Calc. Christoffel Symbols of Hiscock Coordinates

In summary, the conversation discusses the calculation of the Christoffel symbol for a new metric, where the coordinates are given by ##\tau## and ##r##. The metric is given by ##g_{\tau\tau}=v^2(1-f)^2-1## and ##g_{\tau r}=0##, and the question is whether the Christoffel symbol ##\Gamma^\tau_{\tau\tau}=0##. The expert confirms that the Christoffel symbol is indeed zero and suggests using the formula ##\Gamma^a_{bc}=\frac 12g^{ai}\left(\partial_b g_{ic}+\partial_c g_{bi}-\partial_i g_{bc}\right
  • #1
Onyx
121
3
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
Onyx said:
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
PeterDonis said:
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.
 
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  • #5
PeterDonis said:
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)##.
 
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  • #6
Ibix said:
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.
 
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  • #7
Would it be possible to do it with something like Maple?
 
  • #8
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
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 said:
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
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
Ibix said:
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)##?
Ibix said:
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.
 

Related to Calc. Christoffel Symbols of Hiscock Coordinates

1. What are the Calc. Christoffel symbols of Hiscock coordinates?

The Calc. Christoffel symbols of Hiscock coordinates are a set of mathematical quantities used in the study of curved spaces, specifically in the field of general relativity. They represent the components of the connection, which describes how vectors change as they are parallel transported along a curved space.

2. How are the Calc. Christoffel symbols of Hiscock coordinates calculated?

The Calc. Christoffel symbols of Hiscock coordinates are calculated using the metric tensor, which describes the geometry of the curved space. The symbols are derived from the metric tensor using a specific formula involving partial derivatives.

3. What is the significance of the Calc. Christoffel symbols of Hiscock coordinates?

The Calc. Christoffel symbols of Hiscock coordinates are important in the study of general relativity because they allow us to calculate the curvature of a space. They also play a crucial role in the equations of motion for particles and fields in curved spaces.

4. Can the Calc. Christoffel symbols of Hiscock coordinates be used in flat spaces?

Yes, the Calc. Christoffel symbols of Hiscock coordinates can be used in flat spaces as well. In flat spaces, they represent the components of the trivial connection, which describes how vectors change as they are parallel transported along a straight line.

5. Are the Calc. Christoffel symbols of Hiscock coordinates unique?

No, the Calc. Christoffel symbols of Hiscock coordinates are not unique. They depend on the choice of coordinates and can be different for different coordinate systems. However, the curvature of a space is independent of the choice of coordinates and remains the same regardless of which coordinate system is used.

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