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Does EMR emanate in all directions? Does the wave expand like a sphere surface whose radius grows at $c$? I made a commitment to understand special relativity.
Right now I'd like to understand Electro-Magnetic Radiation. What would it "look like"if I could see it emanate. From a book (Simply Einstein) I'm told any movement of an electric charge (e.g. an electron) can start an electromagnetic wave. Now that is cool. So, finally, my question - once a wave is started does it emanate in all directions (a rock thrown in a lake goes in all directions in 2 dimensional lake surface, so would shaking an election start an emf wave that expands as an ever-growing surface of a sphere. That is, a sphere whose radius is growing at the speed of light? If so, where is the photon on this sphere?
| You are right; when an electric charge is forced to vibrate in all directions, it gives off electromagnetic radiation in all directions. You can think of this as consisting of a large number of photons being sprayed in all directions. That radiation then moves away through space as an expanding spherical wavefront, traveling at the speed of light.
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Can I change solid into liquid, gas or solute by cutting again and again until it is a molecule particle? If I have small piece of solid, for example, pure iron powder, fine sand , (at room temperature) then I cut it slowly into half again and again. Every times, after I cut it, I wait for temperature back to room temperature again.
Case 1: If I do this in vaccuum, is it change to liquid or gas ?
I think it might be impossible because intermolecular force will make it form solid again. If it is impossible what is the smallest size (approximately) that I can get and is it look like viscous liquid ?
Case 2: If I do this in water or in polar liquid, is it change to solute ?
Case 3: If I do this in carbon tetrachloride or in nonpolar liquid, is it change to solute ?
| There is no free lunch. The mechanical breaking of bonds by cutting will take as much energy as melting or boiling and the final temperatures should also be the same. It may seem impossible to get a liquid this way, but assuming a perfectly insulated sample, it will. The act of cutting will generate higher energy molecules that are vibrating or moving linearly. This is the definition of heat, so the answer to doing without heat is no, you can't.
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What would qualify as a deceleration rather than an acceleration if speed is unchanged? The instantaneous acceleration $\textbf{a}(t)$ of a particle is defined as the rate of change of its instantaneous velocity $\textbf{v}(t)$: $$\textbf{a}(t)=\frac{\mathrm{d}}{\mathrm{d}t}\textbf{v}(t).\tag{1}$$ If the speed is constant, then $$\textbf{a}(t)=v\frac{\mathrm{d}}{\mathrm{d}t}\hat{\textbf{n}}(t)\tag{2}$$ where $\hat{\textbf{n}}(t)$ is the instantaneous direction of velocity which changes with time.
Questions:
*
*According to the definition (1) what is a deceleration?
*In case (2), when will $\textbf{a}(t)$ represent a deceleration? For example, in uniform circular motion, why is it called the centripetal acceleration and not centripetal deceleration?
| According to this definition, "deceleration" is undefined.
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Is it possible to break any given matter down into protons, neutrons, and electrons and then reorganize those particles into any other form of matter? I was reading about this somewhere but can't find the site now. It was talking about using 3d printers in the future that can intake any matter (dirt, garbage, etc..) and deconstruct it on the atomic level down to just electrons, protons, and neutrons and then use those left over subatomic particles to create anything by rearranging those particles into the needed formation for whatever it is you want to make. This would surely have a huge impact on the way we live since you'd be able to use your own human waste for example to make anything you'd want as long as there were enough particles. But is this theoretically possible?
| Yes, this is theoretically possible. In fact, it has already been done for certain transmutations. But even when turning lead into gold, at present the cost of the process (at least that process) is prohibitive.
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Ultra low level light detection: What measurement principles are used? I was reading up on how Avalanche Photodiodes(APD) are used to detect very low light levels, some are even used in "photon-counting" experiments (in Geiger mode). However, a survey of some commercial APDs such as this indicates that these sensors also have non-negligible dark currents for the large reverse voltages they operate in. The one in the link for example, has a nominal dark current of 50pA, in intensity terms it would be ~1pW (M=100, 0.5A/W sensitivity at 850nm). This is actually pretty high. Even if we treat this as an offset, say, for example in a charge integration based measurement, the noise in this bias current can mask any detection of ultra-low light levels(~1fW). So what techniques are used in signal conditioning? Any links or pointers would be greatly appreciated.
| If background is a problem, there are photomultiplier tubes. But those are bulky and delicate vacuum tubes and require high-voltage power supplies (about 1000 volts).
For solid state detectors, one can use the avalanche photodiodes in an array, where each diode has low area and low dark current. When there is a small flash of light (for example from a particle in a scintillator material), several of the photodiodes in the array will detect a photon at the same time. So adding the output from all diodes together will create a peak that is above the dark current. This kind of "pixels" cannot be used for imaging. The device is called SiPM (silicon photomultiplier).
Another method is gating to only look at the photodiode in the fraction of a microsecond that one knows that there might be a signal (for example, when a laser pulse in the experiment is firing).
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How fast are cyclotrons? I know this question is kind of vague, but I just want have a handle on how fast a average cyclotron can accelerate particles, and what kind of limits there are...
| The cyclotron depends on the fact that the angular frequency is a constant given by $\omega={qB\over mc}$. However, that equation is in the non-relativistic limit.
The correct relativistic equation is $\omega={qB\over mc\gamma}$, so $\omega$ is not a constant when the relativistic parameter gamma increases from its nonrelativistic value of 1. $\gamma$ is related to the energy of the particle be accelerated by the equation $\gamma={T+mc^2\over mc^2}$. This means that the cyclotron will stop working when the kinetic energy T becomes too large. That is, the cyclotron requires that ${T\over mc^2}\ll1$.
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Size of the Universe at end of recombination (~477000 years)? What was the size of the Universe at start of recombination (377000 years) when the Universe started to become transparent and release of trapped photons into what we now know as the CMB and till end of this phase (~477000 years)?
| The redshift at (re)combination is about $z=1100$. The relationship between scale factor and $z$ is $a = (1+z)^{-1}$. So the universe was about 1/1100 of the size that it is now.
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In an experiment, the transducer showed -1.9 Bar pressure. How is it possible? I was doing an experiment on water hammer in pipes. During the experiments, in several cases, the transducer reported negative pressure, as low as -1.9 bar gauge! But I have read that it is not possible for the pressure to practically go beyond absolute zero. I thought it might be due to some calibration issues of the transducer, and recalibrated and double checked the experimental values. I did the same experiment again. But the pressure shot down to 1.9bar or 190 Kpa below gauge, when cavitation occurred inside the pipe.
The pressure transducer i am using is a 0-30 bar transducer (forgot the company name), and i am measuring the values at 50,000 samples per second. the operating pressure inside the pipe in steady flow conditions is 1.7bar gauge and it is being shown correctly.
Can anyone give a possible explanation for the same?
The graph denotes the actual pressure vs time graph obtained during the analysis. the sampling frequency is 100000Hz and the Amplitude shows pressure value in bar, in gauge scale. The time scale is shown in 10microseconds value
| The sensor measures push or pull force on a plate, and, using the area of the plate, translates that into pressure. There are multiple ways to get a "pull" force on the plate without negative pressure. Inertia from moved sensor, internal pressure in the sensor,...
Sensors only perform as expected if they are used within their limitations. Imagine a diaphragm displaced by air under pressure. the amount of displacement somehow is translated into a pressure reading. Now that pressure is removed fast. The diaphragm will bounce back, and without any air to stop it, it will move further than its ususal "0 pressure"-position, for a short time, then resettle on the "0 pressure"-position. This will then be read as "negative" pressure.
The ability to answer to changing pressures, or pressures baove/below certain thresholds is one factor that drives the specialisation of sensors. (Medium, accuracy, drift, etc. are other factors.) Not every sensor will be able to measure every sort of pressure & pressure-change reliably.
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Experiment - does mass of a moving body really increase or is it invariant? Suppose we have a mechanical balance, with two identical particles kept in the two sides. Now the balance does not show any deflection.
Now, one of the particles is given some constant horizontal velocity.
Will the balance show the moving particle to be heavier (that side will move downward )or not?
(There is no friction between the balance and the moving particle)
| Your experiment doesn't actually distinguish between the two definitions, because the two definitions give equivalent dynamics.
In the "rest-mass" framework, the gravitational force between two objects, where one is stationary and the observer's frame and one is moving, is
$$F=\gamma \frac{GMm_0}{r^2}$$
where $m_0$ is the rest mass, because forces perpendicular to the velocity (as is the case here) transform as $F\to \gamma F$ under Lorentz boosts in this framework.
In the "relativistic-mass" framework, we get to keep $F\to F$ under Lorentz transforms, because $m=\gamma m_0$ is no longer Lorentz-invariant. So the force is still
$$F=\frac{GMm}{r^2}=\gamma \frac{GMm_0}{r^2}$$
The difference is merely convention. In one case, you associate the required $\gamma$ with the way that force (or more fundamentally, momentum) transforms under Lorentz boosts. In the other case, you associate $\gamma$ with the way that mass transforms under Lorentz boosts. The two formulations give consistent results.
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First excited state during Bose-Einstein condensate We know that the ground state is macroscopically occupied. What about the first excited state?
| I am writing this answer in continuation from my last answer on your previously related post. As described there, the macroscopic occupation of the ground state is enforced by the quantum statistics by bounding the number density of excited states. Note that consequently the first excited state is also bounded, and therefore there is a limit to which it can be occupied irrespective of the number density of the system. So there is no similar phenomenon of macroscopic occupation of the excited states here, rather the macroscopic occupation pertains only to the ground state which is free to accommodate as many bosons as possible; which isn't true of the excited states.
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Using complex exponential to represent waves in EM Ever since we've been using exponentials to work with electromagnetic waves, I've been confused about the imaginary portion and want to confirm my thinking.
What does the imaginary portion represent? Nothing, right? It's just a side effect of using complex exponentials because they are very easy to deal with algebraically. So, in reality we can completely restructure all the math to be written in terms of cos/sin instead and never let a single imaginary number appear, right?
| Yes, it is out of convenience and to simplify the equations. The implication is that the real part of the complex quantities is taken to get the actual physical value (which gives you some term that is a cosine with some phase).
This is analogous to the way we encode the phase of our quantities in complex alternate current calculations and similarly the amplitude is given by the absolute value of our complex quantity and the phase is given the argument of the complex number.
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London theory, an electromagnetic description? Currently I'm reading "Introducton to Solid-State Physics" by Charles Kittel, 8th edition and about superconductivity. I'm having a bit of trouble getting the whole plot, because as far as I'm concerned, the Meissner effect is an effect closely related to superconductors (let's assume a type II superconductor for the moment).
The magnetic field inside the superconductor, if we let the superconductor be a thin and long one, with axes parallel to the applied magnetic field $B_{ac}$ we can deduce that $B=0$ does not come from electromagnetic theory.
Later on I'm coming to the part where the London equations are actually trying to explain the effect of superconductivity once again, with electromagnetic theory, namely Maxwell's equations.
What am I missing? Where did I lose the plot on this one?
| The first part of the argument is simply saying that the Meissner effect cannot simply be explain as a consequence of perfect conductivity. It is an independent physical phenomenon that has to be explained separately. It is not saying that the entire theory of electromagnetism cannot be applied to superconductors.
The second argument uses the London equations to explain how the Meissner effect does occur. As part of that argument it clearly must, describe the magnetic field. The mathematical description of that field is given, as always, by Maxwell's equations.
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Confusion about condition of dark fringes in YDSE I have read that the condition for dark fringes in YDSE is when the path difference is
$$y_n = \left(n + \frac{1}{2} \right) \frac{\lambda D}{d}$$
where $D$ is the distance between the double slit pane and $d$ is the distance between the two slits. I saw a question about the 5th dark fringe being formed opposite to one of the slits and I had to find out the wavelength of the light used. Pretty simple. I applied the above formula and reached an expression, but found that my answer was in fact incorrect (not even in the options - it was an MCQ) and this formula had been used in the solution:
$$y_n = \left(n - \frac{1}{2} \right) \frac{\lambda D}{d}$$
Now I'm confused. Which of the above formulae is correct?
| Mathematically, both these expressions are equally valid, it only differs in terms of how you label fringes, or rather, how you start counting $n$. One can always redefine a new $n' = n - 1$, start counting from $n' = 0, 1, 2 \ldots$ instead of $n = 1, 2, 3 \ldots$. The maths doesn't change. (Actually, the $n \lambda$ part is redundant, it is only the additional $\pm \lambda/2$ that matters, and one would have destructive interference in both cases, whether plus or minus.) Physically, of course, this variable change is absurd, since you would now have a "zeroth" dark fringe, an absurd title, since the central fringe is bright in this case.
Physically, it may be useful to have the following picture:
In YDSE, the central fringe is bright, let us leave this aside for the time being from the counting procedures. On either side, the path difference $\Delta$ grows to $\lambda/2$ first, and hence we have a dark fringe. Thereafter, $\Delta$ becomes $\lambda$, and we have our first bright fringe, aside from the central one. Still further, $\Delta = 3 \lambda/2$, and we have our second dark fringe.
Thus, if you address this in terms of the physical basis, counting dark fringes as per the condition $\Delta = (m - 1/2) \lambda$, with $m = 1, 2, 3 \ldots$ seems more natural.
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Why do really cold objects evaporate really quick? So I saw a video of LNG (liquid natural gas) and when it got in contact with water, which was room temp the LNG evaporated instantly...why?
Ice takes a while to evaporate like a sec even when hot water is dumped...why do really “cold” liquids evaporate super quick. Is there a name to this phenomenon?
If you can dumb down your answer that would be great...I am only in 9th grade.
| I'll add to the answers above, (though @BobJacobson touched on this). Water has a high heat capacity. Air is very disburse and by contact area, a very very low heat capacity. (Heat capacity is usually measured by mass, not contact area, but I think contact area is important in relation to this question)
Water is able to transfer heat to the LNG and make it boil much faster than it boils in air. The same kind of thing happens to us. A person can sit in a 212 degree sauna for a period of time without being burned. 212 is hotter than most saunas are set, but that temperature is manageable for a healthy person for a few minutes if it's dry air. But put your hand in 212 degree water and you have a first degree burn in 2 seconds. Water, even cold water is able to transfer a lot of heat into LNG or onto a person's skin than air, and as a result, it boils very quickly in contact with water, even what we consider cold water.
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Change of wavefunction due to relativistic speed Imagine a spacecraft which is moving at a speed comparable to the speed of light relative to a reference frame with a hydrogen atom at it's origin. How would the probability distribution function of an electron in 1s orbit look relative to an observer inside the spacecraft?
| Technically, in order to incorporate special relativity with quantum theory, you need Quantum Field Theory.
The Lorentz transformation (in particular, boosts) is not unitary. This means that the wavefunction is no longer properly normalized. This is not the case with rotations, which are unitary. Trying to directly jam relativity into quantum theory results in problems such as infinite negative energy eigenstates, negative probability and so on. In QFT such problems can be resolved.
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Electromagnetic force (EMF) example I am currently studying this example from the book "Introduction to Electrodynamics" - David J. Griffiths.
According to the solution, there is an EMF applied to the metal disk and also there is a current flowing to the resistor.
I was wondering why this happens, since dΦ/dt=0 (there is a uniform magnetic field and the metal disk doesn't change position, it only rotates).
So, E=-dΦ/dt=0
I guess the EMF exists because of the fact that the disk rotates but I can not understand how the flux changes.
| The disc in your figure could be considered as being made up of large number of radial,conuducting ,differential elements rotating with angular velocity ω about the center of the disc.This is same as emf due to rotating rod which comes from motional emf.
Induced emf is the more general term. By Faraday's Law, you get an induced emf whenever there's a changing magnetic flux through a loop. If the changing emf is due to some kind motion of a conductor in a magnetic field, you would call it a motional emf. For example, if a loop moves into or out of a region of field, or rotates, or a bar rolls along a rail, you'd get a motional induced emf. But if the changing magnetic flux were due to, say, an increasing current in a wire, you wouldn't call it a motional emf.
Why does Motional emf change flux?
