# Deriving power from the relativistic energy equation

1. Sep 27, 2011

### Xamien

1. The problem statement, all variables and given/known data
I recently finished a test that asks you to derive
$Power = \frac{dE}{dt} = F \times v$
from the energy equation:
$E^2 = E_{0}^2 + (pc)^2$

2. Relevant equations
$Power = \frac{dE}{dt} = F \times v$
$E^2 = E_{0}^2 + (pc)^2$
$p = \gamma m v$

3. The attempt at a solution
I got there in kind of a messy way but I would like to know how I could have more cleanly shown how to put it together. Here's the way I got to it:
$2E \frac{dE}{dt} = 0 + 2pc \frac{dp}{dt}$
$2(\gamma mc^2) \frac{dE}{dt} = 0 + 2 \gamma mvc \frac{dp}{dt}$
$\frac{dp}{dt} = F \stackrel{and\rightarrow}{} \frac{dx}{dt} = v$
therefore $\frac{dE}{dt} = P = Fv$

Of course, I also realize I may have bungled this, so corrections or at least references to the rules would also be much appreciated. Please, weigh in.

2. Sep 27, 2011

### vela

Staff Emeritus
Your notation is a little confusing because both the force F and the velocity v are vectors, so using the symbol $\times$ looks like you're taking the cross product of the two. What you want to show is$$\frac{dE}{dt} = \vec{F}\cdot\vec{v}$$Use the fact that $\vec{p}^2 = \vec{p}\cdot\vec{p}$ and just basically do what you did, and you'll get that result.

3. Sep 27, 2011

### Xamien

Sorry about the notation confusion. Still getting used to using LateX.

The lightbulb came on with that comparison with the momentum squared. Thank you very much!