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MTW Gravitation problems 6.5, 6.8

  1. Oct 31, 2009 #1
    Hello! I'm stuck on these two problems, any help is appreciated!

    6.5
    1. The problem statement, all variables and given/known data

    Use radar as a distance-measuring device. The radar set measures its proper time [tex]\tau[/tex] between the instant at which it emits a pulse and the later instant when it receives the reflected pulse. It then performs the simple computation [tex]L_{0}=\tau/2[/tex] and supplies as output the "distance" [tex]L_{0}[/tex]. How accurate is the output reading of the radar set for measuring the actual distance L to the object, when used by a uniformly accelerated observer? (L is defined as the distance in the momentary rest frame of the observer at the instant the pulse is reflected, which is at the observer's proper time halfway between emitting and receiving the pulse.) Give a correct forumula relating [tex]L_{0}[/tex] [tex]\overline{=}[/tex] [tex]\tau/2[/tex] to the actual distance L. Show that the reading [tex]L_{0}[/tex] becomes infinite as L approaches [tex]g^{-1}[/tex], where g is the observer's acceleration, as measured by an accelerometer he carries.

    2. Relevant equations

    [tex]gt=sinh(g \tau)[/tex]
    [tex]gx=cosh(g \tau)[/tex]

    3. The attempt at a solution
    We need to find [tex]\tau[/tex] in terms of L.
    Let [tex]x_{1L}[/tex] be the distance the light has traveled at the moment of reflection, and [tex]x_{1R}[/tex] be the distance the spaceship has traveled at the moment of reflection. Then [tex]L=x_{1R} - x_{1L}[/tex]. Setting [tex]\tau_{L}[/tex] to be the proper time at which reflection occurs, and plugging in the "relevant equations", we get
    [tex]\tau_{L} = - g^{-1} arcsinh (g x_{1L} )[/tex]
    so
    [tex]x_{1R} = g^{-1} cosh (arcsinh(g x_{1L}) ) = g^{-1} \sqrt{g^{2} x_{1L}^{2} + 1 }[/tex]
    so [tex]L=x_{1R} - x_{1L}[/tex] solves to
    [tex]x_{1L}= (1-L^{2}g^{2}) / (2g^{2}L)[/tex]

    And then, at reabsorption of the radar beam, the light and the rocket are once again at the same coordinates, so
    [tex]ct - 2x_{1L} = g^{-1}sinh(g \tau) - (1-L^{2}g^{2}) / (g^{2}L) = g^{-1} cosh(g \tau)[/tex]
    From this follows that
    [tex]sinh(g \tau) - cosh(g \tau) = - (1-L^{2}g^{2}) / (g L)[/tex]
    =>
    [tex]e^{-g\tau} = (1-L^{2}g^{2}) / (g L)[/tex]
    i.e.
    [tex]L_{0} = \tau / 2 = g^{-1} ln(\sqrt{ gL / (1-L^{2}g^{2})})[/tex]

    Note that this satisfies the condition that as L->1/g, [tex]L_{0}[/tex] -> infinity.

    However, in the other limiting condition, as L->0, this formula also gives that [tex]L_{0}[/tex] -> -infinity.
    Clearly this is wrong...

    Intuitively, I feel that the answer should take the form
    [tex]L_{0}=L/\sqrt{1-g^{2}L^{2}}[/tex], because this is basically length contraction. I don't know where I went wrong though, and I can't find a mistake in my logic or my math.

    Thanks for any suggestions...

    I'll post my question about 6.8 in the next post (same thread)
     
    Last edited: Oct 31, 2009
  2. jcsd
  3. Oct 31, 2009 #2
    And now, problem 6.8, which is a beast that I am having a lot more trouble wrapping my head around.

    the problem:
    Fremi-Walker plus spatial rotation. Observer transports his orthonormal tetrad in the following fashion:

    [tex]\Omega^{\mu\nu} = a^{\mu}u^{\nu} - a^{\nu}u^{\nu} + u_{\alpha}\omega_{\beta}\epsilon^{\alpha\beta\mu\nu}[/tex]

    where [tex]\epsilon[/tex] is Levi-Civita and [tex]\omega[/tex] is a vector orthogonal to 4-velocity u.

    a) let [tex]e_{0'}=u[/tex]. Show this is permitted by his transport law and that [tex]de_{\alpha'}/d\tau = - \Omega \cdot e_{\alpha'}[/tex]. (easy)

    b) show that the rotational part of the transport law produces a rotation in the plane perpendicular to u and [tex]\omega[/tex] (easy)

    here's where I start having trouble.

    c) suppose the accelerated observer Fermi-Walker transports a second orthonormal tetrad [tex]e_{\alpha''}[/tex]. Show that the space vectors of this first tetrad rotate relative to those of his second tetrad with angular velocity vector equal to [tex]\omega[/tex]. i.e., show that at the moment when the tetrads coincide, show that (in three-dimensional notation, referring to the 3-space orthogonal to the observer's world line):
    [tex]d(e_{j'} - e_{j''})/d\tau = \omega \times e_{j'}[/tex]

    attempt (?) at a solution

    I'm mostly having trouble wrapping my head around what this means. In class, we showed that [tex]d(e_{j})/d\tau = \omega \times e_{j}[/tex]. Does this mean that I simply have to show that [tex]de_{j''}/d\tau = 0[/tex]? If the second tetrad is not rotating, then wouldn't the first tetrad obviously rotate relative to it with angular velocity [tex]\omega[/tex] in the local lorentz frame? Do I need to actually transform coordinates?

    I think I can figure out parts d, e, and f on my own, but I have yet to try to work them - they look straightforward, though. I may post questions about those and/or 7.1 later tonight or tomorrow ... need to study for a Math Methods exam as well...

    Again, thanks for any help you can offer. I'm left a little confused by the question, although all the preceding material in the chapter more or less makes sense. Not asking for a solution, but a nudge in the right train of thought direction would certainly help!
     
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