my question is that we often hear that Lagrangian Dynamics is more General and powerful than Newton's Approach. It's true. But, I want to see for myself that it's true. On this particular problem, which is more difficult than basic Newtonian problems, the solution is hard (?) but still "possible".
(**): The concepts of forced,damped,simple and coupled oscillator are quite clear to me and basic ordinary differential equations as well.
| A good start is to make a free body diagram of all parts. Mark known and unknown forces. Remember that if you have a force $\mathbf{F}$ on one part at the contact point with another part, on the contacting part you have a force $-\mathbf{F}$. Then set up a differential equation for the motion given the total forces on the part(s) of interest.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/330803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Magnets and entropy While teaching thermodynamics our chemistry teacher (physics syllabus covers only first law) told us that a process is reversible if and only if no entropy is created.
Coming back to everyday life while playing with magnets we have observed that some substances stay magnetized for a very very long time (that is to say permanently) while others lose their magnetism at a moments notice. Now since magnetism is induced in some materials permanently I conclude that some change in entropy takes place. Now how exactly does magnetism produce a change in entropy on a micro and macro scale ?
| A system in equilibrium corresponds to a free energy minimum which is a compromise between a minimum of the internal energy with a maximisation of the entropy. When you introduce, say, an unmagnetised ferromagnet to a magnetic field, alignment of the atomic magnetic moments with the external field reduces the internal energy. So if the field is strong enough (i.e. a great enough reduction in internal energy) then magnetisation can occur at the expense of a decrease in entropy.
Note, however, that the total entropy of the universe does not decrease since you need to use a heat engine (e.g. your arm) to move the ferromagnet through the magnetic field which will itself generate entropy.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/330920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is value of speed equal to the magnitude of velocity? For example: - suppose a body covers a semi-circle in $5$ seconds, then distance$=\pi * r$, where $r$ is the radius of the semi-circle. Displacement is $2r$ only. Then the value of speed is $[\pi * r] / 5$ while that of velocity is $2r/5$. Here we can see that magnitude of velocity is different from that of speed. Is it correct?
| Instantaneous speed always equals the magnitude of instantaneous velocity (because the instantaneous displacement is small enough to be regarded as straight-line). In your example, you are comparing average speed and the magnitude of average velocity - and the two of them can be different as you have correctly calculated.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/331262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is Earth part of the system when writing Gravitational Potential Energy? Here is a question which stumped me when teaching high school students.
The Work-Energy equation can be written as:
$$
W_{ext} + W_{non-conservative} = \Delta{U} + \Delta{K}
$$
Here, $\Delta{U}$ refers to the difference in potential energy of the system in consideration. Potential Energy is nothing but the negative of work done by conservative forces.
I would like to talk about one special potential energy, and that is Gravitational Potential Energy.
Now, let's say I have a block of mass $m$. We write the gravitational potential energy for this block as $mgh$. When doing so, we say that this potential energy is the potential energy of the block-earth system. So, we mean that Earth is a part of our system.
Now, if earth is a part of our system, everything on earth is a part of the system. It means if I am standing near this block and apply some force on it, that force will not be external and hence its work done would not be counted in the $W_{ext}$!! This doesn't make any sense.
As a student, I never looked at gravitational potential energy this way. But now, when I look at it, it is mind-boggling to think the whole earth is part of the system.
Please clarify where my logic/reasoning is going wrong.
| The usual approach is to treat the block as the 'system' and Earth as the 'environment.' Then gravity is an external force acting on the system, or stated alternatively: An interaction between the system and Earth. In the energy balance, one may either put gravity in the tally of external work (done on the system), xor introduce a gravitational potential energy to account for it.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/331558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Why do charges on the sphere not fly? Consider a uniformly charged conducting solid sphere. As we know, all of the charge must reside on the surface of the sphere, so let's only take this uniformly charged spherical plane into consideration.
I have color-labelled six representative charge elements ($dq$) on this spherical plane.
Let's take a loot at the red charge element.
The electric field just outside the red charge element (outside the plane) is $\sigma/\epsilon_0$ and that just inside the red charge element (inside the plane) is zero. But, what is the electric field at the red charge element?
Imagine we remove all other charge elements from this spherical plane except the purple charge element (without allowing the purple charge element to change its position on the plane) and compute the field contributed by this purple element at the red dot. If we repeat this process for all of the five non-red charge elements and summate their fields at the red point, we find that the field at the red charge element must be non-zero. So, why doesn't the red charge element just fly off?
| Standard answer for a discountinous field is to take the average of the inside and outside field. So in your case the field on the sphere would be $\sigma/(2\epsilon_0)$. See for example Purcells Electricity and Magnetism.
For your second question: In a conductor the electic charge is only allowed to flow inside the conductor. Normally charge does not flow off the conductor (except if you heat it up a bit); in that case you would loose charge as you mentioned.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/331683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How do varying static point charges exert the same force on each other? If you have two point charges one being 1 Coulomb and the other being 1 Trillion Coulomb, it is said that the electric force from the 1 Coulomb point charge exerted on the 1 trillion Coulomb point charge is equivalent to the electric force from the 1 trillion coulomb point charge exerted on the 1 Coulomb point charge.
How can a 1 coulomb point charge exert the same force as a 1 trillion coulomb point charge?
| Perhaps this analogy will help:
Imagine I have two fans - one with a huge diameter, the other with a tiny diameter. When I put them facing each other, with the huge fan running, I will be able to extract a small amount of power from the tiny fan (because only a tiny fraction of the wind generated by the big fan will intersect with it). Conversely, when the tiny fan is running, almost all its air will be "felt" by the huge fan. But the tiny fan only generates a little bit of air movement...
This is how it is with two different charges (or if you like with two different masses). The same thing (charge, mass) that makes them able to generate a field, makes them susceptible to the field of another (charge, mass). This is indeed a necessary consequence of Newton's third law - the attractive force must be reciprocal (force of A on B must equal force of B on A), so there must be symmetry in the equation describing the force.
If you are OK with the force of Moon on Earth being the same as the force of Earth on Moon, then you should be OK with this. And if those forces were not the same, their would either be crashing into each other, or flying apart...
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/331836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
What is Thermal expansion?
"Thermal expansion arises from the asymmetrical nature of potential
energy curve for atoms in a solid. If oscillators were truly harmonic
separation would not change regardless of the amplitude of vibration."
