Calc. Christoffel Symbols of Hiscock Coordinates

[Onyx]

TL;DR Summary

Calculating the christoffel symbols of Hiscock coordinates.

The Hiscock coordinates read:

$$d\tau=(1+\frac{v^2(1-f)}{1-v^2(1-f)^2})dt-\frac{v(1-f)}{1-v^2(1-f)^2}dx$$

$dr=dx-vdt$

Where $f$ is a function of $r$. Now, in terms of calculating the christoffel symbol $\Gamma^\tau_{\tau\tau}$ of the new metric, where $g_{\tau\tau}=v^2(1-f)^2-1$ and $g_{\tau r}=0$, can I safely assume that $\Gamma^\tau_{\tau\tau}=0$, since $\frac{\partial g_{\tau\tau}}{\partial \tau}=\frac{\partial g_{\tau\tau}}{\partial r}\frac{\partial r}{\partial \tau}$ (in the Jacobi matrix $\frac{\partial r}{\partial \tau}=0)$?

 

[PeterDonis]

in terms of calculating the christoffel symbol $\Gamma^\tau_{\tau\tau}$ of the new metric

What "new metric"? $\tau$ isn't a coordinate, it's proper time.

 

[Ibix]

What "new metric"? $\tau$ isn't a coordinate, it's proper time.

Not in this case, apparently. The new coordinates appear to be $\tau,r$. This appears to be in reference to https://arxiv.org/abs/gr-qc/9707024.

 

[PeterDonis]

This appears to be in reference to https://arxiv.org/abs/gr-qc/9707024.

Where in this paper does the metric shown in the OP appear?

 

[Ibix]

Where in this paper does the metric shown in the OP appear?

There isn't a complete metric in the OP, but the stated $g_{\tau\tau}$ and $g_{\tau r}$ match equation 12 given equation 10 for a definition of $A(r)$.

 

[Onyx]

Not in this case, apparently. The new coordinates appear to be $\tau,r$. This appears to be in reference to https://arxiv.org/abs/gr-qc/9707024.

I apoligize, I should have adopted $dT$ instead to be more clear.

 

[Onyx]

Would it be possible to do it with something like Maple?

 

[Ibix]

I'm struggling slightly to determine the definition of $\tau$ here. The definition you quote, given in the paper I found, is for $d\tau$. By the chain rule that's $\frac{\partial\tau}{\partial t}dt+\frac{\partial\tau}{\partial r}dr$, but I can't make sense of the coefficients in your definition in those terms. Am I missing something here?

 

[Onyx]

This might help:
$d\tau=dt-\frac{v(1-f)}{1-v^2(1-f)^2}dr=dt-\frac{v(1-f)}{1-v^2(1-f)^2}(dx-vdt)=(1+\frac{v^2(1-f)}{1-v^2(1-f)^2})dt-\frac{v(1-f)}{1-v^2(1-f)^2}dx$

 

[Onyx]

This might help:
$d\tau=dt-\frac{v(1-f)}{1-v^2(1-f)^2}dr=dt-\frac{v(1-f)}{1-v^2(1-f)^2}(dx-vdt)$

My instinct is that, since it says that the metric is "manifestly static", $\Gamma^{\tau}_{\tau\tau}$ is indeed zero.

 

[Ibix]

Don't guess, calculate. Generally you know that$$\Gamma^a_{bc}=\frac 12g^{ai}\left(\partial_b g_{ic}+\partial_c g_{bi}-\partial_i g_{bc}\right)$$so in this case$$\Gamma^\tau_{\tau\tau}=\frac 12g^{\tau i}\left(\partial_\tau g_{i\tau}+\partial_\tau g_{\tau i}-\partial_ig_{\tau\tau}\right)$$I think that the metric is diagonal, so the only non-zero $g^{\tau i}$ is the $i=\tau$ one. But since $g_{\tau\tau}$ has no explicit dependence on $\tau$ then the Christoffel symbol is indeed zero.

 

[Onyx]

Don't guess, calculate. Generally you know that$$\Gamma^a_{bc}=\frac 12g^{ai}\left(\partial_b g_{ic}+\partial_c g_{bi}-\partial_i g_{bc}\right)$$so in this case$$\Gamma^\tau_{\tau\tau}=\frac 12g^{\tau i}\left(\partial_\tau g_{i\tau}+\partial_\tau g_{\tau i}-\partial_ig_{\tau\tau}\right)$$I think that the metric is diagonal, so the only non-zero $g^{\tau i}$ is the $i=\tau$ one. But since $g_{\tau\tau}$ has no explicit dependence on $\tau$ then the Christoffel symbol is indeed zero.

I was pretty sure of it was zero. Thank you. One last thing:

For the new metric, is $n_a=(\sqrt{A},0)$?

Don't guess, calculate. Generally you know that$$\Gamma^a_{bc}=\frac 12g^{ai}\left(\partial_b g_{ic}+\partial_c g_{bi}-\partial_i g_{bc}\right)$$so in this case$$\Gamma^\tau_{\tau\tau}=\frac 12g^{\tau i}\left(\partial_\tau g_{i\tau}+\partial_\tau g_{\tau i}-\partial_ig_{\tau\tau}\right)$$I think that the metric is diagonal, so the only non-zero $g^{\tau i}$ is the $i=\tau$ one. But since $g_{\tau\tau}$ has no explicit dependence on $\tau$ then the Christoffel symbol is indeed zero.