Wednesday, December 06, 2017

Relativistic Invariants Rock (modified)

The threshold of reactions provides a great example of the use of relativistic invariants. Consider this problem, from Griffiths’s Introduction to Elementary Particles:

Particle A (Energy E) hits particle B (at rest), producing particles C1C2, …:

A + B → C1, C2,… Cn

Calculate the threshold energy (i.e., minimum E) for this reaction in terms of the various particle masses.

Now one could try to use conservation of four-momentum: pμi = pμf. In that case the four-momenta would have to be calculated in the same frame. The problem is that the lab frame is convenient for the left hand (initial) side of the reaction, while the center of momentum frame is convenient for the right hand side. That is because in the center of momentum frame, the threshold occurs when the particles C1, C2,… Cn are created with no kinetic energy. All their energy, in that case, is in their mass.

Here comes the utility of invariants. The four-momentum squared is also conserved, i.e., the same before and after, but has the added bonus that it is the same in all frames. Therefore we can calculate it in the lab frame for the left hand side, and in the center of momentum frame for the right hand side, and those two must be equal.

In the calculations below we take the speed of light, always denoted by the letter c, to be equal to 1,

c = 1.

This makes the algebra less cumbersome. At the end I'll explain how we undo that assumption so that we get the correct numerical result.

For the left hand side (where particle B is at rest)—and, again, taking the speed of light to be unity,

pμpμ = (E + MB)2 - PA2, where PA is the square of the three-momentum for the incoming particle A.

expanding that out, and using PA2 = E2 - MA2 gives

pμpμ = 2 E MB + MA2 + MB2     (1)

Now we need to calculate the same invariant for the right hand side, which is trivial in the threshold case in the center of momentum where all the final particles are at rest.:

pμpμ = M2     (2)

where M is just the total mass of the final particles, M = M1, M2,… Mn

Even though (1) and (2) were worked out in different frames, invariance means we can equate their answers, leading to the result:

E = (M2 - MA2 - MB2) / 2 M

Of course, this is the total energy of the particle A. The kinetic energy required for particle A, TA is 

TA = E - MA,  (3)

and A's momentum from PA2 = E2 - MA2.

Now, what about this business of letting c, the speed of light, be unity? We easily undo that at the end, for the purposes of calculating real numerical answers, because there is only one way to insert c back into any equation to make it dimensionally consistent. For example, take the Eq. 3 for the kinetic energy. The T and the E are energy, but the M is a mass. But M times c2, as we know, is energy. So the way to make Eq. 3 dimensionally consistent is:

TA = E - Mc2.

It's as easy as that.

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