Q stringlengths 18 13.7k | A stringlengths 1 16.1k | meta dict |
|---|---|---|
How is a Higgs boson created? I have read a lot on Higgs bosons, yet I do not fully comprehend how they are created and how they are "flicked off" the Higgs field. I have also had trouble comprehending why a Higgs boson quickly becomes unstable and decays into more common particles.
How is a Higgs boson created and how and why does it quickly decay?
| A higgs boson is created at an accelerator just like any other particle, by converting energy to mass, according to the famous equation
$$E = mc^2$$
If you take the LHC as example, then protons are accelerated to nearly light speed, having enough energy to create particles as heavy as the higgs.
For a particle to decay it needs phase-space (i.e. the particles that the parent particle decays to need to be lighter than the parent particle) and no conserved quantities that would forbid a decay (e.g. charge conservation).
Since the higgs is heavy and has no quantum number that would forbid a decay, it has a very short lifetime.
| {
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Why does $\ell=0$ correspond to spherically symmetric solutions for the spherical harmonics? In quantum mechanics why do states with $\ell=0$ in the Hydrogen atom correspond to spherically symmetric spherical harmonics?
| One way to understand it is to recognize that for the spherical harmonic $|l,m\rangle$ with $l=0$ (and obviously $m=0$), we have $\hat L_i|0,0\rangle=0$, where $\hat L_i$ is the angular momentum operator in the direction $i=x,y,z$. It is obvious for $\hat L_z$, which eigenvalue is $m=0$, and can be verified for the other two.
Then, the rotation operator $\hat R(\theta)$ around a direction $\vec n$ with angle $\theta$ is given by $$\hat R(\theta)=\exp(i\theta \,\vec n . \vec{\hat L} )$$
from which we clearly see that the state $|0,0\rangle$ is invariant for all rotations : $\hat R(\theta)|0,0\rangle=|0,0\rangle$ and is thus spherically symmetric.
In this formulation, you see that it is the only state like that. You can also show that the state $|l,0\rangle$ is axially symmetric (along $z$), etc. See for instance this nice picture :
| {
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Braiding statistics of anyons from a Non-Abelian Chern-Simon theory Given a 2+1D Abelian K matrix Chern-Simon theory (with multiplet of internal gauge field $a_I$) partition function:
$$
Z=\exp\left[i\int\big( \frac{1}{4\pi} K_{IJ} a_I \wedge d a_J + a \wedge * j(\ell_m)+ a \wedge * j(\ell_n)\big)\right]
$$
with anyons (Wilson lines) of $j(\ell_m)$ and $j(\ell_n)$.
One can integrate out internal gauge field $a$ to get a Hopf term, which we interpret as the braiding statistics angle, i.e. the phase gained of the full wave function of the system when we do the full braiding between two anyons:
$$
\exp\left[i\theta_{ab}\right]\equiv\exp\left[i 2 \pi\ell_{a,I}^{} K^{-1}_{IJ} \ell_{b,J}^{}\right]
$$
see also this paper and this paper.
I would like to know the way(s) to obtain braiding statistics of anyons from a Non-Abelian Chern-Simon theory? (generically, it should be a matrix.) How to obtain this braiding matrix from Non-Abelian Chern-Simon theory?
| How to obtain this braiding matrix from Non-Abelian Chern-Simon theory?
To obtain braiding matrix $U^{ab}$ for particle $a$ and $b$, we first need to know the dimension of the matrix. However, the dimension of the matrix for Non-Abelian Chern-Simon theory is NOT determined by $a$ and $b$ alone. Say if we put four particles $a,b,c,d$ on a sphere, the dimension of the degenerate ground states depend on $a,b,c,d$. So even the dimension of the braiding matrix $U^{ab}$ depends on $c$ and $d$. The "braiding matrix" $U^{ab}$ is mot deterimened by the two particles $a$ and $b$.
Bottom line: physically, the Non-Abelian statistics is not described by the "braiding matrix" of the two particles $a$ and $b$, but by modular tensor category.
| {
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Wave function of a particle in a gravitational field
Suppose we have a particle with mass $m$ and energy $E$ in a gravitational field $V(z)=-mgz$. How can I find the wave function $\psi(z)$?
It should have an integral form on $dp$. Any help would be appreciated.
What I've tried
One way to solve the problem is use of change of variable
$$
x~:=~\left(\frac{\hbar^2}{2m^2g}\right)^{2/3}\frac{2m}{\hbar^2}(mgz-E)
$$
we can reduce Schroedinger equation to
$$
\frac{d^2\phi}{dx^2}-x\phi(x)~=~0
$$
This is a standard equation, its solution is given by
$$\phi(x)~=~B~\text{Ai}(x)$$
where $\text{Ai}$ is the Airy function. But my solution should be (not exactly) like this:
$$
\psi(z)= N\int_{-\infty}^\infty dp \exp\left[\left(\frac{E}{mg}+z\right)p-\frac{p^3}{6m^2g} \right]
$$
| $$
\left[\frac{p^2}{2m}+V(i\hbar\frac{d}{dp})\right]\phi(p)=E\phi(p)
$$
$$
\left[\frac{p^2}{2m}+(-mg)(i\hbar\frac{d}{dp})\right]\phi(p)=E\phi(p)
$$
$$
\frac{1}{i\hbar mg}(\frac{p^2}{2m}-E)\phi(p)=\frac{\phi(p)}{dp}
$$
When integrate we have:
$$
\frac{i}{\hbar mg}(Ep-\frac{p^3}{6m})=Ln\frac{\phi(p)}{\phi(p_{o})}
$$
$$
\phi(p)=\phi(p_{0})e^{\frac{E}{mg}p-\frac{p^3}{6m^2g}}
$$
$$
\psi(z)=\int dp e^{ipz/\hbar} \phi(p)
$$
$$
\psi(z)=\phi(p_{0})\int_{-\infty}^\infty dp e^{i/\hbar \left[ (\frac{E}{mg}+z)p-\frac{p^3}{6m^2g}\right]}
$$
| {
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When is quasiparticle same as elementary excitation, and when is not? Can anyone shed light on the comparison between these two concepts?
| Dictionary for this answer: Excitation = particle; collective excitation = quasi-particle.
Short answer:
Elementary particles are never quasi-particles, by definition of elementary. This does not mean that what now it is thought as an elementary particle could be a quasi-particle of entities to be discovered.
Mathematical answer:
Elementary particles are those that correspond to irreducible representations of the Poincare group.
Physical answer:
Quasi-particles require the existence of an external medium or fields, whereas elementary particles do not. For example, phonons require a solid or a fluid to exist (they are collective modes of the atomic lattice vibration), likewise pions require a quark-antiquark sea. These are not fundamental particles, in the sense that they need the existence of other particles. A closed notion is that of composite particle, for example a molecule is made of atoms which, in turn, are made of a nucleus plus electrons. The difference between quasi and composite particles lies in the fact that quasi-particle are though as collective excitations of many particles (usually of the order on the Avogadro number $\sim 10^{23}$, but there may be far fewer, but not tens), while composite particles are more like building blocks where each constituent may be an elementary particle—such as an electron— or another composite particle—such as an atom— (a molecule is usually made of a few or tens of atoms, an atom usually contains from a few to tens of electrons plus a nucleus). Nevertheless, the difference between both concepts is not sharp; for example, pions are somehow made of quarks and antiquarks, they actually are also collective modes (waves) on the quark-antiquark sea, being quasi-Goldstone bosons of the approximate chiral symmetry.
The difference between elementary and composite particles is tied to the human knowledge at the time. At a certain point, it was though that nuclei were elementary, after that people realized that there were in fact more fundamental constituents (protons and neutrons), and later on quarks and gluons were discovered.
| {
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Does velocity or acceleration cause time dilation? What causes time dilation? Acceleration or velocity?
I've seen multiple comments on this forum that assert velocity is the cause, but that doesn't seem right to me. You can't have velocity without acceleration. It's the inertial force with acceleration that breaks the symmetry. My understanding is that the asymmetry is where the inertial frame changes. Measuring time between two objects with different inertial frames is where you have time dilation. When the acceleration ends, the object is effectively at rest in a new inertial frame and has velocity relative to another object in the original inertial frame.
In other words, acceleration (changing reference frames) is the cause...velocity and time dilation is the effect.
Am I right about this? If there are flaws in my logic I'd like to find and correct them.
| In Special Relativity the time dilation is just a matter of convention in the time measurement between moving frames.
In General Relativity the time dilation is a physical phenomenon which involves a force field (either gravity or acceleration or wathever) that actually slows down the particles of a system. When particles interactions take place with lower frequency, Time (lifetime, aging etc.) is effectively slowed down.
| {
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1 Tesla electromagnet? Is it possible to create a powerful electromagnet at home?
With use of a ferromagnet it seems so...
Using the following formula: $B(Tesla)= k\mu_0nI$. I understand some ferromagnet's like iron could have permeability above 10,000? That would easily boost the field above a 1 Telsa?
Relative permeability of raw iron is: 200,000! There must be something wrong here, see this wikipedia table.
UPDATE: How much input power might be needed?
| Let:
*
*$R$: The radius of the coil, $h$ the height of the coil, $n$: spiral density, ie, the number of spirals per height.
*$r$: The radius of the wire, $A$: The area of the cross section of the wire.
*$L$: The total size of the wiring, $N$: The amount of spirals in the coil.
*$\bar R$: The overall resistance of the coil, $\rho$: resistivity of the material of the wire.
*$B$: The magnetic field (ofc), $k$: The relative magnetic permeability of the core.
*$V$: The voltage difference across the ends of the coil, $I$: The current passing thru the coil
*$P$: Dissipated power by the coil.
With that in mind, some formulas relate our quantities:
$$
B = k\mu_0nI,
\quad n = \frac{1}{2r},
\quad L = 2\pi Rnh,
\quad A = \pi r^2,
\quad P = VI,
\quad V = \bar RI,
\quad \bar R = \frac{\rho L}{A}
$$
These formulas comes from physical laws, or simple geometry. Also, we can relate the amount of spirals $N$ by $N = nh$. Don't forget that we want to minimize dissipated power, and maximize the amount of magnetic field generated. With that in mind, we shall fix $B$, and find $P$.
$$
P =
VI =
\bar RI^2 =
\frac{\rho L}{A}\left(\frac{B}{k\mu_0 n}\right)^2 =
\frac{\rho\cdot 2\pi Rnh}{\pi r^2}\left(\frac{B}{k\mu_0 n}\right)^2 =
\frac{2\rho RhB^2}{nr^2k^2\mu_0^2} =
\frac{4\rho RhB^2}{r k^2\mu_0^2} =
\frac{8\rho RB^2}{k^2\mu_0^2}N
$$
So, your goal is to maximize the most you can all variables in the denominator, and minimize the numerator, for a fixed $B$, in order to minimize the power you require to operate such a thing. I actually find quite interesting that $r$ should be maximized, in order to minimize $P$ for a fixed $B$. Intuition would say otherwise isn't it? You can do the same for voltage:
$$
V =
\bar R I =
\frac{\rho L}{A}\frac{B}{k\mu_0 n} =
\frac{\rho\cdot 2\pi Rnh}{\pi r^2}\frac{B}{k\mu_0 n} =
\frac{2\rho RhB}{k\mu_0r^2}
$$
And for last, to find the necessary current, its a direct relationship with the magnetic field:
$$
I = \frac{B}{k\mu_0n} = \frac{2rB}{k\mu_0}
$$
Hereby, our final results:
$$
I = \frac{2rB}{k\mu_0},\quad\quad
V = \frac{2\rho RhB}{k\mu_0r^2},\quad\quad
P = \frac{4\rho RhB^2}{r k^2\mu_0^2} =
\frac{8\rho RB^2}{k^2\mu_0^2}N
$$
| {
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How to smooth the spectrum of a light source? Could somebody please tell me if there's a reasonably cheap substance or device can I use to smooth the spectrum of a light source?
For example, if the spectrum has spikes as in the blue graph below, is it possible to smear in terms of wavelength (not spatially, as in ordinary diffusers) so that it resembles the red graph more?
The actual shape of the smoothing (convolution kernel) is not really important, but it is preferable that most of the energy is transmitted.
I've searched for "spectrum flattening" filters, but they instead attenuate certain wavelengths rather than spread them out.
I suppose it's possible to rapidly move the light source forward and back so that the Doppler effect would cause a similar shift, but the speeds required would be tremendous!
| The phosphors lining the glass tube of a fluorescent light do a pretty good job of smearing the atomic mercury line emission spectrum into something closer to black body radiation. There could be a phosphor mix that accomplishes what you want...
| {
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What gives an object its colour? My understanding of colour is that atoms in a particular object will absorb certain wavelengths of electromagnetic radiation, and the scattered wavelengths give the object its colour. The absorbed wavelengths contribute to lattice vibrations, increasing the kinetic energy and raising the objects temperature. Is this correct?
What is the sequence of events when a solution of particular atoms is sprayed through a Bunsen burner? I don't understand this. In this case is the colour seen a result of electrons moving up and then back down energy levels?
| In Bunsen burner atoms get heated up which means they absorb energy, and go to some excited states (or even get ionized). And then they de-excite in any way they can which means radiative decay is dominant, as there is almost no other way to dissipate energy. There is no crystal lattice, and compared to luminescence rate collisions are infrequent.
So yes - colour observed in Bunsen burner is related to electronic transitions.
| {
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Work in circular motions Suppose that a satellite circles around a planet that exerts $2000N$ of gravitational force on the satellite.
I understand the fact that since the circular motion and the centripetal force are always perpendicular to each other, the work done by gravity is 0.
However, the satellite is being moved from one point to another, so Intuitively I am thinking "there must be some work done to move the satellite from point A to point B".
But besides the gravitational force, there is no force acting on the satellite.
Can some one explain me the gap with my logic and intuition?
| Centripetal force is always directed towards the center of the circular path and always directed perpendicular to the direction of displacement of particle every where on the circular path.
Therefore angle between centripetal force F and displacement S is 90deg.
Work W= F.S
$$W=FS \cos (\theta)=FS \cos90 = FS\times0 =0$$
| {
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How to calculate velocities after collision? I'm currently writing a program for a particle simulator. One of the requirements is that the particles collide in a realistic way. However, I don't know how to calculate the final velocities.
For each collision, I have the $x$-component and $y$-component of each velocity, as well as the displacement and mass of each particle.
Is it possible to calculate the direction and magnitude of their velocities after the collision? If so, how?
| 2 dimensional collision can be reduced to a 1-dimensional problem in the case of spheres--see here. The $\pm$ you encounter when solving the kinetic energy is likely because there are two solutions and the equations are satisfied by either one. One solution is simply where the particles pass right through eachother, which you can discard.
| {
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Is it possible for bubbles to exist in vacuum? In the case of a bubble, the outside pressure is less then the inside pressure.
If that is the case can bubbles exist in vacuum? I am not sure but this should be true if vacuum has zero pressure
| I disagree. If it's a true vacuum then it can't exist. The pressure equilibrium both inside and outside will be zero, thus the surface tension would collapse it into a "drop" rather than a bubble, as there's no internal pressure to hold it expanded.
Were air to be added inside however, then things would get weird. Since the outer pressure is zero, expansion would be infinite... Now it matters what the surface tension strength of your material is vs. the maximum expansion of air you put inside.
| {
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Is it possible to focus the sun in such way?
Imagine the sliding part of the mirror is controlled by computer and opens on intervals.
