# Apparent distances as seen from within Schwarzschild solution

1. Sep 16, 2008

### Jonathan Scott

I've recently been wondering what the exact relationship is between apparent distances to objects as seen by an observer within a spherical gravitational potential well (described by the Schwarzschild solution) compared with distance as seen by an outside observer.

I've decided that the sort of distance I'm interested in is the one which can be measured locally: the distance by triangulation. That is, you measure the angular difference in the apparent direction of the object from two points then divide the distance between the points by the angle (assumed to be small) to get the distance to the object. Another way of putting this is that it is the spatial radius of curvature of a surface perpendicular to rays arriving from the object.

As a first approximation, one can assume an isotropic metric where the time (dt) factor is approximately (1-Gm/rc^2) and the space (dx, dy and dz) factor is approximately 1/(1-Gm/rc^2), ignoring terms in (Gm/rc^2)^2 and above. In that case, one can simply say that "rulers get smaller" and "time runs slower" deeper into the potential well, and the factor is the same. That has the interesting feature that one would expect any observed quantity of the form frequency/distance (such as m/r terms in a Newtonian gravitational potential when m is expressed in energy or frequency units) to have the same value locally as in the background coordinate system.

The first thing I tried calculating exactly was the apparent distance to the middle of the central mass in the Schwarzschild solution, using simple trigonometry and spherical symmetry. This comes out to the following:

$$\mbox{radius of curvature} = \frac{r}{\sqrt{1 - 2GM/rc^2}}$$

This is obviously the same result as one could have calculated from the simple idea that radial rulers shrink by that factor. However, note that the value of this result is a local observation; it does not depend on the metric. Exactly the same result value should therefore be obtained when calculating the same distance in other coordinate systems, and I've checked that I get the same result for isotropic coordinates, although the trigonometry is somewhat different and the result is not in the same form, showing that the "radial ruler" analogy doesn't hold in this case.

The measured ratio of energy/distance for the central mass as measured by the local observer in this way is exactly equal to the value that would be seen by an external observer using the Schwarzschild coordinate system. Interestingly, this means that there is some physical significance to the use of that coordinate system, as that radial coordinate can be physically defined as the distance calculated using the local radius of curvature divided by the local time dilation factor.

I'd now like to be able to calculate the apparent distances by triangulation as seen by an observer in a potential well to other distant objects, for example in approximately the radial outward direction and tangential directions. I'm assuming that these objects are very distant, and I'm only looking for the overall scale factor that gets applied to such distances, as compared with an observer at the same distance from the distant object but outside a gravitational potential well. However, so far I'm not making much progress, in that I don't have enough symmetry to make it easy, and even my first trigonometric approximations seem to be metric-dependent.

Does anyone know of any standard results in this area, or have any ideas for any reasonably simple approach that might work?

2. Sep 17, 2008

### Mentz114

If we call the two observers on the baseline b1 and b2 and the distant object O, then one can work out the angle that a light ray ( null geodesic) from O to either b makes with the line joining the b's. Assuming a plane triangle the third angle can be got by subtraction. I think I can do this because I've got the required null geodesic ( or at least its tangent vector, which will do ) somewhere for the case where all the points are in the equatorial plane. I'll try and do the calculation when I have more time.

M

3. Sep 17, 2008

### Jonathan Scott

Thanks.

I already tried making use of the standard formula for the deflection of a tangential light ray passing a star, which if I remember correctly is an angle of 4GM/rc^2 where r is the closest approach, so by symmetry the deflection would be half that for the deflection by the position of an observer at the tangent point. The change in deflection angle for a small change in radius then has an effect on the apparent angle to a distant object. In addition, I tried to take into account the fact that rulers in the radial and tangential directions vary compared with the coordinate system used for the metric. However, I got different results for Schwarzschild and isotropic coordinates, which means my calculations must have been wrong.

Incidentally, my previous statement that the apparent distance does not depend "on the metric" was potentially misleading; I meant it doesn't depend on the coordinate system used to define the metric.