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Issues with EF coords.

  1. Nov 23, 2009 #1
    Hi,

    Eddington-Finkelstein coordinates claim to prove that photons can only travel in one direction, towards the central singularity, below the event horizon. My intention here is to put EF coordinates under the microscope and see how that proof stands up to close examination.

    The Eddington Finkelstein coordinates are:

    [tex]dS^2 = 2c\ dr\ dT - c^2\left(1-\frac{2m}{rc^2}\right)dT^2 [/tex] .... (Eq 1)

    (The derivation of this set of coordinates is given in the following post.)

    For the velocity of a photon in these coordinates, both sides are divided by dt^2, dS is set to zero and then the expression is solved for dr/dT like this:

    [tex]\frac{dS^2}{dT^2} = 2c\left(\frac{dr}{dT}\right) - \left(1-\frac{2m}{rc^2}\right)c^2\right)[/tex]


    [tex] \frac{dr}{dT} = \frac{c\left(1-\frac{2m}{rc^2}\right)}{2} [/tex]

    which is the unique unambiguous solution for the velocity of a photon in Eddington-Finkelstein coordinates equal to exactly one half the solution obtained in Schwarzschild coordinates.

    Now a trick is applied. Rather than solve for dr/dT, we solve for dT/dr, so re-using Eq1 and dividing both sides by dr^2:

    [tex]\frac{dS^2}{dr^2} = 2c\left(\frac{dT}{dr}\right) - c^2\left(1-\frac{2m}{rc^2}\right)c^2\left(\frac{dT}{dr}\right)^2[/tex]

    Again for a photon dS is set to zero and the equation is now in quadratic form which has two solutions:

    [tex] \frac{dT}{dr} = 0 [/tex]

    and

    [tex] \frac{dT}{dr} = \frac{2}{c\left(1-\frac{2m}{rc^2}\right)} [/tex]

    Now dT/dr = 0 implies a velocity of plus or minus infinity for a photon in these coordinates, but essentially the velocity is undetermined and we certainly can not say anything about the direction of a photon with velocity dr/dT = 1/0.

    Never the less, we persevere and find another trick to get around this anomaly.

    To plot the light paths, the dT/dr solutions are integrated with respect to r, to give:

    [tex]T = 0[/tex]

    [tex]T = 2c\left(\frac{r}{c^2}+ \frac{2m\ \log(|2m - c^2r|)}{c^4}\right) [/tex]

    This defines 3 paths because we can use +c or -c in the second solution, so we have to select two that suit our purpose of "proving" that photons can only move inwards below the event horizon. Therefore, the first solution is chosen to represent the infalling photon and the second solution using +c is used to represent the outgoing photon. The solution with -c is discarded, for no better reason than it does not suit our purpose here.

    When plotted the following graph is obtained:

    https://www.physicsforums.com/blog_attachment.php?attachmentid=103&d=1259006985 [Broken]

    The outgoing photon path is shown in green. Note that this path has different directions above and below the event horizon but is classified according to its direction above the event horizon. The ingoing photon path is red and although its direction can not be determined, we can define its direction above the event horizon, but this still leaves some uncertainty about its direction below the event horizon and may be in the opposite direction to its direction above the event horizon as is the case for the ingoing photon.

    Now comes the second clever trick. The r axis is tilted clockwise 45 degrees so that the ingoing photon with plus or minus infinite velocity appears to moving like a normal photon, to obtain a graph like this.

    https://www.physicsforums.com/blog_attachment.php?attachmentid=106&d=1259021518 [Broken]

    The arrow showing the direction of the ingoing photon is added without justification. The same graph with more null lines, light cones and the path of a falling particle is shown below. The oblique nature of the r axis is clearer in the following image, but it does not change the fact the light path is still parallel to the r axis and does not have finite velocity despite initial appearances.

    https://www.physicsforums.com/blog_attachment.php?attachmentid=108&d=1259021518 [Broken]

    Now using the same methods as above it can equally be proved that all light paths below the event horizon can only move in the outward (up) direction. The solution with negative c that was discarded earlier is now used for the infalling photon and dT/dr = 0 is used for the outgoing photon. The plotted graph with the r axis tilted in the opposite direction, now looks like this:

    https://www.physicsforums.com/blog_attachment.php?attachmentid=107&d=1259021518 [Broken]

    Clearly, it would seem that Eddington Finkelstein coordinates are inconclusive at best, if the same coordinates that are used to prove that photons can only move inwards below the event horizon, can equally be used to prove that photons below the event horizon can only move outwards.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Nov 23, 2009 #2
    This is the derivation of the Eddington Finkelstein coordinates that I promised in the first post.

    The relationship of the Schwarzschild time coordinate (t) to the Eddington Finkelstein time coordinate (T) is defined as:

    [tex]t = T - \frac{r}{c} -\frac{2m}{c^3} \ log\left| \frac{r c^2}{2m}-1 \right| [/tex]

    The speed of light c has been inserted in the above expression in such a way that ensures each of the terms is consistently in units of time.
    After integration with respect to r the following expression is obtained:

    [tex]dt\ =\ dT-\left[\frac{1}{c}+\frac{2}{c(r-\frac{2m}{c^2})}\right]dr[/tex]

    which simplifies to:

    [tex]dt\ =\ -\frac{dr}{c(1-\frac{2m}{rc^2} )} + dT[/tex]

    The purely radial Schwarzschild metric is given by:

    [tex]dS^2 = \frac{dr^2}{ (1-\frac{2m}{rc^2})} - \left(1-\frac{2m}{rc^2}\right) dt^2 c^2[/tex]

    Substituting the equation for dt given above into the radial Schwarzschild metric gives:

    [tex]dS^2 = \frac{dr^2}{ (1-\frac{2m}{rc^2})}\ -\ \left(1-\frac{2m}{rc^2}\right)\left(-\frac{dr}{c(1-\frac{2m}{rc^2} )} + dT\right)^2 c^2[/tex]

    [tex]dS^2 = \frac{dr^2}{\left(1-\frac{2m}{rc^2}\right) }\ - \left[\frac{dr^2}{ \left(1-\frac{2m}{rc^2}\right) } -2c\ dr\ dT + c^2\left(1-\frac{2m}{rc^2}\right)dT^2 \right][/tex]

    This simplifies to give the ingoing Eddington Finkelstein coordinates:

    [tex]dS^2 = 2c\ dr\ dT - c^2\left(1-\frac{2m}{rc^2}\right)dT^2 [/tex] .... (Eq 1)
     
  4. Nov 24, 2009 #3

    George Jones

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    What did you expect to happen when time runs backwards?
     
  5. Nov 27, 2009 #4
    Hi George,

    Thanks for taking a look and spotting that misunderstanding of mine. I was using c a vector when it is obviously a scalar. However, the end result of using negative c or reversing time, is the outgoing Eddington-Finkelstein coordinates. The outgoing coordinates represent a white hole and that raises questions in itself and I have put those questions in a new thread on advanced EF coordinates here https://www.physicsforums.com/showthread.php?p=2462895#post2462895

    Advanced EF coords do not suffer from some of the ambiguities of regular EF coordinates that I have raised here, but there are still some things I would like to clear up. I hope you take a look at the new thread.
     
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