Is it possible to increase the power of the beam by making it bounce between the mirrors thus going through the lens and then releasing it resulting in beam with more power ?
| Fun fact: You cannot focus the beams from the sun create a hotter area than the surface of the sun itself. This would break the second law of thermodynamics. 6000 °C is hot enough for many spectacular destructive applications, though, a youtube search on "melting steel with sun light" offers hours of entertainment.
If you want to make something that potentially focuses more energy than the surface of the sun, head for a solar pumped laser.
| {
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Is it possible to determine the universality class of phase transitions by just analysing symmetry? Since phase transition is closely connected with symmetry, I am wondering whether it is possible to determine the universality class of phase transitions just by symmetry?
Actually, I found it is quite boring to calculate critical exponent numerically
and I want to find a new method.
| At least for the case of the three "conventional" symmetry classes (Wigner-Dyson classes), the symmetry is directly visible in the structure of your Hamiltonian or, if you do numerics, in the matrix elements of your Hamiltonian matrix. For example, if your tight binding matrix contains only real matrix elements and is symmetric, the corresponding Hamiltonian belongs to the orthogonal symmetry class. Same is true for hermitian matrices with complex matrix elements (unitary symmetry class) and for the symplectic symmetry class, the symmetry relation is $H = \sigma_y H^T \sigma_y$, where $\sigma_y$ is the second Pauli matrix.
From what I know, the universality class also directly classifies the kind of phase transition. Maybe somebody else can shed more light on this topic.
| {
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Holographic Image In a holography set-up, as shown in the figure below, Illumination beam and reference beam both are in phase. The interference pattern generated at the detector contains the whole information about the object. When illumination beam will incident on to the object its phase and amplitude will change and further object beam will recombines with reference beam at the detector. Question is how the Interference pattern provides complete 3D information of the object ?
| I would perhaps answer something like this. Maybe that question helps you also. You have to keep in your mind the proper meaning of whole information; the plate will just contain almost the same information as the one you get from the other side of a window placed there (thorugh it), being its size the same of the plate's one, and provided you are working with the same light source (monochromatic).
Almost comes also from the fact that when reconstructing the stored image you have to know the shape of part of the light field used during the recording process (reference beam). That's what allows you to decode that information.
| {
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What happens when a charged (negative or positive) object touches an insulator? I know that insulators do not conduct electricity because they do not allow the free movement of electrons. Let's assume the object that's going to touch the insulator is negatively charged. Does some charge get transferred to the the insulator or does no charge get transferred at all? I want to understand this more conceptually. I don't really get why insulators are used in electrostatic experiments.
| Well if you don't go into the micro details where the charged sphere may induce extremely low quantity of charge in the insulator before touching and then transfer equally less charge even on touching which would not even be practically noted, NO! There is no charge transfer.
Conceptually you can say the extremely high resistance of the insulators does not give the charge carriers a tempting path and place to go to. As we know that charge normally travels over the paths which offer lower resistance.
For your next question, it would be wrong to say why insulators are used in electrostatics as it sounds like thats onky where they are used, insulators play an important role in electrodynamics (current electricity) just as well but that is not your question. Insulators have a special property, unlike conductors, if you charge an insulator, just by touching it at a particular point with a charged substance (most probably rubbing with something) the charge stays localised, had it been a conductor the charge would have spread over its entire surface and we would observe lesser effects. But since in insulators the charge stays localised, it shows significant effects, which are easily noted.
| {
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Is it possible to "cook" pasta at room temperature with low enough pressure? It is known fact, that boiling point of water decreases by decreasing of pressure. So there is a pressure at which water boils at room temperature.
Would it be possible to cook e.g. pasta at room temperature in vacuum chamber with low enough pressure?
Or "magic" of cooking pasta is not in boiling and we would be able to cook pasta at 100°C without boiling water (at high pressure)?
| Pasta does not really need to be cooked. Instead, it needs to become hydrated. Pasta is just like a dry fruit. If you place it in water for a day or so, it will look at least like it has been cooked.
| {
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Wigner friend experiment Let's supposed we take the Wigner's friend experiment from the metaphysical arena and try to implement it as an actual physical experiment
Assuming Wigner's lab friend is kept as a coherent superposition of states until the external experimenter tries to open the decoherence-insulated lab door, then the following check to detect coherence should work, even after the enclosed physicists has interacted with the dead/alive cat, but before the sealed lab is allowed to interact with the external experimenter:
*
*the sealed lab has two lasers (A and B) inside, that mix at a half-mirror, and then the laser goes out of the lab from the half-mirror. This is the only way the state in the inner lab can affect the external universe. If both lasers are on, only one of the external paths has constructive interference, while the other has full destructive interference.
*if the cat is dead, the sealed lab experimenter turns on laser A and keeps B off. Both branches of the outward paths of the half-mirror give half intensities when the single laser A is on while laser B is off
*if the cat is alive, the sealed lab experimenter turns on laser B and keeps A off. Both branches of the outward paths of the half-mirror give half intensities when the single laser B is on while laser A is off
*After some synchronization time, the external experimenter measures photons from each outward path of the half-mirror, and he sees two possible outcomes:
1) one of the paths has 100% intensity, the other has 0% (coherence is kept)
2) both paths have 50% intensity (coherence was lost previously)
Thoughts? What you expect to be the result of this experiment? why?
| As long as the external experimenter does not attempt to measure the phase of each of the arms - and indeed makes a measurement that does not 'touch' this information - he will observe no contradiction. In each of the definite outcomes of the cat experiment inside the lab, the experimenter inside the lab signals this definiteness by outputting 50/50 intensities out of the lab's two windows.
He also signals out what the outcome was, in the phase of the light, but the external experimenter does not look at this signal. (Whatever pieces of equipment are sensitive to this phase will, of course, be entangled with the inside of the lab.)
Thus, all that the external experimenter can tell is that a definite outcome was obtained inside the lab, by virtue of the light being 50/50 over both windows.
| {
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General relativity in terms of differential forms Is there a formulation of general relativity in terms of differential forms instead of tensors with indices and sub-indices? If yes, where can I find it and what are the advantages of each method? If not, why is it not possible?
| Short answer.
Example: The Palatini action, where the action is a functional of a tetrad/vierbein $e$ and a spin connection $\omega$.
$$S(e, \omega) = \int \epsilon_{abcd} e^a \wedge e^b \wedge \Omega^{cd}$$
where $\Omega$ is the curvature associated to the connection form $\omega$:
$$\Omega = D\omega = d\omega + \omega \wedge \omega$$
Other examples listed here (with references):
Holst action
Plebansky action
Samuel-Jacobson-Smolin action
Goldberg action:
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Increase in Mass with Velocity I just had a confusion. Does the mass of the body actually increase when it is moving with a certain velocity? Or does it only look like the mass has increase to another observer. How can the actual mass of the body increase. Please correct me if I am wrong but I feel that it only seems to the observer that the mass has increased but it does not increase in real.
The increase of actual mass would imply that that its velocity will decrease to comply with conservation of momentum. But how can the velocity decrease if no external force is applied?
| The inertia (resistance to acceleration) increases, and the object's tendency to curve space around it increases as it's velocity approaches the speed of light. But physicists have found it cumbersome to treat mass as a variable, or to keep saying "rest mass" all the time, so we say that the mass-proper stays the same.
Edit:
Some clarification in response to John Rennie's comment is in order. The gravitational field surrounding a small, massive particle, like a ball bearing, is given by the Schwarzschild metric, thus we are no longer considering purely special relativistic effects. If a test mass is moving with respect to the particle (as measured from a great distance), one would consider a boosted Schwarzschild metric.
If one is considering a distribution of mass in motion, the relevant object is the stress-energy tensor, which is the source term for the metric. The $T^{00}$ term is what you would normally call the energy of the particle, and is $\gamma m v^2 \delta(\mathbf{x}-\mathbf{x}_p(t))$ for a single moving particle. Things get a bit murky when you specify exactly what you mean by $\mathbf{x}(t)$, here, and you're best off taking $m$ to be much smaller than the other length scales in your spacetime.
| {
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How can space and time arise from nothing? Lawrence Krauss said this on an Australian Q&A programme.
"...when you apply quantum mechanics to gravity, space itself can arise from nothing as can time..."
Can you elaborate on this please?
It's hard to search for!
| The point is that spacetime can be emergent, i.e., you don't put it in by hand from the start, but it sort of pops out at you along the way.
Sometimes the dynamical degrees of freedom (the variables of the theory, if you like) can be different depending on how you describe the system. The best known example is string theory, where you start with an ordinary conformal field theory in 2 dimensions (no gravity, not really space-time per se), and out pops a 10-dimensional theory of quantum gravity. Oh, and stare at it from another end, and it looks like a theory of infinite dimensional matrices (no space, just time).
The AdS/CFT duality is another, related example, where on one hand you have a field theory in 4 dimensions, but no gravity, and on the other hand you have a theory in 5 dimensions, plus gravity, and you're really just looking at two parts of the same elephant.
There is also loop quantum gravity, causal dynamical triangulations, etc., but these are not as well established nor as successful.
The relevant search terms are induced gravity, emergent gravity, emergent spacetime, etc.
| {
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What exactly is a bound state and why does it have negative energy? Could you give me an idea of what bound states mean and what is their importance in quantum-mechanics problems with a potential (e.g. a potential described by a delta function)?
Why, when a stable bound state exists, the energies of the related stationary wavefunctions are negative?
I figured it out, mathematically (for instance in the case of a potential described by a Delta function), but what is the physical meaning?
| Mathematically, bound states are states that decay sufficiently fast at infinity, so that the probability of finding the particle they describe in far away regions of space is negligible.
It has long been conjectured, based on physical intuition, it is the case for the meaningful quantum mechanical states, such as the eigenfunctions of the Hamiltonian (it is not expected that an atomic electron has a sensible probability of being at infinite distance from its nucleus).
This has been proved mathematically in the eighties, mainly by S.Agmon. Roughly speaking, the result is the following: eigenfunctions of the Schrödinger operator (i.e. corresponding to the discrete spectrum) are exponentially decaying in space. So if $\psi_n(x)$ are such eigenfunctions, $\lvert \psi_n(x)\rvert\leq A e^{-B\lvert x\rvert}$, for some positive constants $A,B$.
| {
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Meaning of the chemical potential for a boson gas My lecturer told me that $\mu$, the Chemical potential, is zero or negative, and in the following example, mathematically it acts as a normalization constant. But is there any physical insight about why boson gas can be zero or negative?
I think it is due to the fact the photon gas can pop up from nowhere (i.e. vacuum fluctuation).
$$
f_{BE}(\varepsilon)=\dfrac{1}{e^{(\varepsilon-\mu)/(k_B T)}-1}$$
| I think you can consider the zero chemical potential to be an effect of popping up of photon. Both chemical potential and the temperature appears as a undetermined multiplier due to conservation of energy and number of particle respectively. For photon, there is no conservation of particle number and hence the corresponding undetermined multiplier, i.e. the chemical potential ($\mu/kT$ to be precise) is also zero for all energy states.
In other way you can say that since the number of particle is indefinite for each energy level, the equilibrium configuration is described by the minimisation of the free energy, $\partial A / \partial N_i =0$. When all the $N_i$ are independent of each other (as a result of spontaneous production and destruction of photon), this implies $\mu_i=0$ for all $i$.
| {
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At what point is the spin determined in a Stern-Gerlach Apparatus Consider a particle with spin that travels through a Stern Gerlach box (SGB), which projects the particle’s spin onto one of the eigenstates in the $z$-direction. The SGB defines separate trajectories for the particles that travel through it depending on their spin.
My Question: At what point is the spin determined when it is in superposition? When the particle starts to feel the magnetic field? Or only when the trajectory is measured in the detector?
This is a similar question, however it does not answer my question.
| The spins are determined when they hit something (i.e. "decoher" by interacting with a macroscopic entity.). You can infer a trajectory for them that extends back to the beginnings of the inhomogeneous field.
| {
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If a Goldstone boson is an excitation moving between degenerate vacua, how do symmetries remain broken? In spontaneous symmetry breaking, moving around the circular valley of the Mexican hat potential doesn’t change the potential energy. These angular excitations are called Goldstone bosons. But doesn't the change of angle implies that the system moves from one vacuum to another because different points on the circular valley represent degenerate vacua?
If Goldstone excitations are like this, how does the symmetry remain broken? Goldstone excitation, by definition (because they represent variations in the coordinate on the circular valley), will then take the system from one vacuum to another. However, this doesn't happen.
| The proposition that the symmetry G is spontaneously broken means that by acting by G on the vacuum configuration, we obtain an isomorphic but different configuration. For the symmetry to be unbroken, the transformations in G would have to map the vacuum configuration onto the same one, not just isomorphic one.
If you reflect the letter R along the vertical axis, you will get Я. This "ya" is isomorphic but it is different, so R isn't left-right-symmetric; the symmetry is broken; it is not ever possible for a symmetry to produce an object that even looks different (isn't isomorphic). It's always isomorphic; the question is whether it is identical. The letter H is mapped to H again so H is left-right-symmetric.
The Goldstone bosons' being nontrivial excitations proves that the action by G is nontrivial so the vacuum is not symmetric under G.
| {
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Why do high current conductors heat up a lot more than high voltage conductors? 120 volts x 20 amps = 2,400 Watts
However, if I increased the voltage and lowered the current, you can also use a smaller wire size (more inexpensive), also have less heat and achieve the same watt Power.
1,000 volts x 2.4 amps = 2,400 Watts
*
*Why doesn't it heat up like current?
*To me this approach seems more efficient and less costly because you don't use as much material, so why isn't this common?
| *
*The point is that the wire ends are not exposed to a potential difference equal to the full voltage. The wire ends differ in potential only by a (small) voltage drop $\Delta V = R\,I$, where $R$ is the resistance of the wire. Dissipation power equals $P = I\,\Delta V = I^2 R = \frac{(\Delta V)^2}{R}$. Now suppose you can transform $(I,V)$ while keeping the product $P_1 = V I$ roughly constant. Then dissipation power is $P = I^2 R = \frac{P_1^2}{V^2} R$, so higher voltage $V$ and, correspondingly, lower current $I$ results in less power dissipation $P$.
*Actually it is common to use high-voltage AC lines for high-power transmissions over long distances. In your home, you will not want to have kilovolt lines within millimeters from your fingertips however, cf. Brandon's comment.
| {
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Does corrosion of iron increase its mass? Does corrosion of a sample of iron increases or decreases its mass? I think that the mass will increase because of additional oxygen atoms.
| The following reactions take place :
$O_2 + 4 e^− + 2 H_2O → 4 OH^−$
$Fe → Fe^{2+} + 2 e^-$
$4 Fe^{2+} + O_2 → 4 Fe^{3+} + 2 O^{2−}$
$Fe^{2+} + 2 H_2O ⇌ Fe(OH)_2 + 2 H^+$
$Fe^{3+} + 3 H_2O ⇌ Fe(OH)_3 + 3 H^+$
$Fe(OH)_2 ⇌ FeO + H_2O$
$Fe(OH)_3 ⇌ FeO(OH) + H_2O$
$2 FeO(OH) ⇌ Fe_2O_3 + H_2O$
(source : Wikipedia )
Now since initially only $Fe$ was present and finally its oxides are present in the sample, there is definitely an increase in the mass of the sample.
| {
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Which of these two different forms of spin-orbit interaction is correct? I am seeing the spin-orbit interaction in two different ways:
*
*$\lambda [\mathbf{p} \times \nabla V]\cdot \sigma$
*$\lambda [\nabla V \times \mathbf{p}]\cdot \sigma$
I don't see how these two expressions can be equivalent though; in (1) there will be derivative operators from momentum acting on the gradient whereas in (2) this will not be the case.
