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Momentum of a photon

  1. Mar 21, 2008 #1
    This thing has troubled me for a long time.. but i think this is eventually starting to get clear to me.

    When we say a photon has a momentum 'p', the momentum isn't it's mass times the velocity.. but it can be interpreted as: If the photon [having frequency [itex]f_1[/itex]] collides with any other particle, and an elastic collision takes place.. and it's frequency now is [itex]f_2[/itex], then the momentum imparted to the particle is: [tex]\Delta p = \frac{h}{c}(f_2 - f_1)[/itex]

    am i right?
     
    Last edited: Mar 21, 2008
  2. jcsd
  3. Mar 21, 2008 #2

    Astronuc

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    For a photon, momentum, p = E/c = h[itex]\nu[/itex]/c, and a change would be given by

    [tex]\Delta p = \frac{h}{c}{(f_2 - f_1)}[/tex], where h/c is a constant. I used nu for frequency f.
     
  4. Mar 21, 2008 #3
    well.. i actually.. said the same thing. I don't get what you were trying to explain. Can u please explain? thanks.
     
  5. Mar 21, 2008 #4

    Astronuc

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    I was explaining that a photon of energy E and a momentum of p = E/c. When a photon scatters (interacts with a particle), there will a change in energy and momentum, and the change in p = (Ei - Ef)/c. This is demonstrated in the Compton effect.
     
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