Relativistic Reaction Kinematics

Consider a particle, with mass m₂ and kinetic energy T_k, incident upon a particle with mass m₁ at rest. The reaction of this two-particle system to create a new two-particle system with masses m₃ and m₄ is expressed with the notation

It is common to call m₂ and m₃ the projectile and ejectile; similarly, m₁ and m₄ are known as the target and recoil. The four-momentum squared is invariant:

So the Lorentz invariant quantity for the initial two-particle system is

.

Algebraic manipulation gives the simplified result,

.

Big accelerators, in pairs

Note that if identical particles were to collide with equal and opposite momenta in the lab frame, the portion of the invariant mass available for reactions (the second term in the previous equation) would go like 4 (mT_k +T_k²) instead of 2mT_k, which is why the biggest particle accelerators like Fermilab and CERN collide opposing beams of particles instead of a single beam on a stationary target. In other words, at energies small compared to the mass of the particles, you get twice the invariant mass for twice the accelerator. But as the energies become comparable to or greater than the mass, the factor of 4T_k² results in big gains. For protons accelerated to 1 GeV, this amounts to a total factor of four for two accelerators compared to one. But at an energy of 1 TeV for protons, this amounts to a factor of two thousand! That's right, it's not 4 for the price of 2, it's 2000 for the price of 2! It's like we can't afford not to build bigger accelerators! (In pairs, of course.)

However, I do nuclear physics experiments, where we play different games than the particle physicists. One is to make precise measurments at low energies that are relevant to astrophysics: for reactions in stars, novae, supernovae, and big bang cosmology. Another is to perform experiments with beams of radioactive nuclei, to study exotic nuclei to further our understanding of nuclear structure. So enough of this particle physics talk about big accelerators!

In the center-of-mass frame, the momenta of m₁ and m₂ are equal and opposite. The invariant s does not change:

Since s is just a constant, this equation determines p_cm.

p_cm is the momentum of both particles in the center-of-mass frame. To determine the rapidity of the center-of-mass frame, we consider that a boost to the target with the cm rapidity gives the taget the momentum p_cm.

A simple fact about the hyperbolic trig functions,

,

allows us to calculate the cm rapidity in a nice closed form:

.

Given the invariant mass and the rapidity in the cm frame, it is simple to calculate the energies of the reaction products as a function of angle. The cm momentum of the reaction products follows from the previous equations, but with m₁ and m₂ replaced by m₃ and m₄:

Given E_k and all the masses, p'_cm is just a number. Applying a boost with the cm rapidity gives the momenta and energies of the products as a function of the cm angle.

The only undetermined parameter is the cm angle. The game we usually play when doing nuclear physics experiments is to determine the probability for a given reaction to happen, either as a function of the cm energy, or as a function of the cm angle, or both.

All measurements are made with detectors in a laboratory environment, so it is important to convert readily between cm and lab quantities.

Inspecting the simplified figure on the right, all the signs for the boost to the ejectile are positive. The ejectile energy and momenta in the lab are

,

, and

.

There is a minus sign to consider for the boost to the recoil. The recoil energy and momenta in the lab are

,

, and

.

That's really all there is to it. Now, if you were deranged and masochistic, you might want to calculate one of the lab energies as a function of the lab angle. This is possible, but quite ugly. To do it, begin by rearranging the momenta so that the cm angle terms are isolated and square them both, so that addition of the squared sine and cosine terms eliminates the cm angle altogether. The remaining equations are quadratic. The form of the quadratic equation is identical for p₃ and p₄:

The quadratic equation gives each momentum as a function of its angle:

These solutions merit some discussion. The quantity under the radical in the second term of the numerator must be greater than or equal to zero, since the momentum must be real. Depending on the kinematics, this could imply a limit on the lab angle. The condition is,

When this condition applies, the positive sign gives the high-energy reaction products, the positive and negative sign solutions become equal at the maximum lab angle (the term under the radical becomes zero), and the negative sign solution gives the low-energy reaction products. This condition usually applies for one of the reaction products in nuclear physics experiments, because beam energies are generally small compared to nuclear masses on the order of many GeV. The lighter reaction products, on the other hand, often have no maximum lab angle.

If the condition does not apply, so that there is no maximum lab angle for the given product, then the positive sign is correct. To see this, imagine that one of the outgoing particles is a photon. Taking the limit as m/p'_cm goes to zero, the numerator becomes

.

Since the magnitude of the momentum can never be negative, in this case the positive sign is correct.

For one last bit of unnecessary suffering, let's calculate the differential cross section as a function of lab angle, assuming an isotropic cross section in the center of mass. To do this, note that

Now to find the derivative, write the cm angle in terms of the lab variables,

,

then be sure to write the result in a convenient form after taking the derivative:

Putting it all together, the result is

This calculation uses the ejectile variables. The same calculation can be made in terms of the recoil variables.

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