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Setting up a extremization problem

  1. Nov 1, 2014 #1

    grandpa2390

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    1. The problem statement, all variables and given/known data
    I am having trouble setting up problems that require me to extremize. I know how to use Euler's equation, but I don't understand how to get what it is that I want to plug in.

    what I need to show is that the the correct path between two points (x,t) and (x2,t2) maximizes the proper time. and that the correct path has a constant velocity v=dx/dt

    2. Relevant equations
    instead of distance, I am supposed to use spacetime interval dτ^2 = dt^2 − dx^2,
    τ is proper time

    3. The attempt at a solution
    I tried to treat this as if it were the same as the distance equation, and the constant velocity v=dx/dt was just a constraint. but I don't know. I don't think that works. I haven't a clue how to set this up.
     
  2. jcsd
  3. Nov 2, 2014 #2

    RUber

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    Have you solved ##d\tau## in terms of dx to remove t from the equations?
     
  4. Nov 2, 2014 #3

    grandpa2390

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    I don't know what you mean. I solved ##d\tau## = sqrt (dt^2 - dx^2) and then simplified it to ##d\tau## = sqrt( 1 - x'^2)dt is that what you are saying?
     
  5. Nov 2, 2014 #4

    grandpa2390

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    Let me try a different tack

    I am trying to understand how to determine this. If I can figure out what equation to put into dy and dg, then I can solve it.

    Maybe if you can explain why other equations were chosen.
    If I want to maximize the area, the equation is ∫dA = ∫y*dx why? I understand that the width varies. that I am integrating between x1 and x2. but isn't y dependent on x? why wouldn't it just be the integral of dx?
    or maybe we are looking at it as ∫dA = ∫y(x)dx ? I guess that would make sense?

    or for length. we said that ds = √((dx)^2 + (dy)^2) that makes sense to me because that is the equation of the hypotenuse. So I feel like I should do the same with dτ = sqrt (dt^2 - dx^2)
    I want to say that f( x' ; t ) = sqrt( 1 - x'^2 ) dt

    that makes sense, but what about the other part? show that the correct path has a constant velocity v=dx/dt . is that a constraint, or is that the result I am trying to prove. how do I handle this?

    do I say v = dx/dt and say g( x' , t ) = x' ?


    edit: I found an example problem that said for a fixed perimeter length and it used k= ∫dl = ∫sqrt (1+y'^2)
    so I guess maybe for a fixed velocity k= ∫dv = ∫ something... maybe dv/dt? dv/dt = x"

    edit: or maybe ∫dv = x'dt

    then maybe I can factor out the dx/dt and solve it that way?
     
    Last edited: Nov 2, 2014
  6. Nov 2, 2014 #5

    grandpa2390

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    tried that. Got x' = (t + C) / \sqrt{1 + (t + C)^2)}

    now to integrate

    edit: I got sqrt ( 1 + (t + C)^2)
    is that right? does that make sense?
     
    Last edited: Nov 2, 2014
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