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I was watching a video about the swarzchild radius and it said that potential energy is $gmm/r$. This cannot be right though because potential energy goes up with distance not down. I'm assuming he meant motion due to potential energy. Also what is the intuition behind the equation $gmm/r$?

BMS
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Ray Kay
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2 Answers2

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The equation you cite is the Newtonian potential energy, and as BMS suggests I'd guess whoever is presenting the video is just being careless with the sign.

Potential energy is not a well defined concept in general relativity. If you do a naive integration of $Fdr$ along a radial line to the event horizon it goes to (minus) infinity at the event horizon. This doesn't mean the potential energy is actually infinite, just that you've encountered a coordinate singularity. The point is that unlike the Newtonian case, in GR you have to me mighty careful that you understand what it is you're calculating.

Re the intuition behind the equation, the potential energy is (minus) the work required to remove the mass from a distance $r$ to infinity. The work to move a distance $dr$ is $Fdr$, so to get the total work we integrate the force from a distance $r$ to infinity:

$$\begin{align} PE &= -\text{total work} \\ &= -\int_r^\infty \frac{GMm}{r^2}dr \\ &= -\frac{GMm}{r} \end{align}$$

John Rennie
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  • What's the $F$ you're integrating? If $F$ is the static force, you don't get $-\infty$ at the horizon. Although you would for $F,\mathrm{d}s$, where $\mathrm{d}s = \mathrm{d}r/\sqrt{1-2M/r}$, or something like that. Maybe some other $F$, too. – Stan Liou Aug 21 '14 at 07:14
  • @StanLiou: the $F$ I'm using is the norm of the four acceleration multiplied by the mass, so it's the Newtonian force multiplied by $1/\sqrt{1-r_s/r}$ and this diverges at $r = r_s$. You're now going to say this isn't particularly meaningful, and I agree - that's my point. – John Rennie Aug 21 '14 at 07:17
  • I'm not going to say that. My point is that integral does not diverge. It's just $\sqrt{1-r_s/r}\big\vert_{r_s}^R$. Hence the suggestion of $F,\mathrm{d}s$ instead, because that would be with respect to a 'radial distance'. – Stan Liou Aug 21 '14 at 07:28
  • Ah, OK, yes ISWYM – John Rennie Aug 21 '14 at 07:29
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In the Schwarzschild spacetime, the Schwarzschild time is a Killing vector field, $\xi = \partial_t$ which generates a conserved quantity for an orbit of a particle with four-velocity $u$: $$e = -g(\xi,u) = \left(1-\frac{2M}{r}\right)\frac{\mathrm{d}t}{\mathrm{d}\tau}\text{.}$$ Additionally, the spacetime is rotationally symmetric, which gives a conserved specific orbital momentum from the $\partial_\phi$ Killing vector field: $$l = r^2\sin^2\theta \frac{\mathrm{d}\phi}{\mathrm{d}\tau}\text{.}$$ Plugging these into the condition that the four-velocity is timelike, $g(u,u) = -1$, one can derive the effective potential $$E_\text{orbital} \equiv m\frac{e^2-1}{2} = \frac{1}{2}m\left(\frac{\mathrm{d}r}{\mathrm{d}\tau}\right)^2 + V_\text{eff}\text{,}$$ $$V_\text{eff} = -\frac{GMm}{r} + \frac{l^2m}{2r^2} - \frac{GMml^2}{c^2r^3}\text{,}$$ which has the same form as the Newtonian effective potential except for the last term.

For radial freefall, $l = 0$, and the effective potential consists solely of the $-GMm/r$ term. Note the different interpretation of these quantities from the Newtonian case, however: $r$ is the Schwarzschild radial coordinate, not the distance from the center, and $\tau$ is the proper time of the orbiting particle, not the absolute Newtonian time, nor the Schwarzschild time.

Stan Liou
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