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123
123
6:05 AM
Hello World..
Hi @JohnRennie Sir
 
John Rennie
John Rennie
@123 hi :-)
 
123
123
Pls share importance, development, limitation the idea of energy.
I have lecture. Need your help. Book and google does not help to answer me.
I found few words about work and energy in K&K where it told . We need force as function of position not function of time. That's why we create mathematical trick to solve this problem to answer the question.
I need more info..
 
John Rennie
John Rennie
6:43 AM
@123 The word energy is used in so many different ways that it doesn't really mean anything.
So there is no useful way to discuss "energy" in general. You need to consider a specific system in whuich the energy is a useful quantity.
@user27286 what question?
 
 
2 hours later…
Noah J. Standerson
Noah J. Standerson
8:49 AM
Hi @JohnRennie . I have an interesting question (or solution rather) . Question : Bag A contains 6 red 4 black balls and Bag B contains 4 red and 6 black balls. One ball is drawn at random from Bag A and placed in Bag B. Then one ball is drawn ramdomly from Bag B and placed in Bag A . If now, one ball is drawn from Bag A, what is the probability that it is red? .
 
John Rennie
John Rennie
Hi Noah :-)
I'm answering a question in another room, but I'll get back to you as soon as I can.
 
Noah J. Standerson
Noah J. Standerson
Hi
Ok no problem
 
user27286
user27286
9:50 AM
@JohnRennie physics.stackexchange.com/questions/276053/… In this question there is something like $\mathbb{I}_A$...I wanted to know what it is? Is it a matrix or vector and what is it's value?
 
John Rennie
John Rennie
@user27286 ah. I can't help. Sorry :-(
 
Nihar Karve
Nihar Karve
@user27286 ACuriousMind told you, it's the identity matrix (if it's acting on a finite-dimensional vector space) - or more generally, the identity operator, which sends a vector to itself.
 
bolbteppa
bolbteppa
10:11 AM
Jefimenko's equations
In electromagnetism, Jefimenko's equations (named after Oleg D. Jefimenko) give the electric field and magnetic field due to a distribution of electric charges and electric current in space, that takes into account the propagation delay (retarded time) of the fields due to the finite speed of light and relativistic effects. Therefore they can be used for moving charges and currents. They are the general solutions to Maxwell's equations for any arbitrary distribution of charges and currents. == Equations == === Electric and magnetic fields === Jefimenko's equations give the electric field E and...
I completely forgot about this Feynman version, if you look at it it's actually beyond stunning, unforgettable comapred to the usual form, the problem with this stuff is it goes on for pages and pages if you naively try to derive this stuff but there's probably a clever quick way if anyone finds it
The usual form being the ones derived here
Liénard–Wiechert potential
The Liénard–Wiechert potentials describe the classical electromagnetic effect of a moving electric point charge in terms of a vector potential and a scalar potential in the Lorenz gauge. Stemming directly from Maxwell's equations, these describe the complete, relativistically correct, time-varying electromagnetic field for a point charge in arbitrary motion, but are not corrected for quantum-mechanical effects. Electromagnetic radiation in the form of waves can be obtained from these potentials. These expressions were developed in part by Alfred-Marie Liénard in 1898 and independently by Em...
 
user27286
user27286
@NiharKarve got it clearly. Thanks
@JohnRennie You did help by bringing this topic.
 
 
4 hours later…
Balarka Sen
Balarka Sen
2:40 PM
"In quantum mechanics, particles with speeds close to that of light can
pop in and out of existence, even from the vacuum"
what does this mean
someone explain
 
Charlie
Charlie
 
Balarka Sen
Balarka Sen
seems to be
 
ACuriousMind
ACuriousMind
@BalarkaSen It's both entirely accurate and nonsense :)
 
Balarka Sen
Balarka Sen
lool
its the first statement of a problem in my special relativity assignment
they want me to compute the lifespan of a muon popping in and out of existence wrt muon frame vs earth frame
 
ACuriousMind
ACuriousMind
It's accurate insofar as indeed particles can be created from nothing more than "energy", but it is nonsense because you will never see a measurable particle pop out from the vacuum.
 
Balarka Sen
Balarka Sen
2:43 PM
i assume $t' = \gamma t$ is what i need to chug but this makes 0 sense lol
 
ACuriousMind
ACuriousMind
The idea that there are "actual particles" going in and out of existence in a vacuum is in the end a misinterpretation of Feynman diagrams as representing actual processes, which they don't.
I've ranted about this in various answers on the site (probably this and this are the most complete ones), and not all other physicists agree to this viewpoint
 
Balarka Sen
Balarka Sen
Thanks, let me see
 
ACuriousMind
ACuriousMind
unfortunately, it seems your special relativity exercise was written by someone with no regard for the subtleties of the notion of "virtual particles" :P
 
Balarka Sen
Balarka Sen
sounds about right
 
ACuriousMind
ACuriousMind
it's also entirely superfluous because a much better demonstration of time dilation would be to do that computation for actual muons produced by scattering of cosmic rays in the atmosphere - without time dilation it is inexplicable why they live long enough to be detected on the ground
 
bolbteppa
bolbteppa
2:56 PM
@BalarkaSen if you're doing SR have a look at that, encapsulates a lot of SR ideas in one go
You can directly relate the muon example to this example to remember how things should go
 
Manas Dogra
Manas Dogra
3:18 PM
I was wondering why people hasn't yet come up with a group theoretical method to find out the Einstein tensor components :) or have they???
 
bolbteppa
bolbteppa
@BalarkaSen For time dilation problems like this you can use pseudo-Riemann geometry to get the answer in one line in 5 seconds, have a think and put those manifolds to use it should take one line from first principles :p
@ManasDogra what do you mean
 
Balarka Sen
Balarka Sen
there's no manifolds, there's a quadratic form
its just linear algebra
i am just confused about the wording physicists use, that's all
 
bolbteppa
bolbteppa
A vector space is a manifold
 
Balarka Sen
Balarka Sen
Lol
if you tell that to a mathematician they'll laugh at you and walk away
 
bolbteppa
bolbteppa
Group theory will tell you the tensor might represent some group, but it's got an explicit definition in terms of derivatives of a metric etc
 
ACuriousMind
ACuriousMind
3:21 PM
@ManasDogra how is any "group theoretical" method supposed to know about the specifics of the metric?
The only groups you could get from a metric I can think of are the holonomy and isometry groups, but that's far coarser information and plenty of different metrics (with different Einstein tensors) have the same groups.
@BalarkaSen that's pretty common among mathematicians trying to physics ;P
 
Manas Dogra
Manas Dogra
We have Wigner-Eckart theorem which lets us evaluate so many matrix elements of tensor operators without having to evaluate all of them individually... Back then symmetry between different values of m was new...

I was just hoping if there is some undiscovered symmetry between the components of the tensor :)
 
Balarka Sen
Balarka Sen
@ACuriousMind fortunately im neither a mathematician nor am i trying to physics
just a guy with a deadline
 
bolbteppa
bolbteppa
Basically from $ds^2 = c^2 d\tau^2 = dt^2 - dx^2 - dy^2 - dz^2$ where the $\tau$ part is the rest frame coordinate(s) of the particle you take a square root then factor out $c d t$ and you get $1/\gamma$ on the RHS and you can solve for $dt$ if you know $d \tau$ but since $\gamma \geq 1$ you can see $dt \geq d \tau$ i.e. it lives longer in the lab frame which is why we see them
 
ACuriousMind
ACuriousMind
@BalarkaSen why aren't you trying to physics :(
 
Balarka Sen
Balarka Sen
i learnt physics people have been doing breakthrough work in percolations recently
 
ACuriousMind
ACuriousMind
3:27 PM
might be, not the kind of physics I know much (or anything) about
 
Balarka Sen
Balarka Sen
you might like lattice gauge theory which is closely related and is relevant to the kind of physics you like
 
ACuriousMind
ACuriousMind
now that I know a bit about
 
Manas Dogra
Manas Dogra
@ACuriousMind Yes... I see, evaluating these kind of things are pretty much difficult without a CAS
 
Balarka Sen
Balarka Sen
how do you think of it? its like a discrete approximation to Yang-Mills?
iirc the simplest model takes a bunch of matrices from SU(2) on the vertices of Z^2 in a way that around the unit square they multiply to I, put an invariant measure on the whole thing, and try to scale
seems like trying to approximate a flat bundle
 
ACuriousMind
ACuriousMind
@BalarkaSen there are several approaches, but yes the common one is discretizing Yang-Mills
you triangulate (or whatever shape the lattice should have) the manifold you're on, and then a discrete gauge field is basically just an assignment of group elements to the edges of the cells
 
Balarka Sen
Balarka Sen
3:34 PM
extremely cool idea
 
ACuriousMind
ACuriousMind
the assigned group elements would be the holonomies/parallel transports of the continuous gauge field along the edges
 
Balarka Sen
Balarka Sen
what other approaches are there
 
ACuriousMind
ACuriousMind
you can have a more traditional lattice and assign a value of the gauge field to each lattice point, like you'd do with any other field
it might be that this is equivalent to assigning elements to the edges if you define the parallel transport/curvature on this lattice correctly, I don't remember, but these approaches are definitely different in implementation since the algorithms you use to compute stuff are different
 
Balarka Sen
Balarka Sen
Ah, ok, vertex vs edge model.
 
