Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
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Two-point correlation function in Peskin's book I am reading Peskin's book on QFT and I reached a part (in chapter 4) where he is analyzing the two-point correlation function for $\phi^4$ theory. At a point he wants to find the evolution in time of $\phi$, under this Hamiltonian (which is basically the Klein-Gordon - $H_0$ - one plus the interaction one). Anyway, when he begins his derivation he says that for a fixed time $t_0$ we can still expand $\phi$ in terms of ladder operators in the same way as we did in the free (non-interaction) case (this is on page 83), i.e. $$\phi(t,\mathbf{x})=\int{\frac{d^3p}{(2\pi)^3}\frac{1}{E_\mathbf{p}}(a_\mathbf{p}e^{i\mathbf{x}\mathbf{p}}+a_\mathbf{p}^\dagger e^{-i\mathbf{x}\mathbf{p}})}.$$ I am not sure I understand why can we do this, for a fixed time. When we wrote this in term of ladder operators for the free case, we used the KG equation in the free case, which resembled to a harmonic oscillator in the momentum space, and hence we got the ladder operators. But now, the equation of motion is different (it has $\phi^3$ term, instead of 0, as before), so can someone explain to me why we can still use the same formula as before, even if the equation of motion is different?
| Klein-Gordon equation only determines the dispersion relation between the energy and the momentum $(p^0)^2-(\vec p)^2=0$ for free scalar field. When the interaction exists in the theory, at any time, the field $\phi$ can be Fourier expanded to the momentum space, with the operators $a$ and $a^\dagger$ explicitly dependent on time $t$ without requiring the above free scalar dispersion relation. This is similar to the situation in the quantum mechanics: any wave function can be expanded using momentum eigenstates, with time-dependent coefficients. The point here is that $a(t)$ and $a^\dagger(t)$ also satisfy the commutation relation at any time $[a_{\vec p}(t),a^\dagger_{\vec p'}(t)]\sim \delta^3(\vec p-\vec p')$ which follows from $[\dot \phi(\vec x ,t),\phi(\vec x',t)]\sim \delta^3(\vec x-\vec x')$, if the interaction involves no derivative of $\phi$.
| {
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Recovering the $ SU(2)$ unbroken limit by taking $m_W, m_Z \to 0$ I'm trying to compute the tree-level scattering amplitude for $e^+e^- \to q \overline{q}$ and $e^+e^- \to W^+W^-$ in unitary gauge. Both processes go through an $s$-channel photon, $Z$ and Higgs while the second process also has a $t$-channel neutrino exchange. I would like to check these results with results in the unbroken $SU(2)$ limit, where the $SU(2)$ gauge bosons are massless. Should I expect to recover the result in the unbroken limit by taking all of the gauge boson masses to zero? Or is there something extra that I have to do?
| Your question (if I understand correctly) can be summarized with:
Are the amplitudes computed in the Standard Model in the limit where the Higg's vacuum expectation value (VEV) $\rightarrow 0$ equal to those computed in the Standard Model without spontaneous symmetry breaking (i.e., either without a Higgs at all or with a Higgs but with the Higgs mass parameter $\mu^2>0$)?
Indeed such amplitudes are equivalent to those in another theory but its not quite either of the theories you proposed. Just removing the Higgs gives the wrong answers for processes involving the longitudinal modes of the vector bosons. Having the Higgs but swapping its mass parameter, doesn't by itself make much sense (what do you mean by the different components of the Higgs then?). However, the amplitudes are equivalent to those computed in the "Goldstone Boson Equivalent" theory.
In this theory you identify the longitudinal modes of the vector bosons with the different components of the Higgs (the ones that get "eaten" in the Unitarity gauge). Then the amplitudes are given by those computed by this ``unbroken symmetry'' theory in the limit that the VEV approaches zero. More formally, the amplitudes are equal to those computed in the full theory up to $ {\cal O}({\rm energy} / m _{W,Z}) $ corrections.
| {
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Is spring force yet another force of tension? I was going through the Hooke's law and studying about Springs in general.
One question which is bugging me is whether "Force by a spring" essentially yet another force of tension (just like in a rope).
I read the following, on the following link, https://www.khanacademy.org/science/physics/work-and-energy/hookes-law/a/what-is-hookes-law
So a bunch questions on which I am seeking clarity,
*
*The coil shape has nothing to do with the "spring" or "spring force"?
*Is spring as good as a rope when considering how it acts when subjected to force? Or is "Elastic force" different than "force of tension".
Also, am I thinking correctly here?
The difference is, ropes can only be pulled and NOT pushed (upon push rope will slack), but springs can be pulled and pushed upon. But fundamentally the force at microscopic level is same in either rope / spring.
| Yes, the forces involved are interatomic (and so fundamentally electromagnetic) in the case of a stretched spring, as for a stretched wire or a stretched rope.
I think, though, that the passage you quote is misleading. If you formed a helical spring out of a metre of steel wire, then the extension of the spring when subjected to equal and opposite forces at either end would be much greater than the extension of the original wire when subjected to the same pair of forces. This is because, for the spring, there is twisting and bending of the wire. Extension of the wire along its length is negligible by comparison. The shape of the spring is indeed magical!
| {
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When is Newton's law of gravity recoverable in GR? Often in a course in GR one can recover Newton's law of gravity under certain assumptions. weak field, slow moving particles etc. Is there a general method to recover Newton's laws of gravity for an arbitrary spherically symmetric, static and asymptotically flat spacetime metric? For example with the following line element
$$ (ds)^2 = A(r) dt^2 + 2B(r) dr dt - C(r) dr^2 -r^2 d\Omega^2, $$
where the functions $A,B,C$ are all determined and the above line element is an exact solution to the field equations.
Or, can we only recover Newton's law of gravity when the components of the metric tensor are of the form $g_{\mu\nu} = \eta_{\mu\nu} + h_{\mu\nu}$ where the $\eta_{\mu\nu}$ are the components of Minkowski's flat space plus a perturbation given by $h_{\mu\nu}$?
| A hint :
\begin{eqnarray}
-A(r) dt^2 + B(r) dr dt - C(r) dr^2 -r^2 d\Omega^2 &=& (dt^2 - dr^2 -r^2 d\Omega^2)\\
&+& (A(r)-1) dt^2 + B(r) dr dt - (C(r) - 1) dr^2
\end{eqnarray}
| {
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Is there a case when it is better to use the integral form of the Maxwell equations rather than the differential form? I was wondering if there is a case where the integral form of the Maxwell equations is preferred over the differential form?
If you could provide with an example for each one of the equations I would really appreciate that.
| The same principles apply as when you normally want the differential form of something vs. the integral form.
It's like asking when do you want to know the total distance and time of a road trip vs. what the speed was at a particular point.
With a uniformly charged sphere, for example, you can use the integral form to get the total flux at some radius, and then use that to infer what the potential is at any point because of the perfect symmetry of a sphere. Wouldn't work with a cube!
The analogy would be if you know you drove the exact same speed throughout an entire road trip, you could calculate what the speed was by dividing total distance / total time. Not so if there was a lot of starting and stopping!
| {
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Peskin and Schroeder spinor high-energy limit (5.26 and A.20) P&S say the high-energy limit of spinor $u^s (p)$ is
$ \sqrt{2E} {1 \over 2} (1-\widehat{p} . {\sigma}) \xi^s $ and similar for the right-handed spinor (formulae 5.26 and A.20). I can't seem to derive this. How do you get this from $\sqrt{p . \sigma } \xi^s $?
| Don't know if anyone is still expecting a reply to this, but here is how I managed to find the result. Peskin&Schroeder found result (3.50) by doing the calculation in a specialized frame (boost along z-axis) which resulted in (3.49). If you apply the high energy limit in that same frame by setting E = p3 and rewrite it in a covariant form (using a dot product with unit momentum vector ^p instead of p3), then you get the desired result which is immediately valid in any frame due to the covariant notation.
To anyone who might be concerned...
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Is this an electromagnetic wave without the magnetic part? Jefimenko's Equations are:
$$
\begin{align}
&\mathbf{E}(\mathbf{r}, t) = \frac{1}{4 \pi \epsilon_0} \int \left[ \left(\frac{\rho(\mathbf{r}', t_r)}{|\mathbf{r}-\mathbf{r}'|^3} + \frac{1}{|\mathbf{r}-\mathbf{r}'|^2 c}\frac{\partial \rho(\mathbf{r}', t_r)}{\partial t}\right)(\mathbf{r}-\mathbf{r}') - \frac{1}{|\mathbf{r}-\mathbf{r}'| c^2}\frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] \mathrm{d}^3 \mathbf{r} \\
&\mathbf{B}(\mathbf{r}, t) = \frac{\mu_0}{4 \pi} \int \left[\frac{\mathbf{J}(\mathbf{r}', t_r)}{|\mathbf{r}-\mathbf{r}'|^3} + \frac{1}{|\mathbf{r}-\mathbf{r}'|^2 c}\frac{\partial \mathbf{J}(\mathbf{r}', t_r)}{\partial t} \right] \times (\mathbf{r}-\mathbf{r}') \,\mathrm{d}^3 \mathbf{r} \\
& \mbox{where the retarded time is: }t_r = t - \frac{|\mathbf{r}-\mathbf{r}'|}{c}
\end{align}
$$
I noticed the following: In the E-field, the second and third terms fall at $1/r$. Meaning, they are radiative terms. That is, electromagnetic radiation. Analogously, In the B-field, the second term falls at $1/r$, thus radiative. However, in the B-Field, there is no radiative term depending on $\partial\rho / \partial t$.
With that in mind, consider the following situation:
$$
\frac{\partial\mathbf J}{\partial t} = 0
\quad\quad\mbox{and}\quad\quad
\frac{\partial\rho}{\partial t} \neq 0
$$
Then, the magnetic field will not be time varying, and won't be radiative. In this situation, the only radiative fields would be electric. Meaning, an electromagnetic radiation without the magnetic part! How is this possible? Is there a mistake somewhere? What prevents it from happening?
An approach was to allow time-varying $\rho$ with charge conservation by means of constant not-null-everywhere $\mathbf J(\mathbf r)$. By continuity equation, one can have a time varying $\rho$ one wants, by simply choosing $\mathbf J$ cleverly. I haven't made further progress with this.
$$
\nabla\cdot\mathbf J + \frac{\partial\rho}{\partial t} = 0
\quad\implies\quad
\frac{\partial\rho}{\partial t} = -\nabla\cdot\mathbf J
$$
| Arguably, it might be, but the sources you've provided cannot be sustained for long: since the current doesn't change in time, you have
$$
\frac{\partial^2\rho}{\partial t^2} = -\nabla \cdot\frac{\partial \mathbf J}{\partial t} = 0,
$$
i.e. if the charge density is increasing with time, then it is doing so linearly with time and without any way to stop, so you will necessarily end up with regions of arbitrarily high charges over sufficiently long times, and that is going to take an unbounded amount of energy.
If you're OK with that, then yeah, you can go on and find out whatever weird properties the emitted field has, but it's not really something that's broadly considered to be a physically allowed situation.
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Is the DC/electro-optical Kerr effect usable in optical quantum computing? It seems from @EmilioPisanty's comments on this question that the AC/optical Kerr effect is impractical for use in optical quantum computers, which leads me to ask: would the DC Kerr effect work?
The major difference between the optical/AC Kerr effect and the electro-optic/DC Kerr effect is that for the DC version to work, one must manually apply the electric field to the medium, whereas for the AC version, the light going through produces the effect itself.
In summary, is it possible to use a medium that exhibits the DC Kerr effect in place of a medium that exhibits the AC Kerr effect in an optical quantum computer?
| To formalize Rococo's comment as an answer: not really, no.
The electro-optic effect can be enormously useful in manipulating the propagation of light in the lab, through devices generally known as electro-optic modulators and their relatives, going all the way from precision spectroscopy down to manipulating the propagation of photons on a chip (example, though as I understand it photonic chips are easier to build around the thermo-optic effect).
However, that is based on external manipulation acting on the system, where the modulations come in through classical fields (either AC or DC) that (i) have very many photons, and (ii) are in completely classical states of the field. For this to be useful for quantum computation, you'd need the modulation to be triggered by one of the single photons in your computation, in a coherent, nondestructive way.
And, as explained in the previous answer, no, that's not currently possible without extensive, dedicated work on very special conditions.
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Why don't free electrons escape from a conductor? The thermal velocity of the free electron in a metallic conductor varies from $10^5\ \mathrm{m/s}$ to $10^6\ \mathrm{m/s}$. In spite of high velocity, free electrons fail to escape from the metallic surface. Why is that?
| If a electron were able to just escape the conductor, then it would be pulled back by the positive charge left behind (An image charge potential, one could say). Due to this it would be pulled back and will not be able to escape until it is able to obtain enough energy to become completely free.
Electrons can attain energy by the photoelectric effect, or by heating the conductor (Thermionic emissions).Two example of Thermionic emissions is the Child-Langmuir Law and The Richardson Law. (This PDF gives you a short intro to both ).
Also, note that if a photoelectric effect were continuously illuminated, and were not connected to a circuit, then after some time the photoelectric emission would stop since the positive charges left behind would accumulate and increase the potential barrier required to escape.
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Are generalised coordinates truly independent? Say we have a system with two generalised coordinates $x$ and $y$. When we solve the equations of motion we find $x=x(t)$ and $y=y(t)$. I can invert one of these solutions to find $t=t(y)$ and therefore get $x=x(t(y))$ which therefore gives me $x(y)$. Does the equation of motion impose a constraint? Are generalised coordinates independent?
| You shouldn't try to make $t$ a coordinate; it's a label for the coordinates, over which the coordinate-dependent Lagrangian is integrated to form the action. (The most obvious problem this causes is that the momentum of time is undefined, viz. $\frac{\partial L}{\partial \dot{t}}=\frac{\partial L}{\partial 1}$.) It would be like trying to change the fields in a theory so one is replaced with the spacetime coordinates that label the fields (note these are integrated over to obtain a field theory's action, which is any fields are analogous to generalised coordinates).
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Is the pressure the same if I heat to the same temperature different closed containers with distinct ratios of water to air? Say I have 3 closed containers of 1L each. By volume, the 1st one is 50/50 air and water, the 2nd is 20% water and 80% air, and the third is 80% water and 20% air. If I heat all of them to the same temperature of 120 C. Are they all under the same internal pressure?
EDIT: They all start at atmosphere pressure (1 atm) and room temperature (20 C) before I close the lid and start heating them.
| I'll show you how to do the calculation for 20% water and 80% air, and you can then repeat the calculations for the other mixtures. We will neglect the initial amount of water vapor in the air at 20 C. So there are 200 cc of liquid water and 0.8 l of air. The specific volume of liquid water at 20 C is 1.002 cc/gm, so the mass of water is 199.6 gm and the number of moles of air is determined from the ideal gas law as [(0.9869)(0.8)]/[(0.0821)(293)]=0.0328 moles = 0.952 gm.
From the steam tables, at 120 C, the equilibrium vapor pressure of water vapor is 1.985 bars, the specific volume of liquid water is 1.060 cc/gm and the specific volume of water vapor is 892 cc/gm. So, if x represents the number of grams of water vapor in the gas space, we have (199.6-x) grams of water in the liquid space. So,we have, $$892x+1.060(199.6-x)=1000$$This gives 0.885 gm of water in the gas phase and 199.6-0.885=198.7 gm of liquid water. So the volume of liquid water at 120 C will be (198.7)(1.060)=211 cc and the volume of gas will be 1000-211=789 cc. Neglecting any air dissolved in the water, the partial pressure of the air in the gas space will be $[(1.0)(800)(273+120)]/[(789)(293)]=1.36\ bars$. So the total pressure in the container will be 1.985+1.36 = 3.35 bars.
| {
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When short circuits are not exactly short circuits?
Here $R1 = 2 \Omega$, $R2= 4 \Omega$ and $R3= 4 \Omega$
Though there looks to be a short circuit in this diagram, my teachers say that this circuit can easily be redrawn into simple parallel circuit. As far as I know a short circuit is an alternate way for current to move where there is no resistance whatsoever. So all of the current will flow through a short circuit if found.