Which one of these is correct?
| Good question. In fact, you can check that $\triangledown V\times \mathbf{P}=-\mathbf{P}\times\triangledown V$. Here is the brief derivation:
$\mathbf{P}\times\triangledown V=(p_y\partial _zV-p_z\partial _yV,p_z\partial _xV-p_x\partial _zV,p_x\partial _yV-p_y\partial _xV)$. Let's take the first component $p_y\partial _zV-p_z\partial _yV$ for example, it's obvious that
$$p_y\partial _zV-p_z\partial _yV=-i\frac{\partial^2}{\partial _y\partial _z}V+\partial _zVp_y-(-i\frac{\partial^2}{\partial _z\partial _y}V+\partial _yVp_z)$$,
but $\frac{\partial^2}{\partial _y\partial _z}V=\frac{\partial^2}{\partial _z\partial _y}V$, thus
$$p_y\partial _zV-p_z\partial _yV=\partial _zVp_y-\partial _yVp_z$$. While the first component of $\triangledown V\times \mathbf{P}$ reads $\partial _yVp_z-\partial _zVp_y$, therefore we arrive at $\triangledown V\times \mathbf{P}=-\mathbf{P}\times\triangledown V$.
| {
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What is the intrinsic colour of the star? Hey I have this question and I am wrestling with it all day and I am completely lost.
The V magnitude of a star is 15.1, B-V =1.6, and absolute magnitude Mv= 1.3 The extinction in the direction of the star in the visual band is av = 1 mag/kpc. What is the intrinsic colour?
Can somebody help me?
Edit: Ok, apparently I have to calculate first the distance and after that the excess colour and then I can calculate the intristic colour.
I guess I need this formula for the distance: V = Mv + 5log(r/10pc) + Av
But if i fill this in and get the R I get a different r and I dont know what the next step if to go to the excess colour.
| A star's color has almost nothing to do with its visual magnitude. The color is a function of size, age, etc. per the OBAFGKMNS classification scale (roughly). The visual magnitude is a measure of the light intensity (in the visual spectrum) which reaches us on Earth, so it depends not only on the star's color but its overall output and distance.
Further, the perceived color is redshifted in proportion to the speed at which the star is receding from us (which in general is a function of distance).
| {
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Non-locality and quanta Quantum mechanics is non-local in that long distance correlations are present, though there is no signalling possible. But QFT is Lorentz invariant and contains quantum mechanics as a special case. I assume this is not a paradox as paradoxes do not exist but I do not understand the details. Can anyone supply a reference or satisfactory explanation?
| Quantum Mechanics is non-local because you are working with particles and interactions all occur instantaneously. This doesn't allow locality. Quantum Field Theory is a framework that instead of working with particles, works with fields. A field has values at each point in space and allows local interactions. However, it doesn't enforce locality. It is only local because we choose equations of motion for our fields that are relativistic. If you wish you can write a non-relativistic Schroedinger equation in QFT and do non-relativistic calculations which will no longer obey locality.
| {
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Delta to Star/Y Conversions and vice versa in Electric Ciruits We all know the basic rules for conversion of $"Delta"$ circuits to $"Star"$ circuits and vice versa. We also know that this is needed for simplification of circuits in complex cases. Can anyone please explain HOW the concept of such conversions came about?
To be clearer, can anyone show the derivation of the Conversions? for both Capacitors and Resistors?
| The concept is a special case of a more general topological notion of graph theoretic duality: see the Wikipedia page for Dual Graph.
Graph theoretic duality is "compatible" with the Kirchoff voltage law (voltages around a loop sum to nought) and charge conservation (currents into a node sum to nought) insofar that nodes in a graph map to loops in a graph theoretic dual, so that we get a meaningful electric circuit for the dual if we swap the roles of voltage and current - the two laws (Kirchoff voltage and charge conservation also swap places). The impedances naturally transform too.
So, with graph-theoretic duality and electrical duality combined, we get the procedure written up in the Dual Impedance Wiki Page. The relationships between the star and its topological dual delta are specifically worked through as an example on this page.
As in Alfred's comment, which references:
http://www.engineersblogsite.com/delta-to-wye-and-wye-to-delta-conversion.html
he says the rules work for any impedance, not simply resistances. The topological reasons given above show why.
| {
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The physical interpretation of Newton's constant $G$ It is well known that the speed of light $c$ can be interpreted as the speed limit for information propagation. Similarly, the Planck's constant $h$ is interpreted as the minimum quantum package of action/entropy. Is there a similar interpretation for the Newton's constant $G$?
| I do not know of any such "limit" interpretation of $G$. The universal gravitational constant $G$ is defined by the Law of universal gravity:
Magnitude of the mutual attractive force two bodies exert on each other is given by
$$
F = G\frac{m_1 m_2}{r_{12}^2},
$$
where $m_1$, $m_2$ are masses of the two bodies and $r_{12}$ is their mutual distance, and is oriented towards the other body. Gravitational constant $G$ is the constant of proportionality in the above equation. In the SI system of units, the constant $G = 6.67\times 10^{-11}$ gives force in newtons, acting on 1 kg body due to another 1 kg body at distance 1 m (this is close to the way $G$ is measured).
| {
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Recovering 4-vector Lorentz transformation from spinor formalism I'm trying to recover the 4-vector transformation laws using spinors. I have defined
$$v^{\dot{a}b} = v^{\nu} \sigma_{\nu}^{\dot{a}b}$$
as usual, with $\sigma_0=1$.
Now with the rules for dotted and undotted spinor indices i get the transformed v for a boost in the z-direction
$$v^{\dot{a}b} \rightarrow v^{\dot{a'}b'}= \left({\mathrm{e }}^{ - \theta \frac{\sigma_3}{2}} \right)^{\dot{a'}}_{\dot{a}} \left({\mathrm{e }}^{ \theta \frac{\sigma_3}{2}} \right)^{b'}_{b} v^{\dot{a}b} = \begin{pmatrix}
{\mathrm{e }}^{- \frac{\theta}{2}}&0 \\ 0&{\mathrm{e }}^{ \frac{\theta}{2}}
\end{pmatrix}
\begin{pmatrix}
v_0+v_3&v_1-iv_2\\v_1+iv_2&v_0-v_3
\end{pmatrix}
\begin{pmatrix}
{\mathrm{e }}^{ \frac{\theta}{2}}&0 \\ 0&{\mathrm{e }}^{ - \frac{\theta}{2}}
\end{pmatrix}$$
where i have used the fact that $\sigma_3$ is diagonal and that ${\mathrm{e }}^A= \begin{pmatrix}
{\mathrm{e }}^{A_{11}}&0\\0&{\mathrm{e }}^{A_{22}}
\end{pmatrix}$ holds for every diagonal matrix A. This gives me the wrong transformation!
It would give me the correct transformation if I had
$$v^{\dot{a}b} \rightarrow v^{\dot{a'}b'}= \left({\mathrm{e }}^{ - \theta \frac{\sigma_3}{2}} \right)^{\dot{a'}}_{\dot{a}} \left({\mathrm{e }}^{ \theta \frac{\sigma_3}{2}} \right)^{b'}_{b} v^{\dot{a}b} = \begin{pmatrix}
{\mathrm{e }}^{- \frac{\theta}{2}}&0 \\ 0&{\mathrm{e }}^{ \frac{\theta}{2}}
\end{pmatrix}
\begin{pmatrix}
v_0+v_3&v_1-iv_2\\v_1+iv_2&v_0-v_3
\end{pmatrix}
\begin{pmatrix}
{\mathrm{e }}^{ - \frac{\theta}{2}}&0 \\ 0&{\mathrm{e }}^{ \frac{\theta}{2}}
\end{pmatrix}$$
But I can't figure out why this should be the case. The only possibility I can think about would be if a relation like
$\ A'_{\nu\mu} = M_{\mu}^{\ \rho}(M_{\nu}^{\ \theta})^{-1}A_{\rho\theta }$ $\rightarrow$ $A'=MAM $ would hold, but i can't find a formula like this. Doing the summation by hand i get the same result as with normal matrix multiplication without using the inverse matrix on the right hand side. Any tip or help would be much much appreciated!
| First thing is that you are missing the imaginary unit in the exponential. The correct transformation matrix should be
$$M=e^{i\frac{\theta}{2}}.$$
Up to this small misprint, the first expression you wrote actually gives the desired transformation of the components $v_1$ and $v_2$. If we do the computation
$$
\begin{aligned}
&\begin{pmatrix}
{\mathrm{e }}^{- \frac{i\theta}{2}}&0 \\ 0&{\mathrm{e }}^{ \frac{i\theta}{2}}
\end{pmatrix}
\begin{pmatrix}
v_0+v_3&v_1-iv_2\\v_1+iv_2&v_0-v_3
\end{pmatrix}
\begin{pmatrix}
{\mathrm{e }}^{ \frac{i\theta}{2}}&0 \\ 0&{\mathrm{e }}^{ - \frac{i\theta}{2}}
\end{pmatrix}=\\
&\begin{pmatrix}
v_0+v_3 & e^{-i\theta }(v_1-iv_2) \\
e^{i\theta }(v_1+i v_2) & v_0 + v_3
\end{pmatrix}.
\end{aligned}
$$
This implies
$$
\begin{aligned}
v_1'-iv_2'&=e^{-i\theta}(v_1-iv_2),\\
v_1'+iv_2'&=e^{+i\theta}(v_1+iv_2),
\end{aligned}
$$
that finally gives the correct transformation.
I'm not sure, may be you made a mistake in multiplying of matrices.
| {
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What form of energy is released in $E = mc^2$? In the famous mass energy equivalence $ E = mc^2$ whenever any form of mass, lets say mass defect in nuclear reactions or any other example of mass is converted to Energy, the released energy is in what form ? Heat ? Light ? Other ?
| See Can we make usable energy from subnuclear particles? for a related discussion. The energy is released both as the kinetic energy of the reaction products and as (usually gamma ray) photons.
| {
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Will density of a metal increases during forging? This question is metallurgical engineering, but I had a similar doubt regarding density of liquids and what causing it.
Forged parts refines defects, dislocations will be moved strengthening the metal. But will the density of forged metal change?
My earlier question was, what causes liquids to have different densities?
| The answer is that in principle yes, the removal of defects will increase the density. The question is by how much.
You specifically ask about forging. The forging process doesn't remove dislocations but instead moves them until they become pinned, which is what causes the work hardening. Well, that's an oversimplification since some defects will be removed but on the other hand fresh defects will be created. So whether forging will change the density, and if so by how much, is a hard question to answer.
The extreme version of removing defects would be to compare the densities of crystalline and amorphous metal. However I'm not sure that the common metals can be prepared in well characterised amorphous forms. If you compare that most common of amorphous solids, silica glass, then the density of the amorphous form is about 10% lower than the density of quartz.
I did have a Google for the effect of forging on density (I'm sure you did too) but I couldn't find anything that looked reliable. I did find this paper that reports a density decrease, but the authors suggest this is due to formation of small cracks. I think it would be a hard experiment to do as there are other variables that are hard to control.
| {
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Does black hole formation contradict the Pauli exclusion principle? A star's collapse can be halted by the degeneracy pressure of electrons or neutrons due to the Pauli exclusion principle. In extreme relativistic conditions, a star will continue to collapse regardless of the degeneracy pressure to form a black hole. Does this violate the Pauli exclusion principle? If so, are theorists ok with that? And if it doesn't violate the Pauli exclusion principle, why not?
|
Does this violate the Pauli exclusion principle? If so, are theorists
ok with that?
The short answers are "yes" and "yes".
Recall that we are talking about what happens inside the event horizon ...
Perhaps the density of states diverges as volume decreases.
However iirc most thinking is around the idea that there is a quark degeneracy limit that has to be overcome like the neutron degeneracy limit.
i.e. those people speculating some process that combines quarks into some boson can pat themselves on the back.
The bottom line is that we don't know enough about how matter/energy behaves in such extreme conditions to be able to do more than speculate.
Also see:
http://www.physicsforums.com/showthread.php?t=600360
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Does a photon instantaneously gain $c$ speed when emitted from an electron? An excited electron loses energy in the form of radiation. The radiation constitutes photons which move at a speed $c$. But is the process of conversion of the energy of the electron into the kinetic energy of the photon instantaneous? Is there a a simple way to visualize this process rather than math?
| MC Physics would suggest that photons are formed from the joining of 2 opposite electro-static charged mono-charges that are emitted from particles or atoms due to excessive vibration. Those joined mono-charges are (almost) instantly accelerated to c by surrounding electric-magnetic forces then any excess force is applied to rotate the real particle to frequency. This is further described at:
https://fs23.formsite.com/viXra/files/f-1-2-9037551_RhNE84zB_MC_Physics-_Model_of_A_Real_Photon_With_Structure_and_Mass.pdf
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Minimum seperation between two Airy disks as a function of the distance between two point sources of coherent light passing through the same objective I have two coherent point sources of light, $A$ and $B$, separated by a distance $L$, which I focus down to the diffraction limit using a high-powered objective (e.g. a $\approx 100x$ objective). If I turn on $A$ and turn off $B$, I have an Airy disk at position $c_1$, and I turn off $A$ and I turn on $B$, I have an Airy disk at position $c_2$. Given that both light sources are sent through the same objective, what is the minimum distance between $c_1$ and $c_2$? Is it simply $L$ scaled down by the objective (i.e. $\frac{L}{100}$)? Or does something odd happen because of e.g. curvature of the lens in the objective?
EDIT: A restatement of this question would be the following - Assuming all of the optics are perfect, if I shine a laser at a point (x,y) on an objective, and then shine the laser at a point (x2,y2), will the peak of the Airy disk move the same distance?
| Having spent rather a lot of years designing and building adaptive optic systems, I've seen this sort of problem in a number of guises.
Certainly you can apply some fitting function to the image of each of your spots separately and determine the centroid of each spot alone. If your detector has sufficient resolution, these two measurements will yield the different locations of the two Airy discs.
The whole deal with the "Rayleigh limit" is simply that the two-spot pattern has no local minimum between the two peaks. In the absence of knowledge of the source, you can't tell whether it's two point sources or a single extended source. If you know in advance that two sources (such as neighbor stars) are producing the image, and you have a good idea of their relative brightness, you can fit the image to a different function, i.e. the sum of two Airy discs, to determine the peaks.
However, remember that all light entering the lens system at a given angle will be focussed to the same location. So it's not moving the laser beam to different positions on the lens but rather aiming them at different entrance angles. (I'm ignoring coma and other off-axis effects of a real optic system here)
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Efflux speed of ideal fluid dependent on diameter? I have a cylinder full of water with diameter $D$ with a round opening on the bottom with diameter $d$. The water is friction-free and incompressible. Now I need a relationship for the efflux speed $v$ with which water exits the cylinder and I shouldn't use the approximation $d \ll D$, but formulate a general relationship.
