Gravitational force and acceleration in General Relativity.

[starthaus]

You know that is not true, as I have asked you many times to derive the proper and coordinate acceleration of a stationary particle from the equations of motion, as you claim to be able to do.
Lets see it then!

For your convenience, here is the start of your derivation;

The first two equations are your own personal hacks, I never wrote them.Let me give you the correct start that solves the problem in three lines:

$$\vec{F}=-grad\Phi$$

where $$\Phi$$ is the potential that you can read straight off the metric. You have two more lines to write, see if you can do it all by yourself, I have already shown you how to do that for centripetal acceleration, you only need to copy the method.

 

[yuiop]

That exact equation is given here http://www.mathpages.com/rr/s6-07/6-07.htm

Do you think that reference is wrong? What do you think the equation should be?

The correct result has already been derived eons ago by using covariant derivatives. Look up Dalespam's post.

The equation given in the reference is for the coordinate acceleration of a free falling particle. It is not wrong. It differs from Dalespam's result, because his result is for the proper acceleration of a stationary particle, which is more relevant to this thread. Both equations are right in their own contexts.

 

[yuiop]

The first two equations are your own personal hacks, I never wrote them.

They are direct copy and pastes, word for word from your posts https://www.physicsforums.com/showpost.php?p=2710570&postcount=2" of this thread.

 

[Tomsk]

Noob question here: why is the lagrangian
$$L = \frac{1}{2}g_{ab}\dot{x}^a\dot{x}^b$$
? Shouldn't it be
$$L = \sqrt{g_{ab}\dot{x}^a\dot{x}^b}$$
since it's the path length we want to maximize?

I tried it that way and got different equations of motion. I'm not sure what to do with them now though! I think I prefer the derivation posted by DaleSpam a couple of pages ago.

 

[yuiop]

I think I prefer the derivation posted by DaleSpam a couple of pages ago.

Just curious. Prefer it to what? So far Dalespam is the only person to give a derivation of the acceleration of a stationary particle, in this thread.

 

[Tomsk]

Just curious. Prefer it to what? So far Dalespam is the only person to give a derivation of the acceleration of a stationary particle, in this thread.

To getting it from the lagrangian. I nearly did it but I was just saying the one with 4-vectors looks much easier.

 

[yuiop]

Let me give you the correct start that solves the problem in three lines:

$$\vec{F}=-grad\Phi$$

where $$\Phi$$ is the potential that you can read straight off the metric. You have two more lines to write, see if you can do it all by yourself, I have already shown you how to do that for centripetal acceleration, you only need to copy the method.

I see you now want to move the goal posts. Back in #2 and #4 you started a derivation using the Lagrangian and "left it as an exercise" for me to complete. That normally implies that you know how complete the derivation starting with the Lagrangian and the equations of motion, but despite many requests you have been unable to do so and now you want to do it using covariant derivatives (already done by Dalespam) or using potentials. Now if you want to do it doing using potentials that is fine and I would be glad to see it because I have never aproached the problem from that angle, but for future reference, if you set people "exercises" you should be able to complete the exercise yourself.

 

[Tomsk]

I managed the lagrangian derivation, the answer I got was
$$\frac{d^2 r}{dt^2} = -\frac{m}{r^2}\alpha\left( 1 - \frac{3}{\alpha^2} \left( \frac{dr}{dt}\right)^2\right)$$
I think that agrees with the other results in this thread, depending on the initial velocity. Does that seem right to anyone? I am not quite sure how to interpret it though. This isn't a proper acceleration is it? So it only applies to free falling particles. But is it equal and opposite to the proper acceleration required to stay a fixed distance from the black hole as well?

 

[yuiop]

I managed the lagrangian derivation, the answer I got was
$$\frac{d^2 r}{dt^2} = -\frac{m}{r^2}\alpha\left( 1 - \frac{3}{\alpha^2} \left( \frac{dr}{dt}\right)^2\right)$$
I think that agrees with the other results in this thread, depending on the initial velocity. Does that seem right to anyone? I am not quite sure how to interpret it though. This isn't a proper acceleration is it? So it only applies to free falling particles. But is it equal and opposite to the proper acceleration required to stay a fixed distance from the black hole as well?

