# Photon's momentum

1. Sep 1, 2010

### Parbat

Photon has momentum but is massless.Doesn't that seem strange?

2. Sep 1, 2010

### Bob S

If you write down the fully relativistic equation for momentum (in pc units) for any particle (including photons) of kinetic energy T, the quation is

(pc)2 = (T + mc2)2 - (mc2)2

= T2 + 2(mc2)T

So, for a bullet use the second term; for a photon use the first (for mass = 0). So a massless photon has momentum.

Bob S

3. Sep 1, 2010

### jcsd

Seems a bit strange of course when we've only encountered Newtonian physics as we're always taught p=mv, but relativirty shows that particles that travel at c can only have zero mass.

4. Sep 1, 2010

### maxverywell

No it's not strange because photon has energy. So its momentum is $$p=E/c=h/\lambda$$

(ok it is little bit strange...)

5. Sep 1, 2010

### Bob S

No.

"Seems a bit strange of course when we've only encountered Newtonian physics as we're always taught p=mv, but relativirty shows that particles that travel at c [STRIKE]can only have[/STRIKE] may have zero mass".

Also, in Newtonian mechanics, p = sqrt(2mE) where E = ½mv2

Bob S

6. Sep 1, 2010

### jcsd

No, not 'may', 'must'. The four-momentum of a particle is tangent to it's wordline for a particle travelling at c this means it's four-momentum is always null. The mass of particle (in the absence of a rest frame to define it) can be taken as the norm of it's four-momentum which is zero for a particle with null four-momentum. Hence all particles travelling at c MUST have zero mass (unless we're going to allow particles to have undefined four-momentum and hence undefined momentum and energy).

In Newtonian mechanics a particles momentum is defined as p(t) = mv(t).

Your definition is clearly incorrect as momentum is a vector quantity whereas your definition defines a scalar quantity.

7. Sep 1, 2010

Staff Emeritus
Exactly.

$$v = c \sqrt{1-\frac{m^2 c^4}{E^2}}$$

8. Sep 1, 2010

### Bob S

You are correct.

My pocket calculator shows that a 3.5 TeV proton in LEP has a v/c =

β = 1-1/2γ2 = 1-3.6·10-8 = 0.999 999 964

For the 1020 eV Oh My God cosmic ray proton

β = 1-1/2γ2 = 1-5·10-23 = 0.999 999 999 999 999 999 999 9 (more or less)

So any particle with mass >0 cannot have a velocity of exactly c. I was not careful enough in rounding off.

Bob S