Proof for Radiation Momentum Transfer to Perfect Absorber?

[Kenny Lee]

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.

 

[Hootenanny]

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

Momentum is given by;
$$p = mc$$

Relativistic mass;
$$E = mc^2 \Rightarrow \frac{E}{c} = mc$$

Combining the equations;
$$p =\frac{E}{c}$$

Using photon equation;
$$\fbox{E = hf \Rightarrow p = \frac{hf}{c}}$$

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

Hope this helps

 

[Kenny Lee]

It helped. Thanks very much.