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Does Big Bang predict the size of the fluctuations in the CMB? From this Ethan Siegel’s article
The temperature fluctuations in the CMB are only 1-part-in-30,000, thousands of times smaller than a singular Big Bang predicts.
Does big bang predict the size of the fluctuations in the CMB?
Do we know what caused the fluctuations in the CMB?
| The big bang model does not predict the size of the CMB fluctuations - rather, the universe we see today is a consequence of those fluctuations since all the structures we see today (clusters of galaxies and superclusters, filaments etc) must have grown from structures that are traced by those fluctuations.
Given that, there was an expectation for the size of the fluctuations before they were measured. One can calculate what size of fluctuations can evolve to produce the present day universe, providing you add the right ingredients (including lots of dark matter, which allows the initial fluctuations to be as small as they are).
The fluctuations themselves arise in two ways. There is a component produced by quantum physics in the initial moments. It is thought that these fluctuations are then amplified in scale by a burst of cosmic inflation in the first fraction of a second. It is this process that predicts the (observable) universe should have a very uniform temperature, since all parts of it were in "thermal contact" in the past. The inflation also predicts a rather flat spectrum to the remaining ripples, with similar power on all spatial scales.
A second component is produced by compressions and rarefactions in the primordial gas as the CMB is formed, about 350,000 years after the big bang. These modify the primordial fluctuations in ways that are characteristic of the cosmological parameters (the Hubble parameter, the matter density in the universe, the amount of dark matter and energy and so on). A measurement of the spatial spectrum of fluctuations therefore enables the inference of the cosmological parameters. Or, you could say that the size and spectrum of the fluctuations ought to be consistent with the cosmological parameters that have been inferred by other methods.
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A Number State of Light through a dielectric If a number state |1> picks up a factor of $e^{-i\phi}$ on going through a certain medium, what factor will a number state |N> pick up? Will it be $e^{-iN\phi}$ or $e^{-i\phi}$ or something else? And what is the justification? Thanks.
| As long as it is a linear optical medium - which is typically the case up to good accuracy - it will act as $a^\dagger\to e^{-i\phi} a^\dagger$ on the creation operators (i.e. in Heisenberg picture). Thus, it acts as
$$
|n\rangle =\tfrac{1}{\sqrt{n!}}(a^\dagger)^n|0\rangle \ \to\
\tfrac{1}{\sqrt{n!}}(e^{-i\phi}a^\dagger)^n|0\rangle=e^{-in\phi}|n\rangle\ .
$$
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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 ?
| It is probably not diffraction unless you can see a separation of colours.
The most likely explanation is scratches on the windscreen (and the covering of dirt) caused by the wiper blades.
More prominent at sunset because the Sun is low in the sky and the windscreen is being hit by a shaft of light.
You will note from the enlargement the absence of colour and the streaks visible adjacent to the shaft of light.
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Only sea water appears blue in color, why this is not happening in river water? Is the salt in the water the reason for scattering sunlight into blue?
| The color of a body of water is usually due primarily to light reflected from the sky and surroundings. Of course the ocean usually doesn't have a lot around it except the sky, but in a place where a lot of light from vegetation reflects off the water, the ocean can appear green.
This is a fresh-water lake; you can see reflections of white sky, blue sky, green trees, and so on.
The colors in the water, especially those in reflections from the sky, tend to be darker and different because the colors in the sky are often polarized. When light glances off a smooth surface, the surface reflects light with horizontal polarization more effectively than light with vertical polarization (see Brewster's Angle), so some skylight can end up darkened - or brightened - relative to other colors depending on the angle of polarization and the angle of incidence.
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Electric field of an infinite sheet of charge I am trying to derive the formula for E due to an infinite sheet of charge with a charge density of $ \rho C/m^2$. I assumed the sheet is on $yz$-plane. I used Coulomb's law to get an equation and integrated the expression that over $yz$-plane. But, I have not succeeded in deriving the correct expression. The answer I am getting is $0$.
Below is the picture of my work. Kindly, have a look and let me know where did I make mistakes.
In actual, E due to a charge sheet is constant and the correct expression is
E $=\rho/2\epsilon$0 aN , where aN is unit vector normal to the sheet.
| Use cylindrical coordinates. The field (on axis) of a ring of charge (radius $R$, charge density $\lambda$) goes like:
$$ E(z) = \frac{1}{2\epsilon_0}\frac{ R z}{(z^2 + R^2)^{\frac 3 2}} $$
then integrate over $R$, using:
$$ \int{\frac{ R z}{(z^2 + R^2)^{\frac 3 2}dR}}=-\frac z {\sqrt{z^2+R^2}}\rightarrow 1$$
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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?
| Sometimes it just means "official" or "standardized" or "really important", but usually it has the more precise meaning "relating to the Hamiltonian formulation of classical mechanics". The canonical momenta are usually first introduced in the Lagrangian framework, but they are the momenta that appear in the phase space of Hamiltonian mechanics. Canonical transformations are symmetries of that phase space that preserve the symplectic structure. Canonical perturbation theory is formulated within Hamiltonian mechanics. The canonical commutation relations are a quantized version of Poisson brackets (as per Dirac's quantization rule).
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Galilean transformations of velocity If I perform a Galilean boost $$x' = x - vt \\ t'=t$$ between two frames $S$ and $S'$, observers in each frame would disagree on the velocity of a particle because
$$ \frac{dx'}{dt'} = \frac{dx}{dt} - v. $$
Well Galilean transformations preserve the spatial intervals $\Delta x$ and time intervals $\Delta t$, so surely they should preserve velocity
$$ u = \frac{\Delta x}{\Delta t}? $$
There is obviously something going wrong here with my reasoning. I know the spatial interval $\Delta x$ is defined at constant time $t$, but if I was in the Galilean boosted frame $S'$ observing an object moving between two points $x_1$ and $x_2$ in $S$, I would observe that the interval $\Delta x$ between these two points would be a constant over time anyway so I could still conclude that the particle travelled a distance of $\Delta x$ in time $\Delta t$. What is going on?
| The confusion is in what $\Delta x$ really is.
Let's say two observers, one in S and the other in S', see a meter stick. They will both agree it is one meter long. So if $\Delta x$ is a set length then yes, this is invariant under the transformation.
Now let's say an observer in S is holding this meter stick. Over any time interval, this observer does not view the meter stick as moving. However, an observer in S' will definitely see the meter stick as moving. So in a given time $\Delta t$, the distance covered by the meter stick ($\Delta x$) will depend on which frame you are in.
Your discussion about velocity should focus on the latter scenario. This is in fact just relative velocity. Different frames will measure different velocities of an object depending on their velocity relative to the object (this is somewhat of a redundant or circular statement). Their measurements are related by their velocities relative to each other.
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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.
|
How does each body know how much to radiate at any particular time?
They don't. If you assume that both bodies are "black", that is they radiate electromagnetic waves due to their temperature as described by the black body radiation equations, they do it because that's the way nature is.
[...] how does the hotter body know [..] the temperature [...] of the cooler body and vice versa?
Again: the bodies are independent of each other with regard to the emitted radiation. If the colder body is hot enough to emit IR radiation at all, it will do so regardless of other bodies around it.
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How can a transformer produce a high voltage and a low current? I understand that in ideal transformers, power is conserved. Because of this the product of voltage and current in the secondary winding is a constant.
This means that voltage and current are inversely related, which seems unintuitive because they are directly related by ohms law.
Shouldn't the emf induced in the secondary winding by the alternating magnetic flux be directly related to the current by some constant, such as the resistance of the secondary winding?
I also came across a term known as impedance that seem to be related to the question. Wondering if it is of any relevance.
| Here is an analogy which might help you grasp how a transformer works.
Think of an electrical transformer as if it were instead a car transmission. The ratio between torque and RPM at the input shaft is altered by the gears to yield a different ratio of torque to RPM at the output shaft.
As an example, in first gear the output shaft turns, say, at 1/4th the speed of the input shaft- but the torque at the output shaft is 4x that applied to the input shaft. the transmission "steps up" the torque and "steps down" the RPM, the torque ratio is 1:4 and the RPM ratio is 4:1.
Now remember that torque is analogous to voltage and RPM is analogous to current. Power is torque x RPM, so since power is conserved in the transmission, torque x RPM at the input shaft = torque x RPM at the output shaft and by analogy voltage x current at the primary winding = voltage x current at the secondary.
This means our transformer analog then steps current down by 1:4 and steps voltage up by 4:1.
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Why not free electrons in atom doesn't radiates em waves\photons? Why not free electrons in atom doesn't radiates em waves\photons, although they move with acceleration? Like 1s electron of Titan, it doesn't emits em waves, yes? Why?
| In quantum mechanics there are no individual electrons which dart about like ping bong balls. But there is an electric charge density which can be derived from the total wave function of the system. In general this charge density is not stationary. To the extent that it is moving, oscillating or accelerating, it will emit or absorb radiation. And that emission or absorption can be calculated exactly by using Maxwell's Equations.
The absoption of light, which occurs as a result of the coupling of the conduction band electrons in a metal to the free electron states outside the metal is, governed by Maxwell's Equations (together with Schroedinger's Equation). The linewidths of the absorption lines in the hydrogen spectrum are governed by Maxwell's Equations (together with Schroedinger's equation). The emission of light from a heated tungsten filament is governed by Maxwell's Equations, as applied to the oscillating charge distributions within that metal filament (which are given by Schroedinger's Equation.)
It is not necessary to talk about photons to explain any of these things, and it is not particularly helpful either.
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Woodworking clamps, does force add up? I was watching a woodworking video about glue, and the guy was clamping two pieces of wood together using a total of 8 clamps. He argued that by doing so he would apply 8 times the maximum force of 150N (a property of the clamp), resulting in 1200N in total.
I think he's wrong. I think the force of 150 N is only working locally where the clamp is and will decline drastically radially from that spot. And so the clamping force on any given spot on the board will never exceed the max. force of the clamp.
Who's right?
| He's right - the forces of the clamps will add up. You seem to be confusing force and pressure. The pressure from each clamp would reduce radially outwards from each clamping point, as you describe (although adding clamps will increase the average pressure acting across the entire length of the planks, thus increasing the force!).
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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.
| A constant force is conservative but not central.
For example: $\vec F=F \hat x$
You can check that the curl of this force is $0$, hence it is conservative. Its potential energy function in 3D space would just be $V(x,y,z)=-Fx+V_0$, where $V_0$ is some constant value.
An example of this is the approximation of gravity near the Earth's surface. In this regime the force is assumed to be constant, and we get the same form as above for the potential energy. Of course gravity on larger scales is a central force for, say, planets in orbit around a central star, which is why I gave the general form first.
Another example of a conservative, non-central force is one that is a superposition (sum) of two central forces.
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Demagnetising $H$ field If we have a permanent magnet we know that we have a $H$ field opposite in direction to the residual magnetic field $B_r$. The $H$ field causes a demagnetisation, but how does it do it? We know that the physical field (the field which causes the Lorentz force) is the magnetic field $B$.
| The H field is caused by the positive and negative poles that develop at the end of the magnet in the opposite direction of the B field. Whether H or B is the real field is somewhat a matter of taste, for neither is directly measurable in any easy way and both are the result of averaging; some even say that within the matter they are both just mathematical fictions. There is one well-known and often referenced experiment by Rassetti: ‘‘Deflections of mesons in magnetized iron,’’ Phys. Rev. 66, 1–5 1944, (note the date!) that claims to verify that the Lorentz force formula does indeed have B in it and not H. Here is a quote from Brown: Magnetostatic Principles in Ferromagnetism
The safest view to adopt regarding the field vectors Band H, as they have been defined at internal points, is that they have definite mathematical meanings in terms of a given magnetization distribution, but that no physical meaning should be attributed to them. Physical interpretations are possible in special cases; for instance, in the case of a homogeneous isotropic toroidal specimen, with a toroidal magnetizing winding, the field vector H is identical with the flux density of the winding alone; but introduction of a gap, or of inhomogeneity or anisotropy, immediately destroys this identity. When physical interpretations are possible, they should be based on direct demonstrations in the relevant specific
cases.
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On Harmonics In Physics I am an IB physics student. I am very confused about the concept of first, second, third, etc harmonics.
My questions are:
*
*How does a wave get from first to second harmonic, and from second harmonic to third harmonic etc? What do we need to do to a wave to get it there? Increase the speed, frequency, wavelength, etc?
*Let's take a free ended rope being oscillated at second harmony. Compared to the fundamental wave, the frequency of the wave now is two times higher, and the wavelength is two times less than the fundamental wave. Does this mean that at second harmony, the speed of the wave increases?
Can you please try to explain this simply, at the level of a someone who still hasn't learnt Calculus based physics? Thanks.
| Harmonics are derived from a mathematical concept called fourier series. If you have a $1\lambda$ standing wave and you have a measuring device in the middle, you'd find a 0, if you then created a standing wave with $2\lambda$ and you measured at the exact same spatial coordinate, you'd find another 0. So, by measuring only one position on the rope, we can't differentiate if this is $1\lambda$, $2\lambda$ or $3\lambda$ and so fourth. This is the nature of harmonics.
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How to derive this expression for the free scalar field in QFT? (Peskin & Schroeder) In the introductory text to quantum field theory by Peskin & Schroeder, they state that in analogy to the simple harmonic oscillator in quantum mechanics, the free scalar field can be expressed as:
$$\phi(\vec{x}) = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{2 \omega_p}} \left( a_p e^{i \vec{p}\cdot\vec{x}} + a^{\dagger}_p e^{-i \vec{p}\cdot\vec{x}} \right) \tag{2.25}$$
In quantum mechanics $\phi$ would be written as:
$$\phi = \frac{1}{\sqrt{2 \omega_p}} \left( a + a^{\dagger} \right)$$
I can see the similarities between the two expressions, as well as the fact that one may expand the free Klein-Gordon field as:
$$\phi(\vec{x},t) = \int \frac{d^3 p}{(2\pi)^3} e^{i \vec{p}\cdot\vec{x}} \phi(\vec{p},t).\tag{2.20b} $$
However I don't get how to reach the final expression given above, especially the exponential with negative sign in the second term. It's probably just a small thing that I am not seeing, but I would be thankful if somebody could give me a hint.
| I'll adopt the abbreviation $kx:=k_0x^0-\mathbf{k}\cdot\mathbf{x}$.
The Klein-Gordon equation $(\square +m^2)\phi=0$ can be solved by a Fourier transform. Writing $\phi(x)=\int d^4ke^{-ikx}\tilde{\phi}(k)$ we get $(k^2-m^2)\tilde{\phi}(k)=0$, i.e. $\tilde{\phi}(k)=\tilde{\varphi}(k)\delta(k^2-m^2)$ for some function $\tilde{\varphi}(k)$. Using$$\delta(k^2-m^2)=\dfrac{\delta(k_0-\omega_\mathbf{k})+\delta(k_0+\omega_\mathbf{k})}{2\omega_\mathbf{k}}$$gives$$\phi(x)=\int\dfrac{d^3k}{(2\pi)^32\omega_\mathbf{k}}(a_+(k)e^{-ikx}+a_-^\ast(k)e^{ikx})$$with$$a_+(k):=(2pi)^3\tilde{\varphi}(\omega_\mathbf{k},\,\mathbf{k}),\,a_-(k):=(2pi)^3\tilde{\varphi}^\ast(-\omega_\mathbf{k},\,\mathbf{k}).$$For real $\phi$, $a_-=a_+^\dagger$, so defining $a:=a_+$ we're done. (You have an erroneous $\sqrt{}$ sign in the Lorentz-invariant integration operator.) Note in particular your $e^{i\mathbf{k}\cdot\mathbf{x}}$ coefficient of $a_k$ is $e^{i(k_0x^0-kx)}$, but I've absorbed the $e^{ik_0x^0}$ factor into my definition of $a_k$ to make the above result's Lorentz-invariance manifest.
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Potential difference across the strips of an inductor The voltage across the ends of an inductor is $L (di/dt)$. Applying the loop law $V = L(di/dt)$, so the voltage across the strips at a distance x measured from the "LEFT END" should be $V \frac xl$ because the inductance of x length is $ L \frac xl$.
