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Gauge Theory

Anonymous Sat 06 Nov 20:55:55 2021 No.13826571 View Reply Original Report

Quoted By: >>13826648 >>13826671

Anybody here know how to derive the "standard" energy momentum tensor of a Yang Mills gauge connection? Applying Noethers theorem for spacetime translational symmetry (i.e the first goto) doesnt give the standard one from literature

Anonymous

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Anonymous Sat 06 Nov 2021 21:05:50 No.13826602 Report

Quoted By:

Have nobody about this but interested. Bump

Anonymous

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Anonymous Sat 06 Nov 2021 21:08:18 No.13826615 Report

Quoted By:

Also no idea

Anonymous

Anonymous Sat 06 Nov 2021 21:20:12 No.13826648 Report

Quoted By: >>13826664

>>13826571

So for reference:

With Noethers thrm:

$T^{\mu}_{\nu}= \frac{\partial \mathcal{L}}{\partial\partial_{\mu}\phi^{a}}\partial_{\nu}\phi^{a} -\delta^{\mu}_{\nu}\mathcal{L}$

So applying this to the YM lagrangian

$\mathcal{L}=\frac{1}{4}\text{tr}F^{\mu\nu}F_{\mu\nu}$
where
$F=\mathrm{d}A + A\wedge A$

yields (for me) (smth like)

$T^{\mu}_{\nu} \sim \mathrm{tr}\left( F^{\mu\alpha}\partial_\nu A_\alpha - \frac{1}{4}\delta^{\mu}_{\nu}F^{\alpha\beta}F_{\alpha\beta} \right)$

but in literature it says (smth like)

$T^{\mu}_{\nu} \sim \mathrm{tr}\left( F^{\mu\alpha}F_{\nu\alpha} - \frac{1}{4}\delta^{\mu}_{\nu}F^{\alpha\beta}F_{\alpha\beta} \right)$

Anonymous

Anonymous Sat 06 Nov 2021 21:26:02 No.13826664 Report

Quoted By: >>13826685

>>13826648
You made a mistake in your derivation, which is obvious by the inclusion of only one partial derivative of the 4-potential rather than the commutator of it leading to the commutator and thus the energy momentum tensor. In fact, your result is simply wrong for the reason of partial derivatives of 4-vectors not being lorentz covariant.

Anonymous

Anonymous Sat 06 Nov 2021 21:28:17 No.13826671 Report

Quoted By: >>13826695 >>13826714 >>13826742

>>13826571
Why are you indulging some other mathematician?

Noether fucked a lot of things up. Symmetries are useful except when they are not. Macroscopic events in the universe tend to not be symmetrical, senso stricto. Check out my jargon, yo.

Anonymous

Anonymous Sat 06 Nov 2021 21:33:55 No.13826685 Report

Quoted By: >>13826731

>>13826664
Oh wow okay indeed using the cov derivative for the derivation of A would indeed fix it perfectly. Totally forgot to consider the Bundle structure and hence the need for it. Thanks !

Anonymous

Anonymous Sat 06 Nov 2021 21:37:12 No.13826695 Report

Quoted By: >>13826742

>>13826671
>Macroscopic events in the universe tend to not be symmetrical, senso stricto.
The fact that the exact symmetry such as time reversal doesn't carry over to the statistical behaviour of some large macroscopic system isn't a disproof of it being there. The dynamics of a box of 10^27 free particles of oxygen are reversible completely and in fact even satisfy the assumption of poincare recurrence. Its an undergrad level point to understand.

Anonymous

Anonymous Sat 06 Nov 2021 21:42:41 No.13826714 Report

Quoted By: >>13826742

>>13826671
Theres a whole lot of symmetry in many macroscopic systems. Just consider the FLRW metrics / Friedmann equations. Now if nature really behaves like that is another point but it very much seems like it

Anonymous

Anonymous Sat 06 Nov 2021 21:50:09 No.13826731 Report

Quoted By: >>13826890

>>13826685
Ill probably do the derivation myself and post it tomorrow, if the thread will still be up.
But really, just sanity check your answers. If you got something which isnt covariant and doesnt agree with the literature, you have made a mistake. It is basically impossible to make a mistake and still meet all constraints of covariance/get the right units/get all the right signs/etc.

Anonymous

Anonymous Sat 06 Nov 2021 21:55:59 No.13826742 Report

Quoted By: >>13826766

>>13826671
>Macroscopic events in the universe tend to not be symmetrical, senso stricto. Check out my jargon.
>jargon

>>13826695
>The dynamics of a box of 10^27 free particles of oxygen are reversible completely
>...in fact...
>Its an undergrad level point to understand

Is non-equilibrium thermodynamics a joke to you?

>>13826714
What is it about the term "jargon" you not understand? Of course there is a whole lot of apparent symmetry. Many non-symmetrical things can agglomerate and produce a virtually symmetrical object...if you think about it in certain ways. And in other ways, it is horribly unsymmetrical.

I mean, could you imagine how insane it would be if your circulatory system were symmetrical? Jesus fucking Christ.

Anonymous

Anonymous Sat 06 Nov 2021 22:07:43 No.13826766 Report

Quoted By: >>13826815

>>13826742
>Is non-equilibrium thermodynamics a joke to you?
Nothing magical happens when you put any amount of atoms in a box, there are no magical laws of thermodynamics not applicable for N atoms, but suddenly applicable for N+1 atoms. Learn some statmech.

Anonymous

Anonymous Sat 06 Nov 2021 22:25:50 No.13826815 Report

Quoted By:

>>13826766
>N atoms, but suddenly applicable for N+1 atoms.
A small joke: a general rule is that there are only numbers 0, 0+epsilon, 1, 2, 3, 4 and infinity, or at least God only knows those numbers. Devil maybe also knows 5, 7 and 8 and 12. Dont take it seriously.

Anonymous

Anonymous Sat 06 Nov 2021 22:54:05 No.13826890 Report

Quoted By:

>>13826731
Would be appreaciated if you did but if not its ok too, i managed to derive it now too.

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