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Geodesic equation in new coordinates question

  • Thread starter PsiPhi
  • Start date
  • #1
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Homework Statement


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


Homework Equations


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

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
[tex]\frac{dx^{\mu}}{ds} = v^{\mu} = \frac{\partial x^{\mu}}{\partial \bar{x}^{\nu}}\frac{d \bar{x}^{\nu}}{ds}[/tex] ; [tex]\frac{dx^{\alpha}}{ds} = v^{\alpha} = \frac{\partial x^{\alpha}}{\partial \bar{x}^{\theta}}\frac{d \bar{x}^{\theta}}{ds}[/tex] ; [tex]\frac{dx^{\beta}}{ds} = v^{\beta} = \frac{\partial x^{\beta}}{\partial \bar{x}^{\phi}}\frac{d \bar{x}^{\phi}}{ds}[/tex]
The metric connection will transform as
[tex]\Gamma^{\mu}_{\alpha\beta} = \frac{\partial x^{\mu}}{\partial \bar{x}^{P}}\frac{\partial \bar{x}^{\theta}}{\partial x^{\alpha}}[/tex][tex]\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}}[/tex]
I substituted these terms into the geodesic equation above, cancelled out partial derivatives on numerators and denominators to obtain:
[tex]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}[/tex]

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

Answers and Replies

  • #2
AEM
360
0

Homework Statement


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


Homework Equations


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

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
[tex]\frac{dx^{\mu}}{ds} = v^{\mu} = \frac{\partial x^{\mu}}{\partial \bar{x}^{\nu}}\frac{d \bar{x}^{\nu}}{ds}[/tex] ; [tex]\frac{dx^{\alpha}}{ds} = v^{\alpha} = \frac{\partial x^{\alpha}}{\partial \bar{x}^{\theta}}\frac{d \bar{x}^{\theta}}{ds}[/tex] ; [tex]\frac{dx^{\beta}}{ds} = v^{\beta} = \frac{\partial x^{\beta}}{\partial \bar{x}^{\phi}}\frac{d \bar{x}^{\phi}}{ds}[/tex]
The metric connection will transform as
[tex]\Gamma^{\mu}_{\alpha\beta} = \frac{\partial x^{\mu}}{\partial \bar{x}^{P}}\frac{\partial \bar{x}^{\theta}}{\partial x^{\alpha}}[/tex][tex]\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}}[/tex]
I substituted these terms into the geodesic equation above, cancelled out partial derivatives on numerators and denominators to obtain:
[tex]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}[/tex]

I was just wondering if my final equation for the geodesic equation in new coordinates [tex]\bar{x}^{\mu}[/tex] 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.
 
  • #3
20
0
Cheers, AEM.

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

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