I don't understand how a temperature increase might not affect the volume of a solid substance if the oscillations were harmonic?
| If the interatomic potential were a perfect parabola, it is true that there would be no thermal expansion. Why? Well, as atoms vibrated more and more, the mean atom position would not change. Sure, it would wobble around more, but it would wobble as much to one side as to the other. The change in average atomic position would be zero. In contrast, if the potential is skewed (as it is for an interatomic potential), the time averaged atomic position will change as the amplitude of vibration increase. On average, it will (usually) move to a greater interatomic distance with increasing temperature.
Of course, this process is neither linear nor necessarily monotonic across all temperatures, as demonstrated by many real materials.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/332524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
One loop diagram in $\phi^4$ theory I know that in the $\phi^4$ theory the tree-level diagram for two in-going and two out-going particles is simply a 'cross' where all the external legs meet at one point.
I'm now interested in a slightly more complicated case where I have 4 rather than 2 outgoing particles and I want not only the tree level diagram but also the one loop diagram. I think I know what the tree-level diagram looks like, but I have some trouble drawing the one loop one. I know that each vertex needs to have 4 lines connected to it. I managed to come up with something but it was just trial and error and I'd like someone to verify whether what I've done is right.
My attempt is:
| Yes, that is the correct one loop topology that appears assuming no snail and/or one particle reducible contributions (inclusion of these gives you a plethora of other diagrams, such as ones where you decorate the tree level contribution with snails etc). With a labelling of the external momenta in place, you can show by simple combinatorics the number of inequivalent permutations of the external legs you have. The contributing diagrams are essentially generated by attaching two legs to each vertex of a triangle.
Naively there are $6!$ permutations of the external legs but to avoid overcounting due to equivalent diagrams related by vertex relabelling we have to thereby divide out by the cardinality of the symmetry group of the triangle which is $|S_3| = 3!$ Now, we also need to divide out by the permutation of two legs at each of the three vertices. So the number of contributing diagrams is $6!/(3! \cdot (2!)^3) = 15$.
The same argument can be applied to e.g the more familiar one loop contribution to $2 \rightarrow 2$ scattering within $\phi^4$ (the so-called dinosaur diagram discussed in many QFT books in the pursuit of renormalisation of the theory at one loop). By exactly the same argument the number of contributing diagrams is $4!/(2! \cdot (2!)^2) = 3$, the $s, t$ and $u$ like channel contributions.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/332649",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Can we theoretically "derive" the mass of a particle? I read a pop sci book on the Higgs which said that particles get their mass due to interacting with the Higgs field. If that is true, could we use first principles to derive the mass of, say, an electron? After all, QED is built on the interactions of particles and fields, right?
| This is the table of the elementary particles of the standard model, SM.
The masses have been measured experimentally and the whole table is part of the postulates/axioms of the standard model. These masses within the SM are generated by the Higgs mechanism. The SM up to now is very successful in describing and predicting data. No theory has come up with a prediction of these postulated masses , while embedding the successes of the SM, so the answer is no for the elementary particles.
All other masses are composites of these elementary particles. For the mass of the hadrons the internal dynamics of QCD have to be used, and there has been a laborious lattice calculation that does give the hadron masses.
In the work presented here, a full calculation of the
light hadron spectrum in QCD,
only three input parameters are required: the light and strange quark masses and the coupling g .
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/332777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Why do papers describe the ytterbium ion as having only a single excited state? Multiple papers describe the ytterbium Yb IV ion as having only two energy levels. Why aren't there more levels in its spectrum?
| There are more levels.
For many experiments, one can choose combinations of laser frequencies such that only a few energy levels are relevant; when reporting those experiments, one can just report a simplified energy-level diagram that only includes the states that are relevant to the experiment.
However, that doesn't mean that the other levels don't exist, and if you do other experiments with other excitation conditions then they can and will be relevant.
The NIST Atomic Spectra Database for energy levels has a good list of the full roster of states for atoms and ions. For Yb III, for example, it starts off as
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/332914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Qualitative description of how an inhomogeneous broadening mechanism can lead to multimode oscillation in a laser cavity I understand that when there is inhomogeneous broadening in a laser cavity, the resultant gain profile is a Gaussian (as opposed to a Lorentzian profile for homogeneous broadening). Due to the wider nature of a Gaussian profile, it is natural that multiple modes fall within the gain bandwidth (FWHM/full width half maximum). This subsequently leads to the possibility of multimode oscillation.
However, I am struggling to understand why exactly this leads to multimode oscillation. Why can't there be multiple modes within the bandwidth, with only one of them oscillating?
To summarise the question: how does this specific broadening mechanism lead to multimode oscillation?
| First of all, in-homogeneous broadening doesn't have to be Gaussian, it happens to be like that only for Doppler broadening (due to Maxwell-Boltzmann distribution of velocities). Other in-homogeneous broadening don't result in a Gaussian gain profile.
In the same way, homogeneous broadening doesn't have to be a Lorentzian, this happens only for life-time broadening.
Second, the mechanism which prevents multiple wavelength from participating in the laser is called "mode competition". It is similar how at ceremonies when someone increases the microphone's volume too much we hear a single frequency oscillating. The gain curve has maximum points which with each round in the cavity are amplified the most until the amplifier gets to saturation. When this happens, because of the homogeneous nature of the broadening, the whole gain curve is lowered so some frequencies stop getting amplified, and this continues on and on until only "the last survivor" remains, a single frequency.
But, when the broadening isn't homogeneous, it means that the different frequencies don't share the amplifying atoms, some atoms amplify more one wavelength and some amplify another, and there is no mode competition. That's why with in-homogeneous broadening there are multiple wavelengths that come out of the laser.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/333004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What is the meaning of "kay effective" $k_{\rm eff}$ in SHM? I am really confused studying for my Physics lectures on oscillations, namely Simple Harmonic Motion. You see, my Professor introduced the topic: when he solved some examples, I noticed that when working with springs, he either:
*
*Uses the actual $k$ of the spring that is given in the wording of the exercise.
*Even though the exercise might give the value of $k$ for the spring, he sometimes works with it to find what he calls $k_{\rm eff}$.
I really do not understand how to tell each case apart and I am really desperate.
One example problem he solved in class is the following. Problem 4 of this problem set:
http://web.mit.edu/8.01t/www/materials/ProblemSets/Raw/f12/ps11sol.pdf
I understand how it solves the problem, but how do I know I need to look for another expression of $\omega$ when, in general, we have that $\omega =\sqrt \frac km$, and k is already given in the problem?
