Orbital velocities in the Schwartzschild geometry

[starthaus]

even though it's late in the game --

i think a point has come up which hasn't been answered very well. So I'll
take a stab at it. You need to be careful when using the calculus of
variations to keep in mind which variables are dependent and which are
independent. In the current case, the only independent variable is the
worldline coordinate (t or s). Everything else is a function of (say t) t.
That means H can be at most a function of t. period. since we know
dH/dt = 0 it is the constant function. meaning a real, honest to goodness,
constant.

If I've expressed myself clearly -- this is enough to prove the result.
The rest is for the formula lovers out there.
S'pose L = L(q_i(t), \dot{q}_i(t)) then by the E-L eqns we know
\frac{d}{dt}\left( \frac{\partial L}{\partial \dot{q}_j} \right) <br /> = \frac{\partial L}{\partial q_j}[/itex]<br /> <br /> Now suppose that L is cyclic in one variable say q_k. That is<br /> \frac{\partial L}{\partial q_k} = 0 &lt;br /&gt; = \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}_k} \right)<br /> <br /> So that \frac{\partial L}{\partial \dot{q}_k} = C<br /> Now suppose to the contrary that C were really a function of the &quot;variables&quot;<br /> ie C = f(q_i).<br /> <br /> Then we get<br /> \frac{\partial f}{\partial q_j} = \frac{\partial^2 L}{\partial q_j \partial \dot{q}_k}&lt;br /&gt; = \frac{\partial^2 L}{\partial \dot{q}_k \partial q_j}<br /> <br /> = \frac{\partial }{\partial \dot{q}_k} \left( \frac{\partial L}{\partial q_j} \right)&lt;br /&gt; = \frac{\partial}{\partial \dot{q}_k}\left( \frac{d}{dt} \frac{\partial L}{\partial \dot{q}_j} \right) = \frac{\partial}{\partial \dot{q}_j}\left( \frac{d}{dt} \frac{\partial L}{\partial \dot{q}_k} \right) = \frac{\partial}{\partial \dot{q}_k}\left( 0 \right) = 0.

<br /> <br /> This is one of the bestr answers I have seen in this thread. Unfortunately, you haven&#039;t taken the time to understand the dispute between me and kev. The derivation that you show above is based on the Euler-lagrange equations, so, your results are valid <b>only</b> <b>only</b> along the arc defined by:<br /> <br /> ds^2=\alpha dt^2-dr^2/\alpha-(rd\phi)^2<br /> <br /> kev is trying to hack his derivation by differentiating as if H and K are constant <b>everywhere</b>. This is obviously not true. To wit, in your example f(r)=r^2\frac{d\phi}{ds} so, \frac{df}{dr} isn&#039;t zero.

 

[starthaus]

yowsa. pot-to-kettle.

There's only ONE independent variable in this problem. The worldine
coordinate s. So what we have are a bunch of functions of 1-variable.
r = r(s), t = t(s)... Now listen up if H were a function it would be a
function of s! (and only s).

When we write
r^2\frac{d\phi}{ds} = H
What we mean is
r(s)^2 \frac{d\phi}{ds}(s) = H(s).

Good, so you just confirmed that kev's

\frac{dH}{dr}=2r\frac{d\phi}{ds} + r^2 \frac{d}{dr}\left(\frac{d\phi}{ds}\right)

is nonsense. Thank you

 

[starthaus]

= -\frac{M}{r^2}<br /> \alpha + \frac{M}{r^2\alpha}\frac{dr^2}{dt^2} <br /> + (r-2M)\frac{d\phi^2}{dt^2}

------------------------------------

Get it now? Will you now stop telling Espen that (57) is wrong?

.

Err, wrong, post 430 gives you the correct (unhacked) solution. As you can see, the correct coefficient is r-3M not r-2M.

 

[yuiop]

LOL. That is an elementary calculus blunder. Shame on you!

