Does Light Really Have Mass?

[yuiop]

...
The mass of zero photons can be nonzero as well. If you take 6 square mirrors of mass m and glue them together to form a cube, then the cube will have a mass of slightly more than 6 m, even if there is only a vacuum inside. This is due to the vacuum energy of the elecromagnetic field inside the cube. So, zero photons can have more mass that a single photon.

I am curious. Does the introduction of one photon to our initially dark vacuum box destroy the vacuum energy? Wouldn't the box have a total mass of box+vacuum+photon?

 

[yuiop]

Oh wait, I didn't think this one statement through. Does a single photon have a zero momentum frame?

I guess another way to define rest mass is total energy minus kinetic energy and for a photon total energy = kinetic energy so its rest mass is unambiguously zero by that definition.

 

[Phrak]

The mass of a single photon is zero. The mass of two or more photons can be nonzero, because mass is the total energy in the zero momentum frame. So, if you have two photons with equal energy moving in the opposite direction, then the mass is twice the energy of a single photon.

The mass of zero photons can be nonzero as well. If you take 6 square mirrors of mass m and glue them together to form a cube, then the cube will have a mass of slightly more than 6 m, even if there is only a vacuum inside. This is due to the vacuum energy of the elecromagnetic field inside the cube. So, zero photons can have more mass that a single photon.

No need to go that far. A single photon standing wave has zero momentum such that m_0 =E / c^2

In fact, it's hard to pin-down the long list of ideal conditions required for massless photons...

 

[Phrak]

On second thought, doesn't

m_0^2 = E^2 - p^2 or more precisely

m_0 = p_\mu

serve in particle physics as the Dirac delta function serves in quantum mechanics; each physical impermissible, but serving as the basis of theory?

The idealization in the first case is an unrealizable photon with energy precisely equal to momenutum, and in quantum mechanics an unrealizable wavefunction with either precise momentum or precise position.

 

[Phrak]

m_0 = p_\mu

I'm compelled to correct this before things get too far along.

m_0 U^\mu = p^\mu

where \ U^\mu is the four velocity, ( \ U^\mu U_\mu = -1 ) so that

m_0^2 = -p_\mu p^\mu