I hope my question is understandable. It might sound like I have no idea, but I feel that I really do not in this new topic and I do not know where to start to understand this.
All help is greatly appreciated.
| $k$ is the spring constant experienced by the body due to actual string. $k'$ effective is the constant in the expression $m\frac{d^2x}{dt^2} = - k'x$ after you've done some rearrangements to your differential equations. For example, supposes you have a mass on two springs of spring constants $k_1$ and $k_2$. Then, your differential equation is: $$ m\frac{d^2x}{dt^2} = -k_1x - k_2x= -(k_2+k_1)x $$ where x is displacement from equilibrium. Then $k' = k_2 + k_1$. So although you might have two springs acting on a system, you get behaviour effectively equivalent to a single spring of spring constant $k'$. Now you can apply this idea to more complicated cases where you have different forces acting on the object. Important point is that if you can rearrange the equations into the $$ m\frac{d^2x}{dt^2} = something * x$$ form, your something is the effective spring constant and you can treat the problem as a mass on a spring.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/333155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Do point charge really exist? Do point charge really exist experimentally?
Am I right with definition of point charge? According to me point charge is a charge having 0 (zero) mass and have 0 (zero) volume.
| In your definition, you have put the mass $=0$ condition. You cannot have a massless charged particle. Permitting them to exist will predict a decay for electrons and we have no evidence that that happens. Of course, the assumption here is that QED is correct and we have no evidence that it is not.
However, if you drop the mass constraint then our humble electron should qualify for all practical purposes. As far as we know, the electron has no spatial extent. A spatial extent does not agree well with QED models.
P.S.:- I must clarify that throughout this post I mean electric charges only. There are other kinds of charges in nature as well (e.g., colour charge).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/333483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 0
} |
What is the total energy stored in the capacitors? Shouldn't the repulsive potential energy stored be included? Is there the repulsive energy stored in the capacitors or that energy is already included in the conventional calculation of the stored capacitor energy?
| Assuming no dissipation, the energy stored in a charged capacitor equals the work done in charging the capacitor.
But the work required to increase the charge $Q$ of a capacitor by $\Delta Q$ increases with $Q$; the more charged a capacitor is, the more work is required to increase the charge by $\Delta Q$.
You might think of this in terms of repulsion, i.e., the more excess electrons there are on a plate, the more work is required to add one more electron to the plate.
So, in this sense, repulsion is included.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/333571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Why does water in a glass ripple but oil does not? When I place a glass of water on the table and hit the table from below I can clearly see ripples in the water with their center in the middle of the glass. However when I do the same with a glass filled with oil there are no ripples. The oil seems much more stable.
Why is this? This is exactly the opposite of what I would expect. Since the water has a higher surface tension than oil I would expect the water to be more stable.
| The oil has a far higher viscosity than the water. Since this is a direct measure of the resistance to gradual deformation from a stress, the oil has far smaller ripples than the water for an equal forces - the ripples will also appear to propagate slower through the oil.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/333883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Field confinement inside superconductor cylinder I'm learning about the Meissner effect, but I can't really grasp an example of
Bleaney's book on electricity and magnetism.
Let's assume a hollow thick cylinder made of superconductor material, with it's axis
in the orientation of an applied field $B_1$. We cool it under it's superconducting transition temperature, so it becomes a superconductor, and as a good one it expells the $B$ lines of its walls interior; there is no $B$ field inside the walls of the superconductor. The field inside and outside the cylinder remains being $B_1$.
After that, we change the applied $B_1$ field to a value $B_2$. The field in the exterior of the cylinder changes from $B_1$ to $B_2$, but the field inside the cylinder remains being $B_1$.
Why the interior field does not change as the exterior field? It would be surely compatible with the expulsion of the field lines, I though that the only constrain were that the $B$ inside the superconductor material must be zero.
| The cylinder is a perfect conductor.
Lenz's law says that when there is a change of flux inside a coil, this will set up an e.m.f. to resist that change. But the moment an e.m.f. is generated, this will create a current in the coil that resists the change in flux.
When the resistance is zero, any change in the flux will immediately be canceled by the change in current around the loop. And so the flux inside the cylinder is "locked" to whatever value it was when the cylinder became superconducting.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/334262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Does a capacitor have a resistance? Does a capacitor have a resistance? And why? When I asked my physics teacher, he said certainly not, but I didn't figure out why. Can anyone please clarify? Thanks in advance.
| I feel, capacitor has infinite resistance, since charge generally does not flow through a capacitor, it stores the charge. It generally has a dielectric medium which does not conduct electricity. Thus its resistance will be same as the resistance of the medium. Very high voltage has to be applied across it so that current flows.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/334385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Tension In A Massive Rope My book says :the horizontal component of the tension in a massive rope is constant.
It does not seem intuitive to me as I don't think that the above approximation (I guess) is possible.I don't know though
Can anyone help me out with this as I am just not getting it.
Any help and hints are appreciated .
| Any section of the rope can be considered as a Free Body. If the section is not accelerating then the horizontal and vertical components of force on it are balanced.
If there are no forces external to the rope acting horizontally on the section, then the forces from the rope at the left and right (ie tension) must be equal. If the rope has mass then there is an external force acting vertically, so the forces on the section from above and below will differ by the weight of the section of rope.
The horizontal component of tension is not constant if the rope is accelerating horizontally - eg oscillating or rotating.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/334498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Electrical conductivity of massless fermions in case of finite temperature I was asked to find conductivity $ \sigma (\omega, T) $ using methods of qft, or more exactly using Matsubara Green's function, for the following system:
qed with massless dirac fermions in case $ T \neq 0 $.
As I understand, coductivity is a linear response coefficient to applying electric field.
$$
j_i (t) = \int_0^{\infty} \sigma_{ij} (\tau, T) E_j (t - \tau) d \tau
$$
So I need to use Kubo formula:
$$
\sigma (\omega, T) = \frac{i}{\hbar} \int_{0}^{\infty} < [\hat{X}(t), \hat{Y}(0)] > e^{i \omega t} dt
$$
I think $ \hat{X} $ should be current operator:
$$
\hat{X_{\mu}} = e \bar{\psi} \gamma_{\mu} \psi
$$
I can write excitation to electromagnetic field as:
$$
\hat{H}_{ext} = i e \bar{\psi} A_{\mu} \gamma_{\mu} \psi
$$
But if I use this as $ \hat{Y} $, I will find linear response to the vector potential, not electric field. Maybe, it is still correct and I just need to recalculate conductivity from this linear response coefficient in someway? So that is the first question: what operator I should use as $ \hat{Y} $?
