Why can't a free electron absorb or emit photon?

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Homework Help Overview

The discussion revolves around the assertion that a free electron cannot absorb or emit a photon, exploring the underlying reasons and implications within the context of special relativity and conservation laws.

Discussion Character

  • Conceptual clarification, Assumption checking, Mixed

Approaches and Questions Raised

  • Participants examine the conservation of energy and momentum in the context of photon absorption and emission by free electrons. Some express confusion about the constraints placed on free electrons, while others reference specific equations and principles from special relativity to support their points.

Discussion Status

There is an active exploration of the topic, with some participants providing mathematical insights and others questioning the validity of the original claim. A few suggest that the article may be misleading or that additional context is necessary to understand the conditions under which free electrons interact with photons.

Contextual Notes

Participants note that the discussion may be influenced by specific scenarios, such as those involving free-electron lasers, where electrons are not truly free but are instead manipulated by external fields.

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Homework Statement


In an article, I am informed that a free electron can neither absorb nor emit a photon. I just want to know why?! Please help me.

Homework Equations




The Attempt at a Solution


I've no idea at all. Photon is the medium boson carrying electromagnetic force, so the relationship between electron and photon is interesting. The above process is said to bef orbidden, and I think some conservation laws must be violated!
 
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look at the conduction electrons in a metal and how they interact with light.
 
Hello, Zeus! I'm so glad to come across your problem. Nice and attractive!
It is essential in special relativity that, a free electron can neither absorb nor emit photons. Hints are given in details below.

1. Absorbing of photon is forbidden.

If a free electron could absorb a photon, then, according to conservation of energy and momentum,
$$
\hbar \omega +mc^2 = \sqrt{p^2c^2 + m^2c^4} (Eq1)
$$
$$
\hbar k = p (Eq2)
$$
where $\omega$ and $k$ are the frequency and wavenumber of the photon, respectively, $m$ the electron's rest mass, $p$ the momentum of the electron after absorbing the photon. Eq2 leads to
$$
p^2=\hbar^2 k^2 = \hbar^2 \frac{\omega^2}{c^2} (Eq3)
$$
Insert Eq3 into Eq2, and square of the left side is
$$
(\hbar \omega +mc^2 )^2 =\hbar^2 \omega^2 + m^2c^4+2mc^2\hbar \omega
$$
when square of the right side is
$$
\hbar^2\omega^2+m^2c^4
$$
So, if Eq1 holds, $\hbar \omega=0$. There is no photon carrying vanishing energy. Hence, absorbing of a photon by a free electron is forbidden.


2. Emitting of a photon is Forbidden
Suppose the intial (before photon emitting) and final (after photon emitting) 4-momentum of the electron are separately $p_\mu$, $p'_\mu $:
$$
p_\mu=(0,0,0,imc), \qquad p'_\mu = \left[p, i\frac{mc}{\sqrt{1-\frac{v^2}{c^2}}} \right] (Eq4)
$$
According to conservation of energy and momentum:
$$
p_\mu=q_\mu+p'_\mu (Eq5)
(p_\mu - p'_\mu)^2 =q_\mu^2 (Eq6)
$$
where $q_\mu$ refers to the 4-momentum of the photon. From Eq5 and Eq6, we have
$$
(p_\mu - p'_\mu)^2 = p_\mu^2 -2p_\mu p'_\mu+p'_\mu^2
=-m^2c^2- 2p_\mu p'_\mu -\frac{mc^2}{1-\frac{v^2}{c^2}} (Eq7)
$$
Recall that
$$
p_\mu p'_\mu = -\frac{mc^2}{\sqrt{1-\frac{v^2}{c^2}}} (Eq8)
$$
Insert Eq8 into Eq7, we have
$$
p_\mu p'_\mu = -m^2 c^2 \left[ 1-\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \right] (Eq9)
$$
Yet $q_\mu^2=0$, so, $p_\mu=p'_\mu$, to ensure Eq5 and Eq6 hold. Thus the state of motion of the electron is not disturbed, with no momentum transported to the photon. Hence, emitting of a photon by the free electron is forbidden.
 
Zeus2nd said:

Homework Statement


In an article, I am informed that a free electron can neither absorb nor emit a photon. I just want to know why?! Please help me.

Homework Equations




The Attempt at a Solution


I've no idea at all. Photon is the medium boson carrying electromagnetic force, so the relationship between electron and photon is interesting. The above process is said to bef orbidden, and I think some conservation laws must be violated!

Sounds like the article is wrong. Or else they are placing constraints on the electrons that are not met in FELs...

http://en.wikipedia.org/wiki/Free_electron_laser

.
 
berkeman said:
Sounds like the article is wrong. Or else they are placing constraints on the electrons that are not met in FELs...

http://en.wikipedia.org/wiki/Free_electron_laser

The electrons in a Free-electron laser aren't actually free in the sense used for the question. They are tightly controlled by an oscillating magnetic field, so that they emit coherent radiation.

An electron can't absorb a photon all by itself, because you can't get conservation of energy and momentum with the electron alone.

Cheers -- sylas
 
I would like to draw your attention to https://www.physicsforums.com/showthread.php?t=308657 , on which another interesting problem on the relationship between photon and electron is discussed. This problem is ----why can an electron - positron pair not be created by a photon in free space(by sheelbe999 )? Here also, mathematical formulation are given in details as a hint.
 
Tian WJ said:
I would like to draw your attention to https://www.physicsforums.com/showthread.php?t=308657...

Oh, beautiful! George Jones [post=2167860]#7[/post] gives a very elegant solution, that works here also.

Imagine that a free electron absorbs a photon. Analyze this in the reference frame where the electron is at rest after the absorption. What is the energy of the absorbed photon in this frame?
 
it is true that a free electron can't emit or absorb a photon but it is misleading to think that a free electron is totally unaffected by the passage of a light wave. the electron will indeed oscillate but it will always emit exactly the amount of energy that it absorbs. the 'net' effect is that it is unaffected.

I think that the light might be scattered in the process. I'm not sure.
 

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