Re: Schwarzschild vs Isotropic Coordinates

carlip-nospam@xxxxxxxxxxxxxxxxxxx wrote:
> Murat Ozer <Murat.H.Ozer@xxxxxxxxx> 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 or the Isotropic coordinates?
>
> You can use either. But you have to be careful that the questions
> you are asking are questions about genuine observables -- that is,
> coordinate-independent quantities. For example: what, exactly, do
> you mean by "impact parameter b"? Do you mean minimum *proper*
> distance from the surface? That's fine. But if you mean "minimum
> value of some radial coordinate," you have to specify which radial
> coordinate you mean.
>
> > This is a relavent question because the equation, the initial
> > conditions, etc. for the trajectory are different in different
> > coordinates.
>
> The equations are. The initial conditions are, too, but only if
> you are careless enough to describe them in terms of a particular
> set of coordinates (in which case, you'd better use those same
> coordinates for the calculation).
>
> > Hence, the distance from the center of the star
> > to the particle at any moment will be different in different
> > coordinates.
>
> "Distance" as measured how? "Any moment" as determined by what
> clock? A particle follows a unique trajectory, which has different
> *descriptions* in different coordinate systems.
>
> (If I want to plot a course from Los Angeles to Shanghai, should I
> use a map based on a Mercator projection or a polar projection? The
> answer is that I can use either one, but that in either case I have
> to recognize that they are maps, with distortions that need to be
> taken into account.)
>
> You might want to look at Boddener and Will, Am. J. Phys. 71 (2003)
> 770, "Deflection of light to second order: A tool for illustrating
> principles of general relativity," which discusses almost exactly
> your question, with detailed computations in Schwarzschild and
> isotropic (and also harmonic) coordinates and a careful explanation
> of their relation and physical equivalence.
>
> Steve Carlip

As everybody will, hopefully, agree with me, it's best to argue on a
realistic case. Suppose a spaceship is approaching the Earth from the
North towards the South. The astronauts determine, at a time of their
choice, the initial impact parameter b (the vertical distance from the
ship to the North axis), and the distance d (the vertical distance from
the ship to the West axis) by sending radar signals to space stations
whose positions are known (Note that the center of the East, North,
West, South axes coincide with the center of the earth.) This way the
initial position of the ship from the center of the earth and the
initial angle Phi can be determined (Phi is 0, and Pi/2 on the East
and North axes, respectively):

position_i = Sqrt(b^2 + d^2); Phi_i = Pi/2 + ArcTan(b/d)

Assume, also, that the astronauts determine the ship's velocity
relative to the earth at the same time, call it v_i.
I will try to determine the position of this spaceship when it just
reaches the South axis at which time the angle Phi = 3Pi/2.
The equations of motion in the Schwarzschild, the Isotropic, and the
Special Relativistic coordinates, respectively, are

u'' + u - m_g/h^2 -3m_g u^2 =0 (Schwarzschild.),

u'' + u - m_g/2*(1+m_g/2*u/h^2*[3(1+m_g/2*u)^2*(1-m_g/2*u)^(-2)
+(1+m_g/2*u)^3*(1-m_g/2*u)^(-3) -2]=0 (Isotropic.),

u'' + u - m_g/h^2 - m_g/h^2*u =0 (Special Relativistic).

Here u = 1/r, m_g = GM/(c^2r) with G and M being the gravitational
constant and the mass of the earth, u'=du/dPhi, and h = r^2 dPhi/ds
is a constant of the motion and can be expressed in terms of the
ordinary angular momentum L as follows:

h = L/mc*( 1-2m_g *u)^(-1) (Schwarzschild)

h = L/mc*(1+m_g/2*u)^6*(1-m_g/2*u)^(-2) (Isotropic)

h = L/mc*(1+m_g*u) (Special Relativistic)
, where m is the mass of the spaceship and cancels out in the
equations.

In the numerical examples that I will present below, the ratio m_g/R_E,
where R_E is the radius of the earth, is much less than 1 so that the
spacetime is very close to being flat in the general relativistic
cases. This enables one to derive the expression for L:

L^2 = m^2v^2/(u^2 + u'^2).

Next, one can write a, say Mathematica , program to numerically solve
for r_{final} = 1/u_{final} when Phi_{final} = 3Pi/2. (Note that I have
been using the same symbols u and r for all the coordinate systems.
Since I treat them separately there is no confusion. ) Here are some
examples:

R |b/R |d/R | v_i(m/s) |Schw./R |Iso./R | Spec.Rel./R |
_________________________________________________________________
R_E | 5 |20 | 10^4 | 20.3335 | 20.3335 | 20.3335 |
| 5 |20 | 10^5 | 2034.29 | 2034.29 | 2034.29 |
| 5 |20 | 10^6 | 213272 | 213277 | 213277 |
_________________________________________________________________
0.5m | 5 |20 | 10^7 | 1.58306 | 1.58049 | 1.5911 |
| 5 |20 | 5x10^7 | 37.5351 | 39.775 | 39.8371 |
| 5 |20 | 10^8 | 129.444 | 159.146 | 159.375 |

In the second set of values, I consider a very compact object, but not
a blackhole, with radius half a meter and mass that of the earth. The
last three columns are the coordinates of the spaceship in the relevant
coordinates in terms of R, the radius of the massive object. These
numbers show very clearly that there is a very strong speed dependence
and for a small but very heavy object the predictions of the
Schwarzschild and Isotropic coordinates may be very different. In all
these cases though, the spacetime is very close to being flat, and
therefore I believe the isotropic coordinates' predictions are the
actual values that the astronauts may measure at Phi_{final} = 3Pi/2
because they are very nearly the same as the (flat) Special
Realativistic predictions.

Murat Ozer

.

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