Using the power rule:

\frac{dH}{dr}=2r\frac{d\phi}{ds} + r^2 \frac{d}{dr}\left(\frac{d\phi}{ds}\right)

You need to prove

r^2 \frac{d}{dr}\left(\frac{d\phi}{ds} \right)= 0

Err, r and \phi are independent variables, so your \frac{d}{dr}\left(\frac{d\phi}{ds}\right) is pure nonsense. You really need to stop winging the calculus rules. Especially when multivariate functions are not covered by the sw package you are using in attempting your differentiations.

Note: I have completely re-edited this post to reflect the reminder by Espen that the anti-derivative of zero is a constant.

If as you claim, r and \phi are independent variables, then it follows that

\frac{d}{dr}\left(\frac{d\phi}{ds}\right) =0 \qquad \qquad (1)

Taking the anti-derivative of both sides with respect to s gives:

\frac{d\phi}{ds} = C_{(r)} \qquad \qquad (2)

where C_{(r)} is a constant with respect to r.

We agree that H is constant with respect to s so we obtain:

\frac{dH}{ds}= \frac{d}{ds}\left(r^2\frac{d\phi}{ds}\right) =0

\Rightarrow \frac{d\phi}{ds}\frac{d}{ds}(r^2) +r^2 \frac{d}{ds}\left(\frac{d\phi}{ds}\right) =0

\Rightarrow \frac{d\phi}{ds}\frac{d}{ds}(r^2) +r^2 \frac{dr}{ds}\frac{d}{dr}\left(\frac{d\phi}{ds}\right) =0 \qquad \qquad (3)

Substituting (2) into (3) gives:

\frac{d\phi}{ds}\frac{d}{ds}(r^2) +r^2 \frac{dr}{ds}\frac{d}{dr}\left(C_{(r)}\right) =0

\Rightarrow \frac{d\phi}{ds}\frac{d}{ds}(r^2) =0

\Rightarrow \frac{d}{ds}(r^2) =0

Taking the anti-derivative of both sides with respect to s gives:

\Rightarrow r^2 = C_{(s)}

where C_{(s)} is a constant with respect to proper time s.

This shows that Starthaus's claims are only valid for circular orbits where r is constant.

 

[espen180]

The antiderivative is ...=constant, but that shouldn't invalidate your proof. In fact, you can use it to prove that H is a constant.

 

[yuiop]

Err, wrong, post 430 gives you the correct (unhacked) solution. As you can see, the correct coefficient is r-3M not r-2M.

In post #430 you give:

\frac{d^2r}{ds^2}=-\frac{m}{r^2}+(r-3m)(\frac{d\phi}{ds})^2

Espen gives:

\frac{ds^2}{dt^2} \cdot \frac{d^2r}{ds^2} = -\frac{M}{r^2}<br /> \alpha + \frac{M}{r^2\alpha}\frac{dr^2}{dt^2} <br /> + (r-2M)\frac{d\phi^2}{dt^2}

Do you not see that the left hand sides of the equations are different and so by elementary algebra we expect the right hand sides to be different?

You are talking about two different quantities. Wrong again.

 

[starthaus]

In post #430 you give:



Espen gives:



Do you not see that the left hand sides of the equations are different and so by elementary algebra we expect the right hand sides to be different?

You are talking about two different quantities. Wrong again.

Err, no. Bottom of post 430 says:

\frac{d^2r}{dt^2}=(r-3m)(\frac{d\phi}{dt})^2-(\frac{\alpha}{K})^2\frac{m}{r^2}+\frac{2m}{\alpha*r^2}(\frac{dr}{dt})^2

 

[espen180]

Err, no. Bottom of post 430 says:

\frac{d^2r}{dt^2}=(r-3m)(\frac{d\phi}{dt})^2-(\frac{\alpha}{K})^2\frac{m}{r^2}+\frac{2m}{\alpha*r^2}(\frac{dr}{dt})^2

Writing out the \frac{\alpha}{K} term, what do you get?

