Negative energy photon emission

In summary: D and 2D/c before that. However, from Observer A's perspective, the lightning bolt struck at exactly the same time as it did from Observer B's perspective. :smile:Your frame of reference depends just as much on your position as your velocity.
  • #1
tickle_monste
69
1
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.
 
  • Like
Likes coolaruns
Physics news on Phys.org
  • #2
Hi tickle_monste! :smile:

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. :smile:
 
  • #3
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
tickle_monste said:
… 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?

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. :smile:
 
  • #5
tiny-tim said:
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. :smile:

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
tickle_monste said:
So, mathematically speaking, there's no difference between these two things?

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! :wink:
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?

Sorry, you've completely changed the subject, and I don't follow what you're saying anyway. :redface:

Frames of reference have nothing to do with positions or times, they depend on velocities.
 
  • #7
tiny-tim said:
Sorry, you've completely changed the subject, and I don't follow what you're saying anyway. :redface:

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

Your frame of reference depends just as much on your position as your velocity. Let's 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
tickle_monste said:
Your frame of reference depends just as much on your position as your velocity. Let's 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 …

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". :wink:

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! :smile:
 
  • #9
But how is adjusting for position any different than adjusting for velocity?
 

1. What is negative energy photon emission?

Negative energy photon emission is a hypothetical phenomenon in which photons, or particles of light, are emitted with a negative amount of energy. This concept is based on the theory of negative energy, which suggests that energy can exist in negative forms.

2. How is negative energy photon emission different from regular photon emission?

Regular photon emission occurs when a particle releases a photon with a positive amount of energy. Negative energy photon emission, on the other hand, would involve the release of photons with a negative amount of energy, which goes against our current understanding of energy as a positive quantity.

3. Is there any evidence for negative energy photon emission?

At this time, there is no concrete evidence for negative energy photon emission. While some theories, such as the Casimir effect, suggest the existence of negative energy, there is no experimental proof of negative energy photons being emitted.

4. What are the potential implications of negative energy photon emission?

If negative energy photon emission were to be proven, it would have significant implications for our understanding of energy and the laws of physics. It could also potentially lead to new technologies and applications, such as harnessing negative energy for power generation.

5. How is negative energy photon emission being studied?

Scientists are currently exploring the concept of negative energy through theoretical calculations and experiments, such as the Casimir effect. However, more research and evidence are needed to fully understand and potentially observe negative energy photon emission.

Similar threads

  • Quantum Physics
Replies
21
Views
813
Replies
1
Views
261
  • Quantum Physics
Replies
15
Views
2K
  • Quantum Physics
2
Replies
38
Views
3K
Replies
6
Views
944
Replies
18
Views
1K
  • Quantum Physics
Replies
9
Views
1K
  • Quantum Physics
2
Replies
47
Views
3K
Replies
7
Views
1K
  • Quantum Physics
Replies
5
Views
747
Back
Top