Orbital velocities in the Schwartzschild geometry

[yuiop]

... I have already uploaded the solution I promised you in my blog, Since you subscribe to my blog, you should have received notification of the update.

I don't remember subscribing to your blog. I occasionally look at it, because for some reason you insist on forcing readers of this forum to look at your blog, rather than answering questions directly in the threads.

Now, explain to me where does your new starting point equation come from?

$$\frac{dr^2}{ds^2} = c^2(K^2-1) +f(r)$$

I have already told you it came from a textbook. Since you want to stall on this point I can also obtain it by going back to my derivation in post #211 of this tread:

Starting with Schwarzschild metric and assuming motion in a plane about the equator such that $\theta = \pi/2$ and $d\theta = 0$

$ds^2=\alpha dt^2-dr^2/\alpha -r^2d\phi^2$ where $\alpha=(1-2m/r)$

Solve for (dr/ds):

$$\frac{dr}{ds} = \sqrt{\alpha^2\frac{dt^2}{ds^2} - \alpha - \alpha r^2 \frac{d\phi^2}{ds^2} }$$

The well known constants of motion are:

$K = \alpha(dt/ds)$ and $H = r^2 (d\phi/ds)$

Insert these constants into the equation for (dr/ds):

$$\frac{dr}{ds} = \sqrt{K^2 - \alpha - \alpha \frac{H^2}{r^2}}$$

Now for radial motion, H=0 so the last equation can be written as:

$$\frac{dr}{ds} = \sqrt{K^2 - \alpha }$$

$$\Rightarrow \frac{dr^2}{ds^2} = K^2 - \alpha$$

$$\Rightarrow \frac{dr^2}{ds^2} = K^2 - (1 -2GM/r)$$

$$\Rightarrow \frac{dr^2}{ds^2} = (K^2 - 1) + 2GM/r$$

which is the same as the equation given by the authors of the textbook.

and if we define a function f such that f(r) = (2GM/r) we recover the form you have quoted (using units of c=1):

$$\Rightarrow \frac{dr^2}{ds^2} = (K^2 - 1) + f(r)$$

Will you now show how you differentiate the above expression wrt (s) without differentiating wrt (r) in an intermediate step, as this is what you have claimed the authors of the book have done?

 

[espen180]

Wrong, the angular velocity gets eliminated between (1) and (2).

Err, you got this backwards, the exercise it to prove that $$H(r)$$ does not depend on $$r$$. Since you stepped in, try proving it, you've been provided (as always) all the tools.
This going to be a good challenge for you, especially considering the fact that $$H(r,\phi)=r^2\frac{d\phi}{ds}$$.

That doesn't change anything. All you end up with is the radial acceleration in terms of H rather than dø/ds.

 

[yuiop]

Let's not waste time and take it for granted that you can obtain:

$$\frac{d^2r}{ds^2} = \frac{-GM}{r^2}\right)$$

by assuming K is contant wrt (s) and differentiating (dr/ds) wrt (s).

Unless you prove a flaw in my calculations in #380, I have demonstrated I can obtain the same result by assuming K is a constant wrt (r) and differentiating (1/2)(dr/dt)^2 wrt (r).

This proves that for radial motion, K is constant wrt (s) AND (r).

This means that your claim in your blog document "General Euler-Lagrange Solution for Calculating Coordinate Acceleration", (which is actually the solution for radial motion only - you messed up the titles) that:

$$\alpha \frac{dt}{ds}= k$$

where, obviously k may be a function of r and possibly t but not of s

is a fallacy, because I have shown k is not a function of r for radial motion. Are you going to correct your blog?


From further qualitative analysis I have come to the conclusion that

for radial motion only when (H = 0) that:

K is contant wrt (s,t and r)

and $(\phi)$ is constant by definition under these conditions,

and for circular motion only. when (K = constant) that:

H is constant wrt (s,t and $\phi$)

and (r) is constant by definition under these conditions.

