Re: Schwarzschild vs Isotropic Coordinates
- From: tessel@xxxxxx
- Date: Thu, 29 Sep 2005 02:15:53 +0000 (UTC)
On Tue, 27 Sep 2005, Murat Ozer wrote:
Suppose that I want to study the trajectory of a test particle that is acted upon by a massive spherical object, say a star. Suppose the test particle comes in from infinity at an initial impact parameter b and is deflected by the gravitational attraction of the star. Assuming that no other forces act on the particle,the equation for the trajectory can be obtained from the geodesic equation sraightforwardly. My question is, which coordinates for the Schwarzschild metric should one use, the Schwarzschild coordinates
ds^2 = -(1-2m/r) dt^2 + dr^2/(1-2m/r) + r^2 (du^2 + sin(u)^2 dv^2),
-infty < t < infty, 2m < r < infty, 0 < u < pi, -pi < v < pi
or the Isotropic coordinates?
ds^2 = -(1-M/2/R)^2/(1+M/2/R)^2 dt^2
+ (1+M/2/R)^4 [dR^2 + R^2 (du^2 + sin(u)^2 dv^2)],
-infty < t < infty, M/2 < R < infty, 0 < u < pi, -pi < v < pi
(where the coordinate ranges indicate the exterior region with approximately a half plane removed, as in a local coordinate patch for ordinary spherical coordinates).
The answer to your question depends on what you are trying to compute. If you do everything correctly you will of course always get the -same- answer for any physically meaningful quantity (compared at the same event!), but one or the other may be more -convenient- for a given purpose.
For example, the Schwarzschild radial coordinate r has a simple geometric meaning: 4 pi r0^2 is the surface area of the surface t=t0, r=r0, happens to be a geometric sphere with Gaussian curvature 1/r0^2. This is often very useful to know.
OTH, the polar spherical isotropic coordinates may be more convenient for studying "strong lensing" since the angles of null geodesics coming into an event are correctly represented, while in the Schwarzschild chart, the light cones do not appear to be "circular" cones, but rather are flattened. This is because in the Schwarzschild chart, "transverse" but not radial distances are not faithfully represented.
(Actually, for strong lensing the Costa chart is even better, but the isotropic chart has still other uses.)
Another possible consideration in studying geodesics is that the geodesic equations in one or the other chart may be -integrable in closed form- for particular "special" values of the parameters. This is in fact the case for tracks of null geodesics in the Schwarzschild chart.
This is a relavent question because the equation, the initial conditions, etc. for the trajectory are different in different coordinates. Hence, the distance from the center of the star to the particle at any moment will be different in different coordinates.
Ah. We've discussed this issue before in this group, but this is a good question since (rather amazingly) I haven't seen many problems in gtr textbooks which ask the student to compute two quantities (in this case lengths) in two charts and checking they give the same answer. This is unfortunate. In particular, in this example, the computation you are asking about raises at least two interesting conceptual issues.
But first, let's set the stage. For concreteness, I will assume you are now talking about a stellar model in which a Schwarzschild fluid (or some other static spherically symmetric perfect fluid model) is matched across a surface of zero pressure (but positive matter density) to an exterior Schwarzschild vacuum region. In this case, "distance to the center" would make sense if understood as the length of a spacelike geodesic arc everywhere orthogonal to the timelike Killing vector @/@t (which is vorcity free, since the solution is static, hence hypersurface orthogonal, and the geodesic is just a "radius" in some spatial hyperslice t=t0). However, our discussion would also work for distance between any two static observers outside the event horizon of a Schwarzschild hole.
In fact, it is probably easier to consider a closely related problem, determining the distance along a "radial arc" (in a slice orthogonal to static observers) from the surface of the star to the world line of one of the static observers, since if you really were interested in the stellar model you should find it easy to find the radial distance between the (very hot) static observer at the center of the star and a static observer on the surface.