But according to the answer given it should be $ V( 1 - \frac xl)$. Why?
What do we mean when we say voltage across strips? Any help is highly appreciated.
| There are 2 different potential differences in this question. One is the PD along each strip in the direction of $l$ - ie horizontal. I shall call this $V_x$. The other is the PD between the 2 strips in the direction of $b$ - ie vertical - at the same horizontal position $x$. I shall call this $V_y$, but note that like $V_x$ it also depends on $x$.
Option D is asking for an expression for $V_y$.
If the current has reached a limiting constant value, which will happen after the battery has been connected for a long time, then the PD $V_x$ along each strip depends only on its resistance not its inductance, because there is no back emf in an inductor if the current through it is not changing. In this case the vertical PD between the strips increases linearly from $0$ to $V_0$ when measured from the RH end where they are shorted : $$V_y=V_0 \frac{x}{l}$$
Before the current becomes constant there is a potential drop $V_R=IR$ across each strip due to its resistance $R$. Initially $I$ is very small so $V_R=IR \ll \frac12 V_0$. At the same time there is a back emf of $V_L=L\frac{dI}{dt}$ while the current increases, where $L$ is inductance. This is also a potential drop across the inductor, and it is linear (proportional to $x$) because inductance $L$ like resistance $R$ is proportional to length. Initially $\frac{dI}{dt}$ is large and $V_L =L\frac{dI}{dt} \approx \frac12 V_0$.
The sum of these 2 PDs equals the PD supplied by the battery, the same as when there is a separate resistor and inductor in series : $$V_R+V_L=\frac12 V_0$$ So the total PD $V_x$ along each strip varies linearly even when the current is changing, and the PD $V_y$ between the 2 strips again increases linearly from the RH end, as in the above equation.
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On String Theory, Supersymmetry and prospects of a theory of everything The fundamental postulate of string theory is that matter is composed of tiny vibrating loops of string, and each vibrational mode of the string corresponds to a different fundamental particle. Now, since there exists infinitely many possible vibrational modes of the string, it follows that there exists infinitely many matter particles (fermions). Since string theory requires supersymmetry to be consistent, it follows that there would exist infinitely many bosons, hence infinitely many forces. Now, would a phenominologically consistent theory of ''everything'' be possible if there exists infinitely many forces ?
| QCD is already a little like this. Mesons and baryons form families of increasing spin and mass ("Regge trajectories "). Nucleons are bound together by exchange of spin-0 pions, but heavier spin-1 vector mesons also play a role, and even tensor mesons of spin-2 and higher. The attempt to develop a theoretical framework which could describe the interactions of such families of particles is exactly how string theory was discovered. In fact, the mathematical structure of interaction (Veneziano amplitude) was found first, then people realized that this is how the excitations of a relativistic quantum string interact.
Also: the higher modes are increasingly heavy (because of the energy they contain), and exchange of superheavy strings corresponds to "forces" felt only over ultrashort distances (before they decay to lighter states). Nonetheless, they are part of what makes string theory different from other forms of quantum gravity at short distances, an aspect of what makes it work. So they're not a bug, they're a feature.
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Is this statement of conservation of charge circular? According to Wikipedia:
A closed system is a physical system that does not allow certain types of transfers (such as transfer of mass and energy transfer) in or out of the system.
According to my textbook, the principle of conservation of charge is:
The algebraic sum of all the electric charges in any closed system is constant.
Isn't this circular logic? In terms of charge, a "closed system" is one in which charge can neither exit nor enter. If the charge neither exits nor enters, then of course the sum thereof stays constant.
Or is the principle saying that the only way for the sum of charge in a system to change is via transfer of charge in or out of the system? (In this case, wouldn't it make more sense to state the principle as "charge can neither be created nor destroyed"?)
| You're right that it's a bit circular as stated. The more rigorous way to state a conservation law is something like:
The rate of change of [quantity] in a bounded system is equal to minus the rate at which [quantity] leaves through boundaries of that system.
A "closed system" is then a system for which both of these rates are zero, i.e., the [quantity] is not moving through the boundary of the system.
The version you propose, "[quantity] can neither be created nor destroyed", is closer to this more rigorous statement. But the rigorous statement is a bit stronger than this. If a charge were to suddenly teleport across the room, without passing through the points in between, this would satisfy your version of the statement; but it would not satisfy the rigorous version of the conservation law above. Moreover, it's perfectly possible in particle physics for charges to be created or destroyed, so long as equal amounts of positive and negative charges are created or destroyed. Your version of the statement would seem to outlaw these events, but the rigorous version does not.
| {
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Is the reason the sunshine is 'extra bright' after rain due to refraction of the additional water in the air? Quite frequently after the sun comes out after rain I experience a 30 minute period where the sunshine is 'unusually bright'. Such that it makes my eyes water.
My question is: Is the reason the sunshine is 'extra bright' after rain due to refraction of the additional water in the air?
| There are some reasons I can think of, and they are not mutually exclusive if true.
1) There is more moisture in the sky. This enables sunlight to scatter more, and so the sky appears to be brighter.
2) The rain is able to clear dust, pollutants, etc. which normally block sunlight. With less of these particles in the air, more light is able to come through the atmosphere.
3) The wet surfaces present after raining are more reflective, and so more light reaches your eyes.
4) If your eyes were used to the limited light during the rain, then perhaps your eyes are more sensitive to the brightness.
In any case, I don't think this has anything to do with refraction though. If refraction properties did change, this just influences the angle at which the light appears to come from, but this would not influence how much light actually reaches your eyes.
| {
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What is the educational path to properly understanding the Higgs Boson? I want to understand the Higgs Boson. Not in terms of analogies and metaphors, but in terms of hard math.
Assume I have a solid background in pre-university maths and physics. (I have many bits and pieces of higher-level stuff, but let's ignore that knowledge because it is not comprehensive or cohesive.)
What sequence of things do I need to learn in order to properly understand the Higgs Boson?
| The basic answer is: learn quantum mechanics, then quantum field theory, then the standard model.
By the way, it's the Higgs field that does all the interesting things - the Higgs boson is, to our knowledge, the most useless aspect of the Higgs field. It does nothing of any importance, and yet, because the Higgs field is so central, the Higgs boson also ends up interacting with most other particles.
There is a question as to why the Higgs boson isn't rendered superheavy by virtual particles, and the specific value of its mass even suggests that it has been finetuned by an unknown mechanism. So the Higgs boson may yet be a window on important undiscovered physics. But for now, it's probably the Higgs field that you want to understand.
| {
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derivative of the electric field along the normal to the surface near the conductor how to derive the formula approves the derivative of the electric field along the normal to the surface near the conductor is inversely proportional to the principal radii of curvature?
$\frac{\partial E}{\partial n}=-E \left(\frac{1}{R_1}+\frac{1}{R_2}\right)$.
I tried to derive it by Gauss's theorem for cylindrical coordinates, but it didn't help
| This formula can be deduced by applying the Conservation of Flux : the flux through any volume above the conductor is zero because it encloses no charge.
Imagine such a volume close to the surface of the conductor. In profile this volume (vastly exaggerated in size) looks like the shaded region ABCD in the diagram below in each of the two principal planes of curvature. The conductor has local radius of curvature $R$ in this plane, and the volume subtends angle $\theta$ from the local centre of curvature. The red arrows represent electric field lines, which are perpendicular to the surface of the conductor and parallel to the sides of the shaded volume.
The lower and upper faces of this volume are approximately rectangular and have areas $$A=R_1\theta_1 R_2 \theta_2$$ $$A'=(R_1+z)\theta_1 (R_2+z)\theta_2=(R_1R_2+[R_1+R_2]z+z^2)\theta_1\theta_2$$ In the limit that $z \to 0$ so that terms in $z^2$ can be neglected, the increase in area is $$\Delta A=(R_1+R_2)z \theta_1\theta_2=(R_1+R_2)z \frac{A}{R_1R_2}=(\frac{1}{R_1}+\frac{1}{R_2})zA$$
The flux through these two faces are the same so $$EA=E'A'=E'(A+\Delta A)$$ $$E-E'=\frac{\Delta A}{A}E'=(\frac{1}{R_1}+\frac{1}{R_2})zE'$$ $$\lim \limits_{z \to 0}\frac{E-E'}{z}=-\frac{\partial E}{\partial z}=(\frac{1}{R_1}+\frac{1}{R_2})E$$
| {
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Intuitively, why do attempts to delay hitting a black hole singularity cause you to reach it faster? In general relativity, proper time is maximized along geodesics. Inside of a black hole, all future-oriented timelike trajectories end at the singularity. Putting these two facts together, we find that any deviation from geodesic free fall decreases the proper time before one hits the singularity, so as Carroll says, "you may as well sit back and enjoy the ride."
[Edit: As Dale points out, the Schwarzchild singularity does not consist of a single spacetime event, so this argument fails in general: one can in fact extend the proper time experienced by a free faller between the event horizon and the singularity to some extent by firing rockets inward. But this cannot occur appreciably in the limiting case where the free fall begins at rest just outside the horizon, which I'll assume to be the case.]
This is of course very much counter to nonrelativistic intuition. In Newtonian gravitation, if you fire your jetback inward, you slow your inward fall and buy yourself more time. Is there any physical intuition for why this isn't the case inside of a black hole (if you start free falling from rest at the horizon)?
| My (very limited) intuition for this is that once you cross the event horizon, the singularity is not so much a distant point in space as it is a moment in future time.
In other words, within the event horizon you're firing your rockets not to avoid some point $(x,y,z)$, but rather to avoid next Thursday. From here, I use my intuition about time dilation and the fact that geodesics are trajectories of maximum proper time.
I'm by no means a GR expert so if this picture is wrong, corrections are more than welcome :)
| {
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What is the value of absolute zero: $-273.15\ \rm °C$ or $-273.16\ \rm °C$? What is the value of absolute zero?
0K=
-273.15 °C
Or
-273.16 °C
It has been confused in different scientific scriptures. The first definition on Dictionary.com, for example.
| According to BIPM (International Bureau of Weights and Measures),
The kelvin, unit of thermodynamic temperature, is the fraction $\frac{1}{273.16}$ of the thermodynamic temperature of the triple point of water.
Hence the triple point of water is $273.16\ \rm K$. Comparing to the $0.01\ \rm °C$ reference (i.e. $273.16\ \mathrm{K}$ is the same as $0.01\ \mathrm{°C}$), we see that absolute zero is $-273.15\ °\rm C$.
BIPM is reliable; see this page on their website about the Kelvin scale. They're the group which publishes the SI brochure.
| {
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Spin glasses overlap Suppose you have a spin glass simulation in which the standard Metropolis MC algorithm is used to sample phase space.
The we calculate the equivalent for the lattice system of the self intermediate scattering function, namely:
$$
C(\tau) = \frac{1}{N}\sum_i^N\left<\sigma_i(t)\sigma_i(t+\tau)\right>
$$
in which $\sigma_I$ is the i-th spin, and the sum is average over all $t$, steps of the MC simulation.
What information can be extracted from this correlation function? Is this the same information we can obtain from the self intermediate scattering function for "normal" (= NON spin) glasses?
Is it correct to say that the long asymptotic long time value is the $q_{EA}$ (Edward-Anderson)?
I don't know how this function is called in the spin glass framework. Do you know how can I find it in the literature and/or have you some paper to suggest about it?
| This is called the Temporal Autocorrelation Function. A bit like spatial correlations, it is related to critical dynamics. For example, in a simple, non frustrated 2d Ising model, time autocorrelation is low in the high temperature phase (which is noisy and therefore forgets quickly), and also in the low temperature phase (which remembers for a long time, but has no variability), but is high in the vicinity of the critical point. In other words, it tells us how much a spin at a given time is related to a spin at a later time. $C$ is obviously a monotonically decreasing function.
I don't know much about actual spin glasses, but your intuition that $C = q$ at infinity seems correct to me, and is reported in these notes. These could also be a starting point to find further literature.
| {
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How can the solutions to equations of motion be unique if it seems the same state can be arrived at through different histories? Let's assume we have a container, a jar, a can or whatever, which has a hole at its end. If there were water inside, via a differential equation we could calculate the time by which the container is empty.
But here is the thing: through the differential equation, with initial condition, I shall be able to know everything about the container: present past and future.
But let's assume I come and I find the container empty. Then
*
*It could have always been empty
*It could have been emptied in the past before my arrival
So this means I am not able to know, actually, all its story. Past present and future.
So it seems there is an absurdity in claiming that the solution of the differential equation is unique. Where am I wrong?
|
So it seems there is an absurdity in claiming that the solution of the differential equation is unique. Where am I wrong?
You seem to be taking that the fact that an equation has a unique solution to imply that that equation is the only one with that solution.
An even simpler example:
The solution of $x=1+2$ is unique - there is only one value of $x$ which satisfies the equation, and that value is $3$.
The solution of $x=4-1$ is also unique, and its also has a unique solution where the value of $x$ is $3$.
Given only the statement that the value of $x$ is $3$, you do not know which equation this was a solution of.
The fact that an equation has a unique solution does not imply that that particular equation is the only one which yields that solution; there will be infinite such equations for which the same state is their unique solution.
| {
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Applying the Heisenberg uncertainty principle to photons The speed of light is a universal constant, so we definitely know the speed of the photons. If we know the speed, then we should not have any information about their location, because of Heisenberg's uncertainty principle. But I'm one hundred percent sure when light goes through my window.
Why is this so?
| The Heisenberg Uncertainty Principle does not involve speed. It involves momentum, and this is one of the places where that distinction is very important. Photons all travel at the same speed, yes, but their momentum can take on any value. As such, the uncertainty in its position and the uncertainty in its momentum are still linked in the same way they would be for ordinary matter.
In fact, due to the fact that photons are traveling at the speed of light, their energy and momentum are related by $E=pc$. Applying the usual Planck-Einstein relation $E=hf$, we can see that an uncertainty in the photon's momentum is also directly proportional to the uncertainty in the photon's frequency, which might be easier to picture (i.e. in the case of photons, the Heisenberg Uncertainty Principle links the uncertainty in their position and the uncertainty in their frequency).
| {
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Why is it much more difficult to horizontally throw a toy balloon than a football? If you horizontally throw a sphere of radius $R$ it will feel, in this direction, a drag force due to air. Assume the drag is given by Stokes law, $F_D=6\pi\eta R v$, where $\eta$ is the air viscosity and $v$ is the horizontal speed. This force cannot "see" the internal structure of a toy balloon, a football or even a metal sphere. However, anyone who ever played with balls and toy balloons noticed that for the same throwing, the ball will have higher horizontal reach for the same time interval. Just think about someone kicking toy balloons and footballs and the distances reached in each case. How is the resistive force considerably greater for the toy balloon?
Even if we consider a quadratic drag, $bv^2$, I suppose the coefficient $b$ would depend only on the fluid and the geometry of the bodies. Again the drag would be equal.
Another way to put this question: How does the density of the sphere contribute for the resistive force?
| Given about the same size of a balloon and a football, the main difference in how far they advance horizontally is their kinetic energies, not air resistance.
Since the velocity of an arm is limited, the initial velocity of a balloon would be similar to the initial velocity of a football and, since the football is much heavier, its kinetic energy will be much greater and it'll take much longer to lose (going against air resistance) than for the balloon.
The reason a balloon appears to have greater air resistance than a football is because air resistance is a dominant force that has to be overcome when throwing a balloon, while the dominant force for a football goes against its "inertial resistance", i.e., is used to speed up its mass.
So, for the football, air resistance represents a small faction of the total "resistance" and, therefore, is almost imperceptible.
| {
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Difference between pressure and temperature If I am given the average kinetic energy of the molecules of a gas or a liquid, how can I tell if the fluid will burn me/crush me/both if I immerse my hand in it?