After that, I want to use Wick theorem to express $ <[\hat{X}(t), \hat{Y}(t)]> $ through Matsubara Green's function. Can I use free Green's function $ S(i\omega_n, p) = \frac{1}{i\omega_n \gamma_0 + \bar{p} \bar{\gamma} + i \epsilon} $, or I need to calculate first loop correction?
| Conductivity can be written as polarization operator divided by frequency (what is known as Kubo formula). As soon as you calculate the polarization operator (one-loop diagram where the Matsubara green functions do enter) at finite T you obtain the conductivity.
Please have a look at the following papers where this very problem is solved for 2+1 dimensional fermions
https://arxiv.org/abs/1608.03261,
https://arxiv.org/abs/1111.3017
However, in the QED in 3+1 dimensions you will have infrared divergences in the calculation which have to handled somehow.
good luck,
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/334604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
The Langevin equation - nature of the forces The Langevin equation is given by:
$$m\ddot x=-\gamma \dot x+f(t)$$
where $f(t)$ is some stochastic force. I know that the first term on the RHS is to do with viscous drag and the second term is to do with random fluctuations in the number particle collisions. I have also heard it said it is due to random density fluctuations - is this the same? Lastly what is the main difference between the origin of the two forces, since they are both fundamentally due to particle collisions?
| The first term on the RHS arises from the assumption that the particle is moving through a viscous fluid (in this case, with no net flow velocity). The collisions from the front deliver a higher impulse than collisions from the rear, on average, which leads to a coherent net force directed opposite the particle velocity.
The second term arises from the fact that the viscous fluid is actually made up of a huge number of tiny particles which randomly kick the particle from all directions. In addition to the coherent drag mentioned above, these discrete kicks can also make the particle randomly jiggle in various directions, which results in an easily observable diffusion effect.
In essence, the first term encapsulates the coherent influence of the tiny fluid molecules which tends to reduce the average particle velocity to zero, while the second term represents the incoherent influence of the tiny fluid molecules which causes the particle to bounce around and diffuse rather than just sit perfectly still.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/334739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What's the observer in double slit experiment? I watched a video on youtube about double slit and it said that when we try to observe electrons it starts to act like a particle but what are these observers that they use? they're shooting one electron at a time so how can they detect them while they're moving in front of the observer?
| An observer is you or someone looking at the electron. It can also be a detector or camera taking pictures etc. it is wrong to think that the mind of a conscious observer will cause an observer effect. The thing is if you look at an electron as it's traveling through the slits you will affect it's a trajectory because you physically interfered with it. In order to see the electron photons need to interact with it and they will affect its path causing the overall interference pattern to be disturbed.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/334997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Intuitive reason for the $T^4$ term in Stefan Boltzmann law The Stefan Boltzmann Law gives a relation between the total energy radiated per unit area and the temperature of a blackbody. Specifically it states that, $$ j= \sigma {T}^4$$ Now using the thermodynamic derivation of the energy radiated we can derive the above relation, which leads to $T^4$. But is there any intuitive reason for the $T^4$ term?
| If you know Quantum Mechanics, you know that you can set length to have dimensions of the inverse of energy. This means that $j$ must have dimensions of energy to the four. If you consider that the only variable with energy units is the temperature, then the energy density must be proportional to $T^4$.
If you consider additional constants with energy dimensions, like a mass, then the derivation is no longer valid.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/335122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 1
} |
What is an operator times one of its eigenstates? I am trying to get a hold of caluclating with matrix elements. I have a Hamiltionian $\hat{H}$ in a two-dimensional Hilbert space, having eigenstates $\psi_1$ and $\psi_2$. My professor wrote down these equations:
$$\hat{H} \psi_1 = H_{11} \psi_1 + H_{12}\psi_2 \\
\hat{H} \psi_2 = H_{21} \psi_1 + H_{22}\psi_2 $$
He said these are alternative way to write the matrix elements of $\hat{H}$. However, I fail to see why is this. I tries to look it up in a linear algebra textbook, maybe this is a special property of matrix multiplications.
I have tried the following:
$$H_{11} = \psi_1^* \hat{H} \psi_1 \\
H_{11} \psi_1 = \psi_1^* \hat{H} \psi_1 \psi_1 $$
I did the same thing with $H_{12}$, added the equations together, but still can not see anything that resembles the original system.
Where do those two equations come from?
| I highly doubt that there being "alternative ways to write the matrix elements" mentioned by the professor has any conceptual significance. After choosing a basis of the two-dimensional Hilbert space, operators can be expressed as matrices,
$$H=\begin{pmatrix}H_{11} & H_{12}\\ H_{21} & H_{22}\end{pmatrix}$$ where $H_{11}, H_{22}$ are real and $H_{12}^*=H_{21}$.
The equations you cite simply mean that we have chosen the basis to be $\{\psi_1, \psi_2\}$. This is an orthonormal basis. In particular for a self-adjoint (as observables must be) matrix, there is always an orthonormal basis of eigenvectors. So the two eigenvectors in this basis are represented as $\psi_1=\begin{pmatrix}1\\0\end{pmatrix}$ and $\psi_2=\begin{pmatrix}0\\1\end{pmatrix}$. Note that given an orthonormal Basis of an n-dim. Hilbert space $\{\psi_1, \dots, \psi_n\}$ we may express the Matrix elements of any operator as scalar products:
$H_{ij}:=\langle \psi_i, H\psi_j\rangle=\langle\psi_i|H|\psi_j\rangle$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/335186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to show that Chern number gives the amount of edge states? When talking about topological insulator and talking about bulk-edge correspondence, it seems to be widely accepted conclusion that the band Chern number (winding number) is equal to, when the boundary becomes open, the amount of edge states. But why? For example, SSH model or Graphene (treat $k_x$ as parameter of a 1D chain and $k_y$ as the wave vector along this 1D chain). When the 1D chain is closed, in the topologically non-trivial phase, Chern number can be calculated to be, say, one. And after cutting the 1D chain open, there is one pair of gapless state living on the two edges. Why is Chern number equal to the amount of gapless edge state pairs? Is there any proof of that?
| Yes, there is a proof of that. The first one appeared by a beautiful paper of Hatsugai in 1993 https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.71.3697 for a particular special classes of models of the IQHE. More general proofs ensued, but it does turn out that the proofs rely on some non-trivial math, the most simple way to present it seems to be rooted in the context of some basic facts of complex analysis, as for example is presented in this paper: https://journals.aps.org/prb/abstract/10.1103/PhysRevB.83.125109
There you see for example that essentially the Cauchy integral formula lies at the heart of the bulk-boundary correspondence. More general proofs rely on more sophisticated math, e.g., K-theory or Fredholm theory.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/335269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Does temperature of boiling solution stay constant? Let us suppose we have a liquid with non-volatile solute, and the solution is raised to its boiling point by applying heat.
