# Momentum of a photon

1. Jan 17, 2006

### Moneer81

Hello,

I am trying to teach myself quantum mechanics so take it easy on me.

In 1905, Einstein proposed that light is made of photons and that each photon carries energy E = hf (f is frequency) and momentum p = h/w (where w is wavelength)

My question is how did he go from the definition of momentum which is mass times velocity to the aforementioned p = h/w. Is there an easy straightforward answer to this or is it something derived from classical thermodynamics?

thanks

2. Jan 17, 2006

### Dense

I don't know the answer, but...

I don't think the word "momentum" refers to precisely the same thing in the relativistic sense Einstein was using it, as it does in newtonian mass*velocity.

Or perhaps the newtonian equation is just a simplified "good enough" version of what's really going on, and is not the true equation? Like, maybe there's an equation of momentum that doesn't refer to mass, so that the momentum of massless particles with energy can be calculated as well as the momentum of massive particles.

Interested to see what the answer is, myself.

3. Jan 17, 2006

### zekise

I understand it goes like this:

p = mv = mc = mcc/c = E/c = hf/c = hc/cw = h/w.

I think this is what my textbook "Quantum Mechanics" by Bransden & Joachain page 39 is saying.

4. Jan 17, 2006

### Moneer81

yeah I'll be interested to figure this out a little better too.
You also raised an interesting question, if the photon is a massless particle that what does it mean for it to have momentum?

For me momentum is a property of mass but I guess you might be right, the momentum Einstein was talking about might be something different that the classical momentum that we are familiar with.

5. Jan 17, 2006

### Moneer81

ok well that makes mathematical sense to me

6. Jan 17, 2006

### Perturbation

The Newtonian formula for momentum is incorrect, it is accurate for non-relativistic motion (velocity much less than that of light) but for relativistic cases (velocity comparable to that of light) it is inadaquate. A photon is an example of relativistic speeds and thus Newtonian mechanics fails, as Einstein showed.

The correct equation for the energy of a body of rest mass m in the relativistic case is

E²=(mc²)²+(pc)²

Where p is the momentum of the body, which is not mv. For a photon m=0, so we get E=pc. According to Planck, the energy E of a photon is hf, so hf=pc. Rearranging we get p=h/wavelength. Notice also that when p=0 (i.e. in the rest frame) we obtain E=mc².

The derivation given by someone else in here doesn't work as it relies on the classical definition of momentum.

7. Jan 17, 2006

### Meir Achuz

I just want to clarify the history.
1. Planck suggested E=h\nu to understand the high frequency spectrum of black body radiation.
2. Einstein used Planck's suggestion to understand the photo effect.
3. The connection with momentum was first made by deBroglie to give a wave understanding of the Bohr hypothesis, leading to QM.
4. It had been known since Maxwell that light had both energy and momentum densities, connected by E=pc.

8. Jan 17, 2006

### Staff: Mentor

To put things together:

$$p = \frac {E}{c} = \frac {h \nu}{\lambda \nu} = \frac {h}{\lambda}$$