Here if the current flows through that part of the wire ACB then after that R2 and R3 being the same it will get confused which way to go and even if it goes both sides then some part of the current is bouncing back and moving from the negative side of the cell to the positive. This is just a conjecture.
So can anyone please describe why there is no short circuit in the circuit though it seems to be? And also how to understand by looking at any circuit that there is actually no short circuit though there seems to be one.
|
As far as I know a short circuit is an alternate way for current to move where there is no resistance whatsoever. So all of the current will flow through a short circuit if found.
This is correct. The part that seems to be confusing you is that this isn't a short circuit.
As stated in the other answer, you have shorted the circuit between A and B using wire C; it will pass around R1 without issue.
Once you reach B, the current is no longer free to move without encountering a resistance. This means that the whole circuit is not shorted.
Comparing it to a circuit without wire C, you could say that C has shorted R1 and R2; but the total circuit has some resistance and is therefore not a short circuit.
It's only a short circuit if it can travel through the entire loop without encountering any resistance.
I believe this may also depend on the terminology you choose. It's very common to say you've made a "short circuit" between A and B; but I can see the logic behind saying the entire system itself is not a short circuit.
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Why does the Lamb shift only affect s-states? The shift is due to interactions between atomic electrons and vacuum state fluctuations, so I would think that there would be an associated energy shift for all states, and not just s-states.
Also, is there any physical relationship between the Lamb shift and the Darwin correction for hydrogen? The explanation for the Darwin term is usually "Zitterbewegung", which is often in turn justified by the same processes contributing to the Lamb shift.
| The s-states are the only orbitals that do not vanish at the origin. This is where the coulomb field of the proton is strong enough to produce measurable vacuum polarization or Lamb shift. Hence the Lamb shift only affects s-states.
| {
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Why does sugar dissolve faster in hot water compared to cold water? Why does sugar dissolve faster in hot water compared to cold water?
| I am going to have to respectfully disagree with Pranjal Rana.
The reason that sugar dissolves more quickly in hot water than in cold water has to do with the kinetic energy of the particles.
The faster (hotter) the water molecules are moving, the more energy they give to the sugar molecules when they collide with them.
Think about it this way: a sugar crystal is just a lot of sugar molecules stuck together. You dissolve the crystal you need to hit the molecules with enough force to break them off of the rest of the crystal.
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Why is a nonlinear crystal necessary to stimulate quantum fluctuations that entangle photons? I've been reading about spontaneous parametric down conversion (SPDC). The Wikipedia article on it says:
A nonlinear crystal is used to split photon beams into pairs of photons that, in accordance with the law of conservation of energy and law of conservation of momentum, have combined energies and momenta equal to the energy and momentum of the original photon and crystal lattice, are phase-matched in the frequency domain, and have correlated polarizations... SPDC is stimulated by random vacuum fluctuations, and hence the photon pairs are created at random times [...]
Why is a crystal necessary for these fluctuations to occur, and how do the fluctuations entangle the incoming photons?
| In order for photons to become entangled they must have interacted at some point. Photons don't interact in free space, but they do interact in nonlinear crystals. That is indeed why these special crystals are called nonlinear, they support some form of multi-photon interaction.
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Does phononic "Superconductivity" exist? Can solids be cooled to a sufficiently low temperature such that sound can travel through them without acoustic impedance?
It seems that if I had a solid in the shape of a torus, sitting in ambient vacuum, cooled to absolute zero, and then decided to introduce a compression along a circular slice (rest of the solid still at 0), then this wave should propagate indefintely looping around the torus.
Can this behavior occur at $T_c > 0$ if one takes into account Bose-Einstein condensation of phonons?
| To some extent phonons are already condensed. Usually they are thought of as the Goldstone modes of broken translational symmetry (ignore optical modes for simplicity). This means the superflow of phonons is constant movement of the crystal itself. In other words, the crystal can move at a constant velocity indefinitely (assuming no outside forces at work). In the torus geometry, the entire crystal will rotate indefinitely without dissipation.
However, I believe what you are really interested in is Bose condensation of the nuclei of the lattice themselves. This is actually what Ultracold atom people achieve regularly. However in their case they are dealing with gases, not solids!
But can you actually see this in a normal solid (say Lithium?)
The answer is probably yes in theory, but in practice it is no. As a rule of thumb for bose condensation, you need to have the thermal de Broglie wavelength of the atoms be at least on the order of the atomic spacing itself. For electrons, this is achieved trivially even at extremely high temperatures because of their tiny mass. However for nuclei, which have masses that are thousands of times heavier, this temperature would need to be in the nano kelvins at the highest (pico kelvin in reality). At that point, you could possibly get in the regime of atomic condensation, but usually you need to get even colder.
Currently, the coldest you can get a solid is in the milli to (high) micro kelvin regime. Getting into nano or pico kelvin is completely out of reach for an ordinary solid. Perhaps it will be possible in the future someday.
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In QFT, can Field Operators at different points in Space-time always be expressed as unitary Transformation of each other? Given an operator valued field $\Phi(x)$, and two points in spacetime, $x$ and $y$, can I always write down something like:
$$
\Phi(y) = U_{x,y}^{-1} \Phi(x) U_{x,y}
$$
With $U_{x,y}$ being an unitary Operator?
| Depends what you call a QFT. If it is an object satisfying the definition commonly known as the set of Wightman Axioms then the property is built into it, i.e., it follows from Axiom W2 together with the trivial transitivity of the action of the Poincaré group on spacetime.
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How can I use a candle to heat glass/quartz to a specific temp Excuse my ignorance of physics. Is it possible to suspend glass or quartz over a candle at a specific distance (in. or mm.) in order to heat the object to a certain temperature, say 300 C. Is there a graph anywhere for something like this? The convection heat transfer above a candle flame? I'm trying to find the temperatures ABOVE a flame but can only find the temperature OF a flame.
Or say I put a funnel upside down over a candle, leaving room for oxygen at the bottom, will that force the heat up out of the top of the funnel and create more temperature control in regards to my first question?
Any help would be appreciated, thanks!
| You can estimate the plume temperature above a fire using correlations from Heskestad or McCaffrey, listed respectively below:
$$T_0=25(\dot Q ^{2/5}_c/(z-z_0))^{5/3} + T_ \infty$$
$$T_0=22.3(\dot Q ^{2/5}_c/z)^{5/3} + T_ \infty$$
Where $T_0$ is the centreline plume temperature, $\dot Q_c$ is the convective heat release rate of the fire, $z$ is the height above the fire source, $z_0$ is the virtual origin and $T_ \infty$ is ambient temperature (all temperatures in Celsius). Virtual origin can be estimated (Heskestad) by:
$$z_0/D=-1.02+0.083 \dot Q^{2/5}_c/D$$
Within the flame region, one typically assumes a constant temperature (the flame temperature of say 800 °C). This was all derived from work with heptane, methane burners.
Some issues you may encounter:
*
*This is for undisturbed plumes, which is impractical in "the real world".
*You can't put an item in a plume at 300°C and expect it to heat to 300°C - there is one more major factors at play: heat gains and losses to the quartz plate both by convection and radiation. You will need to determine your energy gains and losses to predict your plate temperature.
Another interesting piece of work which may help you has been carried out by NIST (http://fire.nist.gov/bfrlpubs/fire05/PDF/f05141.pdf) which characterises heat fluxes at various locations above a candle flame.
| {
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Are the first order phase transitions always associated with a latent heat? Is the first order ferromagnetic transition below the critical temperature associated with latent heat?
For example, the transition of ferromagnetic configuration with all its spins aligned up to a ferromagnetic configuration with all its spins aligned down at $T=T_F<T_c$, when the magnetic field changes from $H\to -H$, is a first order transition. Is this also associated with latent heat?
If yes, how do we calculate it? $Q=T_F\Delta S$, but I think, $\Delta S=0$ because both up-aligned and down-aligned configurations have same entropy. I'm not quite sure that $\Delta S=0$ because if $\Delta S=0$ how can it be discontinuous? But for this transition to be first order, entropy should be discontinuous according to the Ehrenfest criterion.
| I agree with @user8153 that I don’t think entropy has to be discontinuous for the phase transition to be first order. The free energy for this case is $F=-MdH$ since the other parameter temperature is constant. At the phase transition, the first order derivative of free energy, $M = -\frac{\partial F}{\partial H}$ is discontinuous.
| {
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Obtaining Euler-Lagrange equation from action with constraint - Witten's topological sigma model The action of Witten's topological sigma model (defined on a worldsheet, $\Sigma$, with target space an almost complex manifold denoted $X$) takes the form
$$
S=\int d^2\sigma\big(-\frac{1}{4}H^{\alpha i}H_{\alpha i}+H^{\alpha }_i\partial_{\alpha}u^i+\ldots\big), \tag{2.14}
$$
as shown in equation 2.14 of his paper. The auxiliary fields $H^{\alpha i}$ also obey the "self-duality" constraint
$$
H^{\alpha i}={\varepsilon^{\alpha}}_{\beta} J{^{i}}_jH^{\beta j}, \tag{2.5}
$$
where $\varepsilon$ and $J$ are respectively the almost complex structures of $\Sigma$ and $X$.
Now, the Euler-Lagrange equation for $H_{\alpha}^{ i}$ is given in equation 2.15 as
$$
H^i_{\alpha}=\partial_{\alpha}u^i+\varepsilon_{\alpha\beta}{J^{i}}_j\partial^{\beta}u^j. \tag{2.15}
$$
How does one show this? I have tried including the "self-duality" constraint in the action via Lagrange multipliers, but I have not been able to obtain the correct Euler-Lagrange equation in this way.
| Isn't the trick to take
$$
H^{\alpha i}={\varepsilon^{\alpha}}_{\beta} J{^{i}}_jH^{\beta j},\tag{2.5}
$$
and note
$$
H^{\alpha i}=\frac{1}{2}\left(H^{\alpha i} + {\varepsilon^{\alpha}}_{\beta} J{^{i}}_jH^{\beta j}\right)
$$
then plug this into the action to rewrite it as
$$
S=\int d^2\sigma\left(-\frac{1}{4}H^{\alpha i}H_{\alpha i}+\frac{1}{2}\delta_{ik}\left(H^{\alpha i} + {\varepsilon^{\alpha}}_{\beta} J{^{i}}_jH^{\beta j}\right)\partial_{\alpha}u^k+\ldots\right),
$$
where I am using the fact that contraction over Latin indices is done using the "Worldsheet metric" which is (by conformal invariance) equal to $\delta_{ij}$.
| {
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What is the Planck scale magnetic field strength? Using the constants $\mu_0$ (or $\varepsilon_0$), $c$, $\hbar$, $e$ and $G$, it is possible to define two quantities with units of magnetic field :
\begin{align}
B_1 &= \sqrt{\frac{\mu_0 c^7}{\hbar G^2}} \equiv \sqrt{\frac{c^5}{\varepsilon_0 \hbar G^2}} \approx 8 \times 10^{53} \, \mathrm{T}, \tag{1} \\[12pt]
B_2 &= \frac{c^3}{G e} \approx 3 \times 10^{54} \, \mathrm{T}. \tag{2}
\end{align}
Which one is really the Planck magnetic field?
While $B_2$ is simpler, I suspect it should be $B_1$, because it doesn't use the electric charge unit. $e$ is not exactly as universal as $\mu_0$. $B_1$ uses the Planck constant, so it's consistent to call it a Planck "unit", while $B_2$ doesn't use that constant. Also, because of the square root, $B_1$ is a bit more of the same shape as the Planck length :
\begin{equation}\tag{3}
L_{P} \equiv \sqrt{\frac{\hbar G}{c^3}}.
\end{equation}
The Planck units are presented on wikipedia: https://en.wikipedia.org/wiki/Planck_units but it doesn't tell anything about the magnetic field.
We could also argue that $B_1$ is the answer because we can find it by equating the magnetic field energy density with the Planck density (dropping all the dimensionless constants) :
\begin{equation}
\frac{B_1^2}{2 \mu_0} = \frac{M_P c^2}{L_P^3}.
\end{equation}
But then, we could also find $B_2$ by equating the Planck cyclotron angular frequency with the Planck energy :
\begin{equation}
\hbar \omega_{\text{cyclotron}} \equiv \hbar \, \frac{e B_2}{2 M_P} = M_P c^2.
\end{equation}
Both methods are arbitrary.
So what is the Planck magnetic strength?
| Planck units are found simply by multiplying together powers of certain constants; one does not consider specific physical laws to get them, which is equivalent to motivating specific multiplicative constants. (We don't do it this way because setting each Planck unit to $1$, the ultimate goal of having Planck units, would be impossible on a law-based approach.)
Coulomb's constant $k_C=\frac{1}{4\pi\varepsilon_0}=\frac{\mu_0 c^2}{4\pi}$, which appears in an inverse-square law the same way $G$ does, is a Planck unit just like $G$. Thus in Planck units $\frac{\mu_0}{4\pi}=1$, so the Planck unit you want is $\frac{B_1}{\sqrt{4\pi}}$. It definitely isn't $B_2=\frac{c^3}{G\sqrt{\alpha}q_P}$, with Planck charge $q_P=\sqrt{4\pi\varepsilon_0 c\hbar}=\sqrt{\frac{c\hbar}{k_C}}$.
| {
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How can a monochromatic X-Ray tube produce a spectrum in XPS? I thought that if we use monochromatic source, we can only get one peak, if it exists.
Because Photoelectric Effect allows only the electron that have the corresponding frequency(energy level) that can be excited.
appreciate your help to rectify my understanding.
| Because the X-rays, in the Photoelectron spectroscopy device, excite the electrons from the individual core levels out and as we know the electrons from the individual core levels reside in different energy levels. Thus the amount of energy left after the electrons overcome the individual energy barrier of each level is different since they derive all their energy from a single source (The single source). The equation goes like this
$$KE=h\nu -E_f-E_b$$
Where $E_b$ is the binding energy of every electronic level under consideration and $E_f$ is the Fermi level height of the material under scrutiny with respect to vacuum. Since $E_b$ is different for different electronic levels, hence $KE$ of the electrons are different which result in different peaks.
| {
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Expression for angular friction Consider say a door rotating about its axis. Is there, in general, any expression for the frictional hindrance to its motion? I was thinking in line with the coefficient of friction for linear motion on a surface.
| Friction only appears to stop the motion. It will not point radially but rather backwards on the sliding area. That is, it will point opposite to the door swinging direction and not towards the door hinge. Because nothing is moving/sliding in the radial direction. So, the concept of an "angular friction" is nothing more than usual "linear" friction. The only difference is that this usual "linear" friction then causes a torque that counter-acts the swinging - but neither the kinetic friction formula $f_k=\mu_k n$ nor the torque formula $\vec \tau=\vec f_k \times \vec r$ are any new particular "angular" friction expressions.
Since this usual "linear" kinetic friction $f_k=\mu_k n$ does not depend on sliding speed and also not on contact area, then for a rigid door, it will be a constant force regardless of swinging speed or size of hinge. This force will then be applied at some averaged position. If the whole bottom of the door slides, then applied half-way; if only the hinge slides, then applied half-way across the hinge.
So, friction itself does not change, but if you can concentrate it near the rotation point (the hinge), then the counter-torque caused by this friction will be small and not have a large influence.
| {
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Definition of symmetrically ordered operator for multi-mode case? As I know, Wigner function is useful for evaluating the expectation value of an operator. But first you have to write it in a symmetrically ordered form. For example:
$$a^\dagger a = \frac{a^\dagger a + a a^\dagger -1}{2}$$
For single mode case where there is only one pair of creation and destroy operator the symmetrically ordered operator is defined. But for multi-mode case,how is it defined? For example, how would we write
$$a_1^\dagger a_1 a_2^\dagger a_2$$
in a symmetrically ordered form (such that we could easily evaluate its expectation value using Wigner function)?
| OK, I assume you are comfortable with the rules of Weyl symmetrization where you may parlay the $[x,p]=i\hbar$ commutation relation to the $[a,a^\dagger]=1$ one... the combinatorics is identical provided you keep track of the is and the ħs, etc...
So your surmise is sound. Since different modes commute with each other,
they don't know about each other, and you Weyl-symmetrize each mode factor separately.
