Gargantua violates the EP?
Jorrie said:
Let's use Kip Thorne's "Gargantua" from his "Black holes and time warps", sporting 15 trillion solar masses and set up the lab at 1.0001 Rs (like Kip), where the g-force is still 10g, but the tidal forces are negligible. The horizon circumference is 29 light years, so our test arena size should also be negligible by comparison. Since Gargantua is spinning very slowly, the spacetime around it is almost perfectly Schwarzschild and pervect's equations should be valid in Schwarzschild coordinates, i.e. by an observer at 'infinity'.
With the event horizon practically flat underneath us and the gravitational field practically homogeneous, set up Cartesian coordinates with the x-axis normal to the radial and the y-axis parallel to it. We can now determine the instantaneous Cartesian acceleration when a ball is dropped without and with horizontal velocity (must be far below the orbital velocity, which is v_o= 0.707c in this case). Corrective note: the v_o = 0.707c is an error; there are no orbits at r = 1.0001Rs. One can use any local velocity < c.
...
Eq. 5
<br />
\frac{d^2 r}{d\tau^2} = -\frac {m}{r^2(1-2m/r)^{1.5}}\left(1 - \frac {2m}{r} + 2r^2v_\phi^2\right) <br />
Since nobody raised an issue on this equation (based on pervect's derived radial acceleration), I did put in some numbers and was quite surprised! Firstly, I rewrote it in local Cartesian (x,y) coordinates (as defined in the above quote):
Eq. 6
<br />
<br />
\frac{d^2 y}{d\tau^2} = -\frac {m}{y^2(1-2m/y)^{1.5}}\left(1 - \frac {2m}{y} + 2(1-2m/y)v_x^2\right) <br />
<br />
The last term is multiplied by (1-2m/y) to convert the local horizontal velocity (v_x) back to the Schwarzschild tangential velocity rv_\phi required by the quoted equation 5.
This acceleration was numerically integrated over one second, starting at y=1.0001Rs, first for a static particle (v_x = 0) being dropped and then for Kev's particle with v_x = 0.8c (horizontal) velocity, over the same drop in the static x,y frame. I found that the first particle dropped a vertically height of 51 meters in the one second, at a practically constant acceleration of -101.717 m/s^2 (~10g, as Kip used). It obviously reached a speed (v_y) of -101.717 m/s in the one second, not relativistic at all.
The particle moving at v_x = 0.8c dropped the 51 meters in 0.66 seconds, also at a practically constant acceleration of -231.916 m/s^2, reaching a vertical speed (v_y) of -153.06 m/s, still not relativistic. The significance of this is that one can ignore the effect of v_y on dtau and on the vertical acceleration. This particle traveled a horizontal distance of some 158,400 km (0.53 ls), insignificant relative to the 29 ly circumference black hole. We are still in an insignificantly small region relative to the hole (practically infinitesimal).
So, if these calculations are correct, Galileo was wrong - a horizontally moving particle does fall faster than the particle dropped from rest. Before everyone jumps in and flame the calculations, consider this: the test is performed at r=1.0001Rs, far inside the "photon radius", where no stable orbits are possible and any increase in tangential velocity (kinetic energy) drags the object faster into the hole. Qualitatively, there is no problem here - horizontal velocity makes things fall faster in that region and it becomes more severe the closer to the event horizon, where even light's trajectory becomes purely vertical.
Quantitatively, doing the experiment at the photon radius r = 3m with v_x = 1 indeed gives a circular orbit with a gravitational acceleration of -2.349 m/s^2. At r=4m, using v_x = 0.707, the orbit is circular with a gravitational acceleration of -0.82 m/s^2, corresponding to the local centripetal acceleration for a circular orbit at that radius (ca = v_x^2/(r-2). I also found that the increase factor in the acceleration did not change if I increased the mass of black hole without limit. Sobering…
This result is probably controversial and may be wrong, but I'll be very interested to get comments on the validity of the equations for the scenario described, rather than just that it violates the equivalence principle.
-J