 

[starthaus]

and I have proven:

\frac{d}{dr}\left(\frac{d\phi}{ds}\right) \ne 0

kev,

I understand that you did not understand calculus very well and you are desperately trying to cover your mathematical blunder. So, let's try an "intuitive" explanation, what you "proved" above is that the angular speed depends on radius. Do you grasp the enormity of your error?

 

[starthaus]

Writing out the \frac{\alpha}{K} term, what do you get?

What is the puropse of this nonsensical request? The derivation is based on the fact that (see post 430)

\frac{ds}{dt}=\frac{\alpha}{K}

 

[espen180]

What is the puropse of this nonsensical request? The derivation is based on the fact that (see post 430)

\frac{ds}{dt}=\frac{\alpha}{K}

Nonsensical? Write it out (also write out ds/dt using the metric) and then compare the expressions.

 

[yuiop]

kev,

I understand that you did not understand calculus very well and you are desperately trying to cover your mathematical blunder. So, let's try an "intuitive" explanation, what you "proved" above is that the angular speed depends on radius. Do you grasp the enormity of your error?

You are wrong. For a particle in freefall with non zero angular momentum and dr/dt not equal to zero, the angular velocity is indeed a function of r.

 

[starthaus]

You are wrong. For a particle in freefall with non zero angular momentum and dr/dt not equal to zero, the angular velocity is indeed a function of r.

Prove it.

 

[espen180]

Prove it.

Mercury's orbital period is less than Pluto's. QED.

 

[starthaus]

Mercury's orbital period is less than Pluto's. QED.

This is certainly true along the geodesics, where r=r(\phi). This is certainly not true for H=r^2\frac{d\phi}{ds} where r and \phi are independent of each other . See my post 481 to qbert.

 

[espen180]

This is certainly true along the geodesics, where r=r(\phi). This is certainly not true for H=r^2\frac{d\phi}{ds} where r and \phi are independent of each other . See my post 481 to qbert.

When did we stop discussing geodesics?

 

[starthaus]

Nonsensical? Write it out (also write out ds/dt using the metric) and then compare the expressions.

You can do it all by yourself, you seem very good at these "turn the crank" exercises.

 

[espen180]

This is certainly true along the geodesics, where r=r(\phi). This is certainly not true for H=r^2\frac{d\phi}{ds} where r and \phi are independent of each other . See my post 481 to qbert.

This got me thinking.

Starthaus, do you view H and K as initial conditions do you look at their change along a geodesic?

 

[starthaus]

When did we stop discussing geodesics?

We didn't. What you keep missing is the following: there are cleaner ways of deriving the equations of motion that do employ the hacks that kev is using. See for example the derivations of the equations of motion for arbitrary planar orbits (post 53 for derivatives wrt proper time and post 430 for derivatives wrt coordinate time).
We have the geodesic method and its perfect equivalent, the lagrangian method. We don't need any hacks that borrow at will the solutions of the lagrange equations and mix them with methods of differentiation that defy the basic rules of calculus.

 

[starthaus]

This got me thinking.

Starthaus, do you view H and K as initial conditions

You know ODE's , right? You know what the constants signify, right?

When I wrote the Euler-Lagrange equation:

\frac{d}{ds}(\alpha*\frac{dt}{ds})=0

and I put down the obvious solution:

\alpha*\frac{dt}{ds}=K

what did I claim K was?When I wrote the Euler-Lagrange equation:

\frac{d}{ds}(r^2\frac{d\phi}{ds})=0

and I put down the obvious solution:

r^2\frac{d\phi}{ds}=H

what did I claim H was?

All I have been telling kev, for hundreds of posts is "Don't borrow the above solutions only to stick them into expressions derived from the metric followed by your hacky ways of differentiating. There are much cleaner ways of deriving the solutions."

 

[espen180]

You know ODE's , right? You know what the constants signify, right?

