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B 4-momenta question

  1. Feb 24, 2017 #1
    The 4-momenta before a photon collides with a stationary election [itex]P_m=(mc^2,0,0,0)[/itex] or is it [itex]P_m=(mc,0,0,0)[/itex]

    Which is the difference between the two?

    Thanks.
     
    Last edited: Feb 24, 2017
  2. jcsd
  3. Feb 24, 2017 #2

    Dale

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    The four momentum is usually written in units of momentum, so (mc,0,0,0). However, it is even more common to use units where c=1, so that energy momentum and mass are all the same unit.
     
  4. Feb 24, 2017 #3

    jtbell

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    Units of energy versus units of momentum. I was trained in experimental particle physics, so I prefer units of energy ##(E, p_x c, p_y c, p_z c)## but I recognize that others prefer units of momentum ##(E/c, p_x, p_y, p_z)##. Whichever units you prefer, be consistent within a calculation!
     
  5. Feb 25, 2017 #4

    Dale

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    Interesting, does the particle physics community still call it the four momentum if they commonly use energy units?
     
  6. Feb 25, 2017 #5

    Orodruin

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    We do not differentiate between the two since we use units where c=1.
     
  7. Feb 25, 2017 #6

    jtbell

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    When I was a grad student it was general practice to talk of momentum in units of MeV and GeV.

    But we nevertheless called it (3-)momentum or 4-momentum as appropriate.
     
  8. Feb 25, 2017 #7
    Thanks for the replies. So using (mc,0,0,0) is just as valid as using (mc^2,0,0,0) ?
     
  9. Feb 25, 2017 #8

    vanhees71

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    Sure, you only must be consistent with your convention or, most elegantly, set ##c=1##, as the particle physicists do (see the postings above). If I keep ##c \neq 1##, then I usually prefer to use the convention that the four-vector quantities have the dimension of the spatial parts, i.e., for momentum
    $$(p^{\mu})=\begin{pmatrix} E/c \\ \vec{p} \end{pmatrix},$$
    where ##E=c \sqrt{m^2 c^2+\vec{p}^2}## is the relativistic energy (i.e., kinetic + rest energy) of the particle. Then the energy-momentum relation (often called "on-shell condition") can be written in manifestly covariant form as
    $$p_{\mu} p^{\mu}=m^2 c^2.$$

    For the photon of momentum ##\vec{k}##, following thie convention you have
    $$(k^{\mu}) = \begin{pmatrix} |\vec{k}|,\vec{k} \end{pmatrix}.$$
    Because it's massless the energy of the photon is ##E_{\gamma}(\vec{k})=c |\vec{k}|##.
     
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