So, e.g., for your example,
$$
a_1^\dagger a_1 a_2^\dagger a_2= a_2^\dagger a_2 a_1^\dagger a_1 = \frac{a_1^\dagger a_1 + a_1 a_1^\dagger -1}{2} ~ \frac{a_2^\dagger a_2 + a_2 a_2^\dagger -1}{2},
$$
etc.
All you need to recall is that each mode is in a separate (Fock) space of a tensor product, so all operations factor out into a tensor product of Weyl-symmetrized factors.
This is the reason all multimode/higher-dim phase space generalizations of all these distributions functions are essentially trivial, and most subtleties are routinely illustrated through just one mode.
Edit in response to comment on entangled states. One of the co-inventors of the industry, Groenewold, in his monumental 1946 paper, Section 5.06 on p 459, details exactly how to handle entangled states--in his case for the EPR state. The entanglement and symmetrization is transparent at the level of phase-space parameters (Weyl symbols): the quantum operators in the Wigner map are still oblivious of different modes. What connects/entangles them, indirectly, are the symmetrized δ-function kernels involved, even though this is a can of worms that even stressed Bell's thinking. The clearest "modern" paper on the subject is Johansen 1997, which, through its factorized Wigner function and changed +/- coordinates, reassures you you never have to bother with the quantum operators: the entangling is all in the Wigner function and phase-space, instead! (Illustration: 351884/66086.)
| {
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Could a computer unblur the image from an out of focus microscope? Basically I'm wondering what is the nature of an out of focus image. Is it randomized information? Could the blur be undone by some algorithm?
| The goal with a camera lens system (whether a microscope or not) is to deliver light from one point on the object to one point on the sensor. This however cannot be perfectly achieved for several reasons.
*
*Light will diffract off the apeture, the smaller the apeture the more the diffraction.
*The system will only be perfectly in focus for an infinitely thin plane. The larger the apeture the faster the focus will drop off with distance from that plane.
*The lenses themselves will be imperfect.
All of these effects act to attenuate high (spacial) frequency components in the image. We call this blurring.
To reverse blurring we must first characterise the blurring. Characterising the blurring requires making some assumptions about what is in the image and/or the characteristics of the optical elements causing the blurring.
Once we have a mathematical function discribing the blurring process we can calculate an inverse of that function and apply it to the image.
However there are limits to how much we can mitigate the impact of blurring. Firstly because our mathematical model of the blurring is only an approximation and secondly because the inversion process inevitablly amplifies noise in the image.
| {
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Calculating magnetic fields by electric fields Is it possible to calculate the magnetic field due to moving charges by considering a reference frame in which they are stationary, finding electric field and then somehow relating it to the magnetic field in the reference frame where they are moving?
| Yes. To do this, you need to know the Lorentz transformations between the "lab frame" and the "charge frame", as well as how electric and magnetic fields transform between these reference frames.
To get more specific, we will denote quantities measured in the "lab frame" with primes, and those in the "charge frame" without primes. We can also assume that these frames agree on the orientation of their spatial axes (i.e., they think that the $x$-, $y$- and $z$-axes point in the same directions), and that the charge frame is moving in the $+x$ direction at speed $v = \beta c$ relative to the lab frame. Under these assumptions, then the coordinates transform as
\begin{align*}
c t &= \gamma (c t' + \beta x') \\
x &= \gamma (x' + \beta ct') \\
y &= y'\\
z &= z'
\end{align*}
while the fields transform as
\begin{align*}
E'_x &= E_x & E'_y &= \gamma(E_y + \beta c B_z) & E'_z &= \gamma (E_z - \beta c B_y) \\
B'_x &= B_x & B'_y &= \gamma(B_y - \beta E_z/c) & B'_z &= \gamma (B_z + \beta E_y/c)
\end{align*}
If all the charges are at rest in the "charge frame", then we have $\vec{B} = 0$. In principle, then you can calculate $\vec{E}(x,y,z,t)$ in the charge frame, and then use the above transformations to find $\vec{E}'(x', y', z', t')$ and $B(x', y', z', t')$.
If you want to see how the field transformations are derived, I highly recommend Purcell & Morin's Electricity and Magnetism (3rd ed.; previous editions are by Purcell only.) In Chapter 5, this text presents a beautiful argument that magnetic forces must exist given the existence of electric forces and the principles of special relativity. The full transformation laws for the electric and magnetic fields are derived a bit later, in Section 6.7.
| {
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Why are the metric and the Levi-Civita tensor the only invariant tensors? The only numerical tensors that are invariant under some relevant symmetry group (the Euclidean group in Newtonian mechanics, the Poincare group in special relativity, and the diffeomorphism group in general relativity) are the metric $g_{\mu \nu}$, the inverse metric $g^{\mu \nu}$, the Kronecker delta $\delta^\mu_\nu$, and the Levi-Civita tensor $\sqrt{|\det g_{\mu \nu}|} \epsilon_{\mu \nu \dots}$. (Note that $\delta^\mu_\nu = g^{\mu \rho} g_{\rho \nu}$, so only two of the first three invariant tensors are independent). This result is extremely useful for constructing all possible scalar invariants of a given set of tensor fields, but I've never actually seen it proven. How does one prove that there are no other invariant tensors?
| The rotation group ($O(3)$ for Euclidean space, $O(1,3)$ for Minkowski signature) is defined as the group of all linear transformations that preserve the metric tensor. If there were other independent tensors, this would give additional restrictions, so it would necessarily define a subgroup of the rotation group.
Demanding that the volume form (~ the Levi-Civita) is also preserved indeed restricts to the special rotation group.
| {
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How could I measure the colour spectrum of a light bulb and investigate how closely it matches a black body radiation curve? Here is my research question:
What is the colour/spectrum produced by each globe type? What is the temperature equivalence? How closely does a globe match a black body radiation curve?
I will be testing this on halogen, compact fluorescent, and LED bulbs. How could I measure the colour specturm of the bulb? I have been suggested to take photos of the glowing bulb then use photoshop to analyse the colour. Is that a possible solution?
How could I match it to a black body radiation curve? Would I have to plot it out? From research it doesn't seem like a curve that could be hand-drawn, especially CFL and LED:
|
I have been suggested to take photos of the glowing bulb then use photoshop to analyse the colour. Is that a possible solution?
No. The power spectrum of a light bulb is a continuous function $f(\lambda)$ where $\lambda$ is a particular wavelength of light and $f(\lambda)$ is the intensity of the light at just that one wavelength.
A camera throws most of that information away, and reduces a color to just three numbers. That's because your eyes have just three different "color receptors," and almost every color that you are able to distinguish can be mimicked by combining different levels of a red, a green, and a blue light source. The camera, therefore, records the level of light coming through a red filter, a green filter, and a blue filter at every pixel position.
Instead of directly photographing your different light bulbs, you could try photographing them through an inexpensive spectroscope
| {
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Quantifying the baryon asymmetry requires normalizing the baryon number density w.r.t comoving entropy density or photon density-Why? This question has been edited a little to clarify the confusion I have.
The matter-antimatter asymmetry of the Universe quantified by the baryon asymmetry as
$$Y_B=\frac{n_B-n_{\bar{B}}}{s}=\frac{n_B}{s}$$
i.e., the difference in the number densities of baryons $n_B$ and antibaryons $n_{\bar{B}}$ normalized w.r.t the comoving entropy density $s$. Sometimes the baryon assymetry is also expressed by normalizin the baryon number density w.r.t the comoving photon density $n_\gamma$.
What is the significance of normalizing the baryon asymmetry w.r.t $s$? Does it have anything to do with the fact that $sa^3$ is conserved (where $a$ is the scale factor)?
Addendum Why is it that people don't quote simply $(n_B-n_{\bar{B}})$ i.e., the number density of baryons (more precisely, that of baryons minus antibaryons) as the baryon asymmetry? Why do we have to normalize it w.r.t either the photon density $n_\gamma$ or the entropy density $s$?
I understand that as the Universe expands, this number density dilutes. Does it mean that the normalized quantity $Y_B$ don't dilute? If yes, is this the reason for normalizing it?
| You have to normalize it to something because $n_B - n_\bar B$ is just the number of baryons in the universe.
I more frequently see the baryon asymmetry normalized by the number of CMB photons (e.g. this PDG table.
The Particle Data Group's review of Big-Bang cosmology says on page 12
For photons [...]
$$
d(sR^3)/dt = 0.
$$
For radiation, this corresponds to the relationship between expansion and cooling, $T \propto R^{-1}$ in an adiabatically expanding universe.
Note also that both $s$ and $n_\gamma$ scale as $T^3$.
I interpret this to mean that your entropy-scaled baryon asymmetry and my photon-scaled baryon asymmetry are simply proportional to one another, for just the reason that you guessed in your question.
| {
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Mass connected to spring on a frictionless surface vibrating out of control I am not good at Physics so please bear with me. I am trying to understand harmonics and in order to do that I thought its best to ensure my understanding of the equations of motion is correct. To do this I tried to model a mass vibrating on a frictionless surface with restoring force from a spring when acted on by a force F. The parameters are as follows:
M = 5 kg (Mass)
F = 10 kN (Initial force)
k = -1kN/m (Spring constant)
I expected the plot of displacement against time to be uniform i.e without dissipation (or no damping). However, the result shows that the displacement are increasing. Have I made a mistake, if not, could someone explain the phenomena.
The plot is shown below
Algorithm is as follows:
F = 10kN
m = 5kg
k = -1kN/m
initial acceleration, a0 = 10/5 = 2 m/s
Assuming displacement is positive to the right
At t = 0,
u = 0
v = 0
s = 0
dt = 1
a0 = 2
ar = 0
anet= a0-ar
While (t<100),
ds = u*dt + 0.5anet*dt^2;
s = s + ds;
v = u + anet*dt;
t = t + dt
acceleration by restoring force, ar= (k*s/m);
updating net acceleration, anet = anet + ar; //K is already negative
Next;
Update:
I plotted the same graph at different time steps.
TimeStep=0.1s
TimeStep=0.005s
| When working with a situation like this (i.e. time-varying force), it's best to start with the differential equations
\begin{align}
\frac{{\rm d}x}{{\rm d}t}&=v \\
\frac{{\rm d}v}{{\rm d}t}&=F/m
\end{align}
rather than trying to use positions and velocities from kinematics.
The above can be re-witten using a time-stepping technique called leapfrog integration, in which you update positions and velocities at offset intervals:
\begin{align}
x^{n+1} &= x^n + v^{n+\frac12}\Delta t \\
a^{n+1} &= F(x^{n+1}) \tag{1}\\
v^{n+\frac32}&=v^{n+\frac12} + a^{n+1}\Delta t
\end{align}
Here, the fractional $n$ can be thought of the "cell wall" value (e.g., $x_{i+\frac12}=\frac12(x_i+x_{i+1})$) but in time instead of space.
Or you can use a modification called verlocity verlet. This turns (1) into a multi-step process:
\begin{align}
a_1 &= F\left(x^n_i\right)/m \\
x^{n+1} &= x_i^n + \left(v_i^n + \frac{1}{2}\cdot a_1\cdot\Delta t\right)\cdot\Delta t \tag{2}\\
a_2 & =F\left(x^{n+1}\right)/m \\
v^{n+1} &= v_i^n + \frac{1}{2}\left(a_1+a_2\right)\cdot\Delta t
\end{align}
Both integration methods are reasonably simple to implement, requiring definitions for forces and some vectors. The latter of the two is more often the algorithm-of-choice due to it's symplectic nature.
The implementation for (2) is basically
while t < t_end
a1 = Force(x) / m
x += (v + 0.5 * a1 * dt) * dt
a2 = Force (x) / m
v += 0.5 * (a1 + a2) * dt
t += dt
output t, x, v
where dt is your time-step (ought to be small, 0.005 should suffice for your values).
| {
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How is an operator applied to a wavefunction in quantum mechanics? If you have the Hamiltonian operator written as such:
$$\hat H = -\frac{\hbar}{2m}\frac{1}{r}\frac{\partial^2}{\partial r^2}r \tag{1}$$
then to apply the Hamiltonian operator to a wavefunction, do you apply the separate operations in order from right to left, as in:
$$\hat H \Psi = -\frac{\hbar}{2m}(\frac{1}{r}(\frac{\partial^2}{\partial r^2}(r \Psi))) \tag{2}$$
and so on?
At first I thought you applied each operation to the wavefunction separately and then multiplied all the sections together but this seems rather clunky.
Forgive me for asking such a clueless question but I cannot find the answer anywhere!
| It is the same as matrix mathematics. In general quantum mechanics is linear algebra in funny hats.
That is, suppose I want to compute $\frac 12 \langle \hat X \hat P + \hat P \hat X\rangle,$ the closest Hermitian observable to the moment $\langle x p \rangle$ in classical mechanics. The relation that $[\hat X, \hat P] = i\hbar$ is usually taken to mean that in the position basis, $\hat X = (x\cdot)$ while $\hat P = -i\hbar \frac{\partial}{\partial x},$ so the first part of this integral is:$$\langle \hat X \hat P\rangle = -i\hbar \int_{-\infty}^\infty dx~\Psi^*(x)\cdot x\cdot\frac{\partial\Psi}{\partial x},$$ while the second part is $$\langle \hat P \hat X\rangle = -i\hbar\int_{-\infty}^\infty dx~\Psi^*(x)\cdot \frac{\partial}{\partial x}\big(x \cdot \Psi(x)\big).$$
The juxtaposition $\hat X~\hat P$ is really a sort of operator composition, just like how the matrix multiplication between two matrices forms a composition of the transforms that they describe.
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Can it happen that angular momentum is conserved about some points but not others? So angular momentum is conserved about a point if no external net torques act about that point. But is there any occasion when this is only true about certain points? In other words: can it happen that angular momentum is conserved about some points in a system but not others?
| As you already said, angular momentum about a point is conserved if and only if the net external torque about that same point is zero. This is always true.
It might happen though that the net external torque is zero about one point and non zero about some other. In this case angular momentum is conserved about the first point whereas it is not about the second. As an example, consider a particle in free fall. The torque due to gravity about any point in the vertical line containing the particle vanishes and therefore the angular momentum about that point is conserved. For the same system, consider a point belonging to another vertical line which does not contain the particle. The torque is non zero so the angular momentum is not conserved.
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What would a Helmholz coil-like mass configuration look like? (produces locally uniform gravity field) A Helmholtz coil is an arrangement of two circular coils that produces a magnetic field in the center which is locally uniform in direction and magnitude, or at least nearly so. The configuration is optimal when the radius of each coil is equal to the separation between coils.
If the coils were replaced with massive rings, would this also produce a locally uniform gravitational field in both direction and magnitude? Or would a different diameter to separation be better?
Is there a well-recognized name for this configuration of masses?
| Fields scale as $M/r^2$, so scaling a structure's dimensions up by $\alpha$ and decreasing its mass density by $\alpha$ leaves the field unchanged. So the field at the center of a cylinder whose diameter increases linearly along its length (I initially incorrectly suggested exponential) and whose mass-density is decreasing inversely with length will have the same constant non-zero field everywhere along its length.
A linear change with length means, of course, that its a simple cone. We can expand the angle at the vertex until the cone becomes a sphere. So if you take a planet with a hole through the middle that happens to have a mass density inversely proportional to radius, then the field will be constant and reverse sign at the center. In fact, the hole doesn't have to go through the middle. The field everywhere has a constant magnitude, only its angle changes. Actually I read somewhere that the earth's density has a slight tendency towards a 1/r dependence and the field in the first 1000 miles or so doesn't change much, but I don't have a reference.
If we talk about a conical shell then the area mass density has to be constant for the constant field scaling to apply. So next time you eat an ice cream cone please be aware that its gravitational field along its axis is roughly constant.
