# How does EM wave geometrical attenuation affect atomic absorption?

• I
• Leureka
In summary, an electromagnetic wave from a point source in a vacuum has a constant energy that attenuates according to the inverse square law. An absorber atom at a certain distance from the source may absorb some of the wave's energy, which is transferred in discrete amounts called photons. However, this does not mean the wave is made up of photons. The rest of the wavefront is largely unaffected by the interaction, and total energy is conserved within small volumes of space. The inverse process of atomic emission is a quantized transfer of energy to the EM field, resulting in a single-photon state that does not resemble a propagating wave. This process cannot be fully understood without studying quantum electrodynamics. The direction of EM wave emission is
Leureka
TL;DR Summary
How does photon absorption happen in the context of spherically propagating EM waves?
Let's say we have a point source of an EM wave in a vacuum of total energy E, and an absorber atom at some distance from this source, whose first excited state is at the energy B, with B < or = E.

The energy of the wave is constant as a whole, but at each point around the source the energy geometrically attenuates according to the inverse square law.

Does the atom ever absorb the incoming radiation? Or does it absorb only at the right distance where E=B? If that's the case, what happens to the concept of photons having constant energy E=hf?

Also, if the radiation is absorbed, what happens to the rest of the wavefront? The total energy of the wave decreased, but is this reflected in all other points of the wave? Or does the loss of energy only happen in that particular spot of the wavefront?

Leureka said:
Does the atom ever absorb the incoming radiation? Or does it absorb only at the right distance where E=B? If that's the case, what happens to the concept of photons having constant energy E=hf?
The relation ##E=h\nu## is correct but not especially relevant here. We have an electromagnetic wave expanding outwards from the central point, it encounters an atom, and that interaction may transfer some of the EM field's energy and momentum. It turns out that these transfers always happen in discrete amounts, and when a such a transfer happens we we say that "a photon was absorbed" or something similar. But that does not mean that the wave is made up of photons the way a water wave is made up of water molecules.

It is true that we can write the state of the electromagnetic wave as sum of single photon states, and doing so facilitates calculating the probability that the interaction will happen. But that's a mathematical representation chosen for computational convenience.
Also, if the radiation is absorbed, what happens to the rest of the wavefront? The total energy of the wave decreased, but is this reflected in all other points of the wave? Or does the loss of energy only happen in that particular spot of the wavefront?
Total energy within any small volume of space is conserved. Thus in the immediate vicinity of the particle there's a bit less energy in the EM field and a bit more in the particle (which is what we'd expect when the interaction transfers energy from one to the other) while the rest of the wavefront is largely unaffected.

Nugatory said:
The relation ##E=h\nu## is correct but not especially relevant here. We have an electromagnetic wave expanding outwards from the central point, it encounters an atom, and that interaction may transfer some of the EM field's energy and momentum. It turns out that these transfers always happen in discrete amounts, and when a such a transfer happens we we say that "a photon was absorbed" or something similar. But that does not mean that the wave is made up of photons the way a water wave is made up of water molecules.

It is true that we can write the state of the electromagnetic wave as sum of single photon states, and doing so facilitates calculating the probability that the interaction will happen. But that's a mathematical representation chosen for computational convenience.
Total energy within any small volume of space is conserved. Thus in the immediate vicinity of the particle there's a bit less energy in the EM field and a bit more in the particle (which is what we'd expect when the interaction transfers energy from one to the other) while the rest of the wavefront is largely unaffected.
So how does this translate to the inverse process, i.e. atomic emission? Is that also subject to the inverse square law, or is it directed emission? When is it relevant to talk about photons with constant energy as opposed to a field disturbance propagating in 3D?

On this matter, what determines if a source will only direct EM waves linearly as opposed to in all directions? My initial intuition tells me it has something to do with conservation of angular momentum, such as the case of a circularly rotating charge; is this the case?

Leureka said:
So how does this translate to the inverse process, i.e. atomic emission? Is that also subject to the inverse square law, or is it directed emission?
Neither. The inverse process is a quantized transfer of energy from the atom to the electromagnetic field, and the resulting single-photon state of the EM field bears no resemblance to anything that we would recognize as a propagating electromagnetic wave. Unfortunately there's no way to properly understand this without going through a serious textbook on quantum electrodynamics, and that's not something that can be done in an I-level thread. You might, however, give http://www.physics.usu.edu/torre/3700_Spring_2015/What_is_a_photon.pdf a try - it cuts some corners to get down to an I-level treatment.
On this matter, what determines if a source will only direct EM waves linearly as opposed to in all directions? My initial intuition tells me it has something to do with conservation of angular momentum, such as the case of a circularly rotating charge; is this the case?
No. The direction of EM wave propagation is determined by the initial conditions of the time-varying EM field and is a classical phenomenon based on interference. Start with the Huygens-Fresnel principle and, for a practical working example, look at how phased-array radars aim their beams.

• Quantum Physics
Replies
54
Views
3K
• Quantum Physics
Replies
18
Views
2K
• Quantum Physics
Replies
21
Views
1K
• Quantum Physics
Replies
38
Views
3K
• Quantum Physics
Replies
1
Views
648
• Atomic and Condensed Matter
Replies
23
Views
2K
• Quantum Physics
Replies
36
Views
2K
• Quantum Physics
Replies
1
Views
737
• Quantum Physics
Replies
5
Views
2K
• Quantum Physics
Replies
2
Views
446