I too completed the "lagrangian derivation" and sent copies of my derivation to Dalespam and DrGreg to check over for me. I will post the derivation here in a day or two, but I thought I would first give Starthaus the opportunity to show that he is not full of "it" and show he can complete the derivation he started and left for us to complete as "an exercise".

 

[starthaus]

This is very wrong. You need to start with the metric:$$ds^2=\alpha dt^2-\frac{1}{\alpha}dr^2-r^2d\phi^2$$

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

From this you construct the Lagrangian:

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

From the above Lagrangian, you get immediately the equations of motion:

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

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

whre h,k are constants.

The correct Lagrangian (see above) is derived directly from the metric.
From the Lagrangian, you get the equation of motion (see above).

There is a third Euler-Lagrange equation, that I have left for last:

$$2\frac{d}{ds} (1/\alpha \frac{dr}{ds})-[\frac{2m}{r^2}(\frac{dt}{ds})^2-(\frac{dr}{ds})^2 \frac{d}{dr}(1/\alpha)-2r(\frac{d\phi}{ds})^2]=0$$

The above, for $$\frac{dr}{ds}=0$$ produces:

$$\omega^2=\frac{m}{r^3}$$

i.e. Kepler third law.

For a complete derivation of acceleration in rotating frames, see the other thread.

On the other hand, if the motion is radial, i.e. $$\frac{d\phi}{ds}=0$$ the equation of motion becomes:

$$2\frac{d}{ds} (1/\alpha \frac{dr}{ds})-[\frac{2m}{r^2}(\frac{dt}{ds})^2-(\frac{dr}{ds})^2 \frac{d}{dr}(1/\alpha)]=0$$

 

[yuiop]

The correct result has already been derived eons ago by using covariant derivatives. Look up Dalespam's post.

You mean the correct result obtained by Dalespam that agrees with the result I gave in #1 right?

...

On the other hand, if the motion is radial, i.e. $$\frac{d\phi}{ds}=0$$ the equation of motion becomes:

$$2\frac{d}{ds} (1/\alpha \frac{dr}{ds})-[\frac{2m}{r^2}(\frac{dt}{ds})^2-(\frac{dr}{ds})^2 \frac{d}{dr}(1/\alpha)]=0$$

Your derivation is incomplete. You have not shown that you can apply the equations you have mindlessly copied from a textbook to a physical situation. You should be able to answer the questions posed in the OP and give answers in Newtons for force or meters/second^2 for acceleration.

... we should be be able to answer simple questions like "what does General Relativity predict that the weight of a 1kg mass on the surface of a very massive gravitational body to be?" or "what is the initial acceleration of 1kg mass when released from a short distance above a very massive gravitational body?"

I can answer those questions unambiguously. Can you?

 

[starthaus]

You mean the correct result obtained by Dalespam that agrees with the result I gave in #1 right?

Physics is not putting in results by hand. You need to learn how to derive them, kev. For this, you need to learn.

Your derivation is incomplete.

I am trying to teach you how to derive results (instead of putting them in by hand, as you have been doing), so I left the easy part for you, as an exercise. You still have no clue how to derive the equations of motion from the Lagrangian, eh?

 

[yuiop]

Your derivation is incomplete. You have not shown that you can apply the equations you have mindlessly copied from a textbook to a physical situation. You should be able to answer the questions posed in the OP and give answers in Newtons for force or meters/second^2 for acceleration. I can answer those questions unambiguously. Can you?

Physics is not putting in results by hand. You need to learn how to derive them, kev. For this, you need to learn.

I am trying to teach you how to derive results (instead of putting them in by hand, as you have been doing), so I left the easy part for you, as an exercise. ...

