# Momentum of light

1. Oct 16, 2012

### silvercrow

Since momentum = mass x velocity

Therefore for light

P = mv
P = 0 x 3x10^8
P = 0 m/s

Can anybody explain ?

2. Oct 16, 2012

### Pengwuino

Yes. The formula $p = mv$ does not hold for massless objects. The momentum of light is given by $p = {{h}\over{\lambda}}$ where h is Planck's constant and $\lambda$ is the wavelength of the light.

3. Oct 16, 2012

### cepheid

Staff Emeritus
Sure. You are using the definition of momentum from classical (Newtonian) physics. In special relativity, we learn that momentum is actually given by a different expression:

p = γmv

where the Lorentz factor γ is defined as $$\gamma = \sqrt{\frac{1}{1 - v^2/c^2}}$$

In the limit where v << c, this reduces to the Newtonian expression p = mv. (Try plugging in some everyday speeds for v, and you'll find that the difference between the the results given by the two equations is negligible). So the Newtonian formula isn't totally correct: it's just an approximation that works well at low speeds where you don't have to worry about relativistic effects. Anyway, what if you have a particle that travels at the speed of light? Well, as v → c, the Lorentz factor approaches infinity. So, a particle with mass cannot travel at speed c, because to do so would require infinite energy and momentum. Particles with mass are confined to speeds less than c. What if you have a *massless* particle though? Then suddenly it's not so clear what this expression for the momentum would yield. It could be it could take on a finite value, since although γ → ∞, m → 0. So it turns out that we have a sort of "loophole". The exact loophole is that special relativity says that massless particles can travel at speed c, and only at that speed (no lesser, no greater). A photon is an example of just such a particle. If you combine p = γmv along with the expression for the relativistic energy, E = γmc2, you find the following result (a relation between energy and momentum in special relativity):

E2 = p2c2 + m2c4

With things written in THIS form, it's easier to see exactly what happens when m = 0. In that case, you just have E = pc, or

p = E/c

So this (above) is the expression for the momentum of massless particles. For a photon, quantum mechanics says that the photon energy is given by the expression E = h$\nu$ where $\nu$ is the frequency and h is Planck's constant. Therefore, the photon momentum would be:

p = h$\nu$/c = h/λ

with Lambda being the photon wavelength. So the momentum of a photon is proportional to its frequency (or inversely proportional to its wavelength).

4. Oct 17, 2012

### silvercrow

Thank you guys . :D

5. Oct 17, 2012

### TheAbsoluTurk

I am a physics novice, but in the original post it says 0 m/s but it should be kgm/s or Ns

Also Cepheid gave an excellent and straightforward explanation.

6. Oct 17, 2012

### silvercrow

Yes sorry for wrong unit . I frankly didn't notice that .

7. Oct 17, 2012

### Nguyen Quang

Ns can be used for common calculation but kgm/s is more accurate and systematic. You should always use kgm/s

8. Oct 17, 2012

### HomogenousCow

The electromagnetic field also has momentum, even tho it has no mass.
The momentum density of the EM is given by (epsilon-zero)ExB.

9. Oct 17, 2012

### TheAbsoluTurk

Officially, derived units are to be avoided then I presume?

10. Oct 17, 2012

### cepheid

Staff Emeritus
Huh? No, I have no idea what Nguyen Quang is getting at. These are just two two different ways of expressing the same unit that are algebraically equivalent to each other, so how can one possibly be better or "more accurate" than the other? I don't know what that would even mean. Use whichever is more convenient for your particular purpose.