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Schwarzschild geometry unstable orbit GR question

  • Thread starter Allday
  • Start date
  • #1
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1
Hey folks,
working problems in Hartle's GR book and having trouble with this one. Chapter 9 discusses the simplest physically relavent curved geometry, that of Mr. Swarzschild
[tex] ds^s = -(1 - \frac{2 G M}{r}) dt^2
+ (1 - \frac{2 M}{r})^{-1} dr^2
+r^2(d\theta^2 + sin^2\theta d\phi^2) [/tex]
In this geometry the derivative of the effective potential goes to zero at two places. one is a stable equilibrium and the other is unstable.
[tex]
V_{eff} = -\frac{M}{r} + \frac{l^2}{2 r^2} - \frac{M L^2}{r^3}
[/tex]

the unstable r value for circular orbits occurs at
[tex]
r_{max} = \frac{l^2}{2 M}( 1-\sqrt{1-12 (\frac{M}{l}) ^2})
[/tex]
the goal is to show that when this orbit is perturbed in the radial direction that the perturbation grows exponentialy
[tex]
\delta_{r} \propto e^{\tau / \tau_*}
[/tex]
where [tex] \tau_* [/tex] is some constant
we also have a constant related to the conserved energy per unit rest mass to take advantage of
[tex]
\epsilon=\frac{1}{2}(\frac{dr}{d\tau})^2+V_{eff}(r)
[/tex]
i know that the derivative above will determine how this pertubation will grow and that
[tex] \epsilon - V_{eff} = 0 [/tex]
for circular orbits at the unstable r. I've tried expanding the potential around this unstable equilibrium point which yields derivatives of V multiplied by the deltar under a square root for the derivative but I cant see how to connect this to the exponential form of the proper time.
 
Last edited:

Answers and Replies

  • #2
Physics Monkey
Science Advisor
Homework Helper
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If you expand the effective potential around the unstable equilibrium radius [tex] r* [/tex] you find the following form for the effective energy equation:
[tex]
\epsilon = \frac{1}{2}\left(\frac{dr}{d\tau} \right) + V_{eff}(r*) + \frac{1}{2} \frac{d^2 V_{eff}(r*)}{dr^2}(r-r*)^2.
[/tex]
Let me ask you, what is the key observation that tells us the orbit is unstable?

To find an expression for [tex] \tau_* [/tex] and obtain the exponential dependence, try differentiating the effective energy conservation equation with respect to [tex] \tau [/tex] to obtain the equation of motion. What is the solution of the resulting differential equation?
 
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  • #3
164
1
Hi Physics Monkey,
the fact that we are looking at an unstable equuilibrium means the first derivative of the potential is 0 and that the second derivative is negative. I've done the expansion and now differentiated wrt to tau. Assuming that the energy remains constant during the perturbation the left hand side will go to zero leaving,
[tex]
\frac{d^2\delta_r}{d\tau^2} + \frac{d^2V}{dr^2}\frac{d\delta_r}{d\tau}=0
[/tex]
this is a differential equation for r (or the perturbation delta r) with an exponential solution. solving for the constant [tex] \tau^* [/tex] I get that it is equal to
[tex]
\frac{1}{\tau^*}=-\frac{d^2V}{dr^2}
[/tex]
which is positive because the derivative itself is negative. this is reasonable and leads to plunging in for negative pertubations and escaping to infinity for positive pertubations. thanks for the help.
 
  • #4
Physics Monkey
Science Advisor
Homework Helper
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Hi Allday,

You're on the right track but you still have a few mistakes. When differentiating the energy equation and finding the equation of motion, you should obtain
[tex]
\frac{d^2 \delta}{d\tau^2}\frac{d \delta}{d \tau} + \frac{d^2 V}{d r^2} \delta \frac{d \delta}{d \tau} = 0,
[/tex]
so there is a small mistake in your equation. Cancelling the velocity from both terms, you find the equation of motion
[tex]
\frac{d^2 \delta}{d\tau^2} = -\frac{d^2 V}{d r^2}\delta},
[/tex]
which you correctly indicate has exponential solutions. However, you have the time constant wrong. Think about the dimensions of [tex] 1/\tau_* [/tex] and [tex] d^2 V/dr^2 [/tex] (remember that we are using geometric units here and V is dimensionless). You're very close however.
 
  • #5
164
1
[tex]
\epsilon = \frac{1}{2}\left(\frac{dr}{d\tau} \right) + V_{eff}(r*) + \frac{1}{2} \frac{d^2 V_{eff}(r*)}{dr^2}\delta^2_r
[/tex].
[tex] \frac{d\epsilon}{d\tau}=0 [/tex]
Because [tex] \delta_r [/tex] = [tex]r - r^* [/tex]
that means [tex] r = \delta_r + r^* [/tex]
and that means the only difference between [tex] \delta_r [/tex] and r is the constant [tex] r^* [/tex] and either will do in the derivative.
[tex] \frac{d}{d\tau}\frac{d\delta_r}{d\tau} = \frac{d^2\delta_r}{d\tau^2} [/tex]
V and its derivatives evaluated at [tex] r^* [/tex] are constants and so dont contribute to the derivatives.
using the product rule,
[tex] \frac{d\delta^2_r(\tau)}{d\tau} = 2 \delta_r \frac{d\delta_r}{d\tau} [/tex]
therefore it looks like
[tex] \frac{d^2\delta_r}{d\tau^2} = -2 \delta_r \frac{d^2V(r^*)}{dr^2} \frac{d\delta_r}{d\tau} [/tex]
where am i going wrong?
 
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  • #6
Physics Monkey
Science Advisor
Homework Helper
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You forgot the 1/2 in front of the second derivative of the potential. Also, the derivative of [tex] \frac{1}{2}\left( \frac{d \delta }{d \tau} \right)^2 [/tex] with respect to tau is [tex] \frac{d^2 \delta}{d\tau^2}\frac{d \delta}{d\tau} [/tex] (use the chain rule).
 
  • #7
164
1
Ahhhhh got it

hey physics monkey, finally got around to finishing up. I realized what you were saying about the units (havent quite gotten used to checking dimensionality in geometrized units but now that I've looked at it, it seems much simpler...almost everything has the same units!) looks as if the time constant should be inversly proportional to the square root of the second derivative of the potential giving units of 1/time on the left and 1/length on the right which are measured in the same units. pretty sure this is right, thanks for your help

Gabe
 

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