Equivalently, what is the difference between heat transfer and momentum transfer at a molecular level?
| Both pressure and temperature can be thought of as forms of kinetic energy density, but they are divided over different quantities. Pressure is proportional to kinetic energy per unit volume, while temperature is proportional to kinetic energy per particle. The conversion factor between the two measures (per-volume vs. per-particle) is the number density (particles per unit volume). A gas can be at high temperature and low pressure if it has low number density; likewise, a gas can be at low temperature and high pressure if it has high number density. The relationship between these three quantities (pressure, temperature, density) is contained in the equation of state for the material. For ideal gases, this equation of state is the ideal gas law, $P=\frac{N}{V}kT$.
| {
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If an object moves at constant speed, does it necessarily have constant velocity? If an object moves at constant speed, does it necessarily have constant velocity?
| No, it doesn't. Velocity is a vectorial quantity, it has magnitude (speed) and direction. Uniform circular motion can be given as an example in order to help you with your question. Even though the speed is constant in this particular example, the direction changes all the time.
The velocity of an object changes when the net force acting on it is not zero. When the net force is perpendicular to the velocity, it does no work and the kinetic energy of the object remains constant, so the force only changes the direction of velocity not the magnitude (speed).
| {
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Half-life of $W$ and $Z$ bosons $W$ and $Z$ bosons should decay through weak interaction. But their half-life is around $\tau = 10^{-25} s$ which is a typical value for particles decaying through strong force (instead of a $10^{-12}-10^{-6} s$ for a weak interaction decaying particle).
Why this can be?
| Both are weak decay processes but the intermediate vector boson decay occurs at very much higher energies of 80-90 GeV compared to 1 GeV for beta decay.
| {
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Confusion about how an electron gun works I'm a little unclear about the charge balance aspect of an electron gun. Referring to this diagram and similar diagrams I've seen, what I don't get is wouldn't the target of the electrons have to be connected to the positive anode so that the electrons fired at a target can be recycled if the electron gun is needs to operate continuously? Is the target generally placed on the anode opening so it's connected to the positive?
| The cathode source frees up lots of thermal electrons continuously and the electric field between cathode and anode accelerate them toward a specific direction.
and until these two is working electron beam continues to exist with no problem.
| {
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How to identify an MRI artifact on Fourier space? I am trying to find the frequency of the artifact on the MRI image of the knee below both manually and with ImageJ:
As you can see the artifact results in a bar pattern extending horizontally along the image - i.e. a spike artifact.
After transforming to Fourier space, there are a couple of dots along the x-axis that seem to stand out in their intensity (yellow circles), and are therefore potential culprits for the artifact:
at frequencies $5.02\text{ pixels/cycle}$ and $2.4\text{ pixels/cycle},$ but the frequency that I calculate visually (and painfully) on the $256 \times 256\text{ pixel}$ image corresponds to $\approx 53 \text{ dark vertical bars},$ which would amount to
$$\frac{256}{53}=4.8\text{ pixel/cycle}$$
This is close enough to the the higher frequency dot in Fourier space ($5.02 \text{ pixels/cycle})$. Is this the explanation for the artifact?
Is there a contribution from the second dot that should be considered?
Here is the complete analysis of both dots:
$$\small\begin{align}\text{Freq}&=5.019\text{ pix/cycle}\\
\text{Direction}&=181.12^°\\
\text{Phase }&= \arctan(68.263/-87.982)=-0.6598^°\\
\text{Magnitude}&=\sqrt{(-87.982)^2 +(68.263)^2}=111.36
\end{align}$$
$$\small\begin{align} \text{Freq}&=2.438 \text{ pix/cycle}\\
\text{Direction}&=181.091^°\\
\text{Phase }&= \arctan(10.977/-5.43)=-1.11^°\\
\text{Magnitude}&=\sqrt{(-5.43)^2+(10.977)^2}=12.25
\end{align}$$
Great answer on ImageJ forum.
| I agree that 4.8 is pretty close to 5.02. There is nothing else in that neighborhood that grabs my attention.
The higher frequency spike looks like the next harmonic of that first spike frequency. It probably represents some additional structure to the artifact.
Note, unless it is windowed funny, this is NOT a classic spike artifact. In the lower left corner of the image there is a small section of background air in the image. The artifact does not appear (with this windowing) to go into the background as spikes do. Therefore, this artifact is a result of MR signal, not an external spike signal. It may be insufficient spoiling or something similar.
| {
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Degeneracy of anisotropic oscillator I was working on the 3D isotropic harmonic oscillator and I found that the energies are given by:
$$E=\hbar\omega(n_x+n_y+n_z+3/2)$$
Which has a degeneracy of $\tfrac12(n+1)(n+2)$. However, when dealing with the anisotropic case, I'm not sure if there's a degeneracy in energies. For example, for the 2D anisotropic harmonic oscillator with frequencies $\omega$ and $3\omega$, I found the energies to be:
$$E=\hbar\omega (n_x +3n_y +2)$$
For example here, for $n=1$ we can either have $n_x=1$ so $E_1=3\hbar\omega$ or $n_y=1$ so $E_2=5\hbar\omega$. For $n=2$, we can have $n_x=2$ so $E_2=4\hbar\omega$ or $n_y=2$ so $E_2=8\hbar\omega$ or $n_x=1,n_2=1$ so $E_2=6\hbar\omega$. So there seems to be no degeneracy, but I'm not sure as this is the first time I've found this problem.
| The states
*
*$n_x=3$, $n_y=0$
*$n_x=0$, $n_y=1$
are degenerate in energy.
$\tag{QED}$
More generally, any harmonic oscillator of the form
$$
E = \hbar \omega_1 n_1 + \hbar \omega_2 n_2
$$
will be degenerate if $\displaystyle \frac{\omega_1}{\omega_2} \in \mathbb Q$. It is an important exercise to prove that that is the case and to calculate the degeneracies in both 2D and 3D.
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What is a complex phase shift? In a complex methods course I am taking, we were given an equation for a particular driven harmonic oscillator where the driving force is trigonometric. I have worked out the math and obtained an equation that tells me that the driving frequency at resonance is the natural frequency multiplied by i. My tutor tells me that this is a 90 degree phase shift, but I don't really understand why. Isn't a phase shift obtained by adding or subtracting 90 degrees? And how can a frequency, which is a measurable physical value, take on imaginary values? I would understand if we were talking about velocity. Because velocity has a direction, addition or scalar multiplication by a real value would not describe a 90 degree rotation of the vector. But frequency is a scalar quantity. What does it mean to have an imaginary frequency?
| There is an article here: (the optimal driving force is shown to be $90^{\circ}$ out of phase of the motion)
Phase difference of driving frequency and oscillating frequency
Also any vector like $4j + 3i$ can be expressed in phasor form as $5 \angle 41^{\circ}$ or in complex form $4+3i$. Adding $90^{\circ}$ is just a vector of same amplitude at $90^{\circ}$ to the original.
| {
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Speed of sound in a gas and speed of a typical gas molecule Why is speed of sound in a gas less than the average velocity of the gas molecules? Is there an intuitive way to explain this?
| Gas molecules moving with an average speed and sound propagation are two different things. Something you have to keep in mind is that sound is not transmitted by the net motion of particles. Sound is transmitted via pressure waves (or you can just say sound is pressure waves). Furthermore, that the gas molecules have an average speed does not mean that the molecules have a net movement in any direction.
So I would say there is not really a direct relationship. To see this, think about solids. The molecules in a solid are pretty much confined to a lattice-like structure, but the speed of sound in solids tend to be much greater than the speed of sound in fluids.
| {
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How are RF Waves transmitted? What is the mode of transmission for RF waves at 1800 MHz. Is it ground wave propagation, Line of Sight Propagation or Atmospheric reflection (from ionosphere).
What are the different ways for different frequencies of RF waves?
| here are some general rules.
Low-frequency EM waves (up to about 500kHz) propagate mostly by ground wave.
Medium-to-high-frequency EM waves (500kHz to 30mHz) propagate by ground wave and by ionospheric reflection.
High frequency waves (above roughly 50mHz) propagate by line-of-sight.
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Why does work depend on distance? So the formula for work is$$
\left[\text{work}\right] ~=~ \left[\text{force}\right] \, \times \, \left[\text{distance}\right]
\,.
$$
I'm trying to get an understanding of how this represents energy.
If I'm in a vacuum, and I push a block with a force of $1 \, \mathrm{N},$ it will move forwards infinitely. So as long as I wait long enough, the distance will keep increasing. This seems to imply that the longer I wait, the more work (energy) has been applied to the block.
I must be missing something, but I can't really pinpoint what it is.
It only really seems to make sense when I think of the opposite scenario: when slowing down a block that is (initially) going at a constant speed.
| Think of a unit of work as what you do when you lift a kilogram to a height of a metre. How do you do 2 units of work? You lift that kilogram 2 metres. 3 units? 3 metres. Etc. If you lift something the work you do is the force you exert times the distance traveled (height raised).
For your example of pushing a body in a vacuum you are only doing work while you are actually pushing, and so making it go faster. If you let it coast you are not doing work because the force during this time is zero.
The work you put in is the change of the energy of the object. When you lift the weight the work you have done becomes potential energy (stored energy). When you push the object in space the work you do becomes kinetic energy (energy of movement). You can change the potential energy to kinetic energy by letting the thing fall, and obviously it will be falling faster after it has fallen for a greater height.
| {
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Why doesn't a charged particle moving with constant velocity produce electromagnetic waves? A charged particle moving with an acceleration produces electromagnetic waves. Why doesn't a charged particle moving with a constant velocity produce electromagnetic waves? As far I understand, the electric and magnetic fields in space will still be time-dependent, if a charged particle is moving with constant velocity, so they could have given rise to electromagnetic waves, but they don't.
Also, why do accelerating charged particles produce electromagnetic waves? What is Nature's intention behind this phenomena?
| Ok so if you take maxwell equations and manipulate them a little you can get $$\begin{aligned}
\frac{\partial B}{\partial t} \quad & = -\quad \nabla\times E, \\[5pt]
\frac{\partial E}{\partial t} \quad & = \frac{1}{c^2} \nabla\times B - \frac{1}{\epsilon_0}J \\[5pt]
.
\end{aligned}$$
You see the left hand side guarantees you that a varying magnetic field will generate an electric field, and a varying (means it changes in time) electric field will produce a magnetic field. The definition of electric field is given by the force a test charge feels from another charge. more exactly,
$$ E = k q_1 \frac{\vec{x} - \vec{x_1}}{|\vec{x} - \vec{x_1}|^3} $$
where q1 is the charge that gives you the electric field in point x, and this q1 charge is situated at the point x1.
nvm, ignore everything...
Intuitive approach: You sit on an electron, you see Electric field spreading around you but nothing else. The thing that Newton taught us is that you can't tell the difference between standing still or moving with constant velocity. Thus if you can't tell you're moving at all, from Maxwell eq, you can't have B, which is the magnetic field. If you can't have B, you can't have varying B, you can't have varying E, thus you cant have EM field, you will only detect that electrostatic field if you sit on an electron (that moves with constant velocity).
fml I'm the worse explainer ever.
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How does a particle know how to behave? How does a particle know it should behave in such and such manner?
As a person, I can set mass is so and so, charge is so and so - then set up equation to solve its equation of motion but who computes that equation of motion for a particle in real life?
I, as a person, employ smart 'tricks' such as principle of superposition to avoid having to calculate super complicated situation (calculating electrical force by a shape where large circle is hollowed out in the off-center) but if I were to calculate this in a brute force manner, this would take long time for me to calculate. However, nature doesn't seem to face these types of problems.
Given a school of fish, the ones at the edge will sense threat and gives signal to those near them and so on but this analogy doesn't seem to make sense for physical objects generally considered in general physics problems. Am I asking the wrong type of question? Would appreciate input on this.
|
How does a particle know how to behave?
Already the title is about metaphysics, a consciousness is attributed to the particle by the verb "know".
Physics is about modelling observational measurements with mathematical formulae which can predict future behavior. The "knowledge" is collective and comes from an accumulation of an enormous number of observations.
There are metaphysical models which attribute consciousness to particles. A particular one I read in my metaphysical era ( and was always careful to separate physics from metaphysics) is the "units of consciousness" model of Janer Roberts who was channeling Seth ( you cannot get more metaphysical than this :) ).
In this model all that exists is units of consciousness, which, like many dimensional cosmic strings exist from -infinity to + infinity, building up nature as we observe it. In that metaphysical frame, the question has an answer.
There is no answer within Physics theories and models.
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String theory and background independence I have read that string theory assumes strings live in spacetime defined by general relativity which make the theory background dependent (although general relativity is a background independent theory). Background independence dictates that spacetime emerge from more fundamental ingredient than spacetime. Quoting Brian Greene, “Then, the theories ingredients - be they strings, branes, loops, or something else discovered in the course of further research - coalesced to produce a familiar, large-scale spacetime” (Greene, The Fabric of the Cosmos, 2004: 491).
My question: why couldn’t spacetime just be spacetime, a fundamental entity? If so, string theory, based on general relativity, would not be a “great unsolved problem” facing string theory.
| The idea of presenting gravity as a standard quantum field of interacting gravitons is naive, because it does not explain how spacetime becomes curved. All other fields do not affect the background like pens writing on a sheet of paper. The sheet may be flat or curved, but writing on it doesn't change the curvature. GR defines the curvature and thus cannot be presented as a standard QFT.
Quantum gravity is not just about quantizing gravity. It is about building a theory on a background independent of spacetime. A theory built on this background can define spacetime as flat or curved as appropriate, like a projector projecting different images of reality. Then all other fields must be defined on the same background to be able to interact. So quantum gravity is about redefining both gravity and QFT to unite them on a new background.
| {
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Velocity of separation and velocity of approach Can I have a clear cut dimension or difference between velocity of approach and velocity of separation?
In just simple 1D motion.
Considering two rigid spherical masses of different masses and moving with different velocities.
| The problem is that the velocity of approach is a term which introduces ambiguity.
In one dimension, if bodies $A$ and $B$ are travelling at velocities $\vec v_{\rm A}$ and $\vec v_{\rm B}$ the velocity of approach can be said to be either $\vec v_{\rm A} - \vec v_{\rm B}$ or $\vec v_{\rm B}-\vec v_{\rm A}$.
Taking the magnitude of the subtraction does not work as one could envisage a negative velocity of approach where the bodies are moving away from one another.
The approach which removes these ambiguities is as follows.
Let $\vec v_{\rm A} = a\hat i$ and $\vec v_{\rm B} = b\hat i$ where $a$ and $b$ are components of the velocity vectors in the $\hat i$ direction as shown in the diagram below.
As drawn $A$ is trying to get closer to $B$ and $B$ is trying to get further away from $B$ so the "velocity" of approach is $(+a) + (-b) = a-b$ and using similar ideas the "velocity" of separation is $(-a) + (+b) = b-a$.
You will note that $(a-b)=-(b-a)$ which is to be expected as the velocity of approach is the opposite of the velocity of separation.
As an example consider an elastic collision where it can be shown that instead of using the conservation of kinetic energy on can say that for such a collision the "velocity" of approach is equal to the "velocity" of separation.
The example I have chosen is Worked example 6.5: Elastic collision where the solution is obtained via the direct application of the conservation of kinetic energy.
An object of mass $2 \,{\rm kg}$, moving with speed of $12\,{\rm m/s}$, collides head-on with a stationary object whose mass is $6 \,{\rm kg}$. Given that the collision is elastic, what are the final velocities of the two objects? Neglect friction.
Using the velocities of approach and separation one might proceed as follows.
$x$ and $y$ are the components of the final velocities in the $\hat i$ direction.
Equating approach and separation velocities gives $(+12) + (+0) = (-x)+(+y) \Rightarrow 12 = -x+y$
Conservation of momentum gives $2\times 12\hat i + 6 \times 0\hat i = 2 \times x \hat i + 6 \times y \hat i \Rightarrow 24 = 2\,x+3\, y$
Solving these simultaneous equation gives $x=-6$ and $y=+6$ and so the final velocities as $-6 \hat i$ and $+6 \hat i$.
However you may have used the following diagram for this collision "knowing" that the velocity of the $2\,\rm kg$ mass will be reversed after the collision.
