Physics, asked by kalpeshraikar6979, 1 year ago

Nice expressions for geodesics in Penrose diagram for Schwarzschild spacetime?

Answers

Answered by Anonymous
0

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For Minkowski space, it's trivial to find parametric descriptions of the geodesics, which are, e.g., in the case of a timelike geodesic, of the form U=f(α+(1+β)t)U=f(α+(1+β)t), V=f(α+(1−β)t)V=f(α+(1−β)t).

Now consider the case of the Schwarzschild spacetime. Suppose we start with Kruskal-Szekeres coordinates (T,X)(T,X), so that U=f(T+X)U=f(T+X) and V=f(T−X)V=f(T−X), and

ds2=32re−r/2g(U)g(V)dUdV+r2dΩ2,ds2=32re−r/2g(U)g(V)dUdV+r2dΩ2,

where g=(f−1)′g=(f−1)′.

Answered by Ashi03
0
For Minkowski space, it's trivial to find parametric descriptions of the geodesics, which are, e.g., in the case of a timelike geodesic, of the form U=f(α+(1+β)t)U=f(α+(1+β)t), V=f(α+(1−β)t)V=f(α+(1−β)t).

Now consider the case of the Schwarzschild spacetime. Suppose we start with Kruskal-Szekeres coordinates (T,X)(T,X), so that U=f(T+X)U=f(T+X) and V=f(T−X)V=f(T−X), and

ds2=32re−r/2g(U)g(V)dUdV+r2dΩ2,ds2=32re−r/2g(U)g(V)dUdV+r2dΩ2,

where g=(f−1)′g=(f−1)′.

hope helps you

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