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Non-static Newtonian Limit?

  1. Apr 9, 2013 #1
    Hey all,

    I was wondering if anyone could help me clear up the following conceptual problem I'm having:

    When going to the Newtonian limit, authors tend to throw out the terms in the geodesic equation that involve derivatives of the spatial coordinates

    [tex]\frac{d x^i}{d \tau}[/tex]

    because they are smaller than the derivative of time. Now, that seems fine if you're working to leading order. But then, they keep just the term

    [tex]\frac{d^2 x^\mu}{d \tau^2} + \Gamma^\mu_{00} (\frac{dt}{d\tau})^2 = 0[/tex]

    and find that

    [tex]h_{00} = \frac{2 G M}{r}.[/tex]

    Now, by the virial theorem, one finds that [itex]h_{00}[/itex] is then actually second order in [itex]v[/itex]. But then we were not working to leading order at all! So, it seems to me that throwing out said terms in the geodesic equation is not allowed, since the derivatives of the spatial coordinates are only one order of [itex]v[/itex] smaller than the derivatives of time:

    [tex]\frac{d x^i}{d \tau} = \frac{d x^i}{dt} \frac{dt}{d\tau} = O(v) \frac{dt}{d\tau}.[/tex]

    I tried working it out, keeping the terms, and it seems to me there are some that do not vanish to second order. Am I missing something here?
     
  2. jcsd
  3. Apr 10, 2013 #2
    Does anyone have any insight? For example, when I calculate

    [tex]\Gamma^i_{00}[/tex]

    to second order in [itex]v[/itex] (since you end up with [itex]g_{00}[/itex] to second order, I'm keeping everything to that order), I find not only the standard term, but also a term

    [itex]g_{i0,0}[/itex]

    which is second order as long as you keep [itex]g_{i0}[/itex] first order.
     
  4. Apr 10, 2013 #3

    pervect

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    If you're keeping only first order terms, you throw out the second order terms in v - because they are second order, and you're only keeping first order terms.
     
  5. Apr 11, 2013 #4
    Yes, that's the argument most books give, but I don't understand it. Once you're done with the calculation of the Newtonian limit, you find that [itex]h_{00}[/itex], which you kept, is actually second order in [itex]v[/itex] by the virial theorem, so if you were expanding to first order, you should have thrown that out too.
     
  6. Apr 11, 2013 #5

    atyy

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  7. Apr 11, 2013 #6
    It does seem like cheating. However, I wouldn't be so worried except for the fact that when going to post-Newtonian order, e.g. Maggiore takes this static Newtonian limit and uses it to assume that there are no first-order metric components in the expansion in [itex]v[/itex], which doesn't seem to have any other justification. I read elsewhere that these first-order terms actually drop out of the geodesic equation, but I tried my hand at the calculations and some first order terms seem to remain. Does anyone know how this is resolved? I'm guessing it is resolved, since Weinberg (1972) already doesn't worry about it. He doesn't provide a better answer, though, his argument is "from experience with the Schwarzschild metric" and I don't know what that means.
     
  8. Apr 11, 2013 #7

    pervect

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    I don't see the problem. To first order in x, 1+x^2 = 1. It's perfectly fine to want to do a second order expansion, but then you need to include all terms of order x^2, and it's no longer a first order expansion.
     
  9. Apr 11, 2013 #8
    Yes, my point is that in the "first order" expansion, i.e. the Newtonian limit, you actually keep a term [itex]h_{00,i}[/itex], which is second order, not first order at all. To see this, consider the matching to Newtonian gravity, which gives you

    [tex]h_{00} = \phi = \frac{GM}{r}[/tex]

    where I neglected some constant prefactors. By the virial theorem,

    [tex]\frac{GM}{r} \sim \frac{Mv^2}{r}[/tex]

    so [itex]h_{00}[/itex] is second order in [itex]v[/itex].

