Flat Space - Christoffel symbols and Ricci = 0?

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unscientific
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Homework Statement


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(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)##.

ricci1.png

Homework Equations

The Attempt at a Solution



Part(a)
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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:
[tex]\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)[/tex]

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}##.
 
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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.
 
TSny said:
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.

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:
Looks right.
 
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TSny said:
Looks right.

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