4
$\begingroup$

What is the physical meaning of the Eddington-Finkelstein coordinates? I want to see a some physical process (experimental) that could explain the many transformations of coordinates into this mathematical procedure. (really two transformations, but i think that is a big number here)

| cite | improve this question | | | | |
$\endgroup$
6
$\begingroup$

Eddington–Finkelstein coordinates use the same position coordinates as Schwarzschild coordinates, only the time coordinate is transformed, so first consider how to define Schwarzschild coordinates in a physical way. This pdf explains a way of defining the position coordinates in section 9.1.1:

• We may assign a practical definition to the radial coordinate $r$ by

  1. Enclosing the origin of our Schwarzschild spacetime in a series of concentric spheres,

  2. Measuring for each sphere a surface area (conceptually by laying measuring rods end to end),

  3. Assigning a radial coordinate $r$ to that sphere using Area = $4\pi r^2$

• Then we can use distances and trigonometry to define the angular coordinate variables $\theta$ and $\phi$.

Similarly, p. 159 of the book Relativity, Gravitation and Cosmology by Robert Lambourne, viewable online here, mentions that the radial Schwarzschild coordinate is constructed so that a sphere of radius $r$ will have a proper circumference of $2\pi r$, meaning the distance measured by placing a series of short rulers end-to-end along a great circle on the sphere.

I suppose this methods presupposes that you know how make sure the surfaces you've constructed are actually non-rotating spheres, but I suspect (though I'm not sure) that this would be the case as long as observers at fixed positions relative to the surface all saw the apparent size of the black hole's event horizon (which blocks out light from things behind it) as being the same (and constant with time), all felt the same proper acceleration as measured by an accelerometer, and all measured the tidal forces to be the same in small regions around themselves. Incidentally, once you have determined that you are at a fixed position in Schwarzschild coordinates with radius $r$, then if you also have measured the proper acceleration $a$ at your location, then since proper acceleration in Schwarzschild coordinates is given by $a = (1 - 2M/r)^{-1/2} M/r^2$ (as mentioned on p. 152 of Robert Wald's Relativity, online here; note this book uses units where $G=c=1$, so if you want to include those constants I believe the equation would become $a = (1 - (2GM/rc^2))^{-1/2} GM/r^2$ ), you can use this to solve for the mass $M$ of the black hole.

Once you have fixed position coordinates and the mass of the black hole worked out, then to measure time in Schwarzschild coordinates you can put standard clocks at fixed positions, and have a second "Schwarzschild time coordinate" display separate from the clock's display (the clock's display is counting intervals of proper time, distinct from coordinate time), with the two linked by a computer programmed to simply increment the Schwarzschild time coordinate forward by $1/\sqrt{1 - (r_0/r)}$ units of time every time the standard clock ticks forward by 1 unit of time; here $r$ is the radius of the clock and $r_0$ is the Schwarzschild radius $r_0 = 2GM/c^2$ (this is why it was important to determine the mass $M$ of the black hole). This will ensure that an interval of coordinate time $dt$ is related to an interval of proper time $d\tau$ by the correct formula for gravitational time dilation in Schwarzschild coordinates, $d\tau = dt \sqrt{1 - (r_0/r)}$.

The last step in a physical construction of Schwarzschild coordinates would be to make sure that not only are the Schwarzschild time coordinate displays at each location ticking at the correct rate, but all the clocks are also properly synchronized. This can be done with something analogous to the Einstein clock synchronization method, but for clocks at constant position in Schwarzschild coordinates rather than inertial clocks; the idea is for clock A to send a light signal to clock B when A reads $t_0$, then when B receives the signal at $T$ he immediately sends a light signal back to A, and if A receives the return signal at $t_1$, the two clocks are defined as "synchronized" if $T$ is exactly halfway between $t_0$ and $t_1$. The fact that this method can be used in Schwarzschild coordinates is mentioned on p. 186 of Astrophysical Concepts by Martin Harwit, viewable online here. Keep in mind that the times above are meant to be times on the Schwarzschild time coordinate displays, not the proper time of the standard clocks.

So that's a physical method of constructing the Schwarzschild coordinate system. As mentioned here, in relation to Schwarzschild coordinates, "ingoing Eddington–Finkelstein coordinates are obtained by replacing the coordinate $t$ with the new coordinate $v = t + r^*$ ", where $r^*$ was defined earlier as being related to the Schwarzschild coordinate $r$ by $r^* = r + 2GM \ln | (r/2GM) - 1 |$. So, instead of having a "Schwarzschild time coordinate" display next to each physical clock you can have a "ingoing Eddington–Finkelstein time coordinate" display, with the computer first calculating the Schwarzschild time coordinate $t$ using the method described before, and then calculating $v = t + r + 2GM \ln | (r/2GM) - 1 |$ and displaying that $v$. That's it! Coordinate systems are really quite arbitrary in GR, any smooth transformation from one physically meaningful coordinate system is another physically meaningful coordinate system, and the Einstein field equations will hold in all of them. So as long as you have a physical method of implementing one type of coordinate system, a physical method of constructing any other coordinate system related to the first by a known transformation is just to have computers at each point which take as input the physically-instantiated coordinates in the first system, calculate the transformation, and display as output the corresponding coordinates in the second system. It's possible there may be some more "direct" way of constructing the second coordinate system, but it's not really necessary.

