# Need help reproducing geodesic DE from a paper

1. Jul 28, 2012

### andrewkirk

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 $\phi$ are constant along its path. The paper claims the following DE determines the geodesic.
$0= \frac{d^2t}{du^2} - (\frac{dr}{du})^2 \frac{a(t)a'(t)}{1-kr^2}$ [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:
$g_{00} = 1;\ \ g_{11} = -\frac{a(t)^2}{1-kr^2}$
It has been argued elsewhere that, along the geodesic, $d\theta = d\phi = 0$, and since it is a null geodesic, this gives:
$0 = ds^2 = \frac{a(t)^2 dr^2}{1-kr^2}-{dt^2}$

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.

Last edited: Jul 29, 2012
2. Jul 29, 2012

### fzero

I agree with your [12']. It seems consistent with the decay $E\sim 1/a$ during expansion, where $E = dt/du, p=dr/du$ are the energy and momentum satisfying $E^2-p^2=0$.

3. Jul 29, 2012

### George Jones

Staff Emeritus
I, too, don't think that you have made a sign mistake.
(12') seems to lead to (13'),

$$a \frac{dt}{du} = A,$$
Using this, I have recalculated the steps, and I still arrive at (22).

4. Jul 29, 2012

### qbert

i also agree with 12'

different derivation:
$$ds^2 = \left(\dot{t}^2 - \frac{a(t)^2}{1-kr^2}\dot{r}^2\right) du^2$$
dot = d/du.
$$S = \int ds = \int \sqrt{ \dot{t}^2 - \frac{a(t)^2}{1-kr^2}\dot{r}^2} du = \int L(t, \dot{t}, r, \dot{r}) du$$
Equations of motion are E-L equations,
$$\frac{d}{du}\left(\frac{\partial L}{\partial \dot{t}}\right) = \frac{\partial L}{\partial t}$$
The standard trick of varying L2 instead of L yields
$$\frac{d}{du}\left(2 \dot{t} \right) = -\frac{2 a(t) \frac{da}{dt}}{1-kr^2} \dot{r}^2.$$

5. Jul 30, 2012

### andrewkirk

So it does!
George, you are a marvel!
It seems obvious now but for some reason I couldn't see that solution to save my life.

I shall now try to follow your example and reconnect from 13' up to 22. Hopefully, I won't have to come back pleading for help again.

I must say, I am starting to wonder about the standard of peer review at the American Journal of Physics, as that's now two erroneous formulae ([12] and [13]) that appear to have eluded the notice of the reviewers. It was published in 1994. I wonder if things have changed.

6. Jul 30, 2012

### stevendaryl

Staff Emeritus
What is this standard trick? I have for years solved equations of motion involving curved spacetime using a second-order Lagrangian, but I never really understood why it worked. It seemed like a miraculous coincidence.

$\tau =\int \sqrt{g_{\mu \nu} dx^{\mu} dx^{\nu}}$

and maximizing the proper time, I use an "effective Lagrangian"

$L = \dfrac{1}{2} g_{\mu \nu} \dfrac{dx^{\mu}}{d \tau} \dfrac{dx^{\nu}}{d \tau}$

and use the regular Lagrangian equations of motion.

This makes solving for trajectories much more like classical mechanics, but I'm not sure why it works. It appears to work even when you introduce electromagnetic forces, as well; instead of the above expression for $L$, add an interaction term:
$q \ g_{\mu \nu} \dfrac{dx^{\mu}}{d \tau}A^{\nu}$ where $A$ is the electromagnetic vector potential, and q is the charge. (I might have a sign error.)

It appears to work, but I'm not sure why.

Last edited: Jul 30, 2012
7. Jul 30, 2012

### George Jones

Staff Emeritus
Your $L$ is given by

$$L = \frac{1}{2} \tilde{L}^2 ,$$

where $\tilde{L}$ is the standard Lagrangian. Then,
$$\frac{\partial L}{\partial q} - \frac{d}{du} \frac{\partial L}{\partial \dot{q}} = \tilde{L}\frac{\partial \tilde{L}}{\partial q} - \frac{d}{du} \left( \tilde{L} \frac{\partial \tilde{L}}{\partial \dot{q}} \right).$$

Along an affinely parametrized geodesic, both $L$ and $\tilde{L}$ are constant, and thus $\tilde{L}$ can be pulled outside the total derivative (but not outside partials) on the right. Consequently,

$$\frac{\partial \tilde{L}}{\partial q} = \frac{d}{du} \frac{\partial \tilde{L}}{\partial \dot{q}}$$

implies that

$$\frac{\partial L}{\partial q} = \frac{d}{du} \frac{\partial L}{\partial \dot{q}}.$$

If the geodesic is spacelike or timelike, then the converse is also true.

Last edited: Jul 30, 2012
8. Jul 31, 2012

### PAllen

Maybe I'm having a late night brain hiccup, but why can't you just stop here, with a first order separable DE?

$$0= \left(\dot{t}^2 - \frac{a(t)^2}{1-kr^2}\dot{r}^2\right)$$

9. Jul 31, 2012

### andrewkirk

Maybe you can. I wouldn't know: I skipped the maths unit at uni on solving DEs (I did something silly like Group Theory instead), and am now being punished for it by having to try to solve them all from first principles.

However the DE that needs to be solved (and which George solved above) is a bit different because it has a + where the above has a minus.

10. Jul 31, 2012

### andrewkirk

Please don't throw away your working for this yet. I am still trying to establish the path from 13' to 22. Currently I have managed to prove that $1+z=\sqrt{\frac{1-V}{1+V}}$
which is upside-down! I am sure I will find a silly error somewhere in my reasoning that, once corrected, will allow me to turn the formula the right way up. So don't post anything yet, as I will be as pleased as Punch if I can work it out by myself.
But I may need to come back with my tail between my legs asking for help.

11. Aug 4, 2012

### andrewkirk

Turns out there was nothing wrong. My V was negative because the first spatial basis vector in my Lorentz frame at O was pointing away from S rather than towards it. So the negative incoming velocity becomes a positive outgoing velocity, which flips the ratio over and gives the right result.
:rofl:
Thank you everybody for your patience and contributions!

12. Aug 11, 2012

### andrewkirk

Here's my completed rehabilitation of Narlikar's proof that the redshift from cosmological expansion is equivalent to that of a local Doppler redshift. It's a pdf version of a TeX-formatted file.

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62.7 KB
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