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Time measured by free-fall observer near object?

  1. Oct 26, 2012 #1
    What is the time measured by an observer who is in free-fall near an object?

    The Schwarzchild metric is given by:

    [itex] \large c^2 d\tau^2 = (1 - \frac{r_s}{r}) c^2 dt^2 - ( 1 - \frac{r_s}{r})^{-1} dr^2 - r^2(d\theta^2 + \sin^2 \theta d\phi^2) [/itex]

    Now for an observer at a fixed position one uses:

    [itex]dr = d\theta = d\phi = 0[/itex]

    To give an element of proper time as:

    [itex] \large d\tau = \sqrt{1 - \frac{r_s}{r}} dt [/itex]

    This is a time interval experienced by an observer who is stationary in the object's gravitational field.

    But I want the time measured by a free-fall observer so I can't assume that his position is fixed.

    Perhaps this is how to do the calculation.

    I work out the world line for a radial light beam using:

    [itex] d\tau = d\theta = d\phi = 0 [/itex]

    Substituting into the above Schwarzchild metric I find the worldline of a lightbeam given by

    [itex] \large (1 - \frac{r_s}{r}) \ c \ dt = dr [/itex]

    Now a free-fall observer experiences a flat spacetime locally. He must use a time element [itex]d\tau[/itex] such that a light beam's worldline is diagonal so that

    [itex] \large c \ d\tau = dr [/itex]

    Therefore his time element must be:

    [itex] \large d\tau = (1 - \frac{r_s}{r}) \ dt [/itex]

    Does this make sense?
     
  2. jcsd
  3. Oct 26, 2012 #2

    Mentz114

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    The four velocity of the LeMaitre free-faller is
    [tex]
    U=\left(\frac{r}{r-2\,m}, -\frac{\sqrt{2}\,\sqrt{m}}{\sqrt{r}} , 0 , 0\right)
    [/tex]
    which agrees with your result.
     
  4. Oct 26, 2012 #3
    Great - thanks!

    I'm also interested in calculating the correct time interval we should use with the Friedmann Walker metric

    [itex] \large ds^2 = -dt^2 + a^2(t) [ \frac{dr^2}{1-kr^2} + r^2(d\theta^2+\sin^2\theta d\phi^2)] [/itex]

    As co-moving observers we are also in free-fall.

    Thus using [itex]ds = d\theta = d\phi = 0[/itex]

    The worldline of a radial light beam is given by

    [itex] \large \frac{dt}{a(t)} = \frac{dr}{1-kr^2} [/itex]

    Thus the interval of proper time [itex]d\tau[/itex] for a co-moving observer with a locally flat spacetime (assuming k is small) is given by

    [itex] \large d\tau = \frac{dt}{a(t)} [/itex]

    which I believe is called an interval of conformal time.

    I think many people believe we should use cosmological time [itex]t[/itex] as our proper time but according to the above reasoning that is not correct because we are in free-fall (in a local inertial frame) and therefore our proper time is conformal time.
     
    Last edited: Oct 26, 2012
  5. Oct 26, 2012 #4

    Mentz114

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    I don't think that's right. If you divide the line interval by dτ2 and set the spatial velocities to zero you get dt/dτ = 1 for this co-moving frame.
     
  6. Oct 26, 2012 #5
    Could you elaborate a little more?
     
    Last edited: Oct 26, 2012
  7. Oct 26, 2012 #6

    Mentz114

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    I can't eleborate without doing the simple algebra and tex'ing it up which I don't have time for. Get a pencil, paper and eraser and do as I suggested. If you have a problem then you could ask about specifics. Remember that ds=dτ.
     
    Last edited: Oct 26, 2012
  8. Oct 26, 2012 #7
    Ok - this is what I think you did.

    For simplicity starting with the FRW metric with k=0 and only radial spatial component:

    [itex] \large d\tau^2 = dt^2 - a(t)^2 dr^2 [/itex]

    divide through by [itex]d\tau^2[/itex]

    [itex] \large 1 = \frac{dt^2}{d\tau^2} - a(t)^2 \frac{dr^2}{d\tau^2} [/itex]

    For a co-moving frame spatial velocity is zero so

    [itex] \large \frac{dr}{d\tau} = 0 [/itex]

    Therefore

    [itex] \large \frac{dt}{d\tau} = 1 [/itex]

    Is this right?
     
  9. Oct 26, 2012 #8

    Mentz114

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    Your calculation is correct.
     
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