    EDIT:

    Actually, it doesn't even make sense to expand the geodesic equation to first order in [itex]v[/itex], because the accelaration involves two time derivatives, which are each first order.
     
    Last edited: Apr 11, 2013
  10. Apr 11, 2013 #9

    Bill_K

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    The so-called Newtonian limit is really twofold: slow motion and weak field. To say you're working to first order or second order is an oversimplification. Must be more specific. There are two independent small parameters involved: v/c and (for a central field) rS/r. The virial theorem won't hold in general, unless you make an assumption about the relative size of these two parameters.
     
  11. Apr 11, 2013 #10
    Well, that is exactly the assumption made in post-Newtonian theory. It deals with systems held together by gravitational interaction, thus the virial theorem holds.
     
  12. Apr 11, 2013 #11
    Truth of the matter is that pure GR calculations with more than one mass are extremely problematic. In essence GR is a non-local theory as soon as there is more than one mass. This is characterized by the difficulty in attempting to create an initial value for calculations.
     
  13. Apr 11, 2013 #12
    Perhaps someone will recognize my problem if I phrase it slightly differently: the original question I had was why, in the PN expansion of the metric, the [itex]g_{i0}[/itex] components are assumed to be zero to first order. See e.g. page 239 of Maggiore. Weinberg does the same thing in his chapter on Post-Newtonian Celestial Mechanics. Both give barely any justification.
     
  14. Apr 11, 2013 #13

    Bill_K

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    Actually, Weinberg's justification is extremely clear and detailed. Suggest you take another look.
     
  15. Apr 11, 2013 #14
    I was aware that Weinberg performs the calculations in great detail, but the justification for dropping the [itex]g_{i0}[/itex] to first order wasn't clear to me. Having another look at it, I just realized a mistake I was making, however: I was coupling n-th order terms in the Einstein tensor to n-th order terms in the stress-energy tensor, while Weinberg couples them to order (n-2) terms. Doing this obviously makes [itex]g_{i0}[/itex] vanish to first order. I'm assuming the reason for this is the factor of [itex]G[/itex] in the Einstein equations?
     
  16. Apr 12, 2013 #15
    For some reason I can't edit my post, so:

    Just for posterity, my last sentence in the above is of course complete nonsense. A constant cannot be of any order in an expansion parameter other than zero. The real reason is a bit more complicated and has to do with the derivatives of the metric taken in the Einstein equations, as well as with the units of the stress-energy tensor.
     
  17. Apr 13, 2013 #16
    I understood and agree with everything you said there, except I don't understand what you mean by "In essence GR is a non-local theory as soon as there is more than one mass".
    Also, has there been attempts to solve the two body problem in GR? I can hardly imagine what the solutions would look like..assuming that we model the planets as point particles following geodesics, that would couple the EFE to the geodesic equation, the stress energy tensor would be very hard to write down, since we do not know where the bodies are so we cannot decide when ρ is non-zero and finally there is the problem of coordinates...
     
  18. Apr 13, 2013 #17

    Mentz114

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    It's even worse than that. If you're looking for a vacuum solution there is a zero Einstein tensor. Presumably the symmetries of the appropriate SET would be encoded into the Riemann tensor. In order to locate the bodies presumably there would be delta functions or distribution functions in the SET.
     
  19. Apr 13, 2013 #18
    We do not know if analytic solutions with more than one mass exist (except for exceptions like the Schwarzschild lattice closed universe), but even worse, if the Einstein equations are even correct for multiple masses. Right now all we have is solutions for single 'blobs' of mass/energy and vacuum solutions.
     
  20. Apr 13, 2013 #19

    Mentz114

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    There are the Majumdar-Papapetrou spacetimes, vacuums where there can be multiple extremal RN black holes. They are no more physical than the extremal BHs but they are solutions of the EFE.
     
  21. Apr 13, 2013 #20
    I never said anything about the solutions being analytic, but the problem is much the same in electromagnetism, trying solving for the exact motion of two charged spheres. Everything gets coupled to everything else.
     
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