Are you also asking in part about the physical motivation for choosing this particular transform of Schwarzschild coordinates, rather than just how the coordinates of events could be assigned with physical measuring instruments? If so, I believe the main appeal of ingoing Eddington–Finkelstein coordinates is just that they ensure that light rays traveling into the black hole on purely radial paths will all have the same constant coordinate speed throughout their travel, whereas in Schwarzschild coordinates the coordinate speed of an ingoing light ray is continually slowing down as it approaches the event horizon (and in fact they will never reach the horizon at any finite coordinate time in Schwarzschild coordinates, whereas they do in ingoing Eddington–Finkelstein coordinates). Outgoing light rays emitted in a radial direction away from the black hole don't have a constant coordinate speed in ingoing Eddington–Finkelstein coordinates, though there is a separate coordinate system called "outgoing Eddington–Finkelstein coordinates" where they do, but ingoing light rays don't. If you want a coordinate system where both ingoing and outgoing light rays have a constant coordinate speed, try Kruskal–Szekeres coordinates.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ "I suppose this methods presupposes that you know how make sure the surfaces you've constructed are actually non-rotating spheres..." I concur the spheres must be non-rotating. And this is the first time I've ever heard anyone else say this! $\endgroup$ – Colin MacLaurin Feb 10 '18 at 12:02
2
$\begingroup$

Coordinates are not physical. They are entirely up to you. In the Eddington Finkelstein coordinates, the lines of constant $u$ $\theta$ and $\phi$ are null radial lines going out to infinity while $r$ is the coordinate such that the surfaces swept out by the spherical symmetry have an area of $4 \pi r^2$. (i.e., take a point of the spacetime. Apply a spherical symmetery (rotation). Do that with all possible rotations and you will get a sphere. Measure its area, take the square root of that area over $4 \pi$, and call that the coordinate $r$. You can thus label all points in the spacetime with some value for $r$. Label the points on one of those surfaces with $\theta$ and $\phi$ using the standard link between them and the rotations of that sphere. Now choose the pole, $\theta=0$, and construct the orthogonal vectors to that surface at that pole.

The function $\textrm{grad } r$ is a vector orthogonal to each of those surfaces of constant $r$. From any point on that one sphere you labelled with $\theta$ and $\phi$, construct a geodesic with $\textrm{grad } r$ as the tangent vector. Where those geodesics intersect other spheres of spherical symmetry, label those points with the same value of $\theta$ and $\phi$ as on that first sphere. At each point on the surface of each of the spheres you have already labelled with $\theta$ and $\phi$, construct the outward pointing null vector orthogonal to the sphere. (There will be two null vectors at any point which are orthogonal to any tangent vector to the sphere. One of them, will be such that along a line with that vector as tangent, $r$ will increase as you go to the future). Construct the null geodesic with that vector as the tangent. Where it intersects any spheres which you have not already labelled by $\theta$ and $\phi$, use it to give a label of $\theta$ and $\phi$ to the points on that sphere. Spherical symmetry will ensure that this labelling is consistent-- i.e. you will not discover that if you run into a sphere which has already been labelled the values of $\theta$ and $\phi$ assigned to the intersection point will already have been labelled differently. The label $u$ will be chosen such that the surface formed by all of the null geodesics emanating from one sphere will have the same label $u$. So far $u$ is arbitrary. Chose the coordinate $u$ so that the gradient of the function $u$ dotted into the gradient of the function $r$ is unity. This constructs the Eddington Finkelstein outgoing null coordinates. To construct the ingoing ones, carry out the above procedure but instead at each point choose the ingoing null vector ($r$ decreases as you go into the future).

The above defines the Eddington Finkelstein coordinates by a "physical" process. The time translation symmetry of the Schwartzschild metric ensures that it does not matter which point you chose initially to start off the above process. They all produce the same coordinates except for a transformation of adding a constant to the function $u$, and of making a rotation of the $\theta$ and $\phi$ coordinates.

| cite | improve this answer | | | | |
$\endgroup$
  • $\begingroup$ The method you describe doesn't seem to be an experimental process, it's still mathematical. $\endgroup$ – Kyle Kanos Nov 13 '14 at 18:25
0
$\begingroup$

Suppose we start with Schwarzschild coordinates $(t,r,\theta,\phi)$ already given (although we could easily start with less). To choose a correct physical construction, first consider the required theoretical coordinate transformation. For the ingoing, null variant of Eddington-Finkelstein coordinates, the new coordinate transforms from Schwarzschild coordinates as $v=t+r^*$ where $$r^*\equiv r+2M\ln\Big\lvert\frac{r}{2M}-1\Big\rvert$$ Since $v=\textrm{const}$ for radial ingoing photons, we base our construction on this. Pick a sphere of fixed $r=r_0$ and place lasers all around it, pointed straight down / inwards. Then emit pulses of photons which encode numerically the value of $t+r^*$ at time of emission. Then at any event with $r<r_0$, intercept the closest light pulse and take its $v$-value for your coordinate.

Note at $r=r_0$, the transformation becomes simply $v=t+\textrm{const}$. So you could ignore the constant and simply take $v':=t$ as the starting value. If $r_0\gg 2M$, just take the proper time of the laser, assuming they are all synchronised on the sphere! You can construct outgoing, null coordinates similarly.

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.