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Relativistic Gravity

  1. Jan 10, 2015 #1
    I know this topic has come up several times throughout the years, but after reviewing a good number of the threads, I’m still not clear as to whether any kind of consensus on the matter has been reached or not, or if there may be new developments in either theory or observation. If you review the threads below, you will see a divergence of views on the matter. Generally, it’s either 1) yes, a relativistically moving body in frame B will be perceived by a lab frame A as having a greater gravitational pull related to the increase in its “relativistic mass,” 2) there is no effect at all, the relativistically moving frame B has the same gravitational pull as it would in the lab frame of A, and 3) it’s too complicated to figure out because only one of the terms in the stress energy tensor relates to “relativistic mass,” the equations are too hard to figure out, and there’s no observational evidence to compare anything against:

    https://www.physicsforums.com/threads/does-relativistic-mass-create-gravity.68454/page-2
    https://www.physicsforums.com/threads/does-relativistic-mass-induce-gravity.413594/
    https://www.physicsforums.com/threads/relativistic-mass-and-gravity.626128/
    https://www.physicsforums.com/threads/relativistic-mass-and-gravity.675637/

    So, in lieu of observational evidence, most posters in these threads come up with thought experiments to try to gain insight on the manner. I’m going to do the same here with a fresh thought experiment I didn’t see in the other threads and see if anyone can, say, pick it apart some, here goes:


    Earth has a twin planet named “Bearth,” with exactly the same radius and mass in kgs. One day Bearth comes zooming through the solar system and passes Earth at 0.87c (relative to an x coordinate which lines up parallel to the equators of both planets), giving Bearth a relativistic gamma factor of 2 relative to Earth in the x direction. On the planet Bearth there are two twin sisters, Alice and Beth. Both of these girls are 6 feet tall (being twins, of course).


    Alice and Beth are playing a game, the game is for both girls to hold their arms straight out, drop a baseball, and see which one hits the ground first. They are not sure whose ball will, because Alice is standing on the north pole of Bearth and Beth is standing on the equator. Low and behold, though, when they try the experiment/game, they find the balls hit the ground at roughly the same time, and the acceleration of each ball is 10 m/s^2.


    Now, while this going on, Bob on planet Earth is watching Alice and Beth play this game through his powerful telescope and wants to play too. Bob is also 6 feet tall and is standing parallel to Alice on the north pole of the Earth. Looking through his telescope, he sees when Alice and Beth drop their baseballs and then proceeds to drop his own at the exact same instant. What does Bob notice?


    From what I can tell, Bob notices that, while he sees his ball fall to the ground at time t, he sees Alice’s ball fall to the ground at time 2t due to the time dilation gamma factor of 2 described above. Is this not right? In that case, wouldn’t this appear to Bob as if the force of gravity on planet Bearth is less than that of Earth, rather than greater due some putative “relativistic mass.” As far as Beth is concerned, it would seem as though Bob would witness his and her ball hitting the ground at the same time because, while Beth’s ball takes twice as long to hit the ground, she is only half the size as Bob at 3 feet tall due to her length contraction in the x-direction of movement.


    So, at least according to this thought experiment, the relativistic velocity of Bearth would seem to have the effect of making gravity appear to be weaker on Bearth than Earth relative to Bob’s perspective. Is this wrong? If so, please explain why and what theoretical and observational work illuminates this issue.
     
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  3. Jan 10, 2015 #2

    Ibix

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    Note: I'm still learning GR. That said, I think this can be handled with just SR.

    I think you can analyse this quite simply with just Lorentz transforms. In the frame of Bearth, which I shall assume to symmetric and non-rotating with radius R, one ball starts at (t,x,y,z)=(0,R+h,0,0) and finishes at (T,R,0,0). The other starts at (0,0,R+h,0) and finishes at (T,0,R,0). Lorentz tells us that, in the frame of Earth:

    (0,R+h,0,0) → (-γv(R+h)/c2,γ(R+h),0,0)
    (T,R,0,0) → (γ(T-vR/c2),γ(R-vT),0,0)
    (0,0,R+h,0) → (0,0,R+h,0)
    (T,0,R,0) → (γT,-γvT,R,0)

