1. Mar 6, 2015

### davidbenari

I wont go into any details because it is really not important.

In class we handled a problem dealing with an incoming photon that crashes into an electron, the electron gains momentum and a new photon is emitted too.

My first question is about Maxwellian E&M. I understand that according to classical EM an accelerating particle produces an EM wave. And according to the photon theory of light, the wavefront of any such wave is replete with bundles of energy called photons. So in this particle interaction I mentioned above, why has the electron emitted a single photon and not the massive amount of photons that an EM wave would predict?

My second question deals with the mathematical construction of such interactions. It was assumed in this collision that the mass of the system was invariant in the before-after scenarios. This is not the typical assumption for nuclear disintegration, for example. The reason why I think the mass is assumed invariant is because the identity of the particle travelling is known in both cases. Namely, it is an "electron" and electrons have mass $m$. This makes perfect sense, but is it known for a fact that all electrons have the exact same mass?

In the case of protons and neutrons, I guess it is also assumed they all have the same mass (not implying here that m-proton=m-neutron). But I don't like this idea because it suggests that whenever I did problems on nuclear disintegration the mass difference (before vs after) was always a linear combination (with integer coefficients) of the masses of the proton and neutron. I think this is assuming too much, and perhaps not even true when I did in fact do such problems.

Thanks for reading so far! Hehe.

2. Mar 6, 2015

### Staff: Mentor

Your understanding of photons and light is incorrect. If you search the quantum mechanics forum here, you'll find a number of threads discussing how photons interact with matter. (Although if the incoming photon is sufficiently energetic, you may get multiple photons out of the interaction - but that's not what you're describing here).

We know that if all electrons do not have the same mass the difference doesn't show up in the first ten decimal places - we've measured the mass of electrons to that degree of accuracy without seeing any difference. It's also a prediction of quantum field theory that all electrons will have the same mass, and QFT works so well in so many places that it would be bizarre not to trust it here too.

However, this is all a bit of a digression. The energy of the system is conserved across the interaction, not the mass. It just so happens that if the energy of the incoming photon is small compared to $m_ec^2$ then the rest mass of the system before and after will be so close to $m_e$, the mass of the electron, that we can use conservation of mass as a very good approximation.

It's the same general principle - we've measured the masses to nine decimal places without seeing any variation from one proton to another. The QFT argument also applies, although it is a bit more involved because the nucleons are not elementary particles - they're composed of quarks which are.

3. Mar 6, 2015

### davidbenari

Is the thing I said about a linear combinations of the masses of the nucleons (with integer coefficients) a true statement? So that if the mass of the system decreased then necessarily some nucleons disappeared, and its not that some nucleons now "weigh less" ?

I am having trouble seeing how my interpretation of the photon theory of light is incorrect. My textbook (Serway, Modern Physics) says this

""He maintained that the energy of light is not distributed evenly over the classical wavefront, but is concen- trated in discrete regions (or in “bundles”), called quanta, each con- taining energy, hf. ""

I think of this as the EM wavefront but whose energy is distributed over the wavefront in a granular way.

The electron gained momentum, and it accelerated. Why didn't it produce an ordinary Maxwellian wave?

4. Mar 7, 2015

### Orodruin

Staff Emeritus
The nucleons do not weight less, but there is also a binding energy involved. The binding energies are large enough to make a difference.

This is wrong and not how the classical limit of quantum electrodynamics works. The classical limit is more akin to coherent quantum states, i.e., states which do not have a definite number of photons, but are superpositions of states with any number of photons.

Because all of this happened at a level where quantum effects are dominating. At this level, an electron does not accelerate as a classical particle, and it does not emit a classical field in the same way a classically accelerating particle would.

5. Mar 7, 2015

### davidbenari

Just out of curiosity, when is the picture of the Maxwellian wave created by an accelerated charged particle adequate? When is the phenomenon not too quantum?

In class we did a calculation of the number of photons crossing an area $A$ due to a laser light. It assumed a constant intensity, and the energy of the photons as $E=hf$. Was this calculation wrong because it supposes a definite number of photons then?

Thanks.

6. Mar 7, 2015

### Orodruin

Staff Emeritus
While you cannot count photons in this way (just as little as you can know the slit the particle went through in the double slit experiment), you can compute the expectation value of the number of photons.

7. Mar 7, 2015

### Staff: Mentor

Also, for a very very very very large number of photons, such as you get in a typical laser, the variations from the expectation value are very very very very tiny, in percentage terms.