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Einstein Equivalence Principle (EEP) states that

A frame linearly accelerated relative to an inertial frame in special relativity is LOCALLY identical to a frame at rest in a gravitational field.

And also, from this answer, it is stated that the EEP leads to the deduction that a test particle shall follow geodesic.

The reason that the equivalence principle is central to GR is in the fact that you can represent the gravitational field with a metric tensor at all--you can replace a force equation with a geodesic equation for a test mass precisely due to the fact that the geodesic that that test mass follows (or the "acceleration" felt by a Newtonian mass) is independent of the mass of that test particle.

But I fail to see how to infer logically from Einstein Equivalence Principle that a test particle should follow geodesic. Any ideas?

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  • $\begingroup$ related: physics.stackexchange.com/q/24359 (But I think the two questions are different, since this one is about how the geodesic hypothesis follows from the EP, not from GR.) $\endgroup$
    – user4552
    Commented Jun 18, 2018 at 2:17

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From the EEP it follows that if you have a freely falling test particle, then you should be able to define a local cartesian coordinate frame in which:

a) the test particle is at rest. b) the physics is locally described by physics in Minkowski time.

In particular in this frame all christofel symbols must vanish on particle worldline. We trivially conclude that in this local coordinate frame the worldline is parallel transporting the 4-velocity of the test particle. Since the 4-velocity is also tangent to the worldline, the worldline is parallel transporting its own tangent vector, the definition of a geodesic.

The EEP thus implies that test particles must follow geodesics.

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  • $\begingroup$ This seems fine to me. Why the downvote? $\endgroup$
    – user4552
    Commented Jun 18, 2018 at 17:45
  • $\begingroup$ Maybe the question is how the equivalence principle leads to geomety at all. How from requiring (free faling)=(in rest far from gravity sources) we come to the notion of curved spacetime. $\endgroup$
    – F. Jatpil
    Commented Nov 22, 2018 at 17:59
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One way of stating the EP is that Eotvos experiments should give null results. If test particles don't follow geodesics, then it seems likely to give non-null results of Eotvos experiments.

This is all about the EP, not about GR more specifically. We would like general relativity to obey the EP, but this is vague because the EP is hard to define rigorously. Some more rigorous work on this has been done by Ehlers and Geroch, http://arxiv.org/abs/gr-qc/0309074v1 . Without an energy condition, you can't prove geodesic motion of test particles in GR. I wrote up a lowbrow version of their argument in my GR book, section 8.1.3.

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  • $\begingroup$ You mean that it requires an energy condition in order to deduce "test particle follows geodesic" from EEP? $\endgroup$
    – Graviton
    Commented Jun 18, 2018 at 4:04
  • $\begingroup$ I think the energy conditions are only necessary to prove that small bodies always behave like test bodies in the limit that their mass goes to zero. They are, for example, completely unnecessary in proving that black holes follow geodesics in the limit of zero mass. $\endgroup$
    – TimRias
    Commented Jun 18, 2018 at 6:35
  • $\begingroup$ @Graviton: You mean that it requires an energy condition in order to deduce "test particle follows geodesic" from EEP? No, you need an energy condition to deduce this from GR, not to deduce it from the EP. And these are two different meanings of "deduce," since GR is a mathematically well-defined theory, and the EP is not completely well defined. See, Sotiriou et al, "Theory of gravitation theories: a no-progress report," e.g., arxiv.org/abs/0707.2748 . $\endgroup$
    – user4552
    Commented Jun 18, 2018 at 17:39
  • $\begingroup$ @mmeent: They are, for example, completely unnecessary in proving that black holes follow geodesics in the limit of zero mass. This may be true, although it's not at all obvious to me. Black holes don't fit within the assumptions of the proof by Ehlers and Geroch. In their proof, they set up conditions such that in a certain limit, the perturbation to the gravitational field from the test particle vanishes, and you recover the background spacetime. This doesn't work for a black hole, because the topology with the black hole differs from the topology without it. $\endgroup$
    – user4552
    Commented Jun 18, 2018 at 17:43
  • $\begingroup$ The black hole cause is far from trivial, but follows from a match asymptotic expansion analysis. See (for example) some of the recent papers by Adam Pound. $\endgroup$
    – TimRias
    Commented Jun 19, 2018 at 8:53

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