Ok. So what I thought is to equate the Bernoulli law on the top of the cylinder with that on the bottom of the cylinder which gives me $v=\sqrt{2gh}$. Solved. How does the speed of efflux depend on the diameter of the efflux hole? I googled quite a lot and all I could find was the above relationship...
| While calculating velocity of efflux you use bernoulli's theorem as follows :
At the top of container :$ P + \frac{\rho {v_1}^2}{2} + \rho gh_1 = k$
At the efflux : $P_a + \frac{\rho {v_2}^2}{2} + \rho g h_2 = k$
$P - P_a + \rho g (h_1 - h_2) = \frac{\rho ({v_2}^2 -{v_1}^2)}{2} $
Now, in accordance with equation of continuity,
$ v_1 A_1 = v_2 A_2$
$ v_1 (\pi D^2) = v_2 (\pi d^2) $
$ v_1 = v_2 \frac{d^2}{D^2}$
If $ d<<D$ , then $v_1 = 0$.
This gives us :
$v_2 = \sqrt{2gh + \frac{2(P-P_a)}{\rho}}$
Also if top of container is open, or pressure inside container is $P = P_a$ , then $v_2 = \sqrt{2gh}$
If the condition $d<<D$ was not given, we would have to put value of $v_1$ along with $v_2$ in bernoulli's equation.
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Potential energy curve for intermolecular distance
(source: a-levelphysicstutor.com)
I'm trying to understand this curve better, but I can't quite figure out what "negative potential energy" means.
The graph should describe a molecule oscillating between $A$ and $B$, however where I'm stuck in reasoning this is that the PE is equal in $A$ and $B$, but then why does this mean $r$ will increase in $A$ (repel) and decrease in $B$?
| Suppose that two molecules are at distance $B$ and have zero kinetic energy. There's a lower potential energy position in $C$ and therefore the molecules will attract.
They will convert $\epsilon$ potential energy into kinetic energy and reach $C$.
Now, the law of inertia states, and the fact that they have positive kinetic energy indicates, that they will maintain their state of motion at $C$ towards $A$.
Going towards $A$ they will gain potential energy by converting kinetic energy into it (in other words, slowing down).
Once at the distance $A$, they will have gained exactly $\epsilon$ potential energy and will have therefore zero kinetic energy.
At this point the cycle will repeat inverted, they will move towards a distance of $C$ because it's lower energy, surpass it and reach $B$ with zero kinetic energy, which will make the cycle repeat from the start.
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During reflection does the emitted photon have same properties? When light (photon) is reflected the the original photon is absorbed by an electron and then emitted again. Does this "new" photon have the same wavelength, frequency etc. as the original?
| Like any quantum field you can approximate light as either a particle or a wave. However you need to be clear that both are just approximations to the true nature of light and like all approximations they work well in some circumstances and badly in others.
In this case the photon model is a poor way to describe the process of reflection. Reflection doesn't involve photons being absorbed then re-emitted. You're correct that the oscillating light wave of the light interacts with electrons in the reflector, and the resulting oscillating dipoles reradiate light. However, while this is easily described using a wave model it's hard to describe using photons. As a general rule the photon model works well when light is exchanging energy with something so for example it would be a good model if the light was ejecting photoelectrons from the mirror. When we're studying the propagation of light the wave model is a far better approximation.
So I don't think your question can usefully be answered as it's currently phrased. However we can say that the reflected light has approximately the same frequency as the incident light. A say approximately because in principle there are effects that could change the frequency. For example if the mirror is floating in zero-G then the momentum change when the light is reflected will cause the mirror to accelerate by a tiny amount and therefore red shift the reflected light by a tiny amount. In most circumstances we can ignore these small effects.
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How do I explain to a six year old why people on the other side of the Earth don't fall off? Today a friend's six year old sister asked me the question "why don't people on the other side of the earth fall off?". I tried to explain that the Earth is a huge sphere and there's a special force called "gravity" that tries to attract everything to the center of the Earth, but she doesn't seem to understand it. I also made some attempts using a globe, saying that "Up" and "Down" are all local perspective and people on the other side of the Earth feel they're on the top, but she still doesn't get it.
How can I explain the concept of gravity to a six year old in a simple and meaningful way?
| There's a good chance you won't be able to make her understand how it works.
According to developmental psychologist Jean Piaget, children at her age tend to see themselves as the "center of the world", and are incapable of reasoning about it from any other viewpoint. The process is called egocentrism, and is the same reason why a kid gets confused when its mother calls its (the mother's) parents mother/father instead of grandmother/grandfather.
It won't hurt trying to explain, and she may gain some insight, but don't get frustrated if she doesn't.
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Dielectric constant of water I need the dielectric constant of water from $10^{-2}$ Hz to $10^4$ Hz. As stupid as it may seem, I cannot find much info. I've googled for days. All I can find is close to GHz. And the only info close to Hz, ($100$ Hz) shows a great variation. A relative dielectric constant at $100$ Hz of about $4000$. So, I cannot interpolate back in frequency (I put a link to the paper at the end). Does anyone have any info about where I could find this data? I know that for constant current and about $20$ C the constant is $80.1$. What about at $50$ Hz?
I need the complex dielectric constant to test a program. Any lead would be really appreciated.
http://arxiv.org/abs/1010.4089
| There is some data in http://www.nist.gov/data/PDFfiles/jpcrd487.pdf (J. Phys. Chem. Ref. Data, vol. 24, No. 1, 1995, p. 33) See, e.g., Table 2 there. Looks like dielectric permittivity of water is about 78.
| {
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What is the 'resolution' of the reality in pixel terms?
*
*What resolution should a TV screen have so that its image were so faithful as reality as if the TV were a window?
*Also what would happen if Physics could reproduce a 'pixel' of the size $ l_{p}^{2} $ the square of the Planck length? Would be then this the resolution of our reality like watching through a real window?
| It is important to understand that Planck's theory of the existence of length which can be divided by zero is hotly disputed. Einstein famously said about this theory: 'madness is trying the same over and over again and expecting different results'. Theory of continuos space has never been disproven, it works in practice, I think we should stick to it. Planck's length is a fraud, lets call it as such and move on.
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Energy spectrum, mass spectrum, and mass gap In Arnold Neumaier's nice answer to this question:
The energy spectrum in quantum field theory
it is remarked that
'If there is a mass gap (i.e., if no representation of tiny positive mass exists), the states can be restricted to their rest frame, where the spatial momentum vanishes; in this case the energy spectrum agrees with the mass spectrum...'
I wonder why it is that this holds only if there is a mass gap. That is, why could one not look at the mass spectrum as an energy spectrum in a rest frame in the case where there is no mass gap?
| The lack of a mass gap goes in all interesting cases together with the existence of a massless particle. These don't have a rest frame. See, e.g., the section ''What is the mass gap?'' in my theoretical physics FAQ at
http://arnold-neumaier.at/physfaq/physics-faq.html
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How can the commutator operation not be transitive? I noticed the following:
$$[L_{+},L^2]=0,\qquad [L_{+},L_3]\neq 0,\qquad [L^2,L_3]=0.$$
This would suggest, that $L^2,L_+$ have a common system of eigenfunctions, and so do $L^2,L_3$, but $L_+,L_3$ don't. How is that possible?
| NOTE. Since $L_+$ is not normal (normal means $A^\dagger A = AA^\dagger$) it does not admit a basis of orthonormal eigenvectors. However your question can be safely restated replacing $L_+$ for $L_2$ and I will assume it henceforth.
The most elementary case of this phenomenon is given by a triple of normal matrices in $\mathbb C^n$:
$$cI,A,B$$
with $[A,B]\neq 0$ and where $c\in \mathbb C$ is an arbitrarily fixed number. $A$ has a common basis of eigenvectors with $cI$: Every basis of eigenvectors of $A$ is such basis. Similarly, every basis of eigenvectors of $B$ is also a basis of eigenvectors of $cI$. However, though it could happen for some vector, there cannot exist a whole basis of eigenvectors in common with $A$ and $B$, otherwise referring to that basis $A$ and $B$ would be in diagonal form and thus $[A,B]=0$, which is forbidden by hypotheses.
All that is possible thanks to the fact that the eigenspaces of $cI$ are (maximally) degenerate. Two vectors $u$ and $v$ with the same eigenvalue ($c$) of respect to $cI$ remain eigenvectors of $cI$ with the same eigenvalue even if linearly composed: $au+bv$. Nevertheless if $u$ and $v$ are eigenvectors of $A$, in general $au+bu$ is not, but it could be an eigenvector of $B$ (remaining, as said, an eigenvector of $cI$)
The situation is essentially the same when dealing with $L^2$ and $L_2,L_3$.
The eigenspaces $\cal H_l$ of $L^2$ are degenerate and, in each eigenspace, $L^2$ is represented by $l(l+1)I$.
Moreover, as $[L^2,L_i]=0$, each eigenspace $\cal H_l$ is invariant under the action of $L_i$. I mean $L_i({\cal H}_l)\subset \cal H_l$.
Restricting to $\cal H_l$, we find the situation I outlined above: $L$ is represented by $cI$ and $L_2, L_3$ are represented by non commuting operators $A$ and $B$.
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Can a mass matrix be asymmetric? I am developing a mathematical model of a mechanical device consisting basically of coupled harmonic oscillators. It turns out that the system mass matrix is asymmetric. I seem to read somewhere that a mass matrix has to be symmetric, but I am not sure. So I would like to know whether it is possible for a mass matrix in this case to be asymmetric. If it can't, what are the physical implications of an asymmetric mass matrix in this case?
| Whenever the mass is expressed as a real matrix $M$, the mass term or something else that matters is a bilinear expression. For example, the effective mass in condensed matter physics is the mass matrix $M$ so that the kinetic term of the Hamiltonian is written as
$$ E_k = \frac{\hbar^2}{2m} \vec k \cdot M^{-1} \cdot \vec k $$
where $M^{-1}$ is the inverse matrix. One may always divide the latter matrix to the symmetric and antisymmetric part. The antisymmetric part doesn't affect the Hamiltonian (doesn't affect the physics) at all because it is being contracted with the symmetric tensor $k_i k_j$. So without a loss of generality, we may demand the matrix to be symmetric, and everyone does so.
In some contexts, like the mass terms for complex fields in quantum field theory, it is only the Hermitian part of the mass matrix that affects the Lagrangian or the Hamiltonian because the term is $\bar\psi \cdot M \cdot \psi$ with an extra bar (complex conjugation). In that case, we assume that the mass matrix is Hermitian for a reason analogous to the previous paragraph; the anti-Hermitian part doesn't contribute.
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Why does ice form on bridges even if the temperature is above freezing? So with this "arctic blast" continuing, I've noticed that for my area, the temperature drops below freezing just long enough to cause freezing rain, but then the sun comes out and the temperature rises immediately. However, on bridges, ice continues to form.
How can ice form even if the temperature is above freezing?
| The wind's "chill factor" is a surface area effect. In a bridge, the surface area exposed to the wind is at least twice as much as a comparable section of the road, so with the proper conditions, the bridge's surface will be colder than the freezing point while the road is above this point.
For example, assume the still air temperature is 44F and the chill factor -10F. The temperature on the road would be 34F, while the temperature on the bridge would be around 24F.
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light beams of the sun
We receive sunlight on earth surface. What type of light beams are these?
Random/Parallel/Converging/Diverging
I think it should be Diverging as Sun is radiating these beams away. But in one book, answer is given as Random, in another it's Parallel.
| It is difficult to answer this question. An EM wave is generated by vibrating charges and nuclear reactions. Sun is full of vibrating charges and nuclear fusions. Because of this full range of frequencies are emitted. At distances close to sun we observe the directions of waves to be random. But at far away distances the direction of waves seem parallel. Since only parallel waves can have constant separation between them. Converging and diverging waves become distant at longer distances.
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How do photons 'connect' during wireless connection? So wireless router broadcasts a signal and then your device searches. So what actually happens when the photons 'meet' it's kind of like saying, 'ah your one of us, so we will follow you, show us the way'
It's so bizarre, how do photons connect during wireless connection?
| In the sense of routers (transmitters and receivers) it is preferable to consider the wave-model for the electromagnetic field.
The router has an antenna that, due to how it is shaped and how the current running through it is modulated, creates an electromagnetic field that propagates depending on direction and construction of antenna. For routers the ideal case would be to have an isotropic antenna (one that radiates equally in all directions) but this antenna is difficult to construct (a perfect isotropic antenna would be a metal sphere, but it has to be driven by current so this would destroy the perfect-sphere aspect of it) so wifi-routers are not radiating isotropically, but they're designed to be close to it.
When this electromagnetic wave produced by the router hits a wifi-receivers antenna, a current is induced in it. The phase and amplitude (and possibly frequency but this is rarer) of the electromagnetic field depends on the current running through the transmitting antenna, and this also creates a corresponding current in the receiver. The same current that is running in the transmitter is induced in the receiver, with more noise and lower magnitude however.
Your understanding of it as guided photons is inaccurate, they are not guided at all by each other but they are emitted in a directional pattern. These are waves that are travelling, and as any other wave, they surrender to the theory of superposition so they maybe interfere with each other in different spatial locations, but they are not "drawn" to each other in any way.
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What is average life in radioactivity and what is its significance? By definition, average life of radioactive sample is the amount of time required for it to get decayed to 36.8% of its original amount.
But what is the significance of 36.8% and why has that value been chosen?
| When radioactive element A decays to produce element B, the (infinitesimal) number of decayed elements A, $dN$, that occurs in a small time interval, $dt$, is proportional to the initial population of A, $N$:
$$
-\frac{dN}{dt}\propto N
$$
Assuming the proportionality is a constant, then the above becomes
$$
-\frac{dN}{dt}=\lambda N
$$
which has a known solution:
$$
N(t)=N(0)e^{-\lambda t}=N(0)e^{-t/\tau}
$$
where $\tau$ is the mean lifetime. Now when 36.8% of the initial material remains then, $N(t)/N(0)=0.368$ and we get
$$
0.368=e^{-t/\tau}
$$
Taking the natural logarithm of this to eliminate the exponential, we find that
$$
\ln0.368=-1=-\frac{t}{\tau}
$$
Thus, the value 36.8% signifies the amount of material left over after one mean lifetime has passed.
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Idea of Covering Group
*
*$SU(2)$ is the covering group of $SO(3)$. What does it mean and does it have a physical consequence?
*I heard that this fact is related to the description of bosons and fermions. But how does it follow from the fact that $SU(2)$ is the double cover of $SO(3)$?
| I'd like to add to Josh's answer, because he didn't really explain what a universal covering group is. Essentially, a space $T$ is a covering space of another space $U$ if, for an open subset of $U$, there's a function $f$ that maps a union of disjoint open subsets of $T$ to the subset of $U$. Or, more simply worded, pick a piece of your space $U$, and I'll find you some number of different pieces of $T$ that have a function mapping them onto the piece of $U$. In the case of $T = Spin(3)$ and $U = SO(3)$, there are two disjoint subsets of $Spin(3)$ for every subset of $SO(3)$, so we say that $Spin(3) \cong SU(2)$ is the double cover of $SO(3)$.
Now, the way that this relates to bosons and fermions is where Josh's answer comes in. We want physical states to live in vector spaces that carry (projective) representations of our symmetry groups. The "projective" part means that our states may pick up a phase when transformed to other states -- so, for example, if you rotate a spin-1/2 state 360$^{\circ}$, the state picks up a minus sign. It turns out that, at least in the case of $SO(3)$, we can eliminate the need for the "projective" part of this -- and thus those pesky minus signs -- by considering instead representations of the covering space.
