# Gravitational force and acceleration in General Relativity.

1. May 9, 2010

### 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.

Last edited by a moderator: Apr 25, 2017
2. May 10, 2010

### starthaus

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

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

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.

Last edited by a moderator: Apr 25, 2017
3. May 10, 2010

### yuiop

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

4. May 10, 2010

### starthaus

I derived for you the equations of motion, I left this as an exercise for you to calculate.

5. May 10, 2010

### yuiop

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.

Last edited: May 10, 2010
6. May 10, 2010

### Ich

Last edited by a moderator: Apr 25, 2017
7. May 10, 2010

### starthaus

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.

It is not correct to "set $d\phi = 0$ and forget about it. "The simple reason is that $\phi$ is variable

You are becoming abusive every time you are shown wrong.

Last edited: May 10, 2010
8. May 10, 2010

### starthaus

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 , $$\phi$$.

9. May 10, 2010

### DrGreg

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.

Last edited by a moderator: May 4, 2017
10. May 10, 2010

### 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.

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.

11. May 10, 2010

### starthaus

If you do that, you won't get the second equation of motion:

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

12. May 10, 2010

### starthaus

I checked, the two solutions produce different answers. As to rudeness, look at his tone.

I didn't miss anything.

13. May 10, 2010

### 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 :
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 :
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.

Last edited by a moderator: Apr 25, 2017
14. May 10, 2010

### starthaus

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.

15. May 10, 2010

### yuiop

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?

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?

Last edited: May 10, 2010
16. May 10, 2010

### yuiop

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.

17. May 10, 2010

### starthaus

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.

Last edited: May 10, 2010
18. May 10, 2010

### yuiop

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.

Last edited by a moderator: Apr 25, 2017
19. May 10, 2010

### yuiop

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.

20. May 10, 2010

### starthaus

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.

Last edited by a moderator: Apr 25, 2017
21. May 10, 2010

### LukeD

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))

22. May 10, 2010

### yuiop

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.

23. May 10, 2010

### Staff: Mentor

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.

Last edited: May 10, 2010
24. May 10, 2010

### LukeD

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..

25. May 10, 2010

### starthaus

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.