Relativistic acceleration with integration problem

  • #1
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:

Answers and Replies

  • #2
tiny-tim
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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:
 
  • #3
vela
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Use the definition of acceleration is to express the LHS in terms of v and t.
 
  • #4
Gravitino22
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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?
 
  • #5
ideasrule
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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.
 
  • #6
vela
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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.
 
  • #7
Gravitino22
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Thanks vela,

I was trying that approach earlier but i can't solve that nasty integral, going to keep trying
 
  • #8
vela
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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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