Suppose we want to design an accelerator to produce antiprotons. (If this hadn’t already been done, it would be well worth doing. See reference 1 and reference 2.) The question for today is, how much energy must the accelerator supply? For simplicity, assume we will accelerate a proton and smash it into a target containing a high density of stationary protons (e.g. liquid hydrogen).
There is an easy way to answer this question. This provides a wonderful illustration of the power of vectors in general, and four-vectors in particular. No math is required beyond high-school “Algebra I” plus the rule for taking dot-products of 4-vectors. (See reference 3 for details on what we mean by “vector”.) We will be using the spacetime viewpoint, i.e. the modern (post-1908) viewpoint, as described in reference 4 and reference 5. This viewpoint makes it possible – and indeed easy – to solve the problem without doing any Lorentz transformations.
In order to get started, we need to understand what sort of reaction we are going to use. We have already decided on a proton/proton collision, so that tells us what will be on the left-hand side of the reaction equation:
| p + p ⇒ something (1) |
There are all sorts of reactions that cannot possibly occur, because they would violate fundamental conservation laws such as conservation of charge, conservation of baryon number, or whatever. In particular, the following are ruled out:
| (2) |
where p stands for proton and p– stands for antiproton.
The simplest possible reaction will be one that creates a proton/antiproton pair (and keeps the two protons from the RHS):
| p + p ⇒ p + p + p + p– (3) |
Accelerators are hard to build, and we don’t want to make the accelerator much bigger than it has to be. Therefore, we don’t want to consider all possible versions of equation 3, but only the most energy-efficient versions. The minimum total energy will be achieved in the special case where the products of the reaction have the minimum kinetic energy. That means the products will not be moving relative to each other. A bundle containing the four product particles will come flying out the back side of the target.
Let pi be the energy/momentum four-vector for the incident particle. Similarly pt for the target particle, and pb for the bundle of products.
By conservation of momentum, we have
| pi + pt = pb (4) |
Squaring both sides we get
| (pi + pt) · (pi + pt) = pb · pb (5) |
Expanding we get
| pi2 + pt2 + 2 pi · pt = pb2 (6) |
We know many of the terms in this expression. For starters, we know that
| pi2 = −m2 (7) |
where m is the mass of the incident particle, in accordance with equation 12 as explained in section 2. The correctness of equation 7 is obvious in the frame comoving with the incident particle, and since the gorm of a four-vector is invariant, the value in one frame is the value in all frames.
Similarly pt2 also equals −m2.
Similarly pb2 also equals −(4m)2. Don’t forget that the 4 gets squared.
Collecting results, we find
| (8) |
All the equations to this point have been true in all frames. We now specialize to the lab frame. In the lab frame, the target is stationary, so its four-vector has very simple components:
| pt = [m, 0, 0, 0] @ Lab (9) |
Combining the two previous equations and carrying out the dot product, we see that the timelike (energy) component of pi must be 7m in the lab frame; that is:
| pi = [7m, ?, ?, ?] @ Lab (10) |
That tells us that in the lab frame, the incident particle must have a total energy of 7m. We could calculate the momentum, i.e. the spacelike components of equation 10, but we don’t need to.
Note that the question asks how much energy must be supplied by the accelerator. The incident particle was born with 1m of energy, i.e. its rest energy, i.e. its mass, so the accelerator only needs to supply 6m, in accordance with equation 13.
| KErequired = 6m (11) |
In engineering units, the mass of a proton is about a GeV (.938 GeV) so we must design the accelerator to produce about 6GeV.
Presumably it is possible to solve this problem using Lorentz transformations, without four-vectors, but it would be a lot more work.
Note: The Berkeley Bevatron was in fact designed to produce antiprotons. The design energy was very nearly equal to what we calculated in equation 11. Actually it was slightly less, because the designers were clever enough to not use a hydrogen target. They used copper. Protons in a non-hydrogenic nucleus are not stationary. Exclusion principle, orbitals, blah-de-blah. If you manage to hit a nucleon that is moving toward the incident beam, its kinetic energy contributes maybe 20% of the reaction energy.
This calculation depends on the famous formula:
| (12) |
where p is the [energy,momentum] 4-vector of some particle, and m is the mass of the particle. Equation 12 is true and directly applicable in each and every reference frame. The mass is invariant, i.e. the mass is a Lorentz scalar, i.e. the mass is the same in all reference frames.
In any particular chosen reference frame, we can pick apart equation 12 into its timelike and spacelike parts:
| (13) |
where E is the energy in the chosen reference frame, and pS is the spacelike momentum, i.e. the ordinary 3-vector momentum in that frame.
It must be emphasized that the E in celebrated formula E=mc2 represents the rest energy (not the total energy) of the particle. To say the same thing another way, if you want E to be the total energy, then the formula E=mc2 is valid only in the rest frame of the particle, as you can see by comparing it to the more general formula in equation 13.