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Does GR reduce gravitational force on a planet's surface?

  1. May 16, 2015 #1
    Am I correct in thinking that the force of gravity between 2 test objects at rest on a planet's surface is less than it would be for the same objects at rest in deep space? I understand that this occurs because in GR gravitational potential has a mass value which is lost on the surface, while the mass's themselves still retain the same value. If this is correct, does it require a difficult GR calculation? Thanks
     
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  3. May 16, 2015 #2

    phinds

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    I don't understand your question, but gravity is not a force in GR (as it is in Newtonian classical physics), it is a simple consequence of spacetime geometry.
     
  4. May 16, 2015 #3

    PeterDonis

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    As phinds pointed out, in GR gravity is not a force; it's spacetime curvature. But if spacetime curvature is weak enough, and objects are moving slow enough, we can approximate the effects of gravity as a Newtonian force, which can be measured by a Cavendish experiment. If we consider running such an experiment with two objects in deep space, and then moving the two objects to a planet's surface and running the experiment again, the results would be the same in both cases. So in this sense, I think the answer to your question is "no".
     
  5. May 16, 2015 #4
    Thanks for this. The opinion I received from a prof. of Engineering, who also has a degree in mathematical physics, was that there would be a discrepancy for the reasons I mentioned, so I was just trying to nail that down. It sounded right to me, but I can't find any information on the subject on the internet.
     
  6. May 19, 2015 #5
    Um, but it won't be the same in both cases. The objects will have different potential energy, and that contributes to their gravitational pull. So the answer is "yes".
     
  7. May 19, 2015 #6

    A.T.

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    Gravitational potential energy is a property of the system of objects, not of one particular object. The Equivalence Principle indicates that at least for small separations of the 2 test objects the answer is "no".
     
  8. May 19, 2015 #7
    Thanks georgic and A.T. for these replies. I guessed that the force on the surface might be a little less because if we take the grav. potential energy, which equals the kinetic energy at the escape velocity, and use E=mc^2 to convert it into mass, we can plug the values for a particular example, say the Earth, into the equation V^2/V^2max + m1^2/m^2 = 1. V = 11.2 km/sec, m is a test object's mass, lets say 1kg, m1 is 1kg - .62 micrograms (the mass we get from E=mc^2) and Vmax is the highest value for the velocity. If we solve for Vmax we get c, or near enough (we don't need to use E=mc^2; we could just use the experientially derived value.) Any comments about this?
     
  9. May 20, 2015 #8
    Let's say we are in an accelerating spaceship. We have two massive plates that stick together gravitationally. The force is large enough to keep plate B hanging on plate A when we lift plate A up.

    Now we grab the plate A, and the plate B hanging on plate A, and start climbing down a rope.

    Before we have reached the Rindler-horizon, the plates will have been pulled apart. This is easy to calculate.

    In a homogeneous gravity field things are exactly the same as in an accelerating spaceship.

    Force of gravity between two plates is less when the plates are at a lower position.
     
  10. May 20, 2015 #9

    PeterDonis

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    "Potential energy" relative to what? Relative to infinity, yes; but we're not measuring their gravitational pull relative to infinity. Relative to each other, since by hypothesis they are both at the same altitude in the field, their gravitational pull is unchanged.

    In other words, moving the pair of objects from infinity to some finite altitude in a gravity well does not change the objects locally at all.
     
  11. May 20, 2015 #10

    PeterDonis

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    This doesn't make sense. The plates should be separated horizontally, not vertically; otherwise they are not at the same "gravitational potential", which they must be by the OP's hypothesis, at least as I understand it.

    If the plates are allowed to be separated vertically, then of course you can change their separation by going to a regime where the change in proper acceleration from one plate to the other is large enough (close enough to the Rindler horizon in the flat spacetime case, or deep enough in the gravity well in the curved spacetime case). But that's not changing the "force of gravity" produced by the plates.
     
  12. May 20, 2015 #11
    Thanks for this. Has anyone actually done the experiment comparing the force of gravity between two objects on the surface of the Earth with the force in space? A good test of GR I would think. I can't find any reference to it if it has. The effect would not show up in the orbits of the planets because inertial and gravitational mass would change together.

    Another related matter, the time dilation on the surface of a planet equals the time dilation at that planet's escape velocity - generally considered an odd coincidence. Isn't it also a little odd that we can use the mass energy value of the escape velocity, and without reference to E=mc^2 (we only need to use the experimentally derived per unit mass energy value) calculate a value equal to c? Are there several ways to calculate c other than Maxwell's?
     
  13. May 20, 2015 #12

    PeterDonis

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    I don't know that anyone has done a Cavendish experiment in space, which would be the definitive test.
     
  14. May 20, 2015 #13

    pervect

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    In terms of a local frame of reference, as you descend down the rope, absolutely nothing happens to the force between the plates, but the proper acceleration needed to hold station on the rope increases. So of course the plates separate. How does this show that "gravity is weaker"?

