# 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.