# Issues with advanced Eddington-Finkelstein coordinates

1. Nov 27, 2009

### yuiop

This is a follow on from a previous thread about issues I had with regular Eddington-Finkelstein coordinates. First I show a brief derivation of the advanced EF coordinates which appears to have less issues than the regular EF coordinates and then discuss what I think appear to be the remaining issues.

The purely radial Schwarzschild metric is given by:

$$dS^2 = \left(1-\frac{2m}{rc^2}\right) dt^2 c^2 -\ \frac{dr^2}{ (1-\frac{2m}{rc^2})}$$

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

$$dt\ =\ dT -\frac{2m}{c(rc^2-2m)}dr$$

Substituting this equation for dt into the radial Schwarzschild metric and simplifying gives:

$$dS^2 = \left(1-\frac{2m}{rc^2}\right) dT^2 c^2 - \frac{4m\ dT\ dr \ c}{r} -\ \left(1+\frac{2m}{rc^2}\right)dr^2$$

For a photon, dS is is set to zero and solving for dr/dT gives:

$$\frac{dr}{dT} = -c$$

and

$$\frac{dr}{dT} = c \frac{ \left(1-\frac{2m}{rc^2} \right)}{\left(1+\frac{2m}{rc^2} \right)}$$

The above results show that the equation for an ingoing light ray in advanced EF coordinates behaves nicely at the event horizon with no infinities. However it can also be noted that the coordinate velocity of a outgoing photon is zero at the event horizon in these coordinates and arrives at the event horizon in infinite coordinate time, just as it is in Schwarzschild coordinates.

Integrating dT with respect to r for the outgoing photon gives:

$$T = \frac{r}{c} + \frac{4m\ \log(2m - rc^2)}{c^3}$$

Plotting T versus r gives:

https://www.physicsforums.com/blog_attachment.php?attachmentid=114&d=1259383629 [Broken]

which is identical to the ingoing regular EF coordinates after the r axis has been tilted. Note that these coordinates are still not perfect, as there is some uncertainty about whether or not the path of the outgoing photon is continuous across the event horizon at T = -infinity.

Reversing the time coordinate gives the outgoing advanced EF coordinates as plotted below.

https://www.physicsforums.com/blog_attachment.php?attachmentid=115&d=1259383629 [Broken]

The outgoing coordinates are identical to the regular EF coordinates after the r axis has been tilted in the opposite direction to the r axis of the ingoing regular EF coordinates.

The outgoing coordinates (as I understand it) represent a white hole. One issue that I have here, is that light and particles can fall to the event horizon of the white hole in infinite coordinate time and this translates to particles falling to the white hole event horizon in finite proper time. Is that supposed to happen?

As mentioned above, ingoing and outgoing Eddington-Finkelstein coordinates still have uncertainties about the continuity of null paths as they cross the event horizons at T = plus or minus infinity, where the velocity of the photon goes to zero exactly at the event horizon in these coordinates.

I would like to introduce a new coordinate transformation that appears to remove this uncertainty.

First I define a new time parameter T related to the Schwarzschild time parameter t by the following relationship:

$$dt\ =\frac{dT} {\left(1+\frac{2m}{rc^2}\right) \left(1-\frac{2m}{rc^2} \right)}$$

Substituting this new definition of dt into the radial Schwarzschild metric and simplifying gives:

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

For a photon dS is set to zero and solving for dr/dT gives:

$$\frac{dr}{dT} = \pm \frac{ c}{\left(1+\frac{2m}{rc^2}\right) }$$

Integrating dT with respect to r gives:

$$T = \pm \left( \frac{r}{c} +\frac{2m \log(r)}{c^3} \right)$$

which is the light paths in terms of T and r which can be plotted to give the graph below:

https://www.physicsforums.com/blog_attachment.php?attachmentid=116&d=1259383629 [Broken]

I have called the new coordinates "fast coordinates" as they are sort of the opposite of Finkelstein's "tortoise coordinates". The red curve in the graph represents a typical ingoing light path and the green curve represents a typical outgoing light path.

This new metric has the advantage that light paths are smooth and continuous across the event horizon for both ingoing AND outgoing photons. In these new coordinates, the "singularity" at the event horizon has been transformed away, perhaps too successfully, because the statement that there is "nothing special about the event horizon" now means that is is not even what Finkelstein called a "one way membrane" in these new coordinates. The new coordinates also remove the requirement to have ingoing and outgoing versions of the metric. White holes no longer exist in these new coordinates and black holes no longer have the properties we are used to. Clearly something appears to be wrong. It should be impossible to arrive at different physical conclusions, just by using different transformations, because General Relativity is supposed to be a diffeomorphism invariant theory. Can anybody spot where I made the error in the transformation and what rules govern what are allowable transformations?

I have other “issues”, but I will come back to them after this issue has been addressed.