Thinking of a simple structure more analagous to the Helmholz configuration, the field along the axis of a ring radius R goes as $GMx/(R^2+x^2)^{3/2}$. If we place the point mass, $m$, on the axis at distance, $d$, from the ring, the total field is $GMx/(R^2+x^2)^{3/2} + Gm/(x-d)^2$. We can now choose $m$ and $d$ and a position $p$ to set the first, second, and third derivatives to zero. I had to solve this numerically. The answer comes out to be m/M = 1.75620, d/R = -1.41456, p/R = 0.202622. The resulting field is 0.862275*GM/R^2
| {
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Different expressions of cylindrical EM waves if derived from one dimensional or three dimensional wave equation? Consider electromagnetic cylindrical waves. Cylindrical waves can be derived from the plane waves using energy conservation consideration: since the power must be a constant the amplitude of a cylindrical wave must decrease with $\sqrt{r}$. Therefore a cylindrical wave expression must be
$$\mathbf{E}(r,t)=\frac{\mathbf{E}_0}{\sqrt{r}} \mathrm{sin}(kr-\omega t)$$
The function $\sqrt{r} \mathbf{E}(r,t)$ satisfies one dimensional wave equation
$$\frac{\partial^2\xi}{\partial r^2}-\frac{1}{c^2}\frac{\partial^2\xi}{\partial t^2}=0$$
In complex notation the cylindrical wave becomes
$$\mathbf{E}(r,t)=\frac{\mathbf{E}_0}{\sqrt{r}} e^{j(kr-\omega t)}\tag{1}$$
If we call $\xi$ a generic component of $\mathbf{E}$, the three dimensional wave equation is
$$\nabla^2\xi-\frac{1}{c^2}\frac{\partial^2\xi}{\partial t^2}=\square \xi=0$$
The solution in cylindrical coordinates is
$$\xi (r,\phi , z,t) =\sum_{\omega,n,h} R^{0}_{\omega, n, h } H_n\Bigg(r \sqrt{\frac{\omega^2}{c^2}-h^2}\Bigg) e^{j(n\phi +hz-\omega t)} \tag{2}$$
Where $R^{0}_{\omega, n, h }$ is a (complex) constant and $H_n$ is the Hankel function of order $n$.
Under the assumption of cylindrical symmetry of the wave, that is
$$\frac{\partial \xi}{\partial \phi}=0 \,\,\,\,\,\, \mathrm{and} \,\,\,\,\,\, \frac{\partial \xi}{\partial z}=0$$
the asymptotic approximation of $(2)$ (for $r >> \frac{c}{\omega}$) lead to a field that is the same as $(1)$.
My question is: why (under cylindrical symmetry) is $(2)$ equal to $(1)$ only at large distances?
I always thought that $(1)$ gives the expression of a cylindrical wave in all the circumstances. So is $(1)$ "wrong" for small $r$? Or are $(1)$ and $(2)$ describing two different things? If so, what are the differences?
(I have an identical doubt for spherical waves).
| Also: consider the concepts of near and far field. The exact solution contains both. In the far field, only traveling waves exist ($1/r^2$ power for spherical cases) that asymptotically approach plane waves. Near field falls off faster (hence: "near"), moreover there may be phase differences with the driving oscillation--power can go into to field and back to the antenna, for example. I believe the credit-card chip is a near field device, and is hence more secure than RFID--which radiates your information.
| {
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Is Bose condensation experimentally achieved for a free Bose gas? Bose-Einstein condensation is a phenomenon that theoretically happens for free Bose gases because Bose-Einstein distribution holds for free Bose gases. But in superconductors and superfluids, there are interactions between the bosons.
*
*Are superconductors and superfluids really Bose condensates?
*In reality, is it possible to see the Bose condensation in a system of free Bosons?
| Yes, Einstein considered a purely statistical effect, where condensation occurs although interactions are absent. However, for most people the accumulation of particles in the ground state is the key point of a Bose-Einstein condensate. Therefore, most people agree that atomic, molecular, exciton-polariton and photonic Bose-Einstein condensates have been realized. They adopt the modern definition of a BEC (link).
So, to answer your first question, most BEC experts consider superfluid helium as a BEC. In order to convince yourself, you could read the introduction of the Nobel prize speeches.
The second question is harder to answer, because it depends on your definition of "free bosons". E.g. if you work with an atomic BEC, you can use a so called Feshbach resonance to switch off 2-body interactions. However, since you always have some fluctuations in your B-fields, this won't be perfect. Furthermore, the BEC needs to be confined in space. Hence, even if you switch off all 2-body and higher order interactions, the BEC still won't be truly free.
| {
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Traveling Wave Equation $\sin(kx-wt)$ vs $\sin (wt-kx)$ In my textbook most if the times it uses $A\sin(wt-kx)$, but occasionally there is a problem using $A\sin(kx-wt)$
So i just changed it from $A\sin(kx-wt) \to -A\sin(wt-kx)$
but does the amplitude change to $-A$?
Is the wave going downwards first?(as negative amplitude would imply)
| Using $\sin(kx-\omega t)$ or $\sin(\omega t-kx)$ does not make a difference: it is just a matter of convention. You can very legitimately change $A\sin(kx-\omega t)$ to
$-A\sin(\omega t- kx)$ without causing any damage.
Note that the full solution to the wave equation is of the form $A\sin(kx-\omega t+\phi)$. Expanding:
$$
A\sin(kx-\omega t+\phi)= A\sin(kx-\omega t)\cos\phi+A\cos(kx-\omega t)
\sin\phi
$$
and choosing $\phi=\pi$ transform
$$
A\sin(kx-\omega t+\phi)\to -A\sin(kx-\omega t)=A\sin(\omega t -kx)
$$
Hence, the solution $ A\sin(\omega t -kx)$ is just "shifted" by one half of a cycle compared to $A\sin(kx-\omega t)$: this shift does not change the wave.
Finally, the amplitude $A$ is usually defined to be a positive number, and it is one half of the difference between the maximum and the minimum reached by a wave of the type $\sin(kx-\omega t)$. If the wave is "going down first" then it might be better to have an explicit negative sign in front of $A$.
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Does amplitude affect time period for spring mass system? I know that with the formula $T=2\pi\sqrt{\frac{m}{k}}$ the time period is not related to the amplitude. However, would amplitude matter if i do this experiment in real life. Would a greater amplitude result in more friction of some sort?
| $2$ $pi$ frequency is root of return force/(inertia- displacement). A real spring will also have nonlinear terms in return force. The return force can be expanded in powers of displacement such as $F = ax + bx.x + cx.x.x ..$.Where $x$ is the displacement and $a,b c$ etc are constants. $b < a, c < b$ etc. The nonlinear terms, that is those involving $b,c$ etc can be neglected for small oscillations. $2$ $pi$ frequency then takes the form as the root of $a/m$ . $m$ is the mass. This is constant. Note that as the displacement is made to increase by increasing the amplitude of oscillation terms involving $x$ square, $x$ cube etc become larger and cannot be neglected. The frequency will then depend on amplitude.
| {
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Do massive particles redshift the same way as photons in a gravitational field? Let's assume two observers $A$ and $B$ hovering in a gravitation field. $A$ sends a radio transmission of frequency $f_1$ to $B$. $B$ receives this transmission and finds it has frequency $f_2$.
As as second experiment $A$ sends an electron beam to $B$. They measure the energies of these electrons on emission and reception.
Particles have de Broglie frequency which is proportional to energy.
Will this frequency gets redshifted the same way as photon frequencies in a gravitational field so the rate of the original and redshift frequency/energy will be the same as in the case of photons?
In other words, for example, if a 900 keV photon, fired from $A$, gets red shifted to 850 keV when it arrives at $B$, will any massive 900 keV particle get slowed down to 850 keV after it free falls to $B$? - Given the rest mass is smaller than 850 keV, otherwise it would just fall back and never reach the other observer I guess.
I worked it out in flat spacetime between two accelerating observers that keep fixed distance between them, and $A$ just drops particle and $B$ just catches it. And in that case it seems the rate of the total energy of the received and dropped particle is exactly the rate of acceleration induced time dilation between the two observers. I'm unsure if this only works this way in this specific case or if it works in general in any gravitational or fictitious force fields.
| Case 1: Convert 1 kg of matter-antimatter to energy
Case 2: Lower 1 kg of matter-antimatter to a gravity well, convert it to energy there, beam the energy up to the original position.
From conservation of energy: Energy produced in case 1 = Energy produced in case 2
Energy in case2 = Energy beamed up + energy generated during lowering
Energy beamed = Energy of 1 kg of matter - energy lost in redshift
It must be so that:
Energy lost in redshift = energy generated during lowering
Also: Energy generated during lowering = energy lost, or used, when the lowered thing is lifted back to the original position
So energy lost, or used, when the matter-antimatter is lifted back to the original position = energy lost in redshift
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Absorption of light I have two questions:
*
*When an atom absorbs part of spectrum from white light, why doesn't it radiate the photon back? What happens to the photon?
*How does the wave theory of light explain absorption of light?
| If an atom absorbs a photon, an electron will be raised to an excited state. For the relaxation of the excited state, we have basically two possibilities: (1) The electron jumps back to initial state where it started. Thereby it emits the "same" photon (same wavelength) it absorbed initially. However, it doesn't have to be in the same direction as the initial photon. (2) The electron jumps to a different state. Hence, a photon of different wavelength will be emitted.
For you second question look up Beer–Lambert's law: It describes an exponential decrease of the intensity.
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Why do objects feel heavier when held with an extended arm than with a bent arm? I do realise that this is due to torque and that torque is at maximum when the angle between the direction of the torque and the force acting on the object is 90 degrees. I would like to know if it is the angle between the forearm and the elbow that causes the situation I described in the question or whether if it's due to something else.
| I'd also say that this is due to torque. It is not the angle at the elbow. Take some weight in your hand and let it hand straight down from your shoulder. Then pull it up to your armpit. Holding it with an extended arm will be easiest because you only need to apply force in your hand.
When you have the weight in front of you, same height as the shoulder, stretched arm, it will feel rather heavy. You can rotate your elbow downwards such that it locks. Now you do not need any force around your elbow joint, but your shoulder has to counter all the torque that the weight exerts on you via your arm. The muscles are attached via tendons rather close to the bones, therefore they don't have much leverage. Therefore they need to exert a lot of force to keep the weight up.
Moving the weight closer to your body but keeping its height by closing your elbow will lessen the torque on the shoulder and therefore reduce the force that the muscles there have to apply.
| {
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Short pulse of monochromatic light? (Potentially connected to this question, but could not find the answer to my particular question there.)
The frequency spread and time duration of a pulse are related by:
$$ \Delta \omega \Delta t \approx 2 \pi, $$
from which perfectly monochromatic radiation ($\Delta \omega$ = 0) would require an infinite "pulse", $\Delta t \rightarrow \infty$.
Now: let's think of a (locked) CW laser, emitting a stable frequency with a linewidth of ~10s of kHz. Actually let's even assume 0 linewidth, let's assume it's ideal.
I have a shutter (or some other sort of switch) in the beam path, that goes ON and then OFF in a very short amount of time (100s of µs). Because of the finite duration of the pulse, I now have a spread in frequencies, following the Fourier relations.
So there are photons with a little bit more and a little bit less energy than originally. How? What's the interaction that allowed the reshuffling in energy?
| Well, when you bring photons into the question, you should also bring quantum mechanics. The shutter constrains the photon's position, thereby introducing uncertainty in its momentum [equivalently, energy]. Note that this is just $\hbar$ times your original statement.
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What causes the phenomena of a ring appearing around the shadow of an airplane So I was sitting in an airplane and I saw a ring appearing around the shadow of the plane on the clouds. What causes this phenomena? I've added an edited image, so the effect is more pronounced.
| This is called a glory.
You can tell it apart from a circular rainbow because of the thickness of the colours relative to the radius of the turn, the repetition of the colours, and the way it seems to appear on a cloud. It's much more consistent with pictures of a glory (rainbows have a much wider circle they form due to their mechanism).
It's an interference pattern caused by refraction of light in the small droplets.
I also meant to add this to my answer. What is the explanation of the glory (optical phenomenon)? (Diracology also linked in comments).
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How do magnetic cores (ferrites) guide magnetic field? As far as I know, magnetic materials have magnetic dipole moments which align when they are under influence of the outside magnetic field.
Basically they increase the magnetic field strength. But people also say that they guide magnetic field. Does the magnetic field strength decrease around them compared to the state before they were introduced near a coil? Do they really guide magnetic field, or just increase the magnetic field inside them so that magnetic field around them seems small, even though it hasn't decreased?
| Ferromagnetic materials add their own magnetic field to the external field that causes an alignment of their atomic dipoles. If the external field is from a fixed current in a solenoid, the induced field has no effect on the external external field. If the external field is from another ferro-magnet, the induced field may modify the magnetization in the external magnet. If an external solenoid is part of an AC circuit, then the induced field will change the inductance of the solenoid and that will change the current and the external field.
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Does work done depend on the frame of reference? Suppose I am sitting on a bench and looking at a moving car. Force is applied on the car by its engine, and it makes it displace, hence some work is done on the car. But what if I am sitting in the car and looking at the bench? The bench covers some displacement, but who has applied force to it? Is any work done on it?
| In a friction free world, the car moving with a certain velocity will not have any forces acting on it. If there were forces acting on the car, it would be accelerating.
There can be multiple forces acting on a body, so it's really important to specify which force are we talking about when we are calculating the work.
$\mathcal{Case (i) - }$
When you are sitting on a bench, you will observe the car to be accelerating from your non-inertial frame of reference (frame at rest or moving with constant velocity). Since, the car is accelerating, there must be some force acting on it. Therefore, that force will do some work.
$\mathcal{Case (ii)-}$
Now when you are sitting in the accelerating car, an inertial frame, there will be pseudo force acting on all the objects you observe from your frame of reference. That pseudo force will certainly do work on the bench.
Although the engine will not doing any work on the bench.
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Neither a vector, nor a scalar While I was reading a book on mechanics, when introducing the vector multiplication the author stated that multiplying two vectors can produce a vector, a scalar, or some other quantity.
1.4 Multiplying Vectors
Multiplying one vector by another could produce a vector, a scalar, or some other quantity. The choice is up to us. It turns out that two types of vector multiplication are useful in physics.
An Introduction to Mechanics, Daniel Kleppner and Robert Kolenkow
The authors then examine the scalar or "dot product" and the vector or "cross product" (the latter not shown in the above link; can be seen on Amazon's preview) but seem to make no mention of any other method.
My concern is not about vector multiplication here, but what can be that quantity which is neither a scalar nor a vector. The author has explicitly remarked the quantity as neither a scalar nor a vector.
What I think is that, when we define a vectors and scalars, we propose the definition in terms of direction. In one direction is considered and in another it is not considered. Then, how can this definition leave space for any other quantity being as none of the two?
I would be obliged if someone could explain me if the statement is correct and how is it so. Also it would be great if you can substantiate your argument using examples.
| If you have two vectors $\mathbf{a}$ and $\mathbf{b}$, the inner product $\mathbf{a} \cdot \mathbf{b}$ is a scalar, the cross product $\mathbf{a} \times \mathbf{b}$ is a vector and the dyadic product $\mathbf{a} \otimes \mathbf{b}$ is a matrix. It is defined as
$$\mathbf{a}\otimes\mathbf{b} = \mathbf{a b}^\mathrm{T} =
\begin{pmatrix}
a_1 \\
a_2 \\
a_3
\end{pmatrix}\begin{pmatrix}
b_1 & b_2 & b_3
\end{pmatrix} = \begin{pmatrix}
a_1b_1 & a_1b_2 & a_1b_3 \\
a_2b_1 & a_2b_2 & a_2b_3 \\
a_3b_1 & a_3b_2 & a_3b_3
\end{pmatrix}
$$
It occurs a lot in the formalism of quantum mechanics where it is written as $|a \rangle \langle b|$ (using the so-called bra-ket notation by Dirac).
With regard to direction: if you apply a matrix to a vector, the vector may get stretched / compressed along multiple axes. So in contrast to a vector, a matrix involves multiple directions.
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How does a free electron look like? In a simple atom say hydrogen, there is an electron cloud which is spherical in shape. What about a free electron, how big or small will that cloud be? I think the term cloud here means the likelihood the electron can be found, as for a free electron in absent of other forces does it stay as a point-like particle or a standing wave spreading out across the universe?
| Since it's isolated (and therefore has no potential energy), the quick answer is that it resembles (but is not exactly) a Gaussian distribution on the order of 10 - 50 angstroms wide. You can calculate it yourself from the equation for thermal wavelength:
$$
\Lambda = \sqrt{\frac{h^2}{2\pi m k_BT}}
$$
At room temperature, using the mass of an electron we have $\Lambda = 4.3\times10^{-9}$ meters, or 43 Angstroms.
