Negative energy photon emission

1. Aug 19, 2009

tickle_monste

Let's say we just had a normal atom in energy state E, and one of the electrons jumps down to a lower orbital E'. Of course, E - E' is positive because E' is a lower orbital than E, so the energy of the photon emitted must be positive. But what if E' were a higher orbital than E? Could the electron in the atom spontaneously jump up to a higher orbital and release a photon of negative energy? Virtually? Theoretically? I don't expect that we've observed such a thing, or I'd probably have heard of it, but from what I've read on the matter, Einstein's field equations don't forbid negative energy.

2. Aug 19, 2009

tiny-tim

Hi tickle_monste!

The EFE (which is general relativity), and the quantum field theory equations, are a piece of maths, and you could plug negative energy into them if you wanted …

but they wouldn't correspond to anything physical.

Real photons have positive energy, and since a photon is its own antiparticle, the annihilation of a real (positive-energy) photon can always be substitued in the maths of "collisions" by the creation of a (non-real) negative energy photon.

So if E' is a higher orbital than E, obviously that can be achieved by absorbing a photon, which in the maths behaves exactly like emitting a negative-energy photon. But physically, it is detected as the former, not the latter. The latter doesn't exist.

3. Aug 19, 2009

tickle_monste

So mathematically speaking (but without the mathematics), atom A jumps from a higher energy state to a lower energy state, and releases a photon with energy equal to the difference between the higher and lower energy states, and since it jumped from higher to lower, this difference will be 'positive' and we'll say that the photon has 'positive' energy. But then this photon propagates outward until it hits Atom B, and causes it to jump from a lower energy state to a higher energy state. If we assume that the lower and higher energy states of Atom A are equal to those of B, respectively, then would the whole of the absorption process of the photon by Atom B be considered the mathematical negative (in some sense of the word), of the original emission process of that photon by Atom A? Would it then be mathematically correct to say that at the same time that Atom A emits it's positive energy photon, the atom that will absorb it, Atom B, emits a negative energy photon?

4. Aug 19, 2009

tiny-tim

No, things don't happen simultaneously, they happen in sequence …

and for negative-energy, everything works backwards (with things being created after they are annilhilated) …

so, if in reality atom A emits a photon which is absorbed later by atom B, then the maths is the same as atom A absorbs a negative-energy photon (an "anti-photon"!) which was emitted later by atom B.

5. Aug 20, 2009

tickle_monste

Yea, that's what I was I trying to say. So, mathematically speaking, there's no difference between these two things? I understand from the theory of relativity (at least I think I understand) that points in time are just simply different frames of reference, just like different points in space, and there's no reason to prefer one reference frame over the other. Does physics today reconcile this somehow?

6. Aug 20, 2009

tiny-tim

Yes, there's no mathematical difference between them, but there's the big physical difference that the second description requires things to be destroyed before they're created … which is nonsense!
Sorry, you've completely changed the subject, and I don't follow what you're saying anyway.

Frames of reference have nothing to do with positions or times, they depend on velocities.

7. Aug 20, 2009

tickle_monste

Your frame of reference depends just as much on your position as your velocity. Lets say Observer A is a distance of D from the place where lightning strikes. He will say the lightning bolt struck at time D/c, and Observer B, standing a distance 2D from the place where the lightning strikes, will say that it struck at time 2D/c. Two different frames of reference, which we reconcile by adjusting for the speed of light (in fact, the Lorentz transformations which apply to the relativity for observers traveling at different speeds are just a generalization of the notion of relativity for observers standing at different places at different points of time).

So let's say Atom A emits a positive energy photon at time t=0 that travels in a direction in time that Atom A prefers to call 'forward'. At time t=t'>0, Atom B absorbs this positive energy photon. There is no difference between this and Atom B emitting a negative energy photon at time t=t'>0 which travels in a direction in time that Atom A calls backwards, and is absorbed by Atom A at time t=0. Shouldn't we be able to say that both frames of reference (coordinate systems) are equally valid for formulating the laws of physics, and create some sort of tensor that allows us to generalize this validity to all the coordinate systems that could be formed in this way?

8. Aug 20, 2009

tiny-tim

No, we dont!!

An observer allows for the speed of light when he says what time something happened … for example, when we see a supernova a million light years away, we say "that supernova happened a million years ago".

Observer A has a clock, and he sees the lightning at time D/c, so he will say "the distance is D, so the lightning bolt struck at time 0."

Similarly, Observer B has a clock, and he sees the lightning at time 2D/c, so he will say "the distance is 2D, so the lightning bolt struck at time 0."

A and B have the same velocity, so they use the same frame, and they agree on the time of everything!

9. Aug 20, 2009

tickle_monste

But how is adjusting for position any different than adjusting for velocity?