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Why are temperatures generally hotter in the Middle East than in Europe? How come the average temperature in the middle east (Israel, Saudi Arabia, Sudan or lower) is always so much significantly higher than in Europe (say Germany, England etc.)?
I know that the sun rays pass a greater distance to Europe than the middle east, but is that the only factor influencing? And also, the distance isn't that much greater so how come the sun is weakened so significantly in Europe in comparison to the middle east?
| The distance of the Sun from Europe or the Middle East plays virtually no role. After all, many people on the Northern Hemisphere might be surprised that the Earth is closest to the Sun in January – it was on January 4th, 2014. It was 3 million miles or 3 percent closer than it is in July. Nevertheless, the winter is cold! Moreover, these 3 million miles are much greater than 3 thousand miles between Europe and a place of the Middle East but even 3 million miles are too small to really matter.
The winter is cold and Europe is colder than the Middle East for the same reason: the sun rays are bombarding the Earth's surface from a more "horizontal" angle than in the summer or in the Middle East throughout much of the year. When the angle of the sun rays is $\alpha$ from the vertical direction, the actual energy and heat coming per unit are is
$$ \cos\alpha\cdot P $$
where $P$ is the power you only get if the rays are bombarding the surface from a perpendicular, normal direction. If you substitute $\alpha\to 90^\circ$, the expression above goes to zero.
The values of $\alpha$ are generally smaller in the Middle East than in Europe and $\cos\alpha$ is therefore greater because the Middle East is closer to the equator, it has a smaller "latitude", we say, and the equator is the place where the Sun often illuminates the Earth's surface from a perpendicular direction. On the contrary, the poles are cooler because the solar radiation only "touches" the surface while it moves almost horizontally. Consequently, $\cos\alpha$ is very small. Europe is somewhere between the Middle East and the North Pole so its temperatures are somewhere in between, too.
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is there any relation between resistance and magnetism? i was holding a resistance wire(insulated) coiled up and both of its ends were connected to a Ohm-meter it was showing 18 ohms while the circuit was on i pulled the coil from both ends and made that a straight wire with barely a knot in between and observed that the reading of Ohm meter dropped to 16.5 Ohms. Is there any relationship between resistance and magnetic field? or was it just because the ohm-meter is having errors?
| There are a few possibilities here, Wikipedia shows many types of ohmmeters but I am going to consider only two; 1. Constant voltage and current measurer, 2. Constant current and voltage measurer.
In your case as you stretch the wire two things happen :
1. Resistance increases as length is slightly increased and cross sectional area reduced.
2. Self-Inductance is highly reduced, thus reducing back emf.
For the 1st type of ohmmeter both parts result in showing decrease in resistance. As the resistance of circuit increases and opposition potential drops the current through the circuit increases, thus increase in measured current shows decrease in resistance as the device assumes a fixed potential.
For the 2nd type of ohmmeter, the increase of resistance increases the current in voltmeter arm, while reduced self-inductance decreases the current in voltmeter arm. As the reading of voltmeter depends directly on current through it and you observe reduction in resistance I would say that reduction in self-inductance decreases the current much more than that increased due to change in resistance.
Of course, as you see that your stretching of the wire to make it straight does in fact increase it's resistance, both the ohmmeters have faults ! A more accurate way to measure resistance would be to use a meter bridge or potentiometer.
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Temperature: Why a Fundamental Quantity? Temperature is just an indication of a combined property of the masses of the molecules and their random motion. In principle, we can explain "no effective energy transfer between two conducting solid bodies in contact" via a condition in terms of the masses of the molecules and their speeds such that due to the collisions of molecules of two bodies, net energy transfer between two bodies is zero. But it would be a complex calculative work to derive this condition analytically so we use the temperature scale just as a phenomenological parameter to easily determine the condition of "no net energy transfer between conducting solids" for practical purposes. But it does not denote any fundamentally new property of the body separate from the already known mechanical properties of the same. Then why do we call it a fundamental quantity, e.g. in the SI list of fundamental quantities?
| It is just a scale to get thermal equlibrium problems easily but defined in such a way that it can not be expressed only in the terms of the other fundamental quantities.So it is a fundamental quantity.
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If the k of a wave is negative, is the wavelength negative too? My friend went to an interview for a reputed scholarship program and was asked this question.
A wave has an equation $a\sin(\omega t-kx)$.
Sometimes k surely can become -ve.
We know that $k=\frac{2\pi}{\lambda}$.
So $\lambda$ is -ve?
How can this be?
What he said was that we can write $k=\frac{\omega}{v}$.
Since $\omega=\frac{2\pi}{t}$, and t can't be -ve, v should be -ve.
So it implies that the wave is in -ve x direction.
That is why wavelength has come -ve due to this sign convention.
But they didn't agree to it.
Even I think the above is correct.
Where is the problem then?
| Your equation for the wave is really a vector equation:
$$ \psi({\bf x}, t) = a \sin(\omega t - {\bf k . x}) $$
This tends to be glossed over when students are first taught the equation, and to be fair in 1D the dot product $\bf k.x$ is simply $kx$ or $-kx$ depending on whether $\bf k$ and $\bf x$ point in the same or opposite directions.
Anyhow, $\bf k$ is the wave vector and like all vectors has direction and magnitude. Its magnitude is the wave number $k$, and the wave number is equal to $2\pi/\lambda$. Because $k$ is the magnitude of a vector it is always positive, and therefore $\lambda$ is always positive.
When you have a wave travelling in the $-x$ direction it's not $\lambda$ that changes sign, it's the direction of $\bf k$.
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Is tripleaxis planet possible? Imagine. Our solar system. Our sun. Then earth and moon orbiting it. And you have "powers" to create any planet you want, any size, any density, any weight and any velocity. Would it be possible for you (using all knowledge of earth), to create a natural satellite to moon? Whose trajectory would be almost circular/ellipsoidal?
Question actually goes only as follows: Can moon of moon (actual moon) of moon (earth) of sun exist?
| If things are in close, you have the three-body problem (four, given the sun - but it is distant). Three-body orbits are not uncommon in the universe. Orbit evolution prediction is fundamentally impossible, though abundant special solutions exist,
http://en.wikipedia.org/wiki/Three-body_problem
http://news.sciencemag.org/physics/2013/03/physicists-discover-whopping-13-new-solutions-three-body-problem
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Understanding Triplet And Singlet States We know, that for SU(2) representations $$\textbf{2}\otimes \textbf{2}=\textbf{3}\oplus \textbf{1}$$ where $\textbf{2}$ stands for the fundamental representation of SU(2). This means that we have a spin triplet of states and a spin singlet. Can we regard these states as the spin part of wavefunction for the excited states and the ground state of the deuteron nucleus? What is the difference between the members of a triplet?
| The spin part of the quantum state of any system consisting of two spin-$1/2$ particles (including a Deuterium nucleus) can be described as a general linear combination of the singlet and triplet states.
The the symbolic manipulation $2\otimes 2 = 3\oplus 1$ is telling you that the Hilbert space of the system of two spin-$1/2$ particles, which is simply the tensor product of the spin-$1/2$ Hilbert space with itself, admits an orthonormal basis for which $3$ of the states in the basis, the so-called triplet states, have total spin quantum number $s=1$, while one of the states in the basis, the so-called singlet state, has total spin quantum number $s=0$.
More explicitly, if we denote $|s, m\rangle$ as a state with
\begin{align}
S^2|s,m\rangle = \hbar^2 s(s+1) |s,m\rangle, \qquad S_z|s,m\rangle = \hbar m|s,m\rangle
\end{align}
where $S^2$ is the total spin squared operator, and $S_z$ is the $z$-component of the total spin operator, then the triplet states are
\begin{align}
|1,1\rangle, \qquad |1,0\rangle, \qquad |1, -1\rangle,
\end{align}
and the singlet state is
\begin{align}
|0,0\rangle,
\end{align}
and together they form a basis for the spin Hilbert space of the two-spin-$1/2$ system.
As can be seen explicitly in the notation, the triplet states are distinguished by the value of their "magnetic" quantum number $m$.
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Why isn't quantum entanglement just a lack of information? From this question and answer from joshphysics, I didn't understand one thing, even after reading the comments:
Why should assume that entangled electrons will only "decide" their state after observation?
Isn't it just a lack of information? For instance, knowing that we have a black and white marble and picking one afterwards with your eyes closed, doesn't mean that the marble "decides" to be white or black after the observation, i.e. there is no entanglement.
Therefore, I do not know what makes the states of the electrons so special.
| Suppose you have four marbles, each either black or white. Because you haven't looked at them yet, you're only aware of certain probabilities. One way or another, suppose you know that Marbles 1 and 2 have only a 5% probability of being differently colored. Likewise, Marbles 2 and 3 have only a 5% probability of being differently colored, and likewise for Marbles 3 and 4. Then you can conclude that Marbles 1 and 4 have at most a 15% chance of being differently colored --- because in order for 1 and 4 to be differently colored, it must be the case that either 1 and 2 are differently colored or 2 and 3 are differently colored or 3 and 4 are differently colored.
Now suppose that your conclusion turns out to be wrong --- that actually marbles 1 and 4 in fact have a 95% chance of being differently colored. Then something must have been wrong with your reasoning, and if you work through the various things that could have gone wrong, you'll find the most plausible culprit is your assumption that the marbles started out with well defined colors.
Something very like that happens in quantum mechanics. (In the quantum mechanics version, it is not possible to examine marbles 1 and 3 at the same time, or to examine marbles 2 and 4 at the same time.) And as you can see above, if the only problem were "lack of information", you'd still be able to conclude that your quantum marbles 1 and 4 can be differently colored at most 15% of the time, and then experiments would prove you wrong.
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Does optical fiber preserve the angles of incoming light? I am reading up on optical fibers and there's one thing I haven't understood yet:
*
*Does an optical fiber preserve the angles of incoming light?
For example, if we light two LEDs in front of an optical fiber (within its acceptance cone) and hold the other end of the fiber in front of a paper, would we see two separate light dots on the paper (because the angles of the incoming light were preserved), or just one (because the angles were messed up...)?
(Another way of posing this question: What would I see if I looked into the end of an optical fiber – would I see what's in front (within the acceptance cone) of the other end of the fiber or just a blurry dot? - disregarding I would need extremely good eyesight etc...)
| An endoscope uses a bundle of multimode fibers. The number of fibers determines the resolution of the image transported through the fiber bundle. The angle of light isn't preserved in a single fiber. Usually - depending on the length and bending of a such a multimode fiber - angles/modes will be mixed in some way; therefore you need a bunch of fibers to preserve angular or spatial information of light using fiber technology.
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The difference between $\mathcal{N}=2$ short multiplets and BPS states I have some questions about the construction of $\mathcal{N}=2$ supermultiplets for chiral matter. I know that the supermultiplet should not include spin one states since they are always in the adjoint representation. So my first question is: why are spin-one modes always in adjoint representation?
To avoid confusion, I will denote the four supersymmetry generators as $Q^A_\alpha$, where $\alpha$ is the spinor index and $A=1,2$.
Firstly, massless multiplet or short multiplet.
We know $Q^A_2$ generates states with zero norm. Then we can just focus on $Q^A_1$. The supermultiplet can be constructed in this way:
$$|\Omega_{-\frac{1}{2}}>\\
Q^{\dagger 1}_\dot{1} |\Omega_{-\frac{1}{2}}>, \, \, Q^{\dagger 2}_\dot{1} |\Omega_{-\frac{1}{2}}>\\
Q^{\dagger 2}_\dot{1} Q^{\dagger 1}_\dot{1} |\Omega_{-\frac{1}{2}}>$$
The two spin-zero states in the second line form an $SU(2)$ doublet.
Secondly, BPS states. In the presence of a central charge $Z$, one can write the four generators in terms of $A_\alpha$ and $B_\alpha$ after linear transformation with $\{ B_\alpha, B^\dagger_\beta\}=\delta_{\alpha\beta}(M-\sqrt{2}Z)$ and $\{A_\alpha, A^\dagger_\beta\}=\delta_{\alpha\beta}(M+\sqrt{2}Z)$. When $M=\sqrt{2}Z$, one can get $\{ B_\alpha, B^\dagger_\beta\}=0$. Thus, the multiplet will be:
$$|\Omega_0>\\
A^\dagger_\dot{1}|\Omega_0>, \, \, A^\dagger_\dot{2}|\Omega_0>\\
A^\dagger_\dot{1} A^\dagger_\dot{2}|\Omega_0>$$
Again there are only four states, the same as the short multiplet.
So do the two fermion states in BPS states form some kind of doublet as in the short multiplet? It's not so obvious since the relation between $A^\dagger_\dot{1}$ and $A^\dagger_\dot{2}$ are different from that of $Q^1_1$ and $Q^2_1$.
From the above discussion, is it legitimate to conclude that BPS states are different from the short multiplet in $\mathcal{N}=2$? For instance, in the short multiplet, the two scalars form a doublet while I don't see that in BPS states. But in Alvarez-Gaume's review paper hep-th/9701069 section 2.9, he mentioned that the BPS state belongs to a $\mathcal{N}=2$ short multiplet and the two scalar modes form a doublet while the two fermions are singlet. It's like Alvarez-Gaume was saying the two states are exactly the same. So what's missing in my thinking?
Thanks a lot.
| Spin one fields are in the adjoint representation because of the specific transformation behaviour of such modes under gauge symmetries. This is a symmetry requirement on the lagrangian.
Regarding your second question: the two multiplets are different. In one case we have a central charge and in the other we do not. However, the issue is essentially one of terminology. The references I am most familiar with (see for example "Modern Supersymmetry" by Terning) identify the short multiplet with the BPS state, whereas the "massless multiplet or "short multiplet" you refer to is called the hypermultiplet.
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Why does the windshield of my car freeze even if the outside temperature is above freezing? Under what conditions does the windshield of a car freeze even if the outside temperature is above freezing?
It is not clear whether this is related to the question why bridges freeze with non-freezing outside temperatures (see related topic), as I did not see an explanation in terms of radiation there.
It indeed seems that electromagnetic radiation is the culprit for windshields, which would predict that this phenomenon can only occur with an open sky. Does anybody know whether this is true?
And why does ice only form on top of the windshield and not on the metal surface of the car? Is this because heat is conducted much faster on the metal surface?
| This can happen if the windshield itself has a lower temperature due to earlier freezing. This effect can also be seen in freezing rain: Air producing rain is advected over frozen ground and suddenly the traffic is in trouble.
| {
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Ensuring globally hyperbolic geodesically-complete spacetimes Let's say we have an incomplete spacetime A that is globally hyperbolic, does there necessary exist a globally hyperbolic completion?
My guess is no, in which case what further restrictions can be placed on A to ensure that it can always be extended to geodesically-complete globally hyperbolic spacetime?
| Consider Kruskal manifold (omitting the future and past singularities at $r=0$). It is a globally hyperbolic spacetime as well known and it is a maximally extended manifold. However there are incomplete causal geodesics: These reaching the singularities. It is impossible to complete them. So the answer to your first question is: NO.
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What makes a rainbow happen? A rainbow is formed when a raindrop refracts light, but why then does the whole sky not become a huge rainbow when it rains? Would the light not be dispersed into ordinary white light? What causes it to look as if each end is nearly touching the ground?
| It depends on the position of the sun. A rainbow does not exist at a particular location in the sky. Its relative position depends on the position of the observer and the sun. All raindrops refract sunlight in the same way, but only the light from some raindrops reach the observer's eye. This light is what constitutes the rainbow for that observer. The bow is centred on the shadow of the observer's head, known as the antisolar point and forms a circle at an angle of of about 42° to the line between the observer's head and its shadow.
