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I (desperately) need help with the following: What is the null geodesic for the space time $$ds^2=-x^2 dt^2 +dx^2?$$ I don't know how to transform a metric into a geodesic...! There is no need to start from the Lagrangian. I know that $$0=g_{ij}V^iV^j$$ where $V^i={dx^i\over d\lambda}$ where $\lambda$ is some parameter. But I don't know what that parameter is nor how to find the geodesic. Many thanks!! Please help! |
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As you mentioned a null geodesic implies: $$g_{\mu \nu} U^{\mu} U^{\nu}=\left (\frac{ds}{d \lambda} \right )^2=0$$ where $\lambda$ is some affine parameter. If you take $\lambda=t$, then this implies: $$-x^2+\left ( \frac{dx}{dt} \right )^2=0$$ So $~dx/dt=\pm x$ is a null geodesic. (Take a second time derivative to get the actual geodesic.) |
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The easiest way to solve this is to pretend that it IS a Lagrangian: $$I = \frac{1}{2}\int d\lambda \left(-x^{2}{\dot t}^{2} + {\dot x}^{2}\right)$$ Where both $t$ and $x$ are functions of $\lambda$, and ${\dot x} \equiv \frac{dx}{d\lambda}$.. Take the variation of the action, find the minimum, and then set your constants so that your geodesic is null. |
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