Spring 2009 Physics Qualifier. Problem 18

Spring 2009 Physics Qualifier. Problem 18

Spring 2009 Physics Qualifier. Problem 18

Detailed solution, for viewing on a computer.

18. A. Is the following reaction allowed, according to the standard model of particle physics (if not, cite a specific rule violated and why): p e+ + 

B. Assuming that this reaction does occur (whether forbidden or not) and that the proton is initially at rest, what is the energy of the photon? Use mp=938.8 MeV/c2 and me=0.511 MeV/c2.

C. What is the momentum of the photon?

D. What fraction of the speed of light is the speed of the positron? (Assume perfect accuracy of the input numbers).


  1. This reaction is not allowed because Baryon number is not conserved: there is one Baryon on the left, the proton (Baryon number=1), and no Baryons on the right. The Baryon number of a particle is 1/3 the number of quarks inside - 1/3 the number of anti-quarks.

B. This is a problem in relativistic dynamics. To start, whichquantities pertaining to motion are conserved in this reaction? (see next page)

Conserved quantities are energy and momentum.

Problems in relativistic dynamics often leave the student swimming in a sea of algebra. This sea is not deep, though it is wide. To keep your feet on the ground, it may be helpful to have a geometric picture of the interaction. Here is an attempt to provide such a picture.

A familiar conservation law

Consider a game where you add as many vectors as you like, head to tail, so long as the first vector starts at point A and the last one ends at point B:

There are lots of different ways to do this.

This game can be summarized by saying that there are two “conservation laws”:

where represent the arrows, and each arrow obeys the Pythagorean Theorem:

These conservation laws are an algebraic summary of the geometric principle that if the arrows are added head-to-tail they reach from A to B.

Conservation laws in relativity

In relativity you associate each particle with an arrow:

In a collision or reaction between particles, you still have the conservation laws:

however, if you want to represent the particles as arrows, for which I will now use capital letters in a weird font, like this: , then you have to make up a screwy version of the dot product (ie introduce a screwy way of calculating the magnitude, that is the length of the arrow):

Things that obey this screwy version of the dot product are called four-vectors.

In relativistic dynamics, you think of the particles as four-vectors. Thus if particle A crashes into particle B, and particles C and D are created, you can think of the collision as a picture like this:

You can also make statements like: or .

Then, when you want to calculate something, you take a dot product.

Back to Spring 2009, problem 18, part b

Our reaction is: p e+ + The proton is at rest and we want the photon energy.

Represent this by arrows:

So we know that . Since we want to calculate something, we take a dot product:

In that last line an extension of the rule given above for four-vector dot products was used, namely that .

Notice that some notational sneakiness is occurring, in that the same symbol (a dot) is used to indicate the dot product between three-dimensional vectors, which obeys the rule, and the dot product between four-vectors, which obeys the rule given a few lines ago. As far as I can tell you have to use weird fonts and/or overbars to indicate by context which dot product you mean.

Now we remember that since the proton is at rest (ie has zero momentum), conservation of momentum says the positron and photon come out in opposite directions and have equal and opposite momenta, ie.

We also remember that conservation of energy says that . Now we can eliminate all unwanted quantities:

Thus the photon has slightly less energy than the electron.

Using then numbers given,

Part c.

Part d. Using the numbers given,.

Now we access a different part of our knowledge of relativity to say that , where here , and  is the “fraction of the speed of light” that we are after.

Thus and (six nines).

Review of how to solve simple problems in relativistic dynamics:

  1. Draw a picture of what is happening in three-dimensional space.
  2. Draw another picture, that does not attempt to represent what is happening in three-dimensional space, but that instead shows four-vectors, and allows you to write down an equation involving four-vectors.
  3. Take a dot product of four-vectors. Tip: if there’s a particle you don’t know much about, and don’t want in your equations, put it on one side of the equals sign, and put all the other particles on the other side of the equals sign. That way, when you take the dot product, only the rest mass of the unfamiliar particle enters into the resulting equation.
  4. Use the first picture (the one that shows what is happening in three-dimensional space) to deal with the three-dimensional dot products in your equation.
  5. Use conservation of energy and momentum, if necessary, to get rid of unknowns.
  6. If necessary, use the non four-vector relations between energy, momentum, and rest mass (the ones that involve ) to finish up.

The next level of complexity in relativistic dynamics is problems in which you have to switch reference frames. What changes, what doesn’t change, when you switch reference frames?