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Just a small question regarding collisions. Imagine a head-on collision between a photon and a particle with mass that moves with a non-relativistic speed, the particle was on its ground state, completely absorbs the photon, and moves to its next energy level. Is it always the case that the particle ends up with a non-relativistic speed after the collision?

Something more specific:

To study the properties of isolated atoms with a high degree of precision they must be kept almost at rest for a length of time. A method has recently been developed to do this. It is called “laser cooling” and is illustrated by the problem below. In a vacuum chamber a well collimated beam of Na23 atoms (coming from the evaporation of a sample at 103 K) is illuminated head-on with a high intensity laser beam (fig. 3.1). The frequency of laser is chosen so there will be resonant absorption of a photon by those atoms whose velocity is v0. When the light is absorbed, these atoms are exited to the first energy level, which has a mean value E above the ground state and uncertainty of (gamma). Find the laser frequency needed ensure the resonant absorption of the light by those atoms whose kinetic energy of the atoms inside the region behind the collimator. Also find the reduction in the velocity of these atoms, ∆v1, after the absorption process. Data E = 3,36⋅10-19 J Γ = 7,0⋅10-27 J c = 3⋅108 ms-1 mp = 1,67⋅10-27 kg h = 6,62⋅10-34 Js k = 1,38⋅10-23 JK-1

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    $\begingroup$ "Is it always the case that the particle ends up with a non-relativistic speed after the collision?" There isn't a sharp cutoff between relativistic and non-relativistic speed. $\endgroup$ – Brandon Enright Jan 4 '14 at 21:06
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As a practical---rather than in principle matter---the lightest particle with internal structure would be a positronium atom with a mass of around $1 \,\mathrm{MeV}$ and a maximum excitation energy of about $6.8 \,\mathrm{eV}$. That thing simply can't absorb enough momentum from a photon to make it relativistic.

The calculation is similar for any system bound by the electromagnetic interaction.

So the next thing to do is to consider a light system bound by the strong interaction. The lightest choices are the charged pions with masses of $139 \,\mathrm{MeV}$. The excited states of these things are rhos with masses of $770 \,\mathrm{MeV}$. That excitation requires $(770 - 139 = 631) \,\mathrm{MeV}$ and because it is provided by a photon there is a transfer of $631 \,\mathrm{MeV}$(1) of momentum as well, which is an appreciable fraction of the mass of the resulting state and implies that the resulting meson will by mildly relativistic in the rest frame of the incident meson. A similar calculation can be performed for the neutral pion excited to the neutral rho or the omega-0.

Note that $631 \,\mathrm{MeV}$ photons can be hard to come by.


(1) Strictly $\mathrm{MeV/c}$, but I'm working in natural units.

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Is it always the case that the particle ends up with a non-relativistic speed after the collision?

No - it depends on the total energy of the particle after the interaction, which (given that the particle is initially non-relativistic) depends on the mass of the particle and the energy of the photon. Compare the energy of the photon to the mass of the particle (in equivalent energy units) and this should help you decide how to proceed with your calculation.

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To expand on the comment suggesting there was no firm dividing line, ... one should remember that the momentum of a photon was first articulated by Einstein in the 1905 paper that defined special relativity. The goal of the experimental set up is to slow the atoms, so the energy of the photon should be slightly less that the energy of the transition to the next electronic level. The atoms traveling toward the beam see a blue-shifted stream of photons whose energy is just what is needed for excitation. The absorption transfers momentum to the atom and then radiation from the excited state occurs in a random direction, so there is a net reduction in forward momentum. Without special relativity the momentum interaction would not be well-described.

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