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Link between 'time' component of 4-momentum and energy

  1. Nov 28, 2016 #1
    1. The problem statement, all variables and given/known data

    $$ E = -\vec{v_{obs}} \cdot \vec{p} $$

    Where ## \vec{p} ## is the four momentum, and ## \vec{v_{obs}}## the velocity of the observer.

    2. Relevant equations


    3. The attempt at a solution

    This was a stated result in a GR course. I look through my SR notes and find that I have actually never seen a proof that ## p^{0}## is exactly,definitely the energy.

    I understand that the norm of the four momentum will be invariant etc...
    But the phrase in the notes I have "For a free, non-interacting particle there is no physically meaningful property of "energy"" is making me uncomfortable - have I fundamentally misunderstood something?

    Thanks!
     
  2. jcsd
  3. Nov 28, 2016 #2

    Ray Vickson

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    Actually, if ##c## is the in vacuo speed of light, then energy is ##c p^0##; you need that extra ##c## to make the units come out right.

    Anyway, how you can take an inner product of a 3-vector ##\vec{v_{obs}}## and a 4-vector ##p##?
     
  4. Nov 28, 2016 #3
    Good question - I presume it's the four velocity of the observer (these are directly from my lecture notes).

    I guess another part of my confusion is the fact that we have :

    $$
    p^{\mu} = m \frac{dx^{\mu}}{d\tau}
    \\[2mm]
    p^{\mu} = \bigg(E,\vec{\dot{x}} \bigg) \\[2mm]
    \implies m\frac{dx^{0}}{d\tau} = E
    $$
    Which I find a bit odd/ how is this the same quantity as my familiar 'classical' energy? Even if the units do work out (I have missed a factor of c somewhere).

    Edit - The oddness to me being that we somehow have some kind of rate of change of a time co-ordinate associated with an energy.... so that somehow having higher energy, is associated, in some sense with a higher rate of motion in time?? But that sounds bananas to me...
     
  5. Nov 28, 2016 #4

    Ray Vickson

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    No! For a particle of mass ##m## moving with 3-velocity ##\vec{u}## the 4-momentum is ##p = (m \gamma c, m \gamma \vec{u})##, where ##\gamma = (1 - u^2/c^2)^{-1/2}##, while the (total) energy is ##E = m c^2 \gamma##. Note that ##E = p^0 c##, NOT ##p^0##. Why do you continue to ignore this distinction? If nothing else, look at the units: the units of momentum are ##\text{mass}\times \text{velocity}## while the units of energy are ##\text{mass} \times \text{velocity}^2##. It has been that way ever since the time of Newton.
     
  6. Nov 29, 2016 #5
    Sure, sure - the units are important and I entirely agree that ## p^{0}c ## has the correct units of energy. What I find confusing is not the units, but the idea that somehow a translation in time is related to my classical picture of energy :)
     
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