Flat Space - Christoffel symbols and Ricci = 0?

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1. Feb 14, 2015

unscientific

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

(a) Find christoffel symbols and ricci tensor
(b) Find the transformation to the usual flat space form $g_{\mu v} = diag (-1,1,1,1)$.

2. Relevant equations

3. The attempt at a solution

Part(a)

I have found the metric to be $g_{tt} = g^{tt} = -1, g_{xt} = g_{tx} = 2c, g_{xx} = g^{xx} = 0, g_{yy} = g^{yy} = 1, g_{zz} = g^{zz} = 1$.

The christoffel symbols can be calculated by:
$$\Gamma_{\alpha \beta}^{\mu} = \frac{1}{2} g^{\mu v} \left( \frac{\partial g_{\alpha v}}{\partial x^{\beta}} + \frac{\partial g_{v \beta}}{\partial x^{\alpha}} - \frac{\partial g_{\alpha \beta}}{\partial x^{v}} \right)$$

Since all components of the metric are constants, it means $\Gamma_{\alpha \beta}^{\mu} = 0$ and $R_{\alpha \beta}^{\mu} = 0$.

Part (b)

I'm not sure how to approach this question. I know I have to find the Jacobian $\frac{\partial \tilde{x^{\mu}}}{\partial x^{v}}$. I also know the transformation is $\tilde{g_{\alpha \beta}} = \frac{\partial x^{\mu}}{\partial \tilde{x^{\alpha}}} \frac{\partial x^v}{\partial \tilde{x^{\beta}}} g_{\mu v}$.

2. Feb 15, 2015

unscientific

Any input for part (b)?

3. Feb 15, 2015

TSny

For both the original line element and the usual flat space line element the metric components are constants. This suggests a linear transformation. So, maybe introduce a new time coordinate $\tilde{t} = at + bx$ where you can try to find values of $a$ and $b$ to do the job.

4. Feb 15, 2015

unscientific

Letting $c \tilde t = c t - x$ makes it work because $c^2 d \tilde t^2 = c^2dt^2 - 2ct ~dt dx + dx^2$.

Last edited: Feb 15, 2015
5. Feb 15, 2015

TSny

Looks right.

6. Feb 15, 2015

unscientific

Thanks alot for all your help! I'm just starting out in GR so I'm learning as hard as I can.