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Tests of EFE

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  1. May 29, 2015 #1
    There are several tests of general relativity:

    - gravitional redshift
    - deflaction of light
    - perihelion precession of Mercury
    - Shapiro delay
    - Lense-Thirring precession
    - binary pulsars


    Now comes the part that problems me. These all are based on vacuum solution of Einstein field equations.
    With vacuum solution EFE is 0 = 0, IMO not that good proof of validity of equations.


    So is there any real tests of EFE, other than Cosmology? (which have it's problems as we all know)

    Also looking for good explanations why left side of EFE must equal right side.
     
  2. jcsd
  3. May 29, 2015 #2

    Mentz114

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    The field equations follow from extrememizing an action as in any field theory.

    Your concerns about vacuum solutions are groundless. The vacuum solutions have zero Ricci tensor but still possess non-zero Weyl curvature which
    describes the field. It is very technical and if you study hard for a few years you will understand.
     
  4. May 29, 2015 #3

    Dale

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    You can always simplify any equation to that form if you choose. This is a complete non issue.

    All of the tests you mention are non-trivial solutions to the EFE.
     
  5. May 29, 2015 #4
    Can you then write EFE using only Weyl tensors? Is it so that basically Riemann tensor = Ricci tensor (trace part) + Weyl tensor(everything else)?

    I disagree, the vacuum solution is the Schwarzschild metric which have 0 = 0
     
  6. May 29, 2015 #5

    Mentz114

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    The only way to change the EFE is to change the action being extremized.

    A vacuum solution is created by solving the differential equations
    ##G(g)^{\mu\nu} =0##
    where ##g## is a function of the coordinates. We thus find a valid metric, which gives us ##0=0##, which is what we wanted.

    See here http://en.wikipedia.org/wiki/Deriving_the_Schwarzschild_solution
     
  7. May 29, 2015 #6
    Yes, deriving the schwarzschild solution we assume that rigth side must be zero because vacuum and then find metric that gives also left side zero (R = 0).

    But this 0 = 0 does not prove that EFE is correct hypothesis. One can put anything one wants to the right side, just make sure it is zero (e.g. number of magnetic monopoles).

    I ask again, is there any experimental tests that have rigth side of EFE not zero?
     
  8. May 29, 2015 #7

    Mentz114

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    That is rubbish. You have to put the correct metric in. The metric is the solution. Your understanding of logic and mathematics is flawed.

    The Schwarzschild metric satisfies all the experimental tests that have been tried - so your assumption is flawed in any case.

    The equation ##x^2-2ax+a^2=0## has a solution ##x=a## which gives ##0=0##. What is wrong with that logic ?
     
    Last edited by a moderator: May 29, 2015
  9. May 29, 2015 #8
    Nothing is wrong with that logic.

    You have proved that x2-2ax+a2 = 0 is true (have solution) but you have not proved that x2-2ax+a2 = b when b is not 0.

    I agree that the Schwarschild metric satisfies all the experimental tests that have been tried and those I listed on first post.

    I just don't understand how Schwardschild metric can "prove" EFE if equation reads 0 = 0.
     
  10. May 29, 2015 #9

    Mentz114

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    Well, you need to understand what solving an equation means. It is like a set of scales. The scales still work when they have nothing to weight. They are saying ##0=0##.

    Take the quadratic equation I wrote. I could have written ##x^2-2ax = -a^2##. Now put in the solution ##x=a##. Now the equation reads ##-a^2=-a^2##.

    We know the solution is correct when both sides are the same.

    I don't think I can help any further with this.
     
  11. May 29, 2015 #10

    PeterDonis

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    No, it isn't. The EFE is a second order partial differential equation (more precisely, it's a system of 10 of them in the general case). So the RHS being zero just means it's a homogeneous second order partial differential equation, which certainly has solutions other than the trivial one. See here for an overview:

    http://en.wikipedia.org/wiki/Vacuum_solution_(general_relativity)
     
  12. May 29, 2015 #11
    So, you wrote this for wiki? Cool.
     
  13. May 29, 2015 #12

    PeterDonis

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    If you mean, did I write the Wikipedia page I linked to, no, I didn't. I just linked to it as a reference.
     
  14. May 29, 2015 #13

    Dale

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    So what? This is true of any solution of any equation. You can always simplify it to 0=0:

    Given A=B
    A-B=0 by subtraction
    0=0 by substitution

    It is a completely useless complaint because it is always true for any equation. If you over-simplify you can always get 0=0. Your query about experiments or observations with a non-vacuum metric is a valid question, but this bit about 0=0 is not.
     
  15. May 29, 2015 #14

    atyy

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    Conceptually, these are not tests of the pure vacuum solutions, because in addition to the vacuum solution, geodesic motion is assumed. The geodesic motion is derived by assuming the presence of matter: http://arxiv.org/abs/0806.3293.

    What problems does cosmology have? Are you thinking about dark matter?

