Rest Mass?

I understand photons have no rest mass, but I also learned in my physics class p=h/(lambda), and that mass depends on the frequency, my question is why can't something moving at c have rest mass if it can have... momentum I suppose is the term...

If you define momentum as the ability to deliver an impulse, then light has it. This is experimentally demonstrable. See Compton scattering.

It is a tenet of special relativity that no mass can be accelerated so as to break the light barrier in any frame of reference, therefore the photon is assigned zero rest mass on principle. Experiments to test this set a very low possible mass for the photon.
See the FAQ on the Experimental Basis of Special Relativity.

It may sound contradictory to some, but that's the way it seems to be.

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jtbell
Mentor
The general relationship between rest mass (a.k.a. invariant mass), momentum and energy is

$$E^2 = (pc)^2 + (m_0 c)^2$$

If $m_0 = 0$, then E = pc.

It's possible to show that

$$\frac{v}{c} = \frac{pc}{E}$$

so if E = pc, then v = c.

Reversing this logic, we can see that if v = c, then $m_0 = 0$.

thanks you guys, that helps a lot, but then what is the relationship of the momentum of light to mass of light? because I know classically p=MV

jtbell
Mentor
That depends on which kind of mass you're talking about. If you're talking about "rest mass" a.k.a. "invariant mass", it's zero, so there's no meaningful relationship between it and the momentum. To get the momentum you use $E = pc$.

If you're talking about "relativistic mass" then you can combine $E = mc^2$ and $E = pc$ to get $p = mc$.

You have to distinguish between rest mass and relativistic mass before solving your problem. Rest mass is the measure of inertia of a body with very low velocity ( in fact, at rest ) while relativistic mass is the measure of inertia at certain velocity v. The relationship between these two is relativistic mass=rest mass /(sqaure root of (1-v^2/c^2)). If we put rest mass=a, while a not equal to 0, and v=c(the velocity of light) in the equation, the relativistic mass will equal infinity, which is impoosible. By the method of contradiction, therefore we know that something moves at velocity c must have rest mass a equals to 0. You may also look the question in this way. Someone try to measure rest mass of a photon and therefore try to make that photon at rest. However we know that the photon will be annihilated if we stop it. Therefore photon has no rest mass.
As the momentum is defined as the velocity times relativistic mass ( not rest mass ) in special relativity, p=mc in case of photon ( a quantum particle of light ) where m is the relatvistic mass and c the speed of light.

That depends on which kind of mass you're talking about. If you're talking about "rest mass" a.k.a. "invariant mass", it's zero, so there's no meaningful relationship between it and the momentum. To get the momentum you use $E = pc$.

If you're talking about "relativistic mass" then you can combine $E = mc^2$ and $E = pc$ to get $p = mc$.
Its more simple than that since relativist mass, m, is defined as the ratio of the magnitude of momentum to speed and therefore p = mv by definition. Since v = c then p = mv for a photon.

Pete

...I think I've got the gist of it now, thanks guys