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I am trying to understand the paper 'Spectral shifts in General Relativity' by Narlikar.
The paper considers a light ray emanating from the origin of a FLRW coordinate system in a universe whose hypersurfaces of constant time (in that coordinate system) are homogeneous and isotropic. The light ray is a null geodesic, and it is argued that θ and [itex]\phi[/itex] are constant along its path. The paper claims the following DE determines the geodesic.
[itex]0= \frac{d^2t}{du^2} - (\frac{dr}{du})^2 \frac{a(t)a'(t)}{1-kr^2}[/itex] [12]
In my attempts to reproduce this result, I get the opposite sign on the last term , which prevents further progress. I have checked and re-checked, but just can't see what I'm doing wrong. Here is my working:
In the FLRW coordinate system, all off-diagonal components of the metric tensor are zero, and the two diagonal components that concern us are:
[itex]g_{00} = 1;\ \ g_{11} = -\frac{a(t)^2}{1-kr^2}[/itex]
It has been argued elsewhere that, along the geodesic, [itex]d\theta = d\phi = 0[/itex], and since it is a null geodesic, this gives:
[itex]0 = ds^2 = \frac{a(t)^2 dr^2}{1-kr^2}-{dt^2}[/itex]
The DE for the time component of the geodesic is:
$$0=\frac{d^2x^0}{du^2} + \Gamma^0{}_{kl}\ \frac{dx^k}{du}\ \frac{dx^l}{du}\ \ \ \ \text{[}\textbf{11}\text{]} \\ $$
We calculate the Christoffel symbol's value for i=0 as follows:
\begin{align*}
\Gamma^0{}_{kl} &= \frac{1}{2}g^{0\beta}(g_{\beta k,l}+g_{\beta l,k}-g_{kl,\beta})\ \ \ \ \text{[}\textbf{11a}\text{ - see Schutz 6.32]}\\
&= \frac{1}{2}g^{00}(g_{0k,l} + g_{0l,k} - g_{kl,0})\text{ [since }g_{0\beta} = 0 \text{ unless }\beta = 0]\\
&= \frac{1}{2}g^{00}(g_{00,l}+g_{00,k} - g_{kl,0})\\
&= -\frac{1}{2}g_{kl,0} \text{[since }g_{00} = g^{00} = 1\text{, which is constant]}\\
\end{align*}
Inserting this into the geodesic DE [11] we get:
\begin{align*}
0 &= \frac{d^2t}{du^2} - \frac{1}{2}g_{kl,0}\frac{dx^k}{du}\frac{dx^l}{du} \\
&= \frac{d^2t}{du^2} - \frac{1}{2}g_{00,0}(\frac{dt}{du})^2 - \frac{1}{2}g_{11,0}(\frac{dr}{du})^2\text{ [since }\frac{d\theta}{du}\text{ and }\frac{d\phi}{du}\text{ must be zero]}\\
&= \frac{d^2t}{du^2} - \frac{1}{2}\frac{\partial(\frac{-a(t)^2}{1-kr^2})}{\partial t}(\frac{dr}{du})^2 \text{ [since }g_{00}\text{ is constant at 1] }\\
&= \frac{d^2t}{du^2} + (\frac{dr}{du})^2 \frac{a(t)a'(t)}{1-kr^2}\ \ \ \ \textbf{[12']}\ \
\end{align*}
As you can see, 12 and 12' do not match, because of the different sign of the last term. I am sure my working must be wrong, as it doesn't appear possible to proceed from 12', whereas it is from 12. But I just can't find my error.
I would be very grateful for anybody that can identify the error for me.
Thank you very much.
The paper considers a light ray emanating from the origin of a FLRW coordinate system in a universe whose hypersurfaces of constant time (in that coordinate system) are homogeneous and isotropic. The light ray is a null geodesic, and it is argued that θ and [itex]\phi[/itex] are constant along its path. The paper claims the following DE determines the geodesic.
[itex]0= \frac{d^2t}{du^2} - (\frac{dr}{du})^2 \frac{a(t)a'(t)}{1-kr^2}[/itex] [12]
In my attempts to reproduce this result, I get the opposite sign on the last term , which prevents further progress. I have checked and re-checked, but just can't see what I'm doing wrong. Here is my working:
In the FLRW coordinate system, all off-diagonal components of the metric tensor are zero, and the two diagonal components that concern us are:
[itex]g_{00} = 1;\ \ g_{11} = -\frac{a(t)^2}{1-kr^2}[/itex]
It has been argued elsewhere that, along the geodesic, [itex]d\theta = d\phi = 0[/itex], and since it is a null geodesic, this gives:
[itex]0 = ds^2 = \frac{a(t)^2 dr^2}{1-kr^2}-{dt^2}[/itex]
The DE for the time component of the geodesic is:
$$0=\frac{d^2x^0}{du^2} + \Gamma^0{}_{kl}\ \frac{dx^k}{du}\ \frac{dx^l}{du}\ \ \ \ \text{[}\textbf{11}\text{]} \\ $$
We calculate the Christoffel symbol's value for i=0 as follows:
\begin{align*}
\Gamma^0{}_{kl} &= \frac{1}{2}g^{0\beta}(g_{\beta k,l}+g_{\beta l,k}-g_{kl,\beta})\ \ \ \ \text{[}\textbf{11a}\text{ - see Schutz 6.32]}\\
&= \frac{1}{2}g^{00}(g_{0k,l} + g_{0l,k} - g_{kl,0})\text{ [since }g_{0\beta} = 0 \text{ unless }\beta = 0]\\
&= \frac{1}{2}g^{00}(g_{00,l}+g_{00,k} - g_{kl,0})\\
&= -\frac{1}{2}g_{kl,0} \text{[since }g_{00} = g^{00} = 1\text{, which is constant]}\\
\end{align*}
Inserting this into the geodesic DE [11] we get:
\begin{align*}
0 &= \frac{d^2t}{du^2} - \frac{1}{2}g_{kl,0}\frac{dx^k}{du}\frac{dx^l}{du} \\
&= \frac{d^2t}{du^2} - \frac{1}{2}g_{00,0}(\frac{dt}{du})^2 - \frac{1}{2}g_{11,0}(\frac{dr}{du})^2\text{ [since }\frac{d\theta}{du}\text{ and }\frac{d\phi}{du}\text{ must be zero]}\\
&= \frac{d^2t}{du^2} - \frac{1}{2}\frac{\partial(\frac{-a(t)^2}{1-kr^2})}{\partial t}(\frac{dr}{du})^2 \text{ [since }g_{00}\text{ is constant at 1] }\\
&= \frac{d^2t}{du^2} + (\frac{dr}{du})^2 \frac{a(t)a'(t)}{1-kr^2}\ \ \ \ \textbf{[12']}\ \
\end{align*}
As you can see, 12 and 12' do not match, because of the different sign of the last term. I am sure my working must be wrong, as it doesn't appear possible to proceed from 12', whereas it is from 12. But I just can't find my error.
I would be very grateful for anybody that can identify the error for me.
Thank you very much.
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