Total energy at zero rest mass

[espen180]

According to my physics textbook, the equation E^2=\left(mc^2\right)^2+\left(pc\right)^2 suggests that a particle may have energy and momentum even when it has no rest mass, and that the total energy then is E=pc. This strikes me as odd, since the relativistic momentum of a particle is given by p=\gamma mv, which is zero when the rest mass is zero. But that must mean that for a particle with rest mass zero, E=pc must also be zero, and that a particle with zero rest mass can not have energy and momentum, but this is false, because photons are particles with energy and no rest mass.

If anyone would explain this to me, I would be grateful.

Thank you in advance.

[tiny-tim]

Hi espen180! :smile:

(have a gamma: γ :wink:)
… the relativistic momentum of a particle is given by p=\gamma mv, which is zero when the rest mass is zero. But that must mean that for a particle with rest mass zero, E=pc must also be zero …

Why do you think that γmc = 0 when m = 0?

At the speed of light, γ = ∞, so γmc doesn't have to be 0. :wink:

[jtbell]

the relativistic momentum of a particle is given by p=\gamma mv, which is zero when the rest mass is zero

...and v < c. If m = 0 and v = c, that formula gives p = 0/0 which is undefined. Therefore that formula doesn't apply to photons. The general energy-mass-momentum relationship, on the other hand, has no such trouble because it doesn't include the velocity explicitly.

[Phrak]

Why do you think that γmc = 0 when m = 0?
At the speed of light, γ = ∞, so γmc doesn't have to be 0. :wink:

∞ times 0 is 42, isn't it?

[atyy]

∞ times 0 is 42, isn't it?

Wasn't that 6X9 :confused:

[espen180]

...and v < c. If m = 0 and v = c, that formula gives p = 0/0 which is undefined. Therefore that formula doesn't apply to photons. The general energy-mass-momentum relationship, on the other hand, has no such trouble because it doesn't include the velocity explicitly.

I see. What does the Energy-mass-momentum relationship look like? Isn't that the formula I included above?

[diazona]

You mean this one?
E^2=\left(mc^2\right)^2+\left(pc\right)^2
That's it all right...

[Bob S]

According to my physics textbook, the equation E^2=\left(mc^2\right)^2+\left(pc\right)^2 suggests that a particle may have energy and momentum even when it has no rest mass, and that the total energy then is E=pc. This strikes me as odd, since the relativistic momentum of a particle is given by p=\gamma mv, which is zero when the rest mass is zero. But that must mean that for a particle with rest mass zero, E=pc must also be zero, and that a particle with zero rest mass can not have energy and momentum, but this is false, because photons are particles with energy and no rest mass..
Not quite true.
p = γm0v = βγm0c
so pc = βγm0c2 = (E2 - m0c2)1/2
so pc = E when m0c2=0
[βγ goes to infinity as m0c2 goes to zero]
α β γ δ ε ζ η θ ι κ λ μ ν ξ ο π ρ ς σ τ υ φ χ ψ ω

[Fredrik]

Just to say it a bit more clearly: The formula \vec p=\gamma m\vec v only holds for massive particles, while E^2=m^2c^4+\vec p^2c^2 holds for all particles. It even holds for tachyons (m^2<0), but they probably don't exist anyway, so that's less relevant.

[espen180]

Thanks for clearing it up, guys! :)

EDIT:

But since infinity x zero is undefined, do we have to go via quantum physics and use pc=hf instead?

[ZikZak]

Thanks for clearing it up, guys! :)

EDIT:

But since infinity x zero is undefined, do we have to go via quantum physics and use pc=hf instead?

You can if you wish. You could also simply write down E = pc.