'Poking' the Event Horizon

1. Mar 29, 2009

spikenigma

'Poking' the Event Horizon

Forgive what is probably a ridiculous scenario, I was having a discussion earlier

We have Black Hole A and Black Hole B of completely (or near) equal gravitational field strength.

http://img148.imageshack.us/my.php?image=blackhole1.gif

A civilization has developed a magnetically re-enforced pole.

The pole is made up of smaller segments which produce a magnetic/dimagnetic field of arbitrary strength to hold it together and directly counteract gravity/gravitational distortions/stresses

The energy requirements of each segment can be easily met by the civilization at any time as needed - as long as it is not infinity.

The civilization extends the pole towards Black Hole A and Black Hole B (carefully)

http://img530.imageshack.us/my.php?image=blackhole2.gif

The pole extends past the event horizon of each Black Hole

http://img530.imageshack.us/my.php?image=blackhole3.gif

assume time dilation is not an issue to the experimenters

questions:

* what happens?
* can the civilization theoretically receive information at speed of sound from the pole about the inside of the event horizon?

2. Mar 29, 2009

Staff: Mentor

The pole breaks apart.

3. Mar 29, 2009

skeptic2

Space is increasingly contracted radially as the even horizon is approached. At the event horizon space is contracted to zero. An infinite distance exists between the pole and the event horizon.

4. Mar 29, 2009

Hurkyl

Staff Emeritus
Huh? This makes no sense at all.

5. Mar 30, 2009

skeptic2

Space is contracted by the factor sqrt(1 - Vesc^2 / c^2) where Vesc is the escape velocity. At the event horizon where Vesc = c space is contracted to zero resulting in an infinite distance between any object and the event horizon.

Time dilation is not the only impediment to crossing the event horizon. Just as an infinite amount of time must pass before an object can cross the event horizon, an infinite distance must also be crossed.

6. Mar 30, 2009

Ich

A singularity in the derivative does not necessarily mean a singularity of the function (proper distance as a function of "r") itself. In fact, the distance is finite.

7. Mar 30, 2009

stevebd1

Hi Skeptic2

$dr'=dr\sqrt(1-v^2/c^2)$ only applies to the free-falling object that is falling at $v=-\sqrt(2M/r)\,c$ towards the BH, this is the process that means dr always equals 1 for the object free-falling from infinity (as the length contraction due to velocity (SR) cancels out the length expansion, $dr'=dr\sqrt(1-2M/r)^{-1}$ brought on by gravity (GR)). Only the static observer deep within the BH's gravitation field comes close to experiencing the 'infinite' curvature near the horizon. For the static observer, the distance to the EH is-

$$\Delta r'=\frac{\Delta r}{\sqrt{1-\frac{2M}{2M+\Delta r}}}$$

where Δr is the coordinate distance to the EH (distance as observed from infinity) and Δr' is the proper distance to the EH (distance as observed locally from the object within the gravitational field) and M=Gm/c2.

As it's already been pointed out, even without massive tidal forces near the event horizon, the pole would break simply due to the space-like geodesics that reside on the inside of the event horizon (i.e. no object can remain static at a coordinate distance r inside the EH).

Last edited: Mar 30, 2009
8. Mar 30, 2009

Ich

The exact "proper distance" is, without approximation:
$$S=r_2\sqrt{1-\frac{2M}{r_2}}-r_1\sqrt{1-\frac{2M}{r_1}}+M\ln{\frac{r_2-M+r_2\sqrt{1-\frac{2M}{r_2}}}{r_1-M+r_1\sqrt{1-\frac{2M}{r_1}}}}$$
(I think), which is finite even for r1=2M.