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Mach's Principle and General Relativity

  1. Nov 21, 2008 #1


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    I don't know general relativity yet, but I have a few questions for which I would like Yes/No answers.

    Let S be an inertial reference frame with its origin in a relatively empty region of space. Let S' be a concentric frame which rotates with constant angular velocity. A free particle in this region obeys the law of inertia, and therefore moves in a straight line with constant [tex] v [/tex] in in the S system. The dynamics of this particle in the S' system can be calculated by performing a Galilean transformation from S to S' (assume [tex] v << c [/tex]).

    This is what I know about general relativity:

    1. It's equations work in all frames of reference (or at least all physically reasonable frames such as S and S' above)

    2. The acceleration/gravitational field in any frame of reference is ascribed to the mass distribution.

    Now for my questions:

    1. Now, does this mean that in the S' frame, the acceleration field must be ascribed to the rotating celestial sphere?

    2. If so, under the assumptions above ([tex] v << c [/tex], and no nearby masses), will the resulting dynamics of a free particle be the same as the one calculated by the principle of inertia with Galilean transform above?

    3. Was any assumption about the mass of the celestial sphere or average mass distribution of the universe used in the GR calculation?

    Also, I don't know exactly what Mach's principle is. If it has nothing to do with this, just ignore that part of the title and answer the questions anyway.
    Last edited: Nov 21, 2008
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  3. Nov 21, 2008 #2

    Jonathan Scott

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    These questions do relate to Mach's principle. GR does not have a fully satisfactory answer to this question. One variant of "Mach's Principle" is the idea that it should be possible to describe what happens in a rotating frame of reference EITHER by describing it from a non-rotating frame OR by considering the "gravitomagnetic" influence of the universe rotating around it, and both should give the same result. Unfortunately, GR can't really do that so far.

    In GR, we can certainly calculate the effect of one rotating mass, or of a rotating shell around an object, and that does show the right sort of effect at the right sort of scale. However, if we try to extend that to the whole universe, we run into lots of problems. Firstly, we haven't yet got general two-body solutions in GR, let alone the universe (except by assuming it is homogeneous).

    Also, the effective rotation rate depends on an expression which includes G multiplied by a "sum for inertia" involving M/R terms for everything in the universe, but the effective rotation rate is supposed to exactly match the rate at which the fixed stars appear to revolve in this case (giving a 1:1 match between the rotation of the universe and its induced effect), and it is difficult to see how this GR expression for rotation can adjust to match the current distribution of mass in the universe, as this seems to need the gravitational "constant" G to vary according to location, which would be inconsistent with GR.

    I think the general consensus is that GR loosely conforms to Mach's principle, but only approximately.
  4. Nov 21, 2008 #3


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    In the sense that the calculation hasn't been done?

    If it is only approximate, then does that mean that the acceleration field must be ascribed partly to something other than the distant masses? If so, what?
    Last edited: Nov 21, 2008
  5. Nov 21, 2008 #4


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    Loosely speaking, in GR there are two types of gravitational fields. In GR, one can choose a local Lorentz frame, so that the metric at the exact centre of the coordinate system is the same as in SR. As one goes away from the centre, deviations from the SR metric occur.

    An accelerated observer in a flat spacetime in which mass is absent (we assume the observer's mass is negligible) experiences fake gravity. The uniformly accelerated Rindler frame or the rotating frame you mention represent such observers. In his local Lorentz coordinates, he will have a first order deviation from the SR metric because he is not inertial. The Riemann tensor should show that spacetime is absolutely flat.

    A freely falling observer in a curved spacetime caused by the presence of mass experiences true gravity. In his local Lorentz coordinates, he will have not have a first order deviation from the SR metric because he is inertial, but he will have a second order deviation from the SR metric due to the curvature of spacetime. The Riemann tensor should show that spacetime is absolutely curved.
  6. Nov 21, 2008 #5


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    Not only hasn't been done, but cannot be done, unless inertia and gravitation are seen as local effects emerging from matter's interaction with the space in which it is embedded, according to Einstein. If you read his 1920 Leyden address and the 1924 essay "On the Ether" you will see that Einstein was acutely aware of the conflict of GR with Mach's Principle and he had to reject Mach's Principle. You can find "On the Ether" as Chapter 1 of "The Philosophy of Vacuum".
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