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: Hyperbolic Motion - SR

  1. Mar 11, 2014 #1
    1. The problem statement, all variables and given/known data
    Consider a particle in one-dimensional so called hyperbolic motion
    x(t)=[itex]\sqrt{b^{2}+t^{2}}[/itex]
    where b is a constant.

    a) Find[itex]\gamma[/itex](t).
    b) Find the proper time [itex]\tau[/itex](t). (assume that [itex]\tau[/itex]=0 when t = 0
    c) Find x and v[itex]_x[/itex] as functions of the propertime [itex]\tau[/itex].
    d) FInd the 4-velocity u[itex]^{\mu}[/itex].

    3. The attempt at a solution

    A) ok to begin I took the derivative of x(t) to get velocity. tuned out to be t(b[itex]^{2}[/itex]+t[itex]^{2}[/itex])[itex]^{-1/2}[/itex].
    soo therefor [itex]\gamma[/itex](t) = [itex]\frac{[itex]\sqrt{b^{2}+t^{2}}[/itex]}{[itex]\sqrt{1-\frac{t^{2}}{\sqrt{b^{2}+t^{2}}}}[/itex]}[/itex]

    b) so now [itex]\tau[/itex](0) = [itex]\sqrt{t^{2}-(b^{2}+t^{2}}[/itex]
    [itex]\tau[/itex](0) = [itex]\sqrt{0^{2}-(b^{2}+t^{0}}[/itex] = 0
    [itex]\tau[/itex](0) = [itex]\sqrt{-b^{2}}[/itex] = 0
    so would b = 0?
    this is where i'm getting lost.
    c) x as a function of \tau would be [itex]\sqrt{t^{2}-\tau^{2}}[/itex]=x?
    where does v[itex]_x[/itex] come in? would i solve v(t) for t^2?

    d) I know the 4 vector for u[itex]^{\mu}[/itex] is (u^0,u^1,u^2,u^3) and the roattional lorrentz for hyperbolic is
    |t'| = |cosh[itex]\varphi[/itex] -sinh[itex]\varphi[/itex] |
    |x'| |-sinh[itex]\varphi[/itex] cosh[itex]\varphi[/itex] |

    where tanh[itex]\varphi[/itex]=v
    where cosh[itex]\varphi[/itex]= [itex]\gamma[/itex]

    where do i go from here? Thanks for all the help.
     
    Last edited: Mar 11, 2014
  2. jcsd
  3. Mar 13, 2014 #2

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Try using ##d\tau^2 = dt^2 - dx^2##.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Loading...