Since the solvent is being evaporated, the concentration of solute will increase. Thus, the vapour pressure of solution will decrease and the boiling point must increase. Since we are still applying heat, the temperature will keep increasing as the boiling point increases.
But we have learnt that the temperature of a boiling liquid is constant until all liquid is evaporated. Is this only applicable for pure liquid and not for a solution with non-volatile solute?
So,
*
*Will the temperature of the boiling solution be constant or increasing?
*Will the vapour pressure of the solution stay constant (equal to atmospheric pressure) even if vigorously heated so that the solution evaporates very quickly?
| "The temperature of a boiling liquid is constant until all liquid is evaporated" is true only if the nature of the liquid doesn't change. When you have a changing concentration of a solute (because of the evaporation of the solvent), you don't have the "same" liquid as time goes on ... therefore there is no contradiction.
The second question is slightly harder to answer - but unless "vigorously heated" means "explosively heated" I suppose the answer is still "yes".
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/335510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Why can we not measure the distance to the Sun directly using Radar? My question is specifically WHY CAN WE NOT USE Radar to measure the distance to the Sun? What is the reason for that?
Sorry if this is a lame question, I'm not an expert on these things and just occurred to me why not use radar to measure the distance to the Sun directly rather than go through all the complications of doing it indirectly via Venus and then using trigonometry to work it out. What is the reason for that?
| The strength of the radar signal falls rapidly with distance so for objects within the Solar System we are dealing with very faint reflected signals. That isn't a problem with objects like Venus because with suitable signal processing we can extract the radar reflection from the background noise. The problem with the Sun is that it's a (very) strong emitter of radio waves and this black body background completely swamps the radar reflection.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/335678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
What is the accepted meaning of atomic orbitals and elementary particles in Quantum Field Theory? If elementary particles are represented as oscillations of their respective fields, why are atomic orbitals said to represent the probability of finding an electron at a specific location in the electron cloud or orbitals like it is a solid particle? Isn’t it more plausible or intuitive to think of the atomic orbitals (wave functions, oscillations) as the actual particles (electrons) themselves? Like for example the way a single hydrogen atom is experimentally imaged here (specifically page 13, fig.3, if you don’t want to look through the whole article):
https://link.aps.org/accepted/10.1103/PhysRevLett.110.213001
Why are the results described as probability distributions in the above article?
Or the way electrons are imaged here as something that looks like typical normal modes of oscillation (specifically last four pages of the first article and last page fig.4 in the second one):
https://arxiv.org/ftp/arxiv/papers/0708/0708.1060.pdf
http://portal.research.lu.se/ws/files/2746286/3224376.pdf
I understand that for example in the last two articles the images show the momentum distribution of the electrons, but wouldn't that distribution correlate to actual fluctuations of the substance of whatever the electron field would consist of?
Or if I have a totally wrong understanding please correct me.
| This is what I have found at
https://en.wikibooks.org/wiki/Quantum_Mechanics/Waves_and_Modes
An electronic orbital is a normal mode of oscillation of the electronic quantum field, very similar to a light mode in an optical cavity being a normal mode of oscillation of the electromagnetic field.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/335927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How does index of refraction have direction? I found it in my textbook that there are some quantities like the elastic spring constant and index of refraction that have magnitude as well as direction, but are not called 'vectors'. I wonder how they have directions - I can somewhat make some sense of the spring constant but the index of refraction part sounds ambiguous.
| Almost certainly what's being talked about here is the refractive index of an anisotropic material. In this case, there are preferred directions in the material in question and the refractive index, i.e. the reciprocal of the phase velocity of the wave, depends in general both on the wave's direction and polarization.
In an isotropic medium, the electric (and possibly, but very seldom, magnetic) constants are no longer scalars. For example, the electric constant can be a homogeneous linear map - the dielectric tensor described by a symmetric $3\times3$ matrix $ \boldsymbol{\epsilon}$ - that maps the electric field to the electric displacement:
$$\vec{D} = \boldsymbol{\epsilon} \vec{E}$$
and the electric field and displacement are not in the same directions.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/336153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is there a connection between Bertrands theorem and Chaos theory? Bertrand's theorem states
Among central force potentials with bound orbits, there are only two
types of central force potentials with the property that all bound
orbits are also closed orbits, the inverse-square force potential and the
harmonic oscillator potential.
Especially the notion of "closed orbits" reminds me of Lyapunov stability, a prominent concept of Chaos theory. Is there a connection between Bertrands theorem and Chaos theory? Can Bertrands theorem be derived using methods from Chaos theory?
| Demonstrating Bertrand's theorem has attracted lots of interest, and there is quite a bunch of proofs using various methods. My library of papers feature one article [1] which would qualify I think, but it is in French. Here is the published abstract in English
"When a point mass undergoes a central, attractive, gradient force, there exists a one parameter family of circular periodic orbits. Bertrand's theorem asserts that if all the orbits close to these circular orbits are periodic then the potential is Newtonian (i.e. proportional to $1/r$, where $r$ is the distance to the fixed centre of attraction) or elastic (i.e. proportional to $r^2$) (J. Bertrand. Comptes Rendus 77 (1873), 849–853). Following an idea of Michael Herman, we compute the first two Birkhoff invariants of this system along the circular trajectories for a generic potential; then we show how to derive Bertrand's theorem."
and here is a translation of mine of the phrase introducing the developments in the paper:
"In this demonstration, among the non-newtonians and non-harmonic potentials, what prevents the property of having only periodic orbits, comes either for the $1/r^2$ potential from the existence of a strict Lyapunov function, or, for a generic potential, from the existence of motions with two incommensurate frequencies (one precession frequency and one revolution frequency)"
The phrasing "strict Lyapunov function" is a literal translation which I hope makes sense in English: my knowledge of this field is flimsy!