The reason could be that a potential diffference is maintained through the conductor as long as it moves through the uniform magnetic field.This is due to the electric field produced in the conductor due to movement of charges due to lorentz's force
Additional help from the website:https://webhome.phy.duke.edu/~schol/phy152/faqs/faq11/node2.html
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Einstein field equations to the Alcubierre metric I was wondering how Alcubierre derived the metric for the warp drive? Sources have said it's based on Einstein's field equations, but how did he go from this to the metric?
| The metric for the Alcubierre warp drive was constructed by considering the properties that it should obey, and not the matter source (which is why it's fairly unphysical).
The two ingredients used in it are :
*
*A bump function, so that the warp drive is localized in a specific region (and that bump function moves, so that the inside may move along with it)
*A widening of the lightcone in that bump function, so that, compared to the outside, the speed of light is "larger".
Given these two characteristics, we get the properties we want for a warp bubble. It is possible to also get variants by changing them, for instance the Krasnikov tunnel does not have a travelling bump function, but still has a widening of the light cone. This is why it is "static", compared to the warp drive.
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How to show that apparent velocity V(app) is always greater than -c/2 for an object fleeing away from you? I am currently taking the online course on Relativity and Astrophysics provided by Cornell on EdX.
There, how the apparent velocity of an object moving at a fraction of c might sometimes be greater than c was shown. But the thought experiment only described a situation where the moving object was approaching the observer. And the fact that V(app) for an object moving away from the observer in some fraction of light will always be greater than -c/2 was only stated. I tried to think of a situation where I would be able to visualize this but I couldn't. So, can anyone please describe a thought experiment where it can be seen that V(app) for an object moving away at a fraction of speed of light will be greater than -c/2?
If you want to see the exact thought experiment used in the course. This is the hand out. If you can use the same situation as in the handout, it would be even better. Gracias.
| Imagine you have a craft receding from you at basically $c$. At observer time $t$ after launch, the craft shines a light. We know where the craft is at that point in the observer frame, it's $ct$ distance away. The light returns to the observer, and provides evidence of the craft at that distance. But the light takes another time $t$ to return to the observer.
So the image of the craft at $x=ct$ is received at $t=2t$. The apparent speed of the craft based on that image is $\frac{x}{t}$ therefore $\frac{ct}{2t}$ or $\frac c2$
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Inertia on a rotating disc? If I toss a ball upwards in a train moving with uniform velocity, the ball will land right back in my hand. This is because the ball has inertia and it continues to move forward at the speed of train even after leaving my hand.
Now consider I'm standing on the outer edge of a rotating disc (merry-go-round). If I toss a ball upwards, it doesn't fall back in my hand. Why? Doesn't it have a rotational inertia (is that even a term?) to continue rotating even after I let go of it?
Is the ball going to land on a new location on the disc? Or is it going to fall away from the disc? At least the ball should have inertia of tangential velocity at which I tossed the ball upwards, right? So the ball should fall away from the disc? Can someone describe what happens in this situation?
| This example might help visualize your question:
Water drops are ejected from a rotating nozzle. Once a drop is ejected from the nozzle, it follows a parabolic trajectory (let's ignore air drag) in a vertical plane. The drops fall outside of the vase shape but we don't see them well.
We don't see this trajectory when we take a picture, though. What we see is the locations of all the drops at a given instant ("streaklines"). They seem to have the "rotational inertia" you're talking about even though it's just an illusion.
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Schroedinger equation on the line with non-symmetric double well potential In the 3rd volume of the Landau-Lifschitz text book ("Quantum mechanics") problem 3 after section 50 studies the Schroedinger equation on the line with symmetric double well potential, e.g. $U(x)=g(x^2-a^2)^2$ with $g,a>0$. In particular they estimate the difference between the first two lowest energy levels using the method of quasi-classical approximation. (The final result is not essential to my question, nevertheless see p. 184 in the 3rd edition of the book.)
As far as I understand, the essential point in the argument was that the potential is symmetric, i.e. $U(-x)=U(x)$. Apparently it is even more important that the potential $U$ has the same value in the two local minima (which are also the global minima).
I am wondering what happens in the case of the non-symmetric double well potential which has two local minima and assumes at them different values. For example I would be interested to see a proof of the following result I heard: if the wave function of a particle is localized near the local minimum where $U$ is larger (called a fake vacuum or metastable state) then after a long time the particle will be localized near the other local minimum (the true vacuum) due to tunneling.
I heard that metastable states are studied sometimes using eigen-functions of the Hamiltonian with non-real eigenvalues. A discussion of that would be of particular interest to me as well as a comparison with the standard approach using the usual spectrum of the Hamiltonian.
| Cute and interesting question.
What really happens is that the stable points are the global minima (if more than one) of the Hamiltonian, i.e. the states having the lowest eigenvalues.
The effect of tunneling can be seen by using the WKBJ method (just as is done when estimating the life time of alpha decays).
Thanks for the question, I hope my comment has been useful.
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Can air make shadows? I have read about schlieren photography, which uses the ability of non-uniform air to create shadows. Is it really possible that air makes shadows?
| If air particles are dense enough and or clustered together then yes. Shadows are not created so much as cast, kind of like looking in a mirror. The reflection is exactly that a reflection, if light can't penetrate an object it therefor casts around it. To the observer it looks like a shadow or a reflection.
Simplified version: When does this happen with air? When the particles in air have a greater concentration than light particles.
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
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What is the difference between relative time dilation and absolute time dilation I know special relativity says that traveling at high speeds (or really any speed) causes time dilation; and General relativity says that gravity also causes time dilation. I was wondering if relative time dilation (where two observers each measure the other's time to be slow) was caused not by time dilation, but instead because with the relative velocity difference between them, if they became increasingly far from each other, light would take longer and longer to reach them from the other. This would result in them both observing each other to have a slower time, though neither would necessarily experience the time dilation.
| Time dilation in special relativity does not depend on the distance of the two observers. It depends on their relative velocity $v$ only, according to $\Delta t'=1/\sqrt(1-v^2/c^2)\Delta t$. This shows that $\Delta t'$ is in fact delayed compared to $\Delta t$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Harmonic oscillator in microcanonical ensemble Consider a hamiltonian of a simple classical pendulum
$$H=p^2+\omega q^2.$$
How can quantities such as $\langle p^2 \rangle$ or $\langle q^2\rangle$ can be calculated using the microcanonical measure?
| As far as I understand this average values are
$$
<p^2> = A^{-1} \iint_{-\infty}^\infty p^2\delta(p^2+\omega^2q^2-E)\ dpdq,
$$
and similar for $<q^2>$, where $E$ is fixed energy of a pendulum and
$$
A = \iint_{-\infty}^\infty \delta(p^2+\omega^2q^2-E)\ dpdq
$$
One can calculate integrals with the $\delta$-function directly or one can use the symmetry of the problem. If $x = \omega q$, then due to the $\delta$-function we have
$$
<p^2> + <x^2> = E,
$$
and due to the symmetry we have $<p^2> = <x^2>$. Hence $<p^2> = E/2$.
P.S. I think there should be $\omega^2$ in the Hamiltonian, not $\omega$.
Update. About $<p^2> + <x^2> = E$ equality. Probability density function for the microcanonical distribution has the form
$$
\rho(q,p) = A^{-1} \delta(H(q,p)-E).
$$
Here $A^{-1}$ is a normalization constant.
Because of main $\delta$-function property the only possible states $q,p$ of the pendulum are those, for which $H(q,p)=E$. After change of the variable from $q$ to $x$ we have $p^2 + x^2 = E$ for any possible state. Average of a constant equals to the same constant. Hence $<p^2 + x^2> = E$. Average of a sum is a sum of averages and we get what we need.
About $<p^2> = <x^2>$ equality. It is obvious after change of variables $p \leftrightarrow x$ in the integral.
| {
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Line element in Kruskal coordinates I try to calculate the line element in Kruskal coordinates, these coordinates use the Schwarzschild coordinates but replace $t$ and $r$ by two new variables.
$$
T = \sqrt{\frac{r}{2GM} - 1} \ e^{r/4GM} \sinh \left( \frac{t}{4GM} \right) \\
X = \sqrt{\frac{r}{2GM} - 1} \ e^{r/4GM} \cosh \left( \frac{t}{4GM} \right)
$$
Wikipedia shows the result of the line element.
$$
ds^2 = \frac{32 G^3M^3}{r} e^{-r/2GM} (-dT^2 + dX^2) + r^2d\Omega^2
$$
I tried to calculate the metric tensor using $ds^2 = g_{ij} \ dx^i dx^j$. As $T$ and $X$ show no dependence in $\theta$ and $\phi$, the $d\Omega$ seems to make sense, but the calculation of the first component of $g$ was not working.
$$
g_{tt} = J^TJ = \frac{\partial T}{\partial t} \frac{\partial T}{\partial t} + \frac{\partial X}{\partial t} \frac{\partial X}{\partial t}\\
= \frac{1}{32} \left( \frac{r}{GM} - 2 \right) \frac{ e^{\frac{1}{2} \frac{r}{GM}}}{G^2M^2} \left( \cosh^2 \left( \frac{t}{4GM} \right) + \sinh^2 \left( \frac{t}{4GM} \right) \right)
$$
Is this the right way to compute the line elements?
What would be better way to calculate the line elements (maybe starting with the Schwarzschild-coordinates)?
| I would like to add that, contrary to the other answer, it is possible to find the metric using the Jacobian.
The transformation law for the metric tensor from Schwarzschild coordinates to
Kruskal–Szekeres coordinates is as follows:
$$g_{\bar{\alpha}\bar{\beta}}=\frac{\partial x^\alpha}{\partial x^\bar{\alpha}}\frac{\partial x^\beta}{\partial x^\bar{\beta}}g_{\alpha\beta}$$
where barred indices correspond to Kruskal–Szekeres coordinates.
We need to find the Jacobian matrix $\frac{\partial x^\alpha}{\partial x^\bar{\alpha}}$, as you have already done so. The mistake that you made was leaving out the metric terms $g_{\alpha\beta}$. This is a key part of the solution, as we are not transforming from Cartesian coordinates which have the Kronecker delta as its metric.
For example, the calculation for $g_{TT}$ would be as such:
\begin{align*}
g_{TT}&=\frac{\partial x^\alpha}{\partial T}\frac{\partial x^\beta}{T}g_{\alpha\beta}\\
&=\frac{\partial t}{\partial T}\frac{\partial t}{T}g_{tt}+\frac{\partial r}{\partial T}\frac{\partial r}{T}g_{rr}\\
&=16G^2M^2 e^{-\frac{r}{GM}}\left[-\left(\frac{X}{\frac{r}{2GM}-1}\right)^2 \left(1-\frac{2GM}{r}\right)+\left(-\frac{2GMT}{r}\right)^2 \left(1-\frac{2GM}{r}\right)^{-1}\right]\\
&=16G^2M^2 e^{-\frac{r}{GM}}\left(T^2-X^2\right)\frac{(2GM/r)^2}{1-\frac{2GM}{r}}\\
&=-\frac{32G^3M^3}{r}e^{-\frac{r}{2GM}}
\end{align*}
where I used $\frac{\partial t}{\partial T}=\frac{X}{\frac{r}{2GM}-1}$ and $\frac{\partial r}{\partial T}=-\frac{2GMT}{r}$ together with the Schwarzschild metric components in the third line.
I think you can figure out the rest.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Formula for spring constant $k$ Is there a formula for the spring constant $k$ which satisfies the equation $$F=k\Delta \ell~?$$
A similar question was asked but was answered by transposition of this equation... I'm hoping for an answer addressing the physical meaning of a spring constant.
Unlike the question which is marked as a duplicate, I was hoping for a fixed equation which encompasses all factors, rather than just a list.
| The elastic constant of the spring depends on the geometry of the spring. There are formulas relating spring constant to the spring dimensions for various conditions.
For a helical spring, a first approximation formula is
$$ k= \frac{Gd^4}{8ND^3} $$
where G is the shear modulus, d is the diameter of the wire, N is the number of coils and D is the diameter of the spring.
If the extension of the spring is "large" or the ration d/D is not small enough the formula needs extra terms. You can look up papers discussing the details of different springs.
For helical springs, google "On the spring constant of a close-coiled helical spring"
| {
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Does Frame Dragging exert a force? Can a planet in a stable clockwise orbit slow down into a spiral if orbiting a black hole spinning counter-clockwise?
| In this context you cannot really speak about "forces", since in any case the object will follow a geodesic through the curved spacetime.
However, there certainly is a noticeable effect of frame dragging on orbits. As you may or may not know, there exist a smallest possible circular orbit around any black hole. For a non-spinning black hole, for example, this orbit is at 3 times the Schwarzschild radius.
For spinning black holes the radius of this smallest circular orbit depends on the orientation of the orbit. In particular it is much smaller for prograde orbit than it is for retrograde orbits. You can certainly interpret this as the retrograde orbits being "destabilized into an inspiralling plunge" by the frame dragging effect.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Angle in pair production Assuming a very high energy photon (energy $E$) crosses the atmosphere and produces an electron-positron pair, I would like to know what is the angle between these to leptons produced. I was trying to calculate it by applying the energy-momentum conservation and realized that in this case the angle could be 0 if the momentum $p$ does not need to be conserved.
Question: Does $p$ need to be conserved in the interaction or is it enough that the following relation applies:$$
E^2=2\left(p_\text{e}c\right)^2+2\left(m_\text{e}c^2\right)^2
\,,$$where $p_\text{e}$ is the momentum of the resulting electron/positron and $m_\text{e}$ its mass?
| Momentum and energy have to be conserved in pair production. That's why it only occurs near a nucleus: in a perfect vacuum, the simple 'decay' of a sufficiently high-energy photon into an electron-positron pair doesn't conserve momentum.
The consequence of this is that the exact angle between the particles of the pair is not zero. Ideally, we can approximate this angle to $0$ by setting up conditions to ensure that the nucleus has negligible momentum after the collision.
But to predict the actual angle between the leptons, you'll need to know the other forces acting upon the nucleus, the mass of the nucleus, the exact energy of the photon, and something about the motion of the pair, probably velocity..
| {
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"timestamp": "2023-03-29T00:00:00",
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Why is this cloud blue? I saw these clouds on the horizon, behind a ridge (apologies I couldn't get more pixels):
Why is the front cloud darker than the cloud behind? There were no other clouds that I saw which could've been casting a shadow on the front cloud. What would cause a cloud to reflect less light?
| the cloud is "blue" because the microscopic droplets that make the cloud are bigger. Colors of clouds are determined by size of droplets, which governs the wavelengths of light they reflect. I'm sure others can post answers in more detail.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/408100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Mossbauer effect explanation I need to understand Mossbauer effect. Is there any simple explanation? So far I know, when an atomic nucleus emits a gamma-ray photon, the nucleus must recoil to conserve linear momentum.
Consequently there is a change of frequency of the radiation due to the movement of the source the Doppler effect.
What it is good for, this recoils absorption, or emission?
What are the consequences?
What is so special about Mossbauer spectroscopy?
| Starting from
So far I know, when an atomic nucleus emits a gamma-ray photon, the nucleus must recoil to conserve linear momentum.
is to begin with a classical expectation.
Instead say, "When a system emits a gamma ray the system must recoil", but allow that the system could be the single atomic nucleus or if the atom is part of a solid it could be the whole lump of solid matter.
That wouldn't be possible in classical physics because the recoil would have to be transmitted sequentially from one atom to another by inter-atomic forces, but in quantum mechanics the final state in which the whole crystal recoils coherently is consistent with the initial quantum numbers of the system so. But the Totalitarian Principle, holds that outcomes not forbidden must be possible, so outcomes with coherent recoil must happen. It remains to compute the rate (which is difficult), and determine how that effects the system (recoil energy must be deducted from the photon energy).
Getting the photon energy
Compute the final state for a single photon emitted from a system with mass $M^*$ resulting in mass $M < M^*$ in the rest frame of the initial state, by conserving energy and momentum
\begin{align}
M^* c^2
&= E_M + E_\gamma \\
&= \left(Mc^2 + \frac{{P_M}^2}{2M}\right) + E_\gamma \tag{1}\\
0 &= P_M - p_\gamma\\
p_\gamma &= P_M \tag{2} \;.
\end{align}
Where I have used the Newtonian kinetic energy for the slowly recoiling system.
