Christoffel Symbols of Hiscock Coordinates

  • #1
Onyx
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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)##?
 
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  • #2
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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.
 
  • #3
Ibix
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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.
 
  • #5
Ibix
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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)##.
 
  • #6
Onyx
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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.
 
  • #7
Onyx
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Would it be possible to do it with something like Maple?
 
  • #8
Ibix
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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?
 
  • #9
Onyx
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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##
 
  • #10
Onyx
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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.
 
  • #11
Ibix
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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.
 
  • #12
Onyx
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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.
 

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