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7:26 AM
Hello everyone. Please do not judge. In order to graduate (BS) I have to submit a thesis. Now I have around 1 month before my semester starts. Within these 1 month, is it possible to learn GR enough to tackle basic astrophysics? because I want to choose a project reagrding astrophysics.
I tried watching XylyXylyX youtube videos to learn GR but seems like he is going too deep. If anyone also watches him, could you let me know if he is going too deep (for example he made >50 hrs video on tensors alone) or is it natural to learn GR?
 
7:38 AM
You could certainly learn GR in a month but I would do it from a book like Carroll rather than YouTube videos.
Do you really need to learn GR to do an astrophysics project?
@ACuriousMind is there a simple way to explain why the probability of an electron and positron scattering into two photons is high, but the probability of two photons scattering into an electron and positron is low?
 
8:20 AM
@JohnRennie Think about it like this: As long as the two particles are in "collision range", the probability per unit time for the reaction to happen is the same in both directions. But photons always move at the speed of light, so they'll never spend a lot of time in collision range of each other, and if they manage to react but the products are very slow, the products will usually annihilate again before getting far enough from each other for you to detect them.
 
Thanks :-) I was asking because of this question:
0
Q: Why Electron Quantum Field Wants Little Energy But Photon Field Doesn't

MeltedStatementRecognizingIn this Quora post: https://qr.ae/pv5tac, it states that the electron quantum field "wants" to reduce the energy it has, so when a particle and an anti-particle interact and the charges cancel out (so conservation of charge not violated) the electron field uses the opportunity to get rid of the e...

If I understand you correctly the scattering probability is the same in both directions, and it's just the fact that the conditions are different on the two sides i.e. photons move at c while electrons (typically) move much slower than c.
Then presumably the observed scattering probability decreases with increasing centre of mass energy?
 
Yes, the scattering cross section doesn't change (it's the same Feynman diagram, after all!), but it's harder to build a good photon collider than to build a good lepton collider
@JohnRennie it's not that simple - it depends on specifics of your collider setup, a property called "luminosity"
like, if you're shooting two beams of leptons into each other head on, making them faster probably won't significantly reduce the number of collisions that happen
if you're just crossing two beams, that's a different story
 
You would have thought that it wouldn't be hard to get the same number of particles interacting per unit volume per unit second for both electrons and photons ...
4
Q: Energy dependency of the total cross section for different species

DarioPComparing the plots for the total (inelastic) cross sections as a function of the centre of mass energy for $pp$ and $e^+e^-$ collisions: one notes that the trend at high energy is opposite: the $pp$ cross section increases while the $e^+e^-$ decreases. Is there a (simple) explanation for th...

The electron-positron scattering cross section does indeed decrease with increasing COM energy ...
 
@JohnRennie I think I need it. For example some of the offered projects are "structural analysis of spacetime around a (non)rotating black hole with (no) electrical charge".
 
@JohnRennie yes, that's the effect of "they're faster so they have less time to interact"
you won't see this for photons in the same way, but there's another effect that the higher the photon energies are the more other particle-antiparticle pairs become available for a pair of them to create
 
8:32 AM
@ACuriousMind Thanks :-)
 
so even though the fundamental "idea" of the reaction is symmetric, the two behave very differently in practice
 
@Earman Schwarzschild metric in other words?
If you're just trying to understand the geometry you don't need to know the full mechanics of GR. You just need to know how to do calculations with the metric and that's not hard.
 
Yes
@JohnRennie i see. Thanks.
@JohnRennie so do you have suggestions for that? To understand schwarzschild metric, kerr metric etc but not the full mechanics of GR?
 
That's pretty much what I did. Despite the large number of GR answers I've posted on the Physics SE I really only know how to work with metrics and my grasp of the rest of GR is pretty tenuous.
But I picked up the knowledge over several years, so my approach wouldn't be any good for the limited time you have available.
I'd suggest you grab a copy of Carroll's book, start at chapter 1 and see how you get on.
You can skim the chapters that go too deeply into the maths. If you skim through the book you should be able to get an idea of what parts you need in a day or two.
 
9:00 AM
@JohnRennie on it :)
 
@Earman :-)
You are welcome to post here if you have any questions. As I mentioned I'm not strong on the fundamentals of GR but we have people hereabouts who are.
 
 
2 hours later…
11:24 AM
@Earman I would suggest picking up an astrophysics book, it will usually have a chapter on GR that will be enough, you could spend months on a normal GR book and you likely don't need that much detail
 
 
4 hours later…
3:30 PM
@ACuriousMind i think i found an explanation, tho i can't verify it. the optical activity i'm used to is optical rotation, e.g., corn syrup rotates the plane of polarization
if you work in terms of the Poincare sphere, that's rotation around the axis of right/left circular polarized states
what i'm seeing would be explained by the plastic instead giving rotation around the axis of horizontal/vertically-polarized states
in which case horizontally polarized light and vertically polarized light would both go through unchanged, but diagonally-polarized light would become elliptically polarized
(another explanation would be that the plastic is somewhat depolarizing but only along the diagonal directions. i don't like this explanation as much but i can't falsify it using linear polarizers alone.)
 