You are missing the point. As I said in post 79, I have already derived the equations and Dalespam and DrGreg hold copies of that derivation, as proof that I have done so. I have given you over a week to show you can complete your own "exercises" and prove yourself worthy of being "my teacher" and not just some guy trying to look clever by posting exercises from a textbook that he has no idea how to complete. So far you have not completed any of the exercises or derivations you have posted in any thread, that are not already completed in the textbooks you quote from. I am still waiting and so is Tomsk who would like to see the derivation.

 

[starthaus]

You are missing the point. As I said in post 79, I have already derived the equations and Dalespam and DrGreg hold copies of that derivation, as proof that I have done so. I have given you over a week to show you can complete your own "exercises" and prove yourself worthy of being "my teacher" and not just some guy trying to look clever by posting exercises from a textbook that he has no idea how to complete. So far you have not completed any of the exercises or derivations you have posted in any thread,

LOL

that are not already completed in the textbooks you quote from. I am still waiting and so is Tomsk who would like to see the derivation.

I don't think he's waiting on anything, he knows what to do.

 

[yuiop]

Starthaus: If you don't want want to post the derivation because you are worried that I will miss the learning opportunity of deriving for myself, then you have no need to worry because I have availed myself of the opportunity and done just that. You now no longer have any excuses not to publish the completion of the exercise you set and the I suspect the reason you are stalling, is that you do not know how to to complete your own exercise.

P.S. Tomsk clearly expressed an interest in #78.

 

[starthaus]

Starthaus: If you do want want to post the derivation because you are worried that I will miss the learning opportunity

No, I am not worried at all. It is clear that you don't know what to do with the Lagrangian.

 

[yuiop]

No, I am not worried at all. It is clear that you don't know what to do with the Lagrangian.

I have done the derivation from the initial information (What you call "the hard work" and what I call quoting from a textbook.) that you provided on the first page of this thread. It seems you want to move the goal posts again. All I am asking you to do is complete ANY of the exercises/derivations that have started and show you can end up with the equations given in #1.

 

[starthaus]

I have done the derivation from the initial information (What you call "the hard work" and what I call quoting from a textbook.)

No, you haven't. That would require knowledge of the Euler-Lagrange formalism of which you have none. Otherwise, you would have derived the equations of motion from the Lagrangian I derived for you long,long ago.

 

[yuiop]

No, you haven't. That would require knowledge of the Euler-Lagrange formalism of which you have none.

Yes I have and Dalespam and DrGreg have copies of my derivation to prove it. If you publish your derivation and it is different to mine I will learn from the different approach. If you show your derivation (if you are in fact able to do it) I will post my derivation and if it is wrong wrong or flawed, I will learn from that too. You claim your purpose here is to educate me, but so far I have seen no sign of you trying to be genuinely helpful and your posts are just confrontational and inflamatory as possible. Does the word "troll" mean anything to you?

 

[starthaus]

You claim your purpose here is to educate me,

Education doesn't mean doing all the homework for you, you need to learn how to do it yourself.

 

[starthaus]

your posts are just confrontational and inflamatory as possible. Does the word "troll" mean anything to you?

You mean like this:

Your derivation is incomplete. You have not shown that you can apply the equations you have mindlessly copied from a textbook to a physical situation.

:-)

 

[yuiop]

Education doesn't mean doing all the homework for you, you need to learn how to do it yourself.

I keep telling you I have already done "the homework" and two people on this forum have seen it. I am asking you to show you have the right to be a self proclaimed guru/teacher on this forum by showing you can complete your own homework. We can all say "the completion of the derivation is left as an exercise for the reader". This is your opportunity to show that you have the guru status you are trying to portray and are not just full of hot air.

 

[starthaus]

I keep telling you I have already done "the homework" and two people on this forum have seen it.

You mean, you finally cleaned up the mess from post 1? Good for you.

