Momentum of a Massless Particle

I just read one of your statements in which you said that E=m^2.c^4+p^2.c^2 and for a photon, E=pc as the mass of a photon is 0. what i want to ask is that, if p=m.v and for a photon v=c, then E=pc=(m.v)c=(m.c).c= mc^2? So we again reach E=mc^2 and well, if photons dont have mass then energy is 0..which is not possible, So when we say E=pc, how do we know the momentum of a massless particle, is there any equation or anything for it?
- Anushka (age 16)
India

I've marked this as a follow-up to ones that may answer it. The key is to pay attention to the different uses of the symbol "m", either as rest-mass or as the factor in p=mv.

"How do we know the momentum of a massless particle?"  There are several ways. One is to directly measure the momentum by measuring, for example, the force exerted on a mirror by a stream of photons. Here one uses  that p is conserved and also that the particle number can be determined using universal quantum relation E=hf, where E is energy, h is Planck's constant, and f is frequency.  Another way is to use an argument from Maxwell's equations that requires E=pc if momentum is to be conserved. That gives a momentum density in terms of the electric and magnetic fields.  It can be converted to a momentum per particle again using E=hf.  Another way is to look for the missing momentum in events involving a few massive particles and a photon or two.  Other ways include  using the universal quantum relation |p|=h/λ, where λ is the wavelength.  λ can be measured with diffraction gratings or other methods.

Mike W.

 

 

(published on 06/04/2013)

Our teacher equated E=mc² with E=hv to find the mass of a photon. Now, firstly, how is it possible to take out the mass if m=0. Secondly, if the m here is relativistic mass, then why do we apply E=mc² as it is applied when the particle is stationary. Also, the photon is always moving. Shouldn't we apply E=pc=hv?
- Anushka (age 16)

This is a familiar question. I've put it in the right thread.

Here's the key point to re-emphasize, once you've gone over our old answers. The "m" your teacher got, hv/c² , is the mass that photon would contribute to the total mass of a bag of photons with zero net momentum. That inertial mass is not generally  the same as the invariant (rest) mass for a single particle, but it has a great property. The sum of the inertial masses for a collection of non-interacting particles is the same as the inertial mass of the collection as a whole. That's not close to being true for the invariant masses.

Mike W.

(published on 06/21/2013)

Since light has mass, it can be affected by gravity. How much is it affected? Say we could see the light in vacuum, what would we see from outer space looking at the Earth? Would some light be drawn and curved into Earth? If so, then that means it is compressing onto itself to reach Earth.
- Jess (age 13)
Sydney, NSW, Australia

You can follow the rest of this thread to get the basic physics. Yes, the earth bends light. The maximum angle is roughly gR/c², where g is the gravitational acceleration at the surface and R is the Earth's radius. That comes out to about 10^-9 radians, so it's not a big effect. The bending as light goes near the Sun is much bigger. It was first observed in 1919, fitting the prediction of general relativity.

Mike W.

(published on 07/30/2013)

Light bends around massive objects due to gravitational stuff because of space-time stuff but how is light affected by gravity if light has no mass?
- Hinerangi (age 16)
Auckland, New Zealand

see above

(published on 08/07/2013)

I understand that the "speed of light" is a bad title. We should instead refer to the £cosmic speed limit", which describes how fast something without mass will move. How then can light travel at "the speed of light", since light has mass? Or have I got something wrong?
- Chris Whyley (age 53)
Swansea United Kingdom

Maybe the earlier parts of this thread will help.

Mike W.

(published on 10/17/2013)