# PHY6200 W07

## Chapter 15: Special Relativity

In Taylor, read sections 15.1 - 15.13 for this week and 15.13 to 15.18 next week.

u = γ[(dx/dt), c(dt/dt)] = γ(v, c)
p = mu = (γmv, γmc).

### Energy, the Fourth Component of Momentum

What about the fourth component of momentum? In the non-relativistic limit it reduces to the constant mc. But if we expand this and look at higher order terms, we see that

γmc ≈ mc + ½(m/c)v² + ...

The second term looks like kinetic energy divided by c. And, since we'd like conservation of energy as well as momentum, it is a reasonable step to say that the fourth component of momentum is the relativisitic energy of an object, divided by c.

p = (p, E/c)

where p is the relativisitic version of three-momentum, γmv, and E is the relativistic energy. Notice that relativisitic energy is not zero for a particle at rest, rather it is E = mc². For a particle in motion, we say that its relativistic energy is composed of rest energy mc² plus kinetic energy, T = E - mc² = (γ - 1)mc².

There are a number of useful relations that can be found between momentum, energy, and mass. The square of p is an invariant:

p² = γ²m²v² - γ²m²c² = m²c²γ²(β² - 1) = -m²c².

It is also given by p² - E²/c². Therefore we have an important relation

E² = p²c² + m²c4.

Also, since E = γmc², we can show that |p| = γβmc, and β = |p|c/E. The relationship between E, p, and m can be expressed with a triangle.

### Massless Particles; the Photon

Massless particles don't make sense in classical mechanics; a massless particle has zero kinetic energy and zero momentum, and may just as well be ignored. At first glance, relativity also seems to have no room for massless particles since p = γβmc and E = γmc².

But on closer inspection, it is seen that they can exist, with finite momentum and energy.