This leaves open what happens when a falling particle has both non-zero radial and non-zero angular motion at the same time. My initial hunch is that in relativity, energy and momentum are not individually conserved as in Newtonian physics, but are conserved as a pair in the form of momentum-energy. I think the same thing is happening here. The angular momentum and energy are conserved as a pair.

One thing to recall is that when we take the partial derivatives of L with respect to a given variable, obtain K and H we are by definition treating the "other variables" as constants and so we have not proved anything about what happens when we allow the other variables to vary.

I am not "peddling" anything. :LOL:
And I am not back peddaling.

If you are going to pick on spelling, try and get it right. it should be "pedaling" or sometimes "pedalling" (American?) but not with two d's.

 

[Altabeh]

Incorrect. the correct statement is

$$\frac{d H}{ds}=0$$.

Nonsense. I think now we have to also take care of "semantics". This is just because you're only dealing with basics of physics and an ODE which says nothing about constants of motion but the fact that their derivative with respect to s is zero. The scenario of "Killing vectors" calls the shouts here and they exactly describe what such constants are and why they are "constant". During motion, such quantities remain constant and that's the reason for their name.

This comes from integrating the Euler-Lagrange equation,

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

wrt $$s$$ resulting into


$$H(r,\phi)=r^2\frac{d\phi}{ds}$$.

It doesn't. It comes from "Killing vectors" and the following proposition:

Let $$\xi^a$$ be a Killing vector field and let $$\gamma$$ be a geodesic with tangent $$u^a$$. Then $$\xi_au^a$$ is constant along $$\gamma$$.

Err, no.


$$H(r,\phi)=r^2\frac{d\phi}{ds}$$ (1)

and

$$\frac{d^2r}{ds^2}=-\frac{m}{r^2}+r(\frac{d\phi}{ds})^2(1-\frac{3m}{r})$$ (2)

Eliminate $$\frac{d\phi}{ds}$$ between (1) and (2) and show that the resultant expression for $$H(r)$$ does not depend on $$r$$.
Your challenge is to show that $$H(r)$$ is not a function of $$r$$

Nonsense. It's really the first time that I see someone makes use of hacks like $$K(r,\phi)$$ or $$H(r,\phi)$$. Read the above proposition and show that it holds. Then we can discuss where your big fallacies arise.

AB

 

[starthaus]

That doesn't change anything. All you end up with is the radial acceleration in terms of H rather than dø/ds.

Err, no, you get that $$H$$ is a function of $$r$$. Which means that kev's method of calculating differentials is all wrong, something that we've spent 100+ posts in trying to explain to him.

 

[starthaus]

Let's not waste time and take it for granted that you can obtain:

$$\frac{d^2r}{ds^2} = \frac{-GM}{r^2}\right)$$

by assuming K is contant wrt (s) and differentiating (dr/ds) wrt (s).

Unless you prove a flaw in my calculations in #380, I have demonstrated I can obtain the same result by assuming K is a constant wrt (r) and differentiating (1/2)(dr/dt)^2 wrt (r).

This proves that for radial motion, K is constant wrt (s) AND (r).

This means that your claim in your blog document "General Euler-Lagrange Solution for Calculating Coordinate Acceleration", (which is actually the solution for radial motion only - you messed up the titles) that:is a fallacy, because I have shown k is not a function of r for radial motion. Are you going to correct your blog?

Err, no. It is simply the first Euler-Lagrange equation. You can find it in Rindler, now that you've broken down and bought the book.
I don't think it is productive going another 100 posts just to show you that you don't understand the basic methods. So , I will direct you to (11.29) in Rindler's book.

and for circular motion only. when (K = constant) that:

H is constant wrt (s,t and $\phi$)

and (r) is constant by definition under these conditions.

Err, no again. For radial motion $$H$$ depends on $$r$$

This leaves open what happens when a falling particle has both non-zero radial and non-zero angular motion at the same time. My initial hunch is that i

Physics doesn't work on hunches, you need to proove your assertions. I even narrowed it down for you, espen180 and Altabeh what you have to proove. See second half of post 377.

