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How can I show that G=8pT implies the equivalence principle ?

  1. Nov 9, 2006 #1
    If I am asking this question, this is maybe a proof that I need a strong "back-to-the-basics".
    Could you give the way?

    Thanks,

    Michel
     
  2. jcsd
  3. Nov 9, 2006 #2
    Equivalence of acceleration and gravitational forces? I had the impression that principle is basicly implied, before the EFE (which you quote) is imposed, when you start by modelling your gravitational interaction as a curvature of space-time instead of as a force on top of (flat) space-time. Then the EFE is found so as to match that curvature (in certain limits) quantitatively to accelerations that are measured or have been predicted by Newtonian gravity (though we could instead have found a different field equation, or modified the EFE, and still retain the equivalence principle).

    Perhaps more the answer you want is: If you take the EFE, it implies that mass-energy curves space-time in such a way that masses will accelerate towards one another without any other forces (and therefore there is absolutely nothing locally to distinguish between the acceleration you feel a constant distance from the earth and the acceleration you might feel on a rocket ship in deep space). Indeed, on earth the normal electrostatic force that ground applies to your feet (pushing you up to prevent you falling) is no different to the force that might be produced by a rocket engine's thrust or by the floor of an elevator elsewhere.
     
    Last edited: Nov 9, 2006
  4. Nov 9, 2006 #3

    Garth

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    Consider the space-time around [itex]x_{\mu}[/itex] where the following holds:

    [tex]G_{\mu\nu} = 8\pi GT_{\mu\nu}[/tex]

    which gives

    [tex]G^{\mu}_{\nu;\mu} = 8\pi GT^{\mu}_{\nu;\mu}[/tex]

    As [tex]G^{\mu}_{\nu;\mu} = 0[/tex] [The Bianchi identities]

    therefore [tex]T^{\mu}_{\nu;\mu} = 0[/tex]

    [itex]T^{\mu}_{\nu;\mu}[/itex] describes the force density acting at [itex]x_{\mu}[/itex], in other words the force density acting on a freely falling particle is zero, hence it suffers is no acceleration in the freely falling frame of reference and the space-time is Minkowskian in a sufficiently small area around the point [itex]x_{\mu}[/itex].

    Garth
     
    Last edited: Nov 9, 2006
  5. Nov 10, 2006 #4
    Garth,

    I don't understand why you can conclude this:

    I think that the answer may be simple or even obvious. I thought first that the answer would be obtained by comparing motions in two frames (accelerated without gravitation and another with gravity) and concluding that physics is the same. I also considered that it could be simply related to the invariance of the equations.

    Thanks to help me on that,

    Michel
     
  6. Nov 10, 2006 #5

    Garth

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    Yes it is obvious - if you think about it a bit!

    If there is curvature but there is no force denstiy acting on a 'test particle' at a particular location then there is no force acting on it and therefore its four-momentum is constant.

    If there is no curvature at a particular location the test particle's four-momentum is also constant.

    Hence the EEP.

    Garth
     
  7. Nov 14, 2006 #6

    Demystifier

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    The Einstein equation, which is the equation of motion of the gravitational field, does NOT imply the equivalence principle. Instead, it is the equation of motion of MATTER that implies the equivalence principle, provided that matter is MINIMALLY coupled to gravity (i.e. to the metric tensor).

    The conservation of energy-momentum implies the equivalnce principle only if [tex]T_{\mu\nu}[/tex] is specified and equal to the quantity that corresponds to the minimal coupling of matter to gravity.
     
    Last edited: Nov 14, 2006
  8. Nov 14, 2006 #7

    Garth

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    I do not think that statement makes sense does it?

    What does "equation of motion of the gravitational field" mean?

    The Einstein gravitational field equation
    [tex]R_{\mu \nu} - \frac{1}2{}g_{\mu \nu}R = 8\pi GT_{\mu \nu}[/tex]

    connects the curvature of space-time with the distribution of mass, energy and stress.

