Recognitions:

## Orbital velocities in the Schwartzschild geometry

 Quote by espen180 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 2-4 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.
At the start of section 2, should d(theta)/dt=0 be d(theta)/d(tau) instead?

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

Recognitions:
 Quote by atyy Also, I don't understand the comment "turning to the metric", just before Eqn 15, since the metric is Eqn 1?
I see. It's not the metric alone that you use to get Eqn 15. It's the metric plus the geodesic equation which implies that gab.dxa/dtau.dxb/dtau is a constant on a geodesic (following say Eq 2.4 and 2.5 of http://www.blau.itp.unibe.ch/lecturesGR.pdf; also see Altabeh's post #342)

 Quote by starthaus Err, wrong. The above doesn't even make sense. Basic calculus says that it doesn't follow.
Tell us which basic calculus supprts your nonsense here! LOL! I see you're completely bogged down with basics of calculus. Work hard!

 $$r$$ is a coordinate while contrary to your fallacious claims $$R=R(\phi)$$ is not a "plane polar coordinate" but rather a trajectory. You are in no position to "teach" since you don't even know the difference between coordinates and trajectories. Please stop trolling this thread.
Nonsense. Read the page 196 of D'inverno to see why you were blindly correcting the author.

 "... Then (15.20) can be integrated directly to give $$r^2\dot{\phi}=h,$$ where h is a constant. This is conservation of angular momentum (compare with (15.6) and note that, in the equatorial plane, the spherical polar coordinate is the same as the plane polar coordinate R). " Eq. (15.6) from the book is: "$R^2\dot{\phi}=h,$" and the equation (15.8) exactly shows that "$R=R(\phi)$."
Is you labelling us a "troller" another dead-end escape route towards not standing corrected? If so, then I don't want to say "you stop trolling" because I see that you're double finished by now.

AB

 Quote by atyy At the start of section 2, should d(theta)/dt=0 be d(theta)/d(tau) instead? Also, I don't understand the comment "turning to the metric", just before Eqn 15, since the metric is Eqn 1?
Thanks for pointing out, I've fixed it now.

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).

 Quote by espen180 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 2-4 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.
Good job espen. Way to go!

AB

 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 Euler-Lagrange equations themselves. This will invariably make your life easier. Your Lagrangian density reads: $$\mathcal{L} = \left( 1- \frac{1}{r}\right) \left( \frac{\mathrm{d} t}{\mathrm{d} \lambda}\right)^2 - \left( 1- \frac{1}{r}\right)^{-1} \left( \frac{\mathrm{d} r}{\mathrm{d} \lambda}\right)^2 - r^2 \left( \frac{\mathrm{d} \theta}{\mathrm{d} \lambda}\right)^2 - r^2 \sin^2\theta \left( \frac{\mathrm{d} \phi}{\mathrm{d} \lambda}\right)^2$$ using natural units. It is clear that the vector fields: $$\frac{\partial}{\partial t} \quad \textrm{and}\quad \frac{\partial}{\partial \phi}$$ are Killing, so Noether's theorem integrates up two of the Euler-Lagrange equations for you and gives you two constants of motion: $$\left(1-\frac{1}{r}\right) \frac{\mathrm{d} t}{\mathrm{d}\lambda} = \mathrm{const} \,(=E) \quad \textrm{and} \quad r^2 \sin^2\theta \frac{\mathrm{d} \phi}{\mathrm{d} \lambda} = \mathrm{const} \,(=h) \qquad (*)$$ i.e. on a given geodesic, these quantities remain unchanged. Similarly, since $$\partial_\lambda \mathcal{L}=0$$, we know $$\mathcal{L}$$ 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 $$\theta = \pi/2$$ (validity can be deduced from the $$\theta$$ E-L equation) and using (*) in $$\mathcal{L}=k$$ gives the ODE: $$\left( \frac{\mathrm{d} r}{\mathrm{d}\lambda}\right)^2 = E^2 - \left( 1-\frac{1}{r}\right) \left(k + \frac{h^2}{r^2}\right)$$ If you'd prefer to parameterise your geodesics using $$\phi$$, use the second of the constraints in (*) again and you get: $$\frac{h^2}{r^4} \left( \frac{\mathrm{d} r}{\mathrm{d} \phi}\right)^2 = E^2 - \left( 1-\frac{1}{r}\right) \left( k + \frac{h^2}{r^2}\right)$$ If you'd prefer to do all this using tensors, just apply Noether's theorem in the form: if $$L_V g=0$$ (i.e. $$V$$ is a Killing vector) then $$V^\mu \dot{x}^\nu g_{\mu\nu} = \mathrm{const}$$.

 Quote by Altabeh Good job espen. Way to go! AB
Thanks! :)

And thanks for your hard work!

 Quote by Anthony 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 Euler-Lagrange equations themselves. This will invariably make your life easier. Your Lagrangian density reads: $$\mathcal{L} = \left( 1- \frac{1}{r}\right) \left( \frac{\mathrm{d} t}{\mathrm{d} \lambda}\right)^2 - \left( 1- \frac{1}{r}\right)^{-1} \left( \frac{\mathrm{d} r}{\mathrm{d} \lambda}\right)^2 - r^2 \left( \frac{\mathrm{d} \theta}{\mathrm{d} \lambda}\right)^2 - r^2 \sin^2\theta \left( \frac{\mathrm{d} \phi}{\mathrm{d} \lambda}\right)^2$$ using natural units. It is clear that the vector fields: $$\frac{\partial}{\partial t} \quad \textrm{and}\quad \frac{\partial}{\partial \phi}$$ are Killing, so Noether's theorem integrates up two of the Euler-Lagrange equations for you and gives you two constants of motion: $$\left(1-\frac{1}{r}\right) \frac{\mathrm{d} t}{\mathrm{d}\lambda} = \mathrm{const} \,(=E) \quad \textrm{and} \quad r^2 \sin^2\theta \frac{\mathrm{d} \phi}{\mathrm{d} \lambda} = \mathrm{const} \,(=h) \qquad (*)$$ i.e. on a given geodesic, these quantities remain unchanged. Similarly, since $$\partial_\lambda \mathcal{L}=0$$, we know $$\mathcal{L}$$ 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 $$\theta = \pi/2$$ (validity can be deduced from the $$\theta$$ E-L equation) and using (*) in $$\mathcal{L}=k$$ gives the ODE: $$\left( \frac{\mathrm{d} r}{\mathrm{d}\lambda}\right)^2 = E^2 - \left( 1-\frac{1}{r}\right) \left(k + \frac{h^2}{r^2}\right)$$ If you'd prefer to parameterise your geodesics using $$\phi$$, use the second of the constraints in (*) again and you get: $$\frac{h^2}{r^4} \left( \frac{\mathrm{d} r}{\mathrm{d} \phi}\right)^2 = E^2 - \left( 1-\frac{1}{r}\right) \left( k + \frac{h^2}{r^2}\right)$$ If you'd prefer to do all this using tensors, just apply Noether's theorem in the form: if $$L_V g=0$$ (i.e. $$V$$ is a Killing vector) then $$V^\mu \dot{x}^\nu g_{\mu\nu} = \mathrm{const}$$.
Exactly! This simply is based on the proposition given in post #389 and by this perfect explanation we are done here.

AB

 I'm glad it was of some use (I don't have the will power to find post #389).
 Mentor Blog Entries: 10 It's on page 25, if you are using the forum default of 16 posts per page.