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Proper Reference Frame -Accelerated observer

  1. Aug 21, 2013 #1

    WannabeNewton

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    "Proper Reference Frame"-Accelerated observer

    Hi guys. This is regarding section 13.6 (p.327) in MTW. Here the authors consider an arbitrary accelerated observer in any space-time and construct a set of local coordinates carried along the entire worldline of the oberver with the origin of the coordinates comoving with the observer; they term this the "proper reference frame" of the accelerated observer. Note that they do not assume the observer's spatial basis vectors are Fermi-Walker transported along his/her worldline.

    In ex.(13.14) they consider an accelerated observer and a freely falling particle that at some event is coincident with the origin of the observer's "proper reference frame". They say to show that the 3-acceleration of the freely falling particle relative to the "proper reference frame" at that event is given by ##\frac{\mathrm{d} ^{2}x^{j}}{\mathrm{d} x^{0^2}}e_j = -a - 2\omega \times v + 2(a\cdot v)v## where ##j = 1,2,3##, ##(e_j)## are the spatial basis vectors carried by the observer, ##\omega## is the angular velocity of rotation of the spatial basis vectors, ##v## is the 3-velocity of the freely falling particle, and ##a## is the acceleration of the observer him/herself.

    Because this event is on the worldline of the observer (i.e. the origin ##x^j = 0## of the observer's "proper reference frame"), the non-zero christoffel symbols are all given in (13.69a) and (13.69b). Now the freely falling particle satisfies the equations of motion ##\nabla_u u = 0## as usual. Taking again ##j = 1,2,3## in the above coordinates, we simply calculate ##\frac{\mathrm{d} ^2 x^{j}}{\mathrm{d} \tau^2} = -\Gamma ^{j}_{\mu\nu}\frac{\mathrm{d} x^{\mu}}{\mathrm{d} \tau}\frac{\mathrm{d} x^{\nu}}{\mathrm{d} \tau}\\ = -\Gamma ^{j}_{00}(\frac{\mathrm{d} x^0}{\mathrm{d} \tau})^2 - 2\Gamma ^{j}_{k0}\frac{\mathrm{d} x^{k}}{\mathrm{d} \tau}\frac{\mathrm{d} x^0}{\mathrm{d} \tau} \\= -a^j(\frac{\mathrm{d} x^0}{\mathrm{d} \tau})^2 + 2\omega^{i}\epsilon_{0ijk}\frac{\mathrm{d} x^{k}}{\mathrm{d} x^0}(\frac{\mathrm{d} x^0}{\mathrm{d} \tau})^2\\ = (-a^j - 2(\omega \times v)^j)(\frac{\mathrm{d} x^0}{\mathrm{d} \tau})^2##
    because all other Christoffel symbols vanish as per (13.69a) and (13.69b). I don't get where that third term ##2(a\cdot v)v## is coming from. Could someone point it out? Thanks in advance!
     
    Last edited: Aug 21, 2013
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  3. Aug 21, 2013 #2

    Bill_K

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    I believe the point is that the desired equation contains a d2/dx02, whereas the geodesic equation contains d2/ds2, so you have to use the metric (13.71) to convert one to the other, and this produces the additional term. For example d/dx0 on the a·x in the metric will give you an a·v.
     
  4. Aug 21, 2013 #3

    WannabeNewton

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    Hi Bill, thanks for the reply! I don't know how but I totally forgot that they wanted ##\frac{\mathrm{d} ^{2}x^{j}}{\mathrm{d} x^{0^{2}}}## and not ##\frac{\mathrm{d} ^{2}x^{j}}{\mathrm{d} \tau^2}## where, as before, ##\tau## is the proper time along the worldline of the freely falling particle. That certainly helped but I'm still a bit confused.

    For example, we have ##\frac{\mathrm{d} ^{2}x^{j}}{\mathrm{d} x^{0^{2}}}e_j = \frac{\mathrm{d} \tau}{\mathrm{d} x^{0}}\frac{\mathrm{d} }{\mathrm{d} \tau}(\frac{\mathrm{d} \tau}{\mathrm{d} x^{0}}\frac{\mathrm{d} x^{j}}{\mathrm{d} \tau})e_j\\ = (\frac{\mathrm{d} \tau}{\mathrm{d} x^0})^{2}\frac{\mathrm{d} ^{2}x^{j}}{\mathrm{d} \tau^2}e_j + \frac{\mathrm{d} \tau}{\mathrm{d} x^0}\frac{\mathrm{d} x^{j}}{\mathrm{d} \tau}e_j(\frac{\mathrm{d} }{\mathrm{d} \tau}\frac{\mathrm{d} \tau}{\mathrm{d} x^0})\\ = -a - 2\omega \times v + v(\frac{\mathrm{d} }{\mathrm{d} \tau}\frac{\mathrm{d} \tau}{\mathrm{d} x^0})##

    I just don't get how ##\frac{\mathrm{d} }{\mathrm{d} \tau}\frac{\mathrm{d} \tau}{\mathrm{d} x^0}## equals ##2a_i v^i## using (13.71) or otherwise. The freely falling particle isn't at rest in the "proper reference frame" of the accelerating observer so I don't get how the equality comes about.
     
    Last edited: Aug 21, 2013
  5. Aug 22, 2013 #4

    Bill_K

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    I'd say to start from the d/ds end, and use the fact that d/ds is a directional derivative along the world line:

    d/ds = dt/ds ∂/∂t + dx/ds · ∂/∂x

    When the dust clears, use the fact that this is evaluated at the origin, and set x = 0. So

    d2/ds2 = (dt/ds ∂/∂t + dx/ds · ∂/∂x)(dt/ds ∂/∂t + dx/ds · ∂/∂x)
    = (dt/ds)22/∂t2 + (∂/∂t)(dt/ds) ∂/∂t + dx/ds · ∂/∂x (dt/ds) ∂x/∂t + (dx/ds)22/∂x2

    Then (∂/∂t)(dt/ds) = a·v so the second term is a·v v, and dx/ds · ∂/∂x (dt/ds) = v·a so the third term is a·v v also.

    EDIT: Good illustration of why I don't like MTW! They seem to make up their notation ad hoc as they go along.
     
    Last edited: Aug 22, 2013
  6. Aug 22, 2013 #5

    WannabeNewton

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    Ah ok. That makes perfect sense, thank you again Bill! I appreciate the help :smile:

    And yeah, I'm relatively new to MTW and their notation is making my head spin like crazy haha; I've gotten too used to the notation in Wald it would seem.
     
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