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How do the unit vectors in spherical coordinates combine to result in a generic vector? I might be missing the obvious, but I can't figure out how the unit vectors in spherical coordinates combine to result in a generic vector. In cartesian coordinates, we would have for example $ \mathbf{r} = x \mathbf{\hat{i}} + y \mathbf{\hat{j}} + z \mathbf{\hat{k}}$. But in spherical coordinates, the position vector is actually a multiple of the unit vector $ \mathbf{\hat{e}_{r}} $, since $ \mathbf{r} = r \mathbf{\hat{e}_{r}} $ and not a linear combination of $\mathbf{\hat{e}_{\theta}} $, $ \mathbf{\hat{e}_{\phi}}$ and $ \mathbf{\hat{e}_{r}}$ (attached picture). Do we actually combine all 3 unit vectors in spherical coordinates to obtain a certain vector, or are $\mathbf{\hat{e}_{\theta}} $ and $ \mathbf{\hat{e}_{\phi}}$ just an indication of direction?
| In spherical coordinates, the $\theta$ and $\phi$ are like latitude and longitude on the surface of the Earth. At a given point, if we want to express a "south pointing direction" tangent to the surface, we'd use the $\theta$ basis vector. Likewise, an "east pointing direction" would be expressed using the $\phi$ basis vector. The difficulty you have with "position vectors" is that they are always "from the origin". Better to think of displacement vectors, e.g. the difference between two position vectors on a trajectory. Then all three spherical basis vectors are used to express the direction and length of the displacement from the initial point.
| {
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How does the current remain the same in a circuit? I understand when we say current, we mean charge (protons/electrons) passing past a point per second. And the charges have energy due to the e.m.f. of the power supply.
Now tell me, if a lamp has resistance and you hook it in the circuit, how will the current stay the same? The charges obviously lose energy in the lamp and so become SLOWER, which should mean current decreases, right?
[Edit]
All answers explained a bit of everything, so it was hard to choose one. If YOU are looking for an answer, please check the others too, in case the accepted one doesn't answer your question.
| Here is why the current stays the same going through the bulb:
Think of current as water flowing downstream in a river. The water comes to a dam, and flows over the top, and then falls all the way down to the bottom of the dam, then resumes flowing downstream. Every gallon of water that was flowing in the river will go over the top of the dam. Every gallon of water that flows over the top will fall to the bottom. Every gallon that falls to the bottom resumes flowing downstream. Therefore, every single gallon of water that was originally flowing downstream before coming to the dam will wind up flowing downstream again after leaving the dam behind.
We say that water is conserved. One gallon in, one gallon out. Current through a circuit behaves the same way: One ampere in, one ampere out.
| {
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Normal modes of vibration of a plate vs a membrane I have been studying Chladni patterns but recently I have stumbled on some conceptual questions that I seem to not have an answer. At first I thought that the theory would be the same of a vibrating membrane, with very well defined solutions for given geometries like a square or a circular one like it is described in this document (found here) giving resonance at very well defined and discrete frequencies.. But then I saw the computational work done by kai5z in GitHub, where its script is able to generate a Chladni pattern for any frequency. At first I thought that it would have something to be with the fact that it considers a forced 2D oscillator but that doesn't make sense.
Further investigation showed me that Chladni patterns come from the eigenfunctions of a bi-harmonic operator and the vibration modes for a membrane come from the eigenfunctions of the Laplacian, but why? How are the problems different? What makes it imperative to have two theories for the two problems? Are the membrane solutions an approximation for the plate solutions? Are they somewhat related?
| The difference depends on the physical structures: the tension of the membrane must be imposed by means of external forces whereas that of a plate naturally exists in its interior. In other words the relation between deformations and stress is different.
The small deformations of the structure, in the first case, are dynamically described in terms of a transversal D'Alembert equation and the spatial part of this equation of motion includes the spatial 2D Laplacian. In the second case, the model is more complicated and we have a similar equation of motion of the transversal deformations where however the Laplace operator is now replaced by (minus) its square. I am assuming that the membrane and the plate are uniform and isotropic, otherwise the equations become more complicated.
What the two structures share are the eigenfunctions of the relevant spatial operator: they only depends on the form of the structure (and on the boundary conditions). The eigenvalues are different (in fact the eigenvalues of $\Delta^2$ are the square of those of $-\Delta$). So the Chladni figures should be quite similar in some regime for equal shapes.
| {
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Normalization of the action in Special Relativity The action for a massive point particle in Special Relativity is given as
$$A =-mc^2\int d\tau,$$
Where $\tau$ represents the proper time, and $m$ represents the (rest) mass. From what I could understand, the Action must not change with respect to the reference frame, and hence it can be written as $$something\int d\tau$$ but why must the something need to be proportional to mass?
| The equation you mention is the action of a single point particle.
$$S =-mc^2\int d\tau$$
The unit of action is energy multiplied by time, in the present case the rest energy which corresponds to the mass of the particle multiplied by the proper time of the particle.
This equation refers to the action from the point of view of the reference frame of the point particle along its own worldline (which is parameterized by its own proper time), that means that the point particle is its own observer. From its own point of view, its velocity and its displacement in space is always zero, and also its momentum and its kinetic energy is zero. You can say that the equation describes the aging of the particle, its mass (= its rest energy) is transported through time.
| {
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Can a Black Hole become a normal mass again? It is well known how Black hole forms. But once it is formed are there any circumstances which can lead a black hole to becoming a normal mass again?
| Hawking radiation leads to a complete evaporation of black holes, whether there are remnants and the problems this creates is a matter of ongoing discussion.For example see here for a review.
Forty years after the discovery of Hawking radiation, its exact nature remains elusive. If Hawking radiation does not carry any information out from the ever shrinking black hole, it seems that unitarity is violated once the black hole completely evaporates. On the other hand, attempts to recover information via quantum entanglement lead to the firewall controversy. Amid the confusions, the possibility that black hole evaporation stops with a "remnant" has remained unpopular and is often dismissed due to some "undesired properties" of such an object. Nevertheless, as in any scientific debate, the pros and cons of any proposal must be carefully scrutinized. We fill in the void of the literature by providing a timely review of various types of black hole remnants, and provide some new thoughts regarding the challenges that black hole remnants face in the context of the information loss paradox and its latest incarnation, namely the firewall controversy. The importance of understanding the role of curvature singularity is also emphasized, after all there remains a possibility that the singularity cannot be cured even by quantum gravity. In this context a black hole remnant conveniently serves as a cosmic censor. We conclude that a remnant remains a possible end state of Hawking evaporation, and if it contains large interior geometry, may help to ameliorate the information loss paradox and the firewall controversy. We hope that this will raise some interests in the community to investigate remnants more critically but also more thoroughly.`
| {
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The Calculation of work in different frames Say there is a man running on a road. Friction is applying a force hence he is accelerating, but the friction is static hence does no work. We know that the change in kinetic energy is due to internal work done by his muscles.
Now say that the same man is running on a wooden plank kept on a frictionless surface. As the man moves ahead the plank goes behind. At the point of contact, the friction applies force in the forward direction accelerating the man ahead, but since the plank itself is moving behind, the point of contact is going behind. Hence, work done from ground frame is negative. But if I change my frame to the plank then work is zero again. Thus work is frame dependent.
Is my conclusion correct? And I always think if the calculation for work is frame dependent than when we theorize physics we might be missing out on different types of work which may lead to concepts like dark energy. How to physicists think about this.
| From the ground reference frame, there is no net work done. Friction did displace you by a distance of say $d$. However, the same friction also displaced the wooden board by a distance $d'$. These two perfectly cancel and this can be realised by noticing that the center of mass remains fixed throughout this process.
However, with respect to the wooden board or the man, they have indeed done work. This is because both are non-inertial frames where Newton's laws do not hold. There is no apparent equal and opposite force applied back and it seems that work was done out of thin air. For this reason, a pseudo-force is added to make sense out of non-inertial frames.
The concept of dark energy has nothing to do with the frame of reference and was theorised through a series of experiments to explain the rate of expansion of the universe.
| {
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If magnetic field lines don't exist, what are these iron filings doing around a magnet? Obviously the iron filings can be seen aligning themselves along the virtual magnetic field lines produced by the permanent magnet, the virtual magnetic field line is made of electromagnetic field due to the alignment of electrons in the magnet but why the patterns, why lines? Do these lines have thickness? Are they due to interference pattern?
| Magnetic field lines show the direction of force a magnetic mono-pole would experience if it were free to move under the influence of a magnet, but unfortunately, we all know that mono-poles are just a thing of imagination and don't exist in reality. This is given by Gauss' law for magnetism which states
$$\oint \mathbf B\cdot\text d\mathbf s=0$$
Whenever iron filings are scattered around a magnet, they become magnetized and (for a short time) also behave as tiny little magnetic dipoles. Those little magnets can just rotate a bit and align themselves with the field because they can't move and stick to the bar since the experiment table offers some friction.
Now as more of the filings align themselves with the field generated by the adjacent filings, they form a continuous loop around the magnet and that gives us a very nice visual perception of how the magnetic field lines look like. Of course, the "field lines" don't exist in real life.
| {
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Boyle's Law and hot air balloon A bit dumb question because it is really difficult to imagine it. :-
Pressure is force per area. Talking about gases, the pressure is said to be the force molecules exert on walls of let's say a balloon.
Usually in examples of Boyle's law, our teachers mention hot air balloon, that the size of balloon increases as pressure decreases.
But at as height increases pressure decreases because there are less molecules above us.
How will the pressure of molecules outside the balloon effect the pressure of molecule inside the balloon.
| For the system to be in equilibrium, the pressure outside must be equal to the pressure inside, otherwise the inside will contract (if the pressure outside is higher) or expand (if lower). There is a minor contribution given by the elasticity of the balloon capable of handling some extra pressure but we can neglect that here.
So as the atmosphere gets thinner, the balloon expands a bit.
| {
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Gravitational Wave - What is waving? Two kinds of wave transmission are:
*
*Light waves, where a substance (photon) travels as a wave.
*An attached rope, like at the gym, that is "waved" up and down. Here, no substance travels to a new spot, but adjacent parts transmit the energy to others.
QUESTION: Which method do gravitational waves propagate by?
| Let us clear at first that photons are elementary quantum mechanical particles. That classical electromagnetic waves emerge from a confluence of photons can be shown, but classical electromagnetic theory works with the self propagation of the light energy, carried by electric and magnetic fields.
Two kinds of wave transmission are: 1. Light waves, where a substance (photon) travels as a wave.
Lets ignore the quantum frame, as gravity is not yet definitively quantized.
*An attached rope, like at the gym, that is "waved" up and down. Here, no substance travels to a new spot, but adjacent parts transmit the energy to others.
QUESTION: Which method do gravitational waves propagate by?
Like 1., except what is waving is spacetime itself. This was beautifully seen in the experiment at LIGO .
Now it is expected when gravity is quantized that the gravitational waves will be emergent from a confluence of gravitons ( hypothetical at present) the way electromagnetic waves are a confluence, superposition of wavefunctions, of photons.
| {
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If sound passes through material, vibration is produced. So are electromagnetic waves produced too? Sound means vibration of molecules and vibration produces electromagnetic waves. So, this means that sound produces electromagnetic waves directly.
Is this possible?
| If the vibrating material is charged, then electromagnetic radiation will be produced.
There is a type of microphone called the capacitor or condenser microphone, in which the microphone is a capacitor whose capacitance varies with the acoustic vibrations.
https://en.wikipedia.org/wiki/Microphone#Condenser_microphone
The capacitance can be detected either by measuring the impedance to an RF signal, or by applying a high voltage direct current bias and measuring the current flow in and out of the capacitor as its capacitance changes.
Normally the microphone is connected directly to the amplifier circuit, but it should be possible to couple it indirectly. In this case the electromagnetic radiation produced by acoustic vibration would be detectable.
If the vibrating material is magnetized, then electromagnetic radiation will be produced.
A typical electric guitar pickup consists of several magnets (usually one per string) surrounded by a coil of wire. The magnets induce magnetism in the string. The vibrations of the magnetized string are picked up by the coil of wire. Hence the electromagnic radiation produced by the string is detected.
However, in the general case where nothing is done to make the vibrating material an emitter of electromagnetic radiation, it will be difficult to detect (unless the vibration is so intense as to cause the vibrating material to heat up and emit radiation due to its increased temperature.)
https://en.wikipedia.org/wiki/Pickup_(music_technology)
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Dynamicity inside a stationary water drop I was doing some experiments with water drops on lampblack when I saw this.
You can see the full video here.
Inside a water drop which is perfectly still from the outside, you can see some moving things, which I suppose are some lampblack flakes from the surface. I am not bothered about how those flakes got in there, but I am interested in their motion.
Why are they even moving?
Is it some kind of Brownian motion?
I am amazed by the dynamicity inside such a seemingly inactive water droplet.
If I had not seen this, I would have considered the water drop as a simple sphere for the rest of my life.
This observation, at least for me, raises an important question.
Is anything in the world really NOT dynamic?
| Those are flakes of lampblack. Their wettability in water is unequal around their periphery. This allows surface tension forces to apply a netforce in some random direction on the flake and force it to scoot around, until the wettability difference goes away as the water begins to wet out the previously poorly-wetting spots.
| {
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Direction of average acceleration in circular motion I know that the instantaneous acceleration is always directed towards the center of the circle.But what about average acceleration.
In the above figure my book says place change in velocity along the line that bisects angle $r$ and $r'$ and observe that it is directed towards centre.
my question is that is there any rule that we should place it along the angle bisector between the two given points to get average acceleration direction.
Any help will be appreciated
| The average acceleration over one or more complete revolutions is zero. This can be seen from the fact that when the object is on the opposite side its acceleration is opposite.
If the average acceleration were nonzero, the object would not keep returning to the same location.
| {
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Why does the same proportion of a radioactive substance decay per time period? (half life) Just wondering, if decay is random, why does the activity half every half life, as in, why does it have to reduce by the same proportion in the same time period?
| It's a consequence of the fact that the nucleus doesn't know how many other nuclei are in your lump of material. A 3kg block of uranium has to decay at the same rate as three 1kg blocks of uranium. Which means a 1kg block has to decay at 1/3 of that rate. A 1kg block is the same as 3 1/3kg blocks, so a 1/3kg block has to decay at 1/3 of that rate too.
Now suppose you have a 3kg block of uranium (or whatever-ium), and it takes a year to decay into 1kg of whatever-ium (and 2kg of other stuff - let's pretend you have a system to take that away because we're only talking about the uranium here). Since you have 1kg, it must decay at 1/3 of the rate that it did at the start. It takes the same time to decay 2/3kg from a 1kg block as it does to decay 2kg from a 3kg block. That means after a year, you only have 1/3kg left. And decaying 2/9kg from a 1/3kg block takes the same time as decaying 2/3kg from a 1kg block. So after another year, you have 1/9kg left. And so on.
We say that whatever-ium has a third-life of one year.
We can extrapolate with maths. We know it has a ninth-life (1/3 squared) of two years. We know it has a 57.3%-life (square root of 1/3) of half a year. We know it has a half-life of 0.63092975357 years (you need to use logarithms to work this out).
We measure things in half-lives because it's convenient. We could equally well use third-lives or quarter-lives or fifth-lives or two-thirds-lives.
| {
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What is the general prescription for constructing the quantum mechanical momentum operator conjugate to a given coordinate? Obviously, if $x$ is a Cartesian coordinate, then the corresponding momentum operator is $-i \hbar \partial_x$. But what if $x$ is something more complicated, like some sort of curvilinear coordinate in 3D space?
