Constantly accelerating rocket algebra problem

Click For Summary
SUMMARY

The discussion focuses on solving a problem related to the algebra of a constantly accelerating rocket, specifically proving the equation $$ \frac{dv}{dt} = \frac{dv'}{dt'} (1 - v^2)^{\frac{3}{2}} $$ using the relationship between the instantaneous rest frame (S') and the observed frame (S). The key steps involve differentiating the equation $$ dv = dv' (1 - v^2) $$ with respect to time and applying the chain rule. The participants emphasize the importance of recognizing that velocity (v) is a function of time (t') and the correct application of the chain rule in this context.

PREREQUISITES
  • Understanding of special relativity concepts, particularly time dilation.
  • Familiarity with calculus, specifically differentiation and the chain rule.
  • Knowledge of the relationship between proper time and coordinate time in relativistic contexts.
  • Ability to manipulate and differentiate infinitesimal differentials in physics equations.
NEXT STEPS
  • Study the derivation of time dilation in special relativity.
  • Learn advanced differentiation techniques in calculus, focusing on the chain rule.
  • Explore the implications of velocity as a function of time in relativistic physics.
  • Investigate the mathematical treatment of infinitesimal differentials in physics equations.
USEFUL FOR

Students of physics, particularly those studying special relativity, mathematicians interested in calculus applications in physics, and educators looking for examples of relativistic motion problems.

Fek
Messages
13
Reaction score
0

Homework Statement


  • Rocket is accelerating constantly. Let S' be instantaneous rest frame of rocket and S be frame in which rocket is observed moving at velocity v.

Homework Equations


Given: $$ dv = dv' (1 - v^2) $$

Must prove:
$$ \frac{dv}{dt} = \frac{dv'}{dt'} (1 - v^2)^{\frac{3}{2}} $$

The Attempt at a Solution



So differentiate given equation with respect to t and use chain rule to get in terms of t'

$$ \frac{dv}{dt} = \frac{dt'}{dt} * \frac{d}{dt'}[dv'(1 - v^2)] $$
We also know
$$ \frac{dt'}{dt} = (1 - v^2)^{\frac{1}{2}} $$
as t' is proper time.

We also have:
$$ \frac{d}{dt'} (dv' (1 - v^2) = \frac{dv'}{dt'} (1 - v^2) $$

We have the answer! Except this final step isn't right because v is a function of t' as well and chain rule must be used?
 
Physics news on Phys.org
If you have a functional equation, you can differentiate it. If you have an equation involving infinitesimal differentials, you can't differentiate it. Instead, you can divide by another infinitesimal differential.
 

Similar threads

Replies
10
Views
3K
Replies
30
Views
5K
  • · Replies 4 ·
Replies
4
Views
5K
Replies
1
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
3K
  • · Replies 13 ·
Replies
13
Views
5K
Replies
3
Views
1K
  • · Replies 27 ·
Replies
27
Views
2K
Replies
3
Views
2K