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My question is regarding an arbitrary time-dependent spherically symmetric spacetime with line-element, in co-moving coordinates, to be

$$ds^2 = -f(R) dt^2 + a(t)\bigg\lbrace\frac{dR^2}{f(R)} +R^2d\Omega^2 \bigg\rbrace.$$

Obviously, I'm working with something "Schwarzschild-like" and the $f(R)$ and $a(t)$ can be any user-defined function. I need help verifying if the procedure I do below is correct.

First, the geodesic equation for the $R$-coordinate can be written as

$\frac{d^2R}{d\lambda^2}+\Gamma^{R}_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}=0$

where $\lambda$ is an affine parameter. If I multiply both sides of the equation by $\frac{dR}{d\lambda}$ I then get

$\frac{d^2R}{d\lambda^2}\frac{dR}{d\lambda}+\frac{dR}{d\lambda}\Gamma^{R}_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}=0$

which I could then take to mean

$\frac{1}{2}\left(\frac{dR}{d\lambda}\right)^2 + \displaystyle \int{d\lambda\frac{dR}{d\lambda}\Gamma^{R}_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}} = \textrm{constant}$.

From this, can I read off the term in the integral sign to be the effective potential? Furthermore, since the terms inside the integral have no explicit dependence on the affine parameter, can I further reduce it to

$U_{\textrm{effective}} =\displaystyle \int {\Gamma^{R}_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}dR }$

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It seems to me that this won't work, because if you want to interpret your expression as an effective potential, it should only depend on the coordinates, but your

$$\int {\Gamma^{R}_{\mu\nu}\frac{dx^{\mu}}{d\lambda}\frac{dx^{\nu}}{d\lambda}dR }$$

depends on what geodesic is being discussed.

If you want to see if this can be made to work somehow, one thing to try would be to check whether it gives reasonable results in the special case of the Schwarzschild spacetime, where there actually is a well-defined potential.

For generalizations of the potential, you may want to look at the Hansen potentials.

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  • $\begingroup$ sorry there's something I don't get. why should the effective potential only depend on the coordinates? Also, are the Hansen potentials applicable for non-stationary spacetimes? $\endgroup$ – jboy Apr 6 '19 at 9:40

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