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

  1. May 5, 2009 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant equations


    3. 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!
     
  2. jcsd
  3. May 6, 2009 #2
    look at the conduction electrons in a metal and how they interact with light.
     
  4. May 6, 2009 #3
    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.
     
  5. May 6, 2009 #4

    berkeman

    User Avatar

    Staff: Mentor

    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

    .
     
  6. May 6, 2009 #5

    sylas

    User Avatar
    Science Advisor

    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
     
  7. May 6, 2009 #6
    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.
     
  8. May 6, 2009 #7

    sylas

    User Avatar
    Science Advisor

    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?
     
  9. May 21, 2009 #8
    it is true that a free electron cant 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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