Connection transformation from geodesic equations

[cazlab]

I don't know if the tex is displaying properly. On my computer all I see is the geodesic equations in every tex field. In the past when this has happened, it has been fine for others viewing it, but if it doesn't make sense, I will upload a pdf or something. Thanks.

Homework Statement

The geodesic equations are
0&=\frac{\textup d^2 x^\mu}{\textup ds^2}+\Gamma^\mu_{\alpha \beta}\frac{\textup d x^\alpha}{\textup ds}\frac{\textup d x^\beta}{\textup ds}
Write down the geodesic equations in the coordinates
\bar x^\mu
with connection components
\bar\Gamma^\mu_{\alpha \beta}
Write down a relation between
dx^\mu/ds
and
d\bar x^\mu/ds
Hence obtain a relation between
\Gamma^\mu_{\alpha \beta}
and
\bar\Gamma^\mu_{\alpha \beta}

Homework Equations

Given above

The Attempt at a Solution

Well, I'm assuming that no matter what frame you look at it from, the lines are straight, so the geodesic equations in the new coordinates are
0&=\frac{\textup d^2\bar x^\mu}{\textup ds^2}+\bar\Gamma^\mu_{\alpha \beta}\frac{\textup d\bar x^\alpha}{\textup ds}\frac{\textup d\bar x^\beta}{\textup ds}
We also have
d\bar x^\mu/ds=d\bar x^\mu/dx^\nu dx^\nu/ds[/tex]
I don't see much use in the latter except to remove the
dx^\alpha/ds
and
dx^\beta/ds
I won't write down the different methods I have tried, as I have about a hundred pages where I've tried various ways. I am happy to type out one of the ways if people just want to see that I've done something, but in the mean time, I think I can explain my problems without a lot of maths.

First, I equate the RHS of both equations because they are both zero. In doing so, I lose a small amount of information (i.e. saying they are equal can account for situations where they are non-zero), so I don't know if this is the way to go. Anyhow, there are four terms in this. No matter what I do, I always end up with three terms on the right hand side, with one of the Gammas on the left. I know that the answer only has two terms on the RHS. Second, the answer has different symbols for the two Gammas. That is fine for the alpha and beta because they are dummy indices, but the mu remains upstairs in both Gammas. I don't know how to change it, because Gamma is not a tensor so I can't just transform it like a tensor. If I understand how to resolve those two problems, I should be able to solve it, as it seems like it should just be a rearrangement of the equation that equates both geodesic equations.

Hopefully someone can point me in the right direction, and I'm happy to provide more info or show more workings if need be. Thanks in advance

 

[cazlab]

Okay, I think I've made some progress, but not sure. I've done it backwards, but not sure if all of the operations I've applied are 'legal'. Here's what I have

\Gamma_{\mu^\prime\lambda^\prime}^{\nu^\prime}=\frac{\partial x^\mu}{\partial x^{\mu^\prime}}\frac{\partial x^\lambda}{\partial x^{\lambda^\prime}}\frac{\partial x^\nu^\prime}{\partial x^{\nu}}\Gamma_{\mu\lambda}^{\nu}-\frac{\partial x^\mu}{\partial x^{\mu^\prime}}\frac{\partial x^\lambda}{\partial x^{\lambda^\prime}}\frac{\partial^2 x^{\nu^\prime}}{\partial x^\mu\partial x^\lambda}
<br /> \Rightarrow\frac{\partial x^\mu^\prime}{\partial x^{\mu}}\frac{\partial x^\lambda^\prime}{\partial x^{\lambda}}\Gamma_{\mu^\prime\lambda^\prime}^{\nu^\prime}=\frac{\partial x^\nu^\prime}{\partial x^{\nu}}\Gamma_{\mu\lambda}^{\nu}-\frac{\partial^2 x^{\nu^\prime}}{\partial x^\mu\partial x^\lambda}<br />
<br /> \Rightarrow\frac{\partial x^\mu^\prime}{ds}\frac{\partial x^\lambda^\prime}{ds}\Gamma_{\mu^\prime\lambda^\prime}^{\nu^\prime}+\frac{d^2x^{\nu^\prime}}{ds^2}=\frac{\partial x^\nu^\prime}{\partial x^{\nu}}\Gamma_{\mu\lambda}^{\nu}\frac{\partial x^\mu}{ds}\frac{\partial x^\lambda}{ds}+\frac{\partial x^{\nu^\prime}}{\partial x^\lambda}\frac{d^2x^{\lambda}}{ds^2}<br />
<br /> \Rightarrow\frac{\partial x^\mu^\prime}{ds}\frac{\partial x^\lambda^\prime}{ds}\Gamma_{\mu^\prime\lambda^\prime}^{\nu^\prime}+\frac{d^2x^{\nu^\prime}}{ds^2}=\frac{\partial x^\nu^\prime}{\partial x^{\nu}}\left(\frac{\partial x^\mu}{ds}\frac{\partial x^\lambda}{ds}\Gamma_{\mu\lambda}^{\nu}+\frac{d^2x^{\nu}}{ds^2}\right)<br />
<br /> \Rightarrow\frac{\partial x^\mu^\prime}{ds}\frac{\partial x^\lambda^\prime}{ds}\Gamma_{\mu^\prime\lambda^\prime}^{\nu^\prime}+\frac{d^2x^{\nu^\prime}}{ds^2}=\frac{\partial x^\mu}{ds}\frac{\partial x^\lambda}{ds}\Gamma_{\mu\lambda}^{\nu^\prime}+\frac{d^2x^{\nu^\prime}}{ds^2}<br />
That's a fair bit to type out without making a mistake, and it isn't showing up for me, so there's a good chance that there will be a typo. If it's not clear, let me know and I'll fix it up.

First thing, I realized that - even though the two frames are different - the free indices must match on both sides, so both have to have primed or unprimed nu. This is reflected in my result. The result looks good, but I'm concerned because it seems that I treated Gamma as a tensor in the last line, even though it is not. Does this look right to people? If not, how come I seem to get the correct result? Thanks

EDIT: I just realized that, with the free indices being the same on both sides, the second derivatives cancel on both sides, so it doesn't make sense. I'm really stumped.

 

[cazlab]

Okay, I almost have the solution now. However, some of the indices in the first term should be primed, and none of them are. I've uploaded a pdf of my solution. I'd appreciate it if anyone feels like pointing out where I've gone wrong :)

 

[Tangent87]

So from what I can tell you're happy with writing the geodesic equations in different coordinates except you can't get the relation between the connections in the two coordinate systems? I read your pdf and you seem to be making a right mess of this. Here's the simplest way to do it:

\nabla_a V^b=\partial_a V^b +\Gamma_{ac}^b V^c

First express this in terms of the components in the transformed frame (Call this equation 1). Then (since it's a tensor) express it as:

\nabla_a V^b=\frac{\partial x^{&#039;c}}{\partial x^{a}}\frac{\partial x^{b}}{\partial x^{&#039;d}}\nabla_c^&#039; V^{&#039;d}

Expand that out and call it equation 2. Then equate equation 1 and equation 2 and rearrange and you should get the correct relation between the connections. Let me know if you have any problems

 

[cazlab]

It's alright. I was working on this with a friend today. Interestingly, we were both having a similar problem that was due to the same error. The mistake is in the final step. I erroneously used the chain rule on the double-derivative, which would have a more complicated form if it was done. What I should have done, is note that two of the derivatives in that term are just kronecker deltas and therefore just change the indices in the double derivative. This gives the right solution. Thanks for your advice anyway :)