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four-vectors (was: field transformations)



On 04/03/2003 05:41 PM, Joe Heafner wrote:

"Why _avoid_ doing Lorentz transformations?"

In general I like to avoid doing unnecessary work.

I thought Lorentz transformations *define* four vectors, at least their
transformation properties anyway.

I don't think the Lorentz transformations *define*
four-vectors.

By way of analogy, the ordinary D=3 normed Euclidean
space can be defined just fine without mentioning
the word "rotations". Given the machinery provided
by the axioms of the space (i.e. dot products and
such) you can construct rotations.

Similarly, the usual D=4 Minkowski space can be defined
just fine without mentioning Lorentz transformations.
Given the machinery provided by the axioms of the
space, you can construct boosts (and rotations).

http://www.monmouth.com/~jsd/physics/rapidity.htm

Moving away from axioms to interesting physics, the
canonical example to illustrate the four-vector
approach goes like this:

Q: Suppose we want to design an accelerator to
produce antiprotons. How much energy must the
accelerator supply? In particular, assume we will
accelerate a proton and smash it into a target
containing a high density of stationary protons
(e.g. liquid hydrogen).

A: Using conservation of baryon number, we know
that we can't just create an antiproton; we must
create a proton/antiproton pair. Specifically,
the simplest possible reaction will be
p + p ==> p + p + p + pbar

Accelerators are hard to build, so we don't want to
waste energy. The minimum energy will be achieved
if 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 p_i be the energy/momentum four-vector for the
incident particle. Similarly p_t for the target
particle. Similarly p_b for the bundle of products.

By conservation of momentum, we have
p_i + p_t = p_b

Squaring both sides we get
(p_i + p_t) dot (p_i + p_t) = p_b dot p_b

Expanding we get
p_i^2 + p_t^2 + 2 p_i dot p_t = p_b^2

We know many of the terms in this expression. For
starters, p_i^2 is just m^2, where m is the invariant
mass of the incident particle. This is obvious in the
frame comoving with that particle, and since the norm
of a four-vector is invariant, the value in one frame
is the value in all frames.

Similarly p_t^2 also equals m^2.

Similarly p_b^2 equals (4m)^2. Don't forget that the
4 gets squared.

Collecting results, we find
2 p_i dot p_t = (16-2) m^2
p_i dot p_t = 7 m^2

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:
p_t = [m, 0, 0, 0]

Combining the two previous equations and carrying
out the dot product, we see that the timelike
(energy) component of p_i must be 7m in the lab frame:
p_i = [7m, ?, ?, ?]
That tells us the total energy the incident particle
must have, in the lab frame. We could calculate the
spacelike (momentum) components, 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 mass, so the
accelerator only needs to supply 6m.

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.

(The Berkeley Bevatron was in fact designed to produce
antiprotons. The design energy was slightly less than
calculated above, because they were clever enouth 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
one that is moving toward the incident beam, its kinetic
energy contributes maybe 20% of the energy budget.)

http://www.nobel.se/physics/laureates/1959/index.html