Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Is force of gravity dependent on speed of one of the body?

  1. Mar 29, 2012 #1
    Is there any such relation as per general theory of relativity?
     
  2. jcsd
  3. Mar 29, 2012 #2

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    Yes, definitely, but there is no unique definition. For an isolated gravitating mass, and a test body stationary relative to it, there is a unique definition of 'stationary'. Thus you can ask what the proper acceleration is for a stationary world line near a massive body. This is analogous to Newtonian force of gravity.

    However, for a moving body, there is no unique definition of a 'stationary' world line. Similarly, if you consider the gravitating mass to be stationary, and the test body moving, there is no unique definition of 'straight as if the body wasn't there'.

    However, no unique definition doesn't mean no possible definition. For most any plausible definition, the moving gravitating mass exerts more force on the stationary test body ([edit: produces higher peak proper acceleration of the 'stationary' world line]); and the 'straight' moving test body experiences larger proper acceleration (at closest approach) than a stationary test body.

    Unfortunately, the lack of uniqueness means you can't come up with universally agreed on numbers. Different definitions will lead to different amounts of increase in force [edit: peak proper acceleration of resisting world line] due to relative motion.
     
    Last edited: Mar 30, 2012
  4. Mar 30, 2012 #3

    tom.stoer

    User Avatar
    Science Advisor

    Perhaps one could call this question is force of gravity dependent on speed of one of the body? a Category mistake.

    In GR gravity is not a "force".

    Let's assume we have rather heavy bodies (like stars) and a test body (like a satellite). The energy-momentum density T of the stars plus the metric g of spacetime manifold are defining a solution (g,T) of the Einstein field equations.

    The curvature of the manifold defines geodesics C along which test bodies will move.

    A different energy-momentum density of the stars T' with different positions, different motion comes with a different solution g' for the metric, i.e. a different spacetime with (g',T').

    The geodesics C' of this new spacetime are 'different' in a certain sense. But be careful: in principle you are not allowed to compare C and C' b/c they are defined w.r.t. different (g,T) and (g',T').

    It's like saying that a Depp is less stupid than a Twompe; that's dangerous b/c both terms are defined w.r.t. to two different languages; Depp is German for Twit, Twompe is (Haitian) Creole.
     
  5. Mar 30, 2012 #4

    julian

    User Avatar
    Gold Member

    One answer is that if you are moving your mass is greater so the grav force increases, but...(thought experiment) say you have a speed boat at rest, the wieght (due to GRAVITY) equals the force due to water pressure. But what if the boat is moving at speed v with respect to the ocean? The area is reduced by [itex]\gamma[/itex] because of length conrtraction, plus the mass increases and hence the weight of the boat increases...the conclusion is that the boat will sink....
     
    Last edited: Mar 30, 2012
  6. Mar 30, 2012 #5

    julian

    User Avatar
    Gold Member

    The answer is that ALL transvesre forces change under Lorentz tranformations, not just with gravity. You are in motion with respect to a grav body, your mass and hence weight increases, but your transverse force will 'reduce' at the same time your mass and wieght increase.
     
    Last edited: Mar 30, 2012
  7. Mar 30, 2012 #6

    tom.stoer

    User Avatar
    Science Advisor

    as I said, gravity is not a force!
     
  8. Mar 30, 2012 #7

    julian

    User Avatar
    Gold Member

    Yeah but when the strengh of grav field, speed and masses are so small you got to conceed that such considerations are valid.
     
  9. Mar 30, 2012 #8

    tom.stoer

    User Avatar
    Science Advisor

    it's simple: when the velocity of the test particle changes, its geodesic changes as well
     
  10. Mar 30, 2012 #9

    julian

    User Avatar
    Gold Member

    No offesnse, comming from someone who has asked questions that are difficult to understand, but what are you saying exactly?
     
  11. Mar 30, 2012 #10
    Can we please just agree to call mass and energy mass and energy, not mass and mass.

    Also, you need to be more careful here because gravity does not couple to kinetic energy in the same way as it does to mass. It is not clear to me that your analysis is correct at all. Why would the boat driver see his boat sinking?
     
  12. Mar 30, 2012 #11

    julian

    User Avatar
    Gold Member

    Length contraction (boat length) reduces and the force due to water pressure decreases and the weight of boat increases becauses the mass increases... factor of gamma^2 involved....this can only explained by transverse force changing...
     