    We can then evaluate the coordinate time seen by an observer on Earth:
    • For the ball at the "leading" pole this is γ(T-vR/c2) - -γv(R+h)/c2 = γ(T+vh/c2)
    • For the ball on the "side" pole this is γT - 0 = γT
    From this we can see that the gravitational field isn't spherically symmetric anymore, which isn't surprising since Bearth is length-contracted in its direction of motion. I'm not sure what this says about the "strength" though. I'd be inclined to factor out the motion-induced time-dilation, since we'd be able to observe other processes like clocks ticking slow by that factor. In that analysis, the strength perpendicular to the motion is unaffected, perhaps unsurprisingly. However, the field parallel to the motion has changed - the fall time is no longer T, but T+vh/c2. That means that the fall time no longer has the same functional form that it does in Bearth's rest frame. That isn't something you can easily define as "stronger" or "weaker". It's just different.

    I'm a little surprised that the Earth-viewed fall time is greater than the Bearth-viewed one. I wonder if someone could check my maths. And, you know, the whole methodology.
     
  4. Jan 10, 2015 #3

    phinds

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    As nearly as I can tell, there is a clear consensus (the answer is yes), but it's complicated. I think where you are getting confused is because in some of the posts you have referenced, the emphasis of the discussion is on the fact that calculations of "relativistic mass" (a thoroughly deprecated term by the way) do NOT give the right answer when trying to get a numerical answer to your question. The energy put into the "moving" object causes an increase in speed and thus a concomitant increase in momentum and it is the mathematically complicated relationship between that and the gravitational attraction that causes the confusion. In short, what I am saying is that I think all of the answers are "yes", but some of them are more the more correctly stated "yes, BUT ..." and it is the discussion of the "but" that makes it seem somewhat as though some of the answers are "no".
     
  5. Jan 10, 2015 #4
    That's exactly my point, though. We are talking about how Bob sees the gravitational potential on Bearth, not Alice. We can all agree that, since Alice's clock is running slow compared to Bobs, her experience of gravity will be unaffected by her relative motion. Bob, however, will see Alice's ball fall at a slower rate of acceleration than his ball falls at, and will thus conclude that the force of gravity is therefore weaker on Bearth. At least weaker when measured from the poles of Bearth. You could then say, well this is an optical illusion caused by the Lorentz factor, and that if you matched up the fall rate of Alice's ball with the rate her clock ticks, then everything comes out even, etc. And I'd say you are absolutely right. But then you could say the same thing about the twin paradox. Something different is happening there because, when the traveling twin shows up back on Earth, she is physically younger than her brother. So things don't actually add up as you might think they should.

    In the same way, we can imagine an experiment where Alice, Beth, and Bob repeat the ball dropping experiment 1000 times, all at the exact same rate relative to their rest frame, and then compare their results when say, the planet Bearth makes a turn around sometime later and arrives back near the Earth in a twin paradox-type scenario. Here we are going to find that, while in that time Bob did indeed perform the ball drop experiment 1000 times, Alice was only able to perform it, say 500 times. Why? Because that ball was just falling so darn slowly, she couldn't make it to 1000 drops. Why was it falling slowly? Because the gravitational potential on Bearth was weaker during the trip, at least from Bob's perspective. Alice didn't notice the difference, but Bob did. And it's not an optical illusion because there is a measurable effect. Bob was counting on Alice to perform that experiment 1000 times for their class project and she only accomplished half her assignment. The only excuse she and her lab partner Bob are going to have when they show up for class on Monday is that they couldn't complete the experiment in the time allotted because the gravity on Alice's planet was half that on Bob's planet.

    Again, you could say that Alice was only able to perform the experiment 500 times not because the gravity was weaker on Bearth, but only because her clock was running more slowly. And my reply would be to say, "what's the difference?" You can focus on whatever measured variable you want and use it to explain an anomaly in another. For example, say there were no clocks on the planet Bearth and all Bob had to go on were his visual observations. All he could say about the situation is that Alice's ball fell at a slower rate than his and, thus, it appears that the gravitational potential on Bearth is less than it is on Earth.

    Where am I off here?
     
  6. Jan 10, 2015 #5

    A.T.