 

[starthaus]

The first two equations are your own personal hacks, I never wrote them.Let me give you the correct start that solves the problem in three lines:

$$\vec{F}=-grad\Phi$$

where $$\Phi$$ is the potential that you can read straight off the metric. You have two more lines to write, see if you can do it all by yourself, I have already shown you how to do that for centripetal acceleration, you only need to copy the method.

Ok, I see that you don't want to make the effort to learn the Euler-Lagrange formalism, so we can use the simple method I showed for reading the potential straight off the metric.

Here is the second line:

$$e^{2\Phi/c^2}=1-\frac{2m}{r}$$

You have one line to write in order to find the correct expression of the force.

 

[yuiop]

You mean, you finally cleaned up the mess from post 1? Good for you.

There you go, being all confrontational and aggressive again without looking into where and why we differ and seeing if in fact our results are in agreement and just expressed in different ways.It does not help that it has taken over 80 posts to discover what you think the results should be.

Let me give you the correct start that solves the problem in three lines:

$$\vec{F}=-grad\Phi$$

where $$\Phi$$ is the potential that you can read straight off the metric. You have two more lines to write, see if you can do it all by yourself, I have already shown you how to do that for centripetal acceleration, you only need to copy the method.

Ok, I see that you don't want to make the effort to learn the Euler-Lagrange formalism, so we can use the simple method I showed for reading the potential straight off the metric.

Here is the second line:

$$e^{2\Phi/c^2}=1-\frac{2m}{r}$$

You have one line to write in order to find the correct expression of the force.

OK, straight up, I am not familiar with the formalism you are using, as I do not come across it very often. If you care to state the exact titles and ISBN's of the Gron and Rindler books (or the Amazon links), I will order them and at least then we will be using the same references and notation and perhaps we might understand each other better.

Now for your derivation.

The strong field aproximation metric is:

$$ds^2= \left(e^{\frac{2\Phi}{c^2}}\right)c^2dt^2 - \left(e^{\frac{2\Phi}{c^2 }}\right)^{-1}dr^2 - r^2(d \theta^2 - \sin^2 \theta d \phi^2)$$

and the Regular Schwarzschild metric is:

$$ds^2= \left(1-\frac{2GM}{rc^2}}\right)c^2dt^2 - \left(1-\frac{2GM}{rc^2 }}\right)^{-1}dr^2 - r^2(d \theta^2 - \sin^2 \theta d \phi^2)$$

From the above two forms of the metric it is easy to make the identity:

$$e^{2\Phi/c^2}=\left(1-\frac{2GM}{rc^2}\right)$$

Solve the above for $$\phi$$:

$$\phi = \frac{1}{2}c^2 \log \left(1-\frac{2GM}{rc^2}\right)$$

Now you state:

$$\vec{F}=-grad\Phi$$

(as does Wikipedia here http://en.wikipedia.org/wiki/Force#Potential_energy)

so:

$$\vec{F}= -grad\Phi = -grad\left(\frac{1}{2}c^2 \log \left(1-\frac{2GM}{rc^2}\right)\right)$$

$$\vec{F} = \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)^{-1}$$

Now your result for coordinate force does not agree with the equations I got for force in #1 quoted below:

The proper force acting on the stationary test mass on the surface of the gravitational body (i.e its weight as measured by a set of weighing scales on the surface) is:

$$F_0 = \frac{GMm_0}{r^2}\left(1-\frac{2GM}{rc^2}\right)^{-\frac{1}{2}} = m_0a_0$$

The Schwarzschild coordinate force as measured by an observer at infinity is:

$$F = \frac{GMm}{r^2}$$

Now before you jump on your "Starthaus is right and Kev is wrong" high horse I think we should explore why we get different results and see if we are saying the same thing in different ways. You should bear in mind that several posters have stated that the equations I gave in #1 agree with equations given by textbooks that they have. I am sure your derivation is also from a textbook, so it likely that both results are right but stated in different ways.