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What are the correct initial conditions for the moon (in a simulation)? So I've modeled the interactions between the sun and all the planets (and the interactions between the planets) using Verlet integration.
I've used data from Wikipedia for masses, distance from the sun etc.
I initialized the initial velocities of the planets via the critical velocity equation.
This produces nice stable velocities.
I'm unsure of how to calculate the initial velocity of the moon so that it stays in orbit around the earth.
| The moon orbits the earth with a near circular trajectory relative to the earth. So add earth's orbital velocity (around the sun) to the moon's orbital velocity (around the earth).
This will put the moon into an orbit around the earth, but might make it a bit more eccentric (elliptical). To correct this you can use angular velocity around the sun with respect to the barycenter (center of mass) of the earth moon system. The resulting initial velocity will then be their velocity around the barycenter (since the earth also orbits the moon) plus the angular velocity of the barycenter times their distance from the sun.
Once you have calculated all of this you might also want to set the momentum of the entire system to zero, since otherwise the center of mass of the entire system will keep on moving with a constant velocity.
| {
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Basic Thermodynamics: Quasistatic Adiabatic Process I'm going through the exercises in a Thermodynamics book, just to revise and build my intuition. Right now, I'm working on:
Show that for a quasistatic adiabatic process in a perfect gas, with
constant specific heats:
$$PV^\gamma = \left[\text{constant}\right]$$
with $\gamma = \frac{C_P}{C_V}$
where $P$ is pressure, $V$ is volume, and $C_V$ is the constant-volume heat capacity.
I'm not looking for the answer, just for a hint (I'm stuck and want to find the solution myself).
So those are my thoughts:
*
*perfect gas means: $PV = RT$, ($R$ is the universal gas constant)
*adiabatic means: $\mathrm{d}Q = 0$, ($Q$ for heat)
*since there is no heat exchange, the process is reversible
*reversible means: $\mathrm{d}W = -P \, \mathrm{d}V$, ($W$ is for work)
*heat capacity is defined as $C_\text{V} = \left( \frac{\mathrm{d}Q}{\mathrm{d}T} \right)_V$, respectively $C_\text{P} = \left( \frac{\mathrm{d}Q}{\mathrm{d}T} \right)_\text{P}$
If I draw a $PV$ diagram for this situation, it looks like this:$\hspace{175px}$.
Now I want to show that $PV^\gamma=\left[\text{const}\right]$ by going from $\text{State 1}$ to $\text{State 2}$ in the $PV$ diagram.
I've started like this:
$$
W ~=~ -\int_{V_1}^{V_2}P \, \mathrm{d}V ~=~ -\int_{V_1}^{V_2} \frac{RT}{V} \mathrm{d}V ~=~ RT \ln{\left(\frac{V_2}{V_2}\right)}
$$
This leads me into the wrong direction though. I thought about using $R = C_P - C_V$ here, but it doesn't seem to work. Any suggestions?
Please just give me a hint, not the solution.
| As an alternative to the popular proposal of using $dU = c_V dT$, I find it more intuitive to use the equation for the internal energy of an ideal gas, and its differential:
$$
U = \frac{3}{2} N k T = \frac{3}{2} p V\quad \rightarrow \quad dU = \frac{3}{2} p \, dV + \frac{3}{2} V \, dp
$$
where I've assumed (in using "3") that the gas is monatomic. Then the differential form of the First Law becomes:
\begin{align}
dU &= q + w_\text{on sys}\\
dU &= 0 - p dV \\
\frac{3}{2} \left[ p \, dV + V \, dp\right] &= - p \, dV
\end{align}
Consolidating like variables and integrating leads to the answer.
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Water in vacuum (or space) and temperature in space
*
*So, water in vacuum will boil first and then freeze. I don't know how the freeze happens. As pressure lowers to zero, what happened to freezing point? (I know heat taken by vapor, and the water cool down, but I don't think it will be that cold, will it? In vacuum, boiling point is so low that water shouldn't need so much heat as it does in normal pressure, which means vapor actually takes more heat away under normal pressure than in vacuum, so water under normal pressure would be cooler? (I'm guessing)
*And temperature comes from heat generates by motion of molecules (I guess so), so in vacuum, there is no temp?
*What happen when I heat up a vacuum tube?
*Does heat need a medium to “travel"?
| The phase shift (sublimation) of water from Ice to vapor occurs in a relatively narrow window. If the ice sample could remain at a temp above a certain level it would all boil away in the vacuum of space.The evaporation and the sub zero temps of space actually cool the sample below the phase shift window thus remain as Ice.
At just below freezing, with a constant pressure (Around 500 mTorr) the evaporation/sublimation process will actually cool the sample out of the window and stop the sublimation process as the system seeks to stabilize. In this system left to its own around 30% of the sample would be turned to vapor and the rest retained as ice when the system stabilized.To keep the process going the Temp or Pressure must be raised. In space BOTH go low and stop the process very quickly.
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Integration over the state space I came across the concept of average fidelity
$\int f(\psi)d\,\psi$ where the integration is with respect to the uniform Haar measure on pure states.
I've only seen Haar measures in connection with topological groups, so I don't understand what measure is used in this context. Could anyone provide a reference where this is explained ?
| The group $U(n)$ has a unique Haar measure, both right and left invariant, since it is unimodular as it is compact. Now consider the complex projective space $$\cal{P}(\mathbb C^n)=
\left({\mathbb C^n} / \sim\right) - [0]\quad \mbox{when}\quad v \sim v' \quad \mbox{iff}\quad v = cv'\:, \quad c\in \mathbb C -\{0\} \:.$$ The group $U(n)$ acts smoothly (with respect to the natural real smooth 2n-2 dimensional manifold structure of the projective space) and transitively on $\cal{P}(\mathbb C^n)$, trivially: $${\cal P}(\mathbb C^n) \ni [v] \mapsto [U\psi]\:.$$
${\cal P}(\mathbb C^n)$ can therefore be viewed as a smooth quotient of $U(n)$ and the compact subgroup $H_n$ which leaves fixed a point of ${\cal P}(\mathbb C^n)$.
Under these conditions, taking into account that both $U(n)$ and $H_n$ are compact (Lie) topological groups and thus are unimodular, there exist a $U(n)$-invariant positive Borel measure $\mu$ on ${\cal P}(\mathbb C^n) = U(n)/H_n$, unique up to the normalization. That is your measure.
These results were essentially due to Mackay. An account of the general theory can be found in Ch.4 of fundamental Barut-Raczka's textbook on the theory of group representations and applications World Scientific 1986.
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Is crystal momentum an operator? My teacher has for Bloch waves the notation $\langle \vec{r}|\vec{k} \rangle = e^{i\vec{k}\cdot \vec{r}}u_{\vec{k}}(r)$ and uses it consistently. However, does this not assume that there is an operator that has eigenstates $|\vec{k} \rangle$? If so, how would such an operator be defined?
| Sure, it is certainly possible to define a crystal momentum operator, although I haven't heard of people doing this.
You define it by saying that the eigenstates of this operator are Bloch states, and the eigenvalue of each Bloch state is its crystal momentum (translated into the first Brillouin zone). There is a unique linear operator that satisfies these specifications.
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Boosts are non-unitary! Unlike rotations, the boost transformations are non-unitary. Therefore, the boost generators are not Hermitian. When boosts induce transformations in the Hilbert space, will those transformation be unitary? I think no. If that is the case, what is the physical significance of such non-unitary transformations corresponding to boosts in the Hilbert space?
| On the actual Hilbert space of a consistent relativistic quantum mechanical system, the Lorentz transformations including boosts actually are unitary – which also means that the generators $J_{0i}$ are as Hermitian as the generators of rotations $J_{ij}$.
We say that the Hilbert space forms a unitary representation of the Lorentz group.
What the OP must be confused by is the fact that the ordinary vector representation composed of vectors $(t,x,y,z)$ is not a unitary representation of $SO(3,1)$. The $SO(3,1)$ transformations don't preserve any positively definite quadratic invariant constructed out of the coordinates $(t,x,y,z)$. After all, we know that an indefinite form, $t^2-x^2-y^2-z^2$, is conserved by the Lorentz transformations. So on a representation like the vector space of such $(t,x,y,z)$, the generators $J_{0i}$ would end up being anti-Hermitian rather than Hermitian.
But if you take a Lorentz-invariant theory with a positive definite Hilbert space, like QED, the formula for $J_{0i}$ makes it manifest that it is a Hermitian operator, which means that $\langle \psi |\psi\rangle$ is preserved by the Lorentz boosts! The complex probability amplitudes for different states $c_i$ behave differently than the coordinates $t,x,y,z$ above.
Note that the (non-trivial) unitary transformations of $SO(3,1)$ are inevitably infinite-dimensional. Finite-dimensional reps may be constructed out of the fundamental vector representation above and they are as non-unitary as the vector representation. But that's not true for infinite-dimensional reps. For example, the space of one-scalar-particle states in a QFT is a unitary representation of the Lorentz group. For each $p^\mu$ obeying $p^\mu p_\mu=m^2$, and there are infinitely (continuously) many values of such a vector (on the mass shell), the representation contains one basis vector (which are normalized to the Dirac delta function). The boosts just "permute them" along the mass shell which makes it obvious that the positively definite form is preserved when normalized properly.
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Why does gravity decrease as we go down into the Earth? We all know that gravity decreases as the distance between the two increases. Hence
$$ F = G \frac{Mm}{r^2}. $$
Hence the acceleration due to gravity
$$ g =\frac{F}{m}= G \frac{M}{r^2} $$
increases as $r$ decreases. Then why does it decrease as we go deep into the earth?
| It's actually not entirely true that the strength of the Earth's gravitational field decreases as a function of depth. It is true for certain regions in the Earth, but it's untrue for others because of the non-trivial dependence of the Earth's density on depth.
To see what's going on, assume that the Earth is a sphere whose density is spherically symmetric.
Now consider a mass $m$ at some radius $r$ from the center of the Earth. Using Newton's Law of Gravitation, one can show that that given the spherical symmetry, the gravitational attraction on $m$ of all mass with radii greater than $r$ exert no net force on it. It follows that only the mass with radii less than or equal to $r$ contribute to the gravitational force on $m$, which, by the Law of Gravitation is
\begin{align}
F(r) = G\frac{M(r)m}{r^2}
\end{align}
where $M(r)$ is the mass of stuff at radii less than or equal to $r$. Notice, then, that $F(r)$ will be an increasing function of $r$ (and will decrease as $r\to 0$), provided $M(r)/r^2$ is an increasing function of $r$.
Now, If the Earth were uniformly dense with density $\rho_0$, then the mass within a radius $r$ would be
\begin{align}
M(r) = \frac{4}{3}\pi r^3 \rho_0
\end{align}
namely just the density times the volume of a sphere of radius $r$, and in this case the strength of the gravitational field as a function of radius would be
\begin{align}
g(r) = \frac{F(r)}{m} = G\frac{1}{r^2}\frac{4}{3}\pi r^3\rho_0 = \left(\frac{4}{3}\pi g\rho_0\right) r
\end{align}
So in this case, it would be true that the strength of the gravitational field would decrease with increasing depth.
However, the Earth's density is not constant, and instead has some non-trivial dependence on $r$. See, for example, this PREM table. You can clearly see that there are regions in the Earth where the strength of the gravitational field increases because of the varying density.
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Atomic nearest neighbor notation I recently got a correction to a paper that I am writing. The correction references a section in which I talk about nearest neighbors. The comment says:
Do you mean NN, NNN, etc., or NN, 2NN, 3NN? It's different.
To clarify: here is the (111) face of an FCC crystal with nearest neighbors numbered (sorry about the horrible mouse-written numbers):
Starting from the center, I would denote the nearest neighbors as NN, 2NN, 3NN, etc. just as they are numbered. I think it would be the same to call them NN, NNN, NNNN, etc. Does anyone here know what this reviewer could be talking about? Because this seems like a simple concept to me and I have never encountered any confusion with the notation before.
| hmm this is usually the standard notation with
\begin{align*}
\text{NN} &= \text{NN}\\
\text{NNN} &= \text{2NN}\\
&\ldots
\end{align*}
The only thing I could imagine is some kind of numbering of the nearest neighbors: If you have 5 nearest neighbors like in your example with 1NN, 2NN, 3NN, etc. you could mean the first atom which is NN, the second atom which is NN and so on. Nonetheless that would be really weird, but I dont see any other meaning of these abbreviations in a physics context.
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Definition of Ohm in SI basic units in words One way Wikipedia defines Ohm is (this is also teached in school):
$$1\Omega =1{\dfrac {{\mbox{V}}}{{\mbox{A}}}}$$
They describe this definition in words, too:
The ohm is defined as a resistance between two points of a conductor when a constant potential difference of 1.0 volt, applied to these points, produces in the conductor a current of 1.0 ampere, the conductor not being the seat of any electromotive force.
The definition of Ohm in SI basic units is:
$$1\Omega = 1{\dfrac {{\mbox{kg}}\cdot {\mbox{m}}^{2}}{{\mbox{s}}^{3}\cdot {\mbox{A}}^{2}}}$$
It's really hard for me to get that this definition is correct. It's clear that mathematical calculations confirm this definition. But how do you describe the definition of the SI in words like that paragraph on wikipedia?
Edit:
How would you describe it? Although it is not common to do it that way, I think describing it that way, could be very interesting.
| Not sure whether this is correct, but if you have to do it, I think you can say that it is:
the work done by the conductor per unit charge per unit current through the conductor, or in terms of SI units, $\mathrm{\frac JC\cdot \frac1A}$
which is the same as:
the work done by the conductor per unit current per unit time per unit current, $\mathrm{\frac J{A\cdot s\cdot A}}$
We know that the work done is equal to the dot product of the force and the displacement, so it is:
the electric force multiplied by the displacement of the charge carrier per unit time per unit charge squared, $\mathrm{\frac{N\cdot m}{s\cdot A^2}}$
and we know force has SI units $\mathrm{kg\;m\;s^{-2}}$
So I guess you can say that the ohm is the resistance when one newton of electric force causes a charge carrier to displace one meter in one second with a current of one ampere.
I would go on and say that it is the resistance when a charge carrier of one kilogram accelerates at one meter per second squared, and this acceleration causes the charge carrier to displace one meter in one second, producing a current of one ampere. But I'm not very certain about the "charge carrier of one kilogram" part.
| {
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Drift speed of electrons when the object is grounded So I know that the drift speed of electrons is usually pretty slow. Let's say I have a charged sphere and I would ground it over a wire. How fast would the electrons leave the sphere? Would that drift speed be a lot faster than usual? I just want to get the general idea of the speed.
|
So I know that the drift speed of electrons is usually pretty slow.
Yes, if $10A$ of current is maintained in a conductor of cross-section $10^{-4}m^2$, with number density of electrons equal to $9X10^{28}m^{-3}$, drift velocity of free electrons will be $0.000006ms^{-1}$ (with the centimeter scale in your geometry box, it will be $0.0006cms^{-1}$).
Let's say I have a charged sphere and I would ground it over a wire.
By this, I hope you mean connecting the charged sphere to the ground with the help of a conducting wire.
How fast would the electrons leave the sphere?
We can't always say that electrons leave the sphere, if sphere is positively charged, electrons flow from earth to the sphere. Because, earth is neutral and sphere is positively charged. If sphere is negatively charged, electrons flow from sphere to the earth, in that case drift velocity depends on many factors, such as charge on the sphere, potential difference, etc.
Would that drift speed be a lot faster than usual? I just want to get the general idea of the speed.
I would just say that drift velocity in your case will constantly decrease with the flow of charge. I am unable to say you the exact or approximate value of drift velocity because of unavailability of data. But you can have general idea with the example given in the first paragraph.
| {
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How much power would a space craft's magnetic shield require? I've read over the decades that a magnetic shield might protect a spacecraft from cosmic radiation. Its a fascinating idea that might only be theory or science fiction at the moment. In regards to that here is an article just in case someone isn't sure what I mean:
http://physicsworld.com/cws/article/news/2008/nov/06/magnetic-shield-could-protect-spacecraft
The article even mentions simulations about midway into the article. Does anyone know if research has continued?
My big question though, is how much electrical power would be required to protect a space vessel in this way? How large would the field be in order to be useful?
Better yet, please use some practical objects that anyone can wrap their head around: What if you were using similar technology to protect the Apollo Command Module? Or the Discovery One from 2001 (140 meters long).
In laymans terms, how much electrical power would it take to create a magnetic shield?
| If it is a supercon magnet, it runs without losses until the field is deformed by a solar coronal mass ejection blasting by. It offers no protection against energetic photons.
Consider a one tesla field filling a 200 m diameter bubble. How many joules is 4.2 million m^3 of one tesla field? (Suppose the field collapsed into a solenoid shorted by a resistor.) A quench would be messy. Photosynthesis and mammalian metabolism both go through free radical intermediates. A one tesla field polarizes unpaired spins parallel, a triplet sate, preventing bonding or certainly slowing its kinetics. An hour in an MRI tunnel is not six months. Metabolic discrepancies will accumulate.
$1 T = 1 \mathrm{N{\cdot}s{\cdot}C^{-1}{\cdot}m^{-1}}$
${PV = energy}$, 101.325 J/liter-atm
| {
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Is the $N$ factorial in the Partition function for $N$ indistinguishable particle an approximation? I suspect that the $N$ factorial in the partition function for N indistinguishable particles
$$ Z = \frac{ Z_0^N } {N!} $$
is an approximation. Please someone correct me if I am wrong and why or why not. Thanks.
A simple case:
each particle has two states with energy $0$ and $E$. The partition function for a single particle is
$$ Z_0 = 1 + e^{- \beta E} . $$
If there are only two particles, there is the total partition function
$$
Z = \frac{ Z_0^2 } {2}.
$$
But regarding the whole system consisting of these two particles, we can also write
$$
Z = 1 + e^{- \beta E} + e^{-2 \beta E} .
$$
And it is certain that
$$
\frac{ Z_0^2 } {2} \neq 1 + e^{- \beta E} + e^{-2 \beta E}
$$
| Actually, it's exact. The flaw is "regarding the whole system consisting of these two particles, we can also write" $Z = 1 + e^{- \beta E} + e^{-2 \beta E}.$ Assuming the two particles are distinguishable, we have
$$Z=\sum_ig_ie^{-\beta E_i}=1+2e^{-\beta E}+e^{-2\beta E}=Z_0^2,$$
with the $2e^{-\beta E}$ since the state of energy $E$ is doubly-degenerate.
The additional factor of 1/2 in $Z=Z_0/2!$ accounts for indistinguishability.
EDIT: This seems to be wrong. See other answers.
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Physical distribution of salt anions and cations during electrophoresis If I have a volume of $L$ liters of salt water at a concentration of $\approx N$ mM NaCl and I pour it into an electrophoretic apparatus (like this one: ). Once we turn the apparatus on, and set the power level to some number of volts $V$, we would expect that there would be a Lorentz force on the individual Na+ and Cl- ions, pulling them towards the cathode and anode ends of the device, respectively. What does the physical distribution of Na+ and Cl- ions actually look like in the device as it operates?
Each ion will experience some Lorentz force, but separating the two ion types will have entropic penalties, so it's unclear to mean what the equilibrium distribution will look like with some voltage and current level.
A note on why this might be of interest: The apparatus for "gel electrophoresis" shown below is typically used to drag charged molecules (for example, nucleic acids or proteins) through a gel of some density, allowing for size sorting (where the gel leads to better size discrimination than the typical charge/mass ratio one uses to seperate molecular species for mass spectrometry). Proper molecular structure depends on having specific concentrations of specific anions or cations. Thus, I've always wondered if a gradient distributions of anions and cations over the length of distance between the anode and cathode had any undesirable effects. However, nowhere have I ever seen or heard this mentioned in the literature.
| The ion transport is by convection and diffusion. Only a very small part is by migration. As long as there is no anodic and cathodic reaction I would say the concentration is not changing in any part of the solution.
| {
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Can a car skid while moving in a circle? How can a car skid if it is moving around a circular road? It is understandable that it can topple but I feel it impossible that a moving car can skid. It is hard to imagine that a car would skid while traversing a circular path. The wheels of the car are rotating so it seems like a car would topple rather than skid.
| If you understand that a car can topple while negotiating a sharp turn at high speeds, then you just have to consider the fact that skidding happens when the tyres don't have enough grip to prevent the car from being thrown out the trajectory. Newton first law states that any solid tends to continue in a straight path, so the solid has to be pulled perpendicularly to its velocity in order to go in a circle. On a car this force is created by the friction of the tyres on the ground: it effectively "pushes" on the ground when oriented toward the inside of the turn. Toppling happens because of this force: the centrifugal force acting on the centre of gravity (from the fact the solid can't go straight and has to follow another path), above the ground obviously, yields to a torque about the contact line of the outer set of wheels. If you don't have grip for example on a frozen puddle, the car doesn't topple, it slips. On the road, there is a partial grip that detaches rubber from the tyre.
| {
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Can Flow of time become still?
*
*According to theory of time dilation, flow of time slows down significantly at the speed of light.Is there any conditions practically or theoretically when flow of time is reduced to zero means it comes to still?
*Whether time & space are infinite?
| Time is relative. When it comes to Time Dilation, you actually see dilated time of another observer. So, your own time flow won't get frozen in any case.
Hypothetically, you can see another one's time frozen if she is traveling at speed of light (time dilation by speed) or she is at event horizon of Black Holes (gravitational time dilation). Unfortunately, both are not possible in the framework of Relativistic Physics.
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How to get conductivity from Green function $\mathcal{G}(x_1,x_2,\tau)$ of inhomogeneous system? I'd like to study an inhomogeneous system, i.e., momentum is not a good quantum number therein. Therefore, I tried to calculate temperature Green functions like $\mathcal{G}(x_1,x_2;\tau)$, or its twofold Fourier transformation $\mathcal{G}(p_1,p_2;\tau)$.
But how can I get any transport property, e.g., conductivity, from these Green functions? I checked Mahan's overwhelming book, however, it only deals with the formalism of $\mathcal{G}(p;\tau)$ for homogeneous systems. Thanks in advance for any useful information.
| There are a few ways to extract transport properties from your single-particle temperature Green's functions. By analytically continuing it to real time $t$, one gets information about how a particle propagates in the medium. More exactly, you get the probability of the particle traveling a distance $x = |x_1-x_2|$ in the interval of $t$. From this, you could estimate the diffusion constant $D \sim x^2/t$ and then you relate it to conductivity $\sigma$ via Einstein's relation $\sigma \sim e^2 D N(E_F)$. In this picture, the particle moves like a Brownian dust. You may consult the book Conduction in noncrystalline Materials by N.F. Mott for further treatment.
Another way is to use Kubo formula. This formula expresses conductivity in terms of the electron-hole propagator, which can be written as a product of two single-particle Green's functions like yours. Thence you get the conductivity. This method is accurate but less physically transparent.
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The physics of sound boards As a kid I was bemused at why soundboards worked. A small sound could be demonstrably amplified simply by attaching the source to a surface that is rigid and not too thick. How could the volume increase so much given that there was no extra energy added?
As an adult I kind-of-think I know, but there are still many nagging questions. I assume it has to do with the waves propogating from a vibrating object actually being a compression on one side of the object just as they are a decompression on the other side, and something about that lack of coherence limits the volume. Exactly why remains a mystery to me. Is separating the pocket of compression and decompression so that the boundary along which they meet is quite small part of the issue?
My question is what are the physics that make a soundboard work?
Interesting specifics that would be nice-to-knows would be why does a hollow one (like a violin) work better than a solid one (imagine a filled in violin)? How important are the harmonics of the solid? But the real question is what are the physics that make a soundboard work?
P.S. I am a mathematician, so feel free to wax very mathematical if it is necessary to give a good explanation.
| The comments above that say the sound is louder because the soundboard itself begins to vibrate are correct. This is called resonance. It sounds louder because the motion of the board is mechanically more efficient at converting the energy of the system into sound waves than the string alone. The board is an effective radiator of sound energy. A louder sound wave has a larger amplitude.
If you want a mathematical analysis that shows resonance, try the ODE for a one dimensional SHO being driven by sine function. You'll see the amplitude of your oscillator increases near the natural resonance period of the system. You could also do a 2D analysis on a thin plate such as a drum head.
Strings and thin plates are relatively good sound radiators because relatively low force is required to cause a larger displacement, and so a sound wave of larger amplitude.
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Rms value of rectified output from a half wave rectifier The rms value of an alternating quantity which is the input to a half wave rectifier is $\frac{I_{max}}{\sqrt2}$.
Then the rms value of output should be $\frac{I_{max}}{2} \sqrt 2$. But it is given every where that it is $\frac{I_{max}}{2}$. Please tell me where am I wrong.
| In a half wave rectifier note that the only a half cycle is transmitted In other half cycle, current is (approximately) zero. How will it affect the rms value of output?
$Hint :$ Consider that the rms value in the first half cycle is same but in next half cycle it is zero. How do you find the rms value of current? Will it be the same case if you double the time there?
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Does electron in wave form have mass? I heard from my lecturer that electron has dual nature. For that instance in young's double slit experiment electron exhibits as a particle at ends but it acts as a wave in between the ends. It under goes diffraction and bends. But we don't see a rise in energy. It has to produce 500kev of energy (please correct if my approximation is wrong) according to mass energy equivalence relation. But wave is a form of pure energy and doesn't show properties of having mass as of experimental diffraction. So where is the mass gone?
| *
*mass=energy(actually mass is a form of energy) so everything who has energy(wave) has mass also
*if you squeeze all wave in a very little place you get a solid item(like you).
*if you squeeze a solid(like you again) in a very little place you will get a black hole
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Non-symmetric Lorentz Matrix I was working out a relatively simple problem, where one has three inertial systems $S_1$, $S_2$ and $S_3$. $S_2$ moves with a velocity $v$ relative to $S_1$ along it's $x$-axis, while $S_3$ moves with a velocity $v'$ along $S_2$s $y$-axis.
So I constructed the Lorentz transformation by multiplying the transformation from 1 to 2 with the transformation from 2 to 3, and I obtain the transformation. Now, this is all nice and well, and it is also where the question for which I'm doing this stops. However, it made me think: the resulting Lorentz matrix is not symmetric!
Somehow I always thought that they always were (simply because I hadn't encountered one that was not), which I suppose is naive. Is there any information inherent to the fact that the transformation is not symmetric? Does this in some way mean that there is a rotation happening? This is what seemed most plausible to me, as if you boost 'harder' in one direction than in the other, your axis is essentially rotating. Or am I going about this the wrong way?
| Indeed NowIGetToLearnWhatAHeadIs's comment answers your question:
"Simply because I hadn't encountered one that was not." A rotation is a lorentz transformation which is not symmetric.
Indeed the transpose of a rotation matrix is its inverse, and only trivial rotations or rotations through half a turn are involutary (self inverse). To see this in detail, write a rotation matrix as $\exp(H)$, where $H$ is skew-symmetric and real, then $\exp(H)^\dagger = \exp(H)^T$. More explicitly, partition the transformation matrix into $2\times 2$ blocks and consider a rotation about the $x$-axis:
$$\Lambda = \left(\begin{array}{cc}I&0\\0&R\end{array}\right)$$
where $I$ is the identity and $R$ is a $2\times2$ rotation matrix; the latter's general form is, of course,:
$$\left(\begin{array}{cc}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{array}\right)$$
which is clearly not symmetrix.
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Divergent thin lens producing real images The question is as in the title, can a divergent lens produce a real image, when backed by a convergent lens?
Which are the conditions to be respected?
This was a homework and I have received the solution but I think is wrong.
Basing this on research already done online, the only possibility looks like that a divergent lens should have as object a virtual image. Is it true?
EDIT: In the case of a convergent lens to the left, and a divergent lens to the right, with the light coming from the left, how would be?
| You have two good fundamental answers, but a slightly different take on your answer is that yes, this is done all the time in optical systems to compensate for various distortions, aberrations and errors. A good example is an achromatic doublet, where a convex (converging) lens of one material is put in direct contact with a concave (diverging) lens along a surface of common curvature. Of course, the overall effect is converging. The refractive index as a function of wavelength is such that both materials is different, but they are chosen so that both materials the same index at some wavelength in the middle of the "achromatic band". Therefore, at that wavelength, the contacting surface has no optical effect at all. However, at wavelengths higher than this middle wavelength the refractive index, the converging lens has a higher index than the diverging lens, so that the surface between them has a slight nett converging effect. At wavelength shorter than the middle, the diverging lens has the higher index, and so the surface is nett diverging. This engineered variation in optical power is used to offset the overall variation of the system's optical power with wavelength, i.e. it is a mechanism to correct for chromatic aberration.
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How far could the LHC "fire" a proton into space if we (outside LHC) ignore all interactions but gravity? Very simple question, and frankly quite a silly one, but I'm currently writing a lecture for secondary school kids and I'd love to tell them how far the Large Hadron Collider could fire a proton.
The talk is actually about space, so what I'd really like to know is how would the velocity of the proton would compare to, for example, the necessary escape velocity for the proton to escape the Milky Way?
As the talk is mostly about energetics and kinematics I'd be keen to ignore the effect of the protons charge and just look at the kinetic energy being converted to gravitational potential.
| I think I get what you're after. You're looking for a qualitative handle on just how fast those protons are slinging it...
You might try considering the speed differential between an LHC proton and a photon that is launched at the same time. As others have noted the proton is ultra-relativistic (its kinetic energy is waaay greater than its rest-mass), so working out an actual speed is a bit tricky. Luckily, someone already did it.
Using this value, the proton is moving at a speed of $(1 - 9\times10^{-9})c$ and the photon is moving at $c$. So the proton is 2.7 $ms^{-1}$ slower than the photon. The photon is getting away from the the proton at about the speed of a brisk walk.
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free energy and entropy my understanding of free energy and entropy is that as entropy of a system increases its free energy decreases. As sun has free energy and this energy is being converted to useful work, and its entropy is increasing. Its entropy will continue to increase till its material is exhausted to cease free energy production.
Can permanent magnets be a source of free energy? The question arises because magnets are permanent dipoles that lose their orientation very slowly to become disordered and increase its entropy.
| A large entropy is beneficial to a system, in that it lowers the total energy of the system. So a large increase in entropy would decrease the activation energy. Together with the change in enthalpy, they decide the spontaneity of the reaction. If the expression for the Gibbs free energy is negative at a given temperature, the reaction is spontaneous, meaning that it occurs.
However, just because a reaction occurs, it doesn't mean that the reaction is significantly noticeable. The conformational change of diamond to graphite has a negative ΔG, but I have yet to see any diamonds turn to carbon. This is because the reaction rate is so slow, that it would take forever (in our perspective) to see it happen.
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Enthalpy Change in Reversible, Isothermal Expansion of Ideal Gas For the reversible isothermal expansion of an ideal gas:
$${∆H}={∆U}=0 \tag1$$
This is obvious for the case of internal energy because
$${∆U} = \frac {3}{2} n R {∆T} = 0 \tag2$$
and
$${∆U} = -C_P n {∆T} = 0 \tag3$$
For the case of enthalpy it is easy to see that
$${∆H} = -C_v n {∆T} = 0 \tag4$$
I've also seen
$${∆H} = ∆U + ∆(PV) = ∆U + nR{∆T} = 0 \tag5$$
Now for the part I don't understand.
$$dH = dU + PdV \tag6$$
$$dH = dU + nRT \frac {dV}{V} \tag7$$
$${∆H} = {∆U} + nRT \ln\frac {V_2}{V_1} \tag8$$
$${∆H} = 0 + nRT \ln\frac {V_2}{V_1} = \ln\frac {V_2}{V_1} ≠ 0\tag9$$
Clearly, it is incorrect to make the substitution $ P = nRT/V$ in going from $(6)$ to $(7)$. Why is that? I thought equation $(6)$ was always valid, and integrating such a substitution should account for any change in the variables throughout the process. Why does this not yield the same answer as $(4)$ and $(5)$?
| $$H = U + PV \Rightarrow$$
$$dH = dU +PdV + VdP\tag{6}$$
In other words, equation 6 is missing the $VdP$ term.
$$dH = dU + nRT \frac{dV}{V} + nRT \frac{dP}{P}\tag{7}$$
$$ \Delta H = \Delta U + nRT \ln\frac{V_2}{V_1} + nRT \ln\frac{P_2}{P_1}\tag{8}$$
$$P_1 V_1 = P_2 V_2 \text{ (isothermal)}$$
$$\Delta H = \Delta U + nRT \left( \ln\frac{V_2}{V_1} + \ln\frac{V_1}{V_2} \right) = \Delta U = 0\tag{9}$$
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Bulk flow of air in a long tube between Antarctica and Australia? I have a 5km diameter clear plastic tube which is open at each end and runs from the center of Antarctica to Lake Eyre in Australia. The tube is on the ground where it can be and at sea level on the ocean.
Will there be bulk flow of the air in the tube? If so, which way will the air flow?
| Natural circulation in tubes is driven by pressure gradient between warmer and colder parts of the fluid. The pressure gradient results from density gradient, which is induced thermally by heating/cooling the tube at different points. However, tube dimensions You've set, make it a very complex problem, probably impossible to solve. Normally models have to be validated with experimental installations as it is difficult to accurately predict what will happen. With such long pipe and complex heat exchange conditions my guess is that the air will flow in all possible directions in different parts of the tube, with average net flow equal to 0.
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Taking pivot about an accelerating point Given this question:
A small ball of mass $m$ and radius $r$ rolls without slipping on the inside surface of a fixed hemispherical bowl of radius $R>r$. What is the frequency of small oscillations?
The standard solution is to write Newton's second law for the ball and then take the centre of mass of the ball to be the pivot and write
$$\tau = I \alpha.$$
Only the frictional force contributes to the torque in this case. From Newton's second law, I can express the frictional force in terms of the gravitational force and therefore the frictional force can be eliminated in the equation for torque. I then make small angle approximation and get the equation to be of the form
$$k\theta=-I\ddot{\theta}$$
from which I can find the frequency.
Another approach uses the point of contact of the ball with the sphere as the pivot. It has the advantage that the frictional force adds no torque. Both approaches give the same result.
My question is since both pivots that we have chosen are accelerating, why are not fictitious forces considered? In the first place, can the pivots that we choose when writing
$$\tau = I \alpha$$
be accelerating?
| In fact, you are right in that you should consider the fictitious force in accelerated reference frame. As such, using the point of contact between the mass and the bowl is actually not ideal. In the center of mass reference frame, notice if you consider a fictitious force for the accelerated reference frame, it is applied at the center of mass, and thus it does no torque. This is better.
However, the center of mass and the point of contact between the ball and the bowl are actually not the points of reference that I would use. Instead, I would use the center of curvature of the bowl, because this point of reference is not accelerating and still has the benefit of excluding the normal force from the torque equation.
tl;dr: You can only ignore fictitious force in the center of mass reference frame, but otherwise you are right in needing to include it.
| {
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"timestamp": "2023-03-29T00:00:00",
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Confused about Impulse Encountered a problem that involves impulse while studying for my exam and I'm not sure how to even approach it. I know that momentum is conserved, but I'm not sure how to relate that to avg force. Maybe someone can help point me in the right direction? I know that it's in quadrant III, through intuition, but I can't come up with a provable explanation
Relevant equation: $J=F_{avg}\Delta T$
| This post has some information about impulse that you might find useful.
Homework Question involving Momentum
You will not find conservation of momentum useful here. True, the total momentum of object + wall is unchanged by the collision. But the momentum of the object does change.
Since $\Delta P = J = F_{avg} \Delta t$, the direction of $F_{avg}$ and $\Delta P$ must be the same.
| {
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Orbital angular momentum selection rules for three identical particles I'm trying to figure out if there are selection rules for the total orbital angular momentum for a system of three identical particles, say bosons.
For two identical bosons one can argue that the exchange symmetry implies that the state must be parity-even. From the parity properties of the spherical harmonics, this forces the state to have even orbital angular momentum.
How does this work with a system of three identical particles? We may assign two orbital angular momenta: $\mathbf{L}_{1}$ between two of the particles, and $\mathbf{L}_2$ corresponding to the relative angular momentum between those two particles (as a reduced one-particle system) and the third. The total orbital angular momentum is $\mathbf L=\mathbf L_1+\mathbf L_2$.
I can use symmetry arguments to say that the $\mathbf L_1$ quantum number $\ell_1$ is even. But I don't know how to use exchange symmetry to constrain $\mathbf L_2$ since we're now comparing a single particle and a two-particle state.
I'm hoping to be able to use this to argue that a scattering process in some definite orbital angular momentum, total spin, and $CP$ state going to three identical bosons should be $s$-wave or $p$-wave.
Thanks!
| For 3-particle systems the relevant permutation group is $S_3$, which is significantly different from $S_2$ in that not all representations of $S_3$ are 1-dimensional.
Whereas 2-particle states can always be written so they transform back to themselves under permutation of labels, i.e. always possible to write
$$
\psi_a(x_1)\psi_b(x_2)=\frac{1}{2}\psi^+_{ab}(x_1,x_2) + \frac{1}{2}\psi^-_{ab}(x_1,x_2)
$$
with
$$
P_{12}\psi^{\pm}_{ab}(x_1,x_2)=\pm \psi^{\pm}_{ab}(x_1,x_2)
$$
the symmetric and antisymmetric combinations, there are 3-particle states that cannot be made to transform back to a multiple of themselves. Already this is apparent for the triple coupling of spin-1/2 particles: there are two copies of total spin $1/2$ in the decomposition and you cannot arrange the states in $S=1/2$ to be symmetric or antisymmetric under all permutations of particles.
The situation does not improve for 3 spin-1 particles. The fully symmetric states will have total $L=3,1$, the fully antisymmetric states will have $L=0$, but there are states of mixed symmetry with $L=1,2$ (there are two copies of these mixed symmetry states).
Thus, you have no obvious way of knowing if a state with $L=1$ is symmetric or of mixed symmetry.
Of course the spin-statistics theorem does not forbid spin-states of mixed symmetry: the must be combined with spatial stages (also of mixed symmetry) to produce a properly symmetrized total state.
| {
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Does the equation of continuity hold for turbulent flows? My textbook mainly deals with laminar flows. The book derives the equation of continuity, which states that the cross-sectional area times the velocity of a flow is always constant. But nowhere in the derivation does the textbook explicitly assumes that the flow is laminar. So, does the equation hold for turbulent flows too?
|
The book derives the equation of continuity, which states that the cross-sectional area times the velocity of a flow is always constant. But nowhere in the derivation does the textbook explicitly assumes that the flow is laminar. So, does the equation hold for turbulent flows too?
That is only a special case of the equation of continuity for situation where density is regarded constant and velocity is constant across the cross-section of the pipe. In general, equation of continuity is $\partial_t \rho + \nabla\cdot(\rho \mathbf v) = 0$ and holds true even for turbulent flows if the mass of the fluid is locally conserved.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/101555",
"timestamp": "2023-03-29T00:00:00",
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Does a mirror help a near-sighted persion see at a distance clearer? A near-sighted person without eye-glasses can not clearly see things at distance.
If he takes a photo of the things at distance, he can see the things from the photo much clearer, because he can place the photo much closer to his eyes.
If he turns his back at the things at distance, and holds a mirror close to his eyes in a position so that the mirror reflects the things at distance behind him, will he see the things much clearer than if he looked at the things at distance directly?
| I am suffering from myopia (nearsightedness). I tried the same (with out wearing my lens), result was I could see things clearer in front of me than when I saw the same things in the mirror.
The below figure shows image formation for a person suffering from myopia.
Lens in our eye is more convex at its back than in front. For simplicity lets assume it to be convex lens. As our eye lens is of small focal length, we can assume every object to be at infinite distance. When the object is at infinite distance you get image formation as shown below.
Whether you see the object in a mirror in front of you or see the object far away from you, we can assume light rays to be incident from infinite distance (because of small focal length of our eye lens).
Visible light emitted or reflected from objects around us provides information about the world you see.
When you see things behind you in the good mirror it will be able to reflect only $85-95$% (aprox) of light falling from it. When you see same things directly, you will be able to see greater percentage of reflected light. Thus, you will see sharper image.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/101627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
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} |
x-ray in oil droplets experment In oil droplet experiment, x-ray makes the air molecules negatively charged. How does that work? X-ray carries high energy and ionizes air, doesn't that make air positively charged?
| According to this link, the x-rays ionize the gas molecules in the apparatus, not the oil drops directly. So yes, you are right, the air is positively ionized. The newly freed electrons from the gas then adhere to the oil droplets (probably due to an induced dipole moment), producing negatively charged oil.
| {
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"url": "https://physics.stackexchange.com/questions/101811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What are circles on broth (eating soup) surface? Think about broth in the soup, usually it has circles on its surface. What are their properties? Why there are many of them (not a few big blobs)? Are they depended on liquid's temperature? What needs to be added to water so these kind of circles appear?
I can only guess that it is because of oil in the broth, and shape is due to tensions on the surface; is that correct?
| The circles are droplets of fat (triglycerides: triesters of glycerol and fatty acids). The two phases (water and fat) are immiscible because the water molecules are more attracted (hydrogen bonded) to each other than to the fat molecules.
Absent gravity, energy would be minimized by the droplets being spherical, but gravity flattens the droplets. The surface tensions of each pair of the three phases (air-water, air-fat, and water-fat) would be needed to quantitatively describe the situation.
The droplets being seperate is not the lowest energy state. However, surface tension must be overcome for the droplets to coalesce.
If the soup is refrigerated, the fat droplets usually solidify.
The fat droplets usually come from meat in the soup.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/101862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why doesn't diamond glow when hot? In an answer to this SE question, the respondent explains that heating a perfect diamond will not cause it to glow with thermal blackbody radiation. I don't quite follow his explanation. I think it comes down to: there is no mechanism for diamond to generate light in the visible region of the spectrum.
He mentions that interband transitions are well out of the visual range, so there will be no contribution from that.
He mentions that the Debye temperature for diamond is > 2000 K. I presume that the argument here is that optical phonons will be frozen out, too. (But diamond doesn't have infrared-active phonons, does it?)
So is that why hot diamond doesn't glow?
I suppose that if one considers real (not ideal) crystals, imperfections, impurities, and the existence of surfaces lead to the possibility of emission mechanisms, and thus glow. In fact it might be the case that a finite but otherwise perfect crystal might have an extremely faint glow.
Is this basically the reason that hot diamond does not glow? Further elucidation welcome.
| This is because of Kirchhoff's law of thermal radiation. The corollary from it is that emissivity of a material is equal to its absorptivity.
As diamond is transparent even at large temperatures, which can be seen in this answer, its absorption coefficient is very low. Thus its thermal radiation in that spectral region is also very low.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/101960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Why does my kettle only make a noise when it is turned on Almost as soon as I turn my kettle on it starts to make the familiar kettle noise, yet very shortly after turning off the power the boiling noise stops and the kettle is totally silent. The temperature of the water is (almost) the same as when I turned it off. So why is there only noise when energy is being added to the water?
| Formation and collapse of immature water vapor of micro bubbles with a range of radii caused by a high temperature gradient between the heating plate and the ambient water. The noise-like sound is due to different micro bubble sizes.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/102032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How are velocity and dispersion maps of galaxies made? How can I get the velocity dispersion and velocity maps of galaxies given 3D data cubes obtained using an integral field spectrograph?
| The first thing you have to calculate is the overall redshift of the galaxy. When working at this level there are better and worse lines for doing this — for example, $\mathrm{Mg}$ lines frequently trace AGN outflows, so they'll be biased.
Once you have the galaxy's overall redshift, you can look at individual pixels in a data cube (where you measure the spectrum of each pixel). The lines of each cube, preferably the nebular lines (e.g. $\mathrm{OIII}$ and $\mathrm{H\alpha}$), will then be offset from the overall redshift of the galaxy in a way that indicates motion of what's in that pixel relative to the galaxy. That's how you produce the velocity map… basically - you can measure the velocity using multiple lines and produce a flux weighted mean for each pixel.
The velocity dispersion comes from measuring the width of the lines used to measure velocity. See, you can imagine that each observed line comes from relatively narrow lines that are broadened by the Doppler shift of the speed individual emitters/absorbers are moving at (called Doppler broadening). Granted, you have to have some kind of model for how narrow the lines will be in the absence of velocity dispersion, because they don't have zero width in the absence of it (i.e. the gas doing the emission/absorption could be stationary, but the particle in it will move due to their non-zero temperature, whether star or nebula).
The more advanced techniques discussed in the paper linked by @KyleOman are all, fundamentally, elaborations on how to get the best signal to noise from combining multiple measurements based on these physical facts.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/102102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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} |
How does a half-life work? Carbon-14 has a half-life of 5,730 years. That means that after 5,730 years, half of that sample decays. After another 5,730 years, a quarter of the original sample decays (and the cycle goes on and on, and one could use virtually any radioactive isotope). Why is this so? Logically, shouldn't it take 2,865 years for the quarter to decay, rather than 5,730?
| Suppose you start with two kilograms of C-14. After 5730 years you have one kilogram left. Call that piece A. Now get another kilogram of C-14, call it piece B, and put it next to piece A.
You now have two identical pieces of C-14, and yet one of them (A) is supposed to half-decay in 2865 years and the other (B) is supposed to half-decay in 5730 years? Do you see how that doesn't make sense?
Hopefully this convinces you that the rate at which a radioactive element decays can only depend on how much of it there is at that moment, not on how much of the original sample is left.
This is something I don't think any of the other answers explicitly brought up, but Nick Stauner alluded to it in a comment.
| {
"language": "en",
"url": "https://physics.stackexchange.com/questions/102222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
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} |
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