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Different gravitational fields from different reference points

  1. Feb 10, 2012 #1
    Hello. I am new to the idea of Special Relativity. From what I have read, based on what speed something is traveling it experiences an increase in mass. Let's say I am standing on the earth. I am experiencing a gravitational force of (me(m))/r^2. Now lets say I am the same person that somehow got hurled through space at a rediculous speed and I am about to do a close to touch and go fly by of the atmosphereless(I dont want to burn to death or slow down) earth. From my vantage point, the earths mass has greatly increased and from the earths reference frame my mass has greatly increased and therefore (me(m))/r^2 is much greater than the gravitational force experienced earlier correct? Also, if you divide by m you get the acceleration of the superfast me due to the gravity of earth which turns out to be me/r^2. Since the mass of the earth is greater with respect to the superfast version of myself than with the stationary version, I will actually accelerate towards the earth faster than the stationary version of myself correct? Also, wouldn't that mean that just by increasing my speed I have increased the potential energy of gravity from a certian height?
  2. jcsd
  3. Feb 10, 2012 #2


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    The short answer to your question is that you can't get a relativistic theory of gravity just by combining special relativity with Newton's law of gravity. The minimum you have to do is reinvent general relativity.

    By "reference points" I assume you mean frames of reference. Yes, it is certainly true that the gravitational field is frame-dependent. In the frame of reference of someone inside a free-falling elevator, the gravitational field is zero. This is the equivalence principle: http://en.wikipedia.org/wiki/Equivalence_principle

    One problem with your attempted analysis is that you implicitly assume the existence of a frame of reference that encompasses both the earth and the spacecraft doing the flyby. One of the implications of the equivalence principle is that when gravity is involved, frames of reference are only local, not global.
    Last edited: Feb 10, 2012
  4. Feb 10, 2012 #3


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    Have you thought, analyzed, or read about what happens to a charged test particle moving by a charged more massive reference charge (one that isn't affected by the passing test charge) at a high rate of speed? This is a simpler problem than gravity, but provides some insight.

    In particular it's useful to ask whether the electric field is frame independent of the frame, and if not how it transforms.

    And when it turns out the electric field is frame-dependent, with the components in the direction of motion transforming differently than the components perpendicular to the direction of motion, you'll have a little bit of insight of how and why the "gravitational field" is frame dependent.

    Generalizing to gravity isn't as straightforwards as one might hope, but a short and probably misleading answer is that the "gravitational field" transforms similarly to how the electric field does, but with an extra factor of gamma due to the fact that mass increases while charge does not.

    The short and misleading answer will probably be off by a factor as much as 2:1 if you seriously try to use it directly to figure out the bending of light by a gravitational field.

    A more accurate answer requires an extended discussion of what we mean by "gravitational field". Breifly, we can say that Faraday tensor describes the electromagnetic field. The "gravitational field" turns out not to be well defined in the same familar form (in this context, "well defined" means "transforms as a tensor"). It becomes necessary to think of the gravitational gradient, (in Newtonian terms, tidal forces), to describe gravity by a tensor, and the appropriate tensor for describing gravity is the Riemann curvature tensor. This also leads to the notion of gravity as a space-time curvature rather than a force.
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