When I wrote the Euler-Lagrange equation:

\frac{d}{ds}(\alpha*\frac{dt}{ds})=0

and I put down the obvious solution:

\alpha*\frac{dt}{ds}=K

what did I claim K was?


When I wrote the Euler-Lagrange equation:

\frac{d}{ds}(r^2\frac{d\phi}{ds})=0

and I put down the obvious solution:

r^2\frac{d\phi}{ds}=H

what did I claim H was?

In the equations of motion!
How about giving a straight answer?

 

[qbert]

kev is trying to hack his derivation by differentiating as if H and K are constant everywhere. This is obviously not true. To wit, in your example f(r)=r^2\frac{d\phi}{ds} so, \frac{df}{dr} isn't zero.

Sorry, apparently, I haven't made myself clear.

\frac{d}{ds}\left( r^2\frac{d\phi}{ds} \right) =0
in the present context means

r(s)^2 \frac{d\phi}{ds}(s) = r_0^2 \right. \frac{d\phi}{ds}\left|_{s=0}

The number on the right hand side is a number. it isn't a function of \phi or r
or anything. it's a number. If you take the derivative of a number wrt anything
it is 0. everywhere.


second. What you're doing in the piece about f(r) confuses me.
let me write down explicitly how my brain parses this -- i'll leave
in (s) to remind me something is a function of s:
f(r(s)) = r(s)^2 \dot{\phi}(s) \Rightarrow \frac{df}{dr(s)}(s) = 2r(s)\dot{\phi}(s)
which literally makes no sense. How do you
even define a \frac{d}{dr} derivative in this case?
(Remember r isn't an INDEPENDENT VARIABLE here. By inverting the proposed (or once
we have it, the real) solution r = r(s) to give us s = s(r), we could use
\frac{d}{dr} = \frac{ds}{dr} \frac{d}{ds}
but of course that gives us a 0 anyway.

third. On Hacks. /read this at your own peril -- i won't
debate it but it doesn't get said enough here/

There is a longstanding physics tradition. If it works, it's right. This drives
mathies and mathy-inclinded physicists nuts. But the real truth is, physics
is a way of understanding the real world. There doesn't need to be a
logical derivation if the result is right. it's not math. there are many ways
to skin a cat. especially with something as complicated as a g.r. cat.

just because the derivation is flawed doesn't mean the result is wrong
either : http://xkcd.com/759/.

 

[starthaus]

What you're doing in the piece about f(r) confuses me.
let me write down explicitly how my brain parses this -- i'll leave
in (s) to remind me something is a function of s:
f(r(s)) = r(s)^2 \dot{\phi}(s) \Rightarrow \frac{df}{dr(s)}(s) = 2r(s)\dot{\phi}(s)
which literally makes no sense. How do you
even define a \frac{d}{dr} derivative in this case?

Simple, you need to re-read my post 481 to you and the limitations of your method. The short of it is that it applies only along the geodesics.

There is a longstanding physics tradition. If it works, it's right.

That's bad. I do not ascribe to this philosophy.

just because the derivation is flawed doesn't mean the result is wrong
either : http://xkcd.com/759/.

I didn't say that the results were wrong, I said that kev's methods are wrong.
Now, I enjoy very much interacting with you but I have to convince espen180 that he still has errors in the document that started it all. He has the roight approach but he keeps getting the wrong results. Maybe because he mixes correct mathematics with an occasional kev hack.

 

[starthaus]

Nonsensical? Write it out (also write out ds/dt using the metric) and then compare the expressions.

Ok, since you don't seem inclined to do it, I did it. Here is he summary: your (57) is still wrong. To convince yourself that this is the case, make \frac{d\phi}{dt}=0. You should recover your (correct) equation (44). This is not the case.

 

[espen180]

Ok, since you don't seem inclined to do it, I did it. Here is he summary: your (57) is still wrong. To convince yourself that this is the case, make \frac{d\phi}{dt}=0. You should recover your (correct) equation (44). This is not the case.

But (57) is not the complete expression.

 

[starthaus]

But (57) is not the complete expression.

Then either:

-write out (58) and make \frac{d\phi}{dt}=0

or

-make \frac{d\phi}{dt}=0 in (57) and you'll still see that you got things wrong.

I think your errors start with the mess at eq (51)-(52).

 

[Altabeh]

Err, r and \phi are independent variables

Pure nonsesne. For example, in the equatorial plane, the spherical polar coordinate r is the same as plane polar coordinate R=R(\phi) which you seem to have forgotten from my post #335 where I stress that

Look, H=r^2\dot{\phi} is a "constant of motion" and is to be considered in the equatorial plane so that by definition the spherical polar coordinate r is the same as the plane polar coordinate R=R(\phi).

I even gave you the name of a source, but you didn't manage to learn. Even worse, your basec knowledge could have helped you remember from the equations of the orbits of test particles that we put u(\phi)=1/r(\phi)=u. Why do you insist on your onw nonsense claims and hacks all the time?

AB

 

[Altabeh]

This is certainly true along the geodesics, where r=r(\phi). This is certainly not true for H=r^2\frac{d\phi}{ds} where r and \phi are independent of each other.

Nonsense. H here comes from geodesic equations (if not consider the general approach in getting it using "Killing vectors" and the proposition in post #389), which is only valid in its form along geodesics thus the second part of your post contradicts the first; making a nonsense out of the whole of it. LOL!

See my post 481 to qbert.

There is nothing in that post to be inferred about your "hacky" claim above. Get a clue on where your fallacy arises from.

AB

 

[starthaus]

Pure nonsesne. For example, in the equatorial plane, the spherical polar coordinate r is the same as plane polar coordinate R=R(\phi) which you seem to have forgotten from my post #335 where I stress that

Err, wrong . R=R(\phi) is a trajectory , not a coordinate.
Once you express r as a function of \phi it becomes a trajectory.

Even worse, your basec knowledge could have helped you remember from the equations of the orbits of test particles that we put u(\phi)=1/r(\phi)=u. Why do you insist on your onw nonsense claims and hacks all the time?

Why do you persist in the same basic error of mixing r as a coordinate with r(\phi) which is a trajectory?

 

[Altabeh]

We didn't. What you keep missing is the following: there are cleaner ways of deriving the equations of motion that do employ the hacks that kev is using. See for example the derivations of the equations of motion for arbitrary planar orbits (post 53 for derivatives wrt proper time and post 430 for derivatives wrt coordinate time).
We have the geodesic method and its perfect equivalent, the lagrangian method. We don't need any hacks that borrow at will the solutions of the lagrange equations and mix them with methods of differentiation that defy the basic rules of calculus.

Yeah, I agree! Speaking of your old fallacy, dr/ds=0\Rightarrow d^2r/ds^2=0, which you corrected it serenely in your post #251, and the fact that along timelike geodesics you can introduce a "momentarily at rest" co-observer whose velocity is zero at "each" event but yet his acceleration isn't, which you seem to deny it without even daring to take a look at the sources I provided you with days ago, and now recently another nonsense in your post 495, and above all else persisting on them while knowing that they are nothing but hacks and fallacies, all I can say is that you're just trying to not stand corrected after making mistakes and this will not get us to believe in your nonsense ever. Take a swipe at correcting yourself from time to time.

Err, wrong . R=R(\phi) is a trajectory , not a coordinate.
Once you express r as a function of \phi it becomes a trajectory.

Why do you persist in the same basic error of mixing r as a coordinate with r(\phi) which is a trajectory?

Nonsense. It is ridiculous that you want to correct D'inverno. Go read the page 196 of his "well-known" GR book. Where on Earth do they claim r=r(s) isn't a part of spherical coordinates? You're really clueless about basics of physics.

AB