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Energy equal to time component of 4-momentum?

  1. Dec 31, 2011 #1
    I'm sure this question gets brought up a lot, but I can't figure out why this is true. Everywhere I look, people simply equate the two as though it's some axiom, but never an explanation for why. It seems to me like E≠m/√(1-v^2) in general.
     
  2. jcsd
  3. Dec 31, 2011 #2

    Dale

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    It is correct. Can you think of an example where E≠m/√(1-v^2).
     
  4. Dec 31, 2011 #3

    clem

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    Energy is the time component of the momentum 4-vector. Using that fact, and the fact that v=p/E, you can derive your E equation.
     
  5. Dec 31, 2011 #4
    Sure, if one particle's frame measures another particle's velocity to be c/2 with rest mass of 2 kilograms, then its kinetic energy is 1/2 * 2 kg * (1.5*10^8 m/s)^2, which is not equal to its time component of four-momentum, which is 2 kg / √(3/4). The units don't even match up. Now my understanding of special relativity is completely self taught, so I might be missing something huge here.
     
  6. Dec 31, 2011 #5

    jtbell

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    No. In SR, kinetic energy is not [itex]\frac{1}{2}mv^2[/itex], but rather

    [tex]K = \left[\frac{1}{\sqrt{1 - v^2/c^2}} - 1 \right] m c^2[/tex]

    in which by "m" I mean the "invariant mass" which is often called "rest mass."

    When v << c, [itex]K = \frac{1}{2}mv^2[/itex] is a very close approximation to the equation above.
     
  7. Dec 31, 2011 #6

    jtbell

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    Also, in the definition of the four-momentum, (E/c, p_x, p_y, p_z), E is the "total" energy, i.e. E = K + mc^2, not the kinetic energy K alone.
     
  8. Dec 31, 2011 #7
    The reason to define the momentum 4-vector this way is that this way if you have the momentum 4-vector in one frame and you want to know it in another frame in relative motion to the first frame, you just multiply the momentum 4-vector by the Lorentz transformation matrix. This is why we like 4-vectors: their values in different frames are simply related to each other by the Lorentz transformation matrices.
     
  9. Dec 31, 2011 #8

    atyy

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    There's no reason to this. It is a new definition of energy which explains why the old definition of energy is applicable at low speeds. Why this new quantity is also called energy is simply a matter of convention, mostly having to do with physicists believing that "energy" should be conserved.
     
  10. Jan 1, 2012 #9
    It's more than just a belief. Noether's theorem shows that for any symmetry of a physical system, there is a conserved quantity. For classical situations with time translation symmetry, the conserved quantity is energy in the usual non-relativistic sense. By direct analogy, in relativity we call the quantity conserved as a result of time translation symmetry energy, as well.
     
  11. Jan 1, 2012 #10

    atyy

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    Yes, that's a good way to explain the choice of generalization.
     
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