Undergrad Basic question, harmonic coordinate condition algebra

[binbagsss]

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)

 

[Orodruin]

why has the double derivative of the first term vanished whilst a single derivative of the second does not?

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.

 

[binbagsss]

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]

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

$\partial_\nu=\frac{\partial}{\partial x^{\nu}}$

Obvious typo, also the fix is obvious (the $\lambda$ in the last expression should be a $\mu$).

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

 

[binbagsss]

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]

$\partial x^\mu/\partial x^\nu=\delta^\mu_\nu$?

Right, so is it clear how equation (4.85) follows?

 

[binbagsss]

Right, so is it clear how equation (4.85) follows?

ahhh ofc, thanks.