Momentum of a Photon - Is My Understanding Correct?

[rohanprabhu]

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 f_1] collides with any other particle, and an elastic collision takes place.. and it's frequency now is f_2, then the momentum imparted to the particle is: \Delta p = \frac{h}{c}(f_2 - f_1)[/itex]<br /> <br /> am i right?

 

[Astronuc]

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

\Delta p = \frac{h}{c}{(f_2 - f_1)}, where h/c is a constant. I used nu for frequency f.

 

[rohanprabhu]

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

\Delta p = \frac{h}{c}{(f_2 - f_1)}, 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]

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 = (E_i - E_f)/c. This is demonstrated in the Compton effect.