 

[yuiop]

Err, no, you get that $$H$$ is a function of $$r$$. Which means that kev's method of calculating differentials is all wrong, something that we've spent 100+ posts in trying to explain to him.

The point is that you yet to derive your own solution for acceleration in coordinate time, for radial and angular motion in a single equation and demonstrate that my result is wrong.

 

[starthaus]

The point is that you yet to derive your own solution for acceleration in coordinate time, for radial and angular motion in a single equation and demonstrate that my result is wrong.

My derivation has been in the blog since May 28. You should read it sometimes.

As to your approach, your result coincides with mine but your method is a hack. We've been over this.

 

[yuiop]

Err, no. It is simply the first Euler-Lagrange equation. You can find it in Rindler, now that you've broken down and bought the book.
I don't think it is productive going another 100 posts just to show you that you don't understand the basic methods. So , I will direct you to (11.29) in Rindler's book.

I don't have the book. I just see random bits of it on google.

Physics doesn't work on hunches, you need to proove your assertions. I even narrowed it down for you, espen180 and Altabeh what you have to proove. See second half of post 377.

and you have to prove that $dH/dr \ne 0$.

You have not done that yet.

 

[espen180]

Err, no, you get that $$H$$ is a function of $$r$$.

This is a bare assertion and deserves no merit unless you back it up with a derivation, preferably your own. Bare assertions violate PF rules.

 

[starthaus]

I don't have the book. I just see random bits of it on google.

Then , I recommend you get it or that you read my blog and stop imagining errors where there aren't any.

and you have to prove that $dH/dr \ne 0$.

You have not done that yet.

You have this as an exercise explained in the second part of post 377. I gave you all the necessary hints, you have one substitution to perform.

 

[starthaus]

This is a bare assertion and deserves no merit unless you back it up with a derivation, preferably your own. Bare assertions violate PF rules.

It is an exercise for you and kev, all set up in post 377. See which one of you can finish it first.

 

[espen180]

You have this as an exercise explained in the second part of post 377. I gave you all the necessary hints, you have one substitution to perform.

This isn't the homework section. It doesn't work like that here.

 

[starthaus]

This isn't the homework section. It doesn't work like that here.

Well, the only way to proove you wrong is to get you to perform the computations.

 

[yuiop]

My derivation has been in the blog since May 28. You should read it sometimes.

As to your approach, your result coincides with mine but your method is a hack. We've been over this.

Your blog result is not the solution for acceleration in coordinate time, for radial and angular motion in a single equation and your documnet dated May 28 does not have one single mention of the variables phi or theta and is for radial motion only.

As it stands my solution is the only solution so far provided on this thread for acceleration in coordinate time, for radial and angular motion in a single equation and as far I can tell there is no solution provided in the textbooks or on the internet. When Espen has completed his work we will have an independent solution to check it against. You seem to be implying that you have your independent solution and you are gently guiding Espen and Altabeh towards it, but I doubt you have own solution or derivation, because it is not in the textbooks for you to copy it from.

 

[espen180]

Well, the only way to proove you wrong is to get you to perform the computations.

You're unwilling to substansiate your claim, then. Either that, or you know it won't lead anywhere (a false assertion). Either way, you're breaking the rules.

 

[starthaus]

You're unwilling to substansiate your claim, then. Either that, or you know it won't lead anywhere (a false assertion). Either way, you're breaking the rules.

$$H(r,\phi)=r^2\frac{d\phi}{ds}$$ (1)

and

$$\frac{d^2r}{ds^2}=-\frac{m}{r^2}+r(\frac{d\phi}{ds})^2(1-\frac{3m}{r})$$ (2)

Eliminate $$\frac{d\phi}{ds}$$ between (1) and (2) and show that the resultant expression for $$H(r)$$ does not depend on $$r$$.
Your challenge is to show that $$H(r)$$ is not a function of $$r$$

 

[yuiop]

It is an exercise for you and kev, all set up in post 377. See which one of you can finish it first.