So it would "look" something like this, where red indicates a "denser" part of the "cloud" (though this 2D gaussian doesn't do justice to a 3D function):
This is a 3D gaussian. We have to use cross sections to see the probability density though:
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With respect to what we are saying space is homogeneous or space-time is isotropic? I don't really understand what we are talking about when we say space is homogeneous. What we are measuring? My notion is: it should depend on the entity and with respect to that entity one can decide space is homogeneous or not!
May be I'm asking stupid question, actually the fact is I don't understand the phrase "space is homogeneous" or "space-time is isotropic"
| Suppose we measure the density of matter as a function of position. Then we will come up with some function for the density $\rho(r, \theta, \phi)$ where $r$, $\theta$ and $\phi$ are polar coordinates with ourselves at the origin.
*
*The density is isotropic if it is the same in all directions i.e. it is independent of $\theta$ and $\phi$ and just a function of distance, $\rho(r)$.
*The density is homogeneous if we can move our origin, i.e. move to any other point in the universe and our density function $\rho(r)$ is unchanged.
The simplest such function is if $\rho$ is just a constant since it then automatically the same in all directions and the same at every point.
I've used density as an example, but this applies to any property of the universe that is a function of position.
Our universe is obviously not isotropic and homogeneous because in same places there is a vacuum while in others there are stars. However we know it started out as very close to isotropic and homogeneous from measurements of the cosmic microwave background radiation. And if we take average density on a large enough scale then we believe it is still on average isotropic and homogeneous (though this is a somewhat controversial issue).
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Thermodynamics : Doubt in First Law I just started my study of thermodynamics. I am having problem in uderstanding a concept.
Suppose I drop an ice-cube and a rock of cube shape. Which one will reach the ground first? I know ratio of Work to Heat equals Joules Constant. Therefore any change in potential energy will result in heat which in return will melt ice cube thus mass is changing. Will internal energy of ice cube change till it reaches the ground? If mass of ice cube is changing what will be the ratio of kinectic energy of ice cube to that of rock.
| In your assessment of the first law of thermodynamics, you omitted certain terms in the equation that can sometimes be important. The more complete form of the first law should read $$\Delta U+\Delta (KE)+\Delta (PE)=Q-W$$In your ice cube example, $$\Delta (KE)+\Delta(PE)=0$$So, if the ice cube doesn't exchange significant heat with its surroundings or do significant work on its surroundings while it is falling, $$\Delta U=0$$This means that no ice melts and, even if it was initially below the freezing point, there is no temperature change.
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Should every physical problem formulated as a differential equation have a mathematical solution? I encountered the following statement in Boyce's Elementary Differential Equations and Boundary Value Problems :
Not all differential equations have solutions; nor is the question of existence purely mathematical. If a meaningful physical problem is correctly formulated mathematically as a differential equation, then the mathematical problem should have a solution.
Is this true?
| In the best of all possible worlds we know the laws of physics perfectly and can write them as differential equations (or something similar). But we do not live in that world. Instead we create models of the physical world that may not correspond to the actual laws (due to ignorance or just approximation). Good models give informative predictions: there is a mapping between what happens in reality and the model that is close to a bijection, so we can use the model to predict physical responses. How close it has to be depends on the application.
Now, physics as far as we know never fails to produce a "solution" of what the future state of the world will be. But models clearly can fail at this, even models that are accurate in large domains. Depending on the application this might disqualify them - only use models that use differential equations that have solutions! - or be OK if model failures occur far away from the problems of interest. When the Schwarzschild metric in GR predicts geodesics that end up in a singularity it may not be a problem if one is working on orbits or stuff going on far away from the central mass.
The issue isn't whether physics "runs" on differential equations that always have solutions, but all about what properties a useful model should have in the case at hand. In short, it is not about differential equations but modelling problems well.
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Nature of Work done by Forces I am confused about the possible nature of work that a conservative and non conservative force can do.
*
*Do Non conservative forces like friction only do negative or zero work and not positive?
*Can conservative force also do negative work, I know Gravitational Force is conservative force and it can do negative work, but I read somewhere that net work done by conservative forces is positive.
Can someone please clarify?
| Work done $W$ is the dot product of a force $\vec F$ and the displacement of the force $\vec s$.
$W=\vec F \cdot \vec s$
The work done is positive if the force and the displacement are in the same direction and negative if they are in opposite directions.
You watch a lorry accelerating.
A box on the back of a lorry also accelerates and the kinetic energy of the box increases.
The only horizontal force on the box is friction and so it is the frictional force which is doing positive work on the box.
Ignoring air resistance, if you throw the box vertically upwards, the gravitational force on the box is downwards so the work done on the box by the gravitational force is negative and the kinetic energy of the box decreases.
Wait a little until the box starts falling and the the work done by the gravitational is then positive as the kinetic energy of the box is increasing.
When the box arrives at the place it started from just after it left your hand, the net work done by the gravitational force is zero and it has the same kinetic energy as at the start.
| {
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Why does refraction make something look closer with an incident angle of zero? The gaussian formula for a single spherical surface is defined as
$$ \frac{n}{s} + \frac{n'}{s'} = \frac{n'-n}{R}$$
When the radius of curvature, $R$, is infinite - as in the case of a planar surface, we are delt with the case
$$ \frac{n}{s} = \frac{-n'}{s'} $$
and thus
$$ s' = \frac{-sn'}{n} $$
Which means that if you look at an object in another medium with an incident angle of 0, the object may look closer or further away.
I understand why an object is not where "it seems" when looking into a different medium at an incident angle, but I dont understand why an object would appear farther or closer without and incident angle.
Why would an object appear farther or closer due to refraction when the incident angle is zero?
| Let's say you only have one eye open and you're directly in front of the object. The thing is, your pupil has some size to it. Rays that enter are not only leaving the medium surrounding the object at 0 degrees. Your lens must change shape for the rays entering the edges of the pupil to form an image on your retina, and the particular shape you force your lens into provides feedback on the distance.
Less eye-centric:
An observer will see an image located at the place where diverging rays appear to diverge from. It is common to think of the image existing without the observer, but in the end some type of optical device is needed if one wants to measure the location of an image. An object embedded in a medium will emit rays in all directions. Even if you imagine an observer or other device directly in front of the object, the apparent location can be determined by back tracing two or more diverging rays, because the observer or optical device isn't sampling just a single ray that's at 0 incidence. Lenses have size. It's not just a single ray, as you seem to be assuming. Take all the rays entering the optical device, trace them back, and you'll see the image and object are at different locations.
| {
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Time dilation for non-physicists Apologies in advance, as I'm not a physicist, and may use terms incorrectly.
In the movie Interstellar, the planet Miller has a time dilation of one hour to seven Earth years. This has brought up several questions for me:
*
*At what point would someone (outside the gravitational force surrounding Miller) begin experiencing the time dilation?
*How long would it take to send data from the surface to said someone in space outside Miller?
*Would a one hour audio clip recorded on Miller take seven years to play once transmitted?
| Re 1., gravity has infinite range, so the effect of gravitational time dilation will just grow weaker, but theoretically not go away entirely.
In the idealized case of a single source of gravity, the time dilation factor is given by
$$
\frac t{t_0} = \sqrt{\frac{1-\frac{r_s}{r}}{1-\frac{r_s}{r_0}}}
$$
where the Schwarzschild radius $r_s$ represents the mass of the source, $r_0$ the position of a clock close to the source and $r$ the position of an observer that sees the clock as having slowed down by this factor.
Re 2., transmission times would also increase by the factor above. If you want two-way communication, you also have to take into account the travel time of the signal. While I did the calculation, I'm not sure how interesting that expression would be for you (it's less nice than the term above).
Re 3., depends on your mode of transmission: I imagine you'd have to properly restore any digital signal before it could even be decoded, but once that's done, playback would happen without any slowdown. On the other hand, ignorant person that I am (ie without having looked at how radio actually works), I'd expect a direct playback of an analog radio signal would indeed be slowed by the factor above as well (and possibly otherwise distorted, in particular in case of frequency modulation). Also note that the carrier wave would be shifted towards longer wavelengths.
| {
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If we cool gates in semiconductor transistors, can we reduce the overall energy of electrons? I was reading about Moore's law and was a bit confused about why semiconductors aren't cooled down. If we have an issue with quantum tunnelling, cooling the wafer could reduce the energy level of the electrons. We might need more sensitive materials to achieve similar voltage differences but I'm still curious about why this is not a good idea.
Thanks!
| You are right that reducing the temperature will drive down leakage current, and this would lower energy waste and also increase signal fidelity.
To some extent this is already done, though not in the way you imagine. All transistors dissipate energy, so they tend to heat up. Usually heat sinks, fans, water cooling, etc. work to bring the chip back to room temperature. Otherwise the chip would burn itself out very quickly.
What is not done is to cool to significantly lower than room temperature. This is because cooling requires a lot of energy and is not very efficient, so it would make the computer chip even more energy inefficient overall. However, for some large scale things like servers and supercomputers, cooling is not uncommon because of the sheer scale of things
| {
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When particles are collided some new particles come into existence. Is the field of the new particle was there before just not vibrating? I know that when two particles are collided some new particles come into existence and i know that the new particles weren't inside of the original particles. I understand that the energy of the collision can vibrate other fields to make particles.
What I don't understand is "were those other fields nearby the collision site when the collision happen?"
| Think of tangled up wires on top of each other. Sometimes when you move one of them the right way it moves the other. Now let us say this movement is a vibration. You vibrate one and it vibrates the other. This is a cartoon picture of the mathematics. A disturbance in some field (particle A) can cause a disturbance in another field (particle B appearing seemingly from nowhere). At least according to the math.
| {
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Relation between velocity magnitude and the pitch of the collision sound I'm doing a simulation involving collisions, I'm using glass marbles, based on testing I've noticed that the higher the magnitude of the velocity of the marble hitting another marble at rest the higher the pitch of the collision.
With a small magnitude, let's say you move one marble carefully to touch another marble the sound will be lower as well as the pitch of the sound, all these tests are based on trial and error, there are no math or physics equations involved, my question is, is my claim true? Do any of you know the proper relation between these 2 variables?
| Neat question!
First some setup. The collision of marbles is within the domain of contact mechanics. When two spheres collide their surfaces both deform. This forms a circular contact area between them, with
$$ A \propto F_n$$
The normal force $F_n$ being dependent on the velocity of the collision.
Relying on the geometry of spheres we can also express the contact time $T$ of the two marbles as a function impact velocity, $v_0$:
$$T = k \frac{A}{r \ v_0}$$
In this case $r$ is the radius of the marble, and $k$ is a constant that depends on material properties.
In the end:
shorter contact time $\rightarrow$ higher frequency $\rightarrow$ higher pitch.
One additional note. The sound of the collision is not coming from the resonant modes of the marbles. Their natural frequency of vibration too high to be audible to us. (You can read more about that here: https://physics.stackexchange.com/q/376988)
If you want to read more, here's a great slideshow involving the sound of colliding balls, but in a slightly different scenario. A classic paper on this topic is Richards 1979 On the prediction of impact noise, I: Acceleration noise.
| {
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Why does a calculation to count objects covering a certain area seem to give nonsensical units? Suppose you want to estimate the number of atoms in a rectangular sheet of graphene. You might estimate the sheet to have $10^{7}$ atoms along one edge and $2*10^{7}$ atoms along the other edge. Multiplying while keeping track of units, we get
$$10^{7}\text{atoms} * 2*10^{7} \text{atoms} = 2*10^{14} \text{atoms}^2$$
But obviously, there are $2*10^{14}$ atoms, not $2*10^{14} \text{atoms}^2$. What is wrong with this calculation?
| Atoms are not really a unit. They don't combine in the right way when you multiply. If you are careful, you can make them work in some situations, as both other answers show (+1 to both).
But as you saw, they don't work everywhere. Not like a length would. If the distance from atom to atom is 1 Angstrom, there is no problem with an area of $10^{14}$ Angstrom$^2$.
So what is the unit? A count is dimensionless.
Even so, people often will use it as a unit where it works. You just have to be careful not to use it where it doesn't.
Physicists are sometimes sloppy in ways like this where mathematicians are much more careful. For example, a function can have a value of $0$ most places, but have a tall thin spike near $0$, so the area under the spike is 1. This is useful in some situations. Physicists find they need the spike to be infinitely narrow. So they created the Dirac delta function, which is $0$ everywhere except at $0$. The value at $0$ is infinite. The area under the spike is $1$.
A mathematician would find problems with such a "function" and say it doesn't exist. A physicist is careful to use it where it works.
| {
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Neutrinos always travel at same speed? Neutrinos have mass, and only interact with gravitional and weak force, what make it very difficult to detect.
Neutrinos is known to travel very closer to speed of light and have very small mass altought neither have a precise measure.
Particles with no mass like photons and gluons (altought there is no free gluon) must travel at speed of light (vacuum).
Particles that have mass must travel in any velocity lower than speed of light.
Dark matter can have hot particles (near speed of light) and cold particles (non relativistic velocity).
My questions are:
*
*Does neutrinos have always the same velocity?
*Is it possible to have neutrinos travelling with non-relativistic speeds?
*Why only referred neutrinos as possible hot dark matter and not to cold dark matter?
| The 25 detected neutrinos from supernova 1987A--168,000 light years away--arrived over a span of 13 seconds [Wikipedia, which is never wrong]. Not all neutrinos are this energetic, though.
| {
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What exactly happens in the basic QED Feynman diagram? When a photon is absorbed by an electron, I think that the following things happen:
*
*The electron changes in momentum, angular momentum and energy.
*The phase of the electron wave function changes by a fixed angle given by the coupling constant.
Is this correct? Is this complete?
| A Feynman diagram is not depicting a physical process. Virtual particles do not exist. Strictly speaking that particle is not 'a' photon. It's just there to signify that the interaction is mediated by the electromagnetic gauge field $A^{\mu}$.
Physically, what happens is that the electron is always travelling in an ares where it feels the influence of $A^{\mu}$, because the latter has infinite range. So it is not a free particle, but an eigenstate of the Hamiltonian including both free terms and $A^{\mu}$. Its state will therefore evolve into a different momentum state, and will acquire a phase shift, as if it scattered a single photon.
To calculate the maths of this, you would have to resort to perturbation theory, which is where all the confusion originates from.
Perturbation theory needs you to choose a set of 'unperturbed' states, usually taken as the free electrons i.e. solutions to the Dirac equation. The $A^{\mu}$ is then treated as a perturbation, and energy and wavefunction shifts can be computed to higher and higher level of accuracy with higher order perturbative terms. Each term in this series can be pictorially represented by a Feynman diagram, which is what makes them useful.
| {
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Do x-rays get emitted upon initial acceleration in this setup? Based on the diagram and the paragraph before it, the x-rays are caused by decelerating the moving electrons and thus x-rays are emitted where the diagram clearly shows it (the red arrows pointing downwards labeled "x-rays"). But does this mean that x-rays are also emitted near the cathode where the electrons are initially accelerated?
| The acceleration of the electrons whilst travelling between cathode and anode is very much less than the (negative) acceleration of the electrons when they hit the target atoms.
The higher the acceleration the more energetic are the emitted photons.
So the photons emitted from the interaction of the electrons with the target are much more energetic (in the X-ray region) than the photons emitted when the electrons are traversing the gap between the cathode and the anode.
The text hints at this when it states "X-rays are produced when the electrons are suddenly stopped in the anode" after mentioning the acceleration of the electrons between cathode and anode.
| {
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Where does the magnetic field energy of a charged particle moving with a uniform velocity come from? Consider a charged particle initially at rest with respect to an inertial frame.
Let a force act on it so that it gains a velocity 'v'.
It now produces a magnetic field that has some energy associated with it.
