# Photon absorption

1. Apr 20, 2013

### subsonicman

I recently learned that a free electron can't absorb a photon and derived it by showing it would be impossible to conserve both momentum and energy if that were the case. It seems like the same argument would extend to other fundamental particles. Is it true that no free fundamental particle can absorb a photon?

Edit: Changed mass to momentum

Last edited: Apr 20, 2013
2. Apr 20, 2013

### vanhees71

The correct argument is that you cannot conserve energy and momentum for a single particle and photon together with the on-shell conditions. Mass by itself is not a conserved quantity in relativistic physics!

3. Apr 20, 2013

### subsonicman

Oops I'm sorry, I meant to say momentum, thanks for the catch.

4. Apr 20, 2013

### Bill_K

What happens instead, of course, is that the particle absorbs the photon and then reemits one of a different frequency.

5. Apr 21, 2013

### zincshow

Do you have the calculations handy or a link on the internet to show this?

6. Apr 25, 2013

### Parlyne

It's actually pretty simple. In addition to conservation of energy and momentum, each particle must satisfy $m^2 = E^2-|\vec{p}|^2$. (Note that I'm using natural units where c=1. If you want to use conventional units, multiply every m by c2 and every p by c.) For the photon, which has $m_\gamma = 0$, this reduces to $E_\gamma = |\vec{p}_\gamma|$.

For the final state of the electron (or other massive fundamental particle),
\begin{align} m^2 &= E_f^{\phantom{f}2}-|\vec{p}_f|^2\\ &= (E_i + E_\gamma)^2 - |\vec{p}_i + \vec{p}_\gamma|^2\\ &= E_i^{\phantom{i}2} - |\vec{p}_i|^2 + E_\gamma^{\phantom{\gamma}2}-|\vec{p}_\gamma|^2 + 2E_f E_i - 2|\vec{p}_i||\vec{p}_\gamma|\cos \theta\\ &= m^2 + 2E_\gamma (E_i - |\vec{p}_i|\cos \theta). \end{align}

This is clearly only true if $E_i = |\vec{p}_i|\cos \theta$. But, given the first equation above, this means that
\begin{align} \sqrt{m^2 + |\vec{p}_i|^2} &= |\vec{p}_i|\cos \theta\\ m^2 + |\vec{p}_i|^2 &= |\vec{p}_i|^2 \cos^2 \theta\\ |\vec{p}_i|^2 (1-\cos^2 \theta) &= -m^2\\ |\vec{p}_i| &= \sqrt{\frac{-m^2}{\sin^2 \theta}} \end{align}

But, this is clearly impossible, as it would mean that the initial state electron must have had imaginary momentum, which is certainly not physically realizable.

7. Apr 25, 2013

### Jano L.

This result presumes that the rest mass of the electron cannot change. Under this assumption, there is also this simple explanation:

Imagine the electron and photon are two little balls going towards each other in the frame where their total momentum is zero. Now if the electron absorbed the photon, the former would have to stop its motion and stand still, to conserve zero momentum. As a result, it would lose some kinetic energy. The energy of the photon would be lost too, so clearly energy could not be conserved.

Thus some assumption must be wrong. Usually, it is assumed that the rest mass cannot change, so the conclusion is that the electron cannot just swallow the photon.

However, think for a while that the electron could change its rest energy - there are muons and tauons, which are something like heavier versions of the electron. If the electron mass could increase, then electron could swallow the photon; both momentum and kinetic energy could be conserved in the absorption.

Does anybody know whether it is possible in theory/experiment to change the electron into muon or tauon by irradiating it with a light of proper frequency?

8. Apr 25, 2013

### zincshow

Great example, both the electron and the photon would have to come to a stop from opposite directions... But wouldn't this energy go to the bonding of the photon and electron? Does this mean light will travel through a beam of electrons as if they are not there?

9. Apr 25, 2013

### zincshow

Thank you for sharing this, much appreciated. If I understand this correctly, you are looking for (solving for) the initial momentum of the electron after the collision and finding it is impossible. It does though assume that there is no storage of energy due to the inelastic collision does it not?