[1] Jacques Féjoz and Laurent Kaczmarek, Sur le théorème de Bertrand (d'après Michael Herman), Ergodic Theory and Dynamical Systems 24 (2004), 1583-1589
https://doi.org/10.1017/S0143385704000434
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/336481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
Special relativity: train, dock and inclined beam I am sure there is a flaw in my reasoning; but how could I make it right?
• In a stationary wagon, an experiment is carried out: a photon is sent from point $A$ to point $B$ by reflection at $C$ (equidistant from $A$ and $B$); The values $H$ and $L$ are known.
Two synchronized clocks are operating in $A$ and $B$. In $A$, it is triggered at the emission of the photon; In $B$, it stops at the reception of the photon.
The recorded duration is $\Delta t_w$:
$$\Delta t_w=\frac{2x}{c} \quad{} \Rightarrow \quad{} 4x^2=c^2\,\Delta t_w^2$$
With:
$$4x^2=4H^2+L^2 \quad{} \Rightarrow \quad{}4H^2+L^2=c^2\, \Delta t_w^2$$
• The wagon is now supposed to move at a speed $\overrightarrow{v}$ along a dock where there is a fixed observer equipped with a clock. It triggers its clock at the emission of the photon and stops it at the reception. He sees the photon moving along an elongated triangle due to the moving of the wagon:
The duration recorded $\Delta t_q$ is expressed as a function of the distance $D$ traveled by the wagon and the distance $2y$ traveled by the photon :
$$\Delta t_q=\frac{2y}{c} \quad{} \Rightarrow \quad{} 4y^2=c^2 \,\Delta t_q^2 \quad{}\quad{} \Delta t_q=\frac{D}{v} \quad{} \Rightarrow \quad{} D=v\, \Delta t_q $$
With:
$$y^2=H^2+(\frac{D+L}{2})^2$$
• So we get :
$$4y^2=4H^2+(D+L)^2$$
$$c^2 \,\Delta t_q^2=c^2 \,\Delta t_w^2-L^2+(v\,\Delta t_q+L)^2$$
$$\Delta t_q=...$$
This should be wrong, isn't it?
| The problem with your analysis is that it has more unknowns than it has equations, which means you should not expect it to be able to determine the values of any of the unknowns.
Write (as you have) $\Delta t_w$ and $\Delta t_q$ for the transit time of the light beam, measured in the two frames. Write $L_w$ and $L_q$ for the distance from $A$ to $B$, measured in the two frames.
You have correctly shown that
$$\Delta t_w^2=L_w^2+4H^2$$
A similar analysis will show that
$$\Delta t_q^2=(D+L_q)^2+4H^2=(v\Delta t_q+L_q)^2+4H^2$$
These two equations involve the six quantities $v,L_w,H,\Delta t_w,L_q,\Delta t_q$. You want to take three of these ($v$, $L_w$ and $H$) as given and solve for the other three. But you have only two equations to solve for these three unknowns.
You need another equation, which means you need another thought experiment.
The only thing that saves you in the classic case where $L_w=0$ is that we then feel justified in assuming that $L_q=0$, which gives you the extra equation you need.
So if your question is: "What do I do from here?", the answer is: Find another thought experiment to give yourself one more equation. Then you can solve your system.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/336661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If Hawking radiation is true why isn't the entire universe glowing? It says hawking radiation appears due to a particle and anti-particle pair one of which gets sucked in and other one escapes , these particles are said to appear out of nowhere , appear everywhere in the universe and are opposite of each other.
As such in normal universe away from a black hole why doesn't their recombination and Annihilation lead to emission of energy as light or heat like combination of matter and antimatter should?
Why isn't the universe glowing and all of us dead because of the energy from recombination of these particles everywhere as hwaking suggests?
| Hawking radiation comes from the region near a black hole, it is not produced 'throughout space' as you suggest. In this respect the situation is not very different from a star, where the emitted radiation comes from the outer layers. In the case of the black hole it is the region fairly near to but outside the event horizon that radiates.
Next, Hawking radiation itself is mostly electromagnetic. The numerous popular articles which say that Hawking radiation is particle/anti-particle pairs in which one of each pair gets swallowed by the black hole are misleading. Hawking radiation is almost entirely electromagnetic radiation---that is, photons. You can still use the language of particle/anti-particle, but it is a bit misleading.
Finally, Hawking radiation is extremely dim for black holes of ordinary size. For a black hole small enough to emit brightly, the emission process itself uses up the energy of the black hole and it evaporates. So overall one does not expect much Hawking radiation in total from all the black holes of the universe, compared to other sources of light such as stars (except there remains some room for uncertainty about this since it is not yet known very well what the distribution of small black holes may be).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/336913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Why is $\gamma^5$ used to define the projection operator? The properties of the projection operators are defined as:
$$P_+ = \frac{1}{2}(1+\gamma^5)$$
$$P_- = \frac{1}{2}(1-\gamma^5)$$
where $\gamma^5 = -i\gamma^0\gamma^1\gamma^2\gamma^3$
and their key properties are that $P_+^2 = P_+, P_+P_- = 1, P_-^2 = P_-$.
But since $\gamma^0$ has the same property as $\gamma^5$ that ${\gamma^0}^2=1$, if we replaced $\gamma^5$ with $\gamma^0$ in the definition of the projection operators, the newly defined projection operators would also satisfy all the key properties. And it's not like ${\gamma^0}^2=1$ is basis dependent; to show that ${\gamma^5}^2=1$ we actually need to make use of that fact. Why do we then bother to define $\gamma^5$ at all if we could use $\gamma^0$ to define the projection operators?
|
Why do we then bother to define $\gamma^5$ at all if we could use $\gamma^0$ to define the projection operators?
