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Gravitational force and acceleration in General Relativity.

  1. May 9, 2010 #1
    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:

    [tex] \gamma = \frac{1}{\sqrt{1-\frac{2GM}{rc^2}}} [/tex]

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

    [tex]a= \frac{GM}{r^2}\left(1-\frac{2GM}{rc^2}\right)[/tex]

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

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

    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:

    [tex] F_0 = \frac{GMm_0}{r^2}\left(1-\frac{2GM}{rc^2}\right)^{-\frac{1}{2}} = m_0a_0[/tex]

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

    [tex] F = \frac{GMm}{r^2} [/tex]

    Note that:

    [tex]a_0 = a\gamma ^3[/tex]


    [tex]F_0 = F = m_0 a_0 [/tex]

    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. jcsd
  3. May 10, 2010 #2

    This is very wrong. You need to start with the metric:

    [tex]ds^2=\alpha dt^2-\frac{1}{\alpha}dr^2-r^2d\phi^2[/tex]


    From this you construct the Lagrangian:

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

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

    [tex]\alpha \frac{dt}{ds}=k[/tex]

    [tex]r^2 \frac{d\phi}{ds}=h[/tex]

    whre h,k are constants.
    Last edited by a moderator: Apr 25, 2017
  4. May 10, 2010 #3
    and what do you conclude that the coordinate and proper acceleration should be?
  5. May 10, 2010 #4
    I derived for you the equations of motion, I left this as an exercise for you to calculate.
  6. May 10, 2010 #5
    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 [itex] d\phi = 0[/itex] 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
  7. May 10, 2010 #6


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  8. May 10, 2010 #7
    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 [itex] d\phi = 0[/itex] and forget about it. "The simple reason is that [itex] \phi[/itex] is variable

    You are becoming abusive every time you are shown wrong.
    Last edited: May 10, 2010
  9. May 10, 2010 #8
    From the first equation, we can get immediately:

    [tex] \frac{dt}{ds}=\frac{k}{1-\frac{2m}{r}}[/tex]

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


    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 , [tex]\phi[/tex].
  10. May 10, 2010 #9


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    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

    [tex]\frac{M}{r^2\sqrt{1 - 2M/r}}[/tex]​

    (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
  11. May 10, 2010 #10
    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 [tex]h[/tex]), we can in fact just set [tex]\mtext{d}\phi = 0[/tex]. This is because if we start on a geodesic with [tex]\frac{\mtext{d}\phi}{\mtext{d}\tau} = 0[/tex], this remains true for all time. It is a boundary condition on our motion that imposes no extra forces.
  12. May 10, 2010 #11
    If you do that, you won't get the second equation of motion:

  13. May 10, 2010 #12
    I checked, the two solutions produce different answers. As to rudeness, look at his tone.

    I didn't miss anything.
  14. May 10, 2010 #13
    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 :

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

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

    [tex]a_0 \;\; = \;\; \frac{GM}{r^2} \frac{0}{0} [/tex]

    which is indeterminate.
    Last edited by a moderator: Apr 25, 2017
  15. May 10, 2010 #14
    I think you meant [tex]r=2GM[/tex], not [tex]r=2GM/r[/tex].
    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 [tex]r>3GM[/tex] (or, in my notation , [tex]r>3m[/tex]), so the above is a non-issue. You can never get infinite or indeterminate proper acceleration.
  16. May 10, 2010 #15
    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?

    Here is an attempt to read your mind:

    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
  17. May 10, 2010 #16
    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.
  18. May 10, 2010 #17
    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
  19. May 10, 2010 #18
    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
  20. May 10, 2010 #19
    You can if you are lucky and it looks like I was. :wink:

    Actually it wasn't luck. I just refer to conclusions that are rigorously derived by other people that know what they are doing.
  21. May 10, 2010 #20
    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
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