4:13 AM
@bolbteppa ok finally wrote all this up. maybe youd be interested in this chat on subj hope you can drop by with open mind. @Secret + @DanielSank think youd find it of interest + relevant too :) chat.stackexchange.com/transcript/message/56826061#56826061

3 hours later…
6:56 AM
@Slereah Can I ask for some SR help? (acceleration in SR) ... Having some trouble understanding stuff :/

sure

Thanks one moment
Pg 79
Fig i
How did they get that?

7:30 AM
Or do you want me to paste snippets of the relevant pages?

what do you want me to explain, exactly
It's a diagram

So let me summarize my understanding
So in fig f they have basically shown the orientation of the axis for $|v| = .5 c$
And then he uses the spacing of the points to elaborate the change in rapidity
Why is it equally spaced
from point AB?

8:24 AM
*curve AB?
in fig i

3 hours later…
11:18 AM
Neumann's principle states that the physical properties of a crystal must have the same symmetry as the crystal itself. I have only seen this applied to macroscoping properties such as currents and electromagnetic fields. Does anyone know of an application to internal properties? Specifically I want to consider the exchange coupling between localized spins.
The reason I suspect it is different, is that in contrast to currents and electromagnetic fields, I expect the exchange coupling to be positional dependent, not only depend on direction.

11:30 AM
what geometric properties does nonmetricity reflect?
The Riemann curvature reflects the change of a tensor after it is parallelly transported.
what geometric properties does the torsion reflect? If the torsion doesn't vanish, an infinitesimal parallelogram is not closed.

vector norm isn't preserved along a geodesic if the connection isn't metric
in particular this means that geodesics aren't the shortest curves
or at least not generally

11:52 AM

@Bohemianrelativist the same but with the theorem for Lorentz spaces
ie locally a geodesic between two timelike separated points will maximize proper time

12:29 PM
if the affine connection is not Riemann, then the corresponding curvature doesn't indicate the same for the geometry as the Riemann curvature does, right?

what do you mean by Riemann

@Slereah the torsion vanishes and the nonmetricity vanishes.

levi civitta connection, is what it is called
also yes

@Slereah yes.

for a general connection, you can split the curvature into parts
one of which is the Riemann tensor of the levi civitta connection

12:40 PM
@Slereah I have seen that kind of separation in papers which illustrate the teleparallel equivalent to general relativity.
$R\eta=R^{\{\}}\eta+xxx$, then a teleparallel equivalent to GR is formulated by the Lagrangian $-xxx$, where $R^{\{\}}$ is the Riemann curvature.
I found the papers about teleparallel gravity is of a great diversity.
some formulate teleparallel gravity by the Lagrangian with the fundamental variables the tetrad and affine connection and others formulate it by the Lagrangian with the fundamental variable the translational gauge potential.

12:56 PM
Many formulations indeed

are all these equivalent to GR? I don't think so.

I can't vouch for all of them
Haven't looked much into it

1:34 PM
If we're looking at $\phi^4$ loop corrections, and we get an integral (which we've Wick rotated) that looks like: $$\int \tilde d^4k_E \frac{1}{k_E+m^2}$$ for which we then impose a cutoff at $\Lambda$ and note that at large $k_E$, $k^2_E+m^2\sim k^2$. Apparently the integral becomes $$\int^\Lambda \frac{|k_E|^3 d|k_E|}{|k_E|^2},$$ but how come the measure changes here?
what I basically don't see is why $d^4k_E\rightarrow |k_E|^3 d|k_E|$ is true (not the absolute value signs, just why this reduces to a one dimensional integral)

I think that's just a switch to spherical coordinates?
The angular part is symmetric so you just obtain whatever volume of a sphere

ohh you might be right yeah
I guess that's valid as long as $\Lambda$ is large, which it is, ty
oh woops, that's explained 2 lines down lol my bad

2:02 PM
beautiful integral
$$\int \limits_{\gamma} d^4 x \mathcal L + \Gamma r_{\mu\nu}(x_1, x_2, ...)$$

2:19 PM
Profile Pic updated

what is it though?

2:45 PM
what even is $\Gamma r_{\mu\nu}$

3:01 PM
$\Gamma$ is often an effective action term in stuff like that?

1 hour later…
4:03 PM
I just wanted to make the integral look cute!
It just looks good that way,
Tho it means totally meaningless
inspired from this:
2 hours ago, by Charlie
If we're looking at $\phi^4$ loop corrections, and we get an integral (which we've Wick rotated) that looks like: $$\int \tilde d^4k_E \frac{1}{k_E+m^2}$$ for which we then impose a cutoff at $\Lambda$ and note that at large $k_E$, $k^2_E+m^2\sim k^2$. Apparently the integral becomes $$\int^\Lambda \frac{|k_E|^3 d|k_E|}{|k_E|^2},$$ but how come the measure changes here?

in the teleparallel gravity, if the nonmetricity vanishes, a global frame which is achieved by parallelly transporting a frame at a certain point has a constant metric, so this is like an Euclidean metric, but if the nonmetricity doesn't vanish, the metric is not constant, the global frame doesn't look like Euclidean, but the paper says we can choose the global frame to be an orthonormal frame, which has a constant metriic, so why is there such contrdiction?

heck idk

4:42 PM
another puzzle I encounter is that whether there is really a torsion $T^\alpha$ in a teleparallel gravity. If I choose a global frame to be the covariant constant frame, that is, a frame on which $\Gamma^\alpha{}_{\beta\gamma}=0$, then $T^\alpha=D\vartheta^\alpha=d\vartheta^\alpha$, but if $\vartheta^\alpha$ is a global coframe, then it can be chosen to be a holonomic frame globally $\vartheta^\alpha=dx^\alpha$, then $T^\alpha=D\vartheta^\alpha=d\vartheta^\alpha=ddx^\alpha=0$.