Now, from (2) we find $P_M = p_\gamma = \frac{E_\gamma}{c}$, so we can re-write (1) as
\begin{align}
\frac{1}{2Mc^2} E_\gamma^2 + E_\gamma - (M^* - M)c^2 &= 0 \\
E_\gamma^2 + 2Mc^2E_\gamma - 2M(M^* - M)c^4 &= 0 \\
\end{align}
which leads to
\begin{align}
E_\gamma
&= \frac{1}{2} \left[ -2Mc^2 \pm \sqrt{4M^2 c^4 + 8 M (M^* - M)c^4} \right] \\
&= Mc^2 \left[ -1 \pm \sqrt{1 + 2 \frac{\Delta M}{M}} \right]\\
\end{align}
Expanding the square root about zeroto three terms gives
\begin{align}
E_\gamma
&\approx Mc^2 \left[ \frac{\Delta M}{M} - \frac{1}{2}\left(\frac{\Delta M}{M}\right)^2 \right] \\
&= \Delta M c^2 \left[1 - \frac{\Delta M}{2M} \right] \;.
\end{align}
The result is that the larger the system mass, the closer the photon energy is to the total available energy. (Keep in mind that we implicitly assumed $\Delta M \ll M$ when we chose the Newtonian kinetic energy at the start.)
Most sources then go on to make the further approximation of replacing $\Delta M$ by the gamma energy itself.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Does the neutral charged object attract the positive charged object or the negatively charged object? Consider an electrically neutral object:
*
*Is it going to attract a positively charged object or the negatively charged object?
*What is the type of attraction?
*How does it attract or why does not it?
*Why the positive protons of the atom attracts to the neutral neutrons and the negative electrons does not?
| Let's assume that we're talking about macroscopic objects, say particles of soot. The neutral object (call it 'O') will attract either a positively charged object ('P') or a negatively charged object ('N'). Here's why…
If P is placed near O, P will tend to pull the electrons in O towards the side of O nearest P. If O is a conductor, free electrons will move through O; if O is an insulator they will be pulled minute distances within each molecule while the nuclei are pushed minute distances away from P. The result in either case is that O will behave as a dipole with its negative charge closer to P than its positive charge. But P's electric field is stronger the closer we are to P. So P will pull the negative charge of the dipole more strongly than it repels the positive charge of the dipole. So O will experience a net attractive force towards P. And, of course, (appealing to Newton's third law) P will have a net attractive force towards O.
If you go through exactly the same argument, but with N instead of P placed near O, and remembering that N's electric field will be in the opposite direction to P's, (that is N repels electrons and attracts nuclei) you'll find that N, just like P, attracts O, so O attracts N.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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When is it appropriate to solve the time-independent Schrödinger equation? I am currently going through Griffiths over the summer but I am a bit confused by one point and I don't have any instructor to ask, so I was wondering if you could help clarify. In Section 2.3, the harmonic oscillator, he writes: "it suffices to solve the time-independent Schrödinger equation."
Clearly, this is not sufficient in every case. I was wondering how we know a priori that
*
*it is sufficient and
*we are not missing some information by only solving the time independent case.
| If you can solve the time-independent case, it always suffices to do so, since the time-evolution of a stationary state is simply $\psi_n(t)=e^{i\omega t}\psi_n(0)$, and any state can be written as a superposition of stationary states.
You can be sure you are not missing anything because every self-adjoint operator$^1$ (like the Hamiltonian) has a complete orthonormal basis of eigenvectors. This is called the spectral theorem.
On the other hand, the time-independent Schrodinger equation is intractable in all but the simplest of cases, so you are forced to rely on approximations in most cases. This often involves the time-dependent Schrodinger equation.
$^1$There's some fiddly details here for a completely general self-adjoint operator, but they don't really make a difference from a physicist's standpoint.
| {
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If you are ontop of a giant heavy object that is falling from the sky, is it possible to jump and roll at the perfect time to survive? I don't know if gravity or the speed of the drop would prevent us from jumping, but would something like this be possible?
If it's not possible on the ground, is it possible in the ocean?
I'm sorry in advance if I used the wrong tag. I had a hard time figuring out which tag out of the physics tag to choose. If someone can kindly correct me, that would be awesome.
| Damage to a living being with bones, organs, tissues, in a collision
is a transfer of energy into breaking of materials. The more
energy transferred, the more breakage is possible.
So, a large (large volume) heavy object will determine the
terminal velocity of the fall (it acts as a parachute would),
but a parachute is a large light object; a large heavy
object is no parachute, it will have gravity energy
proportional to mass (volume, roughly), and lose energy to
atmospheric drag proportional to area. 'Large' implies the
area is smaller scale than volume (terminal velocity is very high).
So, if the ricocheting fragments of the 'object' and its
target aren't energetic enough to hurt you, if the impact
heating doesn't scorch you, your own personal kinetic energy
will be proportional to the square of the terminal velocity of
the object. No, a leg spasm won't transfer enough energy
to help much.
A grasshopper, on a baseball, would have an excellent chance (but
better if it hopped off the baseball while still at high altitude).
| {
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Are the 7 fundamental SI units able to differentiate between all elementary particles? More specifically, can the 7 base SI units express qualities like quark strangeness (of quarks) and quark color? How do these SI units differentiate between different quarks (charm, up, top...)?
| Properties like the strangeness of a quark are just numbers with no dimension. Quark color is mostly related to a category than to a quantifiable thing, is like saying positive or negative, red or blue. The SI units may help to differentiate between different quarks by measuring the respective rest masses and electric charges (in Joules/c^2 or Kg, and Coulombs, if you wish).
| {
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"timestamp": "2023-03-29T00:00:00",
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What gives mass to dark matter particles? Assuming that dark matter is not made of WIMPs (weakly interacting massive particles), but interacts only gravitationally, what would be the possible mechanism giving mass to dark matter particles? If they don't interact weakly, they couldn't get mass from interacting with the Higgs field. The energy of gravitational interactions alone does not seem to be sufficient to account for a large particle mass. Would this imply that dark matter consists of a very large number of particles with a very small mass, perhaps much smaller than of neutrinos? Or do we need quantum gravity to explain the origin of mass of dark matter?
| There are various ways dark matter could acquire mass that have nothing to do with the standard model weak force. For example, there are theories involving a hidden sector- particles that do not interact with the standard model gauge bosons at all, but have their own interactions.
Note that the Higgs mechanism is not required for all mass generation in the standard model. The massive gauge bosons acquire their mass through the Higgs mechanism, but there are models where the fermionic masses are acquired through different mechanisms. The source of mass for neutrinos in particular is unknown.
Without knowing what dark matter is, it is of course impossible to determine how it acquires mass.
If it has no interactions at all, there's no need for a mechanism to acquire mass. Explicit mass terms in the standard model Lagrangian are only a problem because they break gauge symmetry. If a field doesn't couple to the gauge fields, its mass terms don't break gauge symmetry, and the mass can just be added to the Lagrangian by hand.
| {
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Potential difference across a zero resistance wire So I started off with electrostatics and everything seemed nice and mathematical and justified and then "DC circuits" happened!
I just cannot understand the model of electron flow in electrical circuits. Here are my specific doubts-:
1) If potential difference across a tiny cross section of conducting wire is zero, then why on earth does electron flow across that cross section at all? Never mind potential difference across the whole circuit.
2) Is there a constant electric field across a wire connected to a battery? If yes then how is potential difference across a zero resistance wire constant? Shouldnt it be increasing? Doesnt it violate ohms law? If no, then why do electrons flow at all?
Please take time to consider these doubts and relieve me of my frustration. I havs searched through the net for this but every answer seems like beating around the bush. All of the 4 books I have consulted do not address these facts to my satisfaction.
Frankly I think nobody understands this.
| In ordinary circuit analysis (no superconductors involved) when we say the resistance of a wire is zero, we really mean it's close enough to zero that the voltage drop across it doesn't significantly affect the circuit behavior.
If you want to understand exactly how this could work, you could model each wire as a low-valued resistor, and then take the limit as the resistance value goes to zero.
If you want to know the effect of the actual resistance of the wire, you can easily calculate the resistance from the resistivity of copper (~$1.72\times 10^{-8}{\rm\ \Omega\ m}$) and the geometry (cross-section area and length) of the wire. If you design high power circuits, you will certainly run into situations where wire resistance must be considered in the design.
| {
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Why doesn't magnetomotive force have units of force? Why does it have units of Ampere-turns and not Newtons?
Is it a current, or turns mean metres?
| No, turns doesn't mean meters here. Unfortunately, electromagnetics uses many bizarre units. A force should be a force, not a current. The correct units of magnetomotive force are weber/m or emu/cm, which dimensionally is t/s^2 x t/s, which in a sense means force times t/s, rather than the simple t/s^2 for other forces; this is because the magnetomotive force is actually two-dimensional, not one-dimensional, as are other forces. Study the Reciprocal System and prove it for yourself.
| {
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Operators in infinite dimensions In page 64 of Shankar's Principles of Quantum Mechanics, there are a few lines that leave me doubtful:
It is worth remembering that $D_{xx'} = \delta'(x-x') $ is to be integrated over the second index ($x'$) and pulls out the derivative of $f$ at the first index ($x$).
In the above lines, $D_{xx'} = \delta'(x-x')= \frac{\mathrm d}{\mathrm dx} \delta(x-x')$. Here, $\delta$ is the familiar Dirac-Delta function.
What does "pulls out the derivative of $f$ at the first index ($x$)" mean?
Does it mathematically mean as follows?
$\int_{x-\epsilon}^{x+\epsilon} \langle x|D|x'\rangle \langle x'|f\rangle \mathrm dx' = \int_{x-\epsilon}^{x+\epsilon} \delta'(x-x')\langle x'|f\rangle \mathrm dx'$, which is equivalent to $\int_{x-\epsilon}^{x+\epsilon} \delta(x-x')\frac{\mathrm d}{\mathrm dx'} \langle x'|f\rangle \mathrm dx' = \frac{\mathrm d}{\mathrm dx}f(x) \int_{x-\epsilon}^{x+\epsilon} \delta(x-x')\mathrm dx'$, for the interval of integration is infinitesimal. We are left with the integration of $D_{xx'}$ with respect to the second index.
| Re. "What does "pulls out the derivative of f at the first index (x)" mean?"
The sifting property of the Dirac Delta distribution is
$$\int f(x)\delta(x-a)dx=f(a)$$
and for its derivative
$$\int f(x)\delta'(x-a)dx= -f'(a)$$
or
$$\int f(x)\delta'(a-x)dx= f'(a)$$
So for $\delta'(x-x')$ when its placed under the integral along with $f(x')$ and integrated wrt $x'$ the result will be $f'(x)$. Note that $\delta'(x-x')=-\delta'(x'-x)$ but $\delta(x-x')=\delta(x'-x)$. That's because $\delta'(-x)=-\delta'(x)$.
| {
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Symmetry of the 2d anisotropic Heisenberg model? I am getting confused about the symmetries of the 2d anisotropic Heisenberg model. The Hamiltonian is:
$$H=-\sum_{\langle i,j\rangle}(J_x S_i^xS_j^x+J_yS_i^yS_j^y)\tag{1}$$
I have read (source not publicly available) that this has symmetry $\Bbb{Z}_2$. Which although I agree with - I don't think is the full symmetry. Since as far as I can tell we have the following symmetry generators:
*
*Rotation about $z$ axis by $\pi$.
*Reflection about $xz$ plane
*Reflection about $yz$ plane
which does not appear to be simply $\Bbb{Z}_2$.
My question is therefore: What is the symmetry of the Hamiltonian (1)?
| Short Answer
Yes the symmetry group is larger then $\Bbb{Z}_2$ and is $\mathbf{k_4\cong \Bbb{Z}_2 \times \Bbb{Z}_2}$. But the ground states are only related by $\Bbb{Z}_2$ and it is this symmetry the get's broken in spontaneous symmetry breaking.
Long Answer
Let us look at the individual symmetry groups mentioned in the question and there generators:
*
*Rotation about $z$ axis by $\pi$: This has a single generator given by:
$$\pi_z=\begin{pmatrix} -1 &0 \\ 0 &-1\end{pmatrix}$$
*Reflection about $xz$ plane: This has a single generator given by:
$$ R_y=\begin{pmatrix} 1 &0 \\ 0 &-1\end{pmatrix}$$
*Reflection about $yz$ plane: This has a single generator given by:
$$ R_x=\begin{pmatrix} -1 &0 \\ 0 &1\end{pmatrix}$$
Note that since $R_x=\pi_z R_y=R_y \pi_z$ the total symmetry group is given by:
$$\langle \pi_z, R_y\rangle\cong k_4\cong \Bbb{Z}_2 \times \Bbb{Z}_2$$
where $k_4$ is the Klein four-group.
Now we note that the ground states (say all spins in the $+x$ or $-x$ direction) are related by the generator $\pi_z$ and thus the group $\langle \pi_z\rangle \cong \Bbb{Z}_2$. Whilst the generator $R_2$ leaves the ground states invariant. Thus it is this $\langle \pi_z\rangle \cong\Bbb{Z}_2$ which the above is likely talking about. We could equally view this group as being generated by the rotations $R_x$ since these are related by a transformation that leaves the ground states invariant - namely $R_y$.
As a side note spontaneous symmetry the generator $\pi_z$ (or $R_x$) is broken leaving the system with a symmetry $\langle R_y \rangle \cong \Bbb{Z}_2$.
| {
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What is the connection between Lagrangian symmetry and particle multiplets? I am struggling to see the connection between the symmetries of Lagrangians and particle multiplets. If I had three quark fields arranged in a vector $\Psi(x) = (u(x),d(x),s(x))$, these three fields have a combined Lagrangian
$$ \mathcal{L}= \bar{\Psi}(x)(i\gamma^\mu\partial_\mu-m)\Psi(x) $$
I can see that this Lagrangian has an $SU(3)$ symmetry. When I quantise my theory, I promote my fields $u,d,s$ to quantum operators on an infinite dimensional Fock space. The interpretation of the operators $\hat{u},\hat{d},\hat{s}$ is that they are particle creation operators that create up, down and strange quarks respectively. However, I have also read that quarks live in the fundamental representation of the Lie algebra of $SU(3)$.
My question:
From my studies of quantum field theory, I thought particles live in infinite dimensional Fock spaces, not finite dimensional representation spaces. How do I arrive at the particle multiplets from the symmetry of the Lagrangian?
| In general, a lagrangian can be invariant under more than one transformation so there can be several symmetries and representations related. Specifically, the one particle states furnish representations of the Poincare Group and therefore live in an infinite dimensional space. On the other hand flavor (or color) symmetry does not have anything to do with Poincare, the states live in another space, which is finite dimensional. There will be, for instance, many spaces related to a given quark: The infinite dimensional Fock space associated to momentum and spin, the three dimensional space associated to color and the two (left) and one (right) dimensional space associated to electroweak interaction gives you an explicit example.
| {
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If quantum cloning was possible would entanglement enable faster than light communication? Assume quantum mechanics allowed you to clone particles. How could you use quantum entanglement to communicate faster than the speed of light?
In the 1980s a scientist proposed using quantum cloning to send information between entangled particles faster than light. This led to the discovery that quantum mechanics prohibited quantum cloning.
| Imagine Alice and Bob share a Bell state:
$$
|\psi \rangle = |00\rangle +|11\rangle \ .
$$
Note that this state looks like this in the $X$ basis:
$$
|\psi\rangle = |++\rangle + |--\rangle \ .
$$
As you know if Alice measures her qubit in the $Z$ basis, and gets $|0\rangle$ or $|1\rangle$, Bob's qubit will collapse unto the corresponding state as well. On the other hand if she measures it in $X$ basis, Bob's qubit will also collapse to a state in the $X$ basis.
Now, imagine Alice and Bob have moved to opposite ends of the Milky way, and Bob has made a Billion copies of his qubit. Normally, Bob would be unable to distinguish what basis Alice has measured her qubit, but now Bob can measure half of his qubits in the $Z$ basis and the other half in the $X$ basis. The half that agree with each other shows the basis Alice has measured her qubit in. Therefore, assuming cloning arbitrary quantum states was possible, Alice could instantly send a bit of information to Bob on the other end of the universe.
| {
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Energy conservation on expanding universe Due to the expansion of the universe, the photons emitted by the stars suffer redshift, Its mean that the energy is lowered a little bit. Does this mean that the energy is lost? Does the expansion of the universe violate some conservation principles according to Noether's theorem?
| Redshift happens when the wavelength of the photon is increased, or shifted to to the red end of the spectrum. Energy is never lost but transferred.
