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I was wondering about the things like Compton scattering. As I understand, it is an inelastic scattering of photon on free electron. Inelastic means, that photon changes it's angle and frequency. As this site shows every people confused why electron can't completely absorb photon's energy. The great minds answers, that "because energy and momentum can't conserve simultaneously", and gives math, that, for me, do not explain nothing. So my doubt goes below.

Electron is at rest, it's energy is $m_ec^2$. Here goes photon. Electron absorbs it completely, and start to change it's velocity, BUT, there is an $inertia$ - ability of electron, to save it's velocity constant, if there is no influence on it, and to confront the change in velocity. And then I assume two situations

1.The acceleration process goes into the negative acceleration process, electron looses some energy - new photon. (photon changes it's angle(?) and frequency)

enter image description here

  1. The first one may be bad, because seems, that in transition "the acceleration -> moving constantly" there is no negative acceleration, hence electron doesn't loose anything. So, perhaps, electron emits photon, during the acceleration(how, it's energy increases??)

enter image description here

About mass(or energy, I'm confused already) changing, due to velocity changing.I want to find will the mass of electron change, if speed increases.

electron in rest, v=0

$E=\dfrac{mc^2}{\sqrt{1-v^2/c^2}}$

$E\sqrt{1-v^2/c^2}=mc^2$

$m=\dfrac{E\sqrt{1-v^2/c^2}}{c^2}$

So, electron should change it's mass if speed changes. You probably answer - "you stupid, the velocity change is the derivative of energy change, i.e. electrons energy changes, and then velocity changes". But then it confronts, the fact that electron can't gain energy and momentum.

If my first assumption was correct, then all scatterings are all about a complete absorption and reemitting processes?

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    $\begingroup$ What do you mean by "As this site shows every people confused why electron can't completely absorb photon's energy"? $\endgroup$ – G K Sep 15 '18 at 20:10
  • $\begingroup$ @GK just search "free electron absorb photon", and You'll understand. But You have already offended.. $\endgroup$ – user205628 Sep 15 '18 at 20:13
  • $\begingroup$ Well, you are the one writting it in your post. I wanted you to answer me. $\endgroup$ – G K Sep 15 '18 at 20:14
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    $\begingroup$ In quantum field theories the tree-level description of a scattering event (like Compton scattering) is in terms of an abosorption operator and a creation operator. But you shouldn't understand that as describing a sequence with a well defined time-ordering. Links physics.stackexchange.com/questions/185110/… physics.stackexchange.com/q/162845/520 $\endgroup$ – dmckee Sep 15 '18 at 20:17
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    $\begingroup$ Comments ascribing negative personality traits or states of mind to other users violate the "Be nice." policy and will be removed. $\endgroup$ – dmckee Sep 15 '18 at 20:20
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This is not a full answer, but just to illustrate why a free massive particle (electron) cannot absorb a massless particle (photon). The total energy (squared) of the electron is

$$ E_e^2=m^2c^4+p_e^2c^2 \tag{1} $$

In the rest frame, the initial momentum is zero and the initial energy is $mc^2$. Let's assume that after the collision the electron absorbs the momentum of the photon

$$ p_e=p_p=\dfrac{hf}{c} $$

where $h$ is the Plank constant and $f$ is the frequency. According to $(1)$, the energy (squared) of the electron now should be

$$ E_e^2=m^2c^4+h^2 f^2 \tag{2} $$

However, per energy conservation, the electron absorbs the energy of the photon

$$ E_e=mc^2+E_p=mc^2+hf $$

Taking a square gives

$$ E_e^2=m^2c^4 {\color{red}{+2mc^2hf}} +h^2 f^2 \tag{3} $$

As you can see, the energy $(2)$ derived from the conservation of momentum does not match the energy $(3)$ derived from the conservation of energy. Because both energy and momentum cannot be conserved at the same time, this process is prohibited by the laws of nature. A free electron cannot absorb a photon.

This inequality must be resolved quickly within the timeframe allowed by the uncertainty principle. The extra energy and momentum must be emitted as another photon. Thus the Compton scattering involves an absorption of the initial photon by the electron and an immediate emission of another photon by the electron. Between the events of absorption and emission, the electron becomes virtual. In other words, energy and momentum get out of balance upon the absorption that promopts an immediate emission to restore the balance.

This analysis is valid for elementary particles without internal energy levels. Composite particles, such as atoms, do have internal energy levels and can be excited to these levels to consume the extra amount of energy highlighted in red above. However, the excited state is temporary and this extra energy is soon released by emitting a photon.

The main difference between elementary and composite particles is that the elementary particle (such as the electron) upon absorption of the photon becomes virtual and thus must emit a photon instantly. In contrast, a composite particle (such as an atom) doesn't become virtual on absorption, but only excited. It still must emit a photon soon, but not instantly. This effect is used in lasers where many atoms are excited first and then after a short delay are stimulated to emit a coherent light all together.