There is this question elsewhere on this site. Quantum mechanical analogue of conjugate momentum But it's not clear to me that the accepted answer there is really answering the question I am asking. Nothing in the question or answer specifically addresses non-Cartesian coordinates, certainly not in any detail.
| If an $n$-dimensional configuration manifold $M$ of some physical system is orientable and endowed with a positive volume form
$$\Omega~=~\rho(x)\mathrm{d}x^1\wedge\ldots\wedge\mathrm{d}x^n, \qquad \rho(x)~>~0,$$
we can define a sesquilinear form
$$\langle \phi | \psi\rangle~:=~\int_M \! \Omega ~\phi^{\ast} \psi. $$
The Schroedinger representation of the self-adjoint phase space operators is
$$ \hat{x}^j~=~x^j,\qquad \hat{p}_k~=~\frac{\hbar}{i\sqrt{\rho(x)}} \frac{\partial}{\partial x^k} \sqrt{\rho(x)}, \qquad [\hat{x}^j,\hat{p}_k]~=~i\hbar\delta^j_k \mathbb{1}. $$
See also this related Phys.SE post.
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If I use the Earth as my point of reference on a model, would it be valid if I said that the Sun orbits the Earth? (in my model) What the title says.
If I create a model, and I use the Earth as my point of reference, is it valid to say that the Sun orbits the Earth inside my model? Or is claiming that invalid?
| You are talking about the geocentric model and that model has very complicated orbits of planets around the earth. Here have a look at the orbits of planets around earth:
So why not just use the simpler one in which all planets revolve around the barycenter of the solar system.
Also a point to note is that a model with earth at the center of the solar system is not inertial so Newton's laws of motion wouldn't work without a centrifugal force which only adds to the complexity of the case.
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How many $pep$-related electron-neutrinos $\nu_e$ does the Sun produce? In the sun neutrinos can be produced by the rare proton-electron-proton (pep) reaction:
$$ {}^1_1 H + e + {}^1_1 H \to {}^2_1 H + \nu_e $$
How many pep-related electron neutrinos does the Sun produce?
| The solar pep neutrinos were first detected in 2011 by Borexino collaboration (arXiv:1110.3230). The abstract reads:
We observed, for the first time, solar neutrinos in the 1.0-1.5 MeV energy range. We measured the rate of pep solar neutrino interactions in Borexino to be [3.1$\pm$0.6(stat)$\pm$0.3(syst)] counts/(day x 100 ton) and provided a constraint on the CNO solar neutrino interaction rate of < 7.9 counts/(day x 100 ton) (95% C.L.). The absence of the solar neutrino signal is disfavored at 99.97% C.L., while the absence of the pep signal is disfavored at 98% C.L. This unprecedented sensitivity was achieved by adopting novel data analysis techniques for the rejection of cosmogenic $^{11}$C, the dominant background in the 1-2 MeV region. Assuming the MSW-LMA solution to solar neutrino oscillations, these values correspond to solar neutrino fluxes of [1.6$\pm$0.3]$\times 10^8$ cm$^{-2}$s$^{-1}$ and 7.7$\times10^8$ cm$^{-2}$s$^{-1}$ (95% C.L.), respectively, in agreement with the Standard Solar Model. These results represent the first measurement of the pep neutrino flux and the strongest constraint of the CNO solar neutrino flux to date.
Since you wish to know the rate of pep neutrinos produced, we just need to multiply this flux with area of a sphere with a radius of 1 AU (Earth-Sun distance):
$$
\frac{dN}{dt} = \phi A = 1.6\times 10^{12}\: \mathrm{m}^{-2}\mathrm{s}^{-1} \times 4\pi\times (1.5 \times 10^{11} \:\text{m})^2 \approx \underline{4.5 \times 10^{35} \: \mathrm{s}^{-1}}
$$
So even though it took very long for us to detect them, the number of pep neutrinos created is quite large indeed!
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Due to incompressibility of water same volume of water should flow through pipe of different cross section . is that really true This seems true when I slightly compress the opening of pipe the speed of water flowing through pipe increases and same amount of water flows out but as I further decrease the opening of pipe. Instead of increasing speed of water to maintain same volume flow rate.the amount of water coming out of it decreases and same volume rate stops. Why is it ?
| From the law of mass conservation $$\partial_t \rho + \vec \nabla (\rho \vec v) = 0 $$ follows in the case of a static, incompressible fluid $$\vec \nabla \cdot \vec v = 0,$$ implies, when integrated over a volume $dV$ $$\int dV \; \vec \nabla \cdot \vec v = 0,$$ which, by using the Gaussian theorem we can transform from volume, to surface integral $$\int dV \; \vec \nabla \cdot \vec v = \oint d\vec S \cdot \vec v,$$
and then by taking a flow in a pipe with decreasing diameter, choosing a box with 6 sides as surface around the pipe for $dS$, we get $$ 0=\oint d\vec S \cdot \vec v = \sum_{i=1}^6 \Delta S_i\;\, \vec n_i \cdot \vec v$$, where $n_i$ is the normal vector on each side. Now choose the box such that 4 sides are outside the pipe, those vanish trivially, and what remains is $$ 0 = S_1 v_1- S_2 v_2, \;\;\;\;\;\;\; (1)$$ where the minus comes from taking the correct scalar product with the surface normals.
(1) is now true for any portion of the pipe! So if you know $S_1$, $S_2$ and $v_1$, then you can always determine $v_2$.
Now coming finally to your question, what we ignore in this computation is the viscosity, which wold turn up in the Navier-Stokes equation. Viscosity will couple the velocity with the boundary of your pipe, which can stall the flow in the whole pipe, but still at any time (1) will hold.
It is just that when $v_1$ decreases due to effects of viscosity, then $v_2$ has to decrease accordingly.
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Is there an approximation for the Lorentz factor for very large velocities? I am aware of the approximation generally used for low speeds to calculate the Lorentz factor, that being,
$$\gamma \approx 1 + \frac{1}{2} \left(\frac{v}{c} \right)^2$$
But I need the exact opposite thing -- is there any suitable approximation for when v is extremely close to c?
| This is roughly the simplest you can get it:
$$\gamma = \frac{1}{\sqrt{1-v^2/c^2}} = \frac{1}{\sqrt{1-v/c} \sqrt{1+v/c}} \approx \frac{1}{\sqrt{2}} \frac{1}{\sqrt{1-v/c}}.$$
In other words, if $\Delta v$ is how far the speed is below $c$, then
$$\gamma \approx \sqrt{\frac{c}{2 \Delta v}}.$$
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How is a 25-year-old can of soda now empty without having been opened or poked? I just discovered in my parents' basement a Sprite can from 1995* and also a Coca-Cola can probably from the same year.
Both cans are unopened and have no visible damage or holes. The Coca-Cola can feels "normal", but the Sprite can is empty!
You can hear in my video that there is no liquid sloshing around in it.
I can't find any way that the soda could have escaped.
We also don't see any mess near where the cans were, but I don't know for certain that the cans stayed in the same place for 25 years, so maybe there could have been a mess of liquid leaked out somewhere else if the cans had been stored somewhere else earlier. But nothing feels sticky or looks like there has been a leak of any kind.
What are possible explanations for how a carbonated drink could disappear from a sealed aluminum can?
*I had kept it as a "collector's item" from when the Houston Rockets won their second championship.
| This also happened to me with an unopened 7up can from 1993 which I had stored in my collection. When the can was about 6 years old I found it looking intact but feeling completely empty; underneath and around it was a small puddle of sticky clear goo, hinting that its contents had very slowly leaked through a tiny, indiscernible hole somewhere along the can's bottom. The water probably evaporated during the process leaving just the syrup. It's very likely that the same happened to you, except that in your case the can was untouched for 25 years - surely enough time for the members of some lucky ant colony to loot the fine tasting sugary goo and disappear without a trace.
| {
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Age of Universe from Friedmann Equation - How to actually solve the integral? The Friedmann equation for a flat universe can be written as
$$
H(t)=\frac{\dot{a}}{a}=H_0\sqrt{\Omega_{m,0}\cdot a^{-3}+\Omega_{\Lambda,0}}=H(a)
$$
To calculate the age of the universe, many books jump directly to the result. But there should be some sort of integral in between. I assume one can do the following:
$$
t=\int_0^{t_0}\!\mathrm{d}t=\int_0^{1}\!\mathrm{d}a\frac{\mathrm{d}t}{\mathrm{d}a}=\int_0^{1}\!\frac{\mathrm{d}a}{\dot{a}}
$$
with $\dot{a}$ from above expressin for $H$.
But how is this integral solved? Mathematica did something for hours but did not came up with a result. Most books and wikipedia pages skip directly to the result
$$
t_0=\frac{1}{3H_0\sqrt{\Omega_\Lambda}}\log{\frac{1+\sqrt{\Omega_\Lambda}}{1-\sqrt{\Omega_\Lambda}}}
$$
which leads to the well known result of ~13 billion years (depending on the DM density).
Again: But how is the integral solved?
| for flat universe ($\Omega_m + \Omega_{\Lambda}=1)$
$$H^2 = H_0^2(\Omega_ma^{-3}+\Omega_{\Lambda})$$
or $$\frac{\dot{a}^2}{a^2} = H_0(\Omega_ma^{-3}+\Omega_{\Lambda})$$
which becomes
$$\dot{a}^2 = H_0(\Omega_ma^{-1}+\Omega_{\Lambda}a^2)$$
taking square root
$$\dot{a} = H_0\sqrt{\Omega_ma^{-1}+\Omega_{\Lambda}a^2}$$
let us write in the form of
$$\frac{da}{dt} = H_0\sqrt{\frac{\Omega_m}{a}+\Omega_{\Lambda}a^2}$$
$$\frac{da}{\sqrt{\frac{\Omega_m}{a}+\Omega_{\Lambda}a^2}} = H_0dt$$
$$\frac{da}{\sqrt{\Omega_{\Lambda}}\sqrt{(\frac{\Omega_m}{\Omega_{\Lambda}})(\frac{1}{a}) + a^2}}= H_0dt$$
set $$\frac{\Omega_m}{\Omega_{\Lambda}} = L = 0.44927$$
(I took $\Omega_{\Lambda} = 0.69$, $\Omega_m=0.31$)
By taking $a(t_{now})=1$
$$\int_0^{a(t_{now})=1}\frac{da}{\sqrt{\frac{L}{a}+a^2}} = \int_0^{t_{universe}} \sqrt{\Omega_{\Lambda}}H_0dt$$
by using wolfram the solution becomes,
$$\left.\frac{2}{3}log(a^{3/2} + \sqrt{a^2+L})\right|_0^1 = \sqrt{\Omega_{\Lambda}}H_0t_{uni}$$
or you can use https://www.integral-calculator.com to make a numerical calculation of the integral. In any case we have
$$\int_0^1\frac{da}{\sqrt{\frac{L}{a}+a^2}} = 0.793513$$
Hence,
$$t_{uni} = \frac{0.793513}{0.83066 \times H_0}$$
For $H_0 = 68km/s/Mpc$
$$t_{uni} = 0.9552801 \times H_0^{-1}= 0.9552801 \times 14.39~\text{Gyr} = 13.74 ~\text{Gyr}$$
| {
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Generator of translations in classical mechanics The generator of spatial translations is momentum. In quantum mechanics this makes a lot of sense to me and so we can write the translation operator like this:
$$\hat{T}(\Delta \textbf{r}) = e^{\frac{i}{\hbar}\Delta \textbf{r} \cdot \hat{p}}.$$ Is there an equivalent expression (writing the translation operator as an exponential using the generator) for classical mechanics?
| Considering a translated function:
$$f(x)\rightarrow f(x+\Delta x)$$
We can expand in Taylor series around $x$ and write
$$f(x+\Delta x)=\sum_{n=0}^\infty\frac1{n!}\left. \frac{\partial^n f}{\partial x^n}\right|_{x}\Delta x^n\equiv\sum_{n=0}^\infty\frac1{n!}\left(\Delta x\left. \frac{\partial}{\partial x}\right|_{x}\right)^nf(x)$$
Which formally could be understood as
$$e^{\Delta x \left.\frac{\partial}{\partial x}\right|_x} f(x)$$
From the correspondence $p\rightarrow -i\hbar\frac{\partial}{\partial x}$ you should now see the connection
| {
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Is there a thermodynamic process with constant internal energy but variable enthalpy? Lately I've been trying to wrap my head around those two concepts and the difference between them.
In an ideal gas, $PV$ is said to be an energy related to the kinetic energy of molecules boucing against the boundaries of the system. However, the internal energy $U$ being defined as "the sum of all potential and kinetic energies of each molecule" seems to indicate $PV$ is already part of $U$.
As of my current understanding, I get that $PV$ is counted twice in $H$ since $H=U+PV$ (once in the $U$ as explained above and the remaining $PV$ term). I don't think this makes a lot of sense and is most likely wrong, but I cannot think of a different understanding for $PV$.
If there is a process in whcich $H$ changes but not $U$, I might finally be able to grasp those concepts once and for all.
| $H=U+pV$, so a change in enthalpy will amount to: $dH=dU+VdP+P dV$.
If you want $dU=0$ you can impose $dT=0$. Using the equation of state for an ideal gas we get: $V dP+P dV=nRdT$, and because you want $dT=0$ in your process you have $VdP=-PdV$.
Replacing that in $dH$ you get:
$$dH=0$$ Thus you cannot have a process for an ideal gas in which the internal energy does not change but enthalpy does. But for equations of state other tha an ideal gas it must certainly be possible.
For instance, if you have $P(V-b)=nRT$, you get
$$dH=b dP$$ and $dU=0$
| {
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Why doesn't Rayleigh scattering happen at low altitude in air? An answer I found online was: At high altitude where molecules are far apart, scattered photons can travel without interfering with each other, thus they fill the sky with blue light. But at low altitude, molecules are so close to each other that scattered photons destructively interfere with each other to cancel each other out, that's why we see air as transparent.
However, if they perfectly cancel out, then wouldn't the energy be not conserved?
Another explanation was simply that the effect is only apparent in a very large column of air since the scattering effect is not that noticeable.
So why doesn't Rayleigh scattering happen at low altitude in air?
| There is no problem with energy, and an explanation in terms of waves is more intuitive. How else could one easily understand a wavelength dependence?
First there is dispersion, the increase of the index of refraction at short wavelength, which can be understood with the model of Lorentz oscillators.
In a gas there are fluctuations in the number density on short length scales. That means that the index of refraction of cubes with a size of $\lambda^3$ will fluctuate. This gives more scattering at shorter wavelengths, together the $\lambda^{-4}$ dependence.
I am not sure about the form of the pressure dependence, but I would agree with @planetmaker that your assumption is wrong. Fluctuations are relatively larger at low densities, but the lower the density the less scattering there will be. In the limiting case of the vacuum there is no blue sky.
| {
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When the star formation begins? We can separate the history of the universe in different epochs.
Radiation dominated epoch, matter dominated epoch, and dark energy dominated epoch, and we can divide the epochs in different ways.
For example the radiation dominated epoch can be divided in the epoch before atoms could form and the epoch after the first atoms form (nucleosyntesis) my question is, when the first star form?
What is the aproximated redshift where the stars star to formed?
(I think that is a really high redshift because the observations of the star HD 140283)
| This is the history according to the mainstream model, the Big Bang
After the decoupling of photons at 380.000 years, there is a period marked as "neutral hydrogen forms" that carefully is not given a time to the line of the next stage, which is modern universe. In other plots of timeline for the Big Bang the line is drawn at $10^{{17}}$second
Once matter is mainly neutral, gravity is the main force remaining , and is stronger than the gradual, even if accelerated, expansion. Masses will gravitate statistically , (the hypothesis is), toward the denser regions seen in the cosmic microwave background CMB , thus starting to form the kernels of stars galaxies and clusters of galaxies.
As it is a statistical phenomenon, dependent on the density, which was created with quantum fluctuations during the inflation period, one cannot think that there exists a specific line that separates star formation from hydrogen etc soup in the last interval. There is a probability that stars formed in certain regions very soon after hydrogen became neutral.