My question is where does this energy come from? If it comes from the work done by the force acting on the particle, does it mean that $ W = ΔKE $ is not valid in this case?
| You're right that the work done on the charge is not equal to the change in the charge's kinetic energy in this case. An accelerated charged body (let's assume it's of finite size, to avoid the infinite energy problem of point charges) will have a change in its "near field", and it will send off electromagnetic radiation. The work done by this external agency on the charged body will be equal to the total energy imparted to all three of these things:
$$
W = \Delta KE + \Delta E_\text{near field} + E_\text{radiated}.
$$
It is possible to account for this "deficit" in the resulting kinetic energy of the charge by defining a so-called radiation reaction force, which can in some sense be thought of as the force that the charged body exerts on itself as it accelerates. In this case, you still have $W = \Delta KE$, so long as you define $W$ to be the work done both by the external agency and by the radiation reaction force. However, this force has some weird properties (it's proportional to the time derivative of the acceleration, for one thing), which is why you usually don't hear about it in intro E&M classes.
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Analytical continuation of 2,3,4-point integrals I was reading a paper that gives a nice collection of all scalar integrals that crop up in QCD loop calculations. Such integrals are computed in some kinematic region and then the authors say the results may be analytically continued if so desired. I just wonder how is this analytic continuation done in practice? It's a relatively short paper and the url is https://arxiv.org/abs/0712.1851
The authors state that a particular kinematic region allows for the $i\epsilon$ to be dropped and then one can analytically continue results via the prescription $p_i^2 \rightarrow p_i^2 + i\varepsilon, s_{ij} \rightarrow s_{ij} + i\varepsilon, m_i \rightarrow m_i - i\varepsilon$. I just wonder why this is the case and if the sign choices here are significant?
As a simple example, the analytic continuation of the massive tadpole is given as $I_1^D(m^2) = -\mu^{2\epsilon} \Gamma(\epsilon-1) (m^2-i\varepsilon)^{1-\epsilon}$ but what should I do with this result as it contains an explicit $\varepsilon$ in $m^2-i\varepsilon$?
| This is significant: they are originate from the $+i\varepsilon$ prescription in the Feynman propagators of the original integral. The sign choices are important to obtain the correct sign for the imaginary part of the loop functions.
Note that these prescriptions are relevant only if evaluations are being made on a branch cut. Therefore, for your example, if $m^2 > 0$ (which is usually the case), the $-i\varepsilon$ can be dropped.
| {
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Derivation of Newtons second law of motion from the principle of conservation of energy Is newton's second law a consequence of the principle of conservation of energy? How can we arrive at
net force = rate of change of momentum
using only the law of conservation of energy?
| The connection is provided by the Hamiltonian formalism, if $F = -dV/dx$ is the net force acting on a particle of mass $m$, then the quantity
$$
H(x,p) = \frac{p^2}{2m} + V(x) = E
$$
satisfies the expression
$$
\frac{dp}{dt} = -\frac{\partial H}{\partial x} = -\frac{dV}{dx} = F
$$
| {
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How nuking the poles of Mars would create an atmosphere? Elon Musk said one of the methods they are taking into consideration to create an atmosphere on Mars is to nuke its poles. How nuking the poles would create the atmosphere?
| Please refer here for a similar question.
The idea of nuking the poles of Mars came around when people started to seriously consider the idea of terraforming another planet. If Mars's poles were to be nuked, the Ice Caps, mainly made of carbon dioxide, would vaporize. This vapor of carbon dioxide would be the first step in this process as a small atmosphere would form from this initial step. However, a larger atmosphere is achieved when you consider the greenhouse effect. The carbon dioxide would trap heat from the sun, heating up the ice caps and causing almost the entire Ice Cap to disappear and become a larger atmosphere. With carbon dioxide to warm the planet, trees could then be planted to chemically alter the atmosphere with photosynthesis producing oxygen that we can breathe.
This idea appears to work in theory if there is enough energy in the thermonuclear bombs, however, our current energies of these bombs would require tens of thousand of them to be dropped to release a small amount of carbon dioxide. Therefore this plan may not be feasible.
| {
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What is the physics behind a hard phone cover and a soft phone cover? Does hard plastic distribute forces throughout a device, similar to how the shell of an egg works, much better than a soft cover does? This is the common argument I hear for the hard cover.
How different is it from a soft plastic case? For soft plastic, I have the understanding that it increases time of impact, hence leading to a smaller force on the device when dropped.
| One has a lower Young's modulus than the other. That's it.
| {
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Volovik's argument and superconductivity In Volovik's book he describes the Fermi surface as a vortex in energy+momentum space. Due to a winding number the Fermi surface is topologically protected.
I don't understand how the above topological protection is compatible with superconductivity, which destroys the Fermi surface even for small attractive interactions. If it is a topological phase transition, there should be some type of gap closing, e.g. Fermi surface shrinking to points, which is seemingly not the case. Or is it that the pole in the Green's function still exists in a superconductor, although then I am wondering what Volovik's argument really says about the Fermi surface.
I am familiar with the terminology of Chern numbers, topological insulators etc. I would be very grateful if someone could explain this to me using that language, if possible.
| The picture is compatible with superconductivity. Because in the presence of the pairing term, we rewrite the Hamiltonian in the Nambu basis, where both particle and hole Fermi surfaces should be considered. If the particle Fermi surface corresponds to a vortex, then the hole Fermi surface corresponds to an antivortex. They coincide exactly in the energy-momentum space. The pairing term mixes the particle and hole together, allowing the vortex and anti-vortex to annihilate with each other, which destroys the Fermi surface. However, if we impose the $\mathrm{U}(1)$ symmetry, the vortex-antivortex annihilation will be forbidden and the Fermi surface becomes stable. So Volovik's argument works only in the presence of the $\mathrm{U}(1)$ symmetry.
| {
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Is the top of my ladder really reaching infinite velocity? Here is a classic "related rate" maths problems:
A $10$ ft long ladder rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of $1$ ft/s, how fast is the top of the ladder sliding down?
Let's represent the vertical wall as our $y$-axis oriented from bottom to top, and the ground being the $x$-axis oriented to the right. We call $x$ the distance from the bottom of the ladder to the vertical wall, and $y$ the distance from the ground to the top of the ladder. We are given that $\frac{dx}{dt}=1$ ft/s and we are looking for $\frac{dy}{dt}$.
Since $x^2+2y^2=100$, we have $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$, so $\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}=-\frac{x}{\sqrt{100-x^2}}$. Fine.
But, when the ladder finishes to slide, that is when $x$ is approaching $10$ from the left, $\frac{dy}{dt}$ is decreasing with no bound: $\displaystyle \lim_{x\to 10^-}\frac{dy}{dt}=-\infty$.
This obviously does not match with everyday observation.
What happens when the ladder finishes to slide?
| You are making the false assumption that the ladder remains in contact with the wall. In reality, contact will be lost and the velocity remains finite.
To solve this properly you would compute the rate of rotation of the ladder, and you would notice that there isn't enough torque to keep the ladder turning as fast as you need to keep it touching the wall.
| {
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What causes light to refract? The common answer to this question is that light refracts because its speed changes in different materials. But this means that the photons has internal attraction between each other. Is that the case? Otherwise is something else like density of different materials the reason why light refracts?
please give a more convincing explanation.
| The reason for refraction is the different speed of light in different media. This speed is determined by the refractive index $n$, which in turn can be calculated from the relative permittivity and permeability of the material:
$$n = \sqrt{\varepsilon_r \mu_r}$$
These constants simply define the response of a material to an electric or magnetic field. Light (being made up of $E$ and $H$ fields) isn't interacting with itself, but rather with the material it is transmitted through.
So, here it is more useful to look at light in the classical picture as an electromagnetic wave, and not as photons.
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How to change Pressure and Temperature "The triple point of a substance is the pressure and temperature at which all three phases coexist in equilibrium."
Take water for example.
Let's say I put some water in a rigid container and add some heat.
Then both the temperature and pressure rise because the volume is constant.
If the triple point occurs during the heating, then that implies $V=\frac{nRT}{P}$, where $(T,P)$ is the triple point.
Does that mean the triple point can only exist in that size of a container??
| Really interesting question, the answer is yes -- the triple point exists for closed thermodynamical systems. The key word here is being closed with some volume $V$ and non-interacting with the environment except for some stimuli. Every closed thermodynamical system has a natural thermodynamical equilibrium and each has an triple point unique to it inside the system.
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In quantum mechanics is momentum a vector or scalar? In quantum mechanics, is momentum a vector quantity?
| In quantum mechanics, momentum is a vector operator, but there is a subtlety. If we have a state $|\alpha \rangle$ rotated to $|\alpha' \rangle = \mathcal{D}(R)|\alpha\rangle$, for momentum to be a vector operator we must have that,
$$\langle \alpha | p_i |\alpha\rangle \to \langle \alpha'|p_i|\alpha'\rangle = \sum_j R_{ij}\langle \alpha | p_j |\alpha\rangle$$
under the rotation; in words, the expectation value transforms as an ordinary vector by a rotation matrix $R_{ij}$. It can be shown by considering an infinitesimal rotation,
$$\mathcal{D}(R)=1-\frac{i\epsilon \mathbf{J}\cdot \hat{\mathbf{n}}}{\hbar} $$
that a totally equivalent definition of a vector operator is that it satisfies the commutation relations,
$$[p_i,J_j] = i\hbar \sum_k\epsilon_{ijk}p_k.$$
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Does String Theory predict a particle with twice the mass of the electron? As far as I know, the spectrum of any (?) String Theory is of the form
$$
M^2\propto N
$$
where $N$ is the number operator. The lightest known particle being the electron, I am led to think that we should observe particles with masses $m_e,2m_e,3m_e,\dots$
In more general terms, the spectrum of ST seems to be harmonic, as the operators are always essentially oscillators. Cf. the Veneziano amplitude, with poles at $s=4(n-1)/\alpha'$. This brings me to my question: how does the phenomenology of ST deal with the (obviously?) unobserved tower of particles, with masses in harmonic progression? does the geometry of the extra dimensions have anything to do with the apparently erratic behaviour of low-energy physics?
| It's more complicated than that. To get the Standard Model, a standard technique is to use intersecting brane models, usually many D6s in the compact space, in which the spectrum depends on the angles between the branes and the particles comes from vibrational modes of the open strings stretched between the branes.
Roughly speaking, in superstring theory you will get:
$$\alpha' M^2 = N_B + N_F + f(\theta_i)$$
where $N_B$ and $N_F$ are the oscillator numbers and $\theta_i$ the angles between the branes in the compact spaces (think to factorized tori for simplicity), for some function f. While the angles are not completely arbitrary, due to supersymmetry constraints for instance, you can get a realistic spectrum in this way. The presence of multiple branes in each stack will give the different families of leptons and quarks.
A nice and simple introduction to the topic can be found in the classic book by Zwiebach.
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Can a particle have a decay width approaching its mass? Very short-lived particles have half-lives measured in terms of decay width, with the half-life calculated from the energy-time uncertainly principle.
If a particle has a decay width that approaches the mass of the particle, it never really gets a chance to "exist". No one will ever see it.
But could such a particle (or quasi-particle in condensed matter) "exist" in the sense that it has shows up as a virtual particle in Feynman diagram calculations, with a mass, spin, charge, etc?
The best candidate of such a particle I can think of is the Planck Particle, but quantum gravity isn't renormalizable.
| Well, "no one will ever see it" is excessively harsh. Very broad resonant states are surveyed with partial wave phase-shift analysis, although this sort of thing is more in vogue in nuclear physics these days than in particle physics. However, the fabled scalar σ resonance underlying the eponymous model of chiral symmetry breaking in effective lowest-energy QCD has made its way in and out of the PDG, and, today, it is in .
It has the quantum numbers of the vacuum: it is spinless, neutral, $I^G(J^{PC})=0^+(0^{++})$ :
the celebrated $f^0(500)$ with $\Gamma\approx 400-700$MeV, give or take...
Its Breit-Wigner distribution is now
$$
f(E) = \frac{cM^3}{\left(E^2-M^2\right)^2+M^4}~,
$$
where the normalization constant c is 4π/3, I think, from
$
\int_0^\infty \!\!dE ~~f(E) =1.
$
In these conventions, $f(0)=f(\sqrt{2} M) = f(M)/2$.
In any case, even though the relativistic Breit-Wigner arises from the propagator of the unstable particle, you are not always invited to do perturbation theory with it. For the σ this corresponds to an effective theory of the ferociously non-perturbative chiral symmetry breaking process, normally not even renormalizable. So the urge to dive into the Planck mass for illustration might be misplaced.
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Why are work and energy considered different in physics when the units are the same? There is a question that explains work and energy on stack exchange but I did not see this aspect of my problem. Please just point me to my error and to the correct answer that I missed.
What I am asking is this: Why in physics when the units are the same that does not necessarily mean you have the same thing.? Let me explain.
Please let me use m for meter, sec= second , and kg = kilogram as the units for brevity sake.
The units for work are kg * m/sec^2 * m. The units for kinetic energy are kg* (m/sec)^2. They look that same to me. I need them to be the same so I can figure out the principle of least action. Comments are welcome.
| Perhaps a better analogy than height and width can be found in terms of money. Both the balance in your bank account and amount you pay for, say, your electricity bill are denominated in the same units (dollars where I live), but they represent separate concepts. One is measure of what is stored and the other is a measure of what is transferred.
In most uses 'energy' means an amount available in some system (like the balance in an account) while work and heat represent transfers (like payments).
The analogy can be pushed too far as we often don't care about absolute level of energy and treat negative balances in exactly the same way as positive one while your bank may take a dim view of your maintaining a negative balance.
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Remove lens distorsion with intrinsic or extrinsic parameters? I'm struggling with the theoretics of camera calibration. There are the intrinsic parameters and the extrinsic parameters. I know what they are but what I'm struggeling with is to understand which of these help me getting the lens distortion removed. Not even this article helps me on the question
Hope someone can clarify this?
| The intrinsic parameters describe the relationship between light arriving at the camera from a particular direction, and where it ends up on the sensor / focal plane. As such - once you have calibrated a camera carefully for a given focal length, those parameters are "fixed" for that camera.
But to know where something is, and how big it is etc, you ALSO need to know where the camera is "in the universe" (the coordinate system of the lab, and the relationship to the object in question). This is where extrinsic calibration comes in - it tells you "in what direction the camera is pointing", how it is rotated in space, etc.
You need to know both those things to be able to make sense of the image on the screen. Without the extrinsic calibration, you would not be able to answer the question "am I looking at the front of the object, or at the back?", or "which way is up in this image?". Without the intrinsic calibration, you don't know how to go from "this object looks like it is 1 cm tall on my screen; what is the angle between the rays coming from the top and the bottom as they arrive at the optical center of my camera?"
It follows that lens distortion is addressed with appropriate intrinsic calibration - it's a property of the camera, regardless of where it is in space.
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Simple question about Schrodinger equation (time independent) For a quantum mechanical description of a system (like a small molecule) we can write:
$$\langle\psi|\hat {H}|\psi\rangle = \overline E$$
Question:
Is that energy the same as zero Kelvin energy obtained by statistical mechanics (using $E_n$ energies and partition functions)?
| The average $\langle E\rangle$ you have on the right is not to be understood in the sense of statistical mechanics (and thus as a temperature).
The wavefunction $\psi$ can be a linear combination of states $\psi_n$ of definite energy of your molecule. The $\psi_n$’s are solutions of the Schrodinger equation. Thus,
\begin{align}
\psi(x,t)&=\sum_i a_i \psi_i(x,t)\, ,\qquad \sum_i \vert a_i\vert^2=1\, , \tag{1} \\
\hat H\psi_i(x,t)&=E_i\psi_i(x,t)\, .
\end{align}
With this wavefunction the average energy
$$
\langle E\rangle = \sum_i \vert a_i\vert^2 E_i \tag{2}
$$
is a weighted average of the possible energies of your molecule, with the modulus square $\vert a_i\vert^2$ of the complex amplitude $a_i$ functioning as statistical weight. $\psi(x,t)$ given in (1) is an example of a pure state.
For instance, if you have a hydrogen wavefunction of the form
$$
\psi(r)=a_1\psi_{100}(r)+a_2\psi_{200}(r)
$$
then the average energy of this system is
$$
\vert a_1\vert^2\times (-13.6) + \vert a_2\vert^2 \times (-13.6/4). \tag{3}
$$
Measuring the energy, you will sometimes get the value $-13.6$ (this outcome will occur $\vert a_1\vert^2$ of the time) and sometimes get the value $-13.6/4$ (this outcome will occur $\vert a_2\vert^2$ of the time). The average energy is exactly given by (3).