ACuriousMind
ACuriousMind
I think the first approach is more common when you're interested in global stuff like Wilson lines (especially in 2d where there are no local phenomena and the lattice "approximation" is actually exact), while the second is more common if your gauge theory is coupled to some other fields
(if you're putting fermions on the lattice the gauge field they might be charged under is the least of your worries, usually :P)
@BalarkaSen it is approximating a flat bundle, physicists will prefer to do weird stuff at infinity (or in this case with the boundary conditions) to avoid thinking about non-trivial bundles :P
 
Balarka Sen
Balarka Sen
3:42 PM
Naturally
There's also a natural measure on these discretized models, by putting uniform measure on all configurations on the large boxes of finite size $[-N, N]^2$, and then scaling by $e^{-S}$ where $S$ is some appropriate action -- I think they call it Wilson action? Take some sum of traces over all Wilson loops, IIRC
Do you know the exact construction, and if these measures converge as you scale the lattice finer and finer?
Also, what did you mean by the approximation being exact in 2D?
 
ACuriousMind
ACuriousMind
@BalarkaSen No, but generally "convergence" is a thorny issue for lattice theories - what they're trying to converge against is usually some QFT path integral that doesn't rigorously exist anyway
I'm not an expert there by far, searching for "triviality of $\phi^4$-theory" should probably turn up the simplest of cases where it's a bit unclear what the "convergence" is telling us
 
Balarka Sen
Balarka Sen
I see.
Thanks, I'll have a look there
 
Jake Rose
Jake Rose
This is completely wrong, or at least incredibly bad notation right?
 
ACuriousMind
ACuriousMind
@BalarkaSen In 2d, there are no EM waves, gluons or whatever - a 2d gauge theory has no interesting local solutions to the equations of motion because you lack transverse waves (there are only longitudinal waves when you only have one spatial dimension), the only thing you can study are global properties
and if you look at how the path integral for these works then in 2d you can see that subdividing the 2d lattice cannot change the value of it - making the triangularization finer cannot change the values of your observables.
so any coarse triangularization already suffices to compute whatever you want and turns a lot of the theory into topology where you solve the theory on a pair of pants and build the values for all other manifolds out of the values there and some gluing rules
 
Manas Dogra
Manas Dogra
@bolbteppa Yes and there's a lot of terms even after using Bianchi identities and other symmetry properties...Any tip on how to quickly solve them?
 
ACuriousMind
ACuriousMind
3:53 PM
(needless to say 2d gauge theories are not particularly relevant to describe anything in the real world, and so they're allowed to be neat :P)
 
Balarka Sen
Balarka Sen
Ah I see
 
bolbteppa
bolbteppa
@ManasDogra if you mean something like working out Schwarzschild or Reissner-Nordstrom etc then forms can help make it a bit easier
 
Manas Dogra
Manas Dogra
@bolbteppa I saw Schwarzschild metric is easy in Kruskal–Szekeres coordinates...But I am looking for techniques for a general metric tbh
I just thought that how beautifully group theory solved the problems in QM with Wigner Eckart theorem...and if there is something similar for GTR evaluations...but apparently there can't be :(
 
bolbteppa
bolbteppa
@ManasDogra "These days, infact, one obtains these tensors using well-known software package"
 
Manas Dogra
Manas Dogra
@bolbteppa Hmm
 
Jake Rose
Jake Rose
4:25 PM
Can anybody explain why spherical symmetry implies this about the velocity dispersion?
 
Physics Meta
Physics Meta
1
Q: Choosing a Close Reason

WillOWhat close reason should I choose for a question like this?

discussion questions close-reasons
 
 
6 hours later…
schn
schn
10:03 PM
Reading on Near-field techniques here:
Diffraction-limited system
The resolution of an optical imaging system – a microscope, telescope, or camera – can be limited by factors such as imperfections in the lenses or misalignment. However, there is a principal limit to the resolution of any optical system, due to the physics of diffraction. An optical system with resolution performance at the instrument's theoretical limit is said to be diffraction-limited.The diffraction-limited angular resolution of a telescopic instrument is proportional to the wavelength of the light being observed, and inversely proportional to the diameter of its objective's entran...
It says in the first sentence:
"The diffraction limit is only valid in the far field as it assumes that no evanescent fields reach the detector."
Mathematically, what does this mean?
I'm familiar with the far field of a point charge, the term that goes as 1/r.
Evanescent fields are those that are refracted at an interface for which the incident angle is at or above the critical angle.
 
schn
schn
10:50 PM
Why does the far field assume no evanescent fields are present? Why is this important?
 

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