Consider first the situation represented in the Schwarzchild chart. Say the surface of the star (zero pressure sphere) occurs at r = r1, and that our static observer's world line is
r = r2,u=Pi/2, v =0
where of course r2 > r1 > 2m. (Actually, the surface of any compact object would have to lie somewhat outside r = 2m by Buchdahl's theorem, but never mind that.)
To find the length of the spacelike arc (which is in fact a geodesic arc, an integral curve of the spacelike geodesic vector field sqrt(1-2m/r) @/@r), we need to integrate
ds = dr/sqrt(1-2m/r)
from r1 to r2 > r1. Note that to first order in m, this is
ds = 1+m/r
Plugging in some value, say m=1, into the exact integral, we have some length, which we can convert to meters by plugging in the appropriate factors to convert from geometric units (in which G = c = 1) to kms units.
To compare with the isotropic chart, we proceed similarly, but there are two issues we need to address:
1. Which value of the isotropic radial coordinates R1 < R2 correspond to given values of the Schwarzschild radial coordinates r1 < r2?
2. Which value of the mass parameter M corresponds to a given value of the mass parameter m?
Of course you can anser the first question using the transformation between the two charts (or directly, by considering surface areas of spheres). Then we integrate
ds = (1+M/2/R)^2
from R1 to R2. Note that to first order in M, this is
ds = 1+M/R
which agrees with our previous expression when M = m, and suggests that for weak fields, far from the star the distinction between the two charts is moot (see below).
Now we have expressions in f(r,m) and g(R,M) which should give the same numerical value when we plug in corresponding values for m, M. But we already saw a hint that we should simply identify m,M. Indeed, expanding the Schwarzschild line element to O(1/r) gives
ds^2 = -(1-2m/r) dt^2 + (1+2m/r) dr^2 + r^2 (du^2 + sin(u)^2 dv^2)
and expanding the isotropic line element to O(1/R) gives
ds^2 = -(1-2m/r) dt^2 +
(1+2m/R) [ dR^2 + R^2 (du^2 + sin(u)^2 dv^2) ]
In particular, we see that to O(1/r) and O(1/R)
(1+ m/r) r du ~ (1+M/R) R du
This means that sufficiently far out, R,r essentially coincide, with m,M playing analogous roles. Thus the mass parameters m,M can be identified.
(Even better, you can look up Misner-Sharpe mass in a textbook like Carroll.)
If it were possible to actually measure this distance, which coordinates' predictions would agree with experiment?
Both will, of course, according to gtr.
Here is an explicit example. Say in the Schwarzschild chart we have r1 = 4 and r2 = 8. These correspond to spheres of areas 64 pi and 256 pi respectively. Setting m = 1 and integrating we have
delta s ~ 4.971
Even without working out the transformation to isotropic chart, from A1, A2 we see that r1, r2 correspond to R1 ~ 2.914 and R2 ~ 6.964. Setting M = m = 1 and integrating the expression for radial ds in isotropic chart, we find again
delta s ~ 4.971
That's the beauty of "local diffeomorphism covariance"!
As an exercise, I leave you to consider a model in which two static observers bounce a radar signal off each other and divide the net travel time by two to compute a "light travel time distance". Do you predict that this will give the same result as we just found by laying rulers (in imagination) along a radial spacelike geodesic orthogonal to @/@t? (Call this the "static rulers distance"; unlike the light travel time existence, it clearly only makes sense for hypersurface orthogonal observers). The same result for the inner observer bounding a signal off the outer observer, as for the outer observer bounding a signal off the inner observer? What about two Hagihara observers in stable circular orbits of different radii, both lying in the equatorial plane?
Ambitious students can also think about defining an "optical distance" by measuring through your telescope the visual height of an astronaut (oriented orthogonally to a radial arc) whom you know is exactly 2 metres tall. Does this notion of distance agree with either of the other two?
"T. Essel" (hiding somewhere in cyberspace)
.
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- Schwarzschild vs Isotropic Coordinates
- From: Murat Ozer
- Schwarzschild vs Isotropic Coordinates
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