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Towards a Generic Integral for Tau in Schwarzschild Radial Motion

  1. Sep 29, 2011 #1
    I am trying to formulate an integral representing Tau between two r-values for radial motion in the Schwarzschild solution.

    There are a few possibilities:

    1. Free fall from infinity with zero initial local velocity (v0=0 and r0 -> infinity)
    2. Free fall from infinity with a given local velocity (v0=initial velocity and r0 -> infinity)
    3. Free fall from a certain r-value with a given velocity (v0=initial velocity (including 0) and r0 = r value of the initial velocity)
    4. Free fall from a certain r-value with a given velocity that is negative (v0=initial velocity (including 0) and r0 = r value of the initial velocity)

    I am able to describe all but case 4 when the velocity is directed away from the center of gravity.

    This is the integral I came up with:
    [tex]\LARGE \int _{{\it ro}}^{{\it ri}}-{\frac {1}{\sqrt {{\frac {-rr_{{s}}+r{v_{{0}}}^{2}r_{{0}}+r_{{s}}r_{{0
    }}}{r_{{0}}r}}}}} {dr}[/tex]

    rs = Schwarzschild radius
    ro = Outer radius
    ri = Inner radius
    r0 = Start value of free fall
    v0 = Start velocity of free fall

    Now how do I include the case for a negative local velocity, because by using negative v0 I get complex times (by replacing v02 by v0*|v0|)

    I suspect I need to split up the integral into two parts one for each direction and totaling the results.
    Any help?
     
    Last edited: Sep 29, 2011
  2. jcsd
  3. Oct 1, 2011 #2

    PAllen

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    Science Advisor
    Gold Member

    Can you show some of the work getting from common formulas involving radius of zero velocity as a parameter, to v0 at r0? Maybe no one wants to do this work themselves, but if you show yours, somewhat can spot a suggestion for you.
     
  4. Oct 1, 2011 #3
    I start with

    [tex] \Large \int _{{\it ro}}^{{\it ri}}\sqrt { \left( {E}^{2}-1+{\frac {{\it rs}}{r}} \right) ^{-1}}{dr}[/tex]

    Then since:
    [tex] \Large v_{{r}}= \sqrt{{E}^{2}-1+{\frac {{\it rs}}{r}}}[/tex]

    and thus:
    [tex] \Large E=\sqrt {1-{{\it r_0}}^{-1}+{v_r}^{2}}[/tex]

    Then substituting this in the prior integral we get:
    [tex] \Large \int _{{\it ro}}^{{\it ri}}
    -\sqrt { \left(
    {\frac {{\it rs}}{r}}
    -{\frac {{\it rs}}{{\it r_0}}}
    +{{\it v_0}}^{2}
    \right) ^{-1}}
    {dr}[/tex]

    Which is equivalent, and better readable, to the integral in the first posting.

    For free falling from infinity (e.g. r0-> infinity and v0=0) the integral becomes:
    [tex]\Large \int _{{\it ro}}^{{\it ri}}
    -\sqrt {{\frac {r}{{\it rs}}}}
    {dr}[/tex]

    For free falling from stationary (e.g. v0=0) the integral becomes:
    [tex]\Large \int _{{\it ro}}^{{\it ri}}
    -{\frac {1}{\sqrt {{\frac {{\it rs}}{r}}-{\frac {{\it rs}}{{\it r_0}}}}}}
    {dr}[/tex]
     
    Last edited: Oct 1, 2011
  5. Oct 1, 2011 #4

    PAllen

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    Science Advisor
    Gold Member

    Ok, now I understand your original question. Yes, I think you must use two integrals in many cases for initial velocity in the increasing r direction. When your inner and outer radii don't span the turnaround point, one integral should work; otherwise two are needed, and the turnaround point needs to be an integration limit for both integrals. I gather this is what you were asking, so the answer is a definite yes.

    Note the obvious fact that for outward velocity, inner and outer radius of e.g. 5 and 10 could represent direct motion, for 10 to turnaround of 15 and back to 5. If you limit yourself to not including the turnaround (so one integral suffices), there are still impossible choices for inner and outer radii, which manifest as imaginary integrands.

    OK, I guess the simple case integral requirements could be summed up as: r0=ri, ro (outer) <= turnaround.
     
    Last edited: Oct 1, 2011
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