Momentum of a photon

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Main Question or Discussion Point

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?
 
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Answers and Replies

Astronuc
Staff Emeritus
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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.
 
412
2
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.
well.. i actually.. said the same thing. I don't get what you were trying to explain. Can u please explain? thanks.
 
Astronuc
Staff Emeritus
Science Advisor
18,542
1,684
well.. i actually.. said the same thing. I don't get what you were trying to explain. Can u please explain? thanks.
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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