Astrophysics - Special Relativity

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Homework Help Overview

The problem involves analyzing the motion of a relativistic rocket with a proper acceleration that varies with proper time. The original poster seeks to find the rocket's motion from the perspective of a control tower, starting from rest at a specific position.

Discussion Character

  • Exploratory, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply relevant equations involving rapidity and hyperbolic functions but encounters difficulty in solving for velocity. Some participants suggest using the hyperbolic tangent function in a different form, while others express similar challenges in reaching a solution.

Discussion Status

The discussion is ongoing, with participants sharing different approaches to the problem. There is a recognition of the complexity involved in solving for the variables, and some guidance has been offered regarding the use of hyperbolic functions.

Contextual Notes

The problem includes a hint that relates the proper acceleration to a logarithmic function, which may influence the approach taken. Participants are also noting the limitations of their current methods in deriving a solution.

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Homework Statement



2) If a relativistic rocket has a proper acceleration alpha that
increases with proper time tau according to:
alpha(tau) = 2/[Cosine(tau)^2 - Sine(tau)^2]
find its motion, r(t), from the point of view of a control tower
for whom the rocket is motionless at r(0) = 0.
(Hint: alpha(tau) here is the derivative with respect to tau of
ln[tan(tau + pi/4)] .)



Homework Equations



1. R=Rapidity
2. tanh(R)=β
3. d/dτ(R)=α


The Attempt at a Solution



Using formula #3 and the hint, I have R. Using formula #2 and my TI-89, I got:

(1-β)/2 = cos[t*(sqrt(1-β^2)+pi/4]^2

Using a couple of trig formulas, I have

β-1 = sin(2*t*sqrt(1-β^2))

I'm stuck there. As far as I know, there is no way to solve for β, and thus for the velocity 'v', which means I can't integrate to find r(t).
 
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Try using

[tex]\tanh x = \frac{e^x-e^{-x}}{e^x+e^{-x}}[/tex].
 
vela said:
Try using

[tex]\tanh x = \frac{e^x-e^{-x}}{e^x+e^{-x}}[/tex].

I wind up at the same spot.
 
Do it by hand, and show your work here.
 

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