Orbital velocities in the Schwartzschild geometry

[starthaus]

After reading post #27, the similarity between the Newtonian expression and the one I arrived at makes it very difficult for me to believe it is incorrect. It also predicts that the photon sphere should be at r=3GM. Post #48 seems to confirm my belief here.I tried to use the same approach to derive the coordinate acceleration of a particle dropped from rest at r relative to a stationary observer also at r. The result was
$$\frac{\text{d}^2r}{\text{d}t^2}=-\left(\frac{GM}{r^2}-\frac{2G^2M^2}{r^3c^2}\right)$$

...because you are trying to hack your way to the solution.

which goes to 0 as r approaches the Schwartzschild radius, I'm unsure what to make of that. It also changes sign when r<2GM, so I doubt its validity in that region. Still, it approximates the Newtonian expression at large r, differing only by about 1.4 ppb at Earth's surface.

Does it look correct? If neccesary, I can post my derivation.

This is post #50. You are still struggling with the circular orbits. There is no mention of any radial motion yet. Nor is there any problem statement. I told you to split the threads but you insisted in co-mingling them.

 

[starthaus]

The condition was that that $$\frac{dr}{d\tau}$$ was momentarily zero, not constantly. We are talking about free fall here.

This is post #64. Still no mention of the problem statement, just an attempt to deflect my criticism that you are hacking $$\frac{dr}{d\tau}=0$$ into the general equation of arbitrary orbits. I don't think it is worth continuing.

 

[espen180]

This is post #50. You are still struggling with the circular orbits. There is no mention of any radial motion yet. Nor is there any problem statement. I told you to split the threads but you insisted in co-mingling them.

I'm sorry, "acceleration of a particle dropped from rest at r" must have been a little too vague for you.

 

[espen180]

This is post #64. Still no mention of the problem statement, just an attempt to deflect my criticism that you are hacking $$\frac{dr}{d\tau}=0$$ into the general equation of arbitrary orbits. I don't think it is worth continuing.

Are you saying that a paticular family of geodesics are not a subset of the set of all geodesics?

 

[starthaus]

I'm sorry, "acceleration of a particle dropped from rest at r" must have been a little too vague for you.

Yes, I missed that, you have just embarked on your journey of hacking the general equation. You have just switched to a simpler problem and , since you are hacking, you got the wrong answer.You could have saved a lot of wasted time, we already solved this eons ago, see here. Both radial and circular motion.

So, can you solve the five exerciises I gave you?

 

[Altabeh]

Let's try again, since you have been given that the function $$\frac{dr}{ds}=0$$ FOR ALL VALUES OF s, what can you infer about $$\frac{d^2r}{ds^2}$$?
Keep in mind that $$\frac{d^2r}{ds^2}=\frac{d}{ds}(\frac{dr}{ds})$$.

Same old hack. Doesn't work! Go for another hack!

Irrelevant.

Only in case you don't understand the whole thing!

This is not about boundary conditions. Since when do you plug boundary conditions straight into the ODE? LOL

What?! So what do we do with boundary conditions? As I have noticed earlier several times, such nonsense claims are because of the lack of knowledge in the relevant zones.

Irrelevant in finding the trajectory $$r=r(t)$$ or the velocity $$v=v(t)$$ or the acceleration $$a=a(t)$$.

Again only in case you don't understand the whole thing!

If you think otherwise, using your hack, find these:

1. $$r=r(\tau)$$
2. $$r=r(t)$$
3. $$v=v(\tau)$$
4. $$v=v(t)$$
5. $$a=a(t)$$

The hack is only good only for finding the acceleration at the release point. Worthless for anything else.

Irrelevant and totally nonsense! Consult the sources I provided you with to not go for such hacks as an escape route!

AB

 

[Altabeh]

espen180, did you see my post #132? I have provided a derivation of the orbital velocity of a particle in circular motion from the perspective of a hovering observer.

AB

 

[Altabeh]

I don't think it is worth continuing.

Then stop hacking more. You've done enough of that!

AB

 

[espen180]

espen180, did you see my post #132? I have provided a derivation of the orbital velocity of a particle in circular motion from the perspective of a hovering observer.

AB

Sorry, it must have slipped past me. I'll check it out, thanks.

 

[starthaus]

What?! So what do we do with boundary conditions?

You use them appropriately, the hack that you are supporting (inserting $$\frac{dr}{ds}=0$$ into the ODE) has nothing to do with boundary conditions.

Irrelevant and totally nonsense!

Not really, can you solve the exerciise or not?