Last edited by a moderator: May 4, 2017
2. Nov 27, 2009

### JesseM

From what I've read a Schwarzschild black hole is really more of a "gray hole" from the perspective of an outside observer--he can see both outgoing particles that come from an inner white hole region, and ingoing particles which fall into an inner black hole region, both of which experience only a finite proper time between passing the observer and crossing the horizon. If we imagine that the observer at finite radius sees outgoing particles pass him and ingoing particles pass him at regular intervals (say, one per minute) for all eternity, then each ingoing particle will pass by all the infinite outgoing particles that will subsequently reach the observer before that ingoing particle crosses the horizon in finite proper time (it will pass outgoing particles with a greater and greater frequency as it approaches the horizon), and likewise each outgoing particle will pass by all the infinite ingoing particles that passed the observer before that outgoing particle reaches the observer, so the paths of ingoing and outgoing particles never cross inside the horizon, the outgoing particles coming from a different inner region of spacetime than the ingoing particles fall into. All this is a lot clearer in Kruskal-Szekeres coordinates which show the maximally extended version of the Schwarzschild spacetime, while each version of Eddington-Finkelstein coordinates only covers two of the four regions in the Kruskal-Szekeres diagram.

I'm not confident enough in my understanding of the GR math to address the new coordinate transformation you come up with, but there's got to be a mistake in it if it shows outgoing and ingoing light paths crossing inside the horizon, which isn't possible in the standard coordinate systems (would it be possible to take two paths that cross inside and transform them back into Eddington-Finkelstein coordinates or Schwarzschild coordinates to see if they are really correct null paths?) Perhaps one of the other posters will be able to pinpoint the problem.

3. Dec 1, 2009

### George Jones

Staff Emeritus
Suppose you decide to plunge straight into a black hole. Suppose further that during your entire trip, both inside and outside the black hole, there is a particular star that you can see directly behind (i.e., directly above) you.

While outside the event horizon, you briefly fire a laser in the direction of star (straight up). What is the worldline of the photon?

While on the event horizon, you briefly fire a laser in the direction of star (straight up). What is the worldline of the photon?

While inside the event horizon, you briefly fire a laser in the direction of star (straight up). What is the worldline of the photon?
When defining coordinate transformations via differentials, care must be taken. It looks like your definition is inconsistent.

The completely general coordinate transformation $t = t \left( T , r \right)$ leads to the differential relationship

$$dt = \frac{\partial t}{\partial T} dT + \frac{\partial t}{\partial r} dr.$$.

Your definition doesn't have the second term, i.e., your definition has

$$\frac{\partial t}{\partial r} = 0$$

Partial differentiation with respect to a coordinate treats the other coordinates as constants, so this constraint gives that $t$ is an arbitrary function of $T$ (only), so $t = f \left( T \right)$.

The first term of the general differential relationship combined with your differential relationship gives

$$\frac{\partial t}{\partial T} = \frac{1}{1 - \left( \frac{2M}{rc^2} \right)^2 },$$

so that

$$t = \frac{T}{1 - \left( \frac{2M}{rc^2} \right)^2 } + g \left( r \right),$$

where $g \left( r \right)$ is an arbitrary function of $r$ (only).

These two expressions for $t$ are not consistent.

4. Dec 5, 2009

### yuiop

Hi George,

Thanks for the technical input. It was very helpful to me.

So if I take the radial Schwarzschild metric:

$$dS^2 = \left(1-\frac{2m}{rc^2}\right) dt^2 c^2 -\ \frac{dr^2}{ (1-\frac{2m}{rc^2})}$$

and make $\tau$ constant by setting dS = 0, then the second term of the differential relationship would be

$$\frac{\partial t}{\partial r} dr = \frac{\pm dr}{c\left(1-\frac{2m}{rc^2}\right) }.$$

Is that correct so far?

Now it seems that I am free to define my own time parameter dT, so say for example I choose dT such that

$$\frac{\partial t}{\partial T} dT = \frac{2dT} {\left(1+\frac{2m}{rc^2}\right) \left(1-\frac{2m}{rc^2} \right)}$$

then I can say

$$dt = \frac{\partial t}{\partial T} dT + \frac{\partial t}{\partial r} dr = \frac{2dT} {\left(1+\frac{2m}{rc^2}\right) \left(1-\frac{2m}{rc^2} \right)} \pm \frac{dr}{c\left(1-\frac{2m}{rc^2}\right) }.$$

Substitution of this new expression for dt into the Schwarzschild metric yields:

$$dS =\frac{4\ dT^2\ c^2} {\left(1+\frac{2m}{rc^2}\right)^2 \left(1-\frac{2m}{rc^2} \right)} \pm \frac{4\ dr\ dT\ c} {\left(1+\frac{2m}{rc^2}\right) \left(1-\frac{2m}{rc^2} \right)}$$

Setting dS=0 gives the null paths as

$$\frac{dr}{dT} = \frac{\pm c}{\left(1+\frac{2m}{rc^2}\right)},$$

which is the same as the result and graph I posted at the end of post#1.

