energy - Explain how (or if) a box full of photons would weigh more due to massless photons

I understand that mass-energy equivalence is often misinterpreted as saying that mass can be converted into energy and vice versa. The reality is that energy is always manifested as mass in some form, but I struggle with some cases:

Understood Nuclear Decay Example

In the case of a simple nuclear reaction, for instance, the total system mass remains the same since the mass deficit (in rest masses) is accounted for in the greater relativistic masses of the products per $E=\Delta m c^2$. When a neutron decays and you are left with a fast proton and a relativistic electron. If you could weigh those two without slowing them down, you would find it weighed as much as the original neutron.

Light in a Box

This becomes more difficult for me when moving to massless particles like photons. Photons can transmit energy from one heavy particle to another. When a photon is absorbed the relativistic mass (not the rest mass) of the (previously stationary) particle that absorbs it increases. But if my understanding is correct, the energy must still be manifested as mass somehow while the photon is in-flight, in spite of the fact that the photon does not have mass.

So let's consider a box with the interior entirely lined with perfect mirrors. I have the tare weight of the box with no photons in it. When photons are present the box has an additional quantifiable amount of energy (quantified below) due to the in-flight photons. Say there are $N$ photons... obviously assume $N$ is large.

$$\Delta m = \frac{ E }{ c^2 } = \frac{ N h }{ \lambda c}$$

Interactions are limited to reflections with the wall, which manifest as a constant pressure on the walls. If I hold this box in a constant gravitational field (like the surface of Earth) then there will be a gradient in the pressure that pushes down slightly. Is this correct? Wouldn't there still technically be mass as the photons are in-flight, which would cause its own gravitational field just as all matter does? How is this all consistent with the assertion that photons are massless? Is it really correct to say that photons don't have mass? It seems to be a big stretch.

Please offer a more complete and physically accurate account of this mirror-box.

Answer

The statement that photons are massless means that photons do not have rest mass. In particular, this means that, in units where $c=1$, the magnitude of the photon 3-momentum must be equal to the total energy of the photons, rather than the standard relationship where $m^{2} = E^{2}-p^{2}$.

But, you can create multi-photon systems where the net momentum is zero, since momentum adds as a vector. When you do this, however, since the energy of a non-bound state is always non-negative, the energies just add. So, this system looks just like the rest frame of a massive particle, which has energy associated with its mass and nothing else.

The statement about gravity is a little bit more subtle, but all photon states will interact with the gravitational field, thanks to the positive results of the light-bending observations that have been made over the past century. So you don't even need a construction like this to get photons "falling" in a gravitational field.