1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Setting up a extremization problem

  1. Nov 1, 2014 #1
    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


    User Avatar
    Homework Helper

    Have you solved ##d\tau## in terms of dx to remove t from the equations?
  4. Nov 2, 2014 #3
    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
    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
    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted