# Homework Help: Geodesic equation in new coordinates question

1. Apr 3, 2010

### PsiPhi

1. The problem statement, all variables and given/known data
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.

2. Relevant 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. $$x(s)$$

3. 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, cancelled 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?

2. Apr 4, 2010

### AEM

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. Apr 6, 2010

### PsiPhi

Cheers, AEM.

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