The result is inconclusive as Espen has already explained. If you have a proof, show your hand or shut up.

 

[starthaus]

The result is inconclusive as Espen has already explained. If you have a proof, show your hand or shut up.

$$H(r,\phi)=r^2\frac{d\phi}{ds}$$ (1)

and

$$\frac{d^2r}{ds^2}=-\frac{m}{r^2}+r(\frac{d\phi}{ds})^2(1-\frac{3m}{r})$$ (2)

Eliminate $$\frac{d\phi}{ds}$$ between (1) and (2) and show that the resultant expression for $$H(r)$$ does not depend on $$r$$.
Your challenge is to show that $$H(r)$$ is not a function of $$r$$. Do you think you can do this simple calculation all by yourself? Or is it that you have already done it and you know that it contradicts your claims?

 

[yuiop]

Then , I recommend you get it or that you read my blog and stop imagining errors where there aren't any.

I have already shown that your statement in your blog that for radial motion, K is function of r is a false assertion. In the interests of intellectual honesty, please make it clear that you accept that you made a mistake there.

 

[starthaus]

I have already shown that your statement in your blog that for radial motion, K is function of r is a false assertion. In the interests of intellectual honesty, please make it clear that you accept that you made a mistake there.

We are not talking radial motion, we are talking general orbits.


$$H(r,\phi)=r^2\frac{d\phi}{ds}$$ (1)

and

$$\frac{d^2r}{ds^2}=-\frac{m}{r^2}+r(\frac{d\phi}{ds})^2(1-\frac{3m}{r})$$ (2)

Eliminate $$\frac{d\phi}{ds}$$ between (1) and (2) and show that the resultant expression for $$H(r)$$ does not depend on $$r$$.
Your challenge is to show that $$H(r)$$ is not a function of $$r$$

 

[yuiop]

$$H(r,\phi)=r^2\frac{d\phi}{ds}$$ (1)

and

$$\frac{d^2r}{ds^2}=-\frac{m}{r^2}+r(\frac{d\phi}{ds})^2(1-\frac{3m}{r})$$ (2)

Eliminate $$\frac{d\phi}{ds}$$ between (1) and (2) and show that the resultant expression for $$H(r)$$ does not depend on $$r$$.
Your challenge is to show that $$H(r)$$ is not a function of $$r$$. Do you think you can do this simple calculation all by yourself? Or is it that you have already done it and you know that it contradicts your claims?

Nope, we have done it and seen that the result is inconclusive and does not clearly support either side of the dispute.

If you can provide a clear proof from the above that supports your argument, then please do so, or you are just making unsupported assertions.

 

[starthaus]

Nope, we have done it and seen that the result is inconclusive and does not clearly support either side of the dispute.

No, the result isn't "inconclusive", it is quite conclusive. You have one step to calculate, why aren't you doing it?


$$H(r,\phi)=r^2\frac{d\phi}{ds}$$ (1)

and

$$\frac{d^2r}{ds^2}=-\frac{m}{r^2}+r(\frac{d\phi}{ds})^2(1-\frac{3m}{r})$$ (2)

Eliminate $$\frac{d\phi}{ds}$$ between (1) and (2) and show that the resultant expression for $$H(r)$$ does not depend on $$r$$.
Your challenge is to show that $$H(r)$$ is not a function of $$r$$

 

[yuiop]

We are not talking radial motion, we are talking general orbits.

In your blog you make the clearly make the assertion that K is a function of r in the context of purely radial motion, and I have proved you wrong in that context.

Do you now agree that in the context of purely radial motion, that $K=(1-2M/r)(dt/ds)$ is NOT a function of r, even though the equation for K superficially appears to contain the variable (r)?

 

[starthaus]

In your blog you make the clearly make the assertion that K is a function of r in the context of purely radial motion, and I have proved you wrong in that context.

Do you now agree that in the context of purely radial motion, that K is a function is NOT a function of r?

Err, so you don't understand the challenge or you understand it and you don't want to do a simple calculation? We are talking about $$H$$, not $$K$$ , kev. We are talking arbitrary orbits (that's where your hack is at its ugliest) not about radial motion.

 

[yuiop]

Err, so you don't understand the challenge or you understand it and you don't want to do a simple calculation? We are talking about $$H$$, not $$K$$ , kev. We are talking arbitrary orbits (that's where your hack is at its ugliest) not about radial motion.

I am specifically talking about a false assertion in your blog.

 

[starthaus]

You seem to be implying that you have your independent solution and you are gently guiding Espen and Altabeh towards it, but I doubt you have own solution or derivation, because it is not in the textbooks

True, it isn't in any textbook.

for you to copy it from.

False. Try downloading the second attachment from https://www.physicsforums.com/blog.php?b=1957

 

[yuiop]

@Starthaus: As I said before "you have to prove that $dH/dr \ne 0$"

 

[starthaus]

I am specifically talking about a false assertion in your blog.

1. First of, I assert that "k may be a function of r". In the radial motion it isn't, in the case of arbitrary orbits it is.

2. Second off, we are talking about your hacky method of deriving the equations of motion.

3. While talking about your hacky method, we are unravelling your claim that $$H$$ can be treated under differentiation wrt $$r$$ as if it were a constant. The exercise from post 377 is set up to disprove your assertion.

4. You must have calculated $$H$$ by now and you already know that your method is wrong, this is why you want to divert the discussion.

5. My derivation (two files at https://www.physicsforums.com/blog.php?b=1957 ) does NOT use any assumption about the nature of $$K$$ and/or $$H$$. As such, as opposed to your method, my method is fully rigorous.

 

[yuiop]

False. Try downloading the second attachment from https://www.physicsforums.com/blog.php?b=1957

I have seen it before.

You have a general equation in terms of proper time. - Not what we are looking for,

You have the special case of circular motion in coordinate time. - Not what we are looking for.

You have the special case of purely radial motion in coordinate time. - Not what we are looking for.

We are looking for the general (radial & angular) equation in terms of coordinate time in a single equation as I have produced in my derivation. I think you don't want to produce it because your either not sure how to, or you know that it will be the same as my general result and you won't be able to continue implying my result is wrong.

 

[starthaus]

@Starthaus: As I said before "you have to prove that $dH/dr \ne 0$"

So, I see you don't want to do a simple calculation:

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

Show that the above is not a function of $$r$$.

 

[yuiop]

1. First of, I assert that "k may be a function of r". In the radial motion it isn't, in the case of arbitrary orbits it is.

OK. We have finally established that the presence of the variable (r) in the equation for K does not have to imply that K is a function of (r).

3. While talking about your hacky method, we are unravelling your claim that $$H$$ can be treated under differentiation wrt $$r$$ as if it were a constant. The exercise from post 377 is set up to disprove your assertion.

Your exercise or proof relies on the fact that when H is differentiated wrt (r) the right hand side contains the variable (r). We have now established that that is not sufficient proof that H is a function of (r). Try another tack.

 

[starthaus]

OK. We have finally established that the presence of the variable (r) in the equation for K does not have to imply that K is a function of (r).

For radial motion. NOT for arbitrary orbits. Try reading for comprehension.

Your exercise or proof relies on the fact that when H is differentiated wrt (r) the right hand side contains the variable (r). We have now established that that is not sufficient proof that H is a function of (r). Try another tack.

Post 416 says that you are wrong.

 

[yuiop]

3. While talking about your hacky method, we are unravelling your claim that $$H$$ can be treated under differentiation wrt $$r$$ as if it were a constant. The exercise from post 377 is set up to disprove your assertion.

Your exercise is inconclusive. That leaves the only the rigourous argument that H is NOT a function of r as the one provided by Altabeh in relation to Killing vectors and conserved quantities.

 

[starthaus]

Your exercise is inconclusive.

Post 416 shows plainly that you are wrong.