I’ll work through my thought process…stop me where I make an error:
E=MC[sup]2[/sup] so mass and energy are equivalent.
So…
Take a box (the magical Unobtainium kind that has no mass of its own but is unbreakable and perfectly reflective inside) and put the earth in it.
We cannot see into the box.
At the flip of a switch we can convert all the mass in the box to energy and back at will.
It is my understanding that the gravitational effect of the earth will be felt the same outside the box whether the earth is all energy or all mass.
So, would this mean then that photons must emit gravitons?
Sounds odd to me put that way since gravity is one of the four fundamental forces. I would think if photons emitted gravitons that would make gravity different in an important way.
So then, to be consistent in Quantum Mechanics, photons would have to be said as emitting gravitons right? If not how else do I feel the combined gravity of all those photons?
Once you start talking about gravitons, you’re in the realm of quantum gravity, about which we can say very little. But one of the few things we can say is that, if the virtual particle model of forces can be used with gravity at all (a point which is not established), then gravitons must be able to couple to absolutely all of the other particles.
Yes, this is correct. If it weren’t, you’d be able to build a perpetual motion machine out of it.
Are the four fundamental forces, as described in QM as having a force carrying particle (e.g. a photon), actually not real and just virtual in nature? I thought they were all deemed real and except for the graviton have been spotted by scientists. The graviton being the only holdout because no one has built a detector of sufficient power to see it (but in theory could be if we could build such a device).
So what’s with virtual gravitons?
And is it not at all odd that a “fundamental” force actually emits another fundamental force? I would think “fundamental” would preclude that.
A real physicist will be along shortly. In the mean time, The four fundamental forces are not described by QM. Only three are. There is at this time no theory of quantum gravity.
The virtual part isn’t referring to unobserved particles in the sense that no instrument has ever detected them. I believe Ring was making an analogy to virtual photons which are what an electromagnetic field is made out of.
A real particle can be detected and therefore must carry energy. And this energy would have to come from somewhere. Whereas a virtual particle isn’t on the mass shell (it doesn’t obey Einstein’s relativistic equation that relates energy momentum and mass) and therefore can’t be detected.
As I’m sure you know a virtual particle can only exist within the confines of the energy time uncertainty principle, and there’s even some disagreement as to whether they actually exist or are just mathematical artifacts.
Although technically, a particle that’s detected is really just a virtual particle that was really close to its mass shell, and coupled some particle in the source with some other particle in the detector. A particle is only genuinely real if it propagates an infinite distance.
The thing is, no particle is ever exactly on its mass shell. It’s just that some of them are really, really close, and we often approximate those as real.
Yes, if the graviton has energy, but if it’s a virtual graviton, it doesn’t need to. If you had, say, two perfectly reflective mirrors (you buy those from the same catalog as frictionless tables and massless unbreakable strings) with a pair of photons bouncing back and forth between them, then you’d very occasionally emit real gravitons with positive energy, and the pair of photons really would lose energy (albeit extremely slowly).
Now I’m confused. Doesn’t a particle have to be exactly on its mass shell in order to propagate to infinity? I thought the further from the MS the shorter its life.
Thanks for the response. I don’t understand how a photon can lose energy. It’s has no mass (right?) so the only way (that I know of) for it to lose energy is to slow down but the speed of light is constant. I’m trying to squeeze all this together in my head.
Yes, but does any particle ever actually propagate to infinity, either?
All photons travel at c, but the energy of a photon is proportional to its frequency. A gamma-ray photon, for instance, has an extremely high energy, but a radio photon has a low energy. That’s why gamma rays are more dangerous than radio waves.
And while a system of multiple photons could in principle emit real gravitons, a single isolated photon cannot: There’s just no way to do it while conserving energy, momentum, and angular momentum.
