Can a free particle absorb/emit photons?

Question

As simple as in the title.. I would like to know also some mathematics about it!

Answer 1

It cannot. This is because energy and momentum are not both conserved if a free charged particle (say, an electron) emits a photon. It needs interaction with at least a second charged particle in order to do so (as in Bremsstrahlung). The mathematic involved is that of the energy of a photon $E=\hbar \omega$, energy of a particle $E^2 = m^2 c^4 + p^2 c^2$, momentum of a photon $p = \hbar \omega /c$ and simple trigonometry and basic algebra, very much as in the classical version of Compton scattering.

Answer 2

Rather than thinking in terms of conservation mathematically we can just think of the energy.

Consider the emission case (if we can prove this is impossible, then the absorption case is also impossible if we reverse time).

Let the free particle be at rest - if it is not we can transform to a frame where it is. Then if it emits a photon, the photon's momentum must be opposite the final momentum of the particle to conserve the zero momentum initial state.

But now the electron is moving, so has more energy than its rest energy ($E^2 = p^2 + m^2$ in natural units). The photon also has energy, not that this matters. Key point: the final energy is greater than the initial energy $\implies$ forbidden.

Answer 3

Let's assume it could then $\frac{1}{2} mv^2=\frac{hc}{\lambda}$ also $P=mv=\frac{h}{\lambda}$ by replacing $mv$ in first equation with$\frac{h}{\lambda}$ we get $\frac{1}{2} \frac{h}{\lambda}v=\frac{hc}{\lambda}$ now $v=2c$ which is of course impossible.

Answer 4

I see the consensus above is NO, and also same in this thread Free electron can't absorb a photon

Is this an accepted 'law' of physics?? Because I need to respectfully offer a different solution:

If you are mixing particles and photons you need to include relativism. From the example in the link above (Closed discussion) it is shown that there are no valid solutions to conservation of momentum and energy, thus electron cannot absorb a photon.

There is a solution.

To simplify, we observe the electron initially stationary from the same rest frame, thereby setting $p_1=0$ in the equation:

$p_1+p_ν=p_2,(1)$

where p1=0 and p2 is the momentum of the electron after the interaction and pv is the photon momentum, will then be:

$p_ν=p_2,(2)$

which (2) is incorrect because you are now observing the electron no longer from the same frame, so it must be written as:

$p_ν=γp_2,(3)$

where γ is the Lorentz factor, but (3) is not really helpful yet.

Conservation of energy gives us (still starting with p1=0)

$m_ec^2+p_νc=γm_ec^2,(4)$

, where the right-hand side of (4) is the total energy of the particle. Then subtracting $mc^2$ both sides

$p_νc=m_ec^2(γ-1),(5)$

and the right hand side of (5) is exactly the relativistic kinetic energy of the particle. So not only can the particle absorb a photon, the absorbed photon adds to kinetic energy and will thus remain absorbed until the kinetic energy changes!

The above can be extended to any frame of reference and still holds true.

Answer 5

Assume the contrary, namely consider a free electron at rest fully absorbs a photon that strikes it.

Then by momentum and energy conservation it follows that :

$$\begin{align} p + 0 = \frac{m_0 v}{\gamma} \tag1\\[5pt] pc + m_0c^2 = \frac{m_0c^2}{\gamma} \tag2 \end{align}$$

Solving the above equations yields $v = c$

Only particles with zero rest mass may travel at the speed of light. Electrons, however have rest mass.

Thence a contradiction is reached.