So the answer to your question is NO - No energy is lost.
| {
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Why does a mirror reflect visible light but not gamma rays? Visible light (~500 THz) as well as gamma rays (~100 EHz) are electromagnetic radiation but we can reflect visible light using a glass mirror but not gamma rays. Why is that?
| Reflection is caused by electrons reacting to the electromagnetic field by oscillating at the same frequency. When they do this they emit radiation of the same frequency as the incoming light and this is observed as reflection. This works well if the EM frequency is near the eigen frequencies of the electrons. When the frequency is very high the electrons are simply too massive and the forces retaining them not strong enough - think of a mass on a spring - to follow the electric field. So gamma rays can pass through matter.
| {
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Why does a magnetic field "curl" around a current carrying element? From all the texts I've read, it's always stated that the magnetic field would either curl or warp around the the current flowing within a conductive element, yet, I never was clarified as to why that is. Why would the field "curl" around it? I'm curious if there is an explanation for that nature.
Amazingly, it seems like a perfect circle too, similar to the diagrams that I've studied.
| If you ignore all what you learned about the B-field being a "vector" and you treat it as what really is a 3D skew-symmetric tensor then the mystery goes away. In this view the B-field is a bi-vector, a surface-like quantity whose source is the current element from which it radiates outward, so to speak. Unlike the E-field that has lines of force coming out or ending in charges, the B-field has planes of force coming out or ending in currents and these "planes" form surfaces in which the magnetic action, i.e., attraction-repulsion and torque take place. This is analogous to the way the E-field acts along its lines of force. What conventionally is called the lines of force of the B-field are the orthogonal rays to these surfaces. A uniform current generates a uniform set of planes of action whose orthogonal rays are in fact circles, and they stay so approximately even when the source is curled up into a loop. You can see nice pictures of this in Roche: "Axial vectors, skew-symmetric tensors and the nature of the magnetic field", Eur. J. Phys. 22 (2001) 193–203.
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What would be the charge distribution of a conducting sphere in front of a positive point charge? What would be the charge distribution of a conducting sphere in front of a positive point charge? I mean if it's a positive charge then it should induce negative charge in the near side and positive on the other side. But as it's conducting then it should distribute the charge all over the sphere. So it should make the sphere nutral. Or something extra-ordinary might happen. Assume the sphere is isolated.
| The answer using the method of images is for a grounded sphere, but to correct for this you simply need to add a second image charge
$$
q ^{\prime \prime} = -q^{\prime}
$$
at the center of the sphere. This gives a total "induced" charge of
$$
q_{induced} = q^{\prime}+q^{\prime \prime} = 0
$$
so you have two image charges, one at the center of the sphere with charge $q^{\prime\prime} = q \frac{R}{L} $ and one at $ l = \frac {R^2}{L} $ from the center toward the point charge with charge. $q^{\prime} = -q\frac{R}{L} $.
| {
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As the universe expands, do we have any reason to suspect further separation of the fundamental forces/interactions? At some point, all four forces were one force. (another question: what exactly does that mean?). At some point gravity and the strong force separated out leaving the electroweak force. Then the electroweak force separated out to become the electromagnetic force and the weak force.
I assume we are not done with phase transitions. So are there any theoretical reasons to believe that there won't be any further separations? For example, the electromagnetic force separates into two forces.
How do we know that a force is "fundamental" and not separable?
A related question:
What does it mean to say that "the fundamental forces of nature were unified"?
|
At some point, all four forces were one force.
This is speculative. E.g., it's true in string theory, but string theory is probably wrong.
Others will probably be able to give a more definitive answer, but I think basically the reason the weak force separated out from the electroweak force is that there's a temperature corresponding to the masses of the W and Z bosons, about $10^2$ GeV. (Energy scales can be related to temperatures via the Boltzmann constant.) When the temperature gets below that point, you get a phase transition. If we had been intelligent beings who lived when the universe was hotter than $10^2$ GeV, we could have constructed the standard model, measured the masses of the W and Z, and anticipated that the phase transition would occur. The standard model doesn't have other fundamental bosons whose masses are nonzero but less than those of the W and Z, so I don't think we should expect anything like this to happen.
| {
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A ball attached on a moving string If there is a ball attached on a string and the string's point of hanging is accelerating horizontally at $\vec{a}$, what will be the forces exerted on the ball that is hanging? It is obvious that there will be a gravitational force downwards and a tension force, and there should be another horizontal force on the ball in the opposite direction of the acceleration of the string, but where does that force come from? It should be from the ball's inertia, but how can that be a force?
|
...and there should be another horizontal force on the ball in the opposite direction of the acceleration of the string, but where does that force come from? It should be from the ball's inertia, but how can that be a force?
This apparent force is known as a fictitious force. These forces are not present in an inertial frame, but seem to be there in accelerating reference frames. They are also proportional to the mass of an object.
Imagine that you are doing some experiments with a frictionless table on a train. You set a mass on the table and at some point in time the train begins to accelerate to the right.
You measure that from your reference frame (the train), that the mass accelerates to the left. An external (inertial) observer would see that the mass is stationary and that you, the train, and the frictionless table are all accelerating to the right.
To make $F=ma$ consistent in your frame, you are forced to introduce a bookkeeping or fictitious force on the mass that is directed opposite the train's acceleration.
The same is happening in the car. The car is accelerating in one direction. So if you use a reference frame where the car is at rest, it is a non-inertial frame and the fictitious horizontal force on the ball appears.
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What is $z$ in the Bose gas? I'm studying the ideal Bose gas, and I found this equation for the average occupation number of particles at the fundamental level:
\begin{equation}
\langle n_0 \rangle = \frac{z}{1-z}
\end{equation}
There's no degeneration due to spin, and $z$ is the fugacity defined as
\begin{equation}
z = e^{\mu/kT}
\end{equation}
But this is a problem because this makes $\langle n_0 \rangle$ negative.
Furthermore, I see that $z$ can take values from $0$ to $\infty$, but for $z<1$ that would mean either $\mu<0$ or $T<0$, and that doesn't seem physically possible. How do you define $z$ then?
| $\mu $ is the chemical potential and it is negative in the case of Bose gas.
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Movement within a magnetic field My class went over magnetism and electricity, and my professor repeatedly mentioned how the field moves from north to south, and he would say "they" and "it." I wasn't sure if he meant that the field itself was moving, or particles within the field were moving.
If my professor was referring to the field, then what does he mean? I'm confused as to how the field itself can move from north to south. Shouldn't the field be the general region in which there are magnetic forces, and therefore stationary? Or am I misunderstanding something?
I wasn't sure how to phrase this question, so sorry if the title is misleading.
| It seems that the way he was describing it is confusing you. First off, let's just clarify the assumption that we are dealing with the magnetic field from a fixed source (ie a magnet). Then:
*
*As you correctly stated, the magnetic field is stationary and is just a visual representation of how the magnetic field looks in space.
*The field doesn't "move" from north to south but it does point from north to south along the field lines. So a field line points away from the north side and towards the south side.
*Remember that (stationary) charges do not move in a magnetic field.
*The magnetic field applied a force on moving charges that is perpendicular to both the velocity and the magnetic field direction. You can determine the direction from the rh rule and noting that: $F = qv\times B$.
Let me know if this is not clear to you. I hope this clarifies that he probably mean point rather than move as well as some of the other subtleties around magnetic fields.
| {
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Does the Earth emit gravitational waves? We know about Bohr's model and his vague postulate challenging Rutherford regarding discrete orbits and not emitting electromagnetic waves during this.
Extending this idea to our solar system, does the Earth emit gravitational waves around its orbit of the Sun and if not why not, and if yes, will the Earth fall into the Sun?
| Presuming general relativity is correct, yes the Earth orbiting emits gravity waves. The intensity of this emission is quite low. The intensity of the gravity field the Earth creates is small. The speed the Earth moves around the Sun is small. So the rate of energy being carried away is very small. Thus the stability time of the Earth's orbit (w.r.t. this energy loss) is quite long. Other effects would almost certainly be larger. Tidal drag of the Sun on the Earth, for example. Probably little tugs from the gravity of other planets would be larger than the gravity wave effect. Probably the Earth's orbit will be stable enough for other dire things to happen first, such as the sun getting along in its life span, and drastically changing character. (I'm weaseling on that because I don't recall what is the expected fate of the Sun.)
| {
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Why is voltage described as potential energy per charge? Voltage is often called an electromotive force since it causes a flow of charge. However, it is described in terms of Joules per Coulomb or Potential Energy per Charge.
Question: How does the potential energy associated with charge contribute to its effect on the flow of charge?
High voltage, or high electromotive force, causes high current. So this means charge with high PE will cause high current. This doesn't seem to make sense to me. Why does potential energy affect current?
I know of the PE gradient explanation, but this doesn't make sense to me. In most cases of diffusion, there is an explanation as to why it occurs: particles diffuse from high to low concentration DUE to random particle movement. Things fall from high to low gravitational PE DUE to the force of grav.
Question: Charge moves from high to low PE in a circuit, but why? What is the driving force?
| Charges don't posses potential energy. To be more accurate a system possesses potential energy. In electrostatics, you have to distinguish between potential, potential difference and potential energy.
First of all potential energy of a system refers to the amount of energy spent in assembling the system from infinity. Electric potential at a point refers to change in potential energy of the system when a unit charge is brought to that point. Difference in electric potential at two points is called potential difference. What drives current is potential difference.
Second, in mechanics, force is caused by change in potential energy (more precisely gradient in potential energy). When two points have high potential difference and a charge goes from high potential to low potential, there is tremendous decrease in potential energy. This causes larger force on the charges, since force is proportional to gradient of potential energy, and consequently higher current.
| {
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Sign of work done by friction In Goldstein's classical mechanics (3rd ed.) we read:
"The independence of W12 on
the particular path implies that the work done around such a closed circuit is zero,i.e.
$$\oint \textbf{F}.d\textbf{s}$$
Physically it is clear that a system cannot be conservative if friction or other dissipation forces are present, because $F . d\textbf{s}$ due to friction is always positive and the integral cannot vanish."
My question is: why should the work due to friction be "always positive"? Shouldn't it be nonzero instead?
Also, $F . d\mathbf{s}$ is a typo and should be $\mathbf{F} . d\textbf{s}$ (please let me know if I'm wrong)
| The force of friction, a vector, opposes the actual or impending motion, and the work done by friction can be positive or negative depending on the situation.
We are addressing the mechanical work done by friction on the system $\int_{a}^{b} \vec F \cdot d \vec s$. Typically, physics mechanics assume a rigid body for which there are no "heating" effects (no change in internal energy).
Thermodynamics broadens the definition of work to be "energy transferred between a system and its surroundings, without mass transfer, due to an intensive property difference other than temperature". This broader definition of work includes mechanical work. Work, and heat, can change the internal energy of a system. The work done by the system is the negative of the work done on the system.
For friction, we are addressing the mechanical work done on the system.
| {
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Sound waves produced by drums I am searching for an explanation of how drums produce sound waves. What type are these waves and how do they spread in the air?
| Like all sound waves in air, the waves emitted by a vibrating drum head are called compression or longitudinal waves.
As the drum head vibrates upwards, it pushes on the air next to it, which pushes on the air next to it, and so on. then when the drum head vibrates downwards, it pulls on the air next to it, which then pulls on the air next to it, and so on.
The net result is a train of waves moving away from the drum head in a series of zones of slightly compressed air each being followed by a zone of slightly rarified air. these sound waves move away from the drum head in all directions, becoming weaker in strength as they travel away from their source.
This is a simplified picture of the process. The detailed physics of drum head vibration is more complex than this, and much has been written about it.
| {
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Newton's law requires two initial conditions while the Taylor series requires infinite! From Taylor's theorem, we know that a function of time $x(t)$ can be constructed at any time $t>0$ as $$x(t)=x(0)+\dot{x}(0)t+\ddot{x}(0)\frac{t^2}{2!}+\dddot{x}(0)\frac{t^3}{3!}+...\tag{1}$$ by knowing an infinite number of initial conditions $x(0),\dot{x}(0),\ddot{x}(0),\dddot{x}(0),...$ at $t=0$.
On the other hand, it requires only two initial conditions $x(0)$ and $\dot{x}(0)$, to obtain the function $x(t)$ by solving Newton's equation $$m\frac{d^2}{dt^2}x(t)=F(x,\dot{x},t).\tag{2}$$ I understand that (2) is a second order ordinary differential equation and hence, to solve it we need two initial conditions $x(0)$ and $\dot{x}(0)$.
But how do we reconcile (2) which requires only two initial conditions with (1) which requires us to know an infinite number of initial informations to construct $x(t)$? How is it that the information from higher order derivatives at $t=0$ become redundant? My guess is that due to the existence of the differential equation (2), all the initial conditions in (1) do not remain independent but I'm not sure.
| Power series expansion does not hold for all functions $f(t)$ or for all $t\in\mathbb{R}$, but only for real analytic functions and for $t$ in the radius of convergence. In particular, it does not hold at any point e.g. for functions $C^2(\mathbb{R},\mathbb{R}^d)\smallsetminus C^3(\mathbb{R},\mathbb{R}^d)$. Therefore it is not possible to define any function by giving countably many real numbers $(x^{(n)}(0))_{n\in\mathbb{N}}$.
In particular, Newton's equation may have solutions in $C^2(\mathbb{R},\mathbb{R}^d)\smallsetminus C^3(\mathbb{R},\mathbb{R}^d)$, that therefore do not admit a power series expansion, or in general solutions that are not real analytic for all times, and therefore that do not always admit a Taylor expansion. Nonetheless, these functions are uniquely defined by two real numbers ($x(0)$, $\dot{x}(0)$) and by being solution of Newton's equation (i.e. they are also determined by $m$ and the functional form $F$ of the force).
In case that a solution of the Newton equation is real analytic, then the value of the higher order derivatives in zero is determined uniquely by the solution itself, and thus they also depend only on $x(0)$, $\dot{x}(0)$, $m$ and $F$; no further knowledge is required.
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Relativistic mass of components gives system rest mass? To put it briefly, in the classic thought experiment of a massless box with mirrored insides containing photons, does the relativistic mass of the photons imbue the box with rest mass?
I take it that's the case, because I think that's how baryons are supposed to get their mass, but I'm not really getting how this is happening exactly.
| You can only say that if the components are non-interacting.
As soon as you introduce interactions between the components the mass of the system is neither the sum of the rest masses of the components nor the sum of the "relativistic masses".
I've done a moderately thorough treatment in a previous answer.
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In what ways can entropy of the universe increase? So far, I have been able to come across two scenarios in which entropy of the universe increases. First, when heat flows at a finite temperature gradient and second, during expansion of air particles into a vacuum (Joule's Expansion). Is there any other way in which entropy can increase? I'm assuming that there should be because neither of the above instances explains why entropy should increase during irreversible adiabatic compression. Considering an ideal case and neglecting entropy generation due to turbulence, there must be something $fundamentally$ different about irreversible adiabatic compression that generates entropy. Right?
| Bird, Stewart, and Lightfoot, Transport Phenomena, in Chapter 11, problem 11D.1 (Equation of change for entropy) discuss the fundamental causes of entropy generation/irreversibility. They identify heat flows at finite temperature gradients, viscous dissipation of mechanical energy (associated with viscous stresses resulting from rapid fluid deformation), and, in a later chapter, diffusive mass transfer at finite concentration gradients. The example you gave for Joule Thompson falls under the category of viscous dissipation; this also covers entropy generation associated with rapid adiabatic compression of a gas. For Newtonian fluids (such as air and non-polymeric liquids), the rate of entropy generation per unit volume from viscous dissipation is equal to the viscosity times the square of the "effective" velocity gradient, divided by the absolute temperature.
I recommend that you study the development in this example in Bird et al carefully. I personally found it extremely enlightening, and it cleared up all my gaps in understanding.
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What is the physics behind the movement of the split-finger fastball in baseball? A split-finger fastball, or a splitter, is a baseball pitch thrown like a fastball with an v-shaped split finger grip. When thrown correctly, it moves like a fastball, straight in the air, but suddenly drops as it approaches the home plate. Here is a gif of how the pitch moves.
I've searched various sources that explain the reasons, which aren't very convincing to me:
Reason 1: The ball has a reduced backspin as compared to a regular fastball, and hence falls due to a stronger gravity than the upward Magnus force.
