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Copyright © 2005 jsd

How to Make Antimatter – An Exercise Using Four-Vectors
John Denker

1  Accelerator Design Energy

Suppose we want to design an accelerator to produce antiprotons. There was a time, not so very long ago, when nobody had ever seen an antiproton, and some folks were highly motivated to make some. 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 there will be two protons 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:

p + p  p     ??
p + p  p + p ??
p + p  p + p + p ??
             (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 we started with):

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 18 as explained in section 2. We have chosen to measure things in units such that c=1. That is, our unit of time is consistent with our unit of length.

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 must equal −(4m)2. Don’t forget that the 4 gets squared.

Collecting results, we find

pi · pt = −(16−2) m2         
  pi · pt = −7 m2
             (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 19.

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.

2  Background

Section 1 has already achieved what we set out to achieve. In this section we briefly review some facts about 4-vectors, in order to make this document more nearly self-contained. Section 3 makes some additional remarks, to put the results in context, and to show how they relate to some other things you may already know.

Let x be the four-vector representing the position of a particle. In the frame of some observer Moe, it can be expanded in terms of components:

x =



x0     
x1     
x2     
x3     




 




@ Moe
             (12)

where x0 is the timelike component and the other three components are the spacelike components. We sometimes write t as a synonym for x0.

The four-velocity of the particle is

u := dx / dτ
             (13)

where τ is the proper time. The four-velocity u is not to be confused with the coordinate velocity v which is defined as the spatial part of dx/dt ... where the coordinate time t (not the proper time τ) appears in the denominator.

In a frame comoving with the particle, the components of the 4-velocity are:

u =



1       
0       
0       
0       




 




@ rest
             (14)

which expresses the fact that the particle just sits at a given location and progresses toward the future at a rate of 60 minutes per hour.

The four-momentum is defined quite simply as:

p := m u  
             (15)

The timelike component of the four-momentum is called the energy, and we sometimes write E as a synonym for p0. Accordingly, p is sometimes called the [energy,momentum] four-vector.

The gorm of any 4-vector r can be expanded in terms of components (in any given frame):

r·r = r0·r0 
    r1·r1       
    r2·r2          
    r3·r3         
             (16)

where the minus sign in this equation is almost the only thing that makes the geometry of spacetime different from the geometry of ordinary 3-dimensional Euclidean space.

It should be obvious that the gorm of the four-velocity is

u·u = −1  
             (17)

Since any dot product is a Lorentz scalar, it should be obvious that equation 17 is valid in any frame.

3  Ramifications and Correspondences

3.1  Rest Energy

As a consequence of equation 15 and equation 17 we can easily calculate the gorm of the 4-momentum:

p·p = m2
             (18)

where p is the [energy,momentum] 4-vector of some particle, and m is the mass of the particle. Equation 18 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 chosen reference frame, we can pick apart equation 18 into its timelike and spacelike parts:

m2 = E2 − p12 − p22 − p32                   
  = E2 − p1232                   
             (19)

where E is the energy in the chosen reference frame, and p123 (or equivalently pxyz) is the spacelike part of the momentum, i.e. the ordinary 3-vector momentum in that frame.

We can rewrite this in the form:

E2 = m2 + p1232                   
E = 
m2 + p1232
                    
             (20)

In equation 19, if we happen to be in a frame comoving with the particle, the momentum term is zero. We can then take the square root to obtain the celebrated formula

Erest = m               
Erest = m c2               
             (21)

In the second line we have chosen to write the factors of c explicitly. They drop out if/when our units of time are consistent with our units of length.

Equation 21 is often written in the shorthand form E=mc2, without any subscript on the E, but this is potentially very misleading. The LHS of this formula represents the rest energy of the particle. In any frame other than the rest frame of the particle, mc2 is not the total energy.

3.2  Kinetic Energy

Returning to equation 20, we no longer assume the momentum is zero in our frame, but instead let us merely assume the momentum is small. Then we can expand the square root to first order. That gives us

E = 
m 
1 + p1232 / m2
                        
  = m (1 + ½ p1232 / m2)                       
  = m + ½ p1232 / m                     
  = m + ½ m v2     (to lowest order)
             (22)

where in the last line we have used the fact that the momentum is equal to mu. The spatial part of mu can (for present purposes) be approximated by mv (since the particle velocity v – as measured in our frame – is small). Note that equation 22 is valid to first order in E, and to second order in v and p.

The first term on the RHS of equation 22 is the mass i.e. the rest energy, which is manifestly Lorentz invariant. In particular, it doesn’t change if we change the velocity v. So the part of E that depends on velocity can be written as:

Ekinetic = ½ m v2     (to lowest order)
             (23)

We recognize this as the classical expression for the kinetic energy. We intepret it as nothing more or less than a lowest-order approximation to the full relativistic result. This is an example of the correspondence principle: starting from special relativity and taking the classical limit, we obtain the classical result.

Exercise: Obtain the next term in the power-series expansion. Show that for ordinary non-relativistic velocities, the classical expression for the kinetic energy (equation 23) is a very good approximation. Calculate a bound on |v| (a reasonably tight upper bound) sufficient to ensure that the actual kinetic energy remains within 1% of the classical approximation as given by equation 23. Beware that v is not equal to the spacelike part of u to this degree of approximation.

4  References

1.
Nobel Laurates – 1959
http://nobelprize.org/nobel_prizes/physics/laureates/1959/index.html

2.
Wikipedia article, “Bevatron”
http://en.wikipedia.org/wiki/Bevatron

3.
John Denker,
“Introduction to Vectors”
www.av8n.com/physics/vector-intro.htm

4.
John Denker,
“The Geometry and Trigonometry of Spacetime”.
www.av8n.com/physics/spacetime-trig.htm

5.
John Denker,
“Odometers and Clocks in Introductory Relativity”.
www.av8n.com/physics/odometer.htm
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Copyright © 2005 jsd