# B Confusion on Momentum of Light

1. Aug 11, 2016

### Irfan Nafi

In E2=m2c4+p2c2
how do you prove light has momentum? I've tried but my answer comes to be undefined. Light has m=0, so you're left with p2c2. p=(mv)/sqrt(1-v^2/c^2) When substituting values of light you get that p=0/0. How do you prove light has momentum using math and equations?

2. Aug 12, 2016

### Mr-R

The expression for momentum for photons in QM is $p=\hbar k$. Where $k$ is the wave number ($k=\frac{2\pi}{\lambda}$) and $\hbar$ is the Planck constant divided by $2\pi$ . Or equivalently, $p=\frac{h}{\lambda}$ . Your equation for $p$ is for massive particles with non zero rest mass.
Edit: Also note that in your expression for p if you put v=c (photon) then you get a division by zero! It is good to check for these stuff when you are doing calculations.
Edit 2: Sorry I thought that this was posted in QM section. Apologies.

3. Aug 12, 2016

### andrewkirk

In relativity, p is not necessarily obtained by multiplying mass by velocity.

For a photon, instead we work with the four-momentum, the first component $p^0$ of which, in any reference frame, is the energy of the photon in that frame. The other three components $p^1,p^2,p^3$ are momenta in the directions of the three spatial axes of the frame. We want the four-momentum to be a null vector (vector of zero magnitude), because light travels along null geodesics. So that means that not all the other three components of the four momentum can be zero - since the first component (Energy) $p^0$ is nonzero. If we rotate the frame so that the x spatial axis points in the direction of the photon's travel, we know that the other two momentum components $p^2,p^3$ must be zero, and we conclude that $p^1=p^0$ where $p^0=E=h\nu$ is the photon's energy.

4. Aug 12, 2016

### Orodruin

Staff Emeritus
This is a relation that only holds fot massive particles. You cannot use it for light.

Yes, and so you are done. If light carries energy, then it carries momentum p=E/c.

You don't. This is not how physics works, it is an empirical science. Your theory makes a prediction and you test it in the lab.

5. Aug 12, 2016

### Orodruin

Staff Emeritus
I strongly suspect that this will pass over the head of the OP in a B level thread ...