This does not convince me. This scenario should result in a constant force on the ball throughout it's path of motion, and the descent should be gradual, and not be seen at the end of the pitch.
Reason 2: The slight backspin on the ball forms ripple vortexes as it travels through the air, which eventually leads to the ball developing a topspin, which causes the ball to drop
This reason seems to be more plausible, and can explain why the ball drops suddenly and not gradually. However, I'm not so convinced if this is a very accurate reasoning or not. A pitch known as the knuckleball, thrown with very little spin, utilizes these vortices to make the ball move unpredictably. I'm not certain if the slight backspin allows the pitcher to control the effect these vortices have.
What is the most accurate explanation for why a splitter moves the way it does?
| First off, there is no consistent answer to this question because there is no consistent pitcher, muscle power and slight variations in the manner in which the baseball is thrown will always produce different results from each throw.
It seems to depend on the pitcher releasing the ball closer to the end of the arc of the swing on it than he normally would, he keeps his wrist loose and imparting a slight downward vector just before releasing it.
Based on Splitters, which has stats, (and lots of 'em) and various YouTube videos, particularly Throwing Splitters
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Where does the gravitational energy come from We know that when a mass is pulled by gravitation it's kinetic energy increases.
This is also the case when a photon is being radiated towards a massive body,its wave length decreases and it's energy increases. Since I know that energy is conserved,my question is,where does this extra energy come from?
| It comes from the potential energy from gravity arising from the configuration of the objects.
When an object gains kinetic energy due to the gravitational force, the overall gravitational potential energy of the system decreases by an equal amount.
The graviational field can be thought of as "storing" work energy. You can increase the potential energy of the system by investing energy into re-arranging the system, and you can decrease the potential energy of the system by extracting it as kinetic energy.
You can, for example, imagine taking a book from the floor and placing it on a shelf directly above the original spot. This takes a certain amount of work, $W$. Later, if you decide to remove the shelf and let the book fall back to the floor, the book should accumulate kinetic energy $\frac{1}{2} mv^2$ exactly equal to the original work $W$ you put in.
The "original source" of the energy that the book gains when falling, then, is the work energy that you originally invested in putting it on the shelf. Numerically, they are equal. The gravitational potential took that original energy, stored it, and releases back the same amount when the book falls.
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In general, how are representations used in physics? I want is a basic overview, if there is one, of the meaning (and purpose) of the word representation in general terms. I have looked up sources such as Particle Physics and Representation Theory, but I can't see the general purpose of using representations, beyond its particular applications in particle physics.
In general, what is the point of representation theory?
| You're asking about "representations" in the group-theoretic sense. You can think of a representation simply as a way to make an abstract group concrete, by expressing each group element as a numerical matrix.
This may seem odd because a lot of groups in physics are defined as matrix groups... for example, SU(3) is defined as the group of unitary 3x3 matrices with determinant 1. But it turns out that there are faithful irreducible representations of SU(3) as matrices of an infinite number of other sizes: 6x6, 8x8, 10x10, 15x15, ... (but not 2x2, 4x4, 5x5, 7x7, ...).
So, for example, in an 8-dimensional representation of SU(3), each group element is represented by an 8x8 matrix rather than a 3x3 matrix.
("Faithful" means that the mapping is one-to-one. "Irreducible" means that a representation can't be broken down into smaller representations.)
In particle physics, the color force that binds quarks into protons and neutrons has SU(3) as its gauge symmetry. The 3x3 matrices tell you how the three colors of, say, an up quark mix with each other under a gauge transformation. The 8x8 matrices tell you how the eight gluons mix with each other.
Some groups have non-matrix definitions and can seem extremely abstract. For example, G2 can be defined as the symmetry group of the octonions. But each representation of G2 is just a set of nxn matrices. The basic idea from a mathematical standpoint is that one can represent any abstract group as a set of linear transformations of an n-dimensional vector space, but the specific values of n for which this is possible depend on the group.
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Can X-rays emitted due to bremsstrahlung radiation have frequency matching with other EM waves like visible ones? The continuous X-ray spectrum has x-rays of widely varying frequencies. Since an E-M wave is characterized by its frequency, is it possible for the X-rays coming out of heavy metals due to bremsstrahlung radiation to have the frequency matching with other light waves like visible ones, radio waves, or others?
In short, while producing X-rays, can we produce other types of EM radiation?
| The behaviour of an EM wave as it interacts with matter is frequency dependent. X-rays have high frequencies and show greater penetrability in matter as compared to visible light of lower frequency. If the frequency of EM wave coming out of any source has that of visible light they would not be X rays but visible light.
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How to prove a set of matrices form a representation of Lie algebra? When reading Paul Langacker's The Standard Model and Beyond, I am quite confused on equation 3.29, which says with a set of fields $\Phi _a$, where $a$ goes from 1 to $n$, are chosen to be transformed to it self by Lie algebra generators $T^i$. Therefore we make the following assumption
$$
\left[ T^i , \Phi _a \right] \equiv - L_{a b}^i \Phi _b
$$
It is then said $L^i$ can be easily proven to form a representation of Lie algebra $U_G = e^{- \mathrm{i} \beta ^i T^i}$.
Well, I think if one can prove that $L^i$ and $T^i$ satisfy the same commutation relation, this conclusion is then true. But how do we do that?
Thanks in advance!
| I have the answer now! It is explained in equation 3.36 and 3.37 of the same book.
First, take the adjoint of $\left[ T^i , \Phi _a \right] = - L_{a b}^i \Phi _b$, remember $T^i$ is hermitian, assume $L^i$ is real, which would lead to $\left[ T^i , \Phi _a^{\dagger} \right] = \left( L_{a b}^i \right)^T \Phi _b^{\dagger}$.
Now for a ground state $\left| 0 \right>$, assume $\Phi _a^{\dagger} \left| 0 \right> = \left| a \right>$. Also, the ground state is invariant for now, $T^i \left| 0 \right> = 0$.
Then we would have
$$
T^i \left| a \right> = T^i \Phi _a^{\dagger} \left| 0 \right> = \Phi _a^{\dagger} T^i \left| 0 \right> + \left( L_{a b}^i \right)^T \Phi _b^{\dagger} \left| 0 \right> = 0 + \left( L_{a b}^i \right)^T \left| b \right> = \left| b \right> L_{b a}^i
$$
However, it is true that $T^i \left| a \right> = \left| b \right> \left< b \left| T^i \right| a \right>$, which tells us
$$
L_{b a}^i = \left< b \left| T^i \right| a \right>
$$
Hence $L_{b a}^i$ is a representation of $T^i$ in the particle space.
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Why Does Amplitude Have no Effect on the Energy of a Light Particle? In my high school physics class, I was taught that the energy of light is dependent only on the frequency, as demonstrated in the equation $E = h \cdot \nu$.
My question is, why is amplitude part of the equation? As the amplitude of the light increases, it gets more intense, i.e. brighter (ignoring that more light also makes things brighter), so wouldn't it make sense that light that has a greater amplitude also has more energy? And as an extension of that assumption: that lower amplitudes have less energy?
EDIT: not a duplicate, because I'm asking why amplitude is not part of the energy equation, not if a particle has amplitude.
| As you've said, that's the energy of a light particle, not an arbitrary light wave. As you point out, an electromagnetic plane wave with electric field amplitude $E_0$ has total average energy density $u=\varepsilon_0E_0^2/2$. The formula of energy you have given actually relates this to the number density $n$ of light particles per unit volume, by $u=nh\nu$. As such,
$$n=\frac{\varepsilon_0E_0^2}{2h\nu}.$$
EDIT
The thing to be considered here is, while the amplitude of the wave does matter for its total energy, it has no bearing on the energy of the particles that make it up.
What increases the amplitude of an electromagnetic wave is an increase in the number of photons (light particles, each with energy $E=h\nu$) that are part of this entire wave.
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Calculating an "apparent" speed of a beam in a medium While thinking about relativity, one question sparked my curiosity. If we could observe the trajectory of a light beam by using some partially opaque medium, like smoke, to make it visible, we'd not perceive light as moving at $c$. As the light rays approached the observer, they would appear to move slightly faster than $c$, and as they moved away, slightly slower, only moving at $c$ exactly in the point where the light ray is at a right angle to the line of sight. This is because the reflected light from the smoke would take longer to reach the observer the further from the observer it is emitted.
What would be the exact formula for this apparent velocity of the light ray? I can't see a straighforward way of computing it. It would be a function of the distance of the light front at the moment it is closest to the observer in the trajectory, the time elapsed (positive or negative) from that moment, and the speed of light in the medium.
How can I approach this problem?
| This is a real effect called apparent superluminal motion, with the apparent velocity being
$$v' = \frac{v \sin \theta}{1 - v \cos \theta}$$
in units where $c = 1$. Here, $v'$ is an apparent tangential velocity, defined as $v' = r \omega$ where $r$ is the distance, $\omega$ is the angular velocity of the object in the sky, and $\theta$ is the angle the velocity makes with the radial direction. This isn't quite in the variables that you want (the time and the distance of closest approach) but it should be relatively straightforward to convert it.
For sufficiently large $v$ and $\theta$, $v'$ may exceed the speed of light. Of course it's just "apparent" motion, not a real violation of the speed of light limit, but there was briefly controversy over such observations until somebody figured this out.
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Using Helmholtz Free Energy to Calculate Liquid Density My objective is to find an equation of state (EoS) for density, i.e. density as a function of pressure, temperature and concentration, for aqueous acids, bases and salts.
A StackExchange user suggested that I look into the following paper:
1. Myers JA, Sandler SI, Wood RH. An equation of state for electrolyte solutions covering wide ranges of temperature, pressure, and composition. Ind Eng Chem Res. 2002;41(13):3282–97.
The paper describes the EoS as "the total change in the Helmholtz free energy for forming the electrolyte solution on this path" in which they combine contributions from Peng-Robinson model, Born model, and MSA (Mean Spherical Approximation) model:
$$
A(T,V,\bar n)-A^{IGM}(T,V,\bar n)=\Delta A^{PR}+\Delta A^{Born}+\Delta A^{MSA},
$$
where $T$ is the temperature of the system, $V$ is the system volume, $\bar n$ is the vector of the number of moles of each component of the mixture, and $A^{IGM}$ is the Helmholtz free energy an ideal gas mixture.
In the paper, they included graphs of density vs. molality of aqueous $NaCl, NaBr, CaCl_2$. So it seems like this EoS is what I need, but the paper doesn't explicitly describe how to calculate density from the EoS.
I've seen how taking partial derivatives of Helmholtz free energy results in equations for pressure, entropy, and $\mu$, but not for density. I'm wondering if someone has already figured out how exactly one can calculate density from Helmholtz free energy.
UPDATE: Also, I don't think the Helmholtz free energy I described above (from the paper) is the same as this.
| In the IAPWS 1995 Formulation, the Helmholz Free Energy is used in connection with the density of Water:
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Lie derivative of a vector along itself In Tensorial form, the definition of the Lie derivative for a covariant and contravariant vector are respectively:
$$\mathcal{L}_U V^\mu=U^\nu\nabla_\nu V^\mu- V^\nu\nabla_\nu U^\mu$$
$$\mathcal{L}_U V_\mu=U^\nu\nabla_\nu V_\mu+ n_\nu\nabla_\mu U^\nu$$
At some point of a calculation, I had to calculate $\mathcal{L}_V V^\mu$ that gives zero using that definition. My question is, What is the intuition behind this result?
When foliating a spacetime in Cauchy surfaces as
$$h_{\mu\nu}=g_{\mu\nu}-n_\mu n_\nu$$
where $n^\mu$ is a spacelike normal vector, we get an extra identity of the form
$$n^\alpha \nabla_\beta n_\alpha=0$$
derivating the definition $n^\alpha n_\alpha=1$.
With this in mind, the computation of $\mathcal{L}_n n_\alpha$ gives
$$\mathcal{L}_n n_\alpha=n^\beta\nabla_\beta n_\alpha$$
that is different from zero. Once again, what's the intuition behind this result?
| When you take the Lie derivative of a vector, you are looking at how it changes as you move along integral curves. Now if you look at $L_UU$ you are asking how does $U$ change along its integral curves. But the point of an integral curve is that it’s tangent is always $U$. So $U$ does not change as you travel along the curve (its always pointing ahead)
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Outside electric field due to an off-center charge inside a conducting shell Suppose there is an off-center charge inside a copper conducting shell. We know that we can use the method of images to calculate the charge distribution on the inner surface. We also know that the charge distribution on the outer surface is uniform so as not to cause any unwanted electric fields inside the conductor. However, it is claimed that the electric field outside the conductor is only determined by the charge on the outside of the conductor such that it is as if there was a charge at the center of the conductor. However, I don't see why the inner charges (both placed and induced) should be ignored when calculating the field at some random point outside the sphere. There were meant to cancel exactly inside the conductor but nowhere else.
| One way to explain why the field outside a conducting shell is completely defined by the charge distribution on the outer surface of the shell is to consider that the field inside the conductor is zero.
This means that there are no electric field lines going from the inside to the outside of the shell and, therefore, the field outside the shell due to the induced charges on the outer surface should be identical to the field due to the identically distributed "placed" charges.
In other words, once the re-distribution of charges is completed and the field inside the walls of the conducting shell becomes zero, the internal charges stop contributing to the outside field - their further contribution is "blocked" by the conductor.
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Do the wires having AC current emit EM waves? When ac current is passed through a wire, the electrons in the wire oscillate to and fro in the wire (due to change in polarity of current in the wire). If the ac current is of frequency, say 50Hz, then will the wire emit electromagnetic waves of the same frequency?
| Yes. they do produce an electric field, if we're working with $50$Hz a rough estimate is that the $\lambda$ would be $\frac{c}{f}=\frac{3\cdot 10^8}{50}\approx6000000$m which is a wavelength that's hard to make use of, but yes, em waves are produced.
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How are high-energy detectors able to to distinguish between $m_{e}$ and $m_{\mu}$? I had a teacher pose this interesting question yesterday:
Suppose you're running a high-energy scattering experiment at the LHC. For concreteness, let's suppose it's a 2 to 2 scattering event which involves electrons and/or muons.
The theorist uses QFT to compute some cross-section which comes from the amplitude
$$
\mathcal{A}_{p_{1} p_{2} \to p_{3} p_{4} } = F(s,t,u,m_{e},m_{\mu},\ldots)
$$
The amplitude is a function of the Mandelstam variables $s \equiv (p_1+p_2)^2$, $t \equiv (p_1-p_3)^2$ and $u \equiv (p_1-p_4)^2$, as well as the mass of the electron $m_{e}$ and muon $m_{\mu}$ (and some other stuff).
Because we're running a high-energy experiment we obviously have that $s,t,u \gg m_e,m_\mu$, and for this reason the theorist makes the approximation $m_{e} \approx 0$ and $m_{\mu} \approx 0$.
The Question: How is the LHC able to distinguish between an electron and a muon if the theorist makes the approximation that the electron and muon are both massless?
For some reason, the approximation $m_{e} \approx m_{\mu} \approx 0$ is a bad one and the question is why this is. One idea that a colleague had was that the tracks of the electron and muon look different; because of the cyclotron radius $r \sim \frac{cm}{qB} \propto m$ the magnetic fields used in the machine to track the particles coming out of the collision will see the electron spiral more dramatically than the muon.
Any ideas as to other reasons why?
| Particles as expected to be seen in the CMS experiment at LHC
So electrons and muons leave completely different signatures although because of high energy they may leave a similar track in the tracking detector. Electrons are absorbed in the electromagnetic calorimeter, and muons go through,their electromagnetic interactions minimal due to their large mass, leaving a track signal that can be fitted.
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Have all three flavors of solar neutrino been measured? As far as I know that the sun exclusively produces electron neutrinos ($\nu_e$). When the flux of solar neutrinos ($\nu_e$) is measured on the earth, a depletion is observed in the $\nu_e$ flux i.e., some $\nu_e$'s have "disappeared" in their way from the sun to the earth. As far as I know that this conclusion is drawn, by measuring only$^1$ the $\nu_e$ flux in the detectors. The explanation is that some of the neutrinos get morphed into $\nu_\mu$ and $\nu_\tau$.
