In Taylor, read sections 15.1 - 15.13 for this week and 15.13 to 15.18 next week.
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
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.
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:
It is also given by p² - E²/c². Therefore we have an important relation
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 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