The key observation is that chirality should be preserved under the Lorentz transformation. Hence the $X$ in the chiral projection
$$
P_{\pm} = \frac{1}{2}(1 \pm X)
$$
should commute with 6 Lorentz algebra elements
$$
\gamma^{\mu}\gamma^{\nu}.
$$
Out of all the combinations (15 of them: 4 vectors $\gamma^{\mu}$, 6 bivectors $\gamma^{\mu}\gamma^{\nu}$, 4 pseudovectors $\gamma^{5}\gamma^{\mu}$, 1 pseudoscalar $\gamma^{5}$) of gamma operators, only the pseudoscalar
$$
\gamma^{0}\gamma^{1}\gamma^{2}\gamma^{2}
$$
commutes with ALL 6 Lorentz (antisymmetric) bivectors $\gamma^{\mu}\gamma^{\nu}$, while
$$
\gamma^{0}
$$
does NOT.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/337288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Paradox from mass transfer in a close stellar binary I'm considering a simple binary system made of two stars, of mass $m_1$ and $m_2 < m_1$, on circular orbits around their center of mass. Using Newton's theory of gravitation, it is easy to prove the following formulae for the total mechanical energy and angular momentum of the whole binary ($a \equiv r_1 + r_2$ is the separation of the two masses) :
\begin{align}
E \equiv K + U &= -\, \frac{G m_1 \, m_2}{2 a}, \tag{1} \\[12pt]
L \equiv L_1 + L_2 &= \frac{m_1 \, m_2}{M} \, \sqrt{G M a}. \tag{2}
\end{align}
Suppose that one star is transfering matter to the other star, without any loss. Conservation of matter implies $M = m_1 + m_2 = \textit{cste}$, so $\dot{m}_2 = -\, \dot{m}_1$. Now, many papers/lectures I've found state that angular momentum is still conserved : $\dot{L} = 0$. This gives an equation for the rate of change of the distance $a$ (the orbits are slowly evolving, while remaining approximately circular) :
\begin{equation}\tag{3}
\frac{\dot{a}}{a} = -\, 2 \Big( \frac{\dot{m}_1}{m_1} + \frac{\dot{m}_2}{m_2} \Big).
\end{equation}
See for example these references :
http://www.ast.cam.ac.uk/~pettini/STARS/Lecture18.pdf (see page 8)
http://jila.colorado.edu/~pja/astr3730/lecture32.pdf (see page 12)
Equation (3) implies the following :
\begin{equation}\tag{4}
a(t) = a(0) \Big( \frac{m_1(0) m_2(0)}{m_1(t) m_2(t)} \Big)^2.
\end{equation}
My trouble comes from conservation of energy, if the system is really assumed to be isolated (no loss/gain of wathever). Then the derivative of equation (1) above gives this (from $\dot{E} = 0$) :
\begin{equation}\tag{5}
\frac{\dot{a}}{a} = \frac{\dot{m}_1}{m_1} + \frac{\dot{m}_2}{m_2},
\end{equation}
which is incompatible with equation (3) (unless of course $\dot{m}_1 = -\, \dot{m}_2 = 0$, but then there is no mass transfer).
So what is going on here ? Why angular momentum should be conserved
while energy is not ? Why not the reverse, i.e. that energy is
conserved but not angular momentum ? The authors of the lectures
cited above says nothing about the binary's mechanical energy.
EDIT : About the total angular momentum, if we add the contributions from the size and rotation spin of the stars, we get the true conserved total angular momentum. I'm assuming tidal locked stars, with parallel rotation and revolution axes : $\omega_{\text{rot} \, 1} = \omega_{\text{rot} \, 2} = \omega_{\text{orbital}} \equiv \omega$ :
\begin{align}
L_{\text{tot}} &= L_{\text{orbital}} + S_1 + S_2 \nonumber \\[12pt]
&= \frac{m_1 \, m_2}{M} \, \sqrt{G M a} + (I_1 + I_2) \, \omega, \tag{6}
\end{align}
where $\omega$ is given by Kepler's third law :
\begin{equation}\tag{7}
\omega = \sqrt{\frac{G M}{a^3}}.
\end{equation}
I think that the spin contributions and their time variations are negligible in front of the orbital angular momentum, since $I_1 \propto m_1 \, R_1^2$, and $R_1 \ll a$ (same for star 2).
Any clue on this ?
| One star cannot transfer mass (conservatively) to the other without the material losing specific angular momentum.
In practice what happens is that an accretion disk forms around the accreting star. Viscous processes occur which transfer angular momentum outwards and allow mass to flow inwards. The viscous processes result in heating and energy losses.
When considered as a whole it is reasonable to assume the system conserves angular momentum, but it must lose energy.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/337578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why is light bent but not accelerated? Light is bent near a mass (for example when passing close to the sun as demonstrated in the famous sun eclipse of 1919). I interpret this as an effect of gravity on the light.
However, it seems (to me, at least) that light is not accelerated when it travels directly toward the (bary-)center of the sun. The same gravitational force applies yet the speed of light remains constant (viz. $c$).
What am I missing?
| In Einstein's 1911 paper "The influence of gravitation on the propagation of light", he pointed out that from his generalizrd theory of relativity, the speed of light as viewed from us is different at different locations in a gravitational field. From the Huygens principle, a change in the speed of light causes the wave front to tilt over toward the sun. The exact amount of the deflected angle of star light passing near to the sun is by his calculation 0.85 arcsecond(which he later corrected it to be 1.75 arcsecond). The bending of light path near to the sun is thus caused by the variable speed of light.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/337945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "90",
"answer_count": 6,
"answer_id": 5
} |
Schwarzschild geometry, what is physical meaning of coordinates? A past exam has a question:
For the Schwarzschild metric external to a non-spinning spherical mass, what is the physical significance to the coordinates $t,r,\theta,\phi$?
Not sure how to answer this question, I am thinking there is some obvious canonical answer, but it feels very non-specific.
Is the answer something like, an observer at $r\to \infty$ has that $t$ and $\tau$ are the same, and $r,\theta,\phi$ are all to some extent arbitary?
| $t$ and $\phi$ are adapted coordinates to the generators of unidimensional translation symmetry in time and rotation symmetry in $\phi$. Noether's theorem, in this particular setting the Killing equation, dictates that conserved currents are related to those symmetries. The related charges are the Komar mass and Komar angular momentum. The first one is the canonical mass of the Schwarzschild metric and the latter is zero.
So one "physical" significance of $t$ and $\phi$ are those two conserved quantities.
The typical coordinate choice of $r$ and $\theta$ has the consequence that $g_{\phi\phi}=\sin^2(\theta)g_{\theta\theta}$ and $g_{\theta\theta}=r^2$ which makes the angular part of the metric the canonical metric of a sphere and $r$ is the areal radius. This has been pointed out by @Uldreth.
So in a way the typical coordinate choice is neatly tied to the spacetime symmetries and the metric is relatively simple. That being said depending of the application other coordinate choices (isotropic, Gullstrand–Painlevé coordinates) are better suited.
A decisive answer is difficult because I personaly find that question rather vague because "physical significance" is in my opinion a rather board term and in the end all physics has to independent of coordinate choice.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/338280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Where is a classical computer better than a quantum one? Where is a classical computer better than a quantum computer? Is there any known domain where classical algorithms always beat quantum ones, say, both in terms of time and space complexity?
