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Interpretation of Relativistic Momentum

  1. May 31, 2013 #1
    I understood the derivation of relativistic momentum, but I am uncertain of how to exactly interpret it. One could interpret the arrangement of terms to be relativistic mass times velocity, and this appears to be in agreement with data from particle accelerators (or so I have been led to believe). Alternatively, one could rearrange the terms to describe relativistic momentum as rest mass times proper velocity. In my opinion, parameterizing position with respect to proper time seems much more natural than using coordinate time, much like parameterizing a space curve with respect to length. Proper velocity also retains the properties of classical velocity in that its magnitude ranges from zero to infinity. What is the popular interpretation of relativistic momentum? Which of these two interpretations is correct? Or is there a deeper relationship between the two?
     
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  3. May 31, 2013 #2

    PeterDonis

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    Do you mean 4-momentum? As in, the 4-vector whose components are ##(E, p_x, p_y, p_z)## in an inertial frame?

    They both are, if by "correct" you mean "valid". Consider the components I wrote above; what I wrote is equivalent to ##(\gamma m, \gamma m v_x, \gamma m v_y, \gamma m v_z)##. You can rewrite this as ##\gamma m (1, v_x, v_y, v_z)## or as ##m (\gamma, \gamma v_x, \gamma v_y, \gamma v_z)##. It's just a matter of shuffling around factors.

    Also, you left out one: the one I implicitly used above when I wrote down the components. Under this interpretation, the 4-momentum is often called the "energy-momentum 4-vector", since, as you can see, it includes energy as well as momentum.

    However, I think you have a misconception that these interpretations correspond to "real" things--i.e., that they describe different possible ways that reality can be. That's not correct. The "interpretations" are, as the above makes clear, just different ways of shuffling around factors in the math. They are all valid, but none is more "real" than the others.
     
  4. May 31, 2013 #3

    Dale

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    My prefered interpretation is simply that momentum is a non-linear function of velocity. I don't know why it needs to be interpreted further than that.
     
  5. May 31, 2013 #4

    WannabeNewton

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    The magnitude of 4-velocity is normalized so that it is always (using the ##- + + +## convention) ##-1## (in natural units).
     
  6. May 31, 2013 #5

    robphy

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    While true, "proper velocity" (which isn't a great term.... "celerity" is better) refers to the spatial-component of a 4-velocity. In terms of rapidities, it is ## \sinh\theta##. When divided by the 4-velocity's timelike-component ##\cosh\theta##, one gets the spatial-velocity ##\tanh\theta##. (Of course, as a check: ##-(\cosh\theta)^2+(\sinh\theta)^2=-1##, using your signature convention.)
     
  7. May 31, 2013 #6

    WannabeNewton

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    Oh I see. I thought "proper velocity" meant 4-velocity since we usually refer to 4-acceleration as proper acceleration and stuff.
     
  8. May 31, 2013 #7
    I was referring to the spatial components of 4-velocity as robphy stated. I am currently working through a textbook and I didn't read the section on the energy-momentum 4-vector before posting this. After reading that section and your responses, it has helped to clarify my interpretation of the spatial components of relativistic momentum. Thank you for your responses.
     
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