Of course all this within the model,which is at present mainstream.
edit after discussion:
The process from the time of neutral hydrogen to the present observable universe seems to be under research:
"Exactly how stars and galaxies formed, when the process started and how long it took is currently a major area of research. A simple picture runs like this: about 1 billion years after the big bang the first star forming regions, conglomerates of perhaps 106 to 109 solar masses began to develop. Over the next several billion years, most of these merge to form larger units or are partially destroyed by the energetic supernovae which develop as a natural part of star formation. Within a few billion years most of these have developed into stable configurations of stars and gas and are recognisable as ``galaxies''.
| {
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Does Another Universe Where Time Runs Backwards Revive the Idea of a Cyclical Universe? https://www.pbs.org/wgbh/nova/article/big-bang-may-created-mirror-universe-time-runs-backwards/
This article talks about a new theory where the big bang happened in one direction for us and the other direction where time is going backwards (mathematically). Would that revive the idea of a cyclical universe again? I had this little vision in my head when I read this. A universe chain with a big bang on essentially on either side of the start/end of time and in the next big bang our universe is the one going backward in time and the new one is going forward.
| The problem with mirrors in this situation is the question of energy.
What one person might be doing in a situation might be the mirror image of what another is doing, but is the energy the same? In the situation of a mirror universe colliding with our universe, in theory if you put a mirror in the middle, it should be the same on both sides, but once you cross the line, the rules of energy flows are different. Okay, I've got a degree in philosophy, but one of my university mathematics teachers was the guy who figured out the Black Hole equations, and I remember him saying, "heat is a flow" which means it's dependent on time, so in a collision between universes following different rules of time, which way would the heat flow?
And the amount of energy resulting from a collision between just one star of positive matter and another of antimatter would be so huge that nobody really knows what would happen.
But I think it's an area that should be researched.
| {
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Clarification of the concept "less resistance means less heating" in a wire So my textbook says that the reason cables that are suppose to carry high currents, are thicker that those that are meant to carry lesser current, is that "less resistance (of the wire) means less heating..."? Is this even true?
Isn't CURRENT the reason wires heat up? If we decrease resistance, more current flows, and that should produce more heating!
| Not necessarily. This is because it is not always the current rather a coupling of both which causes the heating effect to occur. What this means is that for wires carrying high currents, lower resistance leads to less dissipated power, according to P=I^2*R, where I is the current in the wire, and R is the resistance. This means that wires which are thicker, which have lower resistance than thinner wires provided other factors remain constant, will not have as much heat produced, since lesser power is dissipated. Therefore, it need not always hold true that higher current leads to more heat being generated.
| {
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Post-measurement $\psi$ in quantum mechanics I have a question regarding the wave function after a measurement. Everything I found online says that this is the following formula:
$\psi = \frac{M_m\psi}{\sqrt{P(m)}}$
Where $P(m)$ is the probability of observing m, the $\psi$ on the left is the wavefunction AFTER the measurement and the $\psi$ on the right is the original wavefunction. However, I cannot find a good definition on how I would go about calculating $M_m$? The Berkley lecture notes say that this is the measurement operator, but how would I go about finding that for my specific problem?
Also the probability function is $P(m) = |<\psi|\omega>|^2$, how would I find $\omega$ in this case? Is it just the eigenstate at that observable?
| You have to think about what a measurement is in QM. Generally speaking, you are going to measure an eigenvalue of an observable. Lets say that you measure $A$ that follows the eigenstate equation $A|a>=a|a>$. Then if you begin with a general state $|\psi>$, after the measurement you will get another state $|\psi'>$ following the equation:
$$|\psi'>=\frac{|a><a|\psi>}{\sqrt{P(a)}}$$
Where $|a><a|$ is called the projector (or measurement operator by your lectures). It basically gives you the part of $|\psi>$ that is in the subspace of eigenstates of eigenvalue $a$.
Physically it means that after a measurement is done, you can be sure that the state of the system is in a subspace with eigenvalue $a$, so that any other measurement of the obsevable $A$ that you make on the system afterwards will spit out the eigenvalue $a$ all the time (if you keep the system unchanged).
So you if you want to calculate $M_m$, which is the projector into the subspace of some eigenvalue, just find the eigenstates of that eigenvalue (lets say $A|a_n>=a|a_n>$ for $n=0,1,...,g$). Then.
$$M_m=\sum_{n=0}^{g}|a_n><a_n|$$
| {
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What happens to oxygen during atmospheric absorption of radiation with a peak at 60 GHz This question discusses atmospheric absorption of radiation (specifically 5G radio waves) due to oxygen at 60 GHz.
I would like to know exactly what happens to oxygen in all its forms (be it air or ozone) when it absorbs radiation at 60GHz.
I'm aware of different types of absorption but not sure what it means for oxygen. Also, the experiments cited in these types of Wikipedia articles seem to look at oxygen atoms or molecules in controlled environments, but I can't quite extract the lay explanations of what happens and how it applies to the air we breathe.
| The oxygen molecule is unusual in having a ground state that has unpaired electrons, and this means it has a net spin of one. Technically the ground state is a triplet. This is unusual because ground states normally have spin zero i.e. they are singlets. For example the $\mathrm{N}_2$ molecule has a singlet ground state.
Anyhow, the non-zero spin of the oxygen molecule causes the ground state to split into a couple of closely separated levels. This type of splitting is known as fine structure. The spacing between these levels is 0.246 meV, which corresponds to a frequency of 59.4 GHz i.e. microwaves with a frequency of 59.4 Ghz have just the right amount of energy to cause transitions between the fine structure levels.
And this is what happens when oxygen absorbs microwaves at 60 GHz. The absorption spectrum is actually very complex because the oxygen molecule changes its rotation as well as jumping between the fine structure levels, so we get a complicated cluster of lines centred around 60 GHz. At atmospheric pressure the lines are broadened by collisions between oxygen molecules and we see only the characteristic broad absorption hump at 60 GHz. To resolve the many lines within this hump requires the absorption to be measured at very low pressures where collisions between the molecules are infrequent.
| {
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Gravity, matter vs antimatter I have a simple question regarding matter-antimatter gravity interaction.
Consider the following though experiment:
If we imagine a mass $m$ and an antimass $m^-$, revolving around a large mass $M$
the potential energy of mass $m$ should be:
$$ U_1=-\frac{GmM}{R} $$
and the potential energy of mass $m^-$ should be:
$$ U_2=-\frac{GmM}{R} $$
or:
$$ U_2=\frac{GmM}{R} $$
depending on the sign of the gravity interaction between matter and antimatter.
If the two particles annihilate to energy, then the gravitational field of $M$ will interact with the emitted photons and will change their frequency.
But, as the interaction between gravity and the photons has nothing to do with the question of the gravity between matter and antimatter, can't we simply use the interaction between gravity and photons, and the energy conservation to establish the nature of the gravity interaction between matter and antimatter?
| This is a perfectly good argument and one of the reasons that all the physicists I know believe that antimatter behaves just like matter in a gravitational field.
It is important to distinguish between antimatter, which is well understood from countless collider experiments, and negative matter (also known as exotic matter), which has never been observed. Antimatter does not have a negative mass. Indeed antimatter is just perfectly ordinary matter - we think it special only because we are made from matter and therefore biased. Negative/exotic matter is very different. If it existed it would cause all sorts of problems with conservation of energy and the stability of the universe.
| {
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The Enigma of Universal Gravitation Forces This is taken from a book called "Physical Paradoxes and Sophisms" by V. N. Lange.
1.22. The Enigma of Universal Gravitation Forces
The law of gravitation can be written $F=\gamma\frac{m_1m_2}{R^2}$.
By analyzing this relationship we can easily arrive at some interesting conclusions: as the distance between the bodies tends to zero the force of their mutual attraction must rise without limit to infinity.Why then can we lift up, without much effort, one body from the surface of another body (e.g., a stone from the Earth) or stand up after sitting on a chair?
| Because in general gravitational attraction is much weaker than electromagnetic repulsion.
For example when you compare gravitational and electrostatic forces between two electrons, both forces are proportional to $R^{-2}$, but if you compare their actual values, you get that gravitation is around $10^{37}$ times weaker.
| {
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What does this notation for spin mean? $\mathbf{\frac 1 2}\otimes\mathbf{\frac 1 2}=\mathbf{1}\oplus\mathbf 0$ In my quantum mechanics courses I have come across this notation many times:
$$\mathbf{\frac 1 2}\otimes\mathbf{\frac 1 2}=\mathbf{1}\oplus\mathbf 0$$
but I feel like I've never fully understood what this notation actually means. I know that it represents the fact that you can combine two spin 1/2 as either a spin 1 (triplet) or a spin 0 (singlet). This way they are eigenvectors of the total spin operator $(\vec S_1+\vec S_2)^2.$ I also know what the tensor product (Kronecker product) and direct sum do numerically, but what does this notation actually represent?
Does the 1/2 refer to the states? Or to the subspaces? Subspaces of what exactly (I've also heard subspaces many times but likewise do not fully understand it). Is the equal sign exact or is it up to some transformation?
And finally is there some (iterative) way to write a product of many of these spin 1/2's as a direct sum?
$$\mathbf{\frac 1 2}\otimes\mathbf{\frac 1 2}\otimes\mathbf{\frac 1 2}\otimes\dots=\left(\mathbf{1}\oplus\mathbf 0\right)\otimes\mathbf{\frac 1 2}\dots=\dots$$
| I should concur with the other answers that there is no substitue for reading up in good texts and WP.
You are right that, in a given basis, there is a similarity (equivalence) transformation implied in the equation of your title: it basically means that the tensor product on the l.h.s. is reducible, by an orthogonal basis change to the r.h.s.; that is, in words,
*
*The Kronecker product of two 2-vectors (spinors; in general you have (2s+1)-dim vectors!) is a 4-vector. But rotations keep two subspaces in it separate: a 3-vector subspace, and a 1-vector (scalar) subspace. However, this is invisible to the naked eye. There is an orthogonal basis change, the Clebsch matrix, which visibly separates these two subspaces, so rotations act on these visibly separately, by block matrix action. (In your singlet case, by no action at all! the rotation matrices are the identity, 1).
Can you find this 4×4 Clebsch matrix $\cal C$ in Problem 4 here for your exact problem? (Hint: mix up just the 2nd and 3rd components by a rotation of $\pi/4$.) The "right is detail-in-left" convention in the tensor product amounts to
$$
\begin{pmatrix} a_1\\a_2\end{pmatrix} \otimes \begin{pmatrix} b_1\\b_2\end{pmatrix} = \begin{pmatrix} a_1 b_1\\a_1 b_2 \\ a_2 b_1\\ a_2 b_2\end{pmatrix} \leadsto \begin{pmatrix} \uparrow \uparrow\\ \uparrow \downarrow \\ \downarrow \uparrow\\ \downarrow \downarrow \end{pmatrix} ,
$$
in the spherical basis notation. The second and 3rd component, then, mix up into $(\frac{\uparrow \downarrow + \downarrow\uparrow }{\sqrt{2}},\frac{\uparrow \downarrow - \downarrow\uparrow }{\sqrt{2}} )$, the triplet component and the singlet component.
The upshot is a direct sum of a 3-vector (components 1,2, & 4) and a singlet (component 3):
$$
\begin{pmatrix} \uparrow \uparrow\\ \frac{\uparrow \downarrow + \downarrow\uparrow }{\sqrt{2} } \\ \downarrow \downarrow \end{pmatrix} \oplus \frac{\uparrow \downarrow - \downarrow\uparrow }{\sqrt{2}} .
$$
Your title formula, however, never picks a basis.
Finally, there are elaborate formulas for recursive compositions of spins, pioneered by Bethe and elaborated by several authors afterwards. Your case is particularly simple, as WP details. I copy the WP formula, which uses dimensionality, instead of spin indices (2s+1 instead of your s), since you can do instant arithmetic checks by ignoring the circles in × and + !
Combining n doublets (your spin 1/2s) nets you
$$
{\mathbf 2}^{\otimes n} = \bigoplus_{k=0}^{\lfloor n/2 \rfloor}~
\Bigl( {n+1-2k \over n+1} {n+1 \choose k}\Bigr)~~({\mathbf n}+{\mathbf 1}-{\mathbf 2}{\mathbf k})~,$$
where $\lfloor n/2 \rfloor$ is the integer floor function;
the number preceding the
boldface irreducible representation dimensionality (2 s+1) label indicates multiplicity of that
representation in the representation reduction. The random walk that takes you there reconstructs the celebrated Catalan's triangle.
For instance, from this formula, addition of three spin 1/2 s yields a spin 3/2 and two spin 1/2s, ${\mathbf 2}\otimes{\mathbf 2}\otimes{\mathbf 2}={\mathbf 4}
\oplus{\mathbf 2}\oplus{\mathbf 2} $; four spin 1/2 s yields two singlets, three spin 1 s, and one spin 2, and so forth.
| {
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Quantization of complex scalar field I'm learning Peskin's qft now and I'm a little confused about problem 2.2 .
Suppose I write the field $\phi(x)$ as:
$\phi(x) =\int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_{p}}} (a_{p}e^{-ipx}+b_{p}e^{ipx})$
I know that $b_p$ should be written as $b_p^\dagger$ because it annihilate antiparticle, otherwise $b_p$ creates particle with negative energy.
However, when I calculate the Hamiltonian, all I got is:
$\int \frac{d^3p}{(2\pi)^3}E_{p}(a^\dagger_{p}a_{p}+b^\dagger_{p}b_{p})$
In my result, the $b$ particles create positive energy as $a$ particle did. I'm not sure if I did something wrong in calculation or there are some other explanation in the result.
| To begin with, Peskin gives the following action:
$$
\mathcal{S} = \int d^4 \left[ \partial_{\mu} \phi^{\ast} \partial^{\mu} \phi - m^2 \phi^{\ast} \phi \right].
$$
Let's begin by considering the classical field theory. The equation of motion is
$$
\left( \partial^2 + m^2 \right) \phi = 0
$$
(and its complex conjugate). A general solution to this equation is given by
$$
\phi(\mathbf{x},t) = \int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 E_\mathbf{p}}} \left( a_{\mathbf{p}} e^{i \mathbf{p} \cdot \mathbf{x} + i E_{\mathbf{p}} t} + b_{\mathbf{p}} e^{i \mathbf{p} \cdot \mathbf{x} - i E_{\mathbf{p}} t} \right),
$$
where $E_{\mathbf{p}} = \sqrt{|\mathbf{p}|^2 + m^2}$. At this point, the quantities $a_{\mathbf{p}}$ and $b_{\mathbf{p}}$ are just some complex functions of momentum. We can also write down the classical Hamiltonian of this system, where the conjugate momentum is
$$
\pi(\mathbf{x},t) = \partial_t \phi^{\ast}(x,t) = - i \int \frac{d^3 p}{(2 \pi)^3} \sqrt{\frac{E_\mathbf{p}}{2}} \left( a_{\mathbf{p}}^{\ast} e^{-i \mathbf{p} \cdot \mathbf{x} - i E_{\mathbf{p}} t} - b_{\mathbf{p}}^{\ast} e^{-i \mathbf{p} \cdot \mathbf{x} + i E_{\mathbf{p}} t} \right)
$$
(and its complex conjugate).
