# Momentum of a photon

1. Mar 21, 2008

### 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: $$\Delta p = \frac{h}{c}(f_2 - f_1)[/itex] am i right? Last edited: Mar 21, 2008 2. Mar 21, 2008 ### Astronuc ### Staff: Mentor For a photon, momentum, p = E/c = h$\nu$/c, and a change would be given by [tex]\Delta p = \frac{h}{c}{(f_2 - f_1)}$$, where h/c is a constant. I used nu for frequency f.

3. Mar 21, 2008

### rohanprabhu

well.. i actually.. said the same thing. I don't get what you were trying to explain. Can u please explain? thanks.

4. Mar 21, 2008

### Staff: Mentor

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.