But to really test this hypothesis, a deficiency in the $\nu_e$ flux is not enough. There must be an experiment where the detector must also measure the $\nu_\mu$ and $\nu_\tau$ fluxes. If adding the fluxes over all three flavors turns out to be equal to the expected flux then only we can be sure that solar neutrinos have undergone oscillation. Has that been achieved in experiments?
$^1$The experiment carried out by Davis et al at the Homestake mines detected $\nu_e$ through the inverse beta decay $\nu_e+^{37}{\rm Cl}\to e^-+^{37}{\rm Ar},$ and found that they were getting about one-third of the number of $\nu_e$ that were predicted from the solar models.
| The SNO experiment was sensitive to all three flavors of neutrinos, and hence provided definitive evidence for solar neutrino oscillations. That's why half of the 2015 Nobel prize for neutrino oscillations went to the director of this experiment, rather than the many other previous experiments.
As usual, there's a bit of confusion depending on whether you talk primarily to theorists or experimentalists. Judging from the model building occurring in the 90's, by 2001 most theorists considered it a given that neutrinos had masses and hence oscillated. But the SNO experiment was the first to really check it for sure.
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Why does this paper use 1/cm for units of frequency? Reading this paper from 1963 $^*$, they use units of cm$^{-1}$ for frequency.
Here is an excerpt:
It doesn't seem like wave number, as they clearly call it frequency. What's going on here?
$^*$ Sievers III, A. J., and M. Tinkham. "Far infrared antiferromagnetic resonance in MnO and NiO." Physical Review 129.4 (1963): 1566.
| If you suppose c=1, then 'centimetre' is the time it takes for light to cross a centimetre, and a count of cycles per centimetre is something that can be done in nature, and also relates to the frequency in cycles per second.
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Global warming and air temperature The layman's question. Is there a maximum temperature that can be reached by air on Earth due to global warming? For example, one that would prevent living organisms from functioning? I read somewhere that with the increase of the sun's brightness by 10%, the average temperature on our planet will be 47 C. This, however, only for about 600 million years. Would global warming enable reaching such a high average temperature?
| Is there a limit? Yes and no.
No, there isn't really a maximum temperature anywhere near what we're used to. It's certainly possible to reach an average temperature of 47°C. That may well trigger a runaway greenhouse effect, when all the oceans evaporate. Look at our neighbour Venus. On Venus, the greenhouse effect leads to a surface temperature of 467°C. That's very far away from where we are now on Earth, and we'll most likely not reach that any time soon, but it's probably not theoretically impossible to reach. On Venus, the partial pressure of CO₂ is around 9 MPa. On Earth, it's closer to 40 Pa, a factor of around 200,000 less.
There is, of course, a maximum temperature; we can't reach $\infty$°C. Venus may be close to the upper bound. But at that maximum temperature, life on Earth will be history.
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Ocean density vs atmosphere density I understand that the density of the oceans on Earth in on average constant regardless of the depth. It is 1020 kg/m^3 at the surface and 1050 kg/m^3 at deep waters.
I understand too that this is not the case with the atmosphere. The density of the atmosphere decreases with height.
Though, I could not find any description in QM, that would describe the difference between the two media, and how they react differently to gravity, and the distance from the center of mass does or does not change their density.
Question:
*
*Is there an explanation why the density of the oceans is (mostly) independent of depth, but the atmospheric density changes with height (is this just because of one is liquid and one is gas)?
*Is there a QM explanation to this (different material) or is this just because of the distance from the center of mass (GR)?
| The incompressibility of liquids is due to the fact that they are made of atoms or molecules of finite radius. The atoms in a liquid are constantly in contact with their neighbors, and increasing the density would require that atoms or molecules overlap, which they typically don't do very readily. as such, liquids have an essentially fixed density.
In a gas, the atoms/molecules are not in contact with their neighbors, so the density can vary widely as the distance between one atom/molecule and its neighbors increases or decreases.
No QM or GR is required here.
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Changin the optical depth (axial resoltuion) on an Optical Cohernce Tomagrpahy system So the axial resolution in a OCT system is given by:
$$I_c=\frac{4ln2}{\pi}\frac{\lambda^2}{\Delta \lambda}$$
My question what would you do to say increase the axial resolution, and how would it effect the image quality?
So looking at the above diagram which kinda answer the first part of my question, that if you move the reference mirror you will either decrease on increase the axial resolution, but how dose the relate to the central wavelength and the bandwidth?
As from the above equation, central wavelength is given by $\lambda$ and the broadband is given by $\Delta \lambda$
| Moving the reference mirror does not change the axial resolution. Its action is to "bring the image in focus", so to speak. Because the light has short coherence length $\Delta\lambda$, the two beams split by the beam splitter do not interfere on the detector. They show clear interference (which is the raw signal) when the optical path difference (OPD = $\int n_{\text{index}} ds_{\text{path}}$) is less than the coherence length, and you need to move the mirror to adjust the OPD.
In analogy, bringing an image in and out of focus is not the same thing as changing the performance (resolution) of the microscope, for the latter, you need to control the numerical aperture.
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Why do materials obey Hooke's law? Why do materials extend proportionally to the force exerted on them (Hooke's law)? I thought that
*
*when materials are compressed or extended under force, their atoms become closer or further apart;
*the inter-atomic forces are essentially electrostatic;
*electrostatic forces (and indeed most other forces) follow an inverse-square law;
So I would naively conclude that springs should follow an inverse square law.
But clearly in most situations the law is linear. Where is my logic flawed?
| When one considers a relatively small segment of a curve, most appear approximately linear when one zooms in far enough. Most materials are not truly linear when stretched a considerable distance, but for small deviations about the equilibrium some are approximately linear. This can be seen when one considers Taylor Series expansions of functions.
The inverse square law for electrostatics can be expanded as a Taylor Series as well, and it contains a linear component as can be seen here:
$$\frac{1}{(r-r_0)^2}=\frac{1}{r_0^2}\left[1+2\left(\frac{r}{r_0}\right)+3\left(\frac{r}{r_0}\right)^2+4\left(\frac{r}{r_0}\right)^3+\cdots\right]\,.$$
For small deviations $\Delta r\approx0$, the inverse square law is approximately linear since the higher-order terms $\mathcal{O}[(\Delta r)^2]$ fall off very quickly. This can be seen using the Binomial approximation $(1+x)^n\approx1+nx$ where $x\approx0$. In this case we have:
$$(\Delta r-r_0)^{-2}=\frac{1}{r_0^2}\left(1-\frac{\Delta r}{r_0}\right)^{-2}\approx\frac{1}{r_0^2}\left[1-(-2)\left(\frac{\Delta r}{r_0}\right)\right]=\frac{1}{r_0^2}\left[1+\frac{2\Delta r}{r_0}\right]\,.$$
Of course this is far too simple to describe the more complex material effects of crystal lattices being stretched from equilibrium, but this is the general idea of how many materials can exhibit linear relationships between force and distance for small deviations from equilibrium.
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Relative motion and time When someone reaches to a speed which is close to the speed of light with respect to earth, will he see the things actually moving faster than when he is in low speeds?
| In SR (special relativity) time is running at a different pace if measured in inertial reference frames in relative motion. A stationary observer (observer in its own rest frame) measures a time $t$ running faster than the proper time $\tau$ of a moving frame. The well known relation is:
$\Delta t = \gamma \Delta \tau$ time dilation
where:
$\gamma = 1 / \sqrt{1 - v^2/c^2}$ Lorentz factor
As the relative velocity $v$ approaches $c$ the Lorentz factor diverges to infinity and the stationary observer measures the clock of the moving frame to tick slower and slower. Of course, the description is symmetric if measured by the moving frame.
To answer to the question, a spacecraft moving at a relativistic speed vs. earth would measure events on the earth occurring slower. That disregarding whether it is going closer or farther from the earth, as the Lorentz factor depends only on the norm of the relative velocity.
Note: A different issue is the relativistic Doppler effect. In that case going closer would mean a blueshift of the radiation, while going farther a redshift vs. the radiation frequency as measured in the earth reference frame.
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Questions about shapes and the reason behind the drag coefficient differences
*
*What is the reason that a streamlined body has a drag coefficient that is lesser than the drag coefficient of a streamlined half-body, when the latter has a completely flat bottom, while the former has bulges on either side of the centerline?
*Why do long cylinders have a lesser drag coefficient than short cylinders? Length means more surface for the flow to move against, but it seems that a shorter object with an otherwise same shape, has more drag to it.
https://img.bhs4.com/26/7/2675e0e869aa5afcf4ea44bd4908acb8248a8a76_large.jpg
| Drag is not friction between the surface of the body and the fluid. The assumption is usually that the fluid is not moving relative to the surface at the surface itself. This is the basis of boundary layer theory. D'alembert's paradox shows that laminar flow does not produce drag. The upshot is that the drag is due to power loss into turbulence - ultimately into heating the fluid. Viscous losses are friction losses, but not friction of the fluid rubbing on the surface. Hence, larger surface area does not in and of itself lead to larger drag.
The streamlined body is designed to produce the low turbulence by balancing the fluid streamlines coming of the back of the body so that they have the same velocity on detaching and do not tend to collide with each other. The half streamlined body has the fluid coming around the top with some downward velocity, while the fluid on the bottom, in principle, is going straight through. The two streams could be expected to merge and create turbulence. It often helps here to recall that the fluid is actually atoms, and think about what they are doing.
Navier Stokes theory is based on the idea of the existence of a velocity field. But, in turbulence, there is no velocity field - as the particles are going every which way and crashing into each other. These collisions lead to creation of entropy, and heating, and at another level, the emission of photons adding to the heat bath.
| {
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What does it mean if the dot product of two vectors is negative? If the dot product gives only magnitude, then how can it be negative?
For example, in this calculation:
$$W = \vec{F}\cdot\vec{r} = Fr\cos\theta = (12\ \mathrm{N})(2.0\ \mathrm{m})(\cos 180^\circ) = -24\ \mathrm{N\,m} = -24\ \mathrm{J}$$
Why is there a negative sign? What does it tell us?
| In your function:
$$W = \vec{F}\cdot\vec{r} = Fr\cos\theta $$
Let's assume that
$$\vec{r}>0$$
There are three cases:
*
*the dot product is 0, this means that the two vectors are perpendicular
*the dot product is >0, this means that the two vectors point approximately in the same direction, that is, their angle is < 90 degrees
*the dot product is <0, this means that the two vectors point in approximately opposite direction, that is their angle is > 90 degrees
| {
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Is the Bose-Einstein condensation a single particle phenomenon? BEC occurs for noninteracting Bosons. Can we conclude that it can be described with a single particle? What is the significance of the number of the particles?
| The Bose Einstein condensate is a QM effect of collective quantum state in which a macroscopic number of particles occupy the lowest energy state and thus is described by a single wavefunction.
All the bosons will be described by the same wavefunction.
So it is not a single particle, but all the particles (their probability distribution) are described by the same wavefunction.
| {
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From where energy comes for the heat released when direction of electric field is changed with a dipole placed in the field
This is an example problem in my book. Forget about the values.
Initial anf final(after heat is released) conditions in this problem are same, ie, dipole moment and electric field are in same direction. We know that energy of a configuration is same irrespective of how it is created. Then how there is release of heat, from where energy is coming in to the system.
| Since the direction of the field changes suddenly, the dipoles rotate quickly and, as a result, when they reach the new orientation, they have some kinetic energy, which will turn into heat, i.e., some energy will be released.
This kinetic energy is equal the difference of potential energies corresponding to the initial orientation of the dipoles and two directions of the electric field.
The energy, naturally, comes from the electric field, which cause the change in polarization.
| {
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How much mass is in gravitational waves? Like photons, I understand gravitational waves to have no rest mass but mass due to their energy. Are gravitational waves a significant part of total mass and what are the main components (black hole collisions, big bang etc)?
| Turning to the very useful cosmic energy inventory by Peebles and Fukugita, they estimate that gravitational radiation from massive black holes make up $10^{-7.5\pm 0.5}$ of the total energy of the universe, stellar binaries contribute the smaller $10^{-9\pm 1}$ and primeval gravitational waves are less than $10^{-10}$.
In short, they make up a rather small component. As comparison, they estimate stars to make up 0.0015 and planets $10^{-6}$ of the energy.
| {
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How does a satellite take pictures when the surface seems to be always cloudy/white? I've just opened ISS video on youtube for the first time, and I must say I'm underwhelmed. There are no oceans/landforms. It's all white everywhere. I'm a bit confused how the satellites can take pictures when the view from space looks like this ? Google maps has clear satellite pictures. Do they use some other frequency light that go straight through clouds ?
| The satellites (optical satellites) utilize the electromagnetic spectrum to bypass the opacity provided by the atmosphere, which are known as atmospheric windows. In these windows the Channels of the sensor can penetrate the atmosphere. For example, in the image below the atmosphere is basically transparent if the bands are selected accordingly.[][http://www.gb.nrao.edu/GBTopsdocs/primer/atmospheric__windows_.htm]
The optical satellites have Red, Blue, Green, Near Infrared channels for observations which are readily transmitted through the atmosphere. Optical satellites cannot see through "clouds" thus proper cloud masks are generated. For example see the image below. Here the clouds obscures the ground information, but you can see the shadow of the cloud, using which the height of the cloud etc can be determined.[][ISRO/NRSC]
Whereas the cloud poses no challenge to the microwave satellites (Synthetic Aperture Radar Satellites) like Sentinel-1(European Space Agency), RISAT-1(Indian Space Research Organization) etc. Hence they are useful in mapping floods during monsoon or cloudy situations where optical satellites are not fruitful. For example, see the image below in which the flooded parts are mapped in blue, obviously microwaves can penetrate the cloud and see beneath.[][Sentinel-1 SAR]
| {
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does tension in the string affect its equilibrium? In my textbook (Sears and Zemansky's University Physics), it is written that the vector sum of the forces on the rope is zero, however the tension is 50 N. Then is tension different than the force? And if not, then why force is zero while tension is not?
A body that has pulling forces applied at its ends, such as the rope in Fig 4.27, is said to be in tension. The tension at any point is the magnitude of force acting at that point (see Fig 4.2c). In Fig 4.27b, the tension at the right end of the rope is the magnitude of $\vec{\mathbf{F}}_{M\ on\ R}$ (or of $\vec{\mathbf{F}}_{R\ on\ B}$). If the rope is in equilibrium and if no forces act except at its ends, the tension is the same at both ends and throughout the rope. Thus if the magnitudes of $\vec{\mathbf{F}}_{B\ on\ R}$ and $\vec{\mathbf{F}}_{M\ on\ R}$ are $50\ \rm N$ each, the tension in the rope is $50\ \rm N$ (not $100\ \rm N$). The total force vector $\vec{\mathbf{F}}_{B\ on\ R}+\vec{\mathbf{F}}_{M\ on\ R}$ acting on the rope in this case is zero!
| Okay, I am a student myself so please re-read it if you don't get at once.
Your textbook says that the Sum of all forces on the rope is zero and yes it is because the rope is in equilibrium.
To understand this first answer these questions,
*
*Is the rope moving? (Hint: No)
*Is the rope pulling the wall? (Yes)
*Is the rope pulling the man? (Yes)
I hope that you know the answers to these questions. (Hint: Read laws of motion)
Now, the man is applying a force on the rope. If there was no force counter acting this then the rope should move right..? Yes, but it is not moving. That means that there is a force on the rope acting in the opposite direction. That is the force exerted on the rope by the wall and that force is equal to the force applied by the man.
So the net force on the wall adds up to zero. I think this is a good answer for your second question, https://physics.stackexchange.com/a/221169/202990.
| {
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Why is Copper(I) Oxide Red? This may appear to be a chemistry problem. But, after reading the Wikipedia article on copper(I) oxide, it seems to have more to do with semiconductor-physics. For example:
… light travels almost as slowly as sound in this medium.
Is that true?
What have Kramers–Kronig relations got to do with it?
To a chemist, who was never brilliant at maths, it takes a bit of understanding.
I know that copper(II) oxide (Mott–Hubbard insulator [semiconductor]) is black because of intervalence charge transfer, giving rise to the generation of a highly polarising Cu(III) species. Similarly, the non-stoichiometric form of nickel(II) oxide (Mott insulator) is black because of a Ni(III) species. Again, silver(I) oxide is black … Ag(III) species.
