Re: Energy:momentum ratio

```To Steve:

Words as "efficiency" and "convert" are only used to not make sentences more
elaborate than they could be, I had not thought the word "efficiency" would
create that much trouble, since one only needs an unit-conversion to make
things right.
It is this same origin that makes the word "convert" not 100% at its place.
Even now, I can't find a single word would be satisfying, the only solution
would be to use sentences like your:

When you take something that has momentum, and you get energy out of it
by slowing it down, you are getting the extra energy it has to
compensate for its momentum (m^2 = E^2 - p^2).  If m is constant, then
when p increases E also increases.

Even then, the words "get energy out of momentum" sounds much like
"convert", "transfer" or "transform" which don't sound very well either.

The derivation is as follows:

The momentum of a particle with velocity v : p=gamma m v
The energy of a particle with velocity v   : E=m c^2 (gamma-1)

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

Simplifying the above gives:

p/E = 1/c {v/(c-Sqrt[c^2-v^2])}

All units are according to S.I.

(When mentioning energy, I only mean kinetic energy)

Timothy

```