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Orbital velocities in the Schwartzschild geometry 
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#523
Jul810, 05:30 PM

P: 1,568




#524
Jul810, 05:45 PM

P: 3,967

If you read Espen's document with any care, you would notice that (57) is derived completely from the Schwarzschild metric and equation (49), without any reference to equations (51)(56). Sloppy and wrong again. 


#525
Jul810, 05:48 PM

P: 3,967

@Starthaus. I see in #523 you have quietly stopped defending your assertion that H is a function of r. Does that mean you are finally convinced that you were wrong about that?
[tex]\frac{d}{ds}(r^2*\frac{d\phi}{ds})=0 \Rightarrow \frac{d}{dr}(H) =0[/tex] but it seems it was not obvious to you (although it is to everyone else), so posted the step by step explanation for you in post #511. 


#526
Jul810, 05:51 PM

P: 1,568




#527
Jul810, 06:43 PM

P: 836

Alright! I have now finished the calculation, thanks to kev's help and support!
Here is the newest document: Download I have removed the sections 24 which dealt with special cases, focusing on the general, unrestricted case. I abandoned the calculation I was currently working on (with ridicculously large expressions etc) for a simpler approach. Please take a look. Edit: Updated with angular accelerations. 


#528
Jul810, 07:39 PM

Sci Advisor
P: 8,792

Also, I don't understand the comment "turning to the metric", just before Eqn 15, since the metric is Eqn 1? 


#529
Jul910, 12:45 AM

Sci Advisor
P: 8,792




#530
Jul910, 02:38 AM

P: 665

AB 


#531
Jul910, 04:24 AM

P: 836

I said "turning to the metric" instead of "from (1) we obtain" since I used the matric in a different form. (spacetime element vs. matrix). 


#532
Jul910, 04:55 AM

P: 665

AB 


#533
Jul910, 05:20 AM

P: 83

I think the OP might be making life difficult for himself. With any form of variational problem, you should try to use Noether's theorem to integrate your equations up, rather than try to go from the EulerLagrange equations themselves. This will invariably make your life easier. Your Lagrangian density reads:
[tex] \mathcal{L} = \left( 1 \frac{1}{r}\right) \left( \frac{\mathrm{d} t}{\mathrm{d} \lambda}\right)^2  \left( 1 \frac{1}{r}\right)^{1} \left( \frac{\mathrm{d} r}{\mathrm{d} \lambda}\right)^2  r^2 \left( \frac{\mathrm{d} \theta}{\mathrm{d} \lambda}\right)^2  r^2 \sin^2\theta \left( \frac{\mathrm{d} \phi}{\mathrm{d} \lambda}\right)^2 [/tex] using natural units. It is clear that the vector fields: [tex] \frac{\partial}{\partial t} \quad \textrm{and}\quad \frac{\partial}{\partial \phi}[/tex] are Killing, so Noether's theorem integrates up two of the EulerLagrange equations for you and gives you two constants of motion: [tex] \left(1\frac{1}{r}\right) \frac{\mathrm{d} t}{\mathrm{d}\lambda} = \mathrm{const} \,(=E) \quad \textrm{and} \quad r^2 \sin^2\theta \frac{\mathrm{d} \phi}{\mathrm{d} \lambda} = \mathrm{const} \,(=h) \qquad (*) [/tex] i.e. on a given geodesic, these quantities remain unchanged. Similarly, since [tex]\partial_\lambda \mathcal{L}=0[/tex], we know [tex] \mathcal{L}[/tex] remains constant, and we set it to {+1,1,0} depending on whether you're interested in timelike, spacelike or nulll geodesics. Call this constant k. Note that all our ODEs are now 1st order. Setting [tex]\theta = \pi/2[/tex] (validity can be deduced from the [tex]\theta[/tex] EL equation) and using (*) in [tex]\mathcal{L}=k[/tex] gives the ODE: [tex] \left( \frac{\mathrm{d} r}{\mathrm{d}\lambda}\right)^2 = E^2  \left( 1\frac{1}{r}\right) \left(k + \frac{h^2}{r^2}\right)[/tex] If you'd prefer to parameterise your geodesics using [tex]\phi[/tex], use the second of the constraints in (*) again and you get: [tex] \frac{h^2}{r^4} \left( \frac{\mathrm{d} r}{\mathrm{d} \phi}\right)^2 = E^2  \left( 1\frac{1}{r}\right) \left( k + \frac{h^2}{r^2}\right) [/tex] If you'd prefer to do all this using tensors, just apply Noether's theorem in the form: if [tex]L_V g=0[/tex] (i.e. [tex]V[/tex] is a Killing vector) then [tex]V^\mu \dot{x}^\nu g_{\mu\nu} = \mathrm{const}[/tex]. 


#534
Jul910, 05:27 AM

P: 836

And thanks for your hard work! 


#535
Jul910, 09:47 AM

P: 665

AB 


#536
Jul910, 05:26 PM

P: 83

I'm glad it was of some use (I don't have the will power to find post #389).



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