This model does not appear to work for copper(I) oxide because the non-stoichiometry, causing the oxidation required for the balancing of charges with the oxide ions, would give Cu(II); which, by definition, is not sufficiently polarising to produce the deep, intense colour observed. Further, the reduced Cu(I) becomes Cu(0), the pure metal.
So, why is copper(I) oxide red?
| Apparently $\rm Cu_2O$ has a band gap of about $2.1\ \rm eV$ (according to the linked wikipedia page). That means it'll absorb photons with a wavelength of less than $590.4\ \rm nm$ (just do the calculation with $E=hf$). For comparison, yellow light has a wavelength in the range $570-590\ \rm nm$. Hence we detect longer wavelengths and $\rm Cu_2O$ looks reddish.
Here's an easy way to think of it: the definition of band gaps tells us that it's the energy difference between possible bands, so if a photon of a lower energy (like the ones in red light) is incident, it can't be absorbed. But yellow wavelengths and higher energy photons (green, blue, and so on) will be absorbed.
| {
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Can Newton’s law of gravitation be derived from Coulomb’s law? I’m casually learning physics and have noticed that Newton’s law of gravitation and the electrostatic force formulas look similar. I’ve asked this question before but would really appreciate another response. Is it possible that the two laws are related? Can the law of gravitation be seen as the macroscopic averaging of Coulomb’s law? So atoms on average have negative charge (positive mass) and thus on a macroscopic scale we observe that two large bodies (eg planets) attract rather than repel. Would it help if we assume that masses can be positive as well as negative? Apologies as I’m not a physicist (rather a data analyst) and these are probably dumb questions.
| What if mass had a sign?
There are (let's keep it simple) Sun, Earth and Moon.
Earth goes around the Sun, so they have different signs.
What about the Moon? If it's attracted to the Earth, it would be repelled by Sun, and vice versa. This is not what happens.
| {
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Zero velocity divergence for incompressible flow is derived from conservation of energy equation or conservation of mass equation? I'm a bit confused about incompressible flow definition. In many textbooks or scientific articles, they simply claim that the incompressibility condition for Navier-Stokes equation is:
$\nabla \cdot \mathbf{u} = 0$
But, nobody says explicitly how to prove that incompressible velocity field should be divergence free.
Here are my findings to derive this equation from basic fundamentals of physics:
For incompressible fluid: from thermodynamics equation of state, we know that density should only depends on equilibrium potentials of pressure and temperature:
$\rho = \rho(P,T)$
If we take material derivative from this equation:
$\frac{D \rho}{D t} = (\frac{\partial \rho}{\partial P})_{T} \frac{D P}{D t} + (\frac{\partial \rho}{\partial T})_{P} \frac{D T}{D t}$
For an isothermal and incompressible fluid:
Incompressible fluid :$(\frac{\partial \rho}{\partial P})_{T} = 0$
Isothermal fluid: $\frac{D T}{D t} = 0$
So finally, these conditions will lead to:
$\frac{D \rho}{D t} = 0$
But, from mass conservation equation (continuity equation), we have:
$\frac{D \rho}{D t} = -\rho \nabla \cdot \mathbf{u}$
As a result: $\nabla \cdot \mathbf{u} = 0$
For compressible fluid: from internal energy balance equation, we know:
$\rho \frac{D e}{D t} = -\nabla \cdot \mathbf{q} + \sigma \cdot (\nabla \otimes \mathbf{u})$
Where $e$ is the internal energy of the system, which is equal to enthalpy at the constant pressure condition, $\mathbf{q}$ is the thermal heat flux, $\sigma$ is the Cauchy stress tensor, which is equal to: $\sigma = -P \mathbf{I} + \tau$, where $P$ is the pressure and $\tau$ is the deviatoric stress.
For isothermal compressible fluid: $\frac{D e}{D t} = 0$ and $\nabla \cdot \mathbf{q} = 0$.
As a result: $\sigma \cdot (\nabla \otimes \mathbf{u}) = 0$.
For a Newtonian compressible fluid, we have: $\tau = \mu (\nabla \mathbf{u} + (\nabla \mathbf{u})^{T}) + \zeta (\nabla \cdot \mathbf{u}) \mathbf{I}$.
Where $\mu$ is the shear viscosity and $\zeta$ is the bulk viscosity.
Finally, the term $\sigma \cdot (\nabla \otimes \mathbf{u})$ could be expanded as:
$\sigma \cdot (\nabla \otimes \mathbf{u}) = -P (\nabla \cdot \mathbf{u}) + \zeta (\nabla \cdot \mathbf{u})^{2} + 2 \mu S \cdot (\nabla \otimes \mathbf{u})$.
Where $S$ is the shear rate tensor, which is defined as: $S = \frac{1}{2}(\nabla \mathbf{u} + (\nabla \mathbf{u})^{T})$.
Finally, we have:
$\sigma \cdot (\nabla \otimes \mathbf{u}) = -P (\nabla \cdot \mathbf{u}) + \zeta (\nabla \cdot \mathbf{u})^{2} + 2 \mu S \cdot (\nabla \otimes \mathbf{u}) = 0$
or
$(P - \zeta (\nabla \cdot \mathbf{u})) (\nabla \cdot \mathbf{u}) = 2 \mu S \cdot (\nabla \otimes \mathbf{u})$
Now, we could argue that at low velocities (low Mach number), the viscous heat dissipation term ($2 \mu S \cdot (\nabla \otimes \mathbf{u})$) is negligble. As a result, we have:
$(P - \zeta (\nabla \cdot \mathbf{u})) (\nabla \cdot \mathbf{u}) = 0$
Finally, we should have:
$P = \zeta (\nabla \cdot \mathbf{u})$
or
$\nabla \cdot \mathbf{u} = 0$
The first equation ($P = \zeta (\nabla \cdot \mathbf{u})$) is contradictory because the thermodynamics pressure $P$ should only depends on equilibrium potentials and not kinetics variables like velocity. As a result, we have:
$\nabla \cdot \mathbf{u} = 0$
So it proves that compressible fluid could be treated as an incompressible flow, when its velocity remains small in comparison to speed of sound (low Mach number).
So my question is that why in classical fluid mechanics textbooks, always people claim the divergence free condition is a direct consequence of mass conservation?! Right now, I show that it could be derived with minimum assumptions from energy conservation equation. Any idea or suggestion is appreciated.
Edition:
Proof of negligible viscous dissipation heat rate:
Full internal energy balance equation:
$\rho \frac{\partial e}{\partial t} + \rho \mathbf{u} \cdot \nabla e = -\nabla \cdot \mathbf{q} + \sigma \cdot (\nabla \otimes \mathbf{u})$
The internal energy will be equal to enthalpy at the constant pressure. As a result we have:
$e = C_{p} \Delta T$
Where $C_{p}$ is the constant pressure specific heat capacity and $\Delta T$ is the temperature difference from reference point.
Also by assuming the Fourier heat transfer law, we have:
$\mathbf{q} = -k \nabla T$
Where $k$ is the heat conductivity.
The internal energy equation could be rewritten as:
$\rho C_{p} \frac{\partial T}{\partial t} + \rho C_{p} \mathbf{u} \cdot \nabla T = k \nabla^{2} T + \sigma \cdot (\nabla \otimes \mathbf{u})$
If we put the expansion of $\sigma \cdot (\nabla \otimes \mathbf{u})$ for a Newtonian compressible fluid, finally we will find:
$\rho C_{p} \frac{\partial T}{\partial t} + \rho C_{p} \mathbf{u} \cdot \nabla T = k \nabla^{2} T -P (\nabla \cdot \mathbf{u}) + \zeta (\nabla \cdot \mathbf{u})^{2} + 2 \mu S \cdot (\nabla \otimes \mathbf{u})$
This equation could be nondimensionalized by taking:
$\theta = \frac{\Delta T}{\Delta T_{0}}$, $t^{'} = \frac{t}{t_{0}}$, $\mathbf{u}^{'} = \frac{\mathbf{u}}{u_{0}}$, $\nabla^{'} = \epsilon \nabla$, $P^{'} = \frac{P}{P_{0}}$, $S^{'} = \frac{\epsilon S}{u_{0}}$
So the above equation could be rewritten as:
$\frac{\rho C_{p} \Delta T_{0}}{t_{0}} \frac{\partial \theta}{\partial t^{'}} + \frac{\rho C_{p} u_{0} \Delta T_{0}}{\epsilon} \mathbf{u}^{'} \cdot \nabla^{'} \theta = \frac{k \Delta T_{0}}{\epsilon^{2}} {\nabla^{'}}^{2} \theta - \frac{P_{0} u_{0}}{\epsilon} P^{'} (\nabla^{'} \cdot \mathbf{u}^{'}) + \frac{\zeta u_{0}^{2}}{\epsilon^{2}} (\nabla^{'} \cdot \mathbf{u}^{'})^{2} + \frac{2 \mu u_{0}^{2}}{\epsilon^{2}} S^{'} \cdot (\nabla^{'} \otimes \mathbf{u}^{'})$
Finally, by taking $\alpha = \frac{k}{\rho C_{p}}$ and its nondimensionalized form $\alpha^{'} = \frac{\alpha t_{0}}{\epsilon^{2}}$, we have:
$\frac{1}{\alpha^{'}} \frac{\partial \theta}{\partial t^{'}} + Pe \mathbf{u}^{'} \cdot \nabla^{'} \theta = {\nabla^{'}}^{2} \theta - \frac{P_{0} u_{0} \epsilon}{k \Delta T_{0}} P^{'} (\nabla^{'} \cdot \mathbf{u}^{'}) + Br_{bulk} (\nabla^{'} \cdot \mathbf{u}^{'})^{2} + 2Br_{shear} S^{'} \cdot (\nabla^{'} \otimes \mathbf{u}^{'})$
Where Peclet, bulk Brinkman, and shear Brinkman numbers are defined as:
$Pe = \frac{u_{0} \epsilon}{\alpha}$
$Br_{bulk} = \frac{\zeta u_{0}^{2}}{k \Delta T_{0}}$
$Br_{shear} = \frac{\mu u_{0}^{2}}{k \Delta T_{0}}$
Finally, for an isothermal fluid: $\theta = \theta_{0} = const.$ and we will have:
$2Br_{shear} S^{'} \cdot (\nabla^{'} \otimes \mathbf{u}^{'}) = (\frac{P_{0} u_{0} \epsilon}{k \Delta T_{0}} P^{'} - Br_{bulk} (\nabla^{'} \cdot \mathbf{u}^{'})) (\nabla^{'} \cdot \mathbf{u}^{'})$
For low Mach number Brinkman numbers (both shear and bulk) are negligible. In fact, Brinkman number should be at least in order of $O(1)$ to consider viscous dissipation heat rate in internal energy equation. For conventional fluids at low Mach number regime Brinkman number is in the order of $O(10^{-3})$, which is negligible.
As a result, we should have:
$\frac{P_{0} u_{0} \epsilon}{k \Delta T_{0}} P^{'} = Br_{bulk} (\nabla^{'} \cdot \mathbf{u}^{'})$
or
$\nabla^{'} \cdot \mathbf{u}^{'} = 0$
Again, we could argue that the first equation ($\frac{P_{0} u_{0} \epsilon}{k \Delta T_{0}} P^{'} = Br_{bulk} (\nabla^{'} \cdot \mathbf{u}^{'})$) could be not true because the thermodynamics pressure should only depend on equilibrium potential and not kinetics variables (e.g. velocity). As a result finally we will find:
$\nabla^{'} \cdot \mathbf{u}^{'} = 0$
or in its dimensional form:
$\nabla \cdot \mathbf{u} = 0$
| Physical meaning of divergence is rate of change of control volume per unit volume. If density is not changing then rate of charge of control volume will be zero this is directly from conservation of mass.
| {
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Do photons violate the uncertainty principle, given that they have a constant speed $c$ with no uncertainty? I have a very basic understanding of quantum physics, but as I understand it the uncertainty principle says that the more precisely you know a particle momentum and the less you know the particle's position.
But I wonder with the photon: given that the velocity is a constant $c$ so there is no uncertainty at all in the speed (and so in the momentum), does that mean for a photon that the uncertainty of the position is "infinite"?
| There is an uncertainty in momentum! Because, for a photon,
$$p = \frac{h}{\lambda} = \frac{h \nu}{c}$$
where $p$ is the magnitude of the momentum, $\lambda$ is the wavelength of the photon and equivalently $\nu$ is the frequency.
So even though photons travel at $c$, their momentum can be uncertain if their frequency is uncertain.
And this ties in exactly with the uncertainty principle: when looking at a wave, be it light, sound, or whatever, how accurately you can know the wavelength depends on how distributed in space (equivalently, in time) the wave is: a brief pulse of sound, say, doesn't have a well-defined frequency at all, while a sound that goes on for a very long time does (or may do). So a photon has a well-defined frequency, and hence a well-defined momentum, only if it is very spread out in space, while a photon which is localised has an ill-defined frequency and hence an ill-defined momentum.
| {
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Why does kinetic energy of elecron on Bohr's orbit not change if zero of potential energy changes? If zero of potential energy is not taken at infinity, total energy and potential energy of electron change but not kinetic energy. Why is that so?
| The zero of PE is at infinity actually. A charge at rest would have 0 energy at infinity. That's why bounded states have negative energy. For the hydrogen atom, the electron at its fundamental level is $-13,6 eV$, negative because it would escape if it had 0 energy, just like any orbit.
Plus, KE depends on velocity, not on PE.
| {
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In a globally-hyperbolic spacetime, does every pair of elements have overlapping light cones? Suppose we have a spacetime $(M,g)$, and denote by $J^+(p)$ the set of points that lie in the causal future of $p$, i.e. $x \in J^+(p)$ iff there is a future-directed timelike curve $\gamma: [0,1]\rightarrow M$ such that $\gamma(0) = p$ and $\gamma(1) = x$.
Question: For any pair of elements $p,q$ in a globally-hyperbolic spacetime, is it always the case that $J^+(p)$ and $J^+(q)$ intersect?
A standard result (see e.g. here, Thm 3.78, page 49) is that every globally hyperbolic spacetime admits some Cauchy surface $\mathcal{S}$, and moreover $M$ is isometric to the product $\mathcal{S} \times \mathbb{R}$.
Without loss of generality we can assume that $p$ and $q$ lie on the same Cauchy surface since the result is trivially true if $p,q$ are timelike/lightlike events, and if $p$ and $q$ lie on different slices, say $p \in \mathcal{S} \times \{t\}$ and $q \in \mathcal{S}\times \{t'\}$ where $t <t'$, then we can just look at some point $p'\in J^+(p)$ that lies on $\mathcal{S}\times\{t'\}$ (such a point always exists since we can pick any inextendible future timelike curve $\gamma$ passing through $p$ and use the definition of a Cauchy surface to conclude that $\gamma$ will pass through $\mathcal{S}\times \{t'\}$) and use that $p<p'$ implies $J^+(p')\subset J^+(p)$.
How can I proceed from here? I was hoping to define some sort of spacelike geodesic connecting $p$ and $q$, and then pick an appropriate element $r \in \mathcal{S} \times \{t+l\}$, where $l$ is the length of the spacelike geodesic connecting $p$ and $q$. Is this the right approach?
|
In a globally-hyperbolic spacetime, does every pair of elements have overlapping light cones?
No. Standard cosmological models of our own universe are a counterexample. There are cosmological horizons, so future light cones do not all overlap, but the universe is globally hyperbolic.
| {
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Discrepancy in general work and pressure-volume work Consider a massless, frictionless piston fit into an airtight container containing an ideal gas. And let us say that the gas undergoes quasi-static isothermal expansion by lowering the pressure applied by the piston slowly.
The piston moves a distance, say $dr$ in an instant in which the force applied on it by the gas is $F$. So, the work done on it by the gas should be $\delta W=Fdr$.
Now, my textbook states that work done on the gas by the piston should be equal to the negative of the work done on the piston by the gas, saying that it is a consequence of Newton's third law of motion.
The problem is, according to Newton's third law of motion, the mutual forces of interaction between the gas particles and the piston should be equal and opposite to one another. But for the corresponding work done to be negative of one another, each gas particle must be displaced in the opposite direction with same displacement $dr$, which is not necessarily true. So, how come does this hold true?
EDIT: Can anyone prove it mathematically that work done on the gas by the piston should be equal to the negative of the work done on the piston by the gas?
| Work done is defined as $dW = \vec{F}.d\vec{r}$. Force on the piston is in direction of it's displacement, so according to definition of work done it will be positive. On the other hand, force on gas is opposite to the displacement of gas, so work done on gas comes out to be negative due to the $cos(\theta)$ term.
For proving that work done by piston on gas should be equal to negative of the work done by gas on piston, you can use work-energy theorem. The theorem states that for all forces net work done on the system should be equal to total change in kinetic energy of the system.
To apply the theorem, for simplicity choose your initial and final states when the piston is at rest and for gas we can assume average velocity is zero at these two states. Now change in kinetic energy is zero which is equal to net work done by the forces. Net work done is summation of work done on piston plus net work done on gas. So this proves work done on gas is negative of the work done on piston. Mathematically,
$\Delta KE = WD_{pg} + WD_{gp}$
Since $\Delta KE=0$
$WD_{pg} = -WD_{gp}$
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Why do lights appear like straight lines on a windshield of a car? (becomes more prominent at sunset and night)
i think thats the Diffraction of light caused by the scratches in the glass. or the wiper blade leave a pattern of dirt and grime in an arc. i hope someone can confirm my logic.
my seconde question is why the light line become curved at bottom of windshield ?
| The streak is caused by refraction (&/or reflection) from scratches on the glass or from streaks of oil, grime, wax, etc., on the surface. As Farcher said, it is not caused by diffraction. The reason the streaks are curved at the bottom of the windshield is because the windshield is curved, and/or because the wiper blade is not rigidly attached to the wiper arm, so the scratches are not precisely circular.
If you look more deeply into this phenomenon, you'll find that it's possible to "draw" 3D images using scratches like those that you've observed, by controlling the center(s) of curvature of the scratches at each point on a surface.
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Electric field between 2 coaxial cylindrical charged tubes Griffith's book on electrodynamics says the electric field between 2 coaxial metal tubes of charge $+q$ and $-q$ is found by using Gauss's law, where the gaussian surface is a cylinder with radius between the outer and inner tube. But that doesn't take into account the electric field created by the outer tube, it will only find the elextric field created by the inner tube, Gauss's law only take into account the enclosed charge.
So my question is, shouldn't the electric field between the 2 tubes be a superposition of the electric field created by each tube?
| Gauss's law always holds. Imagine for a moment that the inner tube is the only tube present. Then, using a symmetry argument, you can calculate the electric field using Gauss's law. Now imagine adding some kind of charge distribution outside of your Gaussian volume. While the total electric flux through the boundary of your volume remains unchanged, its distribution might change. In other words, it might go up in some places and go down in other places. So the symmetry argument no longer works, because the charge distribution outside the Gaussian volume is no longer symmetrical.
Fortunately, in your case the outer charge distribution is symmetrical in exactly the same way that the inner charge distribution is. So the symmetry argument still goes through. In fact, you can do the calculation with only the outer tube, in which case you find that the electric field inside the tube is 0. So the electric field for the space between the tubes is the sum of the electric field contributed by each tube, just as expected.
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Meaning of the word 'canonical' in physics I often encounter the term canonical in my study of physics. What does it mean? There is canonical momentum, canonical transformations and I have even heard the phrase 'proving something more canonically'. What does the word mean in each of these contexts?
| Even in physics, the term canonical requires a disambiguation for clarity. In the contexts you were citing, it means that it is a more general form. E.g. if you are dealing with momentum, then the canonical momentum refers to $p = p + q \bf{A}$, however, momentum in a Newtonian physics course would most certainly refer to $p=mv$, thus, a professor might call momentum canonical to clarify that he does not mean the more simple version, but the more general version.
For other uses of canonical, see Wikipedia's disambiguation below:
https://en.m.wikipedia.org/wiki/Canonical
Hope this clears up your answer!
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How can I say whether a Hamiltonian is integrable or not? The transverse field Ising Hamiltonian $$ H = J\sum_{i=0}^{N}\sigma_{i}^{z}\sigma_{i+1}^{z}+h_{x}\sum_{i=0}^{N}\sigma_{i}^{x} $$ is integrable because it can be exactly solved using Jordan Wigner transformations. But the tilted field Ising Hamiltonian $$ J\sum_{i=0}^{N}\sigma_{i}^{z}\sigma_{i+1}^{z}+h_{z}\sum_{i=0}^{N}\sigma_{i}^{z}+h_{x}\sum_{i=0}^{N}\sigma_{i}^{x} $$ is a non-integrable Hamiltonian. As Jordan-Wigner transformation is a non-trivial transformaiton, just by looking at the initial hamiltonian of a system, how can I say whether it is integrable or not?
| I don't think that level spacing is "enough" to determine a system is "integrable" or not. (of course it depends on how one defines integrability.) The level spacing idea is called Berry-Tabor conjecture, and it is not proven that Poissonian distribution is intrinsic in the case of quantum integrability.
To me, the existence of extensively many conserved charges (with local or quasi-local densities) suffices the "quantum integrability". (or equivalently, the existence of Yang-Baxter equation in the system) Many systems like Lieb-Liniger model and Heisenberg XXZ chain are solved by Bethe Ansatz, while some others are solved using Yangian symmetry, e.g. long-range Haldane-Shastry model.
Of course, if a model after some transformation becomes a free model, as in the case of transverse Ising model, it is integrable. (scattering in free model is trivial and infinitely many conserved charges with local densities are easy to construct.) In general, there is no a priori way to determine whether an interacting quantum system is "integrable" or not.
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In a vacuum can a cooler body radiate Infrared radiation to a warmer body? I mentioned vacuum, because I want to discount the effects of conduction or convection. I simply want to know if some of the infrared radiation(IR) goes from the cooler body to the hotter body? How does each body know how much to radiate at any particular time? I assume that it ultimately comes down to temperature difference but how does the hotter body know what the temperature is of the cooler body and vice versa? We all know that both bodies will radiate IR at the 4th power of its temperature and obviously they will be eventually in equilibrium with each other, each of them then radiating an equal amount to each other.
| I am agree with answers by @Ricky Tensor and @V.F., but more detailed answer is "there are several scenarios". It will depend on:
*
*How much total radiation emits each of bodies
*How much each body absorbs of enother body's radiation (how close they are, how reflective they are)
*What is the rate of cooling for each body (depends on their thermal transfer properties from inside to the surface)
For example if body A emits less radiation than absorbs radiation from hotter body B, them body A will not be cooling. Its temperature will be rizing instead until its radiation become equal to absorbed radiation.
Also if body A is cool but loses its temperature slowly, and body B is very hot but cools fast - there can be a moment than body B becomes cooler than A.
But in every case both bodies A and B lose their temperature slower than they do in absense of other body.
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Are all fields in the universe we know of quantum fields? Are all fields in the universe we know of quantum fields?
Do all fields that exist must be inherently quantum in nature?
How about fields that are yet to be discovered (ie. a new field like Higgs field) , do they all have to be quantum fields?
| Currently all fundamental fields are quantum, except for gravity. For this reason Quantum Gravity is a hot area of research, but the full Quantum Gravity theory has not been developed yet. Why not?
The challenge is not just technical, but conceptual. On one hand, the Quantum Field Theory cannot consistently co-exist with any classical theory. If the Quantum Field Theory is correct, then gravity must be quantum. On the other hand, gravity cannot be just another quantum field theory, because gravity bends the space and time ("the background"), on which the Quantum Field Theory is based, and this creates unreasonable challenges (time is steady and independent in QFT, but depends on the field and is dynamic in GR) that technically result in non-renormalizability of quantization.
The only logical way to resolve this contradiction is to admit that both theories, General Relativity and Quantum Field Theory, are approximations of another unknown yet theory that in itself is neither General Relativity nor Quantum Field Theory. As mentioned in the comment of @Prahar above, chances are that gravity will be quantized in some way, but Quantum Gravity will not be a standard Quantum Field Theory.
Other possibilities also have not been ruled out, such as that gravity may not have a quantum nature or have a nature that would change our understanding of "quantum" and what exactly we mean by it. Thus the answer to your question is that no one knows yet.
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Orthogonality of Scattering states The scattering states solution ($E>V_0$) to the time independent Schrodinger equation for a finite square barrier ($V_0$ ) in an otherwise free region has the form:
$$\psi(x)=\begin{cases}e^{i k x} + r(k) e^{-i k x} &\textrm{for } x<0\\
a(q) e^{i q x} + b(q) e^{-i q x} &\textrm{for } 0<x<L\\
t(k) e^{i k x} &\textrm{for } x>L
\end{cases}$$
where $a,b,r$ and $ t$ can be found by applying conditions of smoothness and continuity of $\psi(x)$ over all $x$.
Question: I need to find whether these wavefunctions $\psi$ are mutually orthogonal or not? I need it to find the scattering amplitude between two such states through an interaction potential.
| If they are orthogonal, then their dot product or in this case the probability must be 0. To check whether they are orthogonal or not, just integrate it all over space. If it comes 0 then they are orthogonal.
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Do speakers have non-radiating nearfield? Antenna nearfield contains energy that is not radiated away,does speaker or any other acoustic device nearfield also have this non-radiating energy element that exist in space and doesnt travel away?
| In the electromagnetic case, we say that a component of the field is non-radiating if the field strength of that component decreases with distance faster than $1/r$. Since the intensity of radiation is proportional to the square of the field strength, it follows that the intensity of a non-radiating component decreases faster than $1/r^2$. The total power radiated through a shell of radius $r$ is proportional to the product of intensity and area (which is $4\pi r^2$), so any intensity decreasing faster than $1/r^2$ leads to a decreasing radiated power with distance, limiting to zero power radiated at infinity.
The same argument can be applied to the acoustic case, as long as we make the right identifications between concepts. Sound intensity is analogous to electromagnetic intensity (as they're both power per unit area), and, since acoustic radiated power is proportional to the square of the sound pressure, the pressure plays the role of the electromagnetic field strength. The only thing left to demonstrate is the existence of components of the pressure field that decrease faster than $1/r$.
For an acoustic point monopole source oscillating at frequency $\omega$ with strength $F$, the pressure as a function of distance is
$$p=\frac{F}{4\pi r}e^{i\omega(t-r/c)}$$
which is exclusively proportional to $1/r$. (Just as in electromagnetism, the real part of this complex pressure is the actual sound pressure.) So an acoustic point monopole source has no non-radiating component.
For an acoustic point dipole source of dipole moment $\vec{F}$ and frequency $\omega$, however, the pressure (retaining the same notation) is
$$p=\frac{ik\cos\theta}{4\pi r}\left(1+\frac{1}{ikr}\right)\vec{F}e^{i\omega(t-r/c)}$$
which has a radiating component, proportional to $1/r$, and a non-radiating component, proportional to $1/r^2$. In the near field, the non-radiating component dominates. Similarly, a point acoustic quadrupole source has, in general, a radiative term proportional to $1/r$, and two non-radiative terms, proportional to $1/r^2$ and $1/r^3$, respectively.
As such, acoustic sources with non-radiating components exist, and they're actually quite common. Any acoustic dipole has a non-radiating component.
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If a drop of water hits something at bullet speed, will it damage it? I can't find this anywhere. If a drop of water hits a glass for example, at the speed of a bullet, will it damage it? Or will it just make the same as a raindrop?
| Wet steam erosion is a big problem in steam turbine design, as you can see from this image:
So yes, water droplets with enough energy can certainly cause damage to even objects made of steel. With enough energy, it could certainly cause damage (or destroy) your glass.
Water jet cutting is another good example of high-energy water being used to damage (cut) something:
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All central forces are conservative forces, but are all conservative forces central forces? I have just been introduced to the concept of central forces, and to the fact that they are per definition conservative forces. I have looked up several examples of central forces (gravity, electric, and spring), but they cover just about all the conservative forces I have ever heard about. Are there any conservative forces that are not central?
There must be, because otherwise there would not be any point in having a subcategory for central forces, yet I cannot find any examples anywhere.
| Paul's answer is great. But I just found out an error in the background information you mentioned: central forces aren't necessarily conservative forces. I'm writing it down so you may have a clearer understanding of the logic relationship between a 'central' force and a 'conservative' force.
For example we may take $$\vec F = x \cdot \hat r$$ and with a bit of calculation we have$$\frac{\partial F_y}{\partial x}-\frac{\partial F_x}{\partial y}= \frac{y^3+yz^2}{(x^2+y^2+z^2)^{3/2}}-\frac{-yx^2}{(x^2+y^2+z^2)^{3/2}}=\frac y r \neq 0$$
Since that's the $\hat z$ term in $\nabla \times \vec F$, we can tell that the curl is not zero, hence the force being nonconservative.
For a central force to be conservative, it must also be spherically symmetric, namely its magnitude must be a function of distance $r$ only. With that we can express $F$ as the gradient of some scalar $T$
$$\vec F = f(r)\cdot \hat r = \nabla T$$
with T being the indefinite integrationof $f(r)$
$$T = \int f(r)$$
Since the curl of a gradient is always zero,that gives us $\nabla \times \vec F = 0$, the proof of $F$ being conservative we're looking for.
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Which is more efficient in stopping a bullet, small cubes or fine grains of sand Recently, I was presented with the following problem, relating to pressure: which would be more efficient in stopping a bullet, small cubes or fine grains of sand.
Using my intuition, I would say the small cubes, since it has a larger surface area as compared to the fine grains of sand, and would likely have a strong structural integrity that is difficult for the bullet to penetrate. I tested this out by punching my Ricola candies (cuboid shaped) as opposed to the sand on the beach, and I discovered that my hypothesis was correct to a certain extent, since it was much more painful punching the candies.
I could not therefore understand why in the war, sandbags are used to block bullets? My teacher also disagreed with me by raising the fact that sand is highly compact and it is thus very hard for a bullet to penetrate. I would like to ask, is my idea correct?
| When you hit sand, the sand has time to move out of the way. It behaves like a fluid in that regard - dipping your finger in water isn't painful. But doing a belly-flop off the high dive into the water is extremely painful, the reason being that the water has no time to get out of the way, so it behaves more like a solid.
Sand is the same way. When a bullet hits sand, the sand has no time to clear out of the way of the bullet, so it takes the brunt of the impact and behaves like a solid. Sandbags are therefore somewhat useful against bullets - the problem is that after a few rounds, sand starts to leak out of the bag. Sand's greatest utility is the fact that it is cheap and abundant, and it doesn't take a long time to place down sandbags in front of your position.
It is also notable that sand is essentially a bunch of small cubes. Your candies won't stop a bullet because they are too big - they won't be able to compact against each other like sand can, and thus the bullet will just pass right through.
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Would a conical siphon be able to lift water greater than 10.1 m? In a discussion with my father, he argued that a siphon can only pull water up to a limit of ~33 ft (10.1 m). I understand that this limit would apply to water with a density of 1.0 g/cm at sea level (101,325 Pa) being pulled with a perfect vacuum (0 Pa) and ignoring the water's vapor pressure. I feel this limit only applies to cylindrical tube siphons, and that the limit could be increased by using a conical siphon with the wider portion towards the top and narrow portion towards the bottom. Is this true and if so, how can I explain it to my father?
My reasoning is that in theory, the water's surface area at the interface with the vacuum does not determine how far water will travel up the siphon, but instead how much water mass the siphon will support. Being that a cone's volume is 1/3 that of a cylinder with the same diameter and height, I would hypothesize that a conical siphon would be able to lift water nearly 3x as high as a standard siphon.
Being that a conical siphon isn't exactly useful and easy to use in everyday applications, I believe this could actually be accomplished by carefully calculating gradually increasing stepped cylindrical tubes as you increase height to perform the same task. That is start with a tube diameter of x, length 5 m, connect to another tube diameter 2x length 5 m, which is again connected to another tube diameter 4x, length 5 m. This 15 m setup, filled with water, would contain the same mass of water as 1 * 5 m + 1/4 * 5 m + 1x/16 * 5 m = 6.5625 m of the top section of tube, and thus could work as a siphon, as it's well under 10.1 m.
| Your father is correct in this case. The pressure of a fluid in hydrostatic equilibrium does not depend on the shape of the container, only the depth, the density, and the pressure at the surface. If a fluid is exposed to atmospheric pressure on one side and vacuum on the other side then the vacuum side will only rise high enough for the fluid pressure to equal the atmospheric pressure. Again, this height is independent of the shape.
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