Crossing the event horizon of a black hole

[Hurkyl]

Those claiming there is no singularity at the event horizon should be asked the following question:

If your head was inside the event horizon and your feet were outside, could you wiggle your toes?

Yes -- but the event horizon will have passed by your feet before the electrical impulse from your brain would have reached them.

Unless you meant that your feet were undergoing sufficiently strong acceleration to remain outside of the event horizon -- in which case they would have already been ripped off of your body.

Yes, there are coodinate transformations of the interior metric that make it look rather normal, but the coordinates do not join up smoothly with those we are coming from on the outside.

Thus the term "coordinate singularity": you've merely given one explanation of why Schwarzschild 'coordinates' are unsuitable for talking about the event horizon.

 

[Hurkyl]

Yes, it is absurd, but not stupid. It may be absurd AND correct.

Since your previous quotation is misleading, allow me to recall that this is all in reference to aranoff's claim:

Therefore, the event horizon is geometrically the same as a straight line.​

The event horizon, for example, is not one-dimensional, so it's clearly not correct.

(Incidentally, if the statement was correct, then it would not be absurd)

 

[PaulDent]

Sure, the signal sent from your brain would reach your toes, and assuming your head is falling into the event horizon after the toes (rather than being kept at a constant distance above the horizon by a rocket or something, which would mean your body would get ripped apart) then sensory signals from the toes could get back to the brain too. The only events on the worldline of the brain that do not have events on the worldline of the toes in their future light cone are events from when the brain is so close to hitting the singularity that there is no time for light signals from the brain to reach the toes before they, too, hit the singularity.

I think you must be talking about an observer in free fall into the BH.

However, consider a BH of mass 1.237e43 Kg, which is still only about 1e-15 the mass of our universe. The gravitational acceleration at the event horizon is c^4/GM, which you will find is our regular 9.81 m/s^2 with the above mass. That means hovering above it with rocket power is no big trick. It would be exactly the same as hovering above the surface of a lake or ocean on earth.
Now I can decide to descend slowly headfirst and dunk my head into the event horizon, and the rockets on my feet hold me there without catastrophic stretching. Now can I wiggle my toes, given that no signal can cross the event horizon in the direction inside to outside?

 

[PaulDent]

Since your previous quotation is misleading, allow me to recall that this is all in reference to aranoff's claim:

Therefore, the event horizon is geometrically the same as a straight line.​

The event horizon, for example, is not one-dimensional, so it's clearly not correct.

(Incidentally, if the statement was correct, then it would not be absurd)

What I was saying is that, from a certain perspective, the event horizon can appear one-dimensional; if you imagine its area appearing to be zero (instead of a sphere) while its thickness or depth appears to be infinite instead of zero, then it morphs to a straight line in the radial direction. Aranoff is not alone in coming to the conclusion that it might appear to look like this from some perspective. I imagine it might TEND to appear that way as you approached the event horizon closer and closer.

But these things are hard to imagine, and my reason for corresponding here is to find help in getting my head round it, so thank you for your take on it.

 

[JesseM]

However, consider a BH of mass 1.237e43 Kg, which is still only about 1e-15 the mass of our universe. The gravitational acceleration at the event horizon is c^4/GM, which you will find is our regular 9.81 m/s^2 with the above mass.

Where did you get that formula for gravitational acceleration? Assuming it's correct, I would guess that it's talking about something like the coordinate acceleration for a freefalling observer in some specific coordinate system like Schwarzschild coordinates. The proper acceleration needed to maintain a constant height above the event horizon (i.e. the G-force felt by someone who is maintaining that height) always approaches infinity as you approach the event horizon--see pervect's post #4 on this thread.

Now I can decide to descend slowly headfirst and dunk my head into the event horizon, and the rockets on my feet hold me there without catastrophic stretching. Now can I wiggle my toes, given that no signal can cross the event horizon in the direction inside to outside?

You most definitely will experience catastrophic stretching if you try to keep your feet at a constant height above the horizon while your head is below it.

