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Why E/c = p ?

  1. Mar 1, 2016 #1
    • Moved from a technical forum, so homework template missing
    Hi people, I studying electromagnetic waves (intermediate) and
    I don't understand how the expression for linear momentum of a wave is obtained, if the wave doesn't carry any mass.
    In particular, I have to explain why the radiation pressure on a perfect absorber is half that on a perfect reflector

    So, I do this:

    P_rad= pressure
    \vec{p}= momentum of wave
    A= transversal area
    Volumen= ctA
    p_den= density of momentum =p/vol
    P_rad=\frac{Δp}{Δt *A} =\frac{ p_den *(c *t*A)}{ t*A} = p_den *c
    but i the books're saying that E/c =p ???? Why?
    Finally, p_den *c = I *c/ c² = I/c to case of absorbent surface.
     
    Last edited: Mar 1, 2016
  2. jcsd
  3. Mar 2, 2016 #2

    DrClaude

    User Avatar

    Staff: Mentor

    In a relativistic context, which you must use when considering light, the energy is given by
    $$
    E^2 = p^2 c^2 + m^2 c^4
    $$
    Therefore, a massless particle has momentum ##p = E/c##.

    As for the second question, think about the difference in the before and after pictures of the collision and conservation of momentum.
     
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