@Semiclassical Isn't that exactly what you'd expect simple uniaxial birefringence ("one axis is faster than the other") to do?
your plastic is essentially a very bad quarter-wave plate
 
probably? i feel like i've forgotten this stuff if i ever knew it well
 
yes, I think that's exactly what you'd expect: if you replace the plastic by a quarter-wave plate you'd expect the same - the plate has two main axes and if the incoming light is aligned with either of them then nothing happens, if not then the result is ellipical polarization
and wave-plates typically are just uniaxial birefringent material
 
nice. my only caveat is that it wouldn't be a quarter-wave plate here, just a generic phase retarder
probably somewhere between quarter- and zero-
 
sure, would be very lucky if it's actually close to being quarter-wave
but in order to test that you'd have to see "how circular" the result at 45° really is
 
3:46 PM
well, if it was circular then you'd expect that the intensity wouldn't change as you rotate the analyzing polarizer
and with mine it does
 
ah, yes
 
that's not enough to quantify how close it is to being circular tho
would be fun to find a household plastic which is accidentally a good quarter-wave plate
("good" meaning approximately---i don't think you'd get a precise one by luck)
 
4:03 PM
Quick question: if the components of a matrix are written like $a^i_j$, is the upper index the row or the column index? I know it's just a matter of choice but I suppose there is a well established convention about this so you don't get crazy over reading different books
 
lol
there are at least two well-established conventions ;)
 
I met both and it is so annoying
On the one hand, you can say contravariant vectors are generally regarded as column vectors, in this sense the upper index being the column index would be more reasonable, on the other hand it looks more natural to read the row index first
Now I get why my professor said he hated working with matrices
 
 
2 hours later…
6:31 PM
@ACuriousMind I do wonder what in the manufacturing process determines the orientation of these axes; they seem to coincide with base and reflection axes of the protractor
 
I'm afraid I don't really understand anything about why/how plastic is birefringent :P
 
Same
Something something long parallel polymers?
 
wiki says something about stress that gets "frozen in" as it cools
but that also doesn't really tell me a lot :P
 
Yeah. And I don’t see much stress coloration when using my polarizers
 
oh
it might be long polymers + an extrusion process
I could believe that extrusion aligns the polymers in a certain direction
 
6:42 PM
Yeah
I’m a little surprised I don’t see more stress effects near the edges tbh. I can find a little of it, but only at what seem like defects (eg the edge bulges out slightly)
In particular I don’t really see it at all on the curved edge
 
7:21 PM
Why is the entropy of entanglement for a system in contact with a reservoir just $S=\mathrm{Tr}e^{\beta H}\mathrm{ln} e^{\beta H} = \mathrm{Tr} -\beta \hat{H}e^{\beta H} = -\beta <E>$ i.e. why is it just the average energy of the subsystem?
Is this just a cooincidence?
That also means as $\beta\rightarrow 0$ (or $T\rightarrow \infty$), $S\rightarrow 0$ i.e. the system detangles with the reservoir
 
7:37 PM
@DIRAC1930 This isn't a quantum phenomenon - in the classical canonical ensemble, you also have $S = \beta E - F$
if you consider the macroscopic definition of entropy in terms of heat (i.e. work "wasted" into internal energy), this should not be surprising
 
fqq
@DIRAC1930 the entangleent entropy is not a measure of entanglement at finite temperature
it only works if you consider the global state (system+environment) in a pure state and explicitly derive the reduced state
 
7:52 PM
Ah okay thanks
 
fqq
also your formula is wrong, you forgot the normalisation which gives the extra term (free energy) from ACM's message
and the entropy is not zero in the infinite temp. limit but $\log d$ where $d$ is the dimension of the hilbert space
 
8:16 PM
Ah okay thanks
 
8:28 PM
So for macroscopic systems, is considering the entanglement entropy not really a useful thing?
 
why do you keep calling it "entanglement entropy"
it's just von Neumann entropy, you can define it for arbitrary $\rho$ without any reference to entanglement
 
Because I'm trying to figure out how entanlged a system becomes with it's environment
Which is defined by the Von Neumann entropy of the reduced density matrix for a system
 
fqq already has objected to that interpretation above
 
Yes I know
 
so instead of asking them why it's wrong or might not work you just decided to ignore that? :P
 
8:40 PM
No I was thinking that maybe you could have the environment and system initially isolated so that the system is in a pure state and then open something
But I'm guessing that wouldn't work
 
9:12 PM
@bolbteppa yes, I realized that later. In fact I took another book along Carroll's one called "Black Holes, White Dwarfs, and Neutron Stars: The Physics of Compact Objects" which has only a chapter about general relativity. And even though I find the approach of Carroll quite smooth, I will probably not follow it after ch1 since this astrophysics book explicitly said that they only wrote Einstein's eqn down but we don't need it (in full generality).
 
fqq
@DIRAC1930 hm, it's complicated. What you wrote is not wrong in principle
y
 
 
1 hour later…
10:29 PM
@fqq Okay thanks
 

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