    The equation of motion of matter is derived from this equation.

    viz: from the Bianchi identities

    [tex]G^{\mu}_{\nu ;\mu} = 0[/tex] therefore

    [tex]T^{\mu}_{\nu ;\mu} = 0[/tex]

    which can then be solved for particular cases such as that of a perfect fluid:

    [tex]T_{\mu \nu} = \rho u_{\mu}u_{\nu} + p(g_{\mu\nu} + u_{\mu}u_{\nu})[/tex]

    Further more, the statement "matter is MINIMALLY coupled to gravity (i.e. to the metric tensor)" would imply the metric does not influence the paths of matter at all.

    Garth
     
  9. Nov 14, 2006 #8

    Demystifier

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    There are many many other solutions besides the perfect-fluid solution. Einstein equation is obtained by varying action with respect to metric. To obtain the matter equation of motion, you need to vary action with respect to matter degrees of freedom. The energy-momentum conservation is NOT the equation of motion.

    Think also about the following. One of the solutions of the energy-momentum conservation equation is the "cosmological constant" solution
    [tex]T_{\mu\nu}(x)=g_{\mu\nu}(x)\lambda .[/tex]
    Does it obey the equivalence principle?
     
    Last edited: Nov 14, 2006
  10. Nov 14, 2006 #9

    Demystifier

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    Did you ever heard about the notion of "minimal coupling"?
    In the case of gravity, minimal coupling means that the only way the nontrivial metric enters the equations is through replacing the ordinary derivatives by covariant derivatives.
     
  11. Nov 14, 2006 #10
    Demystifier,

    You are right about the Einstein's equations, I realised that shortly after posting.
    Clearly I have to check the EP on the equations of motion.
    Would you have a reference on the web that shows that explicitely?

    Thanks,

    Michel
     
  12. Nov 14, 2006 #11

    Demystifier

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    Not really.
    Nevertheless, if you need a pedagogic introduction to general relativity, I recommend
    http://arxiv.org/abs/gr-qc/9712019
     
  13. Nov 14, 2006 #12

    Garth

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    Agreed, that is why I said "such as"
    Agreed, varying the action in 4D space-time necessarily results in the conservation of energy-momentum because the method is true for generalised coordinates.
    In which case, in the case of no external forces acting, you disagree with Wald "General Relativity" see Eq.4.3.6 and the discussion on page 73.
    Obviously in GR it does locally.

    Could you clarify what you meant by "the equation of motion of the gravitational field"? Thank you.

    Garth
     
    Last edited: Nov 14, 2006
  14. Nov 14, 2006 #13

    Garth

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    I am familiar with the expression being used referring to a scalar field, in which case if the scalar field is minimally coupled then its presence affects only the curvature of space-time and does not perturb test particles from their geodesic world-lines through that space-time.
    I have never used "minimal coupling" in that sense and I find its use a little over-redundant. Which authors do use it as such?

    Garth
     
    Last edited: Nov 14, 2006
  15. Nov 14, 2006 #14

    Demystifier

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    1. Of course, but I was talking generally, not about the special case without external forces.

    2. I meant the equation that one obtains by varying the gravitational field in the total action.
     
  16. Nov 14, 2006 #15

    robphy

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  17. Nov 14, 2006 #16

    Demystifier

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    This terminology perhaps is not common in GR, but is common in gauge theories. And you certanly know that GR can be viewed as a gauge theory.
     
  18. Nov 14, 2006 #17

    Garth

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    Agreed, and [itex]G_{\mu\nu}[/itex] describes that curvature.

    Garth
     
  19. Nov 14, 2006 #18

    robphy

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    [itex]G_{\mu\nu}[/itex] describes part of that curvature.

    The point I was making is that the Einstein Field Equations are not required for the so-called Equivalence Principle.
     
  20. Nov 14, 2006 #19

    Garth

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    Agreed, the Riemannian fully describes the curvature.
    Yes indeed, the WEP is satisfied in Newtonian theory and the Brans-Dicke Field Equations equations satisfy the SEP, but noting G varies with position in that theory.

    The OP, however, was asking the question the other way round, that is, does the EFE imply, not is it required for the EEP.

    Garth
     
    Last edited: Nov 15, 2006
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