Deriving power from the relativistic energy equation

Click For Summary
SUMMARY

The discussion focuses on deriving the power equation, Power = dE/dt = F × v, from the relativistic energy equation E² = E₀² + (pc)². The user initially approached the derivation in a convoluted manner but sought clarity on a more straightforward method. Key insights included the relationship between force and momentum, where dp/dt = F and dx/dt = v, leading to the conclusion that dE/dt = P = Fv. Notation confusion regarding vector representation was also addressed, emphasizing the use of the dot product.

PREREQUISITES
  • Understanding of relativistic energy equations, specifically E² = E₀² + (pc)²
  • Familiarity with the concept of momentum p = γmv
  • Basic knowledge of calculus, particularly differentiation
  • Proficiency in vector notation and operations, including dot products
NEXT STEPS
  • Study the derivation of relativistic energy equations in detail
  • Learn about the relationship between force, momentum, and velocity in relativistic contexts
  • Explore vector calculus, focusing on dot products and their applications in physics
  • Practice using LaTeX for clearer mathematical notation in physics
USEFUL FOR

Students and educators in physics, particularly those studying relativistic mechanics, as well as anyone interested in mastering the derivation of power from energy equations.

Xamien
Messages
11
Reaction score
0

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.
 
Physics news on Phys.org
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!
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
546
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
14
Views
2K
  • · Replies 27 ·
Replies
27
Views
2K
Replies
15
Views
1K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
2K
  • · Replies 5 ·
Replies
5
Views
998