Basic question, harmonic coordinate condition algebra

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
6 replies · 2K views
binbagsss
Messages
1,291
Reaction score
12
where ##□=\nabla^{\mu}\nabla_{\mu}## is the covariant D'Alembertian.

##□x^{\mu}=0##

##g^{\rho\sigma}\partial_{\rho}\partial_{\sigma}x^{\mu}-g^{\rho\sigma}T^{\lambda}_{\rho\sigma}\partial_{\lambda}x^{\mu}=0##

So this line is fine by subbing in the covariant derivative definition and lowering index using the metric.

The notes say that it is crucial to realize that when we take the covariant derivative that the four functions ##x^{\mu}## are just functions, not component of a vector. And I guess this is what he is using when he gets:

##g^{\rho\sigma}\partial_{\rho}\partial_{\sigma}x^{\mu}-g^{\rho\sigma}T^{\lambda}_{\rho\sigma}\partial_{\lambda}x^{\mu}=-g^{\rho\sigma}T^{\lambda}_{\rho\sigma}## ,
where ##T^{\lambda}_{\rho\sigma}## is the connection.

Excuse my stupid questions to follow, but I have no idea what is going on here:
- why has the double derivative of the first term vanished whilst a single derivative of the second does not?
- how has , I think he must have used that## \partial_{\lambda}x^{\mu}=1##, where has this came from?

All I can think of is some conditions imposed on the functions such as each four function is made to depend only upon one of ##x^\mu## and normalization.

Thanks in advance.

(Sean Carroll Lecture notes on GR, equation 4.85)
 
Physics news on Phys.org
Orodruin said:
What is ##\partial_\nu x^\mu##?

Your equation is also wrong, the right side has ##\lambda## as a free index while the left does not.

Equation 4.85, http://arxiv.org/pdf/gr-qc/9712019.pdf, that's what I thought.

##\partial_\nu=\frac{\partial}{\partial x^{\nu}}##
 
Orodruin said:
Obvious typo, also the fix is obvious (the ##\lambda## in the last expression should be a ##\mu##).

So what is ##\partial x^\mu/\partial x^\nu##?

I thought he may have renamed ##\lambda## and ##\mu##. The fix wasn't so obvious for me, but ta...

##\partial x^\mu/\partial x^\nu=\delta^\mu_\nu##?
 
Orodruin said:
Right, so is it clear how equation (4.85) follows?

ahhh ofc, thanks.