In quantum statistical mechanics one introduces the concept of a mixted state. For mixed states one cannot define a wavefunction for the system as in (1). The system is described using a matrix that represents a statistical mixture of wavefunction that is not of the form given in (1); the average energy for this statistical mixture is also not of the form (2) since the latter comes from (1), which does not exist for mixted states.
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What is Tsirelson's bound for the original bell's inequality As its well known, quantum correlations break Bell's inequalities only to a certain limit called the Tsirelson's bound. The bound was written for the CHSH inequality.
My question is what is the Tsirelson bound for the original case that Bell dealt with in his article, and how it is proven.
| Tsirelson's bound for CHSH works only with matrix sum. If one uses Kronecker sum instead then the eigenvalues are -4,-2,0,2,4. So the average of CHSH can reach 4 in tensor quantum mechanics.
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Understanding the equation for Potential Energy I am having a hard time understanding why Potential Energy can be calculated in the following way:
$$ \Delta U = U_f - U_i = -\int_{x_i}^{x_f} F_x dx $$
In particular, I don't understand why there is an integral in that equation. That is to say, why is it integrating the force in a system.
| Why an integral?
The integral comes from the work formula. When pushing something over a distance, you are applying work. Work is your force $\vec F$ times the displaced distance $\vec x$:
$$W=\vec F\cdot \vec x$$
If the force (or distance vector) changes along the way, then you have a problem and can't know which $\vec F$ (or $\vec x$) to insert into the formula. So you will have to calculate the work done before and after this change separately, because only here are the $\vec F$ and $\vec x$ constant.
$$W=\vec F_1\cdot \vec x_1+\vec F_2\cdot \vec x_2$$
For many, many changes, you must sum up smaller and smaller pieces of work done. We can invent the sum symbol $\sum$ for many terms summed - and the integral symbol $\int$ when they are essentially infinitely many (indicated by the infinitesimal symbol $d$):
$$
\begin{align}W=&\vec F_1\cdot \vec x_1+\vec F_2\cdot \vec x_2+\vec F_3\cdot \vec x_3+\vec F_4\cdot \vec x_4+\cdots\\
W=&\sum \vec F\cdot \vec x\\
\downarrow\\
W=&\int \vec F\cdot \vec dx\\
\end{align}$$
Why is this potential energy?
The work that a conservative force will do when released is called potential energy. That's all. So before it is released there must be
$$U=-W=-\int \vec F\cdot \vec dx$$
of energy stored. The negative sign indicates that the energy you provide as work to lift the book to the shelf corresponds to negative work done by the conservative force gravity.
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Can a theory gain symmetries through quantum corrections? It is well known that not all symmetries are preserved when quantising a theory, as evinced by renormalising composite operators or by other means, which show that quantum corrections may alter a conservation law, such as with the chiral anomaly, or 'parity' anomaly of gauge fields coupled to fermions in odd dimensions.
However is the reverse possible: can a theory after quantisation gain a symmetry? Or if not, can it gain a 'partial symmetry'?
(For example invariance under $x\to x+a$ for any $a$ is translation symmetry, and invariance under $x\to x+2\pi$ would be said to be a partial symmetry. My question concerns whether a theory can gain a full symmetry, or a partial one at least through being quantised.)
| Maybe not the answer you are looking for but, remember that (Wilsonian) QFTs are defined at a certain scale $\mu$.For example we can take Yang-Mills theory with various matter fields added with a certain set of coupling constants/masses $a_i$. Classically this theory can be made to have conformal symmetry by choosing the couplings in such a way that all the coupling constants are dimensionless. For concreteness let us take $SU(N)$ Yang-Mills theory with 6 scalars in the adjoint representation with a general quartic potential and 4 Dirac fermions with general Yukawa couplings. It is well known that conformal symmetry is broken by quantum effects generically. But it is also known that at a point in parameter space, $\partial_\mu a_i=0$, this theory is superconformal at the quantum level. So it can indeed happen that quantum/loop corrections conspire among each other to enhance symmetries. Another example is ABJM which appears to only have $SU(2)\times SU(2)$ flavor symmetry but actually has $SU(4)$ or even $SO(8)$ symmetry depending on the ranks of the gauge group.
| {
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How the Hertzsprung-Russell Diagram allows us to calculate distance to stars I understand how to interpret a H-R diagram, in the sense that I know that the upper right top corner is occupied by cool stars, but they are very luminous so they must be big; and the bottom left corner are hot stars, not luminous, so they are small in size.
However, I have tried to read a textbook and look online, but have yet to understand how from this information, we can measure the distance to stars of undefined distance.
| You can use an HR diagram along with calibrated evolutionary models to find the distance (and in some cases, mass and age) of individual stars. The method is known as spectroscopic parallax. This is a confusing name because it is not a parallax measurement at all.
The technique is to use spectroscopy, or less precisely the colours, of a star to estimate it's effective temperature and surface gravity. This can normally be used to get a good idea of what kind of star (dwarf, giant and spectral type). This is enough to locate it in the HR diagram and determine the absolute luminosity. From there, the apparent brightness of the star yields its distance.
The technique works best for main sequence stars. Other types have quite age-dependent positions in the HR diagram and since this isn't known and also because the gravity is not usually accurate enough to pin down the exact HR diagram position (although it is usually good enough to distinguish a main sequence star from a giant or subgiant), it doesn't work as well.
Another simple description of this technique, with examples, is given here.
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What does 'massive' scalar field theory mean? I understand what scalar field theory is, it's just a theory that studies scalar fields.
But what does the 'massive' mean?, this might be a trivial question but I just wanted some clarification. Is it just a really big scalar field? If so why does the size really matter a field is presumed to cover all space anyways?
| The Lagrangian density $\frac{1}{2}\partial_\mu\phi\partial^\mu\phi-V(\phi)$ for real $\phi $ has equation of notion $\partial_\mu\partial^\mu\phi=-V'(\phi)$. When the $\phi^2$ coefficient in $V$ is $\frac{m^2}{2}$ with $m\ge 0$, $\phi$ has mass $m$. For a complex field the Lagrangian $\partial_\mu^\ast\phi\partial^\mu\phi-V(\phi,\,\phi^\ast)$ obtains the same eom, but this time we look for $V$ to include $m^2\phi^\ast\phi$. A coupling to a vector field complicates matters, but more generally the squared mass of a real (complex) field is $\frac{\partial^2 V}{\partial^2\phi}$ ($\frac{\partial^2 V}{\partial \phi \partial \phi^\ast}$).
| {
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Use of the Kronecker Delta in Translations A translation in special relativity is, as I understand, a kind of Lorentz transformation given by:
$x^{\mu} \rightarrow \delta^{\mu'}_{\mu}(x^{\mu} + a^{\mu})$ where $ \delta^{\mu'}_{\mu} = 1$ if and only if $\mu' = \mu$.
What I fail to understand is the use of the Kronecker delta here. Why would you need to multiply by $1$ if the two coordinates are the same and $0$ if they aren't? Isn't that the whole point of the transformation, i.e., if two coordinate systems are the same, why have a translation? In other words, what is the point of using the Kronecker delta and what purpose does it serve?
| You are entirely correct; the $\delta$ in this circumstance is entirely superfluous.
It was probably chosen to emphasize a common form for all Poincaré transformations, which can be understood as a pair of a translation and a Lorentz transformation. In this particular context it means "Translations are embedded in the Poincaré group as the pair $(T, I)$ where $T$ is the translation and the Lorentz transformation is the identity transformation." In other words, they are explicitly calling out, for ill-defined reasons, that the "Lorentz" part of the transform doesn't do anything but perhaps relabel an index.
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What is the point of the reaction mass in the LIGO mirror suspension system? I learned from the LIGO official website that the LIGO mirror suspension system consists of a "main chain" and a "reaction chain", and there are small electric motors gently pushing the masses on the main chain (i.e., the mirror and other suspension masses) against their counterparts on the reaction chain to "keep them in place". Here is a link to the material (see the second paragraph under the schematics of the suspension system).
I am wondering why this is a good strategy to achieve better isolation. It appears to me that the whole point of isolation is to keep the masses free (at least in some frequency band and along some direction), and the idea of installing motors seems to be ruining it. Also, I do not understand how these motors are controlled to distinguish between unwanted disturbances which should be corrected for, and true gravitational wave signals which should not be corrected for.
| The vibration control systems are actuating in all the degrees of freedom of the test masses and subtracting motion from them, primarily ground motion at low frequencies.
Only when all these systems are active are the test masses "free" to the degree that the noise in these systems allows.
You can think of it as the noise cancelling system in a pair of headphones: a microphone peaks up background noise and an actuator feeds a control signal to the speakers so that you get a silent playback. The LIGO mirrors are free or "quiet" in a similar way when all the control systems are online.
The reaction mass itself is another, rather big and sophisticated, actuator. It controls the longitudinal degree of freedom of the test mass via electrostatic force.
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Does light heat transparent mediums? I was thinking that if a photon is travelling in the vacuum at $c$, then enters a transparent medium so it's speed becomes $0.8c$, the photon has lost energy. What is that energy transformed into? Is the surface being heated?
| The energy carried by light is entirely independent of the speed it travels at - it is simply given by the light's intensity, which is equal to the photon flux (number of photons carried per second) times the photon energy (equal to Planck's constant times the frequency). If the light is slowed down in a medium, then at constant energy flux the local energy density will increase (the energy is 'squeezed together', if you will), but that is all that happens.
In a transparent medium, no light is absorbed and the intensity does not change: the photon flux remains constant, and the frequency is not altered. Thus, there is no energy loss, either by the light beam as a whole nor from the individual photons that comprise it.
| {
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Hawking Radiation: Attraction of negative mass particles? My understanding of Hawking Radiation is this:
All the time, small amounts of matter and antimatter pop into existence, find each other, and annihilate. When this happens near/on the edge of a black hole, the matter particle slips out into space, while the antimatter particle falls into the black hole and reduces it's mass.
That's fine. My question is; How is the thing with negative mass being pulled in by gravity, while the thing with positive mass is escaping? shouldn't it be opposite of that?
| As far as we know, antimatter does not have a negative mass, it just has an opposite electric and quantum charge (although scientists at CERN’s alpha institute are looking into the idea of antimatter having negative mass). When these two virtual particles are created, either can fall into the black hole, its not always the antimatter. Stephen Hawking noticed is that if these two particles are created at the event horizon of the black hole, one could fall in. This would defy one of the important concepts of quantum mechanics - the conservation of information - which states that these two virtual particles should be able to exchange information with each other if they were created at the same time. If this was the case, than information would have to leave that black hole which is impossible according to relativity as the information would have to break the speed of light. Hawking proposed that black holes can emit radiation, Hawking radiation, that would allow the particles to still share information without breaking the speed of light. This unfortunately brings up a few paradoxes such as the fact that black holes should have a shell of radiation that we haven’t directly observed.
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Spring Constant Dependence on Contact Area I'm actually modeling a contact situation between an elastic cuboid (in reality it is an elastomer) and a non elastic sphere ($c_c\ll c_s$). The spring constant $c$ is defined with the young's modulus $E$, the area $A$ and the height of the cuboid $h$ with
$$c=\frac{A\cdot E}{h}$$
However, the contact area in the described model changes with the indentation depth. So in my opinion the equation in this model should depend on the indentation depth $x$
$$c(x)=\frac{A(x)\cdot E}{h}$$
Am I right, or do I mess something up?
| It is somewhat more complicated than this. To get it right, you'll need to look at the theory of elasticity. There is a model, called the Hertzian model which deals with the force between things like two elastic spheres or a sphere and an infinite space (https://en.wikipedia.org/wiki/Contact_mechanics).
In the case of an a hard sphere and an infinite elastic space which has Young's modulus $E$ and is incompressible $\nu=1/2$, the force of indentation can be shown to be
$$
F= \frac{16E}{9}R^{1/2}\delta^{3/2},
$$
where $\delta$ is the indentation depth, and $R$ is the radius of the sphere.
Now, you have a finite thickness elastic material, Dimitriadis et al. have shown be shown that if $\chi=\sqrt{R\delta}/h$ is small,
$$
F=\frac{16E}{9}R^{1/2}\delta^{3/2}\left[1+0.884\chi+0.781\chi^2+O(\chi^3)\right].
$$
For details, here is the reference:
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC1302067/pdf/11964265.pdf
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EMF generated by a rotating rod I am confused by the following example in my textbook.
Question:-
A metallic rod of length $l$ is rotated with a angular velocity $\omega$, with one end hinged at the centre and the other end at the circumference of a circular metallic ring of radius $l$, about an axis passing through the centre and perpendicular to the plane of the ring. A constant and uniform magnetic field $B$ is present everywhere and is parallel to the axis. What is the emf between the centre and the metallic ring ?
Solution :-
The area of sector traced by the rod in time $t$ is $\dfrac12 l^2 \theta$, where $\theta = \omega t$. Therefore $\varepsilon = \dfrac{B\omega l^2}{2}$.
The math in the solution is not hard to understand. I have trouble understanding the physics.
I know $\varepsilon = \dfrac{\mathrm d\ \phi_B}{\mathrm d \ t} = \dfrac{\mathrm d\ (\bf B\cdot A )}{\mathrm d \ t}$ and in the question it is given that ${\bf B}$ is constant and the angle between $\bf A$ and $\bf B$ is $0$. So we are just left with $\varepsilon = B\dfrac{\mathrm d \ A}{\mathrm d \ t}$.
Why the area taken in the solution is the area of the sector traced by the rod in time $t$ ?
Since the magnetic flux is passing through the whole metallic ring, shouldn't the area be the area of metallic ring ? Which is contant. So the emf should be zero.
| There are two possibilities of how you close the loop.
(Actually several possibilities, but lets consider two for now )
I have shown them in the picture.
In one case, there will be no EMF generated.
In other case, there will be EMF generated.
Two cases shown in picture below.
In both cases, the black shows initial position and blue shows the final position.
On LHS, the voltmeter and wires move with the rod and no EMF is generated.
On RHS, the voltmeter does not move and wires stretch. In this case, there will be EMF generated as the area is changing.
But in this second case, the math involved is not very easy as we have to find the rate of change of the area of triangle.
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Are the Christoffel symbols all zero in gravity-free space? I was looking at the geodesic equation,
$$\ddot{x}^\mu + \Gamma^\mu{}_{\nu\rho} \dot{x}^\nu \dot{x}^\rho = 0,
$$
and thinking about how to identify gravity-free spaces by looking at the Christoffel symbols $\Gamma^\mu{}_{\nu\rho}$. For example, if they were zero, under certain limits I could check what kind of expressions/signs metric derivatives take. Can this be used to show that the space is gravity-free?
| No, not necessarily.
Flat space (which is what I assume you mean when you say gravity-free space) is special because it's possible to choose a global coordinate system in which all of the Christoffel symbols vanish. However, it's easy to make a choice of coordinate system (for example, spherical coordinates) for which the Christoffel symbols are generically non-zero, even in the absence of spacetime curvature.
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Is it strange that there are two directions which are perpendicular to both field and current, yet the Lorentz force only points along one of them? By "strange" I mean 'Is there a reason for this, or is it something we accept as a peculiarity of our universe?'
I see no reason why if magnetic field is in the $+x$ direction and a charge's velocity is in the $+y$ direction, that the force experienced by the charge can't be in either the $+z$ or $-z$ direction, both are perpendicular to $x$ and $y$. The equation for Lorentz force just tells us that it goes in the $+z$ direction, but it seems equally valid that the force would be in the $-z$ direction (I mean there's nothing to distinguish $+z$ from $-z$ anyway, you can turn one into the other by swapping the handedness of your coordinate system). It seems as if the universe is preferentially selecting one direction over the other.
As a practical example, if a current carrying wire is in a magnetic field and it experiences an upward force, why shouldn't it experience a downward force?
| You just discovered something pretty fundamental about the universe. It is conceivable that there would be a "mirror image universe" in which the laws of physics are exactly the other way around. But that's not the universe we live in.
