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Orbital velocities in the Schwartzschild geometry 
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#1
Jun1010, 09:39 AM

P: 836

I'm trying to use the tensor formulation of GR to calculate the velocity of a particle in a circular orbit around a black hole.
Here is the work I have done so far. What concerns me is that I end up getting zero velocity when applying the metric to the differential equations I get from the geodesic equation. I wonder if I have made a miscalculation, but I am unable to find any, so maybe there is a misunderstanding on my part. Any help is appreciated. 


#2
Jun1010, 10:18 AM

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#3
Jun1010, 10:47 AM

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There seems to be a problem with signs in (13), because both terms are negativedefinite.
Using (12), and choosing t=0 to coincide with [itex]\tau=0[/itex], you can set [itex]t=\beta \tau[/itex], where [itex]\beta[/itex] is a constant. Let's also set [itex]\omega=d\phi/der t[/itex]. Then (13) gives [itex]\omega=\pm i\sqrt{2m/r^3}[/itex], where [itex]m=r_s/2[/itex]. This seems sort of right, since it coincides with Kepler's law of periods. However, it's imaginary due to the sign issue. It would also surprise me if Kepler's law of periods was relativistically exact when expressed in terms of the Schwarzschild coordinates, but maybe that's the case. If you can fix the sign problem, then you seem to have the right result in the nonrelativistic limit of large r. You might then want to check the result in the case of r=3m, where I believe you should obtain lightlike circular orbits. 


#4
Jun1010, 12:10 PM

P: 1,568

Orbital velocities in the Schwartzschild geometry
[tex]L=(1r_s/r)\frac{dt^2}{ds^2}r^2\frac{d\phi^2}{ds^2}[/tex] 


#5
Jun1010, 02:57 PM

P: 836

@George Jones , bcrowell
Thanks for pointing that out, and thank you for the reference. I traced the sign error back to a differentiation error when calculating the Christoffel symbol. As for writing [itex]t=\beta\tau[/itex] I don't see how that will help since t does not appear in the other equations. I must be missing something. I will try to arrive at a result and do the "reality checks" you mentioned. @Starthaus Thanks for your input. I am afraid I don't know how to arrive at or what to do with the Lagrangian. From it's appearance it looks just like the metric, so L=1 here, I imagine? EDIT: I arrived at [tex]\frac{\text{d}\phi}{\text{d}\tau}=\frac{c}{r}\sqrt{\frac{r_s}{rr_s}}[/tex]. The units match, but I doubt this is correct, since letting r=3m gives [tex]\frac{\text{d}\phi}{\text{d}\tau}=\frac{c}{\sqrt{2m}}[/tex] while my intuition tells me it should be c. 


#6
Jun1010, 05:53 PM

P: 1,568

[tex]\frac{d^2t}{ds^2}=0[/tex] and [tex]\frac{d^2\phi}{ds^2}=0[/tex] Hint: in your writeup you made [tex]dr=d\theta=0[/tex], remember? You need to think what that means. 


#7
Jun1010, 06:02 PM

P: 836

[tex]\frac{c^2r_s}{r^2}\left(\frac{\text{d}t}{\text{d}\tau}\right)^2r\left(\frac{\text{d}\phi}{\text{d}\tau}\right)^2=0[/tex] which, when I used the substitution [tex]\left(\frac{\text{d}t}{d\tau}\right)^2=1+\frac{r^2}{c^2}\left(\frac{\te xt{d}\phi}{\text{d}\tau}\right)^2[/tex] which I got from the metric, gave me [tex]\frac{c^2r_s}{r^2}+(r_sr)\left(\frac{\text{d}\phi}{\text{d}\tau}\right)^2=0[/tex] which solves to [tex]\frac{\text{d}\phi}{\text{d}\tau}=\frac{c}{r}\sqrt{\frac{r_s}{rr_s}}[/tex] So now I am unsure about where I made my mistake. 


#8
Jun1010, 06:50 PM

P: 1,568

The third Lagrange equation, exists only if r is variable and its correct form would have been: [tex]\frac{r_s}{r^2}\frac{dt^2}{ds^2}2r\frac{d\phi^2}{ds^2}=0[/tex] But you made [tex]dr=0[/tex] (this is why I gave you the hint), so the third equation does not exist. This is the root of your errors. 


