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Invariance of Acceleration in Inertial Reference Frames

  1. Oct 13, 2015 #1
    Claim: The acceleration (both direction and magnitude) for any object is the same in any inertial reference frame.

    Is this claim true? I think it is, but someone mentioned to me that time may be an issue as it's not agreed upon in all inertial reference frames.

    I'd appreciate any references, if available.
     
  2. jcsd
  3. Oct 14, 2015 #2
    Why don't you look up the equations in special relativity and calculate it yourself? My calculation suggests that this statement is wrong, but don't take my words for it, my math is rusted.
    $$a{\rm{'}} = \sqrt {1 - {{{v^2}} \over {{c^2}}}} {(1 - {{vu} \over {{c^2}}})^{ - 3}}a$$
     
  4. Oct 17, 2015 #3
    Thanks Xu Shuang for your reply. It's been a long time ago since I studied special relativity. But from your equation, it seems if the velocities are all much less than the velocity of light, then the statement would be valid, correct?
     
  5. Oct 17, 2015 #4

    Nugatory

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    It is true for proper acceleration but not true for coordinate acceleration.
    Wars have been fought and cities razed to the ground because someone made a statement about acceleration without qualifying whether they meant coordinate or proper acceleration, so it's not surprising that you're finding some disagreement.

    The concept of proper acceleration is usually introduced when you first study special relativity, so this thread maybe belongs in the relativity forum.
     
  6. Oct 18, 2015 #5

    Orodruin

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    Since this thread was originally not in the relativity forum, I think it is worth pointing out that the acceleration is the same in all inertial frames in classical mechanics. It is only when relativistic effects become important when this is no longer true. But as Xu said, you should not take anyone's word for this either, you can simply derive the relation yourself based on the Galilei transformations.
     
  7. Oct 21, 2015 #6
    Is this only for the x-component or for all 3 components?

    Does this mean that the direction of the acceleration vector doesn't change?
     
  8. Oct 27, 2015 #7
    This is just x component. 4-dimensional space-time is too much for my math.
     
  9. Nov 2, 2015 #8
    I think it should be

    $$a{\rm{'}} = (\sqrt {1 - {{{v^2}} \over {{c^2}}}})^{3} {(1 - {{vu} \over {{c^2}}})^{ - 3}}a$$
     
  10. Nov 2, 2015 #9
    Are you sure? It was a full paper of calculation.
     
  11. Nov 15, 2015 #10
    Quite sure.
     
  12. Nov 15, 2015 #11
    I've been meaning to take a closer look at this for a while, it seems to be fairly exhaustive and correct (I went through it once without taking it in properly).
     
  13. Nov 17, 2015 #12
    I did it again, yours is right.
     
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