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Radiation momentum

  1. Mar 19, 2006 #1
    My textbook states without proof that p, the momentum transferred by a wave onto a perfect absorber (normal incidence), is U/ c, where U is the total energy delivered in some time interval.

    Is there a simple proof for this? Just something to help me rationalize. I tried equating kinetic energy to U. Then I set mv equal to p. But what I get is 2U/c, which is the p for a perfect reflector.

    Any help is greatly appreciated.
  2. jcsd
  3. Mar 19, 2006 #2


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    Are you talking about a light wave? If you are the derivation is quite simple;

    Momentum is given by;
    [tex]p = mc[/tex]

    Relativistic mass;
    [tex]E = mc^2 \Rightarrow \frac{E}{c} = mc[/tex]

    Combining the equations;
    [tex]p =\frac{E}{c}[/tex]

    Using photon equation;
    [tex]\fbox{E = hf \Rightarrow p = \frac{hf}{c}}[/tex]

    I think this is the answer your textbook is looking for. However, you can go further;
    [tex]f = \frac{c}{\lambda} \Rightarrow p = \frac{hc}{\lambda c}[/tex]
    [tex]p = \frac{h}{\lambda}[/tex]

    Hope this helps :smile:
    Last edited: Mar 19, 2006
  4. Mar 19, 2006 #3
    It helped. Thanks very much.
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