Now we go to the quantum theory, where the fields $\phi$ and $\pi$ (and therefore the coefficients $a_{\mathbf{p}}$ and $b_{\mathbf{p}}$) become operators. The starting point of quantization is the equal-time canonical commutation relations:
$$
[\phi(\mathbf{x},t),\pi(\mathbf{y},t)] = [\phi^{\ast}(\mathbf{x},t),\pi^{\ast}(\mathbf{y},t)] = i \delta^3(\mathbf{x} - \mathbf{y})
$$
We cannot assume at this point that $a_{\mathbf{p}}$ and $b_{\mathbf{p}}$ are bosonic annihilation operators, we need to actually calculate their commutation relations from the above equations. Inverting the above formulae, we have
$$
a_{\mathbf{p}} = e^{- i E_{\mathbf{p}} t} \int d^3 x \, e^{- i \mathbf{p} \cdot \mathbf{x}} \left( \sqrt{\frac{E_{\mathbf{p}}}{2}} \, \phi(\mathbf{x},t) - i \frac{1}{\sqrt{2E_{\mathbf{p}}}} \, \pi^{\ast}(\mathbf{x},t) \right),
$$
$$
b_{\mathbf{p}} = e^{- i E_{\mathbf{p}} t} \int d^3 x \, e^{- i \mathbf{p} \cdot \mathbf{x}} \left( \sqrt{\frac{E_{\mathbf{p}}}{2}} \, \phi(\mathbf{x},t) + i \frac{1}{\sqrt{2E_{\mathbf{p}}}} \, \pi^{\ast}(\mathbf{x},t) \right).
$$
Now we can compute the commutators directly. They are
$$
[a_{\mathbf{p}},a^{\dagger}_{\mathbf{p}'}] = - (2 \pi)^3 \delta^3(\mathbf{p} + \mathbf{p}'), \qquad [b_{\mathbf{p}},b^{\dagger}_{\mathbf{p}'}] = (2 \pi)^3 \delta^3(\mathbf{p} + \mathbf{p}').
$$
As you can see, one of these commutators is the wrong sign compared to the usual bosonic commutation relations, and as a result the construction of the usual bosonic Fock space proceeds differently. (In particular, $[a^{\dagger} a, a^{\dagger}] \propto - a^{\dagger} $ so $a^{\dagger}$ is actually a lowering operator.) To get the usual commutation relations, we should define
$$
\tilde{a}_{\mathbf{p}} = a^{\dagger}_{\mathbf{p}}, \quad \tilde{a}^{\dagger}_{\mathbf{p}} = a_{-\mathbf{p}}, \quad \tilde{b}_{\mathbf{p}} = b_{\mathbf{p}}, \quad \tilde{b}^{\dagger}_{\mathbf{p}} = b^{\dagger}_{-\mathbf{p}}.
$$
This gives us the usual result. In textbooks they often write down the initial expansion in such a way so that the answer comes out correctly, but in practice one needs to go through the above procedure (until you've done it enough times that it has become automatic).
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Where does energy go in joining capacitors of different capacitance charged by different potential, hypothetically assuming no resistance in circuits? I don't understand why there is any change in initial and final energy since we have already assumed a perfectly conductive circuit. I mean, theoretically at least, there should be no change in energy.
Now, considering there is a change in energy at all, is it because electrons accelerate in moving from one capacitor to another, so energy gets dissipated in the form of electromagnetic radiation?
Image source: NCERT Physics Textbook for Class XII Part I, page 82
| Yes, even assuming a perfectly conductive circuit, there would be some inductance (you may not have studied this yet) which would create an oscillating LC circuit, which would radiate. Some early spark gap radio transmitters worked like this.
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Geometry of Young's experiment for optical path length I am currently studying the textbook Modern Optical Engineering, fourth edition, by Warren Smith. When presenting the concept of optical path length, the author says the following:
With reference to Fig. 1.13, it can be seen that, to a first approximation, the path difference between $AP$ and $BP$, which we shall represent by $\Delta$, is given by
$$\Delta = \dfrac{AB \cdot OP}{D}$$
I'm having difficulty understanding how the mathematics $\Delta = \dfrac{AB \cdot OP}{D}$ corresponds to the figure. I suspect that there is some use of trigonometry and/or geometry that I am not seeing. I would greatly appreciate it if someone would please take the time to explain this to me.
| The assumption is that AB is very small compared with D. Then the two triangles are similar and the sine of the small angle is OP/D = Δ/AB
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Is there work done on a mass from matter-antimatter pairs? Imagine I have a vaccuum at very low temperature and I put a single neutron in, then gamma rays interact to form matter-antimatter pairs within this vaccuum and assume that this happens extremely close to the aforementioned neutron. Since the matter-antimatter pair has mass, there would be a small amount of curvature in spacetime. Would the neutron follow the geodesic created? If so, this means that there was work applied onto the neutron from no initial force. Which would mean a violation in the conservation of energy/force. What is the flaw in this situation?
If the mass of the neutron >>> mass of electron/positron is the issue, then replace the neutron with an electron. Here there would also be either an attractive or repelling electromagnetic force depending on orientation, which still leads to my point.
I am still in high school, so I have a very limited understanding of these topics, sorry for my naivety. But so far the teachers I have asked have not given me a satisfactory response yet.
| It is not possible to have the topology you describe. In order for a gamma ray to create a particle antiparticle pair, it needs the electromagnetic field of a nucleus for energy and momentum conservation to work in the center of mass of the e+e- pair.
In this feynman diagram Z is the nucleus in whose electromagnetic field the pair production can happen.
So you cannot have a neutron and a particle pair only.
As for gravitational interactions, yes, as long as there is an energy momentum tensor the solutions of the General Relativity are followed by all particles . Zero mass particles follow the geodisics, massive ones the solutions of the particular boundary condition problem.
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How do we know not all photons are absorbed? Only those of specific energies? When a photon hits an electron in an atom, its energy has to be equal to the difference in energy between the current shell and a shell with a higher energy level, otherwise it is not absorbed at all.
How do we know not all photons are absorbed? Wouldn't at least some energy of the photon be absorbed since it is an oscillation in the EM field?
|
How do we know not all photons are absorbed?
You can shine monochromatic light on a sample of a material and see whether it is absorbed or not.
For example if you shine different wavelengths of light through a chamber containing a pure gas, and you'll see that only wavelengths that match transitions of that gas molecule are absorbed. Wavelengths that are either too long or too short are not absorbed.
Wouldn't at least some energy of the photon be absorbed since it is an oscillation in the EM field?
It's not possible to absorb "some" of the energy of a photon. The whole concept of a photon is that it is the quantum of energy in the EM field. That means that when something exchanges energy with the EM field, it does so in units of whole photon energies, not fractions of the photon energy.
| {
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Conservation of crystal momentum I am trying to convince myself that crystal momentum is conserved in a periodic lattice modulo a reciprocal lattice vector.
Consider a Hamiltonian $H$ which is periodic under translations of a Bravais lattice vector. The canonical momentum operator $\mathbf{P} = (P_x,P_y,P_z)$ is the generator of translations, so I can write my translation operator as
$$ T(\mathbf{a}) = e^{i \mathbf{a} \cdot \mathbf{P}}, \quad \mathbf{a} \in \mathbb{R}^3.$$
However, for a periodic Hamiltonian, the full symmetry is broken down to translations within the Bravais lattice only. I would express this symmetry as $[ T(\mathbf{a}) , H] =0$ for any Bravais lattice vector $\mathbf{a}$. Now, substituting my translation operator into the commutator, I find
$$ \mathbf{a} \cdot[ \mathbf{P} , H] = 0$$
If my system had the full translation symmetry, I could factor out the $\mathbf{a}$ to conclude that each component of the momentum is conserved: $[P_i, H] = 0$. However, as we are restricted to the Bravais lattice, I can only conclude that $ \mathbf{a} \cdot \mathbf{P}$ is conserved and I would rename $\mathbf{P}$ as the crystal momentum.
I am unsure how I arrive at the fact that the crystal momentum is conserved modulo a reciprocal lattice vector. I imagine it has something to do with assuming I can bring down the exponent in the commutator. I can see why the exponential does not define momentum uniquely, however if I had full translational symmetry, I would be able to say the exponent is conserved. What is different here?
| There is no need to expand the exponential at all. Let the lattice have basis $\mathbf{a}_i$. The fact that
$$[e^{i \mathbf{a}_i \cdot \mathbf{P}}, H] = 0, \quad [e^{i \mathbf{a}_i \cdot \mathbf{P}}, e^{i \mathbf{a}_j \cdot \mathbf{P}}] = 0$$
indicates that we can simultaneously diagonalize the $e^{i \mathbf{a}_i \cdot \mathbf{P}}$ and $H$. Since $e^{i \mathbf{a}_i \cdot \mathbf{P}}$ is unitary, its eigenvalues are pure phases, so we may define
$$e^{i \mathbf{a}_i \cdot \mathbf{P}} |\psi \rangle = e^{i \phi_i} |\psi \rangle.$$
Now, because the $\mathbf{a}_i$ form a basis of $\mathbb{R}^3$, there exist vectors $\mathbf{k}$ so that
$$e^{i \phi_i} = e^{i \mathbf{a}_i \cdot \mathbf{k}}.$$
We can then call $\mathbf{k}$ the "crystal momentum". The reason that $\mathbf{k}$ is only defined up to multiples of reciprocal lattice vectors is because we have not specified $\mathbf{k}$ anywhere in this argument, only its exponential. Indeed, if we add a reciprocal lattice vector $\mathbf{b}_j$, then the phases change by $e^{i \mathbf{a}_i \cdot \mathbf{b}_j} = e^{2 \pi i \delta_{ij}} = 1$ by the definition of the reciprocal lattice.
For full translational symmetry, you can take $\mathbf{a}$ infinitesimal and Taylor expand the exponential, giving $[\mathbf{a} \cdot \mathbf{P}, H] = 0$, and then since $\mathbf{a}$ is arbitrary we have $[\mathbf{P}, H] = 0$. But for the lattice translations, expanding the exponential isn't really clean, and it's not necessary either.
| {
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Outside temp of vessel of is 40 Deg C and Water inside it is 86 Deg C.. Why? I was boiling water in a steel utensil.
Used Fluke's IR Gun (59 Max) to measure the temperature of Water inside AND the surface temp of vessel outside.
The boiling water read 86 Deg C whereas the outside surface temp of vessel read just 40 Deg C. Why is that? If I touched the outside of the vessel it would 'feel' extremely hot but measurement shows just 40 Deg C. So what's going on here?
| How these thermometers works is based on the principle of blackbody radiation and the Stefan-Boltzmann law. There is an IR sensor housed inside the thermometer that senses the total power radiated in a window of IR. And this power is related to the temperature of the emitter. But there is also a dependence on emissivity which is governed by the material properties of the emitter like the reflectivity and such. From the wiki page for Infrared Thermometer:
Most surfaces have high emissivity (over 0.9 for most biological surfaces)[citation needed], and most IR thermometers rely on this simplifying assumption; however, reflective surfaces have lower emissivity than non-reflective surfaces. Some sensors have an adjustable emissivity setting, which can be set to measure the temperature of reflective and non-reflective surfaces. A non-adjustable thermometer may be used to measure the temperature of a reflective surface by applying a non-reflective paint or tape, with some loss of accuracy.
So maybe you can modify the emissivity to the right value and see if you get the right temperature.
| {
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Why is the speed of light in vacuum a universal constant? While getting familiar with relativity, the second postulate has me stuck. "The speed of light is constant for all observers". why can't light slow down for an observer travelling the same direction as the light?
| It is the speed of causality. If there was no limit in how fast information can travel all would happen at once. The universe is a causal place because information needs time to travel a distance. Your question is probably much deeper than you intended it to be.
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Can the real-time Green's function be written in the form of path integral on the real axis? In every textbook, the path integral of the Green's function is written in imaginary-time. I wonder whether we could write real-time green function in the path integral form.
| PS: I notice that if we can distort the time integral line (to include imaginary part), then it is possible to establish the path integral formulation of real-time green function (at least for fermions). So I edited my question so that the time integral has to be defined on real axis.
—————
All right after discussing with professor, I’ll answer the question myself.
The crucial point is that the real-time green function is defined on zero temperature, so the contribution to the two point function will only come from the ground state, and the factor $e^{-\beta H}$ will be thrown away. As a result, the remaining time ordered operator will appear as some evolution operators not in a time ordered place. (Without considering time contour as in @Vadim answer or other more complicated situations) So it can not be written as a path integral.
| {
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Do reflection and diffraction occur at the same time?
When waves pass through a gap as the above, diffraction occurs.
As the upper and lower part of the plane wave (indicated by yellow colour) are striking the barriers, does reflection occur at the same time? Is this the cause of the decrease in amplitude when diffraction occurs?
| You can design the barriers such that wave hitting them is absorbed other than reflected (Key word: impedance matching). The reason why decrease in amplitude in the diffracted wave happens, is not because of reflection, but because of the portion of the wave that passed the opening now has to occupy wider wave front than before.
| {
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Decay of electron? Have we detected any decays of electrons to an electron neutrino and $W$-boson in Fermilab or in CERN? Are neutrinos the only possible stable leptons inside an electroweak field?
|
Have we detected any decays of electrons to an electron neutrino and W- boson in Fermilab or in CERN?Are neutrinos the only possible stable leptons inside an electroweak field?
This cannot happen because energy and charge conservation are absolute laws. Particles can decay to other particles when the sum of the masses of the decay products is smaller than the mass of the original particle, (and also various quantum number consrvations laws are obeyed). See the table
The electron is a charged lepton, leptons can decay to other leptons, but not the electron since the other two charged leptons are heavier.
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Why complex numbers are used in electronics? The impedance of a capacitor or an inductor is imaginary. How do we know these quantities are imaginary?
| An RLC circuit satisfies$$L\ddot{I}+R\dot{I}+C^{-1}I=\dot{V}.$$To solve this with AC voltage such as $V=V_0\cos\omega t,\,V_0\in\Bbb R$, it's convenient to take the real part of a complex choice of $I$ for the case $V=V_0\exp j\omega t$. Substituting $I=I_0\exp j\omega t,\,I_0\in\Bbb C$ gives$$I_0=\frac{j\omega V_0}{C^{-1}-\omega^2L+j\omega R}.$$The special case $C^{-1}=L=0$ gives $I_0=\frac{V_0}{R}$. The general case gives capacitance (inductance) an effective resistance of $\frac{C^{-1}}{j\omega}$ ($j\omega L$), so that $I_0=\frac{V_0}{Z}$ with a complex impedance $Z=R+j(\omega L-\frac{1}{\omega C})$. If $\omega^2LC\ne1$, the phase of $Z$ causes the oscillating expressions for $V,\,I$ to have a phase difference, i.e. in the real case constants $A,\,\phi$ exist with $I=A\cos(\omega t-\phi)$. Since $\omega^2LC\ne1\implies\phi\ne0$, any definition of resistance we give for an RLC circuit has to use complex numbers to represent the LC parts' phasing effects. This is why in general impedance is complex.
| {
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Why can vector components not be resolved by Laws of Vector Addition? A vector at any angle can be thought of as resultant of two vector components (namely sin and cos).
But a vector can also be thought of resultant or sum of two vectors following Triangle Law of Addition or Parallelogram Law of Addition, as a vector in reality could be the sum of two vectors which are NOT 90°.The only difference here will be that it is not necessary that components will be at right angle.
In other words why do we take components as perpendicular to each other and not any other angle (using Triangle Law and Parallelogram Law).
| This is mainly because the X and Y axes are also perpendicular to each other and therefore to measure quantities along these axes the vectors are resolved into X and Y components.
| {
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Does the effect of gravity get slower at high speeds? If a sentient object was vibrating at near light speed and was some distance off the ground. Compared to being relatively stationary to the ground, would it take a longer while for them to fall down onto the floor from their perspective?
| You are basically asking whether an oscillating clock would run slower then a stationary clock on the surface. The answer is yes, because the velocity of the oscillating clock causes time dilation.
However we can answer your question very easily because both effects 1 and 2 make the falling clock run more slowly, so overall the falling clock must run more slowly that the clock on the surface.
Does a clock oscillating in a friction-free hole through the center of a planet run slower than a stationary clock on the surface?
Now you are asking about the duration of this trip to the ground viewed from the frame of the oscillating clock. From the oscillating clock's frame the trip takes very short time.