Does that seem reasonable?

5. Dec 16, 2009

### yuiop

The silence is deafening. :tongue:

If it helps, the main issue I am having is that dr/dt for a light path can be positive or negative in Schwarzchild coordinates and only negative in EF coordinates. Why is the result from EF coordinates treated as a final arbiter of truth and the Schwarzchild result ignored?

I am not sure why you say this. Schwarzschild coordinates, EF coordinates, Lemaitre coordinates and KS coordinatres all show light paths crossing below the event horizon. Perhaps you meant that if light paths cross above the light horizon then they should not cross again below the event horizon and vice versa? If that is the case, then the new coordinates comply with that rule too. Although I have not shown a calculation for light paths crossing above the EH in the new coordinates, I think it would be easy enough to arrange that by the addition of a suitable constants of integration.

6. Dec 16, 2009

### JesseM

With light the meaning of "ingoing" and "outgoing" is perhaps ambiguous since light has no proper time, so it would be clearer to talk about ingoing and outgoing particles with timelike worldlines. If we set the direction of increasing proper time along different particles' worldlines so that they all agree in the region outside the horizon (i.e. if two particles cross paths at some point outside the horizon, both of their proper times increase in the future light cone of that point), then we can distinguish between "ingoing" particles whose proper time is maximized at the point their worldline intersects a singularity, and "outgoing" particles whose proper time is minimized at the point their worldline intersects a singularity. In this case, as long as we restrict our attention to particles which spend part of their worldline in the region outside the event horizon in our universe (as opposed to the parallel universe in the maximally extended Schwarzschild spacetime shown in KS coordinates), then ingoing and outgoing particles never cross paths inside the horizon, the outgoing particles come from an inner "white hole region" while the ingoing particles fall into a separate inner "black hole region" (again this is shown clearly in KS coordinates).

But even if we're just talking about light, the diagram you show here doesn't make seem to make sense:

https://www.physicsforums.com/blog_attachment.php?attachmentid=116&d=1259383629 [Broken]

When light paths cross inside the horizon, then the two paths should form the sides of a future and past light cone from the crossing point. If we look at the crossing point of the red and green paths that are supposed to represent light paths above, and we drew a timelike worldline that also went through the crossing point, its slope would have to be closer to horizontal than the slope of the two light paths at that point, so that the horizontal black line that goes through the crossing point would be timelike inside the horizon (if instead you assumed timelike worldlines were closer to vertical than the slope of the two light paths, then part of the future light cone of that crossing point would lie outside the horizon since the green line escapes the horizon, and part of the past light cone would lie outside the horizon since the red line came from outside the horizon, so there'd be timelike worldlines that dipped below the horizon and then came back out without ever hitting the singularity). But then if you imagine a bunch of timelike worldlines which go through that event and extend them backwards through the past light cone into the region outside the horizon, that would seem to imply that the radial coordinate is always timelike even outside the horizon, and that even for crossing points outside the horizon timelike worldlines will have a slope closer to horizontal than the light worldlines, which I don't think was your intention.

Last edited by a moderator: May 4, 2017
7. Dec 16, 2009

### yuiop

Thanks Jesse,
It might surprise you that actually agree with everything you say here. Although I have probably not made it clear, the point of this thread is to try and establish what the rules are, as to what constitutes a valid transformation to construct an alternative but legitimate coordinate system. GR allows many equally valid coordinate systems but obviously they can not be entirely arbitrary or we could come to all sorts of strange conclusions as demonstrated by the problems you have pointed out with these new coordinates. They obviously break some rules, but what are those rules?

In my "Relativity for Idiots" book it states in a derivation of the regular EF coordinates that two null coordinates

u = t - r* and v = t + r*,

are used in the construction where u and v are new ingoing and outgoing time coordinates respectively, t is the regular Schwarzchild time coordinate and r* is a function of the regular Schwarzschild radial coordinate (r). The book does not explain why these relationships must hold.

In the derivation of the original EF coordinates, r* is know as the "tortoise coordinate" and is defined as:

$$r* = r + 2m \ln\left(\frac{r}{2m}-1\right)$$

and then takes the derivative of the above expression wrt r and obtains:

$$dr* = \frac{dr}{(1-2m/r)}.$$

Again, without explanation it states

dt = dv - dr*

$$dt^2 = dv^2 - \frac{2\ dv\ dr}{(1-2m/r)} + {dr^2}{(1-2m/r)^2}$$

which after substitution into the Schwarzschild coordinates gives the original EF coordinates.

It seems that using the same rule (dt = dv - dr*) can be used to construct the advanced EF coordinates except this time

$$dr*=\ \frac{2m}{c(rc^2-2m)}dr$$

and using dt = dv - dr*

$$dt\ =\ dv -\frac{2m}{c(rc^2-2m)}dr.$$

Does anyone know if dt = dv - dr* is a rule that can be applied generally and if so what is the reasoning or logic behind it?

Last edited: Dec 16, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook