Gravitational force and acceleration in General Relativity.

[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?

 

[yuiop]

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]

Your equation for the coordinate acceleration gives a value that is larger than the proper acceleration .

You don't know what you are doing. It is really simple:

a_{proper}=\frac{d\Phi}{dr'}=\frac{d\Phi}{dr}*\frac{dr}{dr'}
a_{coordinate}=\frac{d\Phi}{dr}

If \frac{dr}{dr'}=\sqrt{1-\frac{2GM}{rc^2} by what factor do the two accelerations differ?

and that is exactly the opposite to the generally perceived wisdom

LOL

 

[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)?

You still have a mistake:

\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}

You can't get anything right.

 

[Shackleford]

Man, tough crowd. LOL!

 

[yuiop]

You still have a mistake:

\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}

You can't get anything right.

LOL. Great coming from someone who made a complete hash of their calculations in this thread https://www.physicsforums.com/showthread.php?t=403978&page=5 and when I clearly pointed out your errors in #71 you continued arguing the mistake was mine until finally admitting your error in #81.

There is no error in my calculation above, unless you are making a petty reference to inconsistent use of explicitly stating G and sometimes using G=1. In that case you have made the a similar error of explicitly stating c^2 in the final term and using implied c^2 =1 in the middle term. We are all big boys here and occasionally lapse between using explicit c or G and sometimes using units of G=c=1. Get over it. We all do it. I suspect this is yet another red herring to distract us from a major blunder in your work.

 

[starthaus]

There is no error in my calculation above.

Of course not, only about several errors still linger around, it's excusable, you just learned how to do the correct calculations. Unfortunately, you still have no clue what you have calculated even after I took you through the derivation step by step.

 

[starthaus]

LOL. Great coming from someone who made a complete hash of their calculations in this thread https://www.physicsforums.com/showthread.php?t=403978&page=5 and when I clearly pointed out your errors in #71 you continued arguing the mistake was mine until finally admitting your error in #81.

There is no error in my calculation above, unless you are making a petty reference to inconsistent use of explicitly stating G and sometimes using G=1. In that case you have made the a similar error of explicitly stating c^2 in the final term and using implied c^2 =1 in the middle term. We are all big boys here and occasionally lapse between using explicit c or G and sometimes using units of G=c=1. Get over it. We all do it. I suspect this is yet another red herring to distract us from a major blunder in your work.

Nah, you are mixing R with r. There is no R. But worse, you have no clue what you have calculated, even after I took you through the derivation step by step.

 

[yuiop]

Nah, you are mixing R with r. There is no R. But worse, you have no clue what you have calculated, even after I took you through the derivation step by step.

Yep, I was right, you were being petty. Sorry for inadvertently using capital R instead of lowercase r. FoR a minute theRe, I thought you had something impoRtant to say.

The important issue is why your equation for coordinate acceleration is so wrong.

 

[starthaus]

Assuming Schwarzschild geometry (uncharged non-rotating body) and defining the gravitational gamma factor as:

\gamma = \frac{1}{\sqrt{1-\frac{2GM}{rc^2}}}a_0 = a \gamma^3

Nope, this is just as wrong as the rest. It is only true for \gamma(v)=\frac{1}{\sqrt{1-(v/c)^2}} (and only when \vec{v} and \vec{a} are co-linear)There is no reason for it to be true for

\gamma = \frac{1}{\sqrt{1-\frac{2GM}{rc^2}}}

Numerology and physics are two very different fields.

 

[starthaus]

The important issue is why your equation for coordinate acceleration is so wrong.

Because you don't know what you are doing, even after someone leads you through the derivation step by step.
You really need to take a break from numerology and reflect on what I explained to you in posts 107 and 114.

 

[Al68]

Your equation for the coordinate acceleration gives a value that is larger than the proper acceleration .

You don't know what you are doing. It is really simple:

a_{proper}=\frac{d\Phi}{dr'}=\frac{d\Phi}{dr}*\frac{dr}{dr'}
a_{coordinate}=\frac{d\Phi}{dr}

If \frac{dr}{dr'}=\sqrt{1-\frac{2GM}{rc^2} by what factor do the two accelerations differ?

So are you claiming coordinate acceleration is greater than proper acceleration or not? Your response here seems to agree with kev's statement.

 

[starthaus]

So are you claiming coordinate acceleration is greater than proper acceleration or not? Your response here seems to agree with kev's statement.

That's what the derivation shows.

 

[yuiop]

Because you don't know what you are doing, even after someone leads you through the derivation step by step.
You really need to take a break from numerology and reflect on what I explained to you in posts 107 and 114.

I can do a derivation based on the exact Schwarzschild solution that gives equations that agree exactly with equations in #1 and disagree significantly with your conclusion based on a field aproximation.

 

[pepeherborn]

Why don' you two guys do the following: You must arrive to Newtons law of Gravitation in the weak field also. Suppose two spherical masses of 1 and 2 kg at a distance of 1m.For example made of iron with density 7.9 g/cm^3.
Write on the left hand side :
m1 = 2 kg
m2 = 1 kg
r1 = 1m
G = 6,67 *10^-11*m^3/(kg*s^2)

F = G* m1*m2/r^2

F = 1.334*10^-10 N

Now write down on the right hand side every single step you need to calculate to get this single number in General relativity including conversions for G = c = 1 etc. So this would settle this discussion (hopefully!)

 

[Dale]

I can do a derivation based on the exact Schwarzschild solution that gives equations that agree exactly with equations in #1 and disagree significantly with your conclusion based on a field aproximation.

Starthaus, I must say that this is the part that troubles me most about the potential approach. It is indeed more elegant, but this field approximation step makes it very suspicious to me. Particularly in light of the results for the rotating reference frame where we found that obtaining the correct answer with the potential approach depended critically on whether you were using a strong-field approximation or a weak-field approximation. We should have been able to tell, from the beginning, that we needed to use the strong-field approximation.

Can you explain more about that step? When is the weak field approximation safe, under what conditions is the strong field approximation required, and when would even the strong field approximation fail? Without that information it seems that we have to do the brute-force approach anyway, just to check if the field approximations were good.