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Are angle measurements rank 0 tensors?

  1. Jun 24, 2010 #1
    If I measure an angle in one reference frame to be 90 degrees, would it be 90 degrees with respect to all other reference frames? That is, is angle measurement a rank 0 tensor? I'm assuming all other reference systems are at non-relativistic velocities.
  2. jcsd
  3. Jun 24, 2010 #2


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    In flat metric, yes. You can look at an angle as inverse cosiine of the dot product, and dot product of two Lorentz vectors is Lorentz-invariant.

    Edit: The part that's implied here is that the coordinate system is inertial. If it's accelerated, that messes with your metric.
  4. Mar 29, 2011 #3
    What if the accelerating coordinate system is far below relativistic speeds. Would invariance still exist for the angle measurement to a good approximation?

    Or does speed not have anything to do with it, only acceleration. If this is the case, if a coordinate system accelerated only at a small fraction of g and far below relativistic speeds, would invariance still exist for the angle measurement to a good approximation?
  5. Mar 29, 2011 #4
    Disregarding relativity, yes, the angle between two vectors is invariant. An accelerating observer will see the same angle as a nonaccelerating observer. A rotating observer will see the same angle as a nonrotating observer. The angle between two vectors is as constant as things get.

    When you throw Einsteinian relativity into the mix, the angle between two vectors is NOT invariant. Let's declare you to be at rest. You draw a square on the ground. Pick a corner. One side coming out of that corner is called "side A" and the other is "side B." Now draw a diagonal line from your corner to the opposite corner. Call this line "diagonal D." Now the angle between side A and diagonal D is 45 degrees, and so is the angle between B and D.

    I get in my rocket and move at, um, 87% of the speed of light in a direction parallel to side B. Then when I look at the square, I'll see Lorentz contraction: side A will look the same to me as to you, but side B will appear only half as long. So from my perspective, the square is now a rectangle where the angle between A and D gets smushed to 27 degrees and the angle between B and D stretches to 63 degrees, but you still think it's a square and that the angles are 45 degrees each.

    But again, if we ignore Einsteinian relativity, lengths and angles do not appear to change just because the observer is moving or accelerating.
  6. Mar 30, 2011 #5
    Ok, I understand, due to Lorentz contraction, how a moving observer at relativistic speeds would observe a different angle measurement than an observer in the frame where the angle measurement is at rest. I also understand at non-relativisitc speeds, exact invariance would still not exist, but the difference would be so trivial we could ignore it.

    Its the acceleration that I'm not sure of. Lets say initially the frame is at rest. An observer in that frame is making his angle measurement. A second observer would agree with the measurement that observer is making because since the frame is at rest, both observers are in the same frame.

    Now, at some point in time, the frame begins to accelerate with respect to the second observer at a huge acceleration rate. Now we compare the measurements of the two observers 1 pico second later from the time the frame began its acceleration, such that the speed of the frame hasn't had the time to acquire any significant speed and is far below relativistic speeds. Lets say relative to the second observer its instantaneous speed is 1 centimeter per second.

    Invariance of the angle measurement for all practical purposes would still exist because of the non-relativistic speed of the frame, but because the frame is accelerating at that instant at a huge rate, would invariance break down-- soley due to the acceleration alone?
  7. Mar 30, 2011 #6
    No, instantaneous acceleration never makes a difference, even when we take relativity into account. Only your instantaneous velocity will affect angle measurements.
  8. Mar 30, 2011 #7
    Thanks for clarifying this problem for me.
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