The transformation under rotations is standard, sincepis a vector, whileEis rotation-invariant.
Under a boost in the 1-direction, by a velocityv, we have
E′ = γ(
E−vp^1)p′^1 = γ(
p^1 −vE/c^2)p′^2 = p^2
p′^3 = p^3 (21.36)whereγ= (1−v^2 /c^2 )−^1 /^2. If this result is to agree with the non-relativistic limit where|v| ≪c,
andγ∼1, we must havep′^1 =p^1 −mv, and this requires that asv→0, the energy of the particle
remain finite, and given by its rest mass,
E=mc^2 (21.37)which is Einstein’s famous relation.
A fully relativistic relation between energy and momentum is obtained by using the fact thatpμ
is a 4-vector, and hence the “square”pμpμmust be Lorentz invariant. Since its value is independent
of the frame where it is being evaluated, we may evaluate it inthe frame where the particle is at
rest, i.e. its momentum is zero, and its energy ismc^2. This gives,
pμpμ=−E^2 /c^2 +p^2 =−m^2 c^2 (21.38)or alternatively written as a formula for energy,
E^2 =m^2 c^4 +p^2 c^2 (21.39)Energy-omentum conservation is then simply the statement that the 4-vectorpμis conserved.
21.5 Particle collider versus fixed target experiments
Suppose we have two possible experimental set-ups for the collisions of particle os massmonto one
another;
- Fixed target: one incoming particle has energyE, the other is at rest.
- Collider: the particles have opposite momenta, and each has energyE/2.
The question is now in which experiment one gets the most “bang for the buck”. Translated into
scientific terms, what this amounts to is the largest total energy in the rest frame.
For the Collider, the center of mass energy available is simplyE, the sum of the energies of the
two particles.
To compute the center of mass energyEcmfor the fixed target experiment, we use the fact
that the total momentumpμhas the same square in the fixed target frame as in the center ofmass
frame where, by definition, its momentum vanishes. This reads as follows,
pμpμ=−(E/c+mc)^2 −p^2 c^2 =−(Ecm/c)^2 (21.40)