  13. Mar 30, 2012 #12

    Mentz114

    User Avatar
    Gold Member

    Are you sure about this ? After a proper acceleration the geodesic certainly does change but a body on a geodesic may have changing velocity.
     
    Last edited: Mar 30, 2012
  14. Mar 30, 2012 #13

    tom.stoer

    User Avatar
    Science Advisor

    sorry for the confusion, the formulation is missleading; what I want to say is that test masses at one spacetime point P starting in one direction but with different initial velocities v, v', v'', ... will follow different geodesics C, C', C'', ...
     
  15. Mar 30, 2012 #14

    Mentz114

    User Avatar
    Gold Member

    Sure thing. Sorry to nit-pick ...
     
  16. Mar 30, 2012 #15

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    In a sense, yes.

    One way to answer this question is to look at things from the perspective of a parameterized post-Newtonian (PPN) formalism. Here gravity is still a "force" but it isn't quite Newtonian gravity. In addition to good old F=GMm/r2, there are some first order perturbation terms that result from general relativity in such a formalism. These perturbative effects on Newtonian gravity involve velocity.

    This approach leads to a very accurate description of the relativistic precession of Mercury. This is also how JPL, the Russian Academy of Sciences, and the Paris Observatory (the organizations that produce the three leading ephemerides) now model the solar system.
     
  17. Mar 30, 2012 #16
    Do you know if the corrections are in first order what you would expect by naive thinking, ie. F->F(1+v2/c2/2), or something different?
     
  18. Mar 30, 2012 #17

    tom.stoer

    User Avatar
    Science Advisor

    I am not sure whether this is what you are looking for, but here's a paper discussing Einstein’s paper “Explanation of the Perihelion Motion of Mercury from General Relativity Theory”: http://www.wbabin.net/eeuro/vankov.pdf

    You will recognize the simplest post-Newtonian terms (for this special case).

    I am not sure if this is what you are looking for; I guess instead of "slow motion + strong gravitational field" you are interested in "fast motion + weak gravitational field"; anyway - the starting point is always the geodesic e.o.m.
     
  19. Mar 30, 2012 #18

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    They're a bit more complex (well, more than a bit more complex) than that. See equation 8.1 in http://iau-comm4.jpl.nasa.gov/XSChap8.pdf [Broken], for example (this is what JPL uses), or see equation 10.12 in http://www.iers.org/nn_11216/SharedDocs/Publikationen/EN/IERS/Publications/tn/TechnNote36/tn36__151,templateId=raw,property=publicationFile.pdf/tn36_151.pdf [Broken].
     
    Last edited by a moderator: May 5, 2017
  20. Mar 30, 2012 #19
    So basically yes, except it's hard to decide what the coefficient in place of 1/2 is because they are using vectors :) The equation in IERS document is pretty simple: first you have the correction from Schwarzschild metric, then two terms which look like v^2/c^2, and then corrections O(c^-4) which contain angular momentum and then it gets properly confusing, but we can drop those for now. The last term looks like some kind of angular momentum correction from the Sun, so lets drop that one too because we don't care about that... Finally, lets just be lazy and say we're on a circular orbit so [itex] \mathbf{r} \cdot \mathbf{\dot{r}} = 0 [/itex] and that it's general relativity so [itex] \gamma = 1 [/itex] and we're left with just
    [itex] \Delta |\mathbf{a}| = \frac{GMv^2}{c^2 r^2} [/itex] which is actually twice as big as you'd expect. Is that about right?
     
    Last edited by a moderator: May 5, 2017
  21. Mar 30, 2012 #20

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    That's off by a factor of three.

    Those "corrections which contain angular momentum" are frame dragging (Lens-Thirring effect), and they are small -- as is the solar influence (and where's the Moon?)