 

[starthaus]

Then stop hacking more. You've done enough of that!

AB

What hacks? I got the same exact results as you did, in 1/3 of computations and six days (6/19 vs. 6/25) and 80 posts (post 53 vs. post 132) ahead of you.

 

[espen180]

Is the following correct? I'm working on the problem of pure radial motion. I have confidence in my calculation but want to confirm it before I continue.

$$\frac{d^2r}{dt^2}=\frac{d}{dt}\left(\frac{dr}{dt}\right)=\frac{d\tau}{dt}\cdot\frac{d}{d\tau}\left(\frac{d\tau}{dt}\cdot\frac{dr}{d\tau}\right)=\frac{d\tau}{dt}\left(\frac{d\tau}{dt}\cdot\frac{d^2r}{d\tau^2}+\frac{dr}{d\tau}\cdot\frac{d}{d\tau}\left(\frac{d\tau}{dt}\right)\right)=\left(\frac{d\tau}{dt}\right)^2\left(\frac{d^2r}{d\tau}+\frac{dr}{d\tau}\cdot\frac{d^2\tau}{dt^2}\right)$$

If this is correct, the only obstacle I still have to counter is the term $$\frac{d^2\tau}{dt^2}$$.

 

[starthaus]

Is the following correct? I'm working on the problem of pure radial motion. I have confidence in my calculation but want to confirm it before I continue.

$$\frac{d^2r}{dt^2}=\frac{d}{dt}\left(\frac{dr}{dt}\right) =\frac{d\tau}{dt}\cdot\frac{d}{d\tau}\left(\frac{d\tau}{dt}\cdot\frac{dr}{d\tau}\right)=\frac{d\tau}{dt}\left(\frac{d\tau}{dt}\cdot\frac{d^2r}{d\tau^2}+\frac{dr}{d\tau}\cdot\frac{d}{d\tau}\left(\frac{d\tau}{dt}\right)\right)$$

Correct.

$$=\left(\frac{d\tau}{dt}\right)^2\left(\frac{d^2r}{d\tau}+\frac{dr}{d\tau}\cdot\frac{d^2\tau}{dt^2}\right)$$
.

Incorrect

 

[Al68]

at that x...but NOWHERE ELSE in the domain of definition of f(x). This is the part that you, kev, Al68, Altabeh seem unable to grasp.

LOL. Every time someone specifies "at that x", you misread it as "at every x", then claim they are "unable to grasp" that something is true "at that x" but not "at every x"? Why do you insist on saying such illogical nonsense?

And I already know I'm a troll that can't even spell cawkyoulous, so you don't need to bother yourself with telling me again.

 

[espen180]

Incorrect

I see the mistake. Then,

$$\frac{d^2r}{dt^2}=\left(\frac{d\tau}{dt}\right)^2\frac{d^2r}{d\tau}+\frac{dr}{d\tau}\cdot\frac{d^2\tau}{dt^2}$$

 

[Altabeh]

You use them appropriately, the hack that you are supporting (inserting $$\frac{dr}{ds}=0$$ into the ODE) has nothing to do with boundary conditions.

Refer to the sources I provided you with. The lack of awareness of "boundary conditions" in your language is completely felt. For another example of boundary conditions and how they can be introduced in non-linear DE see last few posts in https://www.physicsforums.com/showthread.php?t=402515".

Not really, can you solve the exerciise or not?

I don't know since when "nonsense" has been translated into "exercise" but at least I know this has something to do with your leaky logics!

What hacks? I got the same exact results as you did, in 1/3 of computations and six days (6/19 vs. 6/25) and 80 posts (post 53 vs. post 132) ahead of you.

First off, you've not proven anything nor have gotten you any result that I got in my post #132. In the post #53 I see just a couple of wishy-washy equations that do not by any means seem to be giving us the same result I gave for the orbital velocity of a particle in circular motion so don't attach my formula to your nonsense equations.

Second off, the hack that you're following here is the following:

$$dr/ds=0\Rightarrow d^2r/ds^2=0$$

As was given a counter-example to, I don't see any reason to take into account your nonsense claims in support of that.

Third off, if your work was really of any help to this thread, we wouldn't see this thread get stretched to this page.

You better quit your hacks now and rather stick to the derivation given in post #132. It seems like each time you make mistakes, all you do is to find a escape route (read dead-end) to get out of the pressure we impose on you to get corrected. When you feel like you've made a mistake, you are supposed to stand corrected not to be coming at us for why we put our finger at your mistake.

AB

 

[Altabeh]

Is the following correct? I'm working on the problem of pure radial motion. I have confidence in my calculation but want to confirm it before I continue.

$$\frac{d^2r}{dt^2}=\frac{d}{dt}\left(\frac{dr}{dt}\right)=\frac{d\tau}{dt}\cdot\frac{d}{d\tau}\left(\frac{d\tau}{dt}\cdot\frac{dr}{d\tau}\right)=\frac{d\tau}{dt}\left(\frac{d\tau}{dt}\cdot\frac{d^2r}{d\tau^2}+\frac{dr}{d\tau}\cdot\frac{d}{d\tau}\left(\frac{d\tau}{dt}\right)\right)=\left(\frac{d\tau}{dt}\right)^2\left(\frac{d^2r}{d\tau}+\frac{dr}{d\tau}\cdot\frac{d^2\tau}{dt^2}\right)$$

If this is correct, the only obstacle I still have to counter is the term $$\frac{d^2\tau}{dt^2}$$.

You donn't need to do these calculations when there are tons of counter-examples to the fallacious claim that supports $$dr/ds=0\Rightarrow d^2r/ds^2=0.$$

AB

 

[espen180]

You donn't need to do these calculations when there are tons of counter-examples to the fallacious claim that supports $$dr/ds=0\Rightarrow d^2r/ds^2=0.$$

AB

Ah, that's not why I'm doing the calculation. I was able to calculate $$\frac{d^2r}{d\tau^2}$$ for a particle in radial motion and want to calculate $$\frac{d^2r}{dt^2}$$.

 

[starthaus]

First off, you've not proven anything nor have gotten you any result that I got in my post #132.

Sure I did, you are just 6 days and 80 posts late. Or, should I say, 15 days and 130 posts late? Your derivation rediscovers my posts 4 and 6.

In the post #53 I see just a couple of wishy-washy equations that do not by any means seem to be giving us the same result I gave for the orbital velocity of a particle in circular motion so don't attach my formula to your nonsense equations.

It is trivial to get the orbital equation in much fewer steps than you needed.

Second off, the hack that you're following here is the following:

$$dr/ds=0\Rightarrow d^2r/ds^2=0$$

No, I am not using this hack in any of my derivations, I am just pointing out that you shouldn't be using it.

Third off, if your work was really of any help to this thread, we wouldn't see this thread get stretched to this page.

It is not my fault that certain participants (including you) have such a hard time admitting that they are using hacks. <shrug>
Practically, this thread should have ended at post 53 where I gave the general solution. It got stretched because of your failure to undestand that putting in $$\frac{dr}{ds}=0$$ by hand into the ODE describing the equation of motion is the hack. Looks like bot espen180 und kev understood, why do you have such a hard time understanding?

You better quit your hacks now and rather stick to the derivation given in post #132.

LOL, or what? Your post 132 is nothing but my post 53 only 6 days late.

 

[starthaus]

I see the mistake. Then,

$$\frac{d^2r}{dt^2}=\left(\frac{d\tau}{dt}\right)^2\frac{d^2r}{d\tau}+\frac{dr}{d\tau}\cdot\frac{d^2\tau}{dt^2}$$

Better$$\frac{d^2r}{dt^2}=\left(\frac{d\tau}{dt}\right)^2\frac{d^2r}{d\tau^2}+\frac{dr}{d\tau}\cdot\frac{d^2\tau}{dt^2}$$

 

[espen180]

Alright. I have made an attempt to derive an expression for $$\frac{\text{d}^2r}{\text{d}t^2}$$.

Please see section 4 (Pages 5&6) in this document for the derivation and result.
Download

I'm guessing the expression is not fully simplified yet.

EDIT: I made a dumb error (missing an exponent) which propagated and ruined the derivation.

 

[Altabeh]

Seeing that again you make use of your hacks like $$v_p=r\frac{d\phi}{d\tau}$$ is the proper speed, is not weird at all. Such hacks that suffer not having a "physical mold" are to blame for my derivation being long in parts. You first tell us how your hack here "proper speed" is derived. Then I can argue which way is more useful. (Of course your hacks don't leave a room for a comparison.)

The only way to get out of this mess is to use the method introduced in
http://www.astro.umd.edu/~miller/teaching/astr498/lecture10.pdf at page 5. And this makes it 10 times more complicated than my method in post #132. Such fallacious claims, though are giving the same result, are to be considered as a shortcut in your leaky logic. But you shoud read sometimes good papers which provide you with the knowledge required for the issue you're involved with.

AB

 

[starthaus]

Alright. I have made an attempt to derive an expression for $$\frac{\text{d}^2r}{\text{d}t^2}$$.

Please see section 4 (Pages 5&6) in this document for the derivation and result.
Download

I'm guessing the expression is not fully simplified yet.

(35) is correct (you already knew that).
Starting from (37) is incorrect.

 

[Altabeh]

Ah, that's not why I'm doing the calculation. I was able to calculate $$\frac{d^2r}{d\tau^2}$$ for a particle in radial motion and want to calculate $$\frac{d^2r}{dt^2}$$.

Oh my bad! I thought you were effected by that fallacious result.

AB

 

[espen180]

I corrected the error I made in my previous attempt and made a new derivation from scratch.

Please see section 4 (Pages 5&6) in this document for the derivation and result.
Download

It is still not complete (there is a $\frac{dr}{d\tau}$-term in k).

 

[starthaus]

I corrected the error I made in my previous attempt and made a new derivation from scratch.

Please see section 4 (Pages 5&6) in this document for the derivation and result.
Download

It is still not complete (there is a $\frac{dr}{d\tau}$-term in k).

(30) is wrong.
(37) is as wrong as before.

 

[yuiop]

... the lagrangian method has produced the same exact equation of motion as the geodesic method, as it should.

Your attempt at the Langragian method does not produce the same result as the geodesic method. You messed up.

... Your post 132 is nothing but my post 53 only 6 days late.

Your post #53 might be earlier but it is simply wrong. You got the wrong result.

This is what Espen got from the geodesic method in https://www.physicsforums.com/showpost.php?p=2769019&postcount=56" :

$$d^2r/ds^2 + m\alpha/ r^2 (dt/ds)^2 - m/(\alpha r^2)(dr/ds)^2 - r\alpha (d\phi/ds)^2 = 0$$

What you got in https://www.physicsforums.com/showpost.php?p=2768235&postcount=53" was:

$$d^2r/ds^2 + m\alpha/ r^2 (dt/ds)^2 - 3m/(\alpha r^2)(dr/ds)^2 - r\alpha (d\phi/ds)^2 = 0$$

Your (dr/ds)^2 differs from that in Espen's solution by a factor of 3. Your solution is not the same and not correct.

From the Schwarzschild solution, the following relation can be directly obtained:

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

Inserting this expression for (dt/ds)^2 into the two solutions gives the following simplified, but still fully generalised solutions:

Espen solution:

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

Starthaus solution:

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

It is well known result that when purely radial motion is considered ($d\phi/ds=0$) that the following is true:

$$\frac{d^2r}{ds^2} = -\frac{m}{r^2}$$

The Starthaus solution fails this test.

Now it can also be seen from either solution that for purely radial motion that the above equation is true when dr/ds=0. Now you claim that $d^2r/ds^2=0$ when dr/ds=0 so according to your claim, the following must be true at the apogee:

$$0 = -\frac{m}{r^2}$$

Clearly this is not true. This is yet another proof that your claim that $$dr/ds=0\Rightarrow d^2r/ds^2=0$$ is false.

You are clearly wrong on two major issues in this thread and yet you still maintain that you are the tutor and everyone else in this thread is your student. Clearly you have as much credibility as the England football manager or the linesman in the England-Germany game.

 

[yuiop]

Since Starhaus and Espen have obtained two different solutions to the same problem a tie braker is required.

Here is a simple alternative derivation.

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

$$ds^2=\alpha dt^2-\frac{1}{\alpha}dr^2-r^2d\phi^2$$

$$\alpha=1-\frac{2m}{r}$$

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:

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

and

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

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

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

Differentiate both sides with respect to s:

$$\frac{d^2r}{ds^2} = -\frac{m^2}{r^2} + \frac{H^2}{r^4}(r-3m)$$

Re-insert the full form of H back into the equation:

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

This is the same as the result obtained by Espen and is valid for radial or orbital motion.

 

[starthaus]

Your attempt at the Langragian method does not produce the same result as the geodesic method. You messed up.

Sure it does, the fact that you can't do a simple differentiation produces all your confusion.

Your post #53 might be earlier but it is simply wrong. You got the wrong result.This is what Espen got from the geodesic method in https://www.physicsforums.com/showpost.php?p=2769019&postcount=56" :

$$d^2r/ds^2 + m\alpha/ r^2 (dt/ds)^2 - m/(\alpha r^2)(dr/ds)^2 - r\alpha (d\phi/ds)^2 = 0$$

What you got in https://www.physicsforums.com/showpost.php?p=2768235&postcount=53" was:

$$d^2r/ds^2 + m\alpha/ r^2 (dt/ds)^2 - 3m/(\alpha r^2)(dr/ds)^2 - r\alpha (d\phi/ds)^2 = 0$$

Your (dr/ds)^2 differs from that in Espen's solution by a factor of 3.

Err , no. Why don't you show your calculations step by step and I'll show you where you made your mistake.

Your solution is not the same and not correct.

Err, no. Like I said, you need to learn how to calculate differentials. Using software packages is not going to cut it. You need to use the hint I gave you in post 136.

 

[starthaus]

Since Starhaus and Espen have obtained two different solutions to the same problem a tie braker is required.

Nope, we obtained the same exact equation of motion, the fact that you can't recognize it is a problem created by your inability to differentiate correctly.

Re-insert the full form of H back into the equation:

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

This is the same as the result obtained by Espen and is valid for radial or orbital motion.

A. It DOES NOT work for radial motion (make $$\frac{d\phi^2}{ds^2}=0$$ and you'll see why)
B. It is also clearly wrong for circular orbits. Try figuring out $$\frac{d\phi}{ds}$$ and you'll quickly see why your derivation is wrong. I'll let you figure out where you made the errors. You can always find the correct equation in post 139.

 

[yuiop]

Re-insert the full form of H back into the equation:

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

A. It DOES NOT work for radial motion (make $$\frac{d\phi^2}{ds^2}=0$$ and you'll see why)

This is just silly. When $$\frac{d\phi^2}{ds^2}=0$$ the result is obviously:

$$\frac{d^2r}{ds^2} = -\frac{m^2}{r^2}$$

This is the same as a result you quoted in #220 of https://www.physicsforums.com/showthread.php?t=402135&page=14.

It is also the same as the result you quoted in #128 of the current thread:

If yyou do it right, you should get:

$$\frac{d^2r}{ds^2}=-\frac{m}{r^2}$$

It is also the result quoted by mathpages http://www.mathpages.com/rr/s6-07/6-07.htm (first equation.)

B. It is also clearly wrong for circular orbits. Try figuring out $$\frac{d\phi}{ds}$$ and you'll quickly see why your derivation is wrong.

OK. Start with:

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

Solve:

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

For circular motion the radius is constant and so the radial acceleration must be zero $(d^2r)/(ds^2)=0$ so in this limited case:

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

When r=3m (the photon orbit):

$$\frac{d\phi}{ds} = \infty$$

This is the expected result because for a particle moving at the speed of light, ds=0.

For r<3m the result is an imaginary number, indicating that it is not possible to have a circular orbit below r=3m. This is also a well known result.

 

[starthaus]

This is just silly. When $$\frac{d\phi^2}{ds^2}=0$$ the result is obviously:

$$\frac{d^2r}{ds^2} = -\frac{m^2}{r^2}$$

...which is dead wrong

This is the same as a result you quoted in #220 of https://www.physicsforums.com/showthread.php?t=402135&page=14.

It is also the same as the result you quoted in #128 of the current thread:
It is also the result quoted by mathpages http://www.mathpages.com/rr/s6-07/6-07.htm (first equation.)

Nope, look again. Check your math.

 

[starthaus]

OK. Start with:

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

Solve:

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

For circular motion the radius is constant and so the radial acceleration must be zero $(d^2r)/(ds^2)=0$ so in this limited case:

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

Even worse than your answer for radial motion. It is time you stopped your hacking and you started studying.

 

[espen180]

Since Starhaus and Espen have obtained two different solutions to the same problem a tie braker is required.

Here is a simple alternative derivation.

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

$$ds^2=\alpha dt^2-\frac{1}{\alpha}dr^2-r^2d\phi^2$$

$$\alpha=1-\frac{2m}{r}$$

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:

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

and

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

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

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

Differentiate both sides with respect to s:

$$\frac{d^2r}{ds^2} = -\frac{m^2}{r^2} + \frac{H^2}{r^4}(r-3m)$$

Re-insert the full form of H back into the equation:

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

This is the same as the result obtained by Espen and is valid for radial or orbital motion.

You may have made a mistake somewhere, since you have the mass squared instead of just the mass.

 

[starthaus]

You may have made a mistake somewhere, since you have the mass squared instead of just the mass.

This is not the only mistake, the whole "method" is invalid, resulting into bogus results. Can you figure out at what step, in his "derivation", does kev introduce his error? It is pretty gross, so it shouldn't be difficult to spot.