All you need to do is set up two equations in a consistent manner.
Velocities gives $(+12) + (+0) = (+x)+(+y) \Rightarrow 12 = +x+y$
Momentum gives $2\times 12\hat i + 6 \times 0\hat i = 2 \times x (-\hat i) + 6 \times y \hat i \Rightarrow 24 = -2\,x+6\, y$
Solving these simultaneous equation gives $x=+6$ and $y=+6$ and so the final velocities $+6 (-\hat i)+ = - 6 \,\hat i$ and $+6 \hat i$ as before.
| {
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Triangulating Sol's position from another star Let me pose this question as a hypothetical.
Your ship makes a warp/jump through space. There's a malfunction. You're definitely not at your intended destination. You've warped to an unknown star, but you have managed to land you and your crew on a habitable planet orbiting this star. Survival is not an issue.
You need to know where Earth is. How would you go about triangulating Sol's position?
What tools and knowledge would you need?
| If the star is in our galaxy, then you can in principle triangulate your 3d position in the Galaxy from the positions of M31 and the Magellanic clouds.
With access to more specialised equipment (telescopes, spectrographs etc.) it should be possible to identify a set of well-known globular clusters to obtain more accurate Galactic coordinates.
If you have really specialised equipment - like a Gaia satellite (!) - you can use the positions of a network of distant quasars to pinpoint the shift in position from the Sun, using the minute parallax changes due to the differing perspective.
If you weren't too far away from the Sun and had a radio telescope, then the positions of radio pulsars should also give the game away. However, move too far away from the Sun and many pulsars will not be visible because of their narrow beaming angles.
All this assumes you have shifted in space, but not time.
| {
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How can four-momentum be conserved in every frame in an elastic collision? I have this problem:
A particle B is standing still while another one, A, is moving towards it with initial 4-momentum $(E,p,0,0)$. Calculate the change in particle A's 4-momentum as viewed from the particle B's rest frame, in terms of the initial energy E and the scattering angle $\theta$.
I am a bit confused about the 4-momentum conservation. Initially we have $p^i_A=(E,p,0,0)$ and $p^i_B=(m_B,0,0,0)$ finally we should have $p^f_A=(E_f, p_f \cos(\theta),p_f \sin(\theta),0)$ and $p^f_B=(m_B,0,0,0)$. To get the change in momentum I would do $p^f_A-p^i_A$. But the total momentum should be conserved in any frame, but I am not sure how does that work here. In order to conserve it, we would need $E=E_f$ and $\theta=0$ but then the problem would be trivial and also physically you can obviously have angles other than 0. What am I doing wrong?
| I think you've been (understandably) misled by the wording of the problem.
...as viewed from the particle B's rest frame...
That's badly written because there are two different reference frames that could reasonably be called "particle B's rest frame": the one in which particle B is at rest before the collision, and the one in which it's at rest after the collision. Assuming there is a nontrivial collision, these are not the same reference frame.
You've expressed $p_B^i$ in the former frame and $p_B^f$ in the latter frame, but what you should do is express them both in the same frame, since in order to apply conservation of momentum, you need every momentum to be expressed in the same reference frame. (Either one should work.) For example, in the frame in which B is at rest before the collision, it won't be at rest after the collision, and so $p_B^f$ in that frame is going to be something other than $(m_B, 0, 0, 0)$.
| {
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How does particle physics use deep neural networks to find particles? Does anyone use deep learning: RNN, CNN or any other architecture of deep neural networks to asses the standard model or to detect new or unseen particles? What's the status these days in this frontier?
| The answer is no, for using it with the theoretical models, as the other answer says.
BUT
Neural networks are used as tools in deciding kinematic states in particle detectors, assigning momentum and energy to possible new particles and thus identifying them by using a neural network "fit" even by the time I retired in 2000
A multivariate analysis based on the neural networks technique has been used to do the first inclusive measurement of the charmless semileptonic branching ratio of B-hadrons B → Xulvl in the ALEPH experiment at LEP.
In this link the use in high energy experiments is reviewed.
Artificial neural networks are the machine learning technique best known in the high energy physics community . Introduced in the field in 1988, followed by a decade of tests and applications received with reticence by the community , they became a common tool in high energy physics data analysis. Important physics results have been extracted using this method in the last decade.
| {
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Why can we see the cosmic microwave background radiation? This radiation (CMBR) is said to have its origin at the surface of last scattering that exposed itself when the big bang universe had expanded for less than a million years.
In order to see radiation from a source, one has to be on its future light cone. In a universe that is flat and open, which our Universe is asserted to be at the large scale, we are not on the future light cone of this radiation, but almost maximally remote from it. One can also say that the surface of last scattering is not on our own past light cone.
How is this visibility to be understood within standard big bang cosmology?
(This question is different from an earlier one with the same wording.)
| The answer by anna v is in line with what I have seen previously, but is it clear enough?
The "light cones" I mentioned can be shown in a graph in which one axis represents time and the other any spatial dimension. The scaling of the axes can be chosen so that a light cone is represented by a pair of lines inclined at +/- 45 degrees from the time axis.
The future light cone of the radiation starts close to the zero-point of the time axis and is directed forwards.
Our past light cone starts at the time axis much later and is directed backwards.
I see no way of connecting two widely separated points close to the time axis by a 45 degree line without introducing closure or reflection.
| {
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What happens to a radioactive material's atom when it disintegrates? Suppose you initial had radioactive $2^n$ atoms (where $n$ is an integer). Now after a number of halflives the number of left out atoms becomes 1. Now what will happen to it will it disintegrate and the leftover would be half an atom? Now if the reaction stops then the statement "The decaying radioactive atom would never end" then it'll be wrong.
| Let us suppose that you are indeed left with one unstable nucleus with a half life of $\tau$.
The half life is the time interval during which the probability that the nucleus will decay is $\frac 12$.
So you start the clock at time $t=0$ and wait for one half life when the time is $t=\tau$.
The probability that the nucleus will decay in that time is $\frac 12$ and the probability that the nucleus will not decay is $1-\frac 12 = \frac 12$.
So they deacy could happen between a time $t=\tau$ and $t=2\tau$.
Again in that interval of time the probability of a decay is the same as that of not decaying, $\frac 12$.
The probability of the nucleus not decaying between time $t=0$ and $t=\tau$ and then decaying between $t=\tau$ and $t=2\tau$ is $\frac 12 \times \frac 12 =\frac 14 =\frac {1}{2^2}$
The probability of the nucleus not decaying between time $t=0$ and $t=2\tau$ and then decaying between $t=2\tau$ and $t=3\tau$ is $\frac 14 \times \frac 12 =\frac 18=\frac {1}{2^3}$
$...$
The probability of the nucleus not decaying between time $t=0$ and $t=n\tau$ and then decaying between $t=n\tau$ and $t=(n+1)\tau$ is $\frac {1}{2^n} \times \frac 12 =\frac {1}{2^{(n+1)}}$
When one adds up all the probabilities of decaying between $t=0$ and $t=\tau$; $t=\tau$ and $t=2\tau$; $t=2\tau$ and $t=3\tau$ etc which is $$\frac 12 + \frac 14+ \frac 18 + .....=1$$
as expected ie if you wait an infinite time the nucleus will have decayed at some time.
So in the need you may have to wait a very long time for the one remaining nucleus to decay and the probability of decay in the first half life is $\frac 12$.
The exponential decay function is only a ((very) good) approximation if you are dealing with large numbers of nuclei and such that the statistical fluctuations in the rate of decay are very small compared with the rate of decay.
The Phet Alpha Decay simulation is worth a look at?
| {
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What happens to gravity and spacetime when mass turns to energy? What will happen to the distorted space and time around a mass when it is converted into energy?
Will it go back to its original configuration (i.e. with $0$ gravity)?
Or does space time oscillate? Or is there something else that happens?
| To answer this, lets use a single atom as an example. Essentially, the energy contained in that atom (bonds between the sub-atomic particles in the atoms nucleus) radiates away at the speed of light away when released, mostly in the form of Gamma radiation. the gravitational field would radiate out in exactly the same way, since mass is essentially energy stored in the bonds between the particles that make up the nucleus of the atom.
This means the mass really was just stored energy, which is released when those bonds break. It would cause a ripple in the gravitational field. Of course, there being nothing were the atom used to be, there's nothing to attract anything there.
| {
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Work done by friction on a body which is rolling on an inclined plane Why is the work done by friction zero during translational motion but nonzero when the body is rolling on an inclined plane?
| The work done by friction is zero in both cases. This is easiest to see in the formula for power:
$$P=F\cdot v$$
Friction acts at the point of contact where the velocity of both the body and the ground is zero, so P is always zero, so no work is done by friction.
On the other hand, gravity acts at the center of mass where the velocity is nonzero. So P can be nonzero as long as there is a component of velocity in the same direction as gravity. This occurs on an incline but not on level ground.
| {
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Formally from $Z=\mathrm{tr} \,e^{-\beta H}$ to $Z=\sum_{n_1, n_2} e^{-\beta (E_1 + E_2)}$ for two non-interacting quantum harmonic oscillators Suppose we have a Hamiltonian of two noninteracting quantum harmonic oscillators. Then the Hamiltonian can be written
$H = H_1 \otimes I_2 + I_1 \otimes H_2$.
When I start with $Z=\mathrm{tr}\, e^{-\beta H}$ I know that end result for the partition function is $Z=\sum_{n_1, n_2} e^{-\beta (E_1 + E_2)}$, however, how does one formally, i.e., taking all the tensor products into account arrive at this result?
| Let $\{|1, n_1\rangle\}$ be an orthonormal basis for the Hilbert space $\mathscr H_1$ of oscillator 1, and let $\{|2, n_2\rangle\}$ be an orthonormal basis for the Hilbert space $\mathscr H_2$ of oscillator 2, then the set of all tensor products $\{|1, n_1\rangle\otimes |2, n_2\rangle\}$ is an orthonormal basis for the Hilbert space $\mathscr H_1\otimes \mathscr H_2$. Let us abbreviate:
\begin{align}
|n_1, n_2\rangle \equiv |1, n_1\rangle\otimes |2, n_2\rangle,
\end{align}
Notice that for this non interacting Hamiltonian, one has
\begin{align}
e^{-\beta H}
&= e^{-\beta(H_1 \otimes I_2 + I_1 \otimes H_2)} \\
&= e^{-\beta(H_1\otimes I_2)}e^{-\beta(I_1\otimes H_2)} \qquad (\text{since the summands in the exponential commute})\\
&= (e^{-\beta H_1}\otimes I_2) (I_1 \otimes e^{-\beta H_2}) \qquad (\text{use the power series for the exponential}).\\
&= e^{-\beta H_1}\otimes e^{-\beta H_2}
\end{align}
Thus we have
\begin{align}
\mathrm{tr} e^{-\beta H}
&= \sum_{n_1, n_2}\langle n_1, n_2 |e^{-\beta H_1}\otimes e^{-\beta H_2}|n_1, n_2\rangle \\
&= \sum_{n_1, n_2}\langle n_1, n_2 |e^{-\beta E_{n_1}} e^{-\beta E_{n_2}}|n_1, n_2\rangle \\
&= \sum_{n_1, n_2}e^{-\beta E_{n_1}} e^{-\beta E_{n_2}}\langle n_1, n_2 |n_1, n_2\rangle \\
&= \sum_{n_1, n_2}e^{-\beta (E_{n_1} + E_{n_2})} \\
\end{align}
as desired.
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Hydrostatic force on spillway gate? I'm designing an spillway gate that, as the water exceeds a specific height, then it opens pure mechanically, it means the hydrostatic force compensates for the weight of the gate. As you can see the gate is hinged in point $A$.
If i try to find out how high water can reach before the gate opens, then i implement two methods:
1) Write the torque equilibrium, clearly this method gives the right answer
2) Use the force balance, when the normal component of hydrostatic force is equal to the weight of the gate, then the gate should move upward and water will flow under it, this method is false, and i spend an awful amount of time to figure it out, but so far i can't find any explanation.
| Imagine that the water height is at a point where the force at B is exactly zero but the gate has not yet moved upward.
Part of the weight of the gate will now be supported by the hydrostatic force, and part of the weight will be supported by the hinge at point A.
In other words, the hydrostatic force must only lift a portion of the total gate weight.
Or alternatively, if your gate CG is far enough left, then the hinge at point A will be pushing the gate down.
| {
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Could you have sand pipes like water pipes? It's common knowledge that sand behaves like water when in small grains. So can you make a pipe that carries sand in the same way pipes carry water? If not, is there another way you could?
| Yes. Dense-phase pneumatic conveying does exactly this. A powder is fluidized with air and then flows under gravity. It is widely used in process industries. For example, it is widely used in aluminium smelting to convey alumina (aluminium oxide) with particle sizes similar to sand. The solids flow like water when fluidized. A quick search found this recent paper Pneumatic conveying of alumina - comparison of technologies by Garbe, Hilck and Wolf. This is a well-studied area, but much of the literature is behind paywalls.
There are a number of forms of pneumatic conveying:
*
*lean phase, where the pipe is mainly air with particles "blowing in the breeze"
*dense phase, where the pipe is full of particles with just enough air injected along the length to allow flow
*other intermediate cases
This is a special case. The alumina is dry and has been precipitated from solution in the alumina refinery. The grain size and shape are controlled. Factors that influence the "conveyability" are part of the specification.
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In quantum mechanics, is $|\psi\rangle$ equal to $\psi(x)$? So I'm going through my notes and I think I've confused myself. We often imply
$$
|\psi\rangle \to \psi(x)\\
\langle\psi| \to \psi(x)^*
$$
for instance when we talk about eigenvalue equations we interpret
$$
\hat{H}|\psi\rangle =E|\psi\rangle
$$
as simply
$$
\hat{H}\psi(x)=E\psi(x)
$$
but I don't understand why we say $|\psi\rangle \to \psi(x)$ because if that was the case then
$$
\langle\psi|\psi\rangle=|\psi|^2
$$
when clearly
$$
\langle\psi|\psi\rangle=\int_{-\infty}^\infty |\psi|^2 dx = 1
$$
I'm obviously missing something simple, could anyone point out where I'm going wrong?
| To start, the kets are vectors, which means if we want an explicit realization of them, we would need to write them with respect to some basis. The first basis most people see is the position basis, where the basis kets are the states of definite position. Then, an arbitrary state $|\psi\rangle$ can be written as
$$ |\psi\rangle = \int_{-\infty}^{\infty}dx \ \psi(x)|x\rangle,$$
where $\psi(x)$ is the familiar position space wavefunction. This should help explain why the last line has the integral in it, it comes from this superposition of states of definite position.
Now for rewriting $\hat{H}|\psi\rangle = E|\psi\rangle$, the exact same thing happens, but the reason there aren't any integrals is because the $\hat{H}$ in $\hat{H}\psi(x) = E\psi(x)$ is written as an differential operator acting on position space now, rather than an operator that acts in some way on an arbitrary ket. This might be a slight abuse of notation, but it is usually clear from context what basis the Hamiltonian is written in in a given equation.
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Relativity of Jerk Popular expositions of general relativity start with a thought experiment showing that it is impossible to distinguish a constantly accelerating frame of reference in a free fall from a free floating frame of reference.
Thought Experiment:
Person A is a small closed box, free-falling towards earth. Person B is in a small closed box floating around in space. If they both do the same experiments, they should see the same results. For example, if they have a small ball and toss it inside their box, they would both see that ball travel in a straight line (not curving) towards the wall. They would also both feel themselves floating around as if there was no gravity.
The same thought experiment could be applied to a frame undergoing a constant jerk. Does that lead to a new theory of relativity?
| Gravity happens to measurably correspond to a constant acceleration, not jerk. So, no, this doesn't lead to a new theory of gravity. At least not immediately. On the other hand, strictly speaking, there is actually a slight jerk involved in free fall if an object falls long enough for the increasing gravitational field to become relevant. Whether a model for our world involving relativity of jerk could be built, is something I'd like to know myself...
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Once introduced will an electric and/or magnetic field live for ever? So if generate an electric field or magnteic field, will it live for ever? because whenever you get rid of that field for example getting rid of electric field by discharging a capacitor, it will result in changing megntic field and that will result in changing electric field and that will keep on going on it own. Does it mean then that once introduced electric field or magnetic field will become immortal :)
| Electromagnetic energy can be converted into other forms, like heat or mechanical energy as in the case of motors. So the total energy is conserved, but it need not be in the form of the electric or magnetic field.
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Canonical quantisation: How to find the scalar product? I am trying to understand the canonical quantisation procedure. I understood that one takes the classical field equation and replaces the field by an operator Φ which solves the field equations. Then one imposes the commutation relations on Φ and Π.
Since Φ solves the field equation one can write it as
Φ = Σk uk ak + uk* bk
.
where uk are solutions to the classical field equation and form a orthonormal basis. My question is now how to find the scalar product to which they are orthonormal? Is there a definite way how to find this scalar product? Or do I need to guess one?
| The general answer to your question is given by holomorphic quantization (a good treatment of which can be found in Woodhouse, "Geometric Quantization", especially chapter 5 & section 9.21). The idea is that you can build a scalar product from two ingredients:
*
*a symplectic form $Ω$, which is a non-degenerate antisymmetric real-valued form, related to the Poisson brackets via:
$$df = Ω(X_f,\,\cdot\,)$$
$$\{f,g\} = d_{X_f} g$$
(where $X_f$ denotes the Hamiltonian vector field generated by $f$): this you get for free from the commutation relations;
*a complex structure2, which is a (real-)linear operator $J$ on your phase space (seen as a real vector space) satisfying $J^2 = - \mathbf{1}$: given such a complex structure, any phase space can be turned into a complex vector space, defining the complex scalar multiplication by:
$$\lambda u := \text{Re}(\lambda)\, u + \text{Im}(\lambda)\, J u.$$
If $J$ preserves the symplectic structure, i.e.
$$Ω(J\cdot,J\cdot) = Ω(\,\cdot\,,\,\cdot\,),$$
then we can define a sesquilinear form by:
$$\langle u|v\rangle := 2Ω(u,Jv) + 2iΩ(u,v).$$
[To Be Continued...]
1 Woodhouse focuses mostly on finite-dimensional phase spaces, but he treats the more general case, namely Kähler manifolds, which requires extra integrability conditions on $Ω$ and $J$. The answer above only considers the linear case of phase spaces as "Kähler vector spaces", aka. Hilbert spaces.
2 A complex structure is a special case of what is called a polarization in the context of geometric quantization.
| {
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Question regarding average velocity We know average velocity in its strictest sense, means total displacement over total time taken: $$\frac{X_f-X_i}{T_f-T_i}$$
There's a special case, when a body is moving in a straight line with a constant acceleration. Of course since its acceleration is constant, it has to be a rectilinear motion.
In this case average velocity (over a time interval) is simply
$$V_{avg}=\frac{V_1+V_2}{2} \tag 1$$
It can be proved easily, using the equations of motion.
The important point is, this formula works only when a body is moving with a "constant acceleration". As per my teachers and the books I have.
The problem is, why I'm actually putting this post up, there's another case. If the body moves with a velocity $V_1$ for a time interval $t$, and then it moves with a velocity $V_2$ for the same amount of time $t$. In other words, the body traveling with a velocity $V_1$, takes time $t$ to go from a point $A$ to another point $B$, and then it goes from point $B$ to another point $C$ and again, it takes time $t$, traveling with a velocity $V_2$.
In this case, when time intervals are equal, we calculate average velocity (or average speed) by taking the arithmetic mean of individual velocities, and this formula (as per my teachers and the books I have).
So, in this case :
$$V_{avg}=\frac{V_1+V_2}{2}\tag 2$$
But equation (1) and equation (2) are completely identical. Isn't that strange? And odd? Because equation (1) should be valid 'only' when the acceleration is constant. But in the second case, acceleration of the body is not constant during the course of its motion. It's acceleration is zero as it goes from $A$ to $B$, then its acceleration changes as it velocity changes from $V_1$ to $V_2$, and then its acceleration is constant from $B$ to $C$. Even though its acceleration is non-constant, the formula for finding out its average velocity is exactly the same, as in the first case.
Please explain what's going on here. Because the formula $V_{avg}=\frac{V_1+V_2}{2}$ should be valid if and only if acceleration is uniform. (As per my books and my teachers).
Thanks
| It's incorrect to use the word "only". (1) is always valid if acceleration is constant, but that doesn't exclude the other situation. It still can be valid in certain special situations.
| {
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How are particles in a collision chosen? In synchrotron particle colliders, how are the particles which are collided chosen? For the most part, collisions of different types of particles don't do anything like you might expect in a video game; there is no secret recipe list of cool things, each which require different types of particles. So, what makes certain particles more favorable than others?
| It comes down to a trade-off between collision energy and collision 'purity'. Accelerating lighter particles, like electrons, to high energies is difficult because they have higher synchrotron losses. It is easier to accelerate protons to higher energies, however their collisions are 'messier' - the interactions that happen during collisions are much more complex and harder to predict, and so you actually need a higher amount of energy to see the same kinds of particles.
For certain kinds of experiments, and certain kinds of accelerator designs, it's actually preferable to collide elementary particles instead of protons. And for other kinds of experiments, it's preferable to collide much heavier particles, such as lead nuclei.
| {
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How to measure angles in Minkowsky space, and how do they transform? I want to know how an ordinary angle $\theta$ transforms under a Lorentz boost. For that purpose I consider a 4-vector given by
$$ a ^ \mu = ( t , \cos \theta , \sin \theta , 0 ) .$$
The angle I will analyze is the one between this 4-vector and the $a$ axis, $\theta$. I consider a boost
$$ \Lambda ^\mu _{\ \ \ \nu} = \left( \begin{matrix}\gamma & -\beta\gamma & 0 & 0\\
-\beta\gamma & \gamma & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{matrix} \right) .$$
Thus the transformed 4-vector is $$ a ^{\prime \mu} = ( \gamma (t - \beta \cos \theta), \gamma (\cos \theta - t \beta), \sin \theta, 0 ) $$
Now the angle subtended between $a^\prime$ and the $x^\prime$ axis should be the transformed $\theta^\prime$, i.e.
$$ \theta ^ \prime = \tan ^-1 \left( \frac{\sin \theta}{\gamma (\cos \theta - t \beta)} \right) .$$
Now I have some doubts:
*
*Which should be the correct value of $t$, the time component of the 4-vector $a$ used to define the angle $\theta$? I would say $t=0$ in order to have the angle $\theta$ defined by a space-like vector, the same idea as when you define the proper distance between two points. In this case I would think of a sort of proper angle.
*In the case $t=0$ I find that $ \theta^\prime \leq \theta $... This is not consistent with the length contraction along the $x$ axis... In fact, the $x$ component of $a^\prime$ is greater than that of $a$...
I was expecting to obtain something like this:
but I didn't although I have transformed the vector using the transformation law... What is wrong?
| If you're moving, the angle is going to change. What you want to do is to define the origins of your reference frames $S$ and $S'$ at $t_0 = t_0' = 0, x_0 = x_0' = 0$. That is, $t_0 = t_0'$ is when observers in each frame perform their angle measurements. But how do they make a measurement? There was some event at $(x,y), t < 0$ which created light. In order to see the event, the light must have traveled some distance to enter their eyes. This is why you can't use $t = 0$ as the event time, because if the event happens anywhere away from the origin, there is no time for light to travel to the observers. But since $S'$ is traveling w.r.t. $S$, the distance the light had to travel is not the same.
So let's say the observers are measuring the angle from their position to the Sun. For simplicity's sake, say that at $t_0$, the Sun is located at $(x,y) = (1,1)$. This means the observer at $S$ measures the Sun at a $45^\circ$ angle, but we get another piece of info: the light that observer $S$ measures must have travelled a distance equal to $\sqrt{1^2 + 1^2} = \sqrt{2}$, so we can find out the time that light was emitted,
$$-ct_e = \sqrt{2} \Leftrightarrow t_e = -\frac{\sqrt{2}}{c}$$
Now we can boost $x$ to $x'$ to get,
$$x' = \gamma(-c\beta t_e + x) = \gamma(\sqrt{2}\beta + 1)$$
which is always larger than $x$ (for positive velocity).
Since $y' = y$, the angle $S'$ measures is
$$\tan^{-1}\left(\frac{1}{\gamma(\sqrt{2}\beta + 1)}\right) $$
which is always smaller than $45^\circ$.
| {
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Closed field lines in case of a Bar magnet Field lines in case of charges go from +ve to -ve but incase of magnet, they dont start or stop anywhere. They form closed loops. Is this consequence of the fact that single poles dont exist or something else is going on here?
| Yes. In an alternative universe where north and south charges can exist independently just as positive and negative charges can exist then magnetic field lines could start and end at these charges. In such a universe the alternative maxwells equations become symmetric, moving electric charge generates magnetic field and moving magnetic charge generates electric field. However, in our universe these magnetic charges do not exist, and as such magnetic field lines close up.
| {
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Debye Temperature for Copper I am trying to calculate the Debye temperature, $\theta_D$, of copper using the following:
$$
\theta_D = \frac{\hbar v_s}{k_B} \left( \frac{6\pi^2N}{V} \right)^{1/3}
$$
I have the following values: $\rho = 8900$ kgm$^{-3}$, $v_s = 3800$ms$^{-1}$, atomic mass $ M_a=63.5$gmol$^{-1}$. Now, the speed of sound, $v_s$ is not correct for copper according to online tables, and it seems to be closer to $4600$ms$^{-1}$.
However, I also know that the Debye temp for copper is about $343$K.
Using the fact that,
$$
{N\over V} = \frac{N_A\rho}{M_a} = 8.44\times10^{27}
$$
where $N_A$ is Avogadro's number, I get,
$$
\begin{align}
\theta_D &= \frac{(1.055\times10^{-34})(4600)}{(1.381\times10^{-23})}\cdot \left( 6\pi^2\cdot8.44\times10^{27} \right)^{1/3} \\\\ &=279K
\end{align}
$$
Which just isn't right. And using the value for $v_s$ provided in the question gives an even lower answer of $230$K...which isn't right either.
Am I missing something here?
| Great post. And thanks Anyon for the clarifications. I will post the complete clear solution for that problem with links and numbers and while doing that I will correct a small mistake that Anyon did.
First, concerning the density and molar mass of copper, I have
$$ \rho = 8960 \text{ kg/m}^{3} $$ and $$ M_a = 63.546 \text{ u} $$
which I took from wikipedia
https://en.m.wikipedia.org/wiki/Copper
This gives me a density of
$$ \begin{align}
{N \over V} & = \frac{1000 \cdot N_A \cdot \rho}{M_a} \\
& = \frac{1000\cdot6.023\times10^{28}\cdot8960}{63.546} \\
& = 8.49\times10^{28} \text{ atom per m}^3 \end{align} $$
Now, as Anyon said, we have 2 transverse and 1 longitudinal sound waves, which have the speeds
$$ v_{\mathrm{transverse}} = 2325 \text{ m/s} $$ and $$ v_{\mathrm{longitudinal}} = 4760 \text{ m/s} $$ which I took from
https://www.engineeringtoolbox.com/amp/sound-speed-solids-d_713.html
Now, when applying the formula for $ \theta_D $, one should take an "average" speed of sound. Here is where Anyon made a small mistake. Since we have 2 transverse and 1 longitudinal the average speed is obtained by the formula
$$ \bar{v}_s = 3^\frac{1}{3}\left( \frac{2}{v^3_{\mathrm{transverse}}}+\frac{1}{v^3_{\mathrm{longitudinal}}}\right)^{-\frac{1}{3}} $$
Plugging in the numbers, we get
$$ \begin{align}
\bar{v}_s & = 3^\frac{1}{3}\left( \frac{2}{2325^3}+\frac{1}{4760^3}\right)^{-\frac{1}{3}} \\
& = 2611.69 \text{ m/s} \end{align} $$
Now using the formula
$$ \theta_D = \frac{\hbar v_s}{k_B} \left( \frac{6\pi^2N}{V} \right)^{1/3} $$
And plugging in the numbers we have, we get
$$ \begin{align}
\theta_D & = \frac{1.055\times10^{-34}\cdot2611.69}{1.38\times10^{-23}}\left(6\pi^2\cdot8.49\times10^{28}\right)^{1/3}
\\ & = 342 \text{ K} \end{align} $$
Which is within the expected value of 343 K.
| {
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Why do we describe physical systems in Hilbert space? In quantum mechanics we study physical systems associated with a Hilbert space. Why do we need a Hilbert space to describe the state of a system?
| Although there are many perspectives on this, largely united by the (correct) notion that Hilbert spaces allow for geometric tools to be applied, I'd like to present another overlooked perspective: Hilbert spaces are the tool of choice in QM because they can uniquely mimic probability spaces.
In quantum mechanics, we are given an object that entirely describes the physical system under consideration, which we'll label $\psi$, and are tasked with generating a mapping from the space of $\psi$ functions to $\mathbb{R}\in[0,1]$ assigning probabilities $p(\psi)$ to them.
This process involves defining a probability space $(\Omega,\mathcal{F}, P)$ where $\Omega$ represents all of the different observable $\psi$ (call them $\hat{\psi}$), $\mathcal{F}$ represents the different possible "combinations" of $\hat{\psi}$'s (this is called a $\sigma$-algebra), and $P$ is the mapping that associates a probability $p$ with each $\hat{\psi}$ or "combination" of them.
Although Hilbert spaces are not probability spaces, they are uniquely up to the task to mimic these if we do the following:
*
*Associate all observable $\psi$'s, the $\hat{\psi}$'s , with $\Omega$.
*Equate the inner product of a $\hat{\psi}$ with itself to the measure $P$.
*We can decompose the space into an orthonormal basis, which allow us to define any $\psi$ as the linear combination of a set of the observable states $\hat{\psi}$. This is unique to Hilbert spaces. We can then use these to properly define $\mathcal{F}$.
The fact that we can make this decomposition is what lets us properly define a probability space, and Hilbert spaces are the only ones that can do it (at least straightforwardly) because of the notion of orthogonality they induce. (Note that $L^2$ is the only $L^p$ space that defines orthogonality, precisely because it's a Hilbert space.)
| {
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Uniaxial stress question Let's have a rectangular profiled bar. Let us introduce force $\vec{F}$ which pull the bar apart. In the picture below let us make a virtual horizontal cut $A$.
Well, everything is in the picture. Nothing fancy. But the part I'm stuck with is this:
Let's instead of cut $A$ make a cut $B$ which will be perpendicular to $A$'s normal. That is, $B$'s normal is perpendicular to $\vec{F}$. From my point of view, the force $\vec{F}$ will now be shearing plane $B$. But, of cource, every textbook say that there will be NO stress (neither normal nor tangental) on the plane $B$.
And that's where I'm stuck: My intuition says that $\vec{F}$ will shear $B$, but theory says -- it will not.
I guess my problem lies in the fact that I don't understand why Tractrions(Forces) on cuts with different normals can't add up. But nowhere I've seen any thorough explanation about this inability of comparing tractions on different cuts.
Please, help.
| There will be no shear across plane B. To see why, imagine a very small 'needle' embedded in the solid block, with the needle pointing from left to right in your picture. Also, with the center of the needle residing in the center of the block. I'm imagining the needle penetrating plane B. If there were shear, the needle would have to rotate. But it doesn't rotate, it's just displaced.
Something like this (excuse the poor drawing):
I'm representing plane B as the thin black line, and the needle as the thin red line intersecting it. You can imagine tensioning the block on the top and bottom points of plane B. You can see that the needle doesn't rotate.
Now if you apply force asymmetrically, for example on the top-left corner and the bottom-right corner, then the needle would indeed rotate a bit.
| {
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Expected momentum of ground state hydrogen $$ I am trying to calculate the expected momentum of an electron in the ground state of hydrogen atom. This is the wave function.
So far I have done this:$$\iiint_V \Psi^* (-i\hbar) \frac {d\Psi} {dr} r^2 sin\theta dr d\theta d\phi$$
But the answer I am getting is $$\frac {i\hbar}{a_b}$$ which looks wrong because it is imaginary. What am I doing wrong?
| You simply miscalculated the action of $\vec p$ on spherically symmetric functions f(r).
In actuality, your answer should transform vectorially,
$$
\vec p f(r)= -i\hbar \nabla f(r)= -i\hbar~ \hat x ~\partial_r f(r),
$$
so, then, as @KevinDeNotariis suggests,
$$
\langle \vec p \rangle= -i\hbar \int d^3x ~\psi^* \partial_r \psi(r) ~\frac{\vec x}{r} ,
$$
trivially vanishing upon integrating over all directions.
| {
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Forces between wheels and road for a vehicle Let’s assume we have a front-wheel drive (FWD) car, and we apply a torque (Cf) to the front wheels. At this point, assuming no slip condition, a friction force (Tf) will occur due to friction between the rotating wheel and the road. This force will be from left to right (according to the picture).
Now what’s about the rear wheels? They are pushed forward by the car that is now moving thanks to the Tf and they can only begin rolling if the force between them and the road is backward. Is it correct or am I missing something?
What happens if I have no torque (when I release the throttle) and when I brake?
Thank you
| There is a force to the left on the axle of the back wheels $F$ which is accelerating the rear wheels to the right.
The friction force between the tyre and the ground to the left $F_{\rm friction}$ applies a torque about the axle of the wheel to cause it to increase its rotational speed.
The frictional force is try reduce the relative movement between the tyre and the ground at the point of contact or maintain no relative movement.
With no torque being applied and the wheels not rotating there would be relative movement between the tyre and the ground - a skid has occurred with the friction force on the tyre due to the ground in the backward direction.
That friction force is trying to slow the centre of mass of the vehicle down whilst at the same time trying to get the wheels to rotate ie attempting to reduce the relative movement between the tyre and the ground.
| {
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Uncertainty cannot be calculated? I'm doing an experiment on resonance. The phase difference between the driving force and the one oscillating is given by
$$\varphi=\arcsin\left(\frac{y_1}{y_2}\right)$$
where $y_1$ and $y_2$ are measurements of voltages. At resonance $y_1=y_2$ so $φ=\frac{π}{2}$.
However, I'm trying to calculate the uncertainty of $\varphi$ , I need to take the partial derivative of $\varphi$ with respect to $y_1$ and then $y_2$. In both cases , I end up with something in the form of
$$\frac{\partial\varphi}{\partial y_1}\sim\frac{1}{\sqrt{1-\left(\frac{y_1}{y_2}\right)^2}}$$
multiplied by some other constants which don't really matter. So at resonance, my denominator will be zero. Any thoughts?
| The idea that you can use partial derivatives multiplied by the uncertainty in the independent variable(s) to estimates uncertainty in the dependent variable is just an approximation.
It is related to a Taylor series expansion with an implicit assumption that the terms involving the second derivatives and higher can be neglected.
Obviously, for a sinusoidal function this isn't going to be true near $\pi/2$ since the first derivative becomes small.
A better way to proceed in your case is to do a Monte-Carlo simulation, allow $y_1$ and $y_2$ to be generated at random values determined by their own probability distributions and uncertainties, and then calculate the consequent values of $\phi$ and build up a probability distribution of $\phi$ from which an uncertainty can be calculated.
Edit: in response to your edited function, you now have
$$\frac{y_1}{y_2} = \sin \phi$$
Thus
$$ \frac{\partial y_1}{\partial \phi} = y_2 \cos \phi$$
and the problem I mentioned above clearly applies at $\phi = \pi/2$.
| {
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Relationship between strain energy function and strain or stress How one can get the strain or stress from the strain energy function ?
And if one cannot do it, what is the use of that function ?
| The Strain Energy Deformation Function (SEDF) is a scalar $\psi(\varepsilon_{ij},\xi_k)$ defined in term of some strain tensor $\varepsilon_{ij}$ and possibly some internal variables $\xi_k$ (representing non-reversible or dissipative phenomena). This functions is a thermodynamic potential, that can be interpreted as Gibbs free energy per volum unit. The stress tensor components are related to SEDF by means of:
$$\sigma_{ij} = \frac{\partial \psi}{\partial \varepsilon_{ij}}$$
There are several possibilities for defining it, in terms of what representation is taken for stress and strain. For example, a possible choice is to use the Green–Saint-Venant strain tensor and the second Piola–Kirchhoff stress tensor, or, more commonly, the right CauchyGreen deformation tensor and the second Piola–Kirchhoff stress tensor:
$$S_{ij} = 2\frac{\partial \psi}{\partial C_{ij}}, \tag{1}$$
An important property in the case of materials that can undergo irreversible deformation is that their entropy must increase, which is equivalent to the positiveness of the internal dissipation power:
$$\mathcal{D}_{int} = \sum_{i,j} \left(S_{ij}-2\frac{\partial \psi}{\partial C_{ij}}\right)\frac{\partial{C}_{ij}}{\partial t} + \sum_{k} \frac{\partial \psi}{\partial \xi_k}\frac{\partial \xi_k}{\partial t} \ge 0$$
because of $(1)$ the first term is zero.
| {
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Doppler effect in light (Observer moving away from source) I understand this intuitively and can picture it in my head, but when I do it on paper, the result is a sign difference that I cannot understand
According to this diagram the wavelength = ct-vt = t(c-v)
then the periodic time T = t-(vt/c) which should be t+(vt/c) instead of minus, because the second wavefront takes a longer time to pass the observer, if the error was in the direction, why should the velocity v be negative if the velocity of light in the ---> direction is positive and the observer is moving in the same direction as the light waves? Thanks
| For a stationary source the wavelength is $\lambda=cT$. The wavelength does not depend on the motion of the observer. The position of wavefront n is $x=ct-n\lambda=ct-ncT$.
The position of the observer is $x=vt $. Setting those two equal, the observer receives wavefront n at $vt=ct-ncT$ which gives $t_n=\frac{c}{c-v}nT$. Wave front 0 is received at $t_0=0$ and wavefront 1 is received at $t_1=\frac{c}{c-v}T$.
The frequency is the inverse of that so $f_o=\frac{c-v}{c}f_s$ which has the correct sign for an observer moving away from the source.
| {
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Why does the position uncertainty of a harmonic oscillator not have the expectation value squared term? My question is about the derivation for the uncertainty of a quantum harmonic oscillator in the non-zero ground state energy. In my textbook A modern Approach to Quantum Mechanics by John S. Townsend there is a discussion about the position uncertainty $(\Delta x)^2= \big<x^2 \big>+\big<x\big>^2$ for a harmonic oscillator in the ground state energy.
We have established for a harmonic oscillator
$\hat x={\sqrt {\hbar\over 2m\omega}}(\hat a+{\hat a^\dagger})$ so $\big<x\big>=\big<n \big|{\sqrt {\hbar\over 2m\omega}}(\hat a+{\hat a^\dagger}) \big|n\big>$ which gives us $\big<x^2\big>=\big<n\big|{\hbar\over 2m\omega}(\hat a+{\hat a^\dagger})^2\big|n\big>$.
And in a similar fashion $\big<x\big>^2=\big<n \big|{\sqrt {\hbar\over 2m\omega}}(\hat a+{\hat a^\dagger}) \big|n\big>^2$.
To me this means that $(\Delta x)^2=\big<n\big|{\hbar\over 2m\omega}(\hat a+{\hat a^\dagger})^2\big|n\big> + \big<n \big|{\sqrt {\hbar\over 2m\omega}}(\hat a+{\hat a^\dagger}) \big|n\big>^2$. However, the book seems to drop (with no explanation) $\big<x\big>^2$ and comes up with $(\Delta x)^2=\big<0\big|{\hbar\over 2m\omega}(\hat a+{\hat a^\dagger})^2\big|0\big>$. I know we are talking about ground state so I'm assuming that's why the $n$ was replaced with $0$, however as far as I can tell we just have $\big<x^2\big>$. Why is the other term dropped? Is this something to do with the oscillator being in the ground state?
| ::chuckles::
I remember being baffled by how this works out mathematically myself, and it's one of those "I can't believe it's that simple!" ones.
Three facts:
*
*$\hat{a}$ and $\hat{a}^\dagger$ are the lowering and raising ladder operators: they take a numbered state to one numbered either one less or one more that the starting state.
*The numbered states are a set of eigenstates, so they are orthogonal to one another.
*Expand $\left( \hat{a} + \hat{a}^\dagger \right)^2$, and see why it has a very different character than $\left( \hat{a} + \hat{a}^\dagger \right)$.
| {
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In $E=hf$, can $f$ assume any positive value? (Beginner) The energy of photon is given by the equation $E=hf$, where $h=$ Planck's constant, and f=frequency of radiation. Is f quantized, or can it assume any value?
If it can assume any value, then wouldn't this mean that the energy of photons is not quantized? If f can be any value from a continuous series, this would mean that for any number you imagine, there will always be another number that when multiplied by h gives you that first number. So $E$ could assume any value.
However, we know that $E$ is indeed quantized, so could someone please point out the flaw in my reasoning?
PS: I don't believe this question is a duplicate, since it addresses wether radiation frequencies are discrete in the context of an equation. The answers, therefore, aren't limited to "no, the EM spectrum is not quantized" but also explain that for a given frequency, the photons' energy is quantized. In other words, while the questions are similar, I believe that both the question and the answers have angles that are different enough to justify not being duplicates. For example, if the answers to this question had not addressed the photon energy, I would still be confused as to why so many textbooks say that energy is quantized. This isn't the kind of answer that the duplicates would require, however. Thank you for linking the other posts, though, as they could be useful for someone looking for a different answer.
| Suppose you have an electromagnetic field(which is made up of photons) oscillating with frequency $f$ inside a cavity(resonator like two mirrors). Because of quantization, the field cannot have an arbitrary value of energy but only integer multiples of $hf$(i.e., $nhf$). This is what is meant by energy quantization.
| {
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Lagrange Equation - Basics The basic equation of Lagrange is given by,
$$\frac{\mathrm d}{\mathrm dt} \frac{\partial L}{\partial \dot{q_j}} - \frac{\partial L}{\partial q_j} = Q_j \tag{1}$$
where $T$ is the kinetic energy, $V$ is the potential energy, $L = T-V$. But at any given instant $$T + V = \text{Constant}\tag{2}$$ (Conservation of energy)
I understand that the criteria for Lagrange application is that the coordinates that define the system are to be,
*
*Independent
*Complete
*Holonomic
and for a mechanical system $V \neq f(\dot{q_j})$.
Thus, for mechanical systems, the above relation reduces to
$$\frac{\mathrm d}{\mathrm dt} \frac{\partial T}{\partial \dot{q_j}} - \dfrac{\partial T}{\partial q_j} + \frac{\partial V}{q_j} = Q_j . \tag{3}$$
Thus, as the coordinates are to be independent,
The effect of a force in any particular direction (component) will not give a displacement in the other directions, thus the energies in that particular direction can be expressed by isolated equations that are independent of each other (This is what I understood by the coordinates being independent).
Then, by applying
$$V = Constant - T = C - T\tag{4}$$
the above equation would just become
$$\frac{\mathrm d}{\mathrm dt} \frac{\partial T}{\partial \dot{q_j}} - \frac{\partial T}{\partial q_j} + \frac{\partial (C - T)}{\partial q_j} = Q_j \tag{5}$$
and eventually boil down to
$$\frac{\mathrm d}{\mathrm dt} \frac{\partial T}{\partial \dot{q_j}} - 2\frac{\partial T}{\partial q_j} = Q_j \tag{6}$$
as the derivative of a constant is always zero.
I am new to dynamics and so like to know where I am wrong conceptually.
| Think of the difference between the expression $\frac{1}{2}\,\dot{q}^2 - \cos{q} = 1$, which is a special case of the pendulum's total energy, versus ${q}^2\sin^2(\dot{q}) + {q}^2 \cos^2(\dot{q}) + (1- {q})(1+ {q}) \equiv 1$. In the first case it is not true that
$$\cos{q} = \frac{1}{2} \, \dot{q}^2 - 1$$ for all possible entries $(q,\dot{q})$ while $${q}^2\sin^2(\dot{q})+ {q}^2 \cos^2(\dot{q}) \equiv 1- (1- {q})(1+ {q}) $$ always, for all entries $(q,\dot{q})$.
If you check the derivatives for the first equation, $\frac{\partial }{\partial q} \,\cos{q} = - \sin{q}$ but $\frac{\partial }{\partial q} \, \left(\ \frac{1}{2} \, \dot{q}^2 - 1 \right) = 0$ so they are not equal. For the second identity, the equality will holds.
| {
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Gravitationally-induced slowing-down of the spreading of a wave packet The spreading of a wave packet is very fast in quantum mechanics: for an electron, a gaussian wave packet spreads from one angström to 600km in one second! In his famous QM book, Sakurai mentions that there are numerical evidence that taking into account gravitation as
$$-{\hbar^2\over 2m}\Delta\psi(\vec r)
-{\cal G}m^2\int {|\psi(\vec r')|^2\over |\vec r-\vec r'|}d^3\vec r'\psi(\vec r)=i\hbar{\partial\psi\over\partial t}$$
stops the spreading at distances around 500nm. See for example https://arxiv.org/abs/1105.1921
Why should we take into account the gravitational interaction of the wave packet with itself but not the electromagnetic one? In the case of a free electron for example, the Coulomb repulsion of the wave packet by itself would accelerate the spreading.
| At least in the linked paper they discuss using molecules (such as fluorofullerene C$_{60}$F$_{48}$) as the test particle. The molecules are uncharged.
| {
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What is the experimental evidence that the nucleons are made up of three quarks? What is the experimental evidence that the nucleons are made up of three quarks? What is the point of saying that nucleons are made of quarks when there are also gluons inside it?
| The process that was first used to resolve the internal structure of the proton and neutron is called deep inelastic scattering. Basically, you hit the target hadron with enough energy that the probing particle's wavelength is short enough to make out the details of the internal structure of the proton or neutron.
You are right to question the idea that a proton is "made up of three quarks" when there are gluons inside as well. The gluons are of course force-carrying particles which mediate the strong force and keep the quarks bound to one another, but because the gluons carry color charge themselves, they couple in strongly non-linear ways. The upshot is that most of the mass of a proton comes from the interaction energy of the gluons.
| {
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Why does increasing the volume in which a gas can move increase its entropy? Let's say we have a box with a non-permeable wall separating the box in half. There is gas on the other side of the wall. Now we remove the wall so that the gas can diffuse to the other half of the box.
It is said that the entropy of the gas increases because the molecules now have more room to move, and therefore there are more states that the gas can be in. I can understand this well.
But the change in entropy is also defined as follows:
$\displaystyle \Delta S = \frac{Q}{T}$
Where $T$ is the temperature of the gas and $Q$ is the change in heat of the system. But if we look at this definition, why did the entropy change for the gas inside the box? By just removing the wall, the kinetic energy of the molecules does not change, therefore the temperature does not either. We also didn't add any heat to the system, so $Q$ is zero as well. So why did the entropy change?
| The formula you used, $$\mathrm{d} S=\frac{\mathrm{d}Q_{\mathrm{reversible}}}{T}$$is a simplification for situations where all other factors are constant. Here, you're better off using $$\Delta S=nR\ln\frac{V_2}{V_1}$$
($R$ is the gas constant, and $n$ is the number of moles) It's a more sensible formula since the only thing you're changing is volume: the temperature and energy are both maintained as constants.
Eventually, the entropy did change, but the formula you used wasn't the right one, and hence it suggested that entropy wouldn't change.
The Wikipedia page on entropy has a list of useful formulas, you'll need to choose the appropriate one for different situations, depending on what factors you're maintaining as constants.
| {
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What is possible intuitive explanation of inelastic relativistic collsion? In classical mechanics, we say an inelastic collision happens when some energy is transferred to heat and noise without changing the total sum of momentum. However, in special relativity, every component of 4 momentum is preserved, but not the sum of masses. How can we explain it intuitively like we did in classical mechanics?
| In relativistic collisions, the lost kinetic energy appears as rest masses of products. Or the kinetic energy can increase in, say, a $1 \to 2$ decay process, where rest mass energies are converted into kinetic energy.
| {
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Why does current conservation involve an arbitrary function? In section 6.1 of Peskin's quantum field theory introduction, right after equation 6.3, the four current density $j^{\mu}$ is said to be conserved because for any function $f \left( x \right)$ that falls off at infinity, we have
$$
\int f \left( x \right) \partial _{\mu} j^{\mu} \left( x \right) \mathrm{d}^4 x = 0
$$
I am just so confused on the fact that there is a function $f \left( x \right)$ involved in this evaluation. Is current conservation not just $\int \partial _{\mu} j^{\mu} \left( x \right) \mathrm{d}^4 x = 0$?
Thanks in advance!
| I don't know Peskin's book, so I guess. Maybe the author has defined a quantum field as an operator-valued distribution? Then it would make sense to insert a test function, since to say $T=0$ if $T$ is a distribution means $T(f)=0$ for any test function $f$ and we physicists are used to read $T(f)$ as
$$\int\!T(x)\,f(x)\, {\rm d}^4x.$$
Rather meaningless in several cases$\dots$ E.g. $D$ defined by
$D(f)=(df/dx)_{x=0}$ is a tempered distribution on $\Bbb R$ but in
order to write it as an integral you have to resort to
$$D(f) = -\!\int\!\delta'(x)\,f(x)\, {\rm d}x.$$
| {
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If a galaxy forms from a spherical stationary cloud, how much of the gas will escape? Let's ignore the dark matter legend and stay with Keplerian physics.
Assuming that there is a cloud with $N$ stationary particles with the same size uniformly distributed in a sphere and they condense to form a galaxy.
$$N>>10^{\text{many}}$$
Some particles come to the center. Some will escape. Is there any estimation that how much percentage of particles remain in the galaxy and how much will escape to the infinite space?
| To quote Binney and Tremaine Galactic Dynamics, 2nd et p. 556:
From time to time an encounter gives a star enough energy to escape
from the stellar system. Thus there is a slow but irreversible leakage
of stars from the system, so stellar systems gradually evolve towards
a final state consisting of only two stars in a Keplerian orbit, all
the others having escaped to infinity.
So the fraction remaining is $2/N$. This assumes random motions, which is fairly plausible.
The time to evaporation is on the order of $t_{evap}\approx \frac{14 N}{\log(N)}t_{crossing}$ where $t_{crossing}$ is the typical time to cross the cloud of particles, $\langle r\rangle /\langle v\rangle$.
For actual galaxy models one has to take collision cross sections with stars and the central black hole into account, I think about 10% of the stars tend to accrete.
| {
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Does time dilation mean that faster than light travel is backwards time travel? Ok. So my question is, I've always heard it that Faster Than Light travel is supposedly backwards time travel.
However, the time dilation formula is
$$T=\frac{T_0}{\sqrt{1-v^2/c^2}}$$
And while it is true that speeds greater than $c$ turn the denominator negative, doesn't the whole thing get rendered a complex fraction, rather than negative or backwards time flow, due to the square root of a negative number being a complex one?
Wouldn't this then mean that faster than light travel does something weird, rather than backwards time travel? In other words, wouldn't what happens during faster than light travel be some sort travel in a complex plane and wouldn't that have radically different implications to backwards time travel, depending on the direction one took FTL?
| When using formulas in physics it is important to keep in mind the assumptions that the formula is based on. In this case $T_0$ is the time on a clock in its rest frame. It is doubtful that tachyons exist, but if they do then they are not at rest in any inertial frame, so the time dilation formula simply does not apply.
However, the Lorentz transform does apply. So (in units where c=1) if we had a tachyon which moved at 2 c in our frame then it would have a worldline like $(t,x)=(\lambda,2\lambda)$ where $\lambda$ is an affine parameter and the y and z coordinates are suppressed. Now, if we do a Lorentz transform to a frame moving at 0.6 c relative to our frame then the worldline would be $(t’,x’)=(-0.25\lambda, 1.75\lambda)$.
Note that the worldline in the primed frame has the affine parameter increasing as time decreases whereas the affine parameter increases as time increases in our frame. In that sense it is traveling backwards in time in one frame or in the other.
| {
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IQHE, quantized conductance, and zeeman splitting I've been trying to understand IQHE by reading these lecture notes by David Tong.
Mainly, I was trying to understand the quantized hall resistivity in terms of the number of Landau levels crossing the fermi energy.
Then, I began thinking about why spin induced Zeeman splitting is never really mentioned in the context of IQHE.
The lecture notes say that it's because typically the
Zeeman splitting is very small and it polarizes the spin of the electron.
I think the spin based splitting of energy states still confuses me because in my mind, with the spins of electrons taken into account, you have twice as many energy states crossing the fermi energy.
The filling factor in IQHE is the number of landau levels crossing the fermi energy (as shown in the image below). To me, the spin Zeeman splitting seems to double that number.
| This doubling indeed happens in the "quantum spin Hall effect", but those systems are at zero magnetic field (and moreover enjoy time reversal symmetry, which is key). However, in the usual quantum Hall effect, there is a huge static magnetic field, which polarizes all the low energy electrons (the Zeeman splitting is large because it goes like $|B|$). Therefore, they are essentially spinless. See this review http://www.damtp.cam.ac.uk/user/tong/qhe/qhe.pdf , section 1.4.
| {
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Commutator of Gauge Transformations for Yang-Mills Theory Following the conventions of "Quantum Field Theory and the Standard Model" by Schwartz, we have that for Yang-Mills Theory, an infinitesimal gauge transformation acts like
$$\delta_{\alpha} A = d\alpha - i\left[A, \alpha\right].$$
I am trying to compute the commutator of two gauge transformations, which I expect to give
$$\left[\delta_{\alpha},\delta_{\beta}\right]A = i\delta_{\left[\alpha, \beta\right]}A.$$
However, this is not what I find. Carrying out the computation, I find that
$\left[\delta_{\alpha},\delta_{\beta}\right]A = \delta_{\alpha}\delta_{\beta}A - \delta_{\beta}\delta{_\alpha}A = \delta_{\alpha}\left(d\beta - i\left[A, \beta\right]\right) - \delta_{\beta}\left(d\alpha - i\left[A, \alpha\right]\right) = d\alpha - i\left[d\beta - i\left[A, \beta\right], \alpha \right] - d\beta + i\left[d\alpha - i\left[A, \alpha\right],\beta\right] = d\alpha -d\beta - i\left[ d\beta, \alpha\right] - \left[\left[A,\beta\right],\alpha\right] + i\left[d\alpha, \beta\right] + \left[ \left[A,\alpha\right], \beta \right]$
Some of the commutators can be simplified by realizing that
$\left[d\alpha, \beta\right] - \left[d\beta, \alpha\right] = d\alpha\beta - \beta d\alpha - d\beta \alpha + \alpha d\beta = d\left(\alpha\beta\right) - d\left(\beta\alpha\right) = d\left[\alpha, \beta\right]$.
We can also use the Jacobi identity to see that
$\left[\left[A,\alpha\right], \beta\right] - \left[\left[A, \beta\right], \alpha\right] = \left[\left[A,\alpha\right], \beta\right] + \left[\left[\beta, A\right], \alpha\right] = -\left[\left[\alpha, \beta\right], A\right]$.
Putting everything together, we have that
$\left[\delta_{\alpha},\delta_{\beta}\right]A = d\alpha - d\beta + id\left[\alpha, \beta\right] + \left[A, \left[\alpha, \beta\right]\right] = i\delta_{\left[\alpha, \beta\right]}A + d\alpha - d\beta$.
My question is why has the extra $d\alpha - d\beta$ appeared? Am I carrying out some step in the computation incorrectly, or am I missing something conceptually? As I check, I also computed the commutator by starting with the identity
$e^{i\delta_{\alpha}}e^{i\delta_{\beta}}e^{-i\delta_{\alpha}}e^{-i\delta_{\beta}}A = (1 - \left[\delta_{\alpha}, \delta_{\beta}\right])A + \mathcal{O}(\alpha^2)$.
Here I applied the finite gauge transformations on the left hand side, expanded to second order in $\alpha$ and $\beta$, and matched terms with the right hand side. After doing so I found $\left[\delta_{\alpha},\delta_{\beta}\right]A = i\delta_{\left[\alpha, \beta\right]}A$, as expected, so I'm fairly certain the extra $d\alpha - d\beta$ terms should not be present, but I don't understand where my mistake is when I start from the infinitesimal case.
| Try this:
$$
\left[\delta_{\alpha},\delta_{\beta}\right]A = (A + \delta_{\beta}A + \delta_{\alpha}\left(A + \delta_{\beta}A\right)) - (A + \delta_{\alpha}A + \delta_{\beta}\left(A + \delta_{\alpha}A\right)).
$$
| {
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Relationship between intensity and amplitude of light wave I am confused with the realtionship between intensity and amplitude of wave. My understanding is that energy in a wave is proportional to its intensity; which is proportional to the square of the maximum height of the wave. is that a correct understanding.? If that understanding is correct I have a red right and I increase the brightness what happens.? Do I increase the amplitude of the wave.?
| The maximum height of a wave is also referred to as it's amplitude, and yes, you are correct. The energy of a wave is proportional to the square of it's amplitude, and by consequence, it's intensity depends on the square of $A$ as well. If the intensity were to decay as $1/r^2$ then it's amplitude would decay as $1/r$ as well. Increasing the brightness of your light would give it a larger amplitude and thus a larger intensity as well.
| {
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In textbook thermodynamics, does heat $Q$ have a derivative of volume $V$? I'm just trying to get $U(V_2, T_2) - U(V_1, T_1)$ for an non-ideal gas, given the heat capacity at constant volume $c_V(V,T)$ and the equation of state $P(V,T)$. I know I can start from the equation:
$$ dU = dQ - PdV $$
From here, I want to get this in the form that I could integrate from $V_1$ to $V_2$ and $T_1$ to $T_2$, but taking the total derivative with respect to $V$ and $T$ of this equation is very confusing to me.
I know I want to do something like:
$$ dU = \left ( \frac{dU}{dT} \right )_V dT + \left ( \frac{dU}{dV} \right )_T dV $$
I know that $ \left ( \frac{dU}{dT} \right )_V = C_V(V, T) $, but for $\left ( \frac{dU}{dV} \right )_T$, I'm not sure whether this would just be $-p$ or $\left ( \frac{dQ}{dV} \right )_T - p$. In the latter case, I don't know what $dQ/dV \big |_T $ is... it's not mentioned in the equation of state or the heat capacity.
| You don't have
$$\text{d}U=\delta Q-p\cdot\text{d}V$$
but
$$\text{d}U=T\cdot\text{d}S-p\cdot\text{d}V$$
The first principles assumes
$$\text{d}U=\delta Q+\delta W$$
with
$$W=\int_1^2 \delta W=-\int_{V_1}^{V_2} p_\text{ext}\cdot\text{d}V$$
If the transformation is mechanically reversible, on have $$p_\text{ext}\approx p$$
On the other hand, you have with the Maxwell's Relations
$$\text{d}U=C_V \cdot \text{d}T+\left[T\cdot\left(\frac{\partial p}{\partial T}\right)_V-p\right]\cdot\text{d}V$$
so
$$Q=\int_{V_1}^{V_2} p_\text{ext}\cdot\text{d}V+\int_{T_1}^{T_2} C_V\cdot\text{d}T+\int_{V_1}^{V_2}\left[T\cdot\left(\frac{\partial p}{\partial T}\right)_V-p\right]\cdot\text{d}V$$
| {
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What causes burns when in contact with hot water? As I understand it thermal energy (heat) is simply a measure of the kinetic energy of an object (For example : water).Hot water is simply water with a larger kinetic energy in its molecules, right ?
So how do my hands get burned if I immerse them in hot water ? Do the particles collide with my hand and produce burns ?
PS : I may have a conpletely wrong understanding of how heat works .
| Yes, temperature is generally a measure of the average translational kinetic energy of the molecules of an object.
Skin burns occur when the combination of the temperature on the skin and the duration of the exposure of the skin to that temperature exceed the threshold of burn.
Given a solid, liquid (e.g. water), and gas at the same temperature in contact with the skin, a burn will generally occur first (in time) with the solid, followed by the liquid, and then the gas, due to the relative heat transfer rates (conduction and convection).
So yes your skin can get burned due to exposure to water as well as a gas. It will be less severe if you are in contact for the same amount of time with a gas than than with the water, but more severe if you were in contact with a solid at the same temperature,
Hope this helps
| {
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Can any body be uniform in the universe? If I take any body in the shape of a rod and stretch that, after it reaches breaking stress it breaks at one point.
Even though we apply the same the stress on each and every part of the rod it broke at one point. If it's uniform it should break at all points because breaking stress is same for all the parts of body as it's uniform.
| Does Prince Rupert's drop count?
It will break explosively and reasonably uniformly when damaged due to accumulated stress. This is of course not due to uniformity but rather to its peculiar composition. I wonder if you could redefine uniformity so that it will be covered.
| {
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Is this possible to focus common light (not laser) to a small point in the wall and control it with mirrors? I want to make something like a laser projector for an experiment but with common light.
My question is: is this possible to focus a small point of common light to the wall and control it with mirrors? What combination of lenses I'll need for it?
| An alternative to consider is to use a strong light source for example halogen lamp, which is typically used in fiber optic illuminator.
| {
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Example in which light takes the path of maximum optical length According to the modern version of Fermat's principle,"A light ray in going from point A to point B must traverse an optical path length that is stationary with respect to variations of that path.".Is a maximum optical path length possible ?What if we keep adding deviations to the optical path length?
| I offer you two examples.
A) Draw a semicircle on AB as its diameter. Assume semicircle is
reflecting, and you are looking for reflection from A to B. It obviously
happens at C, midpoint of arc AB. I leave to you to show that for any other point P of arc ACB you have AP + PB < AC + CB.
B) Consider two points A, B, with A in vacuum, B in a medium with
refracting index $n$. The locus of points P such that ${\rm
AP}+n\,\rm BP=const$ is a curve (Cartesian oval). If the oval separates vacuum and medium with index $n$, then all rays APB have the same optical length. Let C be the intersection of the oval with AB. Draw a circle having center on AB, between B and C. If medium is delimited by this circle, then for all P on the circle ${\rm AP}+n\,\rm BP < AB$.
| {
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Do we need a small displacement to create a oscillatory motion on the spring? Do we need a small displacement to create a oscillatory motion on the spring with a mass attached to it? Whats the limit of the displacement that we can give initally to create a oscillatory motion? Is it has to be small or it can large?
| For an ideal spring the force as a function of position is $F=-kx$. Since this is a restoring force for all $x$, there is no limit on what initial displacement will cause oscillations (anything will do).
Even if your spring is not ideal, the force will probably have the form of $F=-f(x)$, where $f(x)>0$ for all $x\neq0$, so that the force is still restorative. Therefore you would still have oscillations for all displacements from equilibrium.
Where this breaks down is in the real world your springs can't stretch forever. Even assuming perfect elasticity, your spring wouldn't be able to stretch past it's full stretched out length. For real springs, there is a point where if you stretch the spring past it, the spring will not restore to its original form. Unfortunately, this depends on the physical spring itself, so at this point I can't keep talking in general.
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"timestamp": "2023-03-29T00:00:00",
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How can a particle in circular motion about a fixed point accelerate, if the point doesn't too? When a particle is performing uniform circular motion attached to a string about a fixed centre, at any instant of time its acceleration is directed towards the centre but the centre has no acceleration. But I was taught in school this is not possible because of the string constraint:
The accelerations of the ends of a string are the same if the string is not slack.
Where am I wrong?
| I suspect what you were actually taught is either that the velocity component along the string is the same on both ends, or that the force / tension is the same. One might think that both of these imply also equal acceleration component, by way of simple $f(x) = g(x) \Rightarrow f'(x) = g'(x)$ consideration or via Newton's second law, respectively. But actually neither is the case!
