# Thomson scattering -- Photons can only scatter from free electrons?

Hello ,

Why does an incoming photon can only scatter from an electron if the electron is not bound top an atom ?
Because from what I know a bound electron can absorb a photon then be excited to a higher energy level and then re-emit a photon while transitioning back to it's previous energy level, why does this process doesn't work when the electron is "alone" ?

My own thinking would be that when the electron is bound there is certain energy in that bonded dipole system of the electron and ion and if the energy supplied isn't strong enough to break the bond it can only change the energy of that bond , energy which is then given back but in an unbounded electron there is no energy in the bond because there is no bond so the electron just responds to the incoming wave/particle much like a ship responds to an incoming wave in the middle of a sea (bad analogy but still).
Is my reasoning sound?

Also is it true that if the frequency of the photon is close or at the resonance frequency of the particle (electron) then the particle doesn't respond/move in phase with the wave but 180 out of phase?
In other words the the outgoing radiation matches in frequency with the incoming one but is at a different angle and phase shifted?

thanks.

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PeroK
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A free electron cannot absorb a photon because energy-momentum cannot be conserved in such a process.

• artis, vanhees71, sophiecentaur and 1 other person
vanhees71
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You can also have scattering of a photon on an atom of course. E.g., in a dielectric electromagnetic waves can go through with very little absorption scattering multiple times on all the bound charged particles making up the dielectric.

@PeroK and that energy momentum can't be conserved because why exactly?
Because an unbound electron has no energy levels to "climb" and deposit that energy into the electron - nucleus bond?
Can't an unbound electron just deposit that energy into it's kinetic energy?

One more thing I did not see in texts, so the incoming photons/radiation wave causes the electrons to accelerate and de-accelerate in time which is called bremsstrahltung and causes the electrons to emit their own radiation which is at a different angle than the incoming radiation but what happens to the incoming radiation?
As the name scattering implies it seems to me that the incoming radiation doesn't get absorbed it just interacts with the electrons and causes them to make secondary emission due to bremsstrahlung so does now the outgoing radiation is both electron emitted and primary together?

Also so far I have considered free electrons generalized as a free electron gas, but thomson scattering happens also with bound electrons like those of solids and gases so is the mechanism there the same as well?
If the incoming radiation frequency is high enough (x rays for example) then the photon energy is large enough so that it can momentarily excite an electron to the point where it loses it's bond to the ion and behaves as it were free? Or should I consider that the energy given to the electron is so high that it becomes so loosely bound to the ion that for all practical reasons it can be considered a free electron?
Because I can't see that we could say that Thomson scattering of solid matter in for example X-ray Crystallography ionizes the sample.

@vanhees71 but isn't the scattering cross section with ions negligible? Due to their much larger mass and lower response to radiation of same wavelength as compared to say electrons?

PeroK
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@PeroK and that energy momentum can't be conserved because why exactly?
Because an unbound electron has no energy levels to "climb" and deposit that energy into the electron - nucleus bond?
Can't an unbound electron just deposit that energy into it's kinetic energy?
Do you have the skills to analyse the collision of a photon and electron yourself? It's not to hard using conservation of energy-momentum to show that an electron absorbing a massless photon is impossible.

• vanhees71
PeroK
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If not, watch this:

• vanhees71
If not, watch this:

The energy of an electron = mc^2 ????? LOL

Q1 - why is it's velocity assumed to be zero?
Q2 - what's up with ½mv^2 ?

• weirdoguy and Motore
The energy of an electron = mc^2 ????? LOL

Q2 - what's up with ½mv^2 ?

It's non-relativistic.

It's non-relativistic.

So the electron has speed 0 and it's a relativistic problem? Isn't the relativistic part already taken into account for the calculations of momentum and energy for a photon since they always travel @ c? Isn't that the point?

If the KE of an electron was mc^2 then it's not an electron anymore - it's a bunch of photons isn't it?

So the electron has speed 0 and it's a relativistic problem?

It is because photons are involved and photons are ultra-relativistic. Treating half of the system non-relativisticly and other half relativisticly won't give you proper resoluts. And I don't even see why one would want to do that in that case.

If the KE of an electron was mc^2 then it's not an electron anymore

I don't know what you mean by that. ##mc^2## is rest energy of electron. Its kinetic energy is ##mc^2(\gamma-1)## (##\gamma## is Lorentz factor). Its total energy is ##\gamma mc^2##.

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• PeroK
But the electron is at rest - how is that a relativistic problem?

But anyway, what if the electron has velocity V? Why is he leaving out MV and ½MV^2 ?

But the electron is at rest - how is that a relativistic problem?

Because you can choose reference frame where it has relativistic velocity. We choose it's rest frame for simplicity. Also, it can have relativistic velocity after scattering so you have to treat it relativisticly from the beginnig.

But anyway, what if the electron has velocity V? Why is he leaving out MV and ½MV^2

Then you have to include its momentum and energy (total, not only kinetic), but you have to use proper relativistic formulas: ##\vec{p}=\gamma m \vec{v}## and ##E_{total}=\gamma mc^2##. These really are basics of relativity.

• PeroK
Then you have to include its momentum and energy (total, not only kinetic), but you have to use proper relativistic formulas: ##\vec{p}=\gamma m \vec{v}## and ##E_{total}=\gamma mc^2##. These really are basics of relativity.

And so what happens when you do - do you still get a bunch of terms = 0 , leading to a senseless outcome once again?

I admit, this leads to a couple of interesting scenarios. It was convenient for him to set the speed of the electron to 0, one instance of which he doesn't need to deal with vectors for momentum.

The easiest is when a photon would approach an electron directly from behind - pretty simple.

But what about head on? Momentum leads to a slowing of the electron but energy has been transferred to it, so there's clearly a big problem there.

Ibix
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How are you defining "head on" with a stationary particle?

And so what happens when you do - do you still get a bunch of terms = 0 , leading to a senseless outcome once again?

Yes. In that case (##p## is electrons initial momentum, ##p'## is its final momentum):

Energy conservation:
##\sqrt{m^2c^4+p^2c^2}+\frac{hc}{\lambda}=\sqrt{m^2c^4+p'^2c^2}##
Momentum conservation:
##p+\frac{h}{\lambda}=p'##

Squaring both we get:
##m^2c^4+p^2c^2+\frac{h^2c^2}{\lambda^2}+\frac{2hc}{\lambda}\sqrt{m^2c^4+p^2c^2}=m^2c^4+p'^2c^2##
##p^2+\frac{2ph}{\lambda}+\frac{h^2}{\lambda^2}=p'^2##

Combining both we have:
##m^2c^4+p^2c^2+\frac{h^2c^2}{\lambda^2}+\frac{2hc}{\lambda}\sqrt{m^2c^4+p^2c^2}=m^2c^4+p^2c^2+\frac{2phc^2}{\lambda}+\frac{c^2h^2}{\lambda^2}##

##\frac{2hc}{\lambda}\sqrt{m^2c^4+p^2c^2}=\frac{2phc^2}{\lambda}##

##\sqrt{m^2c^4+p^2c^2}=pc##
Squaring that we get that:

Sorry if it doesn't correlate with the proof in the video - I didn't watch it.

vanhees71