If yes, could you please give me examples?
If no, could you please provide me with a link to the prove?
| You can simulate a classical computer on a quantum computer (with essentially no overhead), but not vice versa. Here's a link to a list of quantum gates that have been found to be useful:
https://en.wikipedia.org/wiki/Quantum_gate
If you look close to the bottom, you can find an implementation of the quantum Toffoli gate.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/338402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Average induced emf in a rotating coil after rotating by 180 degrees When a coil rotates in a magnetic field an emf is induced, and when the coil starts its motion from the position at which its plane is parallel to the field lines, and then it rotates by 180 degrees, the average induced emf in it is zero. I don't understand that point, and I would appreciate if someone explains it for me.
| For this question we must understand that the EMF produced is directly proportional to the rate of change of Magnetic flux linkage.
As the field is parallel to the plane at first we can write that the magnetic flux density, B=k ( assuming be a constant).
Then the function of the flux through the coil will be Flux,F = BAsin(theta)=kAsin(theta)
Note: We take sin as the function because we know t when field is parallel to coil, flux is zero so thus taking sin function which is zero when theta is zero.
Now EMF= -N(dF/dt)= -N(dF/dtheta )x(dtheta/dt)
Now as you will rotate the coil at a constant angular velocity so dtheta/dt will be constant, assume it to be c( it is essentially angular velocity which remains constant as you are rotating at a constant speed here).
So now EMF=-Nc(d/dtheta(kAsin(theta))=-NckAcos(theta)
So the derivative of sin is cos so in interval 0 to 180, cos changes sign and has equal parts positive and negative so as the EMF is now a function of cos(theta) it's average value over interval from 0 to 180 degrees is 0
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/338512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Does a photon also come to rest after some time while travelling through a medium? If you fire a bullet in air, after some time it comes to rest because of air resistance forces. But what about photons? If they are also particles, then the same thing should happen with them. It does not matter how long it takes but they should come to rest as well while travelling through a medium. Does this happen or not?
| A photon is a particle but it has no mass, that is, mass is zero. Because it's mass is zero, it must always travel at the speed of light: $299,792,458$ m/s. Thus, it is never at rest -- Never.
However, this does not mean that a photon is constantly moving forever. Photons interact with other particles that just get into their way. For example, a photon can strike an electron in an atom be completely absorbed with the result that the electron is knocked into a higher energy level representative of the energy given to it by the photon. The photon itself at that point ceases to exist.
But, usually that excited electron is not stable at that higher energy and it will fall back down to the lower energy level where it was before and another photon of energy will be released and this new photon then continues traveling at the speed of light until it interacts with yet another particle.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/338657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
In the proof of the parallel axis theorem, why is the averaged $x$-coordinate "zero by construction"? According to Wikipedia to prove $I=I_{cm}+md^2$:
$I_\mathrm{cm} = \int (x^2 + y^2) \, dm.$ $$I = \int
\left[(x + d)^2 + y^2\right] \, dm$$
Expanding the brackets yields
$$I = \int (x^2 + y^2) \, dm + d^2 \int dm + 2d\int x\,
dm.$$ The first term is $I_{cm}$ and the
second term becomes $md^2$. The integral in the
final term is the $x$-coordinate of the centre of mass, which is zero by
construction.
I'm confused as to why the third term is 0, probably because I don't understand how the $dm$ mass axis of integration is defined.
| It's zero by definition of the centre of mass.
We define the centre of mass as the unique point $\vec{r}_{CM}$ where $\int dm (\vec{r}-\vec{r}_{CM}) = 0$.
Then we define the moment of inertia about an axis passing through the point $O$ with position vector $\vec{r}_O$ as:
$I_O = \int dm (\vec{r} - \vec{r}_O)^2$
In this expression we add and subtract $\vec{r}_{CM}$ to get:
$I_O = \int dm (\vec{r} - \vec{r}_{CM} + \vec{r}_{CM} - \vec{r}_O)^2$
Relabelling $\vec{r}_{CM} - \vec{r}_O = \vec{d}$ gives $I_O = I_{CM} + Md^2 + 2\vec{d} \cdot \int dm (\vec{r}-\vec{r}_{CM})$
The last integral is the one you had represented as $\int dm x$ Presumably your $x$ is defined relative to the centre of mass, I've just made this explicit by having everything defined in terms of vectors with an arbitrary origin. The last integral vanishes by comparison to the definition of centre of mass at the beginning.
P.S. all the integrals with respect to mass are defined as $\int dm = \int dV \rho $ where both integrals are taken over the body in question and $\rho$ is the mass density in the volume considered. (Or alternatively, they're defined over all space, and the fact that $\rho$ vanishes when you're not in the body causes those contributions to vanish).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/338742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What is Gray, from a physics POV? Quora explains how white and black colors fit into the spectrum of visible light. It explains that white is all colors together while black is the lack of color.
So, where is Gray? Gray is the mix of all-colors and no-color!! What does that mean? Can somebody explain Gray, the mysterious color?
| All colors are only in the mind. Light has a mix of wavelengths, but it doesn't have color until someone sees it.
When light enters the eye, it hits rods and cones in the retina. Cones are color receptors. There are three kinds. Each kind is sensitive to a range of wavelengths. Color is the result of stimulation of the cones, and additional processing in the brain.
The image is from The Color-Sensitive Cones at HyperPhysics. Copyright by C. R. Nave, Georgia State University. A good starting link is the Light and Vision page.
Loosely, the sensors are sensitive to long, medium, and short wavelengths. The ranges overlap. Most light, even single wavelength laser light, stimulates more than one. The graph shows which are stimulated by single wavelength light at different wavelengths. The colors we see are determined by the mix of stimulations. The bottom of the graph gives names of colors for single wavelength light.
Grey is not on the list. Grey requires a mix of wavelengths that stimulate the three types more or less equally. So do black (very little stimulation) and white (more).
There is more to it than that. The perception of color is affected by colors around it. There are photographs where two different patches reflect the same light. But the colors we perceive are different because of the surroundings. For example, see the Checker Shadow illusion.
By Original by Edward H. Adelson, this file by Gustavb [Copyrighted free use], via Wikimedia Commons
Also no wavelength will stimulate only the "Green" cones. They are always stimulated in combination with other cones. I once read it is possible to stimulate them with a probe. The person saw a color he had never seen before. I wish I could find a link. Quora might be a good place to start.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/339130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
} |
Subsets and Splits