    For general relativity with matter, there seems to still be research going in these areas;
    http://www.einstein-online.info/spotlights/hydrodynamics_realm
    http://arxiv.org/abs/1210.4921
    http://arxiv.org/abs/1206.2503
     
  16. May 29, 2015 #15

    PeterDonis

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    Just to clarify, this is referring to geodesic motion of bodies that are not "test bodies", because they have significant self-gravity, correct? For example, geodesic motion of the Earth in the gravitational field of the Sun.

    This "presence of matter" is still very different from the "matter" that is present in cosmological solutions. In the latter, the matter significantly affects the spacetime geometry everywhere. In the former, the matter only significantly affects the spacetime geometry (in the sense of requiring a nonzero stress-energy tensor in the local solution of the EFE) in isolated regions corresponding to the central gravitating body producing the overall geometry (e.g., the Sun), and the self-gravitating bodies undergoing geodesic motion in this geometry (e.g., the Earth). Everywhere else, the solution is a vacuum solution, and that vacuum solution is what is used to predict the results of the classic solar system test experiments. (For example, nobody uses a nonzero stress-energy tensor for the Sun to predict the bending of light by the Sun; they just use the Schwarzschild geometry of the vacuum region exterior to the Sun.)
     
  17. May 29, 2015 #16

    atyy

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    I only meant that there are no test bodies, so although we treat the earth in the gravitational field of the sun as a test body, we do it by assuming a solution in which matter is present, and derive the treatment of the earth as a test body as a very good approximation.
     
  18. May 30, 2015 #17
    First I like to thank to get my thread back. I know my questions are very edge of the this forum rules. I will try not to go over.

    I will try to make my point by some simple examles.

    One can write equation:

    apples = oranges

    This is true if set both apples and oranges to zero. Equation reads 0 = 0. But surely it is not true if there are nonzero items of either.

    One can write equation:

    polar bears = constant * pink elephants.

    Assume we make some experimetal tests on Afrika. Because there is no polar bears and neither pink elephants on Afrika, we are happy that test results agree with our equation which reads 0 = 0. We know there is polar bears on north pole but we have not yet made any tests because north pole is so far away from Afrika.

    Can we now say from our test results that our polar bear/pink elephant equation is correct? No we don't. We have only tested one point (0=0).
    This is the reason I asked is there any experimental test results where we have EFE something else than 0 = 0.
     
    Last edited: May 30, 2015
  19. May 30, 2015 #18

    atyy

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    @CycoFin, as DaleSpam and others have said, your objection makes no sense. Every equation can be written in the form U = 0, including the full Maxwell's equations with charges and currents, and Newton's laws etc. It is true that there are regimes of GR that remain untested. However, you can consider the vacuum solutions of GR analogous to Maxwell's equations without charges. Even in the absence of charge, Maxwell's equations predict interesting phenomena such as electrotromagnetic waves that travel over great distances.
     
  20. May 30, 2015 #19

    atyy

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    In fact, the equation can be rewritten as

    apples - oranges = 0.

    Also, the equation can be true if there are nonzero items of either - it means that the number of apples and oranges is the same.
     
  21. May 30, 2015 #20
    Now I am little bit lost. You say that even we have RHS stress-energy T zero, LHS Einstein tensor G is not???

    I know there are several diffrent vacuum solutions with different metrics but it does not change that in tensor point of view EFE reads 0 = 0 (these are zero tensors, not zeros)
     
  22. May 30, 2015 #21
    Yes but if you make experimental tests when you never have neither apples nor oranges, you can not say if this equation is explainng nature.

    All you can say that there is no apples or oranges on universe.
     
  23. May 30, 2015 #22

    atyy

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    But the vacuum solutions correspond to the case where where apples and oranges are not each zero.
     
  24. May 30, 2015 #23
    Vacuum solutions makes our RHS zero. There is no energy or stress on vacuum (not talking about cosmology here now). So our oranges are zero.

    Then we assume that our LHS side must be zero also. We assume apples zero. We make measurements that agrees our assumption that apples are zero.

    Can we really then say that apples = oranges because they both are zero? What about changing that energy-stress tensor to pink elephant tensor. That is also zero in vacuum.

    So why we don't just use apples = 0 and be happy. And add those oranges once we have tests that contain non-zero numbers of oranges (of course if our apples = oranges is still valid)
     
  25. May 30, 2015 #24

    Nugatory

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    Of course not.
    It's the metric tensor that we're solving for. The EFE supplies constraints on what this unknown metric tensor might be; one of these constraints is that the Einstein tensor calculated from the unknown metric tensor must be zero at points where the stress-energy tensor is zero. Given the complexity of the relationship between the metric and the Einstein tensors, ##G=0## is a seriously non-trivial constraint.
     
  26. May 30, 2015 #25
    Yes, we have G = 0, I don't disagree this and metric explanation of gravity makes sense and this have been very successfully tested.

    But surely there must be some experimental test results so we can write G = constant * T, where T is nonzero energy-stress tensor.

    If not, R = 0, where R is ricci tensor is enough to descripe gravity effects on vacuum.
     
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