The first thing to observe is that your equation for force fails on dimensional analysis because it does not have the units of force. It would appear your equation has units of acceleration, but even then it does not agree with the equation I gave for coordinate acceleration in #1:

...the initial coordinate acceleration of a test mass released at r is (according to the observer at infinity) is:

$$a= \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)$$

I think when you fix your equation so that it has the correct units of force, rather than units of acceleration and identify who makes the measurements, then we might find some agreement.

 

[starthaus]

There you go, being all confrontational and aggressive again without looking into where and why we differ and seeing if in fact our results are in agreement and just expressed in different ways.It does not help that it has taken over 80 posts to discover what you think the results should be.
OK, straight up, I am not familiar with the formalism you are using, as I do not come across it very often. If you care to state the exact titles and ISBN's of the Gron and Rindler books (or the Amazon links), I will order them and at least then we will be using the same references and notation and perhaps we might understand each other better.

Now for your derivation.

The strong field aproximation metric is:

$$ds^2= \left(e^{\frac{2\Phi}{c^2}}\right)c^2dt^2 - \left(e^{\frac{2\Phi}{c^2 }}\right)^{-1}dr^2 - r^2(d \theta^2 - \sin^2 \theta d \phi^2)$$

and the Regular Schwarzschild metric is:

$$ds^2= \left(1-\frac{2GM}{rc^2}}\right)c^2dt^2 - \left(1-\frac{2GM}{rc^2 }}\right)^{-1}dr^2 - r^2(d \theta^2 - \sin^2 \theta d \phi^2)$$

From the above two forms of the metric it is easy to make the identity:

$$e^{2\Phi/c^2}=\left(1-\frac{2GM}{rc^2}\right)$$

Solve the above for $$\phi$$:

$$\phi = \frac{1}{2}c^2 \log \left(1-\frac{2GM}{rc^2}\right)$$

Now you state:

$$\vec{F}=-grad\Phi$$

(as does Wikipedia here http://en.wikipedia.org/wiki/Force#Potential_energy)

so:

$$\vec{F}= -grad\Phi = -grad\left(\frac{1}{2}c^2 \log \left(1-\frac{2GM}{rc^2}\right)\right)$$

$$\vec{F} = \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)^{-1}$$

Good, you finally did the computations. It is not "my result", it is the "correct result". Now, you need to think a little how you calculated the gradient, this will explain why you got the result you got.

Now your result for coordinate force does not agree with equations I got for force in #1 quoted below.

That's too bad, if you end up buying Rindler, you will find out that, contrary to your post 1 (and to your incorrect claims above),

$$\vec{f} = \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)^{-1}$$

is indeed the coordinate force per unit mass. You can multiply by $$m_0$$ all by yourself. Besides, you have now the results derived rather than put in by hand as you did in post 1. I did not mislead you, I guided you to discovering the correct results.

Now, if you could perform one more calculation (exactly one line) you would also get the correct expression for the proper force.

 

[starthaus]

The Schwarzschild coordinate force as measured by an observer at infinity is:

$$F = \frac{GMm}{r^2}$$

Nope, I just guided you through the correct derivation, results put in by hand are not physics. Besides, they are most often likely to be wrong.

 

[yuiop]

$$\vec{f} = \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)^{-1}$$

is indeed the coordinate force per unit mass. You can multiply by $$m_0$$ all by yourself. Besides, you have now the results derived rather than put in by hand as you did in post 1. I did not mislead you, I guided you to discovering the correct results.

Now, if you could perform one more calculation (exactly one line) you would also get the correct expression for the proper force.

I am afraid I have no idea of how to obtain the proper force from your coordinate force equation, as I am unable to relate to the physical meaning of your equation. It differs by orders of magnitude from my equations (which others in this forum say are correct). If you state what you and Rindler claim the proper force to be, (and how it is obtained) I will try and figure out why our equations differ.

 

[starthaus]

It differs by orders of magnitude from my equations (which others in this forum say are correct).

They aren't. The one that you posted in post 1 is wrong, the one that you derived following my hints is correct. You are just a few steps away from getting the correct results.

If you state what you and Rindler claim the proper force to be, (and how it is obtained) I will try and figure out why our equations differ.

I gave you a hint. You need to look at how you calculated the gradient, $$grad \Phi$$ for $\Phi = \frac{1}{2}c^2 \log \left(1-\frac{2GM}{rc^2}\right)$. How did you do it?

 

[yuiop]

I gave you a hint. You need to look at how you calculated the gradient, $$grad \Phi$$ for $\Phi = \frac{1}{2}c^2 \log \left(1-\frac{2GM}{rc^2}\right)$. How did you do it?

Force is the differential of potential so I took the differential with respect to r. Put another way, the differential of a curve is the gradiant of the curve. How does that help?

 

[starthaus]

Force is the differential of potential so I took the differential with respect to r. Put another way, the differential of a curve is the gradiant of the curve. How does that help?

So, you calculated $$\frac{d\Phi}{dr}$$ and you obtained the coordinate acceleration. What do you need to do in oder to get the proper acceleration?

 

[yuiop]

So, you calculated $$\frac{d\Phi}{dr}$$ and you obtained the coordinate acceleration. What do you need to do in oder to get the proper acceleration?

Are you saying that the proper acceleration is:

$$\frac{d\Phi}{dr'} = \frac{d\Phi}{dr/\sqrt{(1-2M/r)}} = \frac{M}{r^2}\left(1-\frac{2M}{r}\right)^{-1/2}$$

(assuming $$dr/dr' = \sqrt{(1-2M/r)}$$ where dr/dr' is the ratio of coordinate length to local (proper) length in the radial direction)?

 

[starthaus]

Are you saying that the proper acceleration is:

$$\frac{d\Phi}{dr'} = \frac{d\Phi}{dr/\sqrt{(1-2M/R)}} = \frac{GM}{r^2}\left(1-\frac{2GM}{R}\right)^{-1/2}$$

(assuming $$dr/dr' = \sqrt{(1-2M/R)}$$ where dr/dr' is the ratio of coordinate length to local (proper) length in the radial direction)?

Almost. You still have errors:

$$dr/dr' = \sqrt{(1-2M/r)}$$

$$\frac{d\Phi}{dr'} = \frac{d\Phi}{dr/\sqrt{(1-2M/r)}} = \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)^{-1/2}$$

 

[yuiop]

Finally.

Good, your equation for the proper acceleration in a gravitational field agrees with the one I gave in #1. That leaves us with the problem that your coordinate acceleration does not agree with the coordinate accleration in #1. Other posters have said that all the equations I posted in #1 are correct, so we should try and find out why we disagree.

I can not find any references that agree with the equation you have obtained for coordinate acceleration, but I can find some that agree with mine. I have also derived the coordinate acceleration from the information you provided in #8 (quoted below) and the result agrees with the result I get in #1. I have no doubt that you got your equations from a good reference, but you may have misinterpreted or applied them in the wrong context.

From the first equation, we can get immediately:

$$\frac{dt}{ds}=\frac{k}{1-\frac{2m}{r}}$$

From the above, we can get the relationship between proper and coordinate speed:

$$\frac{dr}{ds}=\frac{dr}{dt}\frac{dt}{ds}$$

Differentiate one more time and you will obtain the correct relationship between proper and coordinate acceleration. You can do the same exercise for the angular coordinate , $$\phi$$.

Your equation for the coordinate acceleration gives a value that is larger than the proper acceleration and that is exactly the opposite to the generally perceived wisdom.

 

[starthaus]

Good, your equation for the proper acceleration in a gravitational field agrees with the one I gave in #1. That leaves us with the problem that your coordinate acceleration does not agree with the coordinate accleration in #1.

It doesn't leave "us". It leaves "you" with the fact that your equation in post 1 is wrong. I already guided you how to get the correct equation, why are you insisting? Just to waste time?