Light -- Zero Rest Mass and Momentum

[Major_Kong]

How can light have both zero rest mass and momentum? Isn’t rest mass required for momentum? Does light have mass when moving at c? Can light only move at c?

[Whack-a-Mole]

Maybe the link below will help. I’ll post a bit of it so you can see if it would be interesting to you but I don’t want to offend the copyright gods either by posting too much:

Question

My science data book says that the mass of a photon is zero. But how can photon exert radiation pressure, as it occurs in the sun?

Asked by: Suraj

Answer

Actually what your science book says or is trying to say anyway,is that the photon’s rest mass is zero. The concept of the rest mass derives from special relativity. The rest mass is the mass of a particle (in our case the photon) as measured by an observer who sees the particle still and with zero speed. In other words, the particle is at rest as far as this observer is concerned. Thus comes the term REST mass. But according to special relativity, light ALWAYS travels with the light speed c, and is NEVER at rest. And so it has zero REST mass.

SOURCE: http://www.physlink.com/Education/AskExperts/ae180.cfm

[Exapno_Mapcase]

The equation to use is not the classic E = mc[sup]2[/sup], which goes to zero when mass is zero, but

E[sup]2[/sup] = p[sup]2[/sup]c[sup]2[/sup] + m[sup]2[/sup]c[sup]4[/sup], where p is momentum.

With photons having zero rest mass, the equation goes to a simple E = pc.

This is what shows the energy of a zero rest mass particle.

http://www.colorado.edu/physics/phys2170/phys2170_spring96/notes/2170_notes1_4.html

[Francis_E_Dec_Esq]

*Originally posted by Exapno Mapcase *
**With photons having zero rest mass, the equation goes to a simple E = pc.
**

Right. No momentum, then no energy.

Here is another way to think of it.

Photons are just modes of vibration of the electromagnetic field. And we know waves of the EM field have to be able to carry momentum. Jiggle an electron at X, an EM wave travels away from it to Y causing electron at Y to jiggle. The momentum of electron X was transferred to Y via the field. The wave doesn’t have mass, but it does transfer momentum.

A photon has zero rest mass, but that shouldn’t bother you because photons are never at rest. But in one sense they could be considered to have mass.

The relativistic expression for a particle’s momentum is*

p = m[sub]R[/sub]v = mv/(1-v[sup]2[/sup]/c[sup]2[/sup])[sup]1/2[/sup]*

where m is the rest mass and

m[sub]R[/sub] = m/(1-v[sup]2[/sup]/c[sup]2[/sup])[sup]1/2[/sup]

is the relativistic mass. As you can see, the relativistic mass is also equal to p/v, just as it is in Newtonian physics. Although the rest mass of a photon vanishes, the ratio p/v = p/c does not and you can define that as the “mass” of the photon.

This is perfectly valid and quite in accordance with the intuitive notion of inertial mass. The only problem is that the relativistic mass varies with the speed and most people are used to thinking of the rest mass as the mass, but it is really just a matter of definition. In fact, this definition of mass is more in accordance with the OP’s intuition than rest mass is.

[Ring]

*Originally posted by Francis E Dec, Esq *
**
This is perfectly valid and quite in accordance with the intuitive notion of inertial mass. The only problem is that the relativistic mass varies with the speed and most people are used to thinking of the rest mass as the mass, but it is really just a matter of definition. In fact, this definition of mass is more in accordance with the OP’s intuition than rest mass is. **

I wouldn’t say that relativistic mass is completely wrong but it’s certainly considered verboten by a very large majority of physicists.

Francis E Dec, Esq you might want to try this thought experiment. Let’s say that you apply a longitudinal force to a relativistic particle of rest mass m, how would you calculate the relativistic mass so that F = ma (in terms of the rest mass)?

[Ring]

I should probably give the answer to my proposed question.

Longitudinal direction:

*F = Gamma[sup]3[/sup]m[sub]0[/sub]

And for a force applied in the transverse direction:

F = Gamma*m[sub]0[/sub]

Gamma = 1 / (1 - v[sup]2[/sup] / c[sup]2[/sup]) [sup]1/2[/sup]

Instead of a nice invariant scalar quantity we now have a matrix quantity.

[Ring]

Damn I meant to hit preview not submit.

I should probably give the answer to my proposed question.

Longitudinal direction:

*m[sub]r[/sub] = Gamma[sup]3[/sup]m[sub]0[/sub]

And for a force applied in the transverse direction:

m[sub]r[/sub] = Gamma*m[sub]0[/sub]

Gamma = 1 / (1 - v[sup]2[/sup]/c[sup]2[/sup])[sup]1/2[/sup]

Instead of a nice invariant scalar quantity we now have a matrix quantity.

[Francis_E_Dec_Esq]

Thanks, Ring. Very interesting. I’ll have to play with that some more.