#9
Jun1110, 02:42 PM

P: 836

I don't understand why it shouldn't exist. I derived the general case, then assumed dr=0 and substituted that into the equations. dr doesn't even appear in that equation anymore, do I don't see how it has any influence.
EDIT: Where does the factor of 2 come from there? I somehow didn't appear in my calculations. For the velocity, I obtained [tex]v=\frac{\text{d}\phi}{\text{d}\tau}r=c\sqrt{\frac{r_s}{rr_s}}=\sqrt{\frac{2GM}{r\frac{2GM}{c^2}}}[/tex] My intuition says that this is wrong by a factor of [itex]\sqrt{2}[/tex], since then it would give v=c at [itex]r=\frac{3GM}{c^2}[/itex], but I don't see how that factor dissapeared. 


#10
Jun1110, 08:57 PM

P: 1,568

From the correct equation [tex]\frac{d^2\phi}{ds^2}=0[/tex] you should obtain (no surprise): [tex]\frac{d\phi}{ds}=constant=\omega[/tex] The trajectory is completed by the other obvious equation [tex]r=R=constant[/tex] You get one more interesting equation, that gives u the time dilation. Start with: [tex]ds^2=(1r_s/R)dt^2(Rd\phi)^2[/tex] and you get: [tex]\frac{dt}{ds}=\sqrt{\frac{1+(R\omega)^2}{1r_s/R}}[/tex] or: [tex]\frac{ds}{dt}=\sqrt{1r_s/R}\sqrt{1\frac{(R\omega)^2}{1r_s/R}[/tex] The last equation gives you the hint that: [tex]v=\frac{R\omega}{\sqrt{1r_s/R}}[/tex] The last expression is what you were looking for. 


#11
Jun1210, 04:48 AM

P: 836

The only thing I said is that I have constant r. If you have a function, say f(x)=x^{2}, you can say that x is fixed at three, but you can still get the slope at x=3 by differentiating wrt x.
As for the last equation you posted. I don't doubt its validity, but I don't see how it will get me anywhere, since neither v nor [tex]\omega[/tex] are known. In fact, since [tex]v=\omega R[/tex], v cancels on both sides. 


#12
Jun1210, 08:49 AM

P: 1,568

[tex]ds^2=(1r_s/R)dt^2R^2d\phi^2[/tex] So, your Christoffel symbols need to reflect that. They don't. 


#13
Jun1210, 10:09 AM

P: 836

Of course the trajectory is a circle. I imposed that restriction by setting r=constant after deriving the general case geodesic equations. The equations saying d^{2}t/ds^{2}=0 and d^{2}φ/ds^{2}=0 are neccesary consequences. What I am seeking is an expression which gives the orbital velocity as a function of r. I define [tex]v=\frac{\text{d}\phi}{\text{d}\tau}r[/tex], and except for the factor [tex]\sqrt{2}[/tex] my result reduces to the Newtonian formula at large r, which makes me beleive my derivation is valid, the erronous factor [tex]\sqrt{2}[/tex] notwithstanding. 


#14
Jun1210, 10:25 AM

P: 1,568

[tex]\frac{ds}{dt}=\sqrt{1r_s/R}\sqrt{1\frac{(R\omega)^2}{1r_s/R}[/tex] The last equation gives you the hint that: [tex]v=\frac{R\omega}{\sqrt{1r_s/R}}[/tex] since, in GR: [tex]\frac{ds}{dt}=\sqrt{1r_s/r}\sqrt{1v^2}[/tex] 


#15
Jun1210, 10:50 AM

P: 836

And is v, in your case, measured by an observer from infinity? You have to give a definition. [tex]v=\frac{\text{d}\phi}{\text{d}\tau}r[/tex] and [tex]\frac{\text{d}\phi}{\text{d}t}r[/tex], for example, aren't the same, so you have to specify.
Aside from that, how can your equation be used to calculate the orbital period as a function of r only? 


#16
Jun1210, 11:07 AM

P: 1,568

The orbital period is not a function of r. 


#17
Jun1210, 11:20 AM

P: 836

Please link to the EulerLagrange derivation, and I'll do my best to understand it. 


#18
Jun1210, 11:32 AM

P: 1,568




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