Deriving power from the relativistic energy equation

AI Thread Summary
The discussion centers on deriving the power equation from the relativistic energy equation, specifically how to express power as the derivative of energy with respect to time. The user initially presented a messy derivation and sought clarification on a cleaner approach. Feedback highlighted the confusion in notation, particularly regarding the use of vectors for force and velocity, suggesting the dot product instead of a cross product. A comparison with momentum squared helped clarify the derivation process. Overall, the exchange emphasizes the importance of proper notation and understanding vector relationships in physics.
Xamien
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Homework Statement


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

Homework Equations


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

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.
 
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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.
 
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!
 
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