Geodesic equation in new coordinates question

AI Thread Summary
The discussion centers on transforming the geodesic equation into new coordinates, denoted as \(\bar{x}^{\mu}\), using the connection components \(\bar{\Gamma}^{\mu}_{\alpha\beta}\). The transformation of tangent vectors and the metric connection is outlined, leading to a derived equation for the geodesic in the new coordinates. Concerns are raised about the validity of canceling partial derivatives and potential issues with index notation under the Einstein summation convention. A reference to Weinberg's "Gravitation and Cosmology" is suggested as a resource to clarify the transformation properties of the geodesic equation. The conversation emphasizes the importance of ensuring mathematical rigor in coordinate transformations.
PsiPhi
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Homework Statement


Suppose \bar{x}^{\mu} is another set of coordinates with connection components \bar{\Gamma}^{\mu}_{\alpha\beta}. Write down the geodesic equation in new coordinates.


Homework Equations


Using the geodesic equation: 0 = \frac{d^{2}x^{\mu}}{ds^{2}} + \Gamma^{\mu}_{\alpha\beta}\frac{dx^{\alpha}}{ds}\frac{dx^{\beta}}{ds}
where s represents the parameterised curve, i.e. x(s)

The Attempt at a Solution


Now the question asks to move to a new coordinate system (i.e. bar over the terms). So, I began by identifying which terms will be effected by the transformation. Namely, the tangent vectors will transform according to
\frac{dx^{\mu}}{ds} = v^{\mu} = \frac{\partial x^{\mu}}{\partial \bar{x}^{\nu}}\frac{d \bar{x}^{\nu}}{ds} ; \frac{dx^{\alpha}}{ds} = v^{\alpha} = \frac{\partial x^{\alpha}}{\partial \bar{x}^{\theta}}\frac{d \bar{x}^{\theta}}{ds} ; \frac{dx^{\beta}}{ds} = v^{\beta} = \frac{\partial x^{\beta}}{\partial \bar{x}^{\phi}}\frac{d \bar{x}^{\phi}}{ds}
The metric connection will transform as
\Gamma^{\mu}_{\alpha\beta} = \frac{\partial x^{\mu}}{\partial \bar{x}^{P}}\frac{\partial \bar{x}^{\theta}}{\partial x^{\alpha}}\frac{\partial \bar{x}^\phi}{\partial x^{\beta}} \bar{\Gamma}^{P}_{\theta\phi} - \frac{\partial ^{2}x^{\mu}}{\partial \bar{x}^{\theta} \partial \bar{x}^{\phi}} \frac{\partial \bar{x}^{\theta}}{\partial x^{\alpha}} \frac{\partial \bar{x}^{\phi}}{\partial x^{\beta}}
I substituted these terms into the geodesic equation above, canceled out partial derivatives on numerators and denominators to obtain:
0 = \frac{d}{ds}\left(\frac{\partial x^{\mu}}{\partial \bar{x}^{\nu}} \frac{d \bar{x}^{\nu}}{ds}\right) + \frac{\partial x^{\mu}}{\partial \bar{x}^{P}}\frac{d \bar{x}^{\theta}}{ds}\frac{d \bar{x}^{\phi}}{ds} \bar{\Gamma}^{P}_{\theta\phi} - \frac{\partial ^{2}x^{\mu}}{\partial \bar{x}^{\theta} \partial \bar{x}^{\phi}} \frac{d \bar{x}^{\theta}}{ds} \frac{d \bar{x}^{\phi}}{ds}

I was just wondering if my final equation for the geodesic equation in new coordinates \bar{x}^{\mu} was correct?

I suspect a flaw in the mathematical logic when I cancel out the partial derivative in the numerators and denominators. Was this procedure allowed?
Also, I think there may be a problem with my indices with respect to Einstein summation convention.
Thoughts?
 
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PsiPhi said:

Homework Statement


Suppose \bar{x}^{\mu} is another set of coordinates with connection components \bar{\Gamma}^{\mu}_{\alpha\beta}. Write down the geodesic equation in new coordinates.


Homework Equations


Using the geodesic equation: 0 = \frac{d^{2}x^{\mu}}{ds^{2}} + \Gamma^{\mu}_{\alpha\beta}\frac{dx^{\alpha}}{ds}\frac{dx^{\beta}}{ds}
where s represents the parameterised curve, i.e. x(s)

The Attempt at a Solution


Now the question asks to move to a new coordinate system (i.e. bar over the terms). So, I began by identifying which terms will be effected by the transformation. Namely, the tangent vectors will transform according to
\frac{dx^{\mu}}{ds} = v^{\mu} = \frac{\partial x^{\mu}}{\partial \bar{x}^{\nu}}\frac{d \bar{x}^{\nu}}{ds} ; \frac{dx^{\alpha}}{ds} = v^{\alpha} = \frac{\partial x^{\alpha}}{\partial \bar{x}^{\theta}}\frac{d \bar{x}^{\theta}}{ds} ; \frac{dx^{\beta}}{ds} = v^{\beta} = \frac{\partial x^{\beta}}{\partial \bar{x}^{\phi}}\frac{d \bar{x}^{\phi}}{ds}
The metric connection will transform as
\Gamma^{\mu}_{\alpha\beta} = \frac{\partial x^{\mu}}{\partial \bar{x}^{P}}\frac{\partial \bar{x}^{\theta}}{\partial x^{\alpha}}\frac{\partial \bar{x}^\phi}{\partial x^{\beta}} \bar{\Gamma}^{P}_{\theta\phi} - \frac{\partial ^{2}x^{\mu}}{\partial \bar{x}^{\theta} \partial \bar{x}^{\phi}} \frac{\partial \bar{x}^{\theta}}{\partial x^{\alpha}} \frac{\partial \bar{x}^{\phi}}{\partial x^{\beta}}
I substituted these terms into the geodesic equation above, canceled out partial derivatives on numerators and denominators to obtain:
0 = \frac{d}{ds}\left(\frac{\partial x^{\mu}}{\partial \bar{x}^{\nu}} \frac{d \bar{x}^{\nu}}{ds}\right) + \frac{\partial x^{\mu}}{\partial \bar{x}^{P}}\frac{d \bar{x}^{\theta}}{ds}\frac{d \bar{x}^{\phi}}{ds} \bar{\Gamma}^{P}_{\theta\phi} - \frac{\partial ^{2}x^{\mu}}{\partial \bar{x}^{\theta} \partial \bar{x}^{\phi}} \frac{d \bar{x}^{\theta}}{ds} \frac{d \bar{x}^{\phi}}{ds}

I was just wondering if my final equation for the geodesic equation in new coordinates \bar{x}^{\mu} was correct?

I suspect a flaw in the mathematical logic when I cancel out the partial derivative in the numerators and denominators. Was this procedure allowed?
Also, I think there may be a problem with my indices with respect to Einstein summation convention.
Thoughts?

Sorry it took a while to get back to you. If you have it available, look at page 102 of Weinberg's "Gravitation and Cosmology. There he shows that the geodesic equation transforms like a vector.
 
Cheers, AEM.

I do have Weinberg's text, I will have a look.
 
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