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## Gravitational force and acceleration in General Relativity.

I was pondering this thread Correction term to Newtons gravitation law. 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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 Quote by kev I was pondering this thread Correction term to Newtons gravitation law. 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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 Quote by 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: $$L=\alpha (\frac{dt}{ds})^2-\frac{1}{\alpha}(\frac{dr}{ds})^2-r^2(\frac{d\phi}{ds})^2$$ 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?

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## Gravitational force and acceleration in General Relativity.

 Quote by kev 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.

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 Quote by starthaus 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.

Recognitions:
 Quote by kev Not being able to find a clear, simple definitive answer to that question on the internet...

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 Quote by kev 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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 Quote by 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 , $$\phi$$.

 Recognitions: Gold Member Science Advisor Prof Nick Woodhouse of Oxford University shows here (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.

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.

 Quote by starthaus 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.

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 Quote by LukeD 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:

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

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 Quote by LukeD 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.

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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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 Quote by kev 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.

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 Quote by starthaus 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?

 Quote by starthaus 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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 Quote by 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.
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.

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 Quote by kev 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.