# PHY6200 W07

## Chapter 15: Special Relativity

### Reading:

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

u = γ[(d**x**/dt), c(dt/dt)] = γ(**v**, c)

p = mu = (γm**v**, 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, γm**v**, 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 = γ²m²**v**² - γ²m²c² = m²c²γ²(β² - 1) = -m²c².

It is also given by **p**² - E²/c².
Therefore we have an important relation

E² = **p**²c² + m²c^{4}.

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.

© 2007 Robert Harr