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Black Hole : Infalling Observer looking back

  1. Apr 6, 2013 #1

    A.T.

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    What blue shift of distant light would you observe, while free falling into a black hole. It seems that hovering at the horizon (which is not possible), would result in infinite blue shift. But what about the observed blue shift in free fall, as a function of the radial Schwarzschild coordinate?
     
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  3. Apr 6, 2013 #2

    pervect

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    It depends on your energy-at-infinity. If you free fall in to a black hole from zero velocity at infinity (an energy-at-infinity of 1 including your restmass) you see a redshift (not a blueshift) of 2:1 as you look straight back.

    You attribute this redshift to tidal forces.

    I don't recall the thread(s) where I originally worked this out, my recollection is I converted to ingoing Finklestein coordinates. I did find a later thread where I repeated this, and another thread where George Jones did a simlar calculation.

    IN https://www.physicsforums.com/showpost.php?p=4174007&postcount=8 George mentions he got similar results using Paineleve coordinates, which might also be a good choice.

    In ingoing EF coordinated, you can read the shift right off of v(tau), once you compute it, because v is constant along an ingoing light beam falling at rest from infinity.

    https://www.physicsforums.com/showpost.php?p=4185014&postcount=12 and in particular it's succesor https://www.physicsforums.com/showpost.php?p=4185014&postcount=13

    The results below apply to a black hole with mass m=2

    [add]
    to avoid the messy algebra, another approach would be to start with

    [tex]\frac{dr}{d\tau} = \sqrt \frac {2m}{r}[/tex]
    [tex]\frac{dt}{d\tau} =\frac{1}{1-2m/r}[/tex]

    then compute [itex]dv / d\tau[/itex] from the above relations and the defintion of v, using the chain rule:

    [tex]v = t + r + 2\,m\,\ln \left| \frac{r}{2m} - 1 \right|[/tex]

    You'll still have to cancel out infinite terms though.
     
    Last edited: Apr 7, 2013
  4. Apr 6, 2013 #3

    A.T.

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    Thanks pervect, that's interesting.
     
  5. Apr 7, 2013 #4

    pervect

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    You are welcome- I decided to expand on the analysis to handle the case of variable energy-at-infinity, which I call E

    The case of E=1 corresponds to "at rest at infinity".

    We can write the radial ingoing geodesic using the usual formula:
    http://www.fourmilab.ch/gravitation/orbits/

    [tex]\frac{dt}{d\tau} = \frac{E}{1-2m/r}[/tex]
    [tex]\frac{dr}{d\tau} = -\sqrt{E^2 - 1 + 2m/r}[/tex]

    We choose the negative sign on the sqrt because for infalling objects [itex]dr/d\tau[/itex] must be negative.

    To convert to EF coordinates we define the tortise coordinate r*, and use v=t + r*.

    http://en.wikipedia.org/w/index.php?title=Eddington–Finkelstein_coordinates&oldid=538982166

    [tex]r* = r + 2m\;ln \left| r/2m -1 \right| [/tex]
    for both r>2m and r<2m we can write
    [tex]\frac{dr\!*}{dr} = \frac{1}{1-2m/r}[/tex]

    As we agrued previously, v is constant for any infalling light-beam, and can be thought of as the "time of emission at infinity".

    Then the doppler shift is the rate of change of v, the time of emission at infinity, with [itex]\tau[/itex], the proper time.

    Thus, knowing that v = t+r*, we can write

    [tex]\frac{dv}{d\tau} = \frac{dt}{d\tau} + \left( \frac{dr*}{dr} \right) \left( \frac{dr}{d\tau}\right) = \frac{E}{1-2m/r} - \frac{\sqrt{E^2-1+2m/r}}{1-2m/r}[/tex]

    We can see that we have the difference of two infinite quantites at r=2m :-(.

    To get the doppler shift at r=2m, it's convenient to do a series expansion of the above expression around r=2m

    This gives:

    [tex]\frac{dv}{d\tau} \approx \frac{1}{2E} + \frac{1}{16\,m\,E^3} \left(r -2m \right) + \frac{2E^2-1}{64\,m^2\,E^5} \left(r -2m \right)^2 + ....[/tex]

    For values at or near the horizon we can use the series expansion, for values far away from the horizon we should use the original expression (the one which becomes singular at the horizon).

    For E<1, the object can never reach infinity. For the limit of low E, one has a blueshift at the horizon, which makes sense (the object didn't have enough energy to reach infinity, and can be considered to be dropped in from some low value of r). It started out experiencing blueshift, and continues to experience blue shift as it falls in

    For E=1, we see the doppler shift I mentioned of 1/2 at the event horizon. E>1 implies starting out with some velocity towards the black hole at infinity which gives more redshift.

    I believe the computed value for redshift also makes sense if we take the limit as r->infinity, but I haven't looked at that as closely as I might. Clearly the doppler shift is 1 (no doppler shift) for E=1 at r=infinity, which is correct.
     
    Last edited: Apr 7, 2013
  6. Apr 7, 2013 #5
    I do not believe this is correct.

    For instance the observed redshift for a radially free falling (from infinity) observer at the event horizon is indeed 2:1, however this value is independent of the size of the black hole. Tidal forces however depend on the size of the black hole.

    I think the gravitational redshift (or blueshift) is related to the 'sum' of all curvature between origin and destination, popularly called gravitational potential. Add to that the kinetic Doppler effect which is the local velocity wrt a static observer and you have the total red- or blueshift.


    I think the redshift of a signal from infinity for a radially free falling observer using Schwarzschild coordinates is:

    [tex]\left( E+\sqrt {{\frac {{E}^{2}r-r+2\,m}{r}}} \right) ^{-1}[/tex]

    which becomes:

    [tex] \left( 1+\sqrt {{\frac {2m}{r}}} \right) ^{-1}[/tex]

    When E=1

    Correct?
     
    Last edited: Apr 8, 2013
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