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Why can't a free electron absorb a photon? But a one attached to an atom can.. Can you explain to me logically and by easy equations? Thank you..

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  • $\begingroup$ I didn't realize a free electron could not absorb a photon. Or at least I never though about it but I will now. That makes photon / electron interaction a little more interesting when you think about a double slit experiments with electrons and observations. $\endgroup$ – Bill Alsept Dec 23 '15 at 9:28
  • $\begingroup$ Oh yes I have it written on my notebook but I didn't understand my teacher's explanation.. $\endgroup$ – user65035 Dec 23 '15 at 9:35
  • $\begingroup$ I would like to mention that an electron in Bremsstrahlung process is not BOUND to ion but a photon impacts it and changes the electron's path. $\endgroup$ – Amin R. Dec 23 '15 at 11:28
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Can't believe I can't find a duplicate. It is because energy and momentum cannot be simultaneously conserved if a free electron were to absorb a photon. If the electron is bound to an atom then the atom itself is able to act as a third body repository of energy and momentum.

Details below:

Conservation of momentum when a photon ($\nu$) interacts with a free electron, assuming that it were absorbed, gives us \begin{equation} p_{1} + p_{\nu} = p_{2}, \tag{1} \end{equation} where $p_1$ and $p_2$ are the momentum of the electron before and after the interaction. Conservation of energy gives us \begin{equation} \sqrt{(p_{1}^{2}c^{2} + m_{e}^{2}c^{4})} + p_{\nu}c = \sqrt{(p_{2}^{2}c^{2}+m_{e}^{2}c^{4})} \tag{2} \end{equation} Squaring equation (2) and substituting for $p_{\nu}$ from equation (1), we have $$ p_{1}^{2}c^{2} + m_{e}^{2}c^{4} + 2(p_{2}-p_{1})\sqrt{(p_{1}^{2}c^{2}+m_{e}^{2}c^{4})} + (p_{2}-p_{1})^{2}c^{2}=p_{2}^{2}c^{2}+m_{e}^{2}c^{4} $$ $$ (p_{2}-p_{1})^{2}c^{2} - (p_{2}^{2}-p_{1}^{2})c^{2} + 2(p_{2}-p_{1})c\sqrt{(p_{1}^{2}c^{2}+m_{e}^{2}c^{4})} = 0 $$ Clearly $p_{2}-p_{1}=0$ is a solution to this equation, but cannot be possible if the photon has non-zero momentum. Dividing through by this solution we are left with $$ \sqrt{(p_{1}^{2}c^{2}+m_{e}^{2}c^{4})} - p_{1}c =0 $$ This solution is also impossible if the electron has non-zero rest mass (which it does). We conclude therefore that a free electron cannot absorb a photon because energy and momentum cannot simultaneously be conserved.

NB: The above demonstration assumes a linear interaction. In general $\vec{p_{\nu}}$, $\vec{p_1}$ and $\vec{p_2}$ would not be aligned. However you can always transform to a frame of reference where the electron is initially at rest so that $\vec{p_1}=0$ and then the momentum of the photon and the momentum of the electron after the interaction would have to be equal. This then leads to the nonsensical result that either $p_2=0$ or $m_e c^2 = 0$. This is probably a more elegant proof.

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    $\begingroup$ By george, I think he's got it. $\endgroup$ – Digiproc Dec 23 '15 at 10:22
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    $\begingroup$ +1. Does this imply that in Compton scattering, the electron is 'off-mass shell'? I was wondering, how to avoid confusions (like this) between conservation laws of asymptotic scattering (well defined incoming fourvectors), and a virtual processes inside a diagram (it is commonly called absorption too inside a diagram, I believe -- "In compton scattering the electron is absorbed and emitted", people might say). $\endgroup$ – Mikael Kuisma Dec 23 '15 at 10:27
  • $\begingroup$ This answer assumes that the mass of the electron is the same before and after the absorption. What justifies that? $\endgroup$ – WillO Dec 24 '15 at 14:04
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    $\begingroup$ @WillO The rest mass of an electron is invariant. $\endgroup$ – Rob Jeffries Dec 24 '15 at 14:50
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    $\begingroup$ @RobJeffries: I apologize in advance for the fact that if I were better educated I wouldn't have to ask this, but: Why can't the combined electron-plus-photon have a new rest mass that allows both energy and momentum to be conserved? If electrons are defined in such a way that their mass is invariant, then we are not allowed to call this new particle an electron. But I still don't see why it can't exist. $\endgroup$ – WillO Dec 24 '15 at 14:57
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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. (source)

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

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The book is wrong. See Compton scattering.

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    $\begingroup$ In Compton scattering a photon with a different momentum is emitted. $\endgroup$ – CuriousOne Dec 23 '15 at 10:06
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    $\begingroup$ ...but a photon is also absorbed? Maybe the point of the 'lecturers notes' is related to conservation of some quantity (angular momentum?), which can be broken by virtual particles. In any case, "Free electron can't absorb a photon" is a bad way to represent a thing like that. $\endgroup$ – Mikael Kuisma Dec 23 '15 at 10:10
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    $\begingroup$ Mmm maybe he means that it can't be completely absorbed but partially such that the amount absorbed turns as a kinetic energy for the electron, right? $\endgroup$ – user65035 Dec 23 '15 at 10:10
  • $\begingroup$ Didn't know if the OP was asking if it can be absorbed completely. In theory (depending on the observers ref frame) the product photon can have negligible momentum, such that almost all is absorbed. $\endgroup$ – Digiproc Dec 23 '15 at 10:14
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    $\begingroup$ This answer is not complete enough to be useful, in my opinion. $\endgroup$ – Danu Dec 23 '15 at 10:22
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The statement that free electrons can't absorb (any) photon can be disproofed at once. In a particle accelerator how electrons are accelerated? By electromagnetic fields. And a EM field is nothing else as a lot of photons. To conserve momentum and other properties the electron emits back photons. In the case of electron acceleration this emitted photons have less energy than the incoming photons.

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    $\begingroup$ "And a EM field is nothing else as a lot of photons."...this is wrong. A classical EM field is just that, a classical field, and a lot of photons is just that - a lot of photons. Classical EM waves have a certain relation as being "made out of photons", cf. this question, but in general, it is completely wrong to say that the classical field is made out of the quanta of the quantum field. $\endgroup$ – ACuriousMind Dec 23 '15 at 15:05
  • $\begingroup$ @ACuriousMind We are both obsessed from some imaginations. But I prefer to reply to the questions, answers and comments. Argumentation is something different from state something. I will be happy to get disproofed step by step, but read my questions and answers and the replyes. What I hear is that this is not teached. Is this the only argument? Poor discussion. Try to destroy my chain of well known phenomena. $\endgroup$ – HolgerFiedler Dec 23 '15 at 15:50
  • $\begingroup$ it seems that it must emit back photons of same energy. After the first sentence, I was waiting something around the classical math of Cherenkov radiation ... ( not the reverse Cherenkov radiation ) $\endgroup$ – user46925 Dec 25 '15 at 18:08