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  • $\begingroup$ I would call it "reductio ad absurdum" , energy and momentum have to be conserved at all times, you should also add the "not composite with binding energies" in the statement "free massive particle" +1 for doing the algebra $\endgroup$ – anna v Sep 17 '18 at 3:57
  • $\begingroup$ @anna Exactly and this is why the absorption is impossible. $\endgroup$ – safesphere Sep 17 '18 at 4:00
  • $\begingroup$ The thing, that also confuses me, is that, if electron completely absorb a photon, then where does photon goes? It just had to disappear? But we consider a system of electron and photon, and if there is photon, hence there is no system $\endgroup$ – user205628 Sep 17 '18 at 12:13
  • $\begingroup$ @Artur There are matter particles called fermions (e.g. the electron) and "energy" (or field) particles called bosons (e.g. the photon). Fermions have quantum numbers that conserve and normally cannot disappear. Bosons don't have such a restriction (other than for spin or charge). Bosons are like pure energy, they can emerge and vanish in numbers. So we are not concerned with the photon disappearing upon absorption, because it is what the photon is for, its whole life purpose is to transfer energy and momentum (plus spin) from one charged particle to another. $\endgroup$ – safesphere Sep 17 '18 at 19:15
  • $\begingroup$ @anna I have added a section on composite particles. Thanks! $\endgroup$ – safesphere Sep 17 '18 at 19:45
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Energy and momentum conservation are very very basic in physics, as also special relativity and quantum field theory in particle physics.

Electron is at rest, it's energy is $m_ec^2$. Here goes photon. Electron absorbs it completely, and start to change it's velocity.

This scenario cannot happen with the electron because it is not composite. There is no way to absorb energy for the electron in the standard mainstream model of physics.

If a photon hits an atom, which is a composite object, there is a probability that an electron will go to a higher energy state, so the photon will be completely absorbed and the mass of the excited atom will be the invariant mass of the two four vectors that have been added, and the atom, if at rest, will acquire a momentum. (solution 5 in this exam )

There are no constituents in the electron that can go to a higher energy level and absorb a photon, so at best, a lower energy photon may leave the interaction, part of its energy and momentum transferred to the electron. There are no excited electrons.

This is what the compton scattering diagram shows for computing the probability of the scattering crossection. (Feynman diagrams are strict prescriptions for a mathematical formulation)

comton

The classical interpretation you are trying to impose is rejected from energy and momentum conservation; in the microcosm of electrons and photons it is quantum mechanics that can mathematically model the scattering. The electron line is off mass shell after the first vertex, i.e its mass is not the mass of the electron, it is a virtual electron just because of energy and momentum conservation. It recovers its mass at the second vertex where a different energy/momentum photon exits, and a different energy momentum electron, but both on their mass shell, 0 for photon and $m_e$ for the electron.

If you are interested in particle physics you should study quantum mechanics, and quantum field theory. Own speculations are a waste of time , and not appropriate for this site.

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  • $\begingroup$ "Energy and momentum conservation are very very basic in physics" the confusion is not about laws, but about why they can't conserve simultaneously. $\endgroup$ – user205628 Sep 16 '18 at 7:24
  • $\begingroup$ energy and momentum conerve simulatneously in the same inertial system, assumed during any particles interactions. That is why adding the four vector of the photon would increase the mass of the electron and that is not acceptable in special relativity unless the electron is a composite particle that can be excited to a higher level, and this is not the case. The electron has no internal structure . The virtual exchanges of the feynman diagrams take care of this by taking the electron off mass shell, makng it virtual, i.e. not on its measured mass, (the one in the table) $\endgroup$ – anna v Sep 16 '18 at 7:56
  • $\begingroup$ You say composite, energy levels, only a structure can absorb, etc.. What does it mean at all? It's all because of relativity principle? I mean, if we consider an alone electron, there is no bodies, relative to which, we can measure electron's speed(and hence energy), so that's why there is no sense to talk about photon absorption by electron, or what? $\endgroup$ – user205628 Sep 16 '18 at 8:21
  • $\begingroup$ read the paragraph above about the atom and its change in mass when it absorbs a photon $\endgroup$ – anna v Sep 16 '18 at 8:33
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    $\begingroup$ the lorenz boost has nothing to do with the invariant mass. If you cannot understand four vectors better look elsewhere for entertainment. $\endgroup$ – anna v Sep 16 '18 at 15:12
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The most fundamental description we today have of scattering processes is quantum field theory. In quantum field theory, we are allowed to talk about the particles before and after the scattering process, but not about them during the process, the reason being that the intermediate quantum state during the scattering is so unlike the distinct particle states that the classical idea that we could identify individual particles and talk about their acceleration simply does not apply. For more on what we can and cannot say about quantum scattering, see also this answer of mine.

In particular, Feynman diagrams are perturbative expressions that have no intrinsic relation to what's "really" going on - so any explanation involving "virtual particles" (except for those not referring to internal lines in a Feynman diagram, see this answer of mine is as unfounded in the full non-perturbative formalism as the idea that you can track any "acceleration" during the scattering process.

There is neither "emission" nor "absorption", just a quantum scattering process during which are we are not allowed to apply any reasoning relying on identifying individual particles. The (almost) full extent of what quantum field theory says about photon-electron scattering is the Klein-Nishina formula that gives the distribution of possible angles between the incoming and outgoing photon after the scattering process.

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