10. Apr 25, 2013

### Jano L.

Yes, but that would lead to a change of electron's rest mass. It is usually thought impossible.

In the two ball picture, if they are no a head-on collision, no; they will bounce of each other and scatter in some other directions.

11. Apr 25, 2013

### Bill_K

Photons do not bounce. They can only be created or absorbed. And in between, travel in a straight line. You said earlier that an electron cannot just swallow a photon, but that is exactly what does happen. The result is an electron that's off the mass shell, briefly, which then emits a new photon of the same energy.

This is not possible.

12. Apr 25, 2013

### Jano L.

So it is possible, but only for a short time? How long is that? And what prevents the electron from keeping the energy longer?

13. Apr 25, 2013

### Parlyne

I'm going to have to disagree with Bill_K. This should, in fact, be possible; but, it should also be such a highly suppressed process that, in practice, we should never have significant enough statistics to have a reasonable chance of seeing it happen.

The problem is that changing flavors requires a W to be involved. But, since the only external particles are leptons and a photon, the W must appear in a loop; and, given that we're dealing with leptons, the loop will also need to contain neutrinos. But, this is actually where the possibility of the flavor change arises. Because neutrinos mix, it is possible to have an electron enter the W/neutrino loop and a muon or a tau exit (if the right kinematic conditions can be arranged). This process is just the time reverse of the long sought $\mu \rightarrow e\gamma$, which has, as yet, not been observed.

14. Apr 26, 2013

### zincshow

If the energy is somehow accounted for, it could be a long time. For example, if the electron had a capability to store energy on an internal basis (through internal vibration or some kind of periodic internal motion), then the photon would not reappear until the internal process broke down for some reason.

15. Apr 26, 2013

### zincshow

A photon can be absorbed by a free electron. The proof provided in post #6 that it cannot, is a Newtonian view that no energy is absorbed or lost due to the collision. In the words of Newton and Einstein:

A note from Newton's Principles definition #4 in 1687:

Impressed Force - This force conflicts in the action only; and remains no longer in the body when the action is over.

In Einstein's second paper on relativity in 1905, he explicitly concludes:

"Radiation carries inertia between emitting and absorbing bodies". It is important that not only does something receive a "kick" from the momentum of the energy, but the internal inertia (i.e., the inertial mass) of the body is actually increased. (from mathpages.com)

16. Apr 26, 2013

### Bill_K

Thanks, I stand corrected!

17. Apr 26, 2013

### Staff: Mentor

You are misreading the argument in #6 I believe. The mass/momentum/energy relationship implied by that Einstein quote is mathematically represented by the equation in the very first sentence of Parlyne's post, the one that's the start of the derivation.

18. Apr 26, 2013

### zincshow

Sorry, but I disagree. He breaks the energy into before and after the collision, the momentum into before and after the collision, but leaves the mass the same before and after the collision. Einstein's quote addresses exactly that. To quote post #2 "Mass by itself is not a conserved quantity in relativistic physics!"

19. Apr 26, 2013

### Bill_K

The rest mass of an electron IS the same before and after the collision. You're misinterpreting Einstein's quote, which is meant to apply to a macroscopic object that can have internal degrees of freedom, such as thermal energy. The rest mass of an elementary particle cannot be changed.

20. Apr 26, 2013

### Staff: Mentor

No, but mass and energy are together conserved. That Einstein quote refers to the way that adding energy to an object can be interpreted as the object gaining mass while the energy source loses energy, the amounts of each being related by the famous E=mc2.

In the century since Einstein wrote that, serious students of physics have learned that it's often easier to work with the (entirely equivalent) mass/energy relationship that Parlyne started with in #6:
$$E^2 = (m_{0}c)^2 + (pc)^2$$
where m0 is the mass of the particle when it as rest so has zero momentum.

(There are some subtleties if the object in question has internal degrees of freedom so can absorb energy without changing its momentum and kinetic energy, but none of that applies here because we're dealing with electrons and photons - nice simple point particles).