To take the derivative of these velocities, you need to take into account that the direction changes.
$$\begin{align}
\mathbf{v}_0(t) =& 0
\\ \mathbf{v}_1(t) =& r\cdot\omega\cdot\begin{pmatrix}-\sin(\omega\cdot t)\\\cos(\omega\cdot t)\end{pmatrix}
\end{align}$$
The radial component of $\mathbf{v}_1$ is indeed always zero
$$
\mathbf{v}_1(t)\cdot\mathbf{e}_{\mathrm{r}}
= r\cdot\omega\cdot\Bigl(-\cos(\omega\cdot t)\cdot\sin(\omega\cdot t) + \sin(\omega\cdot t)\cdot\cos(\omega\cdot t)\Bigr)
= 0
$$
but the radial component of the acceleration is nonzero. You need to first take the derivative of the velocity vector, then project it into the radial direction, not take the derivative of the radial velocity only.
$$
\frac{\mathrm{d}\mathbf{v}_1}{\mathrm{d}t} = r\cdot\omega^2\cdot\begin{pmatrix}-\cos(\omega\cdot t)\\-\sin(\omega\cdot t)\end{pmatrix}
= -r\cdot\omega^2\cdot \mathbf{e}_{\mathrm{r}}.
$$
With Newton's law you can't argue because this assumes masses. For a true circular motion around a center, you need an infinitely massive central anchor and a finite-mass satellite. The center has then virtually zero acceleration despite handling a significant force. Not so for the satellite.
| {
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Wigner $D$ matrix How to derive symmetry relation of Wigner $D$ matrix? I mean this relation
$$
D_{m',m}^j (\alpha,\beta,\gamma)
= (-1)^{m'-m} D_{-m',-m}^j (\alpha,\beta,\gamma)^*.
$$
I want to derive this, but I don't know how. Anyone derive this, please.
| The Wikipedia page that you linked to explains, at the end of the section “Properties of the Wigner D-matrix”, that this property is a consequence of the commutation of the rotation matrix with the time reversal operator $T$. The derivation is given there.
| {
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What is the relation between physical theory and physical law? Gravitational law was explained by Newtons theory of gravity. So a law was described by a theory. What is the theory for Newton's laws of motion?
| Laws, postulates, principles are the the axioms used in physics mathematical models to pick up from the mathematical solutions those that describe data and measurements an predict new situations successfully, validating the theory. As in mathematics one cannot question the axioms, so in physics mathematical models the laws are assumed axiomatically.
As in mathematics, what is a theorem can become an axiom in a different formulation and the axiom becomes the theorem, this can also happen with the physics models allowing different mathematical formulations, which have to be consistent between them.
| {
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Is there a medium less dense than vacuum, in which light can travel faster than $c$? Is there a medium less dense than vacuum, in which light can travel faster than $c$? If not, can we make it?
| The answer would seem to be "no", because you make a medium less dense by removing material from it. Once you get to a vacuum, you are only left with how good is the vacuum?
An experiment for an undergraduate optics lab would be to build a Michelson interferometer with a gas cell in one arm. As the gas is pumped out, the interference fringes shift. You can actually calculate the change in the effective speed of light at different air pressures, and project how it would change as the pressure declines further and further.
The limiting value is the speed of light in a vacuum.
| {
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What is *diagonal* long range order? I have seen this question about off-diagonal long range order in superfluids.
What’s the difference and the significance between long range diagonal and off-diagonal long range order?
| Found the answer. It means density modulation.
While off-diagonal long-range order is defined as $n^{(1)}(r,r')=\langle \psi^\dagger (r)\psi (r')\rangle$, diagonal long-range order is the special case where $r = r'$. Qualitatively, one can see that when $r=r'$ the argument of the expectation value is $|\psi(r)|^2$, i.e. the probability density of finding the particle in space. Hence, diagonal long-range order $\leftrightarrow$ density profile of the system.
| {
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Is the Hydraulic analogy of a resistor wrong? According to Wikipedia's Hydraulic analogy page
A resistor is analogous to a constricted pipe such at the one in the photo below.
Bernouli's principle tells us that the pressure in both sides would be identical in both of the wide sections of the pipe
and in the narrow section of the pipe the pressure would be lower
it also tells us that the velocity inside the narrow section would be faster than the velocity in the wide parts
But, when a resistor is placed in a circuit, the rate of electron gets lower and stays that way throughout the circuit and causes a voltage drop across it,
so when exiting the resistor, the pressure does not return to the entry pressure like it does in the pipe analogy.
so, in what way is this Hydraulic analogy of a resistor actually analogous to a resistor ?
| You're probably aware of this; but just to cover all the bases, in the hydraulic analogy, pressure represents voltage, flow rate represents current, and as you mentioned, a pipe restriction represents a resistor.
For starters, there's a comment in your question that should be addressed:
Bernouli's principle tells us that the pressure in both sides would be identical in both of the wide sections of the pipe and in the narrow section of the pipe the pressure would be lower
This is true; but for a real flow when the effects of viscosity are considered, the flow restriction will actually create a pressure drop, so pressure at the end will be slightly reduced compared to before the restriction. The more restrictive, the greater the pressure drop. This is analogous to voltage drop across a resistor.
Due to the pressure drop across the restriction, doing this with pipes will decrease the total flow rate through the pipe, the same way a resistor reduces the total current in a line. Because there is a pressure drop in the pipe, that means you either require a greater pressure to get the same outlet flow rate, or by applying the same pressure, the flow at the outlet is less.
This is analogous to what happens with resistors in circuits. The confusion comes from not accounting for the pressure drop across the restriction, and looking too closely at the localized velocity in the restriction, compared to the actual flow rate.
| {
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How to calculate conductivity / electron mobility from theory? Is there a way to make quantitative statements about the conductivity of materials with band theory?
If not I should still be able to get information about the conductivity from Green-Kubo relations of the electron wavefunctions in the material, right?
| What you are asking are transport properties of materials, which represent the response of the system to an external perturbation (such as electric field) so that you cannot obtain just from band structure.
For the transport of the classical particle, you can solve the Boltzmann transport equation, from which you can derive the Drude formula for conductivity. For fully quantum treatment, you can use nonequilibrium Green's function method (or Keldysh formalism), from which you can derive the current formula and transmission coefficient.
From bandstructure, you can obtain mobility, see section3.4 in this paper.
| {
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Why don't all gasses have infinite entropy? Entropy of an ideal gas is defined as the logarithm of the number of possible states the gas can have multiplied by Boltzmann's constant:
$${\displaystyle S=k_{\mathrm {B} }\log \Omega .}$$
In deriving the Maxwell-Boltzmann distribution, we initially start by counting a finite number of states, so this definition of entropy makes sense. But in the end we say that the number of possible states is so high that we can acctually say the distribution is continuous. But if the distribution is continuous, the number of possible states is infinite. So why is entropy not always infinite when a continuous distribution is used?
|
So why is entropy not always infinite when a continuous distribution is used?
If the space of possible states is a continuous region, then the original definition of entropy is not useful (all entropies would be the same, infinity). One may introduce different definition:
$$
S = k_B \ln \Omega
$$
where $\Omega$ is volume of the region (as opposed to number of points in that region). This is sometimes done in statistical physics (pre - quantum statistical physics).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Deriving Friedmann Equations without General Relativity Can we derive the analytic Friedmann Equations without general relativity, starting from completely classical/nonrelativistic arguments? (If we consider sufficiently small volumes.)
| A Newtonian approach is possible but of of course it is not rigorous.
We're in a newtonian approximation now, and we want to describe the motion of a unit mass at a point P on the surface of a sphere.
So let's take this spherical distribution of matter, take a point at a distance $l$ from the origin. The equation of motion of that point is
$$
\frac{d^2l}{dt^2} = - \frac{GM}{l^2} \tag{1}
$$
$$
\Rightarrow \, \, \, \ddot{l}\, \dot{l} = \frac{d}{dt} \frac{\dot{l}^2}{2}= - \frac{GM}{l^2} \dot{l} = \frac{d}{dt} \frac{GM}{l} \tag{2}
$$
from which you can find
$$
\frac{d}{dt} \Big(\frac{\dot{l}^2}{2} - \frac{GM}{l} \Big) =0 \tag {3}
$$
$$
\Rightarrow \, \, \dot{l}^2 = \frac{2GM}{l} + constant \tag{4}
$$
This is the equation of conservation of energy per unit mass.
Now in cosmology you have
$$l= d_c \frac{a}{a_0} = \tilde{D} a \tag {5}$$
where $d_c$ is the comoving distance and $a$ is the usual scale factor. We put this in $(4)$ and remembering that $\tilde{D}$ is not affected by the time derivative and $M = \frac{4\pi}{3} \rho (\tilde{D} a \, ) ^3$, we get
$$
( \,\tilde{D} \dot{a} \, )^2 = \frac{2G \frac {4\pi}{3} \rho (\tilde{D} a \, ) ^3}{\tilde{D}a} \tag{6}
$$
From which we have
$$
\dot{a} = \frac{8\pi G}{3} \rho a^2 + constant \tag{F2}
$$
that is the second Friedmann equation when $constant = -kc^2$.
Now we derive the first Friedmann equation simply replacing $\rho$ with $\rho_{eff} = \rho + \frac{3p}{c^2}$ that is we put a relativistic term in the density by hand. It goes like this, starting from $(1)$
$$
\frac{d^2l}{dt^2} = - \frac{GM}{l^2} = -\frac{G}{l^2} \frac{4\pi}{3} \rho_{eff} l^3 = - \frac{4 \pi G}{3} l \big( \, \rho + \frac{3p}{c^2} \, \big)
$$
Using again $(5)$ one obtains the first Friedmann equation
$$
\ddot{a} = - \frac{4 \pi}{3} G \big( \, \rho + \frac{3p}{c^2} \, \big) a \tag{F1}
$$
You can find this kind of treatment in Coles-Luccin: Cosmology-The origin and evolution of cosmic structure. I don't know in which chapter cause I don't have the book and I used my notes for this answer, but I'm pretty sure my notes were taken from that book.
| {
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Ladder operators and energy levels I am studying how to get the normalization factor algebraically in the exited states of the harmonic oscillator using the beautiful ladder operators technique.
I am stuck at the point where is stated that the product of the ladder operators yields the energy levels (in $\widehat{a}^\dagger\widehat{a}$ case) and the energy levels + 1 (in $\widehat{a}\widehat{a}^\dagger$ case).
$$\widehat{a}\widehat{a}^\dagger\psi_n= (n+1)\psi_n$$
$$\widehat{a}^\dagger\widehat{a}\psi_n=n\psi_n$$
The thing is that I know how to get them mathematically using:
$$\widehat{a}^\dagger|n\rangle=\sqrt{n+1}|n+1\rangle$$
$$\widehat{a}|n\rangle=\sqrt{n}|n-1\rangle$$
But I am not satisfied with this due to the fact that my textbook (Griffiths) does not obtain them like that but using both the Schrodinger Equation and the solution to the exited states of the harmonic oscillator:
$$\hbar\omega\left(a\pm a\mp \pm \frac{1}{2}\right)\psi = E \psi$$
$$\psi_n = A_n (\widehat{a}^\dagger )^n \psi_0$$
But I am not getting them using the stated way. May you please prove it?
I am also wondering why the product of the two ladder operators do yield energy levels without a constant of proportionality. In fact; when just one operator is multiplied by the eigenstates you get a constant of proportionality multiplied by the eigenstates (+1 or -1; depending on the operator we used). But this may be another question so please let me know whether I should post it separately:
| I think you misunderstood what he meant. Griffiths did use one of the equations you stated :
\begin{equation}
\hbar\omega\left(a\pm a\mp \pm \frac{1}{2}\right)\psi = E \psi \ ,
\end{equation}
but the other equation he used (2.61 in my edition) wasn't the form of the eigenstates, but the eigenvalues which are written beside it, and not numbered separetedly:
\begin{equation}
\hbar\omega\left(n +\frac{1}{2}\right) = E_n.
\end{equation}
Substitution of this equation, as well as the Hamiltonian $H=\hbar\omega\left(a_+a_- + \frac{1}{2}\right)$ on the first gives the result immediately. a_+a_- is found by direct inspection and the other equation can be infered, for example, from the commutation relation $[a_+,a_-]=1$
| {
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Formula for potential energy? Conservation of energy? How would we know what formula to use for potential energy?
In my class, $mgh$ was used, but when dealing with a spring, it's ${1\over2}kx^2$. Is that because that's the elastic potential energy formula?
Also, for elastic and inelastic collisions, momentum is conserved. But kinetic energy is conserved only in elastic collisions, what does this really mean?
| For the first part, you have to remember the work done by a spring. Say we have a coil spring squeezed by amount of x cm and locked in this position, hence loaded with a force of F= kx.
But when you release the lock and allow the spring to stretch back to its relaxed length, F gradually tapers off to zero.
So the work done by spring is average of F and 0, or
$ (kx+0)/2 (x) = \frac {1}{2}kx^2$
As for the second part of your question, in an inelastic collision the total moment is conserved, but the kinetic energy is partially lost to vibration of parts after impact which is turned to heat and plastic deformation of them.
| {
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Entanglement in atoms, nuclei and quantized fields Are the particles that are bounded by chemical bonds or nuclei joined by nuclear forces entangled or are they pure states?
In addition to that, are the subatomic particles in the atoms and nuclei, excitations of a single electron field, quark field, entangled or are just pure states in general when measured?
That is, if we don't measure anything, usual postulates of quantum mechanics say that states are in linear superposition, but are they mixed, pure or entangled states in general?
| Quantum systems become entangled through interaction with each other. Entanglement is broken when the entangled particles decohere through interaction with the environment; for example, when a measurement is made.
| {
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"timestamp": "2023-03-29T00:00:00",
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Wave behavior of particles When people say that every moving particle has an associated wave, do they mean that the particles will move up and down physically, for example when we say that a moving electron has a wave associated with it, does the electron physically oscillate? Or is it some other wave, like a probability wave? I really don't understand the latter.
| They do not mean that the particle itself moves up and down in a wavelike way. What is meant by wave/particle behavior or duality is something more subtle; something that many very smart people have spent their lifetimes working on and which my answer will treat in a simplified way which I hope you can grasp.
For objects that are very, very small, it is possible for them to manifest radically different behavior depending on the details of their environment, whether they are singly isolated or part of a large population, and how they interact with the tools we use to detect and study them.
For example, in the case of a single electron that is speeding through space, it's possible to interrupt its path with a detector that registers the impact of the electron as if it were a tiny bullet. It is also possible to interrupt a stream of electrons with a detector that bends their paths just as if they were a train of waves instead of a stream of tiny bullets.
The standard interpretation of the so-called wave equation that describes the propagation of an electron through space is that the probability of finding the electron at a given location along its path can be extracted from that equation, and that this probability varies from point to point in a way that is wavelike: the crests of the wave represent spots where the electron is most likely to be found, and the troughs represent spots where it is least likely to be found.
There is a strict and well-defined mathematical formalism which is used by skilled practitioners to handle questions like this, and there are others here who can furnish you with that if you wish.
| {
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A perfectly fitting pot and its lid often stick after cooking I encountered a question that asks for the reason that a perfectly fitting pot and its lid often stick after cooking when it cools down. The answer in the solution manual was that the pressure decreases when the temperature decreases. I understand that point, but my problem is the following: The pot before being heated during cooking is influenced only by the atmospheric value, and the food inside that pot is the atmospheric pressure also. So this is the initial state for the pressure inside the pot now. When heated the pressure inside the pot increases. When we let the pot cool down to the room tempreture, would the pressure return to its initial value which is the atmospheric pressure? Why does the solution manual consider a vacuum inside the pot?
| You've made the wrong assumption about what a perfectly fitting lid does. Rather than sealing perfectly, it will vent gas when the interior is at a higher pressure (the lid will lift), but seal when the exterior pressure is higher (the lid is pressed down).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/440322",
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