A Roller Coaster Ride through Relativity/Appendix G

From Wikibooks, open books for an open world
Jump to: navigation, search

The relation between Energy and Momentum[edit]

The total relativistic energy E and the relativistic momentum p of a body are given by the following expressions:

E = \gamma M_0 c^2 = {M_0 c^2 \over \sqrt{1-v^2/c^2}}
p = \gamma M_0 v = {M_0 v \over \sqrt{1-v^2/c^2}}

We wish to eliminate v from these equations. First square and multiply across:

E^2(1-v^2/c^2) = M_0^2 c^4
p^2(1-v^2/c^2) = M_0^2 v^2

Now for a diabolically cunning move, multiply the second equation by c2 and subtract!

(E^2 - p^2 c^2)(1-v^2/c^2) = m_0^2 c^4(1 - v^2/c^2)

from which we obtain:

E^2 - p^2c^2 = M_0^2 c^4

An alternative (and in my opinion better) way of writing this equation is:

E^2 - E_0^2 = p^2 c^2

where E0 is the rest-mass energy of the body.

It is instructive to compare this expression with the non-relativistic relation between energy and momentum which is calculated as follows

KE = \tfrac{1}{2} M v^2 ~~~\text{and}~~~p = Mv

so

KE = {p^2 \over 2M}

It is not easy to see, at first, how the relativistic expression will reduce (as it must) to the non-relativistic one when v is small, but it does. Watch!

Since

E^2 - E_0^2 = p^2 c^2

we can write

(E - E_0)(E + E_0) = p^2 c^2

Now (E - E0) is just the relativistic kinetic energy KEr which, at low speeds approximates to the ordinary kinetic energy KE.

At low speeds, the total relativistic energy E and the rest-mass energy E0 are virtually equal and equal to Mc2 so:

KE~.~2 M c^2 = p^2 c^2

from which it is easy to see that

KE = {p^2 \over 2M}

as expected.

Back to the introduction ...

Back to top ...