There is an interesting Feynman lecture one the topic of symmetry - in particular, on the difficulty of trying to explain "clockwise" to someone without any visual aids. See this link
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
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Does providing more heat to a pan of boiling water actually make it hotter? Sometimes my wife has a pan of water 'boiling furiously'. Is the extra heat (wasted in my opinion) actually making any difference, apart from reducing the amount of water in the pan - which could be done by pouring some away?
| Water is not going to have a temperature significantly higher than 100° C, even if you top up the gas (or whatever).
But if there is not just water, the final result can be different. For example, in a Pilaf rice (like a Paella) it is very important to time the thing so that the broth is "finished" just when the rice is cooked --- and you do not want to remove it physically because it has to release the flavors to the rice. So the quantity of heat will have an effect here.
| {
"language": "en",
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How do nonlinear phenomena arise from linear theories? How is it possible that linear theories, for example maxwells equations or the schroedinger equation, produce nonlinear physics?
| One way of getting non-linear results from linear equations is to have interactions that are spread over time and/or space. This will require integration of the linear interaction, which will result in non-linear behavior. For example, if you integrate $dx/dt = kx$, you get $x=\exp(kt)$, which is clearly non-linear in both $k$ and $t$.
In another example, take Hooke's law $F = -k x$ (where $F$ is the force, $k$ is the spring constant, and $x$ is the displacement from equilibrium). You can generalize it to continuous media with stress = constant $\times$ strain. This is clearly linear. However, if you calculate the deflection of a beam of a material of uniform composition and cross-section, you get non-linear effects, for example with deflection with a given force depending non-linearly on the length of the beam. This is because deformation at one point in the beam influences deformation at another point as static equilibrium is achieved. This coupling between different parts of the beam causes the non-linear term to arise.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why is a centered parallelogram not a 2D Bravais cell, but a centered rectangle is? We had a disagreement regarding 2D Bravais lattices during a lecture. The lecturer told us that a centered rectangle forms a Bravais lattice in 2D, but a centered parallelogram isn't:
We couldn't come up with a satisfying definition of a Bravais lattice though. It appears to be a lattice with certain periodic/translation/etc. symmetries; but intuitively, putting a dot in the center of a rectangle symmetry-wise has the same effect putting one in a parallelogram does. Why is the distinction then?
So, our questions are:
*
*Does a centered rectangle form a Bravais cell? (Wikipedia lists it as one.)
*Does a centered parallelogram not? (Wikipedia doesn't list any.)
*Why, and/or why not?
Edit: The question (Is the centered parallelogram a Bravais-cell?) was in a test we discussed, so an exact answer to this is encouraged.
| A centred lattice is defined [1, section 1.3.2.4.] as a sublattice formed by the integral linear combinations of independent vectors ($\renewcommand{\vec}[1]{\boldsymbol{#1}}\vec{a}$ and $\vec{b}$ on your diagrams) plus the translation of this sublattice by a finite number of so-called centring vectors. In your case, we only need one, $(\vec{a}+\vec{b})/2$, which goes from the "starting point" of $\vec{a}$ and $\vec{b}$ on your diagram to the centre of either your rectangle or your parallelogram. So, yes, both of the cases you consider fit with the definition of a centred lattice.
However, in the parallelogram case, the centred and the non-centred cells belong to the same Bravais class because there is no geometrical difference between them: they both have an arbitrary angle, as opposed to the rectangular case where the primitive cell defined by $(\vec{a}+\vec{b})/2$ and $(\vec{a}-\vec{b})/2$ has a different geometry (no right angle). So the statement on your lecture notes is wrong: "not Bravais lattice" should be replaced by "not distinct Bravais lattice".
[1] International Tables for Crystallography (2016). Vol. A, [online access]
| {
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Energy stored in electric field and work done to change a charge configuration I had read a problem where a point charge is placed at the center of a thin grounded conducting shell. The energy for removing the shell off to infinity was asked. It was calculated by using the following expression for energy stored in the electric field, before and after, and subtracting them.
Is the energy in some charge configuration only stored in the electric field? For example, is the energy required to move one positive charge in the presence of another given by the difference in the energy stored in the field before and after?
| It is true that the energy stored in a given system of charges is stored in its electric field.
However, it is difficult to calculate the work done in moving one positive charge in the presence of another in terms of difference in energy stored in the electric field, as the electric field energy of a point charge diverges as we approach it.
| {
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Proof quantum random walk $\sigma \propto n$? In a classical random walk, the standard deviation $\sigma \propto \sqrt{n}$, while in a quantum random walk it goes as $\sigma \propto n$.
Both discrete time and continuous time quantum random walks have been shown to display the above behaviour.
I can prove the classical case, as it is just related to the variance of the binomial distribution, but how does one prove the relation for the quantum random walk? All the papers that I have read either assume it or derive it numerically. Is an analytic proof possible?
| First note that the statement is generally true only in the translation-invariant setting.
For an elementary proof of this in the discrete time setting using Fourier techniques, see the paper by Grimmett et al. in
https://arxiv.org/abs/quant-ph/0309135
There, the weak convergence of $Q_t/t$ is proved for quantum walks on regular lattices, where $Q_t$ is the time-evolved position operator in the Heisenberg picture.
This statement also accounts for the ambiguity of the standard deviation which does not tell the whole story since it doesnt discriminate between different principal axes of the lattice.
| {
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Rotations of eigenstates of $S_z$ I have a question regarding the rotation of spinors in a spin-1/2 system.
We have a Spin generator $\hat{S}$ for rotations of spinors.
A rotation around the axis $\vec{n}$ with the angle $\phi$
is generated by the operator:
$$
D_{\vec{n}}(\phi) = \exp(-i\phi \hat{S}\cdot \vec{n})
$$
This operator can also be written, for a rotation about $z$ e.g. as:
$$D_z(\phi) = \cos\left(\frac{\phi}{2}\right) - i \sigma_z \sin\left(\frac{\phi}{2}\right)
$$
Here, $\hat{S}_i = \sigma_i/2$ and $\sigma_i$ are the pauli matrices.
$\sigma_1 = \begin{bmatrix}0 & 1 \\
1& 0 \end{bmatrix}$ $\sigma_2 = \begin{bmatrix}0&-i\\ i&0\end{bmatrix}$ and $\sigma_3 = \begin{bmatrix} 1&0\\ 0&-1\end{bmatrix}$
Then we have given two states with a spin into the direction of the z-axis:
$\vert S_z= + \frac{1}{2} \rangle =
\begin{bmatrix}1\\ 0\end{bmatrix} = \vert {\uparrow}\rangle $
and
$\vert S_z = - \frac{1}{2} \rangle = \begin{bmatrix}0\\ 1\end{bmatrix} =
\vert {\downarrow}\rangle $
Now my question is:
With which rotation $D_\vec{n}(\phi)$ can the eigenstate $\vert S_x = +\frac{1}{2}\rangle$ be obtained using $\vert\uparrow\rangle $?
How can I calculate that?
| The simplest way to think about it is by thinking of spin as a classical vector.
What kind of rotation would take a vector completely along $\hat z$ to the $\hat x$ axis? Clearly, this would be a rotation in the $xz$ plane, i.e. a rotation about $\hat y$.
The same argument will work for spin. You might care to reflect on the relation between the classical angle and the angle of rotation in spin space, remembering that $\vert +\rangle$ and $\vert -\rangle$ are orthogonal vector in spin space but antiparallel in ordinary space. That’s why $\phi$ is mutiplied by $1/2$ in your expressions.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/364163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Does Humid Air Rise (neglecting the effects of weather turbulence-for example in a closed environment like a sealed room) An internet search will find many people claiming that Humid Air (water vapor) rises. However, I am skeptical because if lighter molecules rose then, it follows that the air would be stratified by molecular weight.
Facts:
Water vapor has molecular weight = 18
Nitrogen has molecular weight = 28
Oxygen has molecular weight = 32
So, Water Vapor has the least mass.
However, If lighter gases rose
then (using the above data)
Nitrogen would rise
Oxygen would fall
and we would be breathing 99% Oxygen (neglecting the 1% of other gasses)
this is not true
So
I believe that gasses do not behave this way
as illustrated by the fact that we breathe 78% nitrogen and 21% oxygen
Therefore, it seems to me that (in a closed environment) gasses mix randomly, with perhaps a very small gravitational contribution.
| There are no sources or sinks of oxygen or nitrogen to speak of, so there are no fluctuations in the ratios of these gasses to speak of. Water vapor, on the other hand, is constantly being added and removed--air can be dried by cooling it to condense out the water, and then reheated, for example. It is quite possible to have humid air interacting with dry air of the same temperature. In that case, until the two air masses mix, the humid one will tend to rise above the dry one.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Feynman diagrams for eigenvalue perturbation theory I posted this question in MathOverflow but was not lucky with the answers, so wil try here.
Suppose I have a matrix given by a sum
$$A=D+\epsilon B$$
where $D$ is diagonal and $\epsilon$ is small, and I want the eigenvalues of $A$ as power series in $\epsilon$. The first two orders in perturbation theory are well known. Third and higher orders are briefly discussed here. However, the equations become horrible.
I hear that Feynman diagrams are an efficient way to formulate perturbation theory, but I can't find an accessible exposition of this approach. Note that I have in mind the simple matrix setting. I don't want vacuum states, quantum field theory, path integrals, many body, etc. Can someone help?
| If you want to find the eigenvalues of a finite dimensional matrix $A$ as a Taylor series on $\epsilon$ there are well known procedures to do that. If your object $A$ is infinite dimensional things become more complicated but can be carried out in principle. (Almost) Everything you want to know on the subject can be found in Kato’s book perturbation theory for linear operators.
The Feynman diagrams are essentially simply a convenient way to write a particular term of the perturbation expansion when your objects are (quantum) field theories. For example, it turns out that, because of the form of the interaction term (your $B$) you can already say that some terms in the perturbation expansion are going to be zero. In general you have a diagrammatic way to write such terms which is useful because it provides an easy way to write down these terms. At the same time it also provides a physical intuition for what these terms do and this is probably even more important.
In a simple matrix setting Feynman diagrams are of no use as they cannot even be defined.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Why is it "bad taste" to have a dimensional quantity in the argument of a logarithm or exponential function? I've been told it is never seen in physics, and "bad taste" to have it in cases of being the argument of a logarithmic function or the function raised to $e$. I can't seem to understand why, although I suppose it would be weird to raise a dimensionless number to the power of something with a dimension.
| The other answers are correct that when you think of it in terms of unit analysis you cannot add quantities that have different units to each other. Even so, formally you can always do something like
$$f\left(\frac{x}{1\operatorname{m}}\right)$$
to get something that works, mathematically.
Where it becomes bad taste/bad practice is that you introduced that denominator, yourself, by hand. In any physical problem that requires you to evaluate some complicated function, like $\sin$, $\ln$, or $\exp$, there will always be some physically relevant quantity with the same units that will allow you to form a unitless quantity. For example, when working with the simple harmonic oscillator we can combine the spring constant, $k$, and the mass, $m$, to produce a quantity with the units of inverse time, $\omega \equiv \sqrt{k/m}$. It is that $\omega$ that allows us to sensibly write $x=A\sin(\omega t)$ to describe the motion of the oscillator.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/364771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "86",
"answer_count": 7,
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} |
Why does $2\to 2$ scattering cross-section have $E_{CM}^2$ in denominator? For $2\to 2$ scattering with equal masses we have
$$\left(\frac{d\sigma}{d\Omega}\right)_{CM} = \frac{1}{64 \pi^2 E_{CM}^2} |\mathcal{M}|^2.$$
(Schwartz's QFT eq. 5.33) Can we make the general statement that total cross section increases as $E_{CM}$ decreases for scattering processes? I'm guessing not because $\mathcal{M}$ may have a complicated dependence on energy/momenta.
As an example, take the case of $e^+e^-\to \mu^+\mu^-$. With $m_e = 0$ we have
$$\frac{d\sigma}{d\Omega} = \frac{\alpha^2}{4 E_{CM}^2} \sqrt{1-m_{\mu}^2/E_{\mu}^2} \left(1+ m_{\mu}^2/E_{\mu}^2 + (1- m_{\mu}^2/E_{\mu}^2)\cos^2\theta\right)$$
in the c.o.m frame (Schwartz 13.78). Here it's pretty clear that total cross section increases as energy decreases.
But I would think that particles scatter more when they are given more energy? The limiting case of course being zero velocity $\to $ zero scattering. I would appreciate any intuition on the relation between $E_{CM}$ and total cross section.
|
Can we make the general statement that total cross section increases
as $E_{CM}$ decreases for scattering processes?
If the considered QFT is unitary (at least perturbatively), there is a constraint on the cross-section following from the unitarity, called Froissart bound. It is the direct consequence of the optical theorem. The latter states that
$$
\sigma_{1+2\to\text{all}} \simeq \frac{1}{s}\text{Im}(M_{\text{forward}}),
$$
where $\sigma_{1+2\to\text{all}}$ is the $1+2\to \text{all}$ scattering cross-section, while $s$ is the invariant mass of the colliding particles. The amplitude $M_{\text{forward}}$ is the forward scattering amplitude bounded by
$$
M_{\text{forward}}\lesssim s\ln^{2}(s)
$$
Therefore, for general $1+2\to\text{all}$ scattering one obtains the Frossairt bound:
$$
\sigma_{1+2\to \text{all}} \lesssim \ln^{2}(s)
$$
But I would think that particles scatter more when they are given more
energy?
The more energy of the particle is, the less corresponding de Broglie wave-length is. Therefore qualitatively it's rather expected that the colliding particles with very large energies will overlook each other.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/364901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What would happen to an isolated block of material I was thinking recently about what might happen if you were to place a block of material in the middle of a complete vacuum. Obviously there's not going to be a way to ever achieve such a scenario but what would happen if you were to put a block of let's say steel at 100C in a vacuum such that the block is not in contact with any material connected to the containment and have it such that outside energy is minimized. I assume the block would lose heat/vibrational energy but what would be the mechanism for such an energy loss and what time scale would it take for the block to reach let's say 0C?
Let me know if there's anything I can add to make the question more clear.
| The amount of energy will depend on the mass of the block and what it is made up of. A small body will lose energy at such a slow rate that it can outlast our sun.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/365019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If I have a graph displaying the equation of elastic potential energy, why does the graph increase quadratically? I was wondering why the graph is quadratic. If one could also link me to an external source to explain this, I would be just as grateful.
| This is an addition to @Steeven's answer.
Your graph plots the distance a marker is catapulted vs. the pull-back distance on the rubber bands. We can find a theoretical prediction for how far your projectiles should go.
Assuming the markers land at the same height at which they are launched, they will travel a distance of
$$d = \frac{v^2\sin2\theta}{g},$$
where $v$ is the launch speed, $\theta$ is the angle of launch (zero begin horizontal), and $g$ is the acceleration due to gravity. All of the marker's kinetic energy comes from the rubber bands, so we can write
$$K = U$$
$$\frac{1}{2}mv^2 = \frac{1}{2}kx^2$$
$$v^2 = \frac{kx^2}{m}$$
where $K$ is the kinetic energy, $U$ is the potential energy, $m$ is the mass of the marker, $k$ is the spring constant of the rubber bands, and $x$ is the pullback distance. Substituting this $v^2$ into the range equation above results in
$$d = \frac{kx^2\sin2\theta}{mg}.$$
So, we expect the range of the marker to be a quadratic function of the pullback distance.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/365155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Where will the Goldilocks zone be when the Sun becomes a red giant? In about 5 billion years, when our Sun expands into a red giant making our planet uninhabitable, where will the new Goldilocks zone be? Could life form on a new planet in the Goldilocks zone? Environment suitable for human life?
| Fortunately, our galaxy won't last that long. The Shapely-Super-Cluster is pulling hundreds of galaxies, including our Milky Way, into a large gravitational anomaly called "The Great Attractor". It's only 220 million years away... Which is pretty darn quick when compared to five billion years. The Sun will be OK for a long time because it doesn't have anywhere near the density/mass to supernova.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/365297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
"answer_count": 4,
"answer_id": 3
} |
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