Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose you have an ideal mirrored box that contains enough photons as to have a relativistic mass equivalent to the [rest mass + kinetic energy] of an electron. In other words, the two systems have the same total energy.

How would I distinguish these two systems? They have equivalent gravitational pulls and I believe they have equivalent inertia. Are charge and spin the only distinguishing factors (I'm assuming the "box" itself has no properties of interest)?

share|cite|improve this question
the box can't have no properties of interest. – Ron Maimon Sep 5 '12 at 4:37
up vote 2 down vote accepted

In practice, the box would be the biggest distinction. After that, the charge and spin differences would make it very obvious. There are a lot of other differences that could be traced back to the charge and spin, e.g. you can't build photon-box circuits, you can't let photon-boxes be captured into atoms, etc. The scattering behavior of a box of photons with some other particle would also be very different from that of an electron, in the sense that electrons scatter off pretty much everything but photons don't. This starts to get pretty speculative, though, because who knows what would be allowed in a universe in which you can build an invisible, massless, substanceless box?

There would actually even be a difference in the gravitational effect, because the stress-energy tensor of an electromagnetic field is not the same as that of a matter field. But that difference would probably be pretty tiny.

share|cite|improve this answer
Great answer! How is gravity different for EM energy vs mass-energy? – Nick Sep 5 '12 at 4:41
As you may know, the source of gravity is the stress-energy tensor, not just mass or energy. Wikipedia describes the electromagnetic stress-energy tensor, but for a single particle at rest, it would be simply the mass in the top left component and everything else zero. (Not counting any quantum gravity corrections that may exist, but we don't know anything about that anyway.) – David Z Sep 5 '12 at 4:51
@DavidZaslavsky: The stresses would have to be nonzero in the photon-containing box to zero out the photon pressure, which is why the effect is to look like a massive particle, and also why the box stresses are essential for reproducing a mass-only stress tensor. – Ron Maimon Sep 5 '12 at 17:50

If the box was magic, massless and such which is common in physics (see massless frictionless pulley) then we might consider the box when it is moving. If the box was moving relative to us, then so are the photons. As such, its speed should be the same no matter how fast we try to move relative to the box. If the box was a massive particle, we could arbitrarily set the velocity of the box relative to ourselves by simply adjusting our own velocity.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.