# Gravitational force and acceleration in General Relativity.

#### yuiop

I was pondering this thread https://www.physicsforums.com/showthread.php?t=401713" when it occurred to me that 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?".

Not being able to find a clear, simple definitive answer to that question on the internet, I decided to do some "back of the envelope" calculations and this is what I came up with (using a combination of hunches, intuition and guesses):

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

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

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

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

The initial acceleration of the test mass when released at r, as measured by a local observer at r is:

$$a_0 = \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)^{-\frac{1}{2}} = a \gamma^3$$

This is equal to the proper acceleration of the test mass when it is stationary in the gravitational field at r, as measured by an accelerometer.

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:

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

Note that:

$$a_0 = a\gamma ^3$$

and:

$$F_0 = F = m_0 a_0$$

just as in the linear analogue in Special Relativity for longitudinal acceleration and force.

Does that seem about right? Can anyone find any solutions in "the literature"?

P.S. I have not shown my working or referenced any sources that I based my calculations on, but I can if anyone is interested.

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#### starthaus

I was pondering this thread https://www.physicsforums.com/showthread.php?t=401713" when it occurred to me that 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?".

Not being able to find a clear, simple definitive answer to that question on the internet, I decided to do some "back of the envelope" calculations and this is what I came up with (using a combination of hunches, intuition and guesses):

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

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

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

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

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

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

From this you construct the Lagrangian:

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

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.

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#### yuiop

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

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

From this you construct the Lagrangian:

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

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.
and what do you conclude that the coordinate and proper acceleration should be?

#### starthaus

and what do you conclude that the coordinate and proper acceleration should be?
I derived for you the equations of motion, I left this as an exercise for you to calculate.

#### yuiop

I derived for you the equations of motion, I left this as an exercise for you to calculate.
I have shown my conclusions, so why don't you show your conclusions and then we will see where we differ? That is of course assuming you know how to calculate the proper and coordinate acceleration from the Lagrangian.

You may not have noticed, but my original post is about particles that are stationary or very nearly stationary (eg a free falling particle near its apogee). I do not know why you always need to ovecomplicate things by introducing things that were not in the original post like horizontal/orbital motion. To get to the crux of the matter, you need to simplify things as much as possible, so set $d\phi = 0$ and forget about it.

Remember I said in the OP that I am looking for a clear simple answer. If you have not got one, you might as well stay out of this thread. There are plenty of general solutions that cover every possible permutation of variables, but I am trying to cut out the crap. I suggest you do the same.

Remember Schwarzschild found the first correct solution to the EFEs by assuming charge = 0 and angular momentum of the gravitational body = 0. That is the way to go. Take the simplest case first.

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#### Ich

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#### starthaus

That is of course assuming you know how to calculate the proper and coordinate acceleration from the Lagrangian.
You don't calculate it from the Lagrangian, you calculate it from the equations of motion. It is a simple exercise in calculus, this is why I left it for you. I did all the heavy lifting.

You may not have noticed, but my original post is about particles that are stationary or very nearly stationary (eg a free falling particle near its apogee). I do not know why you always need to ovecomplicate things by introducing things that were not in the original post like horizontal/orbital motion. To get to the crux of the matter, you need to simplify things as much as possible, so set $d\phi = 0$ and forget about it.
It is not correct to "set $d\phi = 0$ and forget about it. "The simple reason is that $\phi$ is variable

Remember I said in the OP that I am looking for a clear simple answer. If you have not got one, you might as well stay out of this thread. There are plenty of general solutions that cover every possible permutation of variables, but I am trying to cut out the crap. I suggest you do the same.
You are becoming abusive every time you are shown wrong.

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#### starthaus

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.
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 , $$\phi$$.

#### DrGreg

Gold Member
Prof Nick Woodhouse of Oxford University shows http://people.maths.ox.ac.uk/~nwoodh/gr/index.html [Broken] (Section 12.1) that the "acceleration due to gravity", i.e. the proper acceleration of a hovering object, is

$$\frac{M}{r^2\sqrt{1 - 2M/r}}$$​

(in units where G=c=1) although the method he uses requires you to be familiar with covariant differentiation.

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#### LukeD

We did this calculation in my GR class last semester. It took me a bit of time to find my book, but as I suspected (the answer is given on p.261 of Hartle with other relevant formulas scattered throughout the chapters), every formula that you wrote is exactly correct.

You are becoming abusive every time you are shown wrong.
Spoken with the certainty of someone who didn't bother to check if there were any differences between what you two wrote! There aren't of course. You're both absolutely correct. (You may have missed where kev said that he wasn't reproducing his work)
By the way, the Lagrangian is independent of angular momentum, so by conservation of angular momentum (your $$h$$), we can in fact just set $$\mtext{d}\phi = 0$$. This is because if we start on a geodesic with $$\frac{\mtext{d}\phi}{\mtext{d}\tau} = 0$$, this remains true for all time. It is a boundary condition on our motion that imposes no extra forces.

#### starthaus

By the way, the Lagrangian is independent of angular momentum, so by conservation of angular momentum, we can in fact just set $$\mtext{d}\phi = 0$$.
If you do that, you won't get the second equation of motion:

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

#### starthaus

Spoken with the certainty of someone who didn't bother to check if there were any differences between what you two wrote!
I checked, the two solutions produce different answers. As to rudeness, look at his tone.

(You may have missed where kev said that he wasn't reproducing his work)

I didn't miss anything.

#### yuiop

LOL, your right. It seems I end up back where I started every couple of years. Also, I was hoping for a more authorative second opinion than myself! At the time of that old thread I felt the conclusions were inconclusive, but looking back at the old thread all the information is there and I simply had not made sufficient connections to totally convince myself I was on the right track. I think I have a clearer picture now (hopefully).

One thing that always confused me was this statement from http://www.mathpages.com/rr/s6-04/6-04.htm :
At the apogee r = R where dr/dt = 0 the quantity in the square brackets is unity and this equation reduces to

$$\frac{d^2r}{d\tau^2} = \frac{GM}{r^2} \; \; \; (6)$$

This is a measure of the acceleration of a static test particle at the radial parameter r.
I always took that to mean that the proper acceleration of the static test particle is given by (6) but with hindsight I now see that they do not claim that and the correct equation is given later in http://www.mathpages.com/rr/s7-03/7-03.htm :
However, this acceleration is expressed in terms of the Schwarzschild radial parameter r, whereas the hovering observer’s radial distance r' must be scaled by the “gravitational boost” factor, i.e., we have dr' = dr/(1-2m/r)^(1/2). Substituting this expression for dr into the above formula gives the proper local acceleration of a stationary observer

$$\frac{d^2r'}{d\tau^2} \;\; = \;\; \frac{GM}{r^2} \frac{1}{\sqrt{1-2GM/r}}$$

This value of acceleration corresponds to the amount of rocket thrust an observer would need in order to hold position, and we see that it goes to infinity as r goes to 2m.
One thing I would quibble over is that the above equation proves conclusively that the proper acceleration becomes infinite at the event horizon. The acceleration equation can be written in a more general way as :

$$a \;\; = \;\; \frac{GM}{r^2} \frac{(1-2GM/r)}{(1-2GM/r_o)^{3/2}}$$

where $r_o$ is the location of the observer and r is the location of the stationary particle. Setting $r_o = \infty$ gives the coordinate acceleration and setting $r_o = r$ gives the proper acceleration. When r = 2GM the proper acceleration is then:

$$a_0 \;\; = \;\; \frac{GM}{r^2} \frac{0}{0}$$

which is indeterminate.

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#### starthaus

When r = 2GM/r the proper acceleration is then:

$$a_0 \;\; = \;\; \frac{GM}{r^2} \frac{0}{0}$$

which is indeterminate.
I think you meant $$r=2GM$$, not $$r=2GM/r$$.
The more important thing is that it is easy to prove that physically possible solutions for the equations of motion I derived exist only for $$r>3GM$$ (or, in my notation , $$r>3m$$), so the above is a non-issue. You can never get infinite or indeterminate proper acceleration.

#### yuiop

I checked, the two solutions produce different answers. As to rudeness, look at his tone.
The tone you detect is my irritation and frustration at your habit of saying my conclusions are wrong without saying why they are wrong or what your conclusions are.

You always expect me to draw your conclusions for you "as an exercise". That would end up in "circular criticism". If I draw your conclusions for you and then say your conclusions are wrong I would infact be criticising my own conclusions. For me to draw your conclusions for you, requires me to read your mind and that is very difficult at the best of times and even more so over the internet when you may be thousands of miles away and I can not see your facial expressions or body language.

So why not state what YOUR conclusions are and maybe we can have a sensible conversation?

From the first equation, we can get immediately:

$$\frac{dt}{ds}=\frac{k}{1-\frac{2m}{r}}$$
You think the above is the relationship between the coordinate time and the proper time of a stationary particle. Your exercise is to show why the above is not applicable to a stationary particle. For a hint, see my last post.

If that is not what you are thinking, then like I said I am not a mind reader. Why not simply state what you are thinking?

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#### yuiop

I think you meant $$r=2GM$$, not $$r=2GM/r$$.
The more important thing is that it is easy to prove that physically possible solutions for the equations of motion I derived exist only for $$r>3GM$$ (or, in my notation , $$r>3m$$), so the above is a non-issue. You can never get infinite or indeterminate proper acceleration.
Your right about the typo. I should have wrote r = 2GM and have now corrected it that post. Thanks. I usually write r = 2GM/c^2 but we are now using units of c=1.

The equations of motion are valid for r>2m but there are no circular orbits between r=2m and r=3m.

#### starthaus

The tone you detect is my irritation and frustration at your habit of saying my conclusions are wrong without saying why they are wrong or what your conclusions are.
I showed you why your solution was wrong, you can't "fiddle" or "guess" the correct solutions, you need to derive them from base principles. I have done it in the past, every time I have pointed out your errors.

