mass deforms space-time, yielding what we call gravity.
energy and mass are the same thing.
photons do not deform space-time.
is number three right? i’ve never heard of light doing anything to local space-time topology, but then again, this certainly wouldn’t be the first time i’ve been ignorant.
if number three is right, how could this be?
jb
No, number 3 is wrong. Individual photons don’t deform space-time very much, of course, because most of them don’t really have enough energy to do anything interesting, but if you had a high enough energy density of photons somewhere, you’d get appreciable gravity.
But photons have no mass, right? Okay, I know it’s technically zero “rest mass”, and photons are never at rest, so the concept of “relativistic mass” is introduced. But zero multiplied by any number is still zero. So how does this work?
What deforms spacetime is the amount of rest mass in the volume. This means that a single photon, or a group of photons moving in the same direction, won’t deform spacetime, since transforming to the photon’s “rest frame” gets rid of all of their energy. But two photons travelling in opposite directions do have rest mass, so when they approach each other they will deform the space.
Omphaloskeptic,
With respect, I don’t find that to be a very satisfactory answer. I’m not saying you’re wrong, just that I don’t (yet) understand it. If photons have zero rest mass, what’s the mechanism for having however many zeros add up to a non-zero?
Omphaloskeptic are you saying that transforming to a photon’s “rest frame” would not distort spacetime?
The rest mass of a system (the center-of-mass energy) is not necessarily the same as the sum of the rest masses of its components; it can be larger or smaller.
For example, the mass of an atomic nucleus is not quite equal to the sum of the masses of all of the protons and neutrons in the nucleus. The mass of a helium nucleus, for example, is smaller than the mass of two protons and two neutrons. (This difference in mass is what gets converted to energy in fission and fusion.)
Another example: some particles (such as the neutral pion) can decay entirely into photons. This doesn’t change the center-of-mass energy of the system (though of course the resulting photons tend to escape the original region, reducing the spacetime curvature that way instead). In this case, the mass of the pion (and the mass of the system of photons resulting from the decay) is larger than the rest masses of the photons.
The mechanism for extra/missing mass in these two cases is the same as that in the case of two approaching photons: In the system’s center-of-mass frame, the components of the system may have kinetic energy along with their rest masses. This energy can’t simply be transformed away, as it can in the single-particle case, because transforming kinetic energy away from one particle means giving it to another.
This might still be an unsatisfactory answer; I can try again if it is.
I think I begin to see your point. I guess my confusion stems from trying to apply Newtonian physics to quantum events. I suppose f=mv doesn’t work for photons, right?
Of course we can’t just Lorentz-tranform to the photon’s rest frame, since it’s moving at c. But if we transform to a frame moving arbitrarily close to c in the same direction as the photon, we find that the photon’s energy (in this new frame) can be reduced arbitrarily close to zero. This is a hint that there shouldn’t be any spacetime curvature associated with a single particle’s kinetic energy. (For a more dire example, we could transform to frames moving in the opposite direction, and give the photon as much energy as we wanted. We could transform an unsuspecting electron into a particle with so much kinetic energy that it would collapse into a black hole, just because we thought about it!)
But yet, the electromagnetic field surely does gravitate (witness Reissner-Nordstrom black holes, for example). Granted, this is a classical phenomenon. Nevertheless, what is the electromagnetic field but a really huge collection of photons?
And what distortes space-time, incidentally, isn’t the amount of rest mass in the volume per se, right? It’s the amount of stress energy in the volume (and the momentum density, the energy flux, and the momentum flux). Since, as far as I’m aware, one should be able to find a frame wherein some of the above are non-zero for photons, it follows obviously that they act as a source for gravity, despite being massless.
Because the number you’re multiplying by is infinity. Zero times infinity equals whatever you want. The formula is:
relativistic mass = rest mass × gamma
Where gamma is based on your speed. If you’re still (speed = 0) then gamma = 1. If you’re going the speed of light (speed = c) then gamma = infinity.
Omphaloskeptic wrote:
The source for gravitation is the energy-momentum tensor, not just the rest mass. If each individual photon doesn’t deform space-time, how can you possibly get spacetime deformation from a group of them?
All 3 from the OP are at least incomplete. Rewriting:
energy and momentum deform space-time, yielding what we call gravity. Mass is usually by far the largest component here.
energy, momentum, and mass are related by E[sup]2[/sup] = M[sup]2[/sup]C[sup]4[/sup] + P[sup]2[/sup] (where P is the momentum).
photons deform space-time.
E[sup]2[/sup] = M[sup]2[/sup]C[sup]4[/sup] + P[sup]2[/sup]C[sup]2[/sup]
wow. i didn’t know momentum warped space-time! is momentum like the halfway point between energy and matter (or, perhaps an admixture of is more to the point)?
jb
I agree. What I was trying to say is that an individual photon (or a collimated beam) does not cause what’s usually thought of as spacetime curvature. (We can transform to a frame in which the photon has arbitrarily low energy, so the stress-energy tensor is arbitrarily close to zero.) But this is only partly right. The Riemann scalar curvature is zero for a single photon, but the Ricci tensor isn’t. Sorry for my confusion.