Relativistic acceleration with integration problem

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SUMMARY

The discussion focuses on solving a relativistic acceleration problem involving a spaceship that accelerates at a rate equivalent to Earth's gravity (g) for a time period (T). The final velocity (V) is derived using the equation V=c[1+(c/gT)^2]^(-1/2). Participants discuss the integration of the Lorentz factor (gamma) and the relationship between acceleration (a), velocity (v), and time (t), emphasizing the need to use the chain rule for proper integration. Key insights include the necessity of separating variables and utilizing trigonometric substitution to solve the integral.

PREREQUISITES
  • Understanding of relativistic physics concepts, specifically Lorentz transformations.
  • Familiarity with calculus, particularly integration techniques and differential equations.
  • Knowledge of acceleration and its relationship to force and mass in physics.
  • Experience with trigonometric substitutions in integral calculus.
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  • Study the derivation and application of the Lorentz factor in relativistic mechanics.
  • Learn about solving separable differential equations in physics.
  • Explore techniques for integrating functions involving trigonometric substitutions.
  • Review the principles of acceleration in the context of special relativity.
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Gravitino22
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Homework Statement



Consider a spaceship that accelerates so that the passengers feel and acceleration equal to that of gravity of the earth’s surface, g. If the spaceship undergoes this acceleration for a time T, show that the final velocity is given by:
V=c[1+(c/gT)^2]^(-1/2)


Homework Equations



F=(gamma^3)ma where gamma= [1-(v/c)^2]^-1/2 (the Lorentz factor)

The Attempt at a Solution


Since the passengers always feel the acceleration of gravity, you don’t actually feel acceleration you feel the force mg. So at any time the passengers must feel mg so:

mg=(gamma^3)ma
a=g/(gamma^3)

I converted gamma into the function and tried to integrate to obtain V but that’s where I got stuck .

a=g[1-(v/c)^2]^3/2

I trying to integrate with respect to dt but v it self is a dx/dt so I am wierded out by that. I went to the professor and he gave me a hint in which I have to convert adt into another thing using the chain rule (whatever that means).
Somehow I think I have to integrate with respect to dv and have my limit be gT instead of T but iam not sure how to get to there.

Any help is greatly appreciated I have my test this week and I need to get a A :(.
 
Last edited:
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Hi Gravitino22! :smile:

(try using the X2 tag just above the Reply box :wink:)

Your professor is probably thinking of something like
∫ f(v) dt = ∫ [f(v) / (dv/dt)] dv/dt dt = … :wink:
 
Use the definition of acceleration is to express the LHS in terms of v and t.
 
vela said:
Use the definition of acceleration is to express the LHS in terms of v and t.

Thats what i was trying

a=dv/dt then pass dt to the RHS to integrate and which also has a v=dx/dt which is where iam confused.

And for tiny-tim's response

Iam not seeing where your going with that...

=f(v)dtdv? double integration?
 
a=dv/dt, so you can write dv/dt=g[1-(v/c)^2]^3/2. You can integrate that to get a relationship between v and t. It's true that v=dx/dt, but since you're not asked to find x as a function of t, this equation is irrelevant.
 
The differential equation is separable, so you can get all the v's on one side and t on the other and then integrate each side.
 
Thanks vela,

I was trying that approach earlier but i can't solve that nasty integral, going to keep trying
 
Use the trig substitution v/c=sin θ. Then don't make the really stupid mistake I did, and the answer (for the integral) will pop right out.
 

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