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    Since all experiments are slowed down, not just the ones involving gravity, its simpler to describe it as time slowing down. Trying to describe the same effect by tweaking of all possible forces would be more complicated, and conflict with static force measurements.
     
  7. Jan 10, 2015 #6
    I'm aware of that which is why I put it in quotes. But the reason I used it is because I think the topic of this thread is a good way to analyze or expose why it is a deprecated term. On it's face, it would seem as if a "relativistic mass" increase of a body, A, traveling relative to rest body, B, would make it appear to B that A had a greater gravitational potential. This apparently isn't as simple as taking the relativistically corrected mass value and plugging it into Newton's equation. The question is why is this, and what is the actual effect on gravity of a relativistically moving body? Furthermore, I'm not sure how useful it is even to dissociate "relativistic mass" into components of, say, "rest (inertial) mass" plus kinetic energy. Even in this case the added kinetic energy should increase the gravitational potential of the body A relative to B. But is this observed? I don't know, that's what I'm trying to get at here. Are there models of how mass and energy terms are split up in the stress energy tensor to explain how spacetime curvature around relativistically traveling bodies is manifested?

    Also, I'm not an astronomer but couldn't they, say, compare binary star systems of similar mass and orbital-radial arrangements but traveling at different velocities relative to each other and the Earth and get some experimental data that way? Again, I'm way out of loop on how these things are measured but just putting it out there.

    That's a good point, especially considering that Beth at the equator of Bearth would return to Earth's rest frame having actually completed the experiment 1000 times as in Bob's case. Although, she would have also aged half the time Bob had during her journey. Which sort of complicated the picture again..
     
  8. Jan 10, 2015 #7

    Vanadium 50

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    You're starting from the assumption that SR+Newtonian Gravity will give you a realistic theory of relativistic gravity. It won't. That's why GR was developed.
     
  9. Jan 10, 2015 #8

    Nugatory

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    Why? To the extent that relativistic mass is a useful concept, it applies to the three-acceleration of an object in response to applied forces - and that doesn't have much to do with gravity.
     
  10. Jan 10, 2015 #9

    Dale

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    With just SR the balls never hit the ground.
     
  11. Jan 11, 2015 #10

    PeterDonis

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    Why do you think this would happen? Wouldn't the number of experiments completed by Beth have to bear the same relationship to the amount she ages as Alice's does?
     
  12. Jan 11, 2015 #11

    Ibix

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    Peter - you are correct, of course. What I was trying to get at was that it is possible to take a couple of points on the path of each ball without worrying about the details of why the balls are following that path. Then we can just Lorentz-transform those events and end up being able to say something about the behaviour of the force (or "force" in this case) wthout invoking the full machinery of GR.

    I think that argument is ok.
     
  13. Jan 11, 2015 #12

    Dale

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    I don't think that will tell you anything about gravity since it is not a force in relativity. A similar argument could be used to tell you how the EM force transforms, but not gravity.
     
  14. Jan 11, 2015 #13

    PeterDonis

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    This won't work because you can only use a Lorentz transform this way in flat spacetime, and if gravity is present spacetime is not flat.
     
  15. Jan 11, 2015 #14

    Ibix

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    OK - but I thought the whole point of tensors (which the 4-position of the events are) was that they transformed in a straightforward way under a Lorentz boost. I take it that I've been a bit too casual about that. What would (a sketch of) the correct procedure be?
     
  16. Jan 11, 2015 #15

    PeterDonis

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    First, "4-positions of events" are not tensors. Nor are differences between two 4-positions, i.e., displacements. In flat spacetime, you can get away with thinking of displacements as 4-vectors, but this doesn't work in curved spacetime, so it's best not to get into the habit.

    Second, the point of tensors is that they transform in a straightforward way under general coordinate transformations. This is a much wider class of transformations than Lorentz transformations--which, as I said, don't even work in curved spacetime anyway.

    First, you need correct expressions for the worldlines of the two falling objects; you assumed they were straight lines in the Bearth-centered coordinates, but they're not.

    Second, you need a correct expression for the metric in the Bearth-centered coordinates; it's not the Minkowski metric.

    Third, you need a correct expression for the coordinate transformation from Bearth-centered coordinates to Earth-centered coordinates; it's not a Lorentz transformation.
     
  17. Jan 11, 2015 #16

    Ibix

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    D'oh. No meaningful concept of "adding two positions together" means positions do not form a vector space. Although you can get away with it in flat space by using vectors from the origin interchangeably with coordinates, since there's a trivial relationship.

    Ah - Lorentz transforms are equivalent to switching from coordinates derived from one set of rigid rods with Einstein-synchronised clocks and another set in relative inertial motion. And you can't build such a system in a curved space-time. I'm sure there's a differential geometry way of explaining that...

    I need to solve the geodesic equation in a Schwarzchild metric (I noted earlier that I assumed Bearth was not spinning) subject to the initial condition that the particles are at rest far, far above the event horizon. Right?

    I have no idea how to do this. More reading needed, obviously.
     
  18. Jan 11, 2015 #17

    Dale

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    That is also the part that I don't know. I am not aware of a standard metric for a moving spherical mass.
     
  19. Jan 11, 2015 #18

    PeterDonis

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    Right.

    There's a solution called the Aichelburg-Sexl ultraboost that covers this general type of situation:

    http://en.wikipedia.org/wiki/Aichelburg–Sexl_ultraboost

    However, I don't know the transformation between this solution and the ordinary Schwarzschild metric.
     
  20. Jan 11, 2015 #19

    Dale

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    AFAIK, there isn't such a transformation. The Aichelburg Sexl metric is a pp wave spacetime, meaning that it is massless. There is no event horizon.
     
    Last edited: Jan 11, 2015
  21. Jan 11, 2015 #20

    pervect

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    My $.08 on the concept of forces in general relativity

    1) Non gravitational forces are well defined, for instance we can find the force on a charged test particle by comparing it to a reference uncharged test particle. This gives rise to tensor quantities like the Faraday tensor.

    2) If you look up "forces" in a GR book, you'll usually not find much under that heading

    3) If you look up "Christoffel symbols in a GR book, you'll find lots of stuff under that heading.

    4) A subset of the Christoffel symbols can be related to "fictitious forces" in a purely Newtonian context. So if you are in an accelerating elevator accelerating in the z direction, the Christoffel symbols of your connection due to your acceleration (fi.e ##\Gamma^z{}_{tt}##) appear as "fictitious forces" in a Newtonian context.

    5) The idea that we can use the fictitious forces in accelerating elevator as some sort of model for "gravity" has a long history in the development of the equivalence principle in particular and GR in general. This suggests that we look for a relationship between the Christoffel symbols and the idea of "gravitational forces".

    6) Unfortunately, due to point 2), we don't find a lot in the textbooks on this topic. So perhaps we are invited to draw our own conclusions. But we'll have to be careful when talking about them if there isn't any published consensus view to refer to. Perhaps there is some definitive reference - if someone wishes to quote one, it'd be interesting reading.

    7) Christoffel symbols are more than "just" forces - but so is gravity. For instance, we have "gravitational time dilation". This is not a characteristic of any other sort of "force" other than gravity, but is definitely associated with gravity. But Gravitational time dilation is also a Christoffel symbol.

    8) So we, or at least I, am led to the conclusion that Christoffel symbols are the replacement for "gravitational forces" in GR. They are, of course, not tensors. How do we explain the fact that forces in Newtonian mechanics do transform as tensors? I would suggest that the point is that while Christoffel symbols are not guaranteed to transform like tensors, under certain specialized limited sorts of coordinate transformations, they can - it's not prohibited, it's just not guaranteed under general transofmations.

    If we want to replace forces with Christoffel symbols we still have an issue, because there are too many of the later.. But the extra Christoffel symbols (like gravitational time dilation) do have some physical significance.

    So - we (or at least I) can regard the Christoffel symbols as some sort of "superset" of forces - forces + more. And in the context (for instance) determining the equations of motion, we just instead talk about using the Christoffel symbols in the geodesic equation and we note that the equations are formally very similar to the differetial euations we used to get when we had 'forces" in Newtonian theory.

    So my end conclusion is - don't worry so much about forces in GR, instead try to understand thier more-or-less replacement, Christoffel symbols.
     
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