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#### yuiop

I showed you why your solution was wrong. I have done it in the past, every time I have pointed out your errors.
If you are refering to this thread https://www.physicsforums.com/showthread.php?t=397403&page=11 then it clear that atyy, George Jones and Dalespam showed you were wrong and no one supported your argument. If fact Dalespam gave a lengthy, rigorous and very impressive demonstration that you were wrong starting at #165 of that thread and ending at #169.

Actually most of the stuff you say is not incorrect in itself, but for some reason you always seem to be talking about a different subject to the subject of thread and what everyone else is talking about. You are doing that in this thread too.

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#### yuiop

I showed you why your solution was wrong, you can't "fiddle" or "guess" the correct solutions,
You can if you are lucky and it looks like I was. Actually it wasn't luck. I just refer to conclusions that are rigorously derived by other people that know what they are doing.

#### starthaus

If you are refering to this thread https://www.physicsforums.com/showthread.php?t=397403&page=11 then it clear that atyy, George Jones and Dalespam showed you were wrong and no one supported your argument. If fact Dalespam gave a lengthy, rigorous and very impressive demonstration that you were wrong starting at #165 of that thread and ending at #169.
I knew you were going to go there. I don't know why you would do that since there were so many calculus errors that you made in that thread. As to the proof produced by Dalespam, I repeatedly pointed out the error in his approach so, in the end, we agreed to disagree. If you want to discuss more on the subject of transformation of centripetal acceleration in SR, we can do it in the appropriate thread.

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#### LukeD

One thing that always confused me was this statement from http://www.mathpages.com/rr/s6-04/6-04.htm :

At the apogee r = R where dr/dt = 0 the quantity in the square brackets is unity and this equation reduces to

$$\frac{d^2r}{d\tau^2} = \frac{GM}{r^2} \; \; \; (6)$$

This is a measure of the acceleration of a static test particle at the radial parameter r.

I always took that to mean that the proper acceleration of the static test particle is given by (6) but...
Right, this is the radial acceleration of our test particle as measured against the coordinate r. But this isn't the acceleration measured by our test particle. It's what we'd get if we parallel transported the acceleration vector all the way to infinity and then asked the observer at infinity how much acceleration he would measure if he had that acceleration vector.

To get the acceleration measured by our observer at r, you need to measure the acceleration vector in the orthonormal basis used by an observer at r. So the observed acceleration is
$$\frac{d^2 x^\alpha}{d\tau^2} \cdot \hat e^r$$
Where $$|\hat e^r|^2=1$$ and $$\frac{d x^a}{d\tau} \cdot \hat e^r = 0$$. e^r is directed toward larger r value.

It is very important to remember that formula for the dot product involves the metric.
Because our observer is at constant r, just like our observer at infinity, we can choose $$\hat e^r$$ to be in the same direction as the observer @ infinity's radial vector, but
$$\hat e^r$$ has a different length
(remember, a length of a vector is only defined at a point in spacetime. If we put both of these vectors at the same point, then e^r is bigger by a factor of (1-2M/R)^(-1/2))

#### yuiop

I knew you were going to go there. I don't know why you would do that since there were so many calculus errors that you made in that thread. As to the proof produced by Dalespam, I repeatedly pointed out the error in his approach so, in the end, we agreed to disagree. If you want to discuss more on the subject of transformation of centripetal acceleration in SR, we can do it in the appropriate thread.
In that thread it took over 150 posts and a lot of bickering to play the "guess what Starthaus is thinking" game when you could have simply stated what you were thinking in one or two posts and now it seems you want to play that game again in this thread.

So far, in our new "guess what Starthaus is thinking" game, I have guessed in #15 that you are thinking that the relationship between the proper time and the coordinate time of a stationary particle is given by:

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

and you have not denied it. Right there is a mistake in your thinking.

#### Dale

Mentor
As to the proof produced by Dalespam, I repeatedly pointed out the error in his approach so, in the end, we agreed to disagree.
We certainly can go back and do this again if you wish. My derivation and my results are correct, the "error" you pointed out comes straight from the reference (Gron) that you agreed to use, and your error lies in not knowing what proper acceleration is.

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#### LukeD

I checked, the two solutions produce different answers. As to rudeness, look at his tone.
If you got different answers by the 2 methods, then you were doing something wrong. I've checked kev's original answer with a few different sources as well as with my own work, and he made no mistakes at all. I also checked your formulas (the few that you wrote), and they were going in the right direction. You never got very far with them though, and your later comments suggest to me that you probably don't know the correct way to proceed.

In particular, you mention dr/ds as being the proper speed, but this is incorrect. The observer at r has to deal with length contraction in addition to the time dilation that you handled. He measures a dr'/ds, not dr/ds.
Working with 4-vectors instead of the individual coordinates leads to a much cleaner treatment that allows you to transform between observers' measurements easily.

Attacking someone and insisting that they are wrong just because they didn't show their work (which he said he would if you asked) usually doesn't get you a nice response...

Anyway, the question's been cleared up..

#### starthaus

In that thread it took over 150 posts and a lot of bickering
It is not my problem that you had so much trouble with getting a few basic derivatives correct <shrug>. So many errors in your math makes it so painful teaching you.

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