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Finding the proper time

  1. Dec 28, 2005 #1
    Finding the "proper time"

    I'm currently working through the book "A first course in general relativity" by Bernard F. Schutz, and I am kind of stuck on one of the problem on page 56 question 19.

    A body is said to be uniformly accelerated if its acceleration four vector [tex]\vec{a} [/tex] has constant spatial direction and magnitude, say [tex] \vec{a} \cdot \vec{a} = \alpha^2 \geq 0[/tex].

    (a) Suppose a body is uniformly accelerated with [tex]\alpha[/tex]. Find the speed of the body after time [tex]t[/tex], and the distance travelled.

    This one is okay, I manage to work it out.

    [tex] v(t) = \frac{\alpha t}{\sqrt{1+\alpha^2t^2}}[/tex]

    By integrating [tex]v(t)[/tex] with respect to [tex]t[/tex], I get

    [tex]x(t) = \frac{1}{\alpha}\ln(\frac{1}{\sqrt{1+\alpha^2t^2}})[/tex]

    (b) Find the proper time elapsed for the body in (a), as a function of [tex]t[/tex]

    This is the one I'm having problem with. I started with the expression of proper time in terms of the Lorentz invariant

    [tex]-d\tau^2 = -dt^2 + dx^2[/tex]

    I then computed [tex]dx/dt[/tex] and obtain an expression for [tex]dx^2[/tex].

    [tex]\frac{dx}{dt}=-e^{2\alpha x}\sqrt{e^{-2\alpha x} -1}[/tex]

    [tex]dx^2 = (e^{2\alpha x} - e^{4\alpha x})dt^2[/tex]

    substitute this into the Lorentz invariant, with a bit of algebra we get

    [tex]d\tau^2 = (1 - (\frac{\alpha t}{1 + \alpha^2t^2})^2)dt^2[/tex]

    [tex] \tau = \int\sqrt{1 - (\frac{\alpha t}{1 + \alpha^2t^2})^2}dt[/tex]
    I spent quite a lot of time try to evaluate the integral in vain. I tried "normal" substitution, trigonometric substitution. I even tried to do it by integrate by parts by writing as

    [tex]\int(\frac{d\tau}{dt})^{2}dt = \int{1 - (\frac{\alpha t}{1 + \alpha^2t^2})^2}dt[/tex]

    And evalute the LHS and RHS separately to no avail.

    Can anyone help me to evaluate this integral, or point out my mistakes?

    Appreciate it.

    P.S. Even if the physics is wrong, I am still interested in knowing how to evaluate this type of integral.
    Last edited: Dec 28, 2005
  2. jcsd
  3. Dec 28, 2005 #2


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    Homework Helper

    It seems to me like:

    [tex] \tau = \int\sqrt{1 - (\frac{\alpha t}{1 + \alpha^2t^2})^2}dt[/tex]

    should be:

    [tex] \tau = \int\sqrt{1 - (\frac{\alpha t}{\sqrt{1 + \alpha^2t^2}})^2}dt[/tex]

    Then the integral becomes straighforward.
    Last edited: Dec 28, 2005
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