    After dropping the [itex]\vec r \cdot \vec v[/itex], frame dragging, and solar effects we're left with
    [tex]\Delta |\vec a| \approx \frac{GM_E}{c^2r^2} \left( 2(\beta+\gamma)\frac {GM_E}{r} - \gamma \, v^2\right)[/tex]
    For a circular orbit, [itex]GM/r \approx v^2[/itex]. Note: This is exact in Newtonian mechanics; since we're dealing with a perturbation on top of a perturbation, we can consider this to be essentially correct in a PPN formulation as well. Thus the relativistic correction reduces to
    [tex]\Delta |\vec a| \approx \frac{GM_E}{c^2r^2} (2\beta + \gamma)\,v^2[/tex]
    Since β=γ=1 in general relativity, this becomes
    [tex]\Delta |\vec a| \approx \frac{3GM_Ev^2}{c^2r^2}[/tex]
    This is pretty much what you'll see as a given in a simplistic derivation of the relativistic precession Mercury.
     
  22. Mar 30, 2012 #21

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    Maybe I'm blind, but I don't see any O(c^-4) corrections. I definitely agree we can dispense with the J terms for simplicity - consider the hypothetical non-spinning earth. I also agree it makes sense to pretend the sun doesn't exist, for simplicity, so that leaves only the first line.

    Then, using your circular orbit suggestion, I get something different:

    4(GM)^2/(c^2 r^3) - (GM/r^2) (v^2/c^2)



    -------
    One thing I don't understand is that versions of these approximate GR equations of motion I've seen before (like the first of DH references), have acceleration in a term on the right . This version does not. What gives?

    [Edit: I see DH posted first. Fortunately, our comments are similar.]
     
    Last edited: Mar 30, 2012
  23. Mar 30, 2012 #22

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    What gives is that those other references such as my first one are not valid for a satellite in Earth orbit. That equation in the second article I referenced gives the perturbative effect Δa. It does not say how to compute the non-relativistic aspects of the acceleration; that's an exercise left to the reader.

    You can't pretend that the only perturbations to spherical gravity are due to other bodies and general relativity. For an artificial satellite, atmospheric drag is huge for vehicles in low Earth orbit, and the non-spherical nature of Earth's gravity field dominates over relativistic effects out to order 20 or so even out to geosynchronous orbit. Even solar radiation pressure swamps relativistic effects.

    The same goes for orbits about the Earth's moon. Low lunar orbits are very bizarre.
     
  24. Mar 30, 2012 #23

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    Unfortunately, approximations that require that the speed be small (for instance the PPN formalism, i.e. Paramterized Post Newtonian approximations) can't really address the issue of how speed affects gravity, because the speed has already been assumed to be small.

    The biggest pitfall to watch out for in trying to treat gravity as a "force" is how it behaves in different reference frames. I'm not aware of anything that plays the role of "force" in full GR that transforms like a tensor (much less like a 4-vector, which is what one expects a force to transform like).

    The closest candidate I'm aware of are the Christoffel symbols - which are not tensors, and have the wrong rank as well.

    This argument may sound technical, and it is - but not transforming properly ultimately leads to a lot of confusion. It breaks the usual way in which physics summarizes all the different things that different observers might measure into a single, unified, observer-independent framework.

    In fact, I think that trying to understand how gravity transforms between different frames , how it appears to different observers, may be the root of the OP's question.
     
  25. Mar 30, 2012 #24

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    Actually, on review, I see something much more basic. The acceleration terms on the right of general EIH equations are accelerations of bodies other than the 'test' body relative to the barycenter. For satellite and earth, in earth centered coords, this is zero by definitions.
     
  26. Mar 30, 2012 #25

    PAllen

    User Avatar
    Science Advisor
    Gold Member

    The EIH equations do have velocity dependent terms, so they indicate, even for low velocities, that there is v^2/c^ base term (where v is relative velocity of test body and gravitating mass).

    Of course, more generally, I agree there is absolutely nothing in GR that has properties similar to gravitational force. My preferred analog is to match the physical situation. When we think of gravitational force in a Newtonian framework, we mean the force exerted on an object to resist free fall. This concept generalizes to GR except for the major complication that for a non-stationary geometry (e.g. a large mass moving by), there is no unique definition of a stationary world line. Yet, a reasonable answer is to take a world line that maintains a fixed position in Fermi-normal coords of a distant star (assuming asymptotically flat universe). Computing this, I get first order corrections of order gamma^2. In another thread, Bill_k addressed this problem in a different way, yet also got corrections of order gamma^2.
     
    Last edited: Mar 30, 2012
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook