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Orbital velocities in the Schwartzschild geometry

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starthaus
#523
Jul8-10, 05:30 PM
P: 1,568
Quote Quote by kev View Post
My proof in #511 contains no initial assumption that H is a constant with respect to r.
Sure it does, your "proof" is circular.



It has also been demonstrated by a number of people in this thread, that your assertion that [itex]dr/dt=0 \Rightarrow d^2r/dt^2=0[/itex] is false.
Basic calculus says that you are wrong.


I have also proven step by step that your assertion that "equation (57) in Espen's document is wrong", is yet another false assertion on your part.
No, you haven't. (51)-(57) is a mess. If it weren't, then espen180 would have completed (58). He can't , because of the mess he's gotten into following your tips.


You also keep making the false assertion that equation (30) in Espen's document is wrong, simply because you do not understand the context, while everyone else in this thread does.
Err, no, equation (30) is definitely wrong. I even pointed you to the correct equation in my blog but you kept claiming that (30) is correct as is. espen180 (and you) are missing a factor of [tex](3\alpha(r)/\alpha(r_0)-2)[/tex] where [tex]r_0[/tex] is the initial drop distance.


Your contributions to this thread is just a list of false assertions, red herrings and unhelpful vague statements that the work of other contributers are wrong without any indication of why it wrong or how to fix it. Your claim to be the "teacher" of the likes of Altabeh and espen180, when their knowledge of GR, physical intuition and mathematical ability is way beyond yours, is just laughable.
Ad-hominems will not prove your point, quite the opposite.
yuiop
#524
Jul8-10, 05:45 PM
P: 3,967
Quote Quote by starthaus View Post
No, you haven't. (51)-(57) is a mess. If it weren't, then espen180 would have completed (58). He can't , because of the mess he's gotten into following your tips.
Espen's work is an independent work that he is trying to do the derivation without any reference to constants. All he done is ask me to proof read it, which I have done and privately we have now sorted out the problems behind the scenes and succeeded in completing his objective. We would have done it a lot sooner, if it was not for your false leads such as equation (30) and (57) being wrong, when they are not.

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.
yuiop
#525
Jul8-10, 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?


Quote Quote by starthaus View Post
Basic calculus says that you are wrong.
Everyone elses's basic calculus says you are wrong. See the post by George.

Quote Quote by starthaus View Post
Sure it does, your "proof" is circular.
It is only circular if it is obvious 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.
starthaus
#526
Jul8-10, 05:51 PM
P: 1,568
Quote Quote by kev View Post
Espen's work is an independent work that he is trying to do the derivation without any reference to constants. All he done is ask me to proof read it, which I have done and privately we have now sorted out the problems behind the scenes and succeeded in completing his objective. We would have done it a lot sooner, if it was not for your false leads such as equation (30) and (57) being wrong, when they are not.
Sure they are, you need to pay attention.


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.
What I've been telling you is that all equations, starting with (51) and ending with (57) are wrong. This explains why neither you, nor espen180 have been able to complete the trivial equation (58), even after I showed you how to do it exactly 250 posts ago.
espen180
#527
Jul8-10, 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 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.
atyy
#528
Jul8-10, 07:39 PM
Sci Advisor
P: 8,367
Quote Quote by espen180 View Post
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?
atyy
#529
Jul9-10, 12:45 AM
Sci Advisor
P: 8,367
Quote Quote by atyy View Post
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)
Altabeh
#530
Jul9-10, 02:38 AM
P: 665
Quote Quote by starthaus View Post
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!

[tex]r[/tex] is a coordinate while contrary to your fallacious claims [tex]R=R(\phi)[/tex] 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

[tex]r^2\dot{\phi}=h,[/tex]

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:

"[itex]R^2\dot{\phi}=h,[/itex]"

and

the equation (15.8) exactly shows that

"[itex]R=R(\phi)[/itex]."
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
espen180
#531
Jul9-10, 04:24 AM
P: 836
Quote Quote by atyy View Post
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).
Altabeh
#532
Jul9-10, 04:55 AM
P: 665
Quote Quote by espen180 View Post
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
Anthony
#533
Jul9-10, 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 Euler-Lagrange 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 Euler-Lagrange 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] E-L 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].
espen180
#534
Jul9-10, 05:27 AM
P: 836
Quote Quote by Altabeh View Post
Good job espen. Way to go!

AB
Thanks! :)

And thanks for your hard work!
Altabeh
#535
Jul9-10, 09:47 AM
P: 665
Quote Quote by Anthony View Post
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:

[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 Euler-Lagrange 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] E-L 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].
Exactly! This simply is based on the proposition given in post #389 and by this perfect explanation we are done here.

AB
Anthony
#536
Jul9-10, 05:26 PM
P: 83
I'm glad it was of some use (I don't have the will power to find post #389).
Redbelly98
#537
Jul9-10, 06:52 PM
Mentor
Redbelly98's Avatar
P: 12,071
It's on page 25, if you are using the forum default of 16 posts per page.


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