Deriving power from the relativistic energy equation

[Xamien]

Homework Statement

I recently finished a test that asks you to derive
[itex]Power = \frac{dE}{dt} = F \times v[/itex]
from the energy equation:
[itex]E^2 = E_{0}^2 + (pc)^2[/itex]

Homework Equations

[itex]Power = \frac{dE}{dt} = F \times v[/itex]
[itex]E^2 = E_{0}^2 + (pc)^2[/itex]
[itex]p = \gamma m v[/itex]

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:
[itex]2E \frac{dE}{dt} = 0 + 2pc \frac{dp}{dt}[/itex]
[itex]2(\gamma mc^2) \frac{dE}{dt} = 0 + 2 \gamma mvc \frac{dp}{dt}[/itex]
[itex]\frac{dp}{dt} = F \stackrel{and\rightarrow}{} \frac{dx}{dt} = v [/itex]
therefore [itex]\frac{dE}{dt} = P = Fv[/itex]

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.

 

[vela]

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

 

[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!