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Is SR an approximation or a special case?

  1. Apr 18, 2012 #1
    I was having this debate with a friend, and I wanted to know if I was correct. My friend was saying that SR is an approximation for GR (albeit a very good one) with the specific conditions of only inertial reference frames, and I was saying that SR is exactly accurate with GR, and so is directly solvable from GR with those conditions made (doesn't matter if you can actually reduce GR mathematically to SR, just that logically you can).

    Am I right?

    Edit: I guess I can appendix this with another related question: can you mathematically reduce GR to SR?
     
  2. jcsd
  3. Apr 18, 2012 #2

    Dale

    Staff: Mentor

    Actually, the answer to the question in the title is that both are correct. In a region of spacetime small enough to neglect tidal effects SR is a good approximation to GR. SR is also a special case of GR, specifically a solution to the EFE with no stress energy anywhere.

    However, it seems like you may be under the mistaken impression that SR is restricted to inertial frames only. That is not the case. You can analyze accelerated motion or use non-inertial frames in SR just fine. All that you cannot do in SR is treat gravity. See:
    http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html
     
    Last edited: Apr 19, 2012
  4. Apr 18, 2012 #3
    SR is just the case of a flat space-time. Or put mathematically, the curvature of space is zero.

    It is just as exact as general relativity, but like DaleSpam said locally a curved space can be approximated by a flat one. Think of tangent planes to curved surfaces.
     
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