    As long as you use local measurements (local clocks, rulers, and scales), the laws of physics (as modelled by General Relativity, at least) don't depend on your location, or the gravitational potential of your location.
     
  15. May 20, 2015 #14
    I'm so lazy that I copy-paste and edit:

    If the plates are allowed to be separated vertically, then of course you can change their separation by going to a regime where the proper acceleration is large enough (close enough to the Rindler horizon in the flat spacetime case, or deep enough in the gravity well in the curved spacetime case).

    We might say that the "force of gravity" produced by the plates changed. Or we might say the "force of gravity" produced by the acceleration changed.


    I happen to like the first alternative. Because .......... a mountain climber causes a constant tension on a rope, when he climbs down the rope in a uniform gravity field. I mean the anchor does not break away from the rock, no matter how far the climber descends. Right?
     
  16. May 20, 2015 #15

    PeterDonis

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    No, it can only be the second. The first alternative won't work. See below.

    What is a "uniform gravity field"? If by that you mean "proper acceleration does not change with height", then the thought experiment you describe is not a "uniform gravity field"; the Rindler horizon is associated with a congruence of Rindler observers, and their proper acceleration varies with height above the horizon. So the tension in the rope caused by the climber won't be constant.

    If, OTOH, you insist on a congruence of observers whose proper acceleration is constant, then this is the Bell congruence, and they are not at rest relative to each other. (More precisely, the expansion scalar of the Bell congruence is nonzero, which means it is not rigid.) So you can't use that congruence to model a static "gravitational field" in your thought experiment. The rope that the mountain climber is climbing down will be stretched further and further as time goes on, even if the climber is not there.
     
  17. May 20, 2015 #16
    In SR we see time and mass changing together, so if time changes in a gravitational field isn't it worth checking to see if there's some kind of mass change as well? It just seems like due diligence to me. I may be naive about this but 2 test objects at rest in deep space have more energy than the same objects at rest on the surface of a planet. E = mc^2, therefore we have the equivalent of mass loss, and therefore a reduction in the force of gravity between them compared to what it would be in space. By the way, the forum has not addressed my observation about c. A comment or two would be appreciated.
     
  18. May 20, 2015 #17

    PeterDonis

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    More precisely, we see time dilation and relativistic mass changing together. But relativistic mass is not the source of gravity in GR; the stress-energy tensor is.

    Energy relative to what? "Energy" is not an absolute quantity; it's relative. And once again, the "energy" you are talking about is not the source of gravity in GR; the stress-energy tensor is. The stress-energy tensor that describes these "test objects" (technically that's not a good term because that term is usually used to describe objects whose own mass is negligible, so they don't act as sources of gravity) is what determines the gravity they produce, and that's different from the "energy" you are talking about.

    In short, you are reasoning from a mistaken assumption about what property of the test objects determines the gravity they produce. When you use the correct property, the stress-energy tensor of the objects, you get the answer I and others have already given several times in this thread: the gravity between the two objects is the same when they are in a gravity well and when they are out in empty space far from all other bodies.
     
  19. May 20, 2015 #18

    First let's use a clock and a ruler to measure acceleration. The clock is affected by time dilation. So our accelerometer is affected by time dilation.

    Then let's try to probe acceleration at various altitudes using a fishing rod, a fishing line and a weight, the amount of bending of the fishing rod tells us something about the acceleration at the position of the weight.

    The second accelerometer is actually just a force meter. Multiplying the force by the time dilation factor at the position of the weight yields the proper weight of the weight at the position of the weight. Proper acceleration can be calculated from proper weight.

    If we put the part that measures the force at the lower end of the fishing line, then the fishing line accelerometer agrees with the clock and ruler accelerometer about acceleration increasing when moving down in uniform gravity field. The two gravitating disks that break apart at some height is same kind of accelerometer as this one. It is affected by time dilation.


    What if an accelerating spaceship tows a weight with force-meters at both ends of the towing line? Do we agree that the lower force-meter's reading is higher? (towing line is massless, or we subtract its weight)
     
    Last edited: May 20, 2015
  20. May 20, 2015 #19
    Thanks again for the replies. You know vastly more about GR than I do, so I accept what you've told me. But experiment trumps theory, and as I've pointed out, there is a simple experiment available to put GR's interpretation to the test. My feeling is that you agree. Saying that the experiment is unnecessary because we have the GR solution is not good science. Good science puts theory to the test whenever possible. I don't understand why this experiment has never been done, unless the Cavendish experiment lacks enough sensitivity, but it's my understanding that it's extremely sensitive, so it's likely to be able to do the job. Cheers
     
  21. May 20, 2015 #20
    By uniform gravity field I mean the "gravity field" inside an arbitrarily large accelerating spaceship, or the gravity field inside sufficiently small room on a planet. OTOH I use the term correctly. :)

    When a climber climbs down a rope in an uniform gravity field, the force of gravity that he feels increases, because he is affected by increasing time dilation. The force pulling the upper end of the rope is the force felt by the climber divided by time dilation factor at climber's position, so it's constant.
     
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