A very good example for this to understand is the cosmic muon decay. Without this relativistic effect, the muons would decay and never reach the ground.
At rest, muons disintegrate in about 2 x 10-6 seconds and should not have time to reach the Earth's surface given there speed and travel distance. Because they move at close to the speed of light, however, time dilation extends their life span as seen from Earth so they can be observed at the surface before they disintegrate.
https://www.physlink.com/education/askexperts/ae611.cfm
Please note that you are asking whether this changes the effect of gravity, but both oscillating and stationary clocks are affected by gravity.
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Rotation in molecules I am a bit confused about the rotational motion in molecules. Assuming the bond length is constant, the motion can be described as a rigid rotor. In the center of mass frame the energies are given by BJ(J+1) and the wavefunctions are spherical harmonics. However when we measure the energies or the angular momenta, we do it in lab frame. So I am a bit confused. Is the formula for the energy the same both in lab and CM frame? And if not, what is the formula in the lab frame? Also, is the wavefunction the same in both frames or, in other words, is the angular moment of the molecule the same in both frames. Actually I am a bit confused about how is the angular momentum defined in the CM frame. Isn't the molecule stationary in that frame? Yet the wavefunctions in the CM frame (spherical harmonics) do show a clear angular momentum dependence. Can someone help me clarify these things? Thank you!
| Like NewUser mentioned, the motion can always be broken down into motion of centre of mass and motion about centre of mass. In our case, we interact with the system using light. And this causes transitions in the energy levels.
The system factors into one continuous energy spectrum corresponding to the kinetic energy of the free centre of mass and one discrete energy spectrum corresponding to the rigid rotor. The former causes light to scatter and the latter causes absorption. So if you look at absorption spectra, you'll find the signature of the discrete rotational energy levels.
| {
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Why is a Lorentz transformation of a Lorentz transformation also a Lorentz transformation? Why is a Lorentz transformation of a Lorentz transformation ($x''$,$y''$,$z''$,$t''$) also a Lorentz transformation?
| I would say that you want a mathematical proof. You can see in Reign's answer a particular case, but here I will post a more general one.
As you may know, a Lorentz transformation can be written as a 4-dimensional matrix, $\Lambda$, so that $x' = \Lambda x$. This matrix must fulfill this relation,
\begin{equation}
\Lambda^T \eta \Lambda = \eta
\end{equation}
where $\eta$ is the Minkowski metric.
If you consider a Lorentz transformation you are thinking of $x' = \Lambda_1 x$. A Lorentz transformation of a Lorentz transformation is then $x'' = \Lambda_2 x' = \Lambda_2 \Lambda_1 x$. Now, you have a transformation matrix $\Lambda = \Lambda_2 \Lambda_1$ and you would like to know whether this new matrix $\Lambda$ is a Lorentz transformation or not. To do this, take the previous equation and do some calculations:
\begin{equation}
\Lambda^T \eta \Lambda = (\Lambda_2 \Lambda_1)^T \eta (\Lambda_2 \Lambda_1) = (\Lambda_1)^T (\Lambda_2)^T \eta \Lambda_2 \Lambda_1
\end{equation}
Now, you can use that $(\Lambda_2)^T \eta \Lambda_2 = \eta$ because $\Lambda_2$ is a Lorentz transformation, so that you get
\begin{equation}
\Lambda^T \eta \Lambda = (\Lambda_1)^T \eta \Lambda_1
\end{equation}
For the same reason you can use that $(\Lambda_1)^T \eta \Lambda_1 = \eta$.
Finally you get
\begin{equation}
\Lambda^T \eta \Lambda = \eta
\end{equation}
where $\Lambda = \Lambda_2 \Lambda_1$. This proofs that a Lorentz transformation of a Lorentz transformation is a Lorentz transformation.
| {
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Can a wormhole be created if it has not always existed? I know there are solutions to Einstein's field equations that give a wormhole geometry. But they are time independent. They are static. Is there a process where empty flat spacetime can evolve into a wormhole by an appropriate flow of matter and energy and negative energy?
If so, it would change the topology of spacetime. Does General Relativity permit this? How would a hole in spacetime form? What determines where the other mouth of the wormhole would be located?
| The current state of knowledge is basically that we don't know.
The topology change is contentious - it is unclear if it is permitted, but there are also disagreements about why and how. The standard approach is to shout "quantum gravity!" and escape in the confusion.
One classical argument for why making a wormhole would be problematic is the topology censorship theorem. It states:
Every causal curve extending from past null infinity to future null infinity
can be continuously deformed to a curve near
infinity.
Roughly speaking, this says that an observer, whose trip begins and ends near
infinity, and who thus remains outside all black holes, is unable to probe
any nontrivial topological structures.
Now, there are issues with the theorem (since it assumes the null energy condition that wormholes and quantum fields often break, and some topological assumptions). But it seems to be a good reason to suspect wormhole formation or existence is not allowed unless it gets hidden by a topologically spherical event horizon. Except that general relativity on its own seems to be too much of a local theory to be really good defence against non-trivial topology.
There are quantum gravity papers arguing that a cosmic string breaking by tunnelling can produce traversable wormholes and gleefully break the (classical) topological censorship. Here the wormholes show up at the ends of the string, initially next to each other.
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Cup anemometre can measure 3D wind component? As part of my PhD thesis I am looking for information about anemometres.
I have heard that cup anemometre can only measure the x and y component of the wind, the z being left out because it measures the wind thanks to moving part.
I have been unable, so far, to find a source for that affirmation, is it true?
I know that cup anemometre are slower than sonic to measure sudden wind changes (speed and directions) due to the inertia of the mechanics, but nothing about wind component.
Thanks for the help!
| After some more searches I have found two online sources confirming my statement:
*
*https://www.omega.ca/en/resources/anemometers
*https://www.eas.ualberta.ca/jdwilson/EAS327/eas327wind.html
As well as a book "Wind Energy: theory and practise" p40. which says: "cup anemometers are the sensor type most commonly used for for the measurements of near-horizontal wind".
If this helps anyone else in the future!
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What is *physical meaning*? What do we mean when we talk about physical meaning of a quantity, an equation, theory, etc.? Should the physical meaning touch on the relation between the math and the real world? Or does it have more to do with how the equation/theory is used by physicists?
Background
For the immediate background that prompted me to ask this question see the discussion that followed answers to this question.
This forum contains nearly 3000 questions of the type What is the physical meaning of X... but do we know what we are asking?
Opinion
I think the question is important, because it defines the special place of physics among other disciplines. When we ask about a physical meaning of something we really ask how this something is related to the real world, as opposed to purely mathematical reasoning. Mathematicians and biologists do not question mathematical or biological meaning of their objects of study, since it is obvious. Yet, physicists must justify their calculations by basing them on the experimental data and making experimental predictions (as opposed to mathematicians). In the same time physicists cannot do experiments without developing complex mathematical models (unlike biologists or chemists - even though these are often more knowledgeable about complex statistical methods than an average physicist.)
| I understand physical meaning as setting the context.
We can learn a lot about the maths of 2-spin particles, operators, probabilities and expected values for example. But without learning about Zeeman effect and Stern-Gerlach experiment, all the stuff seems coming from nowhere.
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Why do antiferromagnets occur at lower temperature than ferromagnets? The minimal model for describing magnets is the Heisenberg Hamiltonian
$$H = -\frac{1}{2}J\sum_{i,j} \mathbf{S}_i \cdot \mathbf{S}_j$$
Where $i,j$ are nearest neighbors and the factor of $1/2$ is for double counting.
If $J$ is positive, spins will want to align to save energy (ferromagnets), and if it is negative they will anti-align (antiferromagnets). Ultimately $J$ comes about from Pauli exclusion and electrons not wanting to sit in the same orbital (Coulomb repulsion).
But if I look at a table of ferromagnets here, I see transition temperatures up to 1400 K. On the other hand, the highest transition temperature for antiferromagnets is a measly 525 K, with most being below room temperature.
Why do antiferromagnets generally occur at significantly lower temperatures than ferromagnets?
One can argue that maybe $\vert J\vert$ is larger in ferromagnets than antiferromagnets (as one of the current answers does), but this just begs the question. Why should that be the case (assuming it is true)? I don't see an experimentally-verified theoretical basis for asserting $\vert J_{\mathrm{AFM}}\vert < \vert J_{\mathrm{FM}}\vert$.
This question came up in a class I am teaching to talented senior undergraduates.
| This is just speculation, but the excitation spectrum of a ferromagnet is (in general) quadratic, while the spectrum of an antiferromagnet is linear. Possibly this difference in the transition temperature arises from the greater ease of creating excitations in antiferromagnets.
| {
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Is De Broglie's formula $\textbf{p}=\hbar \textbf{k}$ applicable to a discrete wave number system? I don't know if my question has sense at all but while doing my homework there appeared in my mind this question.
Say, for a particle in a box, the confinement makes that the wave number k is discrete, depending on integer numbers $\textbf{n}=(n_{x}, n_{y},n_{z})$.
If De Broglie's formula holds, it doesn't mean that momentum $\textbf{p}=\hbar \textbf{k}$ is also discrete?
| Yes it is applicable. Note that the expectation value of momentum is zero for any stationary solution in a box, so $\langle \psi | \frac{\hbar}{i} \vec \nabla |\psi\rangle = 0$. Still $ \psi | \vec p |\psi = 0$, without the integration over space is the Noether momentum distribution belonging to the Schrödinger lagrangian ${\cal L} = i\hbar \psi^* \dot \psi = -\hbar^2 \vec \nabla \psi^* \cdot \vec \nabla \psi +qV\psi^* \psi$ .
| {
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How do we know that one particular solution for the velocities of a two-body elastic collision is the correct one over the other? Assuming there is a 1-D collision between two bodies, having masses $m_1$ and $m_2$, if we conserve energy and momentum, we get two solutions.
$$
v_{1,i} = v_{1,f} \\
v_{2,i} = v_{2,f}
$$
or
$$
v_{1,i} = -v_{1,f} \\
v_{2,i} = -v_{2,f}
$$
Both of these are valid mathematical solutions under the conservation laws. If so, apart from practical experimentation, how do we decide which one of these is the correct answer? Is there an analysis that we should do locally within the system, rather than just using global laws?
Note: Subscripts i and f denote initial and final states.
| Conservation of Energy and Momentum alone does not give you a decision between the two options; it's the same with normal three-dimensional billiard balls.
Energy and momentum are conserved perfectly fine if the two billiard balls just pass 'magically' through each other and continue their movement unchanged. Still this doesn't happen.
You need to use further information, like the law that no two objects can be at the same place at the same time.
| {
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Explain how scaling of the inverse square law breaks down at a stars surface If the radiation pressure at distance $d>R$ from the center of an isotropic black body star is found to be $$P_{rad}=\large{\frac{4\sigma T^4}{3c}}\left[1-\left(1-\frac{R^2}{d^2}\right)^{\frac{3}{2}}\right],$$
a) How do I show that $P_{rad}$ obeys an inverse square law for $d \gg R$?
b) Why does the inverse square law scaling break down close to the stars surface?
| For the answer to the question a), just use a Taylor expansion in the parameter $x= R/d \ll 1$, so that
\begin{equation}
(1-x^2)^{3/2} \simeq 1- \frac{3}{2} x^2
\end{equation}
and then you obtain a inverse square law in $d$
\begin{equation}
P_{rad}= \frac{2\sigma T^4}{c} \frac{R^2}{d^2}
\end{equation}
| {
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How can a red light photon be different from a blue light photon? How can photons have different energies if they have the same rest mass (zero) and same speed (speed of light)?
| They differ in their energy. Special relativity states that $E=\sqrt{m^2c^4 + p^2c^2}$. For a massive particle, there is a one on one relation between its energy and speed. In the limit $m \rightarrow 0$ this is no longer the case. All massless particles move at light speed, but their energy/momentum can vary.
| {
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Why does frequency remain the same when waves travel from one medium to another? I was reading about reflection and refraction on BBC Bitesize and I can't understand why frequency is a constant in the wave speed equation. I can't visualise the idea of it. I know that wave speed and wavelength are proportional to each other but how can I tell the speed of a wave by looking at a random oscillation?
Here's where I got confused: https://www.bbc.co.uk/bitesize/guides/zw42ng8/revision/2 (the bottom of the page about the water)
| Instead of thinking of a travelling wave, it is better to think of a field, where each point in space and instant of time is associated to an electric and a magnetic field. The EM fields are normal to the direction of propagation. Let's choose a field in the $z$ direction and propagating in the $x$ direction, and suppose the boundary normal to $x$:
$$E_z(x,t) = E_0cos(kx - \omega t)$$
If $k$ is different for each media, it is possible to write for a point in the boundary both equations below, for any given $t$:
$E_z(x_b,t) = E_0cos(k_1x_b - \omega t)$
$E_z(x_b,t) = E_0cos(k_2x_b - \omega t)$
I can always choose the boundary as the origin, $x_b = 0$ and continuity is assured. The idea is that it is possible to have the 2 cosines in phase at the boundary, independent of the time.
But if $\omega$ is different:
$E_z(x_b,t) = E_0cos(kx_b - \omega_1 t)$
$E_z(x_b,t) = E_0cos(kx_b - \omega_2 t)$
It is not possible to assure continuity all the time, only for some t's:
$kx_b - \omega_1 t = kx_b - \omega_2 t + 2\pi n$ =>
$t = 2\pi n / \Delta \omega$
| {
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Why can't a wave travel in a non-elastic medium? Why a wave cannot propagate in a non-elastic medium
We know that wave is a distrubance and carries energy. In this sense let imagine fall of dominoes, which carries disturbance and energy. Here fall of dominoes is non-elastic and we can see that wave propagates.
Can I call it as a wave?
| I think it goes like this.
A mechanical wave is indeed a periodic perturbation the particles of the medium through which it travels, and if the medium is non-elastic, as most media are, then energy will be lost at each 'passing on' of the energy from one particle to the next. This is the cause of the attenuation of mechanical waves over distance. The more elastic the medium, the farther the wave will travel. A wave can travel in a non-elastic medium, but not very far.
Although it is not quite the same thing, consider the speed of sound in air (331 m.s-1), water (1,403 m.s-1), ice (3,838 m.s-1), iron (5,120 m.s-1) & diamond (~12,000 m.s-1). As we progress through that list, the medium becomes more elastic(/brittle), and sound will travel farther.
| {
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Imaginary part of semiconductors index of refraction I understand that the index of refraction is complex and can be expressed as such: $ \widetilde{\eta} = \eta + i \kappa $. However I’ve been searching for a bit and I am unable to find the derivation of why the imaginary part of the refractive index in semiconductors is as follows $$k = \frac{\lambda \alpha}{4 \pi}$$
Can someone demonstrate?
| If $\alpha$ is the attenuation coefficient, such that $|E|^2 \propto e^{-\alpha x}$ it is, by pure identification, the definition of
$\alpha$.
Let's write:
$$ E=E_0 \exp\big(i (n+ik)k_0 x\big) $$
where $k_0=\frac{2\pi}{\lambda}$ is the vacuum wave number.
