Crossing the event horizon of a black hole

[stevebd1]

Yes and no. The expression does hold inside the event horizon, but, inside the event horizon, r has a different meaning than it has outside the event horizon.

Could you possibly explain or demonstrate the implications this has on the equation for the gravity gradient inside the event horizon.

 

[George Jones]

Could you possibly explain or demonstrate the implications this has on the equation for the gravity gradient inside the event horizon.

Once inside the event horizon, r is a timelike coodinate, and passage from the past to the futures implies going from a larger r to a smaller r. This and the expression for tidal force imply that inside the event horizon, tidal forces steadily increase for all observers and eventually become unbounded.

 

[stevebd1]

Thanks for the reply George

While I understand the implications of r becoming a timelike coordinate beyond the event horizon (once past the event horizon, the singularity is in everybody's future), could you elaborate on the last sentence in your post, particularly the part about tidal forces becoming unbound-

This and the expression for tidal force imply that inside the event horizon, tidal forces steadily increase for all observers and eventually become unbounded.

 

[George Jones]

While I understand the implications of r becoming a timelike coordinate beyond the event horizon (once past the event horizon, the singularity is in everybody's future), could you elaborate on the last sentence in your post, particularly the part about tidal forces becoming unbound-

Since r is timelike with smaller r pointing to the future, r must decrease along the worldline of any observer inside the event horizon. Consequently, along the worldline of any observer, 1/r^3 increases, tidal force increases. As r \rightarrow 0 (which it must, since r must decrease), tidal force becomes unbounded.

 

[PaulDent]

I think I missed a 2 in my original post. The Schwartzschild radius is 2GM/c^2.

I do believe the Newtonian formula for Gravitational pull applies down to the event horizon, i.e. acceleration due to gravity = GM/r^2 still. Substitute r=Schwartzschild radius and you get the acceleration at the event horizon. This can't go to infinity if the gravity gradient remains finite - they either both go to infinity or neither does.

 

[George Jones]

I think I missed a 2 in my original post. The Schwartzschild radius is 2GM/c^2.

I do believe the Newtonian formula for Gravitational pull applies down to the event horizon, i.e. acceleration due to gravity = GM/r^2 still. Substitute r=Schwartzschild radius and you get the acceleration at the event horizon.

For which observer?

=PaulDent]This can't go to infinity if the gravity gradient remains finite - they either both go to infinity or neither does.

Again, for which observer?

 

[PaulDent]

For which observer?


Again, for which observer?

I am a bit weak on that question! But it is a stationary observer outside the event horizon, and I can't decide if it is the acceleration that a distant observer decides is being experienced at the event horizon, or if it is the oberver at the event horizon that experiences that acceleration.

 

[stevebd1]

I do believe the Newtonian formula for Gravitational pull applies down to the event horizon, i.e. acceleration due to gravity = GM/r^2 still. Substitute r=Schwartzschild radius and you get the acceleration at the event horizon. This can't go to infinity if the gravity gradient remains finite - they either both go to infinity or neither does.

Basically, what you’re saying is, is why is the Newtonian equation for gravity (g=Gm/r^2) affected by coordinate acceleration (dr'=dr(1-Rs/r)^-1/2, curved space) while the Newtonian equation for tidal forces (dg=2Gm/r^3) remains unchanged as if dealing with flat space. The way I see it is that in close proximity to the event horizon, the curvature of space increases until the velocity induced by the curve is c at the event horizon, an acceleration experienced by the local observer, (for the observer at infinity, the local observer will appear to slow down and freeze at the event horizon due to the curvature having the opposite effect on light sent from near the event horizon, redshifting into infrared and eventually into long wave radio waves). For the local observer, the rate of increase in Newtonian gravitational acceleration remains constant but appears to increase to infinity because the space it's in is curving towards the horizon (hence why gravity is multiplied by coordinate increase to represent proper acceleration). At the event horizon, gravity will appear to diverge when in fact, it's going about business as usual, increasing steadily over r and it is the space it's in that rapidly accelerates (notice how at the event horizon it's the coordinates that diverge and not the gravity) so the equation for tidal forces (the rate that gravity increases) remains correct and unchanged. While this isn't a mathematical answer and there's probably a better way of explaining it, this is a mental picture I have of the situation.

Steve

 

[PaulDent]

Basically, what you’re saying is, is why is the Newtonian equation for gravity (g=Gm/r^2) affected by coordinate acceleration (dr'=dr(1-Rs/r)^-1/2, curved space) while the Newtonian equation for tidal forces (dg=2Gm/r^3) remains unchanged as if dealing with flat space.

What about the force on an object near the event horizon though?

Imagine a huge black hole, with low gravity gradient at the event horizon. I use a rocket to hover just outside the event horizon and use a simple pendulum to measure g at my feet and g at my head. If I did that at different heights above the event horizon, getting

(ghead(1),gfeet(1)) 10000km away, (ghead(2),gfeet(2)) 9000km away...(ghead(10),gfeet(10)) 1000km away,

then I can plot the gravity gradient at 1000km steps down towards the thorizon and then integrate that curve to get the value of g versus distance.

I begin to see a problem with coordinates though. WHO says I am 1000, 2000...10000km from the event horizon? Different observers will come up with different views of that. And is that distance dr what I multiply by gravity gradient in the integration?

In fact, does the event horizon appear to me to recede as I approach it, so that I experience an infinite number of 1000km steos to get there, and my g-integral goes to infinity?

 

[stevebd1]

If you take into account the velocity induced by the curvature (basically v_{rel}=-\sqrt{r_s/r}\ c, zero at infinity, c at the event horizon) then dr remains 1 as in some alternative metrics for static black holes. In this case, the surface gravity is derived from Killing* vectors, which equals-

\kappa=\frac{c^4}{4GM}

but the above only applies when free-fall velocity into the black hole is accounted for and used in conjunction with a global rain frame metric such as http://en.wikipedia.org/wiki/Gullstrand-Painlevé_coordinates" .

An extract from wiki-

'In relativity, the Newtonian concept of acceleration turns out not to be clear cut. For a black hole, which can only be truly treated relativistically, one cannot define a surface gravity as the acceleration experienced by a test body at the object's surface. This is because the acceleration of a test body at the event horizon of a black hole turns out to be infinite in relativity.

Therefore, when one talks about the surface gravity of a black hole, one is defining a notion that behaves analogously to the Newtonian surface gravity, but is not the same thing. In fact, the surface gravity of a general black hole is not well defined. However, one can define the surface gravity for a black hole whose event horizon is a Killing horizon.

The surface gravity κ of a static Killing horizon is the acceleration, as exerted at infinity, needed to keep an object at the horizon.'

Source- http://en.wikipedia.org/wiki/Surface_gravity"


A Hawking Radiation calculator tool which also calculates the Killing surface gravity-
http://xaonon.dyndns.org/hawking/


*Named after mathematician Wilhelm Killing