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Deriving the displacement of an object

  1. Jul 18, 2014 #1
    Hi, I'm trying to derive the displacement of an object with proper acceleration (constant) α.
    t is our laboratory time.

    The goal:
    [tex]x(t) = {1 \over α} [ \sqrt{1+α^2t^2} -1 ] [/tex]

    How to approach this? :(

    Note for anyone who's wondering: no, this is not homework...
     
  2. jcsd
  3. Jul 18, 2014 #2

    WannabeNewton

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    Do you know how to obtain the parametric equations ##x(\tau)## and ##t(\tau)## for a uniformly accelerating object of magnitude ##\alpha##, where ##\tau## is the proper time of a comoving clock? If so then your result is immediate, you just have to use a bit of algebra.

    If not then your first step would be to calculate ##(t(\tau),x(\tau))##. Use ##a^{\mu}a_{\mu} = \alpha^2## and ##u^{\mu}u_{\mu} = -1## to get the differential equations for ##x(\tau)## and ##t(\tau)## and solve them.
     
  4. Jul 18, 2014 #3

    pervect

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    The easiest approach is the textbook approach using 4-velocities, or proper velocities.
    You'll find the following worked out in MTW's "Gravitation", in tensor notation. I've re-written it to use non-tensor notation. I've also chosen to use subscripts in places where MTW uses superscripts for readability.

    Let u be the 4-velocity (proper velocity) of the object, and let ##\tau## be proper time. We wish to solve for ##u(\tau)##, we will take care of the conversion to find x(t) later.

    Then by definition we can write the components of the 4-velocity ##u_0## and ##u_1## as:

    ##u_0 = dt/d\tau \quad u_1 = dx/d\tau##

    where ##x(\tau)## is the displacement in terms of proper time.

    The proper acceleration a is just the rate of change of the proper velocity with respect to proper time, i.e

    ##a_0 = \frac{d }{d\tau} u_0## and ##a_1 = \frac{d }{d\tau} u_1##

    The 4-magnitude of the proper velocity is a constant. With the sign conventions being used, it's -1. Thus ##{u_1}^2 - {u_0}^2 = -1##.

    The 4-magnitude of the proper velocity is a constant. With the sign convetions being used it's just ##\alpha## (no minus sign). Thus

    ##a_1^2 - a_0^2 = \alpha^2##

    Because the magnitude of the proper velocity is always constant, the proper acceleration is always perpendicular to the proper velocity

    ##u_1\, a_1 - u_0 \,a_0 = 0##

    If you wish, you can derive this mathematically and formally from the fact that ##u_1 = \sqrt{u_0^2 - 1}## using the chain rule.

    The simple way to proceed is to note that we have

    ##{u_1}^2 - {u_0}^2 = -1 \quad u_1\,a_1 - u_0\,a_0 = 0 \quad a_1^2- a_0^2= \alpha^2##

    where the last equation comes from the fact that the magnitude of the proper acceleration is constant and equal to ##alpha##.

    We can write down the solution ##a_0 = \alpha \, u_1 \quad a_1 = \alpha \, u_0## which we can easily verify by inspection.

    This is equivalent to the differential equation:

    ##\frac{d}{d\tau} u_0 = \alpha \, u_1 \quad \frac{d}{d\tau} u_1 = \alpha \, u_0## Combining this with the fact that ##u_1^2 - u_0^2 = -1## we immediately write down the solution

    ##u_0 = \cosh (\alpha \, \tau) \quad u_1 = \sinh (\alpha \, \tau)##

    We integrate to find:

    ##t = \frac{1}{\alpha} \sinh(\alpha \tau) \quad x = \frac{1}{\alpha} \cosh (\alpha \, \tau) ##
    This doesn't have x=0 at t=##\tau##=0, which is just a matter of an integration constant. But we'll address this issue later, we'll just note at this point we can add a constant value to x and we will still have a solution, the proper velocity won't be affected by a choice of origin for x.

    To solve your problem, we don't need to write down the proper acceleration, but it's a good check. We see that it's magnitude is constant as desired.

    ##a_0 = \alpha \, \sinh(\alpha \, \tau) \quad a_1 = \alpha \, \cosh (\alpha \, \tau)##

    To find x as a function of t (rather than x and t as a function of tau) we start with
    ##t = \frac{1}{\alpha} \sinh(\alpha \tau) \quad x = \frac{1}{\alpha} \cosh (\alpha \, \tau) ##

    We note that
    ##x^2 - t^2 = \frac{1}{\alpha^2}##

    Solving for x, we have

    ##x = \sqrt{t^2 + \frac{1}{\alpha^2}}##

    The solution won't be affected if we add a constant to x to make it equal to 0 when t=0 as we mentioned previously. Adding this offset, we have

    ##x = \sqrt{t^2 + \frac{1}{\alpha^2}} - \frac{1}{\alpha}##

    which is equal to (though formatted differently) than the expression you wished to find.
     
    Last edited: Jul 18, 2014
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