You get then:
$$ E=E_0 e^{ink_0 x} e^{-k\,k_0\,x}$$
and
$$|E|^2=|E_0|^2 e^{-2k\,k_0\,x}$$
Hence $\alpha=2k\,k_0=
\frac{4\pi\, k}{\lambda}$.
| {
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Why do we need Gauss' laws for electricity and magnetism? The source of an electromagnetic field is a distribution of electric charge, $\rho$, and a current, with current density $\mathbf{J}$. Considering only Faraday's law and Ampere-Maxwell's law:
$$
\nabla\times\mathbf{E}=-\frac{\partial\mathbf{B}}{\partial t}\qquad\text{and}\qquad\nabla\times\mathbf{B}=\mu_0\mathbf{J}+\frac{1}{c^2}\frac{\partial\mathbf{E}}{\partial t}\tag{1}
$$
In an isolated system the total charge cannot change. Thus, we have the continuity equation that is related to conservation of charge:
$$
\frac{\partial\rho}{\partial t}=-\nabla\cdot\mathbf{J}\tag{2}
$$
From these three equations, if we take the divergence of both equations in $(1)$, and using $(2)$ in the Ampere-Maxwell's law, we can get the two Gauss' laws for electricity and magnetism:
$$
\nabla\cdot\mathbf{B}=0\qquad\text{and}\qquad\nabla\cdot\mathbf{E}=\frac{\rho}{\varepsilon_0}\tag{3}
$$
Therefore, the assumption of $(1)$ and $(2)$ implies $(3)$. At first glance, it could be said that we only need these three equations. Also, conservation of charge looks like a stronger condition than the two Gauss' laws (it's a conservation law!), but, as the article in Wikipedia says, ignoring Gauss' laws can lead to problems in numerical calculations. This is in conflict with the above discussion, because all the information should be in the first three equations.
So, the question is, what is the information content of the two Gauss' laws? I mean, apart of showing us the sources of electric and magnetic field, there has to be something underlying that requires the divergence of the fields. If no, then, what is the reason of the inherently spurious results in the numerical calculations referred?
(Also, I don't know what type of calculation is referred in the article.)
| There is a paper linked to the cited statement at wikipedia. In short the system is actually not overdetermined. The authors report that numerical methods, which ignore the divergence-free conditions lead to inaccurate solutions. They show that they are needed to guarantee the uniqueness of the solutions (you have to take account for the boundary conditions).
| {
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Why does the dielectric field not cancel out the capacitor's field? When a conductor is in a region with electric field, free charges will move until they balance out the external electric field. However in dielectrics this does not happen. I know that charges are bounded to the atoms, and there is only a small portion that will be near the surface of the the capacitor, but should we not also consider the small electric fields inside the polarized atoms? They may add up and cancel out the external field.
| Any volume element within the dielectric that's large enough to encompass many molecules, but smaller than any scale of interest, will be electrically neutral regardless of whether or not there is an induced polarization. So the molecules in the bulk do not contribute to the macroscopic electric field.
| {
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Is heat $\delta Q$ an exact differential for an isochoric process (ideal gases)? Generally speaking, heat and work are path-dependent, thus $\delta Q$ and $\delta W$ are not exact differentials. By first law of thermodynamics, we know that $dU=\delta Q - \delta W$
but $\delta W=0$ for an isochoric process, that yields $dU=\delta Q$. Does this make work an exact differential in this specific situation? Am I neglecting something? This sounds weird to me.
| The reason why heat and work are path dependent and therefore not exact differentials is that, unlike internal energy, they are not system properties. While there can be a change in internal energy of a system, there is no “change” in the work or heat of a system because a system does not “possess” work or heat.
Consequently $\delta$ means an amount of energy transfer to or from a system in the form of heat and work, as opposed to $d$ which means a change in the amount of internal energy possessed by the system. The fact that an isochoric process eliminates energy transfer in the form of work does make heat a system property. It simply means the change in internal energy is due solely to energy transfer by heat. Similarly, the fact that an adiabatic process eliminates energy transfer in the form of heat does make work a property of the system. It simply means the change in internal energy is due solely to energy transfer in the form of work.
The selection of a specific path between two states does not make heat or work a system property.
Hope this helps.
| {
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Entropy as a state function - Is it just a postulate of the second principle? I read quite a few questions on this website dealing with the idea of demonstrating that entropy is a state function. None of the answers I read seemed to be fully conclusive. So my question is : is there anything wrong in saying that the second principle postulates the existence of entropy as a state function? Or is there a definitive demonstration showing that from a more restrictive statement of the second principle?
| One complete description of the second law of thermodynamics is:
$\exists \quad S=S(U,V) \quad \textrm{with} \quad dS\geq0$
It indeed contains the existence of entropy as a state function. I for myself don't like the weird historical formulations.
| {
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Why does the range of this integral work out this way? I have a bit of trouble in finding the same limits for the integral in Eq. (17.111) from Peskin & Schroeder. We have something like
$$ \int_0^1 dx' \int_0^1 dz f(x',z) \delta(x-zx').$$
Posing $y=zx'$, I find
$$\begin{align}
\int_0^1 dz \int_0^1 \frac{dy}{z} f\biggl(\frac{y}{z},z\biggr) \delta(x-y)
&= \int_0^1 \frac{dz}{z} 1_{[0,z]}f\biggl(\frac{x}{z},z\biggr)\\
&= \int_0^z \frac{dz}{z} f\biggl(\frac{x}{z},z\biggr).
\end{align}$$
Instead, P&S find
$$ \int_x^1 \frac{dz}{z} f\biggl(\frac{x}{z},z\biggr).$$
I must have overlooked some property of the Delta distribution. Can someone point out my mistake?
| P&S's equation implicitly assumes that $0\leq x\leq 1$:
$$\begin{align}\int_{[0,1]} \!dz &\int_{[0,1]} \!dx' ~f(x',z)~ \delta(x-zx')\cr
~=~&\int_{[0,1]} \!dz\int_{\mathbb{R}} \!dx' ~1_{[0,1]}(x')~f(x',z)~ \frac{1}{|z|}\delta(\frac{x}{z}-x')\cr~=~&\int_{[0,1]} \!\frac{dz}{|z|} 1_{[0,1]}(\frac{x}{z})~f(\frac{x}{z},z)\cr~=~&\int_{[0,1]} \!\frac{dz}{|z|} 1_{[0,1]}(x)~\theta(z\!-\!x)~f(\frac{x}{z},z)\cr~=~&1_{[0,1]}(x)\int_{[x,1]} \!\frac{dz}{z}~f(\frac{x}{z},z).\end{align}$$
| {
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Term for radius and gradient of spacetime distortion? A black hole would distort spacetime time to a greater degree than planet Earth. That is, both the radius and the gradient of the distortion are greater.
Is there a term that combines "greater radius" and "greater gradient" into one?
In layman terms, a black hole distorts spacetime more "aggressively" than planet Earth.
| Spacetime distortion is measured by the Riemann curvature tensor $R_{\mu\nu\lambda\kappa}$. This tensor has 256 components, but various symmetries reduce the number of independent components to 20. So, in general, it takes 20 numbers at each point in spacetime to fully describe how spacetime is distorted.
The simplest way to compare the spacetime curvature of a Schwarzschild black hole versus that of the Earth is to consider a curvature invariant like the Kretschmann scalar. This is just a single number, and it is much larger at the event horizon of a stellar-mass black hole than it is at the surface of the Earth.
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Can pressure inside system ,with small hole in them , build up when heated? Can one build up pressure in system for example kettle with hole sufficent of releasing water vapor ?
Kettle starts with 21c 1atm inside and outside and will be heated as fast as possible, can this system build up pressure inside The kettle even If it has hole that can let the steam escape? And can anyone explain why it won’t or will?
1st scenario:
pressure cannot build up because Steam can escape from hole with as much kinetic energy as it has been heated. Hence pressure cannot build up and pressure stays somewhat consistent, and steam escapes only because it requires more room and has kinetic energy to an velocity to expand and because of hole it only expands in one direction.
2nd scenario:
If hole is small enough it creates some sort of “resistance” and unknown force to me, enables the pressure difference increase inside kettle and this is the reason why steam escapes?
Can one explain which scenario is more correct or explain correctly the phenomena.
| If heating fast enough, one can build a pressure, since it takes finite time for the pressure to equilibrate through a small hole. Let us take, for example, a punched air baloon or a tire - initially the pressure inside is high and the extra air escapes through the hole; it happens quite fast, but not instantly.
An example even closer to the question is the behavior of a pressure cooker - in principle, the valve is never super-tight - it is just a metallic ball on the top of the hole, so there are definitely micro-holes through which the steam can escape. Yet, these holes are too small to slow the pressure equilibration, so that the inside pressure can build.
| {
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Massless string vs massless spring in a mass-spring system
Two masses connected by a massless spring, on a frictionless surface , and a force of $60$N is applied to the 15kg mass such that it accelerates at 2 $\frac{m}{s^2}$. What is the acceleration of the $10kg$ mass?
I came across this question. I first thought that that the $10$kg was constrained to move at the same acceleration. But when I work it out, I get $a_2$ = 3 $\frac{m}{s^2}$. And it is the correct answer according to the book.
What I am unable to understand is, isn’t the $10$kg mass constrained to move at the same acceleration as the $15$kg mass? I thought we could replace the massless spring by (or treat it as) a massless string and results would be the same. Am I making a fundamental mistake?
| a string is rigid so cannot be extended or compressed, its both end would move with same acceleration
in spring, it can extend, if spring extends then it would apply equal forces on both
bodies i.e $kx$ ,towards left for $15 kg$ and towards right for $10 kg$.
the only force moving $10 kg$ is $kx$
now if you apply this concept to string you will get different acceleration on both sides which is not possible, so acceleration in case of strings is constrained
| {
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Can an electron transit within the same $n$-level? Under selection rules for multi-electron atoms in LS coupling, its generally written that there is no restriction on the total quantum number $n$. Does it mean that an electron transit within the same $n$ level, i.e., can $\Delta n =0$?
| As far as i know when you derive the transition rules for some interaction with some interaction hamiltonian $H_{int}$ you investigate the matrix element $\langle f | H_{int} |i\rangle$ between some initial $|i\rangle$ and final $|f\rangle$ state labeles by the appropriate quantum numbers. The allowed transitions are given by the non-zero matrix elements, which constrains the relation between the quantum numbers between the initial and final state. The transition rules are usually formulated as only the constraints, so if it doesn't say anything about $\Delta n$ for example it should not matter. To see a list of transitions in multi electron atoms here.
It might be useful to look at the derivation of the selection rules once again to properly understand the meaning of them.
| {
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What was the mean distance photons travelled before colliding with electrons in the matter plasma before recombination? What was the mean distance photons travelled before colliding with electrons in the matter plasma before recombination?
I have checked other answers close to this but they only mention a mean distance but not what it actually is.
| Since the cat is out of the bag and I calculated this in another (non-duplicate) question, I'll repeat the calculation here.
The baryon density today is about $\rho_0 \sim 4\times 10^{-28}$ kg/m$^3$, then at a redshift of $z\sim 1200$ (just prior to recombination), the number density of free electrons (assuming a fully ionised hydrogen gas) is roughly
$$n_e = \frac{\rho_0 }{m_u}(1+z)^3= 4\times 10^8\ {\rm m}^{-3}$$
The mean free path of a photon in the plasma is $1/(\sigma n_e) = 4\times 10^{19}$m, where $\sigma$ is the Thomson scattering cross-section. Thus the average photon can travel about 4000 light years before being scattered.
Since the "size" of the universe is $\sim ct$, where $t$ is the time since the big bang, and since $t \sim 300,000$ years at $z\sim 1200$, then the universe is effectively opaque to the radiation within it.
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Physics of the trikke tricycle I love my trikke, but I still do not understand what propels it forwards.
It is very clear that the energy comes from my legs and not from my arms (I only have to touch the handle bar ever so lightly), but I do not see how my shifting weight from side to side can result in a forward pointing force.
How is the side to side movement converted into a forward moving force?
(And just to be clear: My trikke is not electric).
| First off, you might as well ask, what makes any vehicle go forward? When you're riding a bike, you're just pushing pedals up and down, so how does that end up making the bike go forward? When you're driving a car, how does the engine make the car go forward? After all, the pistons in the engine just oscillate back and forth.
The answer in all cases is that the forward force ultimately comes from friction with the ground. If a bike, trikke, or car were on perfectly frictionless ice, they all wouldn't be able to go forward. Of course, this doesn't mean that the ground itself is suppling the energy. For bikes or cars, you (or the engine) supply the energy, but the friction force is necessary to convert it into useful forward motion; without friction, the energy just makes the wheels spin in place.
Now specifically, the trikke works just like ice skates and rollerskates work. In the first half of a step on the trikke, you let your body fall down a bit, slightly forward and to the left. If you were on a frictionless surface, this would make your upper body move forward and your lower body move backward, generating no net motion. But because there is friction with the ground, your entire body and the trikke are overall accelerated forward and to the left. In the second half of the step, you move your body back up; that's the part that requires energy input from the legs. Then this is repeated in a step where you fall down forward and to the right, giving a net forward motion.
| {
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Is information lost from black body radiation? If you heat water in a microwave all of the water is heated by a narrow wavelength of photons. As the water cools it emits different photons of radiation in the form of infrared light.
Can you tell how the water was heated from the cooling process? How many photons and their energy.
| In a way the information is in the blackbody radiation: each photon carries $3.897\pm 2.522$ bits of information. It is just scrambled.
The important thing here is that water does not have a very good memory. Certainly there are hydrogen bonds producing a complicated microstructure, but the interactions mix up any information with the whole structure within 100 picoseconds. You send the collected works of Shakespeare into the water as microwave pulses, and you get them distributed into the thermal noise degrees of freedom nearly instantly - they are there, but you will not recoverable just as past sounds are not recoverable from the heat in a room.
One should remember that information in this sense is Shannon information - how many bits does it take to describe the state - rather than semantic information - what does it mean? Noise has maximal Shannon information, but is meaningless.
So the heating history is forgotten quickly, and only the temperature matters.
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Does Lorentz's force and magnetic dipole alignment torque explain same or different forces inside an electric motor? As depicted in the image below borrowed from this page the motion of the rotor in an electric motor can be explained by appealing to the Lorentz force on a current carrying wire (left side) or by the orienting torque of a magnetic dipole in a magnetic field (right side).
Do these two explanations describe two separate forces that are accumulated when calculating the torque of a motor or are they just differently framed descriptions of the same underlying force?
| I had hoped to see answers to your question, but since this is not the case, I allow my own answer here.
The torque in your left image can be explained as follows:
*
*Electrons have a magnetic dipole moment. In a wire without an external magnetic field, the magnetic dipoles of the electrons are randomly oriented.
*When the electrons are in a magnetic field, their magnetic fields get aligned with the external field, and the loop behaves like the magnet in your right picture. It rotates (clockwise) and comes to rest with the wires close to the poles. The magnetization of the wire depends from the susceptibility of the wires material.
*When a small current is switched on, the Lorentz force acts against the magnetic alignment and the wires move slightly away from the vertical position. What you observe is that the movement of the electrons within an external magnetic field is responsible for the lateral movement of the wire. The Lorentz force acts against magnetic alignment.
*When more electrical power is used, the rotor starts to rotate.
So
Do these two explanations describe two separate forces that are accumulated when calculating the torque of a motor ...
is the right answer.
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How to calculate the width of the dark fringes? I am talking about the single slit diffraction experiment.
The width of the central bright fringe is twice as wide as that of the others bright fringes. It can be calculated easily as follows.
\begin{align}
\text{width of other bright fringes} = \frac{\text{wave length}\times\text{distance between screen and slit}}{\text{width of the slit}}
\end{align}
Question
From the intensity plot, it is clear that the width of dark fringes is zero. But when we look at the spectrum the width is not exactly zero. Could you tell me how to find the width of the dark fringes? I think several percentage of the maximum intensity should be considered as dark, right? What percentage is it usually adopted by physicists?
| This is actually somewhat more involved. You can compute the intensity/irradiance of the Fraunhofer diffraction pattern exactly and the result is the so-called Airy pattern:
$I(\theta)=I_0\cdot \left[ \frac{2 \cdot J_1(k \cdot a \cdot sin \theta)}{k \cdot a \cdot sin \theta} \right]^2$,
where $\theta$ is the observation angle, $k$ is the wavenumber and $a$ is the size of your aperture. This gives you the following intensity graph you have shown in your question.
I am referencing the Wikipedia article here. In order to evaluate it you need to be able to compute Bessel functions of the first kind and these are available in most programming languages, such as Python and Fortran, as well as software like Matlab.
The angles where the intensity minima occur are the zeros of these Bessel functions $J_1(x)$. Starting from there you can e.g. compute the angle $\theta$ at which the first intensity minimum occurs:
$sin(\theta) \approx \frac{\lambda}{d}$,
where $\lambda$ is your wavelength and $d$ the width of your aperture.
More zeros can be found on this Mathematica website. Here is the list of the first five zeros of the Bessel function of the first kind for reference:
*
*3.8317
*7.0156
*10.1735
*13.3237
*16.4706
*...
So e.g. for the first intensity minimum you would have to solve $k \cdot a \cdot sin(\theta) = 3.8317$ for $\theta$, which is:
$\theta = arcsin \left( \frac{3.8317}{k\cdot a} \right)$
This also means that the intensity is zero at specific angles and not over a continuum. It only appears so due to the contrast of the image you have chosen.
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Calculating work of an object moving up a slope If an object is being pushed across a horizontal surface, the equation for the work done is $W = F s$, where $s$ is the horizontal displacement.
If an object is being lifted to a height of $h$, the equation for the work done is $W = F h$, where $h$ is the vertical displacement.
If an object is being pushed up a slope, or if a human is moving up a flight of stairs, however, only the vertical displacement is concerned in calculating work, while the horizontal displacement is omitted. Why is this so?
(With hindsight, I think this is a conceptual misunderstanding that only arises because it is quoted out of context!)
| You should understand the fact that (Work)W = F.S
(Work Done)W = FS Cos(theta)
We will not consider Horizontal motion because Weight of the object makes 90 Degrees with Horizontal displacement which gives Work Done in Horizontal Displacement as Zero [Cos(90) = 0] .
| {
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“Bananagrams” under black light? There is a game called “Bananagrams” which includes a bunch of pieces with a letter on each. It seems when I shine a black light flashlight on the letters, the “M” letters glow, but no other pieces do.
All the pieces appear the same under normal lights (except of course the letters on each piece). Why would only the M’s glow, can someone explain what may be happening here?
| Well, obviously either the 'M' pieces are made of some different material which flouresces, or they are painted / coated with some material that does while the other pieces are not.
I would be tempted to mail the makers and say, basically, 'I found this interesting thing' and ask them if they know: they look like they are fairly small and you might therefore get to talk to someone who actually knows about the manufacturing.
| {
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What is global Lorentz transformation and what is local Lorentz transformation? I will consider $\textbf{spacetime}$ as $(M,\eta)$ where $M$ is a four dimensional $\textbf{manifold}$ and $\eta$ the metric which in this coordinates
$$
\begin{align*}
x \colon M &\longrightarrow \mathbb{R}^4\\
p &\mapsto x(p)=(x_0,x_1,x_2,x_3).
\end{align*}
$$
is given by
$$\eta=dx^0\otimes dx^0-dx^1\otimes dx^2-dx^2\otimes dx^1-dx^3\otimes dx^3 \tag1$$
An $\textbf{observer}$ is a worldline $\gamma$ with together with a choice of basis
$
O=v_{\gamma,\gamma(\lambda)} \equiv e_0(\lambda) , e_1(\lambda), e_2(\lambda), e_3(\lambda)
$
of each $T_{\gamma(\lambda)}M$ where the observer worldline passes, if
$$
\eta(e_a(\lambda), e_b(\lambda)) = \eta_{ab} = \left[ \begin{matrix} 1 & & & \\ & -1 & & \\ & & -1 & \\ & & & -1 \end{matrix} \right]_{ab} \tag2
$$
$v_{\gamma,\gamma(\lambda)}$ is the tangent vector of the the curve $\gamma$ at the point $\gamma(\lambda)$
In text books i've found three definition of $\textbf{Lorentz transformation} \quad \Lambda$
*
*$\Lambda \colon \mathbb{R}^4 \longrightarrow \mathbb{R}^4$ is a group of coordinate transformations that leave eq.1 in the same form ,that is $\Lambda \cdot x(p)=y(p)=(y_0,y_1,y_2,y_3)$ such that in this coordinate $$\eta=dy^0\otimes dy^0-dy^1\otimes dy^2-dy^2\otimes dy^1-dy^3\otimes dy^3 $$
*$\Lambda \colon M \longrightarrow M$ a Spacetime diffeomorphism such that $\Lambda_* \eta=\eta$ where $\Lambda_* \eta$ is the pullback of the metric $\eta$
*$\Lambda \colon T_pM \longrightarrow T_pM$ such that $\Lambda O=O'=e'_0(\lambda) , e'_1(\lambda), e'_2(\lambda), e'_3(\lambda)$ satisfy the eq.2 that is
$$
\eta(e'_a(\lambda), e'_b(\lambda)) = \eta_{ab} = \left[ \begin{matrix} 1 & & & \\ & -1 & & \\ & & -1 & \\ & & & -1 \end{matrix} \right]_{ab}
$$
My question is which is of these transformation is global Lorentz transformation and which is local?
| The three definitions are the same. They are ways of saying the same thing. Since you have a manifold $(M,\eta)$ this is a flat, Minkowski, spacetime. The Lorentz transformation is global on Minkowski spacetime.
In a curved spacetime the metric is usually denoted $g$, rather than $\eta$. $g$ is, in general, a function of time and position. At each point there is a Minkowski tangent space, meaning that the manifold is locally Minkowski (to the accuracy of measurement) and that local Lorentz transformations can be applied within a neighbourhood of each point.
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How does one interpret thermodynamic differentials? When we study we usually refer to quantities in terms of differentials, why is this?
Secondly how do we get an intuitive idea of how to deal with the infinitesimal quantities involve? Like what is the right perspective to look at these things from? It is quite different from most physics I have encountered so far.
| I guess most of the time you are interested in understanding how a system's, for example, internal energy change as you change the volume. On the other hand, some quantities such as internal energy or enthalpy are independent of the way you change the system so therefore like gravitational potential energy they only depend on initial and final states of the system so you can write them as $\Delta U$ or $\Delta H$.
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Are antileptons and antibaryons linked? The recent news about the T2K experiment got me thinking: is there any linkage in the Standard Model between the matter and antimatter categories across the families of Standard Model particles? Are antileptons necessarily linked to antibaryons?
As a specific example: In our universe "matter" is made up of electrons $e^-$ and protons $p$. Antimatter particles are positrons $e^+$ and antiprotons $\bar p$.
$p$ and $\bar p$ are obviously a matter-antimatter pair, but is there any theoretical reason the $e^-$ is the same type of matter as the $p$? Could there be a universe in which $p$ and $e^+$ are the "matter" particles and $\bar p$ and $e^-$ are the "antimatter" particles?*
* Besides the fact that obviously that would be a weird universe where you couldn't make atoms.
| This was the first thing Dirac thought of when he produced the Dirac equation which predicts the positron. He thought there might be some hidden loss of symmetry, and that the positron could be the proton. Within a couple of years of the hypothesis, it had been fairly conclusively rejected and Dirac predicted that the anti-electron (as he then termed it) should actually exist. The positron was observed soon afterwards (more strictly, it had already been observed, but not recognised).
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Wavefunction of a photon Does anyone have an explicit closed-form expression for the wavefunction of a single photon from a multipolar source propagating through free space? Any basis is acceptable as long as it is a single photon state.
A reference would also be appreciated, but not essential.
————————
A possible duplicate has been suggested: Does a photon have a wave function or not?
But this question primarily concerns the existence of the wavefunction and is not what I am looking for. None of the answers provide an explicit expression for the wave function, and neither the question nor the answers discuss a multipole source. The multipole source, in particular, is central to my question.
| I posted an answer to a similar question here. Admittedly, many different things may be implied when talking about the wave function of a photon. However, one should keep in mind is that, unlike electrons, photons are classically waves. Quantization neither adds nor subtracts from their wave-like properties, but injects discreteness (i.e. makes them from an electromagnetic field into countable photons). Their wave modes remain the same - plain waves of the electromagnetic field.
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Does Loop Quantum Gravity predict general relativity in semi-classical Limit? Recently i read about Loop Quantum Gravity in Wikipedia and found this below statement here.
Presently, no semiclassical limit recovering general relativity has been shown to exist.
But i also came across a paper in arxiv whose abstract claims that Einstein's equations arise as a result of Semi-Classical Limit from Covariant Loop Quantum Gravity.
In this paper we explain how 4-dimensional general relativity and in particular, the Einstein equation, emerge from the spinfoam amplitude in loop quantum gravity.
So is the Wikipedia article outdated or is the paper's idea falsified?
| About a decade ago, Ashoke Sen did some calculations that show that you can use Euclidean methods compute the logarithmic corrections to the entropy of Schwarzchild black holes in terms of low-energy, classical data. He remarked that these entropies had also been calculated within loop quantum gravity, and that the results do not match the answer. This strongly suggests that LQG does not give the correct semi-classical limit.
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Can Higgs potential provide a cosmological constant? Usually, in particle physics, people do not care about a constant term in scalar field potential. Rather, attentions are paid to the local profile at the minimum. But in the context of cosmology, the absolute value of the potential has a physical meaning; it is a cosmological constant and can cause the Universe to accelerate or decelerate. My impression is that the naive potential for the Higgs field has a negative value at the minima. Do people take it seriously as a negative cosmological constant? Is the dynamical change in the value of the potential at the minimum during EWSB taken into account?
| There are two things to consider here: the late-time accelerated expansion, often discussed within the context of the cosmological constant, $\Lambda$, and primordial inflation. Let's take the latter first.
In primordial inflation, one has a field initially evolving in the false vacuum of some potential energy function. The false vacuum must be very flat in order to get enough inflation and generate the right spectrum of density perturbations. For example,
Cosmologists early on hoped that the SM Higgs, or perhaps a GUT Higgs, might have been what drove primordial inflation but the trouble was that the false vacuum wasn't flat enough. In recent years, this idea has been regained momentum when it was realized that if one coupled the Higgs to gravity (as a non-minimal coupling with the scalar curvature), then one could flatten this region and obtain successful inflation, see https://arxiv.org/abs/0710.3755.
Once the Higgs decays to the true vacuum, primordial inflation ends. If the true vacuum, however, has a nonzero vacuum energy, then it's possible that the Higgs field could contribute to the observed late-time accelerated expansion. As you note, the classical vacuum energy of the Higgs is negative, $\rho_{\rm Higgs, \,vacuum} < 0$, and this is not what we observe. Therefore, it's possible that the effective cosmological constant is something like $\Lambda_{eff} = \Lambda + \rho_{\rm Higgs,\,vacuum}$, but this subtraction would need to be fine tuned to get our tiny observed amount of cosmological constant.
(Image from: https://indico.cern.ch/event/180122/attachments/239069/334713/NExT_2012_Atkins.pdf)
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Escape velocity for electric field As we know gravitational force and electric forces are quite similar I was wondering if there is minimum velocity called escape velocity required to escape gravitational field of earth so is there a minimum velocity to escape electric field of earth since electromagnetic forces are stronger than gravitational forces. If yes then how to calculate it?
| A problem is that the Earth is not just a conducting sphere with charge on its surface as this text Electricity in the Atmosphere illustrates.
However you also need to consider the charge on the object trying to escape which may well be under the influence of other charges which are surrounding it.
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When Is It Appropriate To Use The Ladder Operator Method in Quantum Mechanics? I'm trying to understand when it is intuitively obvious that the ladder method would be best used to tackle a problem in quantum mechanics.
| Two excellent examples of the use of ladder operators can be found in Introduction to Quantum Mechanics (3rd Ed, Griffiths):
1. The 1D harmonic oscillator, in chapter 2 (pg 43) and
2. The spherical harmonics for the total angular momentum operator (pg 159).
In both of these, there is a common theme. If there exists two operators such that one takes an arbitrary eigenfunction (of the operator we are trying to find the eigen configuration for) to the next one and the other takes the eigenfunction to the previous one, then it is possible to use ladder operators. In both cases, Griffiths just gives operator(s) that work as the ladders and there isn’t a “god given” method to find that they are supposed to be, so it can be challenging to find the appropriate ladder operators for a problem.
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Lorentz transformation of four potential In the following, $c=1$ (and so $\beta=v$) and the signature is $(-+++)$.
The four-potential is $A_\mu=(-\phi,\mathbf{A})$. It transforms as $A'_{\mu'}=\Lambda_{\mu'}^{\ \ \ \ \mu}A_{\mu}$ so that, under a boost in the $x^1=x$ direction, it becomes $$A'_{\mu'}=\begin{pmatrix}A'_0\\A'_1\\A'_2\\A'_3\end{pmatrix}=\begin{pmatrix}\gamma&v\gamma&0&0\\ v\gamma&\gamma&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}\cdot\begin{pmatrix}-\phi\\ A_1\\ A_2\\ A_3\end{pmatrix}=\begin{pmatrix}\gamma(-\phi+vA_1)\\\gamma(-v\phi+A_1)\\ A_2\\ A_3\end{pmatrix}$$ Throughout, the unprimed quantities are quantities from before the boost.
From $\mathbf{B}=\nabla\times\mathbf{A}$, it is clear that $$B'_1=(\nabla\times\mathbf{A}')_1=\partial_yA_3'-\partial_zA'_2=\partial_yA_3-\partial_zA_2=B_1$$ i.e. it is unchanged. However, I then find that $$\tag{1}B'_2=(\nabla\times\mathbf{A}')_2=\partial_zA_1'-\partial_xA'_3=-\gamma\partial_z\phi+\gamma v\partial_zA_1-\partial_xA_3.$$
I know the answer is supposed to be $B_2'=\gamma(B_2+vE_3)$, and I also know that $$E_3=(-\nabla\phi-\partial_0\mathbf{A})_3=-\partial_z\phi-\partial_0A_3$$ so that I can write my answer (1) as $$\tag{1}B'_2=(\nabla\times\mathbf{A}')_2=\gamma E_3+\gamma\partial_0A_3+\gamma v\partial_zA_1-\partial_xA_3.$$
I don't really know how to proceed towards the correct answer from here. I suspect I might have made a mistake. Any help would be greatly appreciated.
| It might help in this case to consider that the electric and magnetic fields are the "components" of the electromagnetic tensor. Then one can use the transformation properties of a rank-2 tensor to derive the expression of the transformed fields.
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Why is tension needed to create a wave in a string? Suppose, a long straight string is present in vacuum. I oscillate one end of the string with a certain frequency. Shouldn't a wave be formed? If it is formed, what will be the velocity of the wave?
| Tension provides a restoring force, which is necessary to have oscillations (like a spring or gravity for pendulums). Otherwise any attempt to excite waves will only produce an inelastic deformation.
However, if the gravity is present, there will be tension created due to the non-zero mass of the string which might be sufficient to have waves.
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Mach cone geometry from Mach number Given a Mach number, how would I go about determining the geometry of the associated Mach cone? Apologies, I'm not too well versed in physics.
| A wave in the shock front moves away from its point of origin at the speed of sound. The plane moves away from that point at the Mach number times the speed of sound. In a given time the distance traveled by the plane forms the hypotenuse of a triangle with the distance traveled by the sound forming the side opposite the apex of the cone. One over the Mach number gives the sine of the angle at the apex (measured from the path of the plane).
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What is the term to describe when pressure exerted between two obejcts is balanced? I'm searching for a term here.
All materials compress (some more than others).
Newton's Third law states:
...all forces between two objects exist in equal magnitude and opposite direction: if one object A exerts a force FA on a second object B, then B simultaneously exerts a force FB on A, and the two forces are equal in magnitude and opposite in direction: FA = −FB
So for example, if a rubber ball is placed on top of a sponge, both would feel a "constant" force exerted on each other (in this case due to gravity). Now obviously the sponge would compress more, while the rubber ball would hardly compress.
What is the term to denote that the force applied by and to each of these objects results in a balance of compression?