God why is Tensorflow the industry standard
It's practically non-functional
Gotta spend the whole day just trying to get it working

of course I can choose another coordinate system so that $\vartheta^\alpha$ is not holonomic, but the torsion is a tensor, so if it vanishes in a coordinate system, it should vanish in all coordinate systems.

0

The list of topics for this site on https://physics.stackexchange.com/help/on-topic includes this entry: Experimental designs and results Example: What is needed to claim the discovery of the Higgs boson ? I think what is meant here is questions related to the design of experiments themselves, ...

@Bohemianrelativist Isn't the exterior derivative different if a torsion is present
I seem to recall that there is an extra form in there

@Slereah can you elaborate by writing out the formula? what is that extra form?

5:00 PM
Something like $$D^\omega \psi = d\psi + \omega \wedge \psi$$

@Slereah why not PyTorch?

@NiharKarve AI libraries are usually some unholy mishmash of whatever

What were you using it for

@Slereah what is $\omega$? The formula looks like a covariant differential.

Heck if I know
I don't recall too much wrt torsion

5:05 PM
Hello guys..

5:33 PM
Hey, Figueroa-O'Farill's on MO
big up EMPG

@Bohemianrelativist are there any notes with a simple/easy motivation for even considering that approach

5:45 PM
Quick question, the logarithmic divergence of the one-loop correction in $\phi^4$ theory, doesn't this have two divergences, one as $\Lambda\rightarrow 0$ and one as $\Lambda\rightarrow \infty$? Does this just mean it shows IR and UV divergence?

@Charlie yes

Fair enough
is that just a generic feature of diagrams like that? as the momentum of the intermediate particle approaches zero?
seems like I would almost expect a problem there

it's common to have both kinds of divergences, yes

ok then

@ACuriousMind since your the most mathematical physics person I know. I was wondering if you knew anything about Lieb Robinson bounds? and how would someone with little math background (like me) gain the technical expertise?

5:48 PM
@MoreAnonymous I don't know anything about that, I'm afraid

wikipedia provides several references at the end of the introduction about where information on it can be found en.wikipedia.org/wiki/Lieb-Robinson_bounds

@Charlie I was hoping for something along the lines of a graduate introductory notes atleast and sketch my way from there

Do you already have some experience on the topic or something?
Wikipedia says that a rigorous introduction is given in the book they reference, that's usually a good place to go if one actually wants "expertise"

@Charlie I got curious about it :/
*kinda equivalent
of micro causality from QFT in QM

Reading through the wiki page would be a good start, then check the links they're giving, if it's a fairly obscure topic there may not be that many standard texts that are very comprehensive, you might just end up reading papers
I actually through non-rel qm was entirely non-causal, don't the amplitdues just decay exponentially outside the lightcone

5:56 PM
@Charlie So I'm sure you already know but many body QM is a special case of QFT

I did not know that no

Really?

I haven't looked into mb qm

Well u can use the notes mentioned here if you want
0

Background So basically there is a dictionary/correspondence between first and second quantised quantum mechanical theories. When I suddenly turn on a potential in potential in first quantised theories I can use the sudden approximation and find the probability of the final state. I can use the d...

The intro is super brief but outlines what I'm talking about

I haven't looked into it because it doesn't sound interesting tbh :p

6:00 PM
@Charlie Well each to their own :P
I found it super interesting
since I had a good intuition of QM
But never of QFT

6:14 PM
Intuition is basically synonymous with practice
at first, qft seems like an unholy mess

afterwards, it still seems like an unholy mess, but you've learnt not to expect any better from reality
2

;_;

Argh
Had to reinstall the whole linux
The whole hog

I feel much like today's SMBC on the matter :P

Pondering on a conceptual question. Why is a Weyl gauge always possible, i.e. the electric potential $V=0$?
Is it due to the Helmholtz theorem?
What does it say in layman terms?

6:27 PM
@schn If you look at the gauge transformations, you can always choose the gauge transformation so that that comes out

@ACuriousMind that's a nice answer :)

there's not necessarily a deep reason that a specific choice of gauge is possible - you just have to look at the transformations and figure out whether or not you can solve the equations for the transformed potentials being equal to the desired values in your gauge

Right. So $\mathbf A'=\mathbf A +\nabla \lambda$ and $V'=V-\frac{\partial \lambda}{\partial t}$. Choose $\lambda$ so that $\frac{\partial \lambda}{\partial t}=V$?

yes, exactly

@bolbteppa Do you mean teleparallel approach?

6:35 PM
Yeah

@ACuriousMind Which would just be $\lambda=\int_0^t V dt'$, kind of?

@schn yep (but you shouldn't use the same $t$ for both the integration variable and upper bound)

Right, sloppy (fixed)

@bolbteppa the motivation I read from papers of my MSc supervisor is that the energy-momentum 3-form or even the angular momentum and center-of-mass moment 3-form of the gravity formulated by teleparallel theory is a tensor, rather than psuedotensor.