I'm not even sure if balance is the right word to describe this. I'm trying to describe that the compression of each object will no longer increase or decrease. The closest term I conjured up was "equilibrium of pressure."
| This (when total force on an object or interface is zero, such that there is no net motion) is just called "mechanical equilibrium."
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Why doesn't a backward wave exist? (Huygens principle) Since every point on a wavefront act as a source of secondary wave (wavelets) then why do we get only forward wavefront not backward.
Huygens principal says that amplitude of the backward wave is zero, but why and how it happens?
| Today the principle could be explained as follows.
Every medium has an elasticity and a viscosity. In simple words, the first describes how deeply a body can penetrate or shift the medium over time. The second describes how the medium around is displaced over time.
Imagine a hammer falling lightly on a metal block. The hammer deforms the metal elastically at this point and the metal gives way. In which direction? In all directions. Where the metal is homogeneous, at the same rate. What you get is Huygens (semi)spherical wave.
Note that this wave has a longitudinal and a transverse component. In the direction of the hammer blow, the transverse component predominates (as with sound), and perpendicular to it all around the surface, the longitudinal component predominates.
Absolutely important is that the initial point of the disturbance determines the propagation direction. For a slit, the edges and the wall are the two disturbances. The edges bend the wave behind the edges of the slit, the wall reverse the movement direction.
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What is the possibility for a gamma ray photon to pass through a composite particle like a proton? As I know, the photon can be either absorbed or reflected from a particle, but as baryons or other hadrons are composite particles is there a defined possibility of passing through the volume inside the particle, and if it is small is that because of the relativistic speeds of quarks inside a hadron?
| At the level of gamma ray energies, it is more complicated. Once there is enough energy in the photon for production of particles , scattering of photons on nuclei becomes more complicated. See what a proton looks like here.
See this experiment how many channels with created particles there are .
elastic γp → V , V a vector meson
proton dissociative γp → V N
photon dissociative γp → Gp , G a hadronic state
double dissociative γp → GN
hard non-diffractive γp → X
soft non-diffractive γp → X
In quantum mechanics there will always be a probability of the gamma going through only diffracting
In general it is all a matter of probabilities to be calculated with quantum field theory models.This is a review that goes into some details for high energy photon-proton interactions.
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Is the net torque of an only translating rigid body zero, independent of the point chosen? I know that the Newton-Euler equations can be proven using the center of mass as reference, but I was wondering if this is a special case, or if you can provide a counter-example. We know that when a rigid body is only translating the net torque through the center of mass is zero. Is this true when we evaluate the torque using other points too?
| The net torque on an object can be defined as
$$\boldsymbol{\tau}=\frac{\mathrm d {\mathbf p}}{\mathrm d t}\tag{1}$$
where $\mathbf p$ is the angular momentum. The angular momentum of any body about any general point is
$$\mathbf p=I_{\text{COM}} \boldsymbol{\omega} +m\mathbf r\times \mathbf v$$
where $I_{\text{COM}}$ is the moment of inertia of the body about its center of mass, $\omega$ is its angular velocity, $\mathbf r$ is the position vector of the center of mass, $m$ is the mass of the body and $\mathbf v$ is the velocity of the center of mass of the body. Plugging this into $(1)$, we get
\begin{align}
\boldsymbol{\tau}&= \frac{\mathrm d (I_{\text{COM}} \boldsymbol{\omega} +m\mathbf r\times \mathbf v)}{\mathrm dt}\\
&=\underbrace{\frac{\mathrm d (I_{\text{COM}} \boldsymbol{\omega})}{\mathrm d t}}_{\text{Term 1}} + \underbrace{\frac{\mathrm d (m\mathbf r\times \mathbf v)}{\mathrm d t}}_{\text{Term 2}}
\end{align}
Since the rigid body is only translating, thus the term 1 becomes 0 (Here we have assumed that $I_{\text{COM}}$ is constant). Now applying product rule to differentiate term 2,
$$\boldsymbol{\tau}=m\left(\frac{\mathrm d \mathbf r}{\mathrm d t}\right)\times \mathbf v + m \mathbf r \times \left(\frac{\mathrm d \mathbf v}{\mathrm d t}\right )$$
Here we have assumed that $m$ is constant. Now, since our reference point is stationary, thus, $\mathrm d \mathbf r/\mathrm d t=\mathbf v$ and, also, $\mathrm d \mathbf v/\mathrm d t =\mathbf a$. Thus
$$\boldsymbol{\tau}=m\mathbf v \times \mathbf v+ m \mathbf r \times \mathbf a$$
Since the cross product of a vector with itself yields a zero vector, thus,
$$\boxed{\boldsymbol{\tau}=m\mathbf r \times \mathbf a}\quad \equiv\quad \boxed{\boldsymbol{\tau}=\mathbf r\times \mathbf F}$$
This is your final expression of torque. It is only zero when $\mathbf r$ and $\mathbf a$ are along the same line, or any one of the two of $\mathbf r$ and $\mathbf a$ is zero.
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Why don't we use rapidity instead of velocity? In school we learn that we can add velocities together, and then later on we learn that it's not correct and that there is a speed limit. Why create all this confusion when we could just use rapidity to begin with?
Rapidity is defined as $w = \mathrm{arctanh}(v / c)$, where $v$ is velocity and $c$ is the speed of light in a vacuum. Rapidities can be summed and have no upper bound. At non-relativistic speeds it acts proportional to velocity.
In fact, at non-relativistic speeds, we could substitute $v$ for $wc$ (rapidity times speed of light), and one could hardly tell the difference. The ISS moves rather fast at a velocity of 7660 m/s (27,576 km/h), and has a $wc$ of about 7660.0000016667 m/s. Why can't we just substitute velocity for rapidity in real-world and classroom use, and end the confusion about why there is a speed limit once and for all?
| I think the main reason, as already captured in part by other answers (e.g. Davide Dal Bosco's), is the following: velocity is a physical quantity, it tells us how far something goes in a given time.
Rapidity may be mathematically convenient due to its relativistic addition properties, but what does it tell us?
As an example, the rapidity of light is $w = \textrm{arctanh}(1) = \infty$. Isn't it much more useful to know that light moves at $c=299 792 458 \frac{m}{s}$ through space?
Mathematically, we can transform everything back and forth as we wish to simplify our calculations. But in the end, we will want to know something physical: the velocity.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/547371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
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Is the force of gravity always directed towards the center of mass? This is a pretty basic question, but I haven't had to think about orbital mechanics since high school. So just to check - suppose a [classical] system of two massive objects in a vacuum.
If the density of either object is the same at a given distance from the center, and both objects are spherical, then both objects can be treated as point-masses whose position is the [geometric] center of the original sphere. In the case that either object is not spherical or has an irregular distribution of mass (I'm looking at you, Phobos!), both objects can still be treated as point-masses but the center of mass rather than the geometric center must be used.
Is this correct?
| No. For example, the gravity of a cubical planet of uniform density, which can be computed analytically, is not directed towards its center (or any other single point).
You can also imagine a dumbbell-shaped mass distribution where the two heavy ends are very far apart. If you drop an apple near one end it is going to fall toward that end, not toward the middle of the “neck”.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/547505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
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} |
Can massless particle have effective mass? The effective potential was probably very familiar in many concepts.
However, what about effective mass?
Suppose a massless particle. For simplicity, suppose it's not some superficial particle, i.e. it has observable effect.
Is it possible for such massless particle to gain an "effective mass" through dynamical interaction?
For example, a photon could well obtain a $e^-\sim e^+$ pair in space, but I'm not sure weather it's a meaningful case.
Further, what does it mean to four momentum for such effective mass, if it exists.
| A photon, when propagating in a photonic crystal, will have a nonzero effective mass in all bands, except possibly the point $\vec k=0$ in the lowest band. This effective mass has all the weird properties of that of electron in crystal: anisotropy, varying value in the Brillouin zone, varying sign etc..
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/547610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Is the potential energy stored in a spring proportional to the displacement or the square of it? Suppose a mass of $M$ kg is hanging from a spring in earth. The mass will stretch the spring about $x$ m. So the change in the gravitational potential energy is $mgx$ J (supposing $x$ to be very small compared to the radius of earth).
And this amount of energy will be stored in the spring as potential energy. So,
Change of gavitational energy = $mgx$ = potential energy stored in the spring
And it seems that the potential energy stored in a spring is proportional to displacement $x$. But the potential energy in a spring is $U=\frac{1}{2}kx^{2}$ and so it's proportional to $x^2$, the square of displacement. So surely I am wrong somewhere. But where am I wrong?
| The elementary work $dW$ done on the spring in a elementary displacement $dx$ is:
$dW = Fdx$.
At this point the spring has already been stretched by $x$, so $F = kx$.
$dW = kxdx$.
Integrating from $x=0$ (the unstretched position) until the final $x$:
$W = \frac{kx^2}{2}$
$\frac{kx^2}{2} = mgx$, the loss of potential gravitational energy equals the stored elastic energy.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/548021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Magnetic flux through circular loop Purcell says
if finite current flows
in a filament of zero diameter, the flux threading a loop made of such
a filament is infinite! The reason is that the field $B$, in the neighborhood
of a filamentary current, varies as $1/r$ where $r$ is the distance
from the filament, and the integral of $~B \times \textrm{area}~$ diverges as $\int dr /r $
when we extend it down to $r = 0$.
This page clearly shows that the magnetic flux through a circular loop of radius $R $ is $\dfrac{\mu _0 I}{2R}$. So the flux through it is $\dfrac{\pi\mu _0 I R^{2}}{2}$, which is obviously finite. Where does the contradiction come from?
| $\dfrac{\mu _0 I}{2R}$ is the magnetic field at the centre of the loop, however, the magnetic field is not the uniform across the plane of the loop, getting stronger closer to the loop and exploding as Purcell notes if the filament radius is zero.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/548103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How does cutting a spring increase spring constant? I know that on cutting a spring into n equal pieces, spring constant becomes n times.
But I have no idea why this happens.
Please clarify the reasons
| This happens because spring constant is not really a constant. If you consider any normal elastic material, when a force F is applied, the strech is given by hooke's law :
$$\frac{{F}/{A}}{{\Delta L}/{L}}=Y$$
where $Y$ is the young's modulus of material, which is upto a limit, constant and depend only on the material.
This means the strech
$$\Delta L = \frac{FL}{AY}$$
or
$$F = \frac{AY}{L}\Delta L$$
we see that the proportionality constant for spring is thus $\frac{AY}{L}$. Thus, for a spring of $\frac{1}{2}$ the length, spring constant would be double.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/548186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 4
} |
Mathematically prove that a round wheel roll faster than a square wheel Let's say I have these equal size objects (for now thinking in 2D) on a flat surface.
At the center of those objects I add equal positive angular torque (just enough to make the square tire to move forward).
Of course the round tire will move faster forward and even accelerate (I guess). But how can I mathematicaly prove/measure how better the round tire will perform?
This for my advanced simulator I'm working on and I don't want to just Hardcode that rounds rolls better, square worse, etc.
I know the answer could be very complex, but I'm all yours.
| If you allow an eccentric star gear hub like a Wankle engine then neither has a moving center of mass. The square would wear out faster due to uneven loading (same issue as Wankle engine seals).
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/548519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "64",
"answer_count": 12,
"answer_id": 9
} |
What is the range of Pauli's exclusion principle？ In many introductions to the pauli's exclusion principle, it only said that two identical fermions cannot be in the same quantum state, but it seems that there is no explanation of the range of those two fermions. What is the scope of application of the principle of exclusion? Can it be all electrons in an atom, or can it be electrons in a whole conductor, or can it be a larger range?
| In quantum mechanics, particle interactions can be of two types, scattering interactions and bound states.
What is the scope of application of the principle of exclusion?
The Pauli exclusion principle applies to bound states of electrons in the solutions of potential equations for atoms/molecules/lattices. It will apply to fermions in general , for example no two muons can occupy the same muonic hydrogen energy level.
Can it be all electrons in an atom,
All electrons of an atom have to occupy different energy levels. Energy levels might be degenerate, but they must be different in a quantum number( example spin orientation for example)
or can it be electrons in a whole conductor,
The electrons in a whole conductor are very lightly bound, which means the energy levels they occupy are very close to continuum, i.e. there will always be an available energy level with different quantum numbers to occupy, this is what allows to have more general quantum mechanical models for solids as the band theory of solids.
or can it be a larger range?
So range has meaning for the Pauli exclusion principle only when one is talking of bound states that have energy levels labeled by quantum numbers available for occupation.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/548642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 6,
"answer_id": 3
} |
Do things have colors because their electrons are getting excited when photons hit them? Atomic electron transitions can be caused by absorbing a photon with a certain wavelength. An electron jumps to an higher energy level, then it falls back and a photon is emitted. The perceived color of the photon depends on the energy absorbed by the electron.
Could we say that electrons in the atoms of different objects are excited when white light hits them, and they release photons which in turn causes the object have a color?
| In addition to the other answers one should add color perception. White light contains all frequencies, if some are absorbed the scattered back light will have a different spectrum perceived differently by the receptors in our eyes.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/548900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
In what sense is the equation of motion of a damped oscillator not time-symmetric? Consider the equation of motion of a damped oscillator
$$\frac{d^2 x}{dt^2} + \gamma \frac{dx}{dt} + \omega_0^2 x = 0 \,. $$
Why does the equation of motion not satisfy time-symmetry? Is it related to drag term?
I am new to this area..so can you recommend related articles?
| Indeed, it is the drag term: it is easy to see that $y(t) = x(-t)$ satisfies equation
$$\frac{d^2y(t)}{dt^2} -\gamma\frac{dy(t)}{dt} + \omega_0^2y(t)=0.$$
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/549143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Does the $U(1)$ vector current flip under charge conjugation? The conserved $U(1)$ current of the Dirac Lagrangian is given by $j^\mu = \bar{\psi} \gamma^\mu \psi$, where $\bar{\psi} = \psi^\dagger \gamma^0$. As this is interpreted as electric current I would expect it to flip sign under charge conjugation. Charge conjugation Of a spinor $\psi$ is defined as $\psi^c = C\psi^*$ where $C$ is the unitary charge conjugation matrix that satisfies $C^\dagger \gamma^\mu C = -(\gamma^\mu)^*$ for all gamma matrices.
If I calculate the $U(1)$ current under charge conjugation I find
$$ j^\mu_c = \bar{\psi^c}\gamma^\mu \psi^c \\ = (C \psi^*)^\dagger \gamma^0 \gamma^\mu C \psi^* \\
= (\psi^\dagger)^* C^\dagger \gamma^0 C C^\dagger \gamma^\mu C \psi^* \\
= (\psi^\dagger)^* (\gamma^0)^* (\gamma^\mu)^* \psi^*
\\
= (\bar{\psi} \gamma^\mu \psi)^*\\
= (j^\mu)^* $$
Which hasn’t flipped sign as I thought it would. Have I made an error in my analysis?
Any hints would be appreciated. Thanks!
| For any fermion bilinear we have
$$
\psi^T_\alpha A_{\alpha\beta} \chi_\beta = - \chi^T_\beta A^T_{\beta\alpha}\psi_\alpha\,.
$$
So
$$
\begin{aligned}
(\bar\psi \gamma^\mu \psi)^* &= -\psi^* (\gamma^\mu)^\dagger(\gamma^0)^\dagger\psi
\\&= -\psi^* \gamma^0\gamma^0(\gamma^\mu)^\dagger\gamma^0\psi \\&=
-\bar\psi \gamma^\mu\psi\,.
\end{aligned}
$$
Where I used $(\gamma^0)^2 = 1$ and $\gamma^0(\gamma^\mu)^\dagger\gamma^0 = \gamma^\mu$. In the first line I applied the identity at the beginning with $\psi^T \to \bar\psi^*$ and $\chi \to \psi^*$.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/549272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
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