 

[PaulDent]

No, this is just an artifact of the way Schwarzschild coordinates work, you're free to use a coordinate system where there is no "switch" and the time coordinate continues to be physically timelike inside the event horizon, such as Kruskal-Szekeres coordinates.

I am just getting round to taking another look at the Kruskal extension. The last time I looked, I had many problems with it, at least the way it was presented, as it had quantities under square root signs that became negative. And that wasn't the only problem. It basically transforms only the r and t terms, and leaves the terms in theta and phi as is to get "Mixed Edington-Finkelstein coordinates". However that does not allow construction of a 4d metric that I can recognize as supporting our physics. To be able to unequivocally say that a metric represents a universe like ours, it should be possible to get it in the form

a^2(t)(dx^2 + dy^2 + dz^2) - (cdt)^2

i.e. Minkowski with some tolerable rate of expansion or contraction

 

[PaulDent]

You most definitely will experience catastrophic stretching if you try to keep your feet at a constant height above the horizon while your head is below it.

Then it seems you agree that there IS a singularity at the event horizon, at least to an observer who is not in free fall. And the catastrophic stretching is nothing to do with the rocket thrust, as in my example, that would only have to be g=9.81m/s^2, as here on earth.

 

[PaulDent]

Where did you get that formula for gravitational acceleration? Assuming it's correct, I would guess that it's talking about something like the coordinate acceleration for a freefalling observer in some specific coordinate system like Schwarzschild coordinates. The proper acceleration needed to maintain a constant height above the event horizon (i.e. the G-force felt by someone who is maintaining that height) always approaches infinity as you approach the event horizon--see pervect's post #4 on this thread.

gravitational pull = GM/r^2 at static radius r outside the event horizon

substitute schwartzschild radius r = GM/c^2

and you get c^4/GM as the gravitational pull at the event horizon.

For modest BHs (small M) this is enormous. But for ginormous M, it becomes tolerable.

 

[JesseM]

gravitational pull = GM/r^2 at static radius r outside the event horizon

But that's a Newtonian formula--you can't expect it to work in cases where GR diverges significantly from Newtonian gravity, and the vicinity of a black hole is definitely one of these cases! As I said, see pervect's post #4 here for the actual formula for gravitational acceleration near the event horizon, calculated from GR, which does go to infinity as you approach the horizon.

 

[JesseM]

You most definitely will experience catastrophic stretching if you try to keep your feet at a constant height above the horizon while your head is below it.

Then it seems you agree that there IS a singularity at the event horizon, at least to an observer who is not in free fall. And the catastrophic stretching is nothing to do with the rocket thrust, as in my example, that would only have to be g=9.81m/s^2, as here on earth.

Again, you're mistaken in using Newtonian formulas here, the proper acceleration definitely does have to do with rocket thrust, and the thrust needed to maintain a given height goes to infinity as you approach zero height above the horizon. I don't think most physicists would consider this a physical singularity though, as it is only a theoretical requirement that reaches infinity at the horizon rather than any quantity actually measured by anyone (there are no actual objects maintaining zero height above the horizon)--as an analogy, the theoretical requirement "energy required to increase one's velocity by 0.1c" also goes to infinity at 0.9c, but no one would say that means there's a physical singularity at 0.9c.

 

[stevebd1]

I think PaulDent might be confusing gravity with gravity gradient which, as far as I know, remains unaffected by Schwarzschild coordinates (or any kind of metric), the gravity gradient being a measure of how much gravity increases over 1 metre and in the case of stellar black holes, results in 'spaghettification' at the event horizon. The equation for gravity gradient (or tidal forces) being-

$$\frac{2Gm}{r^3}$$

where the results are in m/s^2/m. Multiply the answer by 2 to get an approx figure for the change in gravity from head to foot. This is why if you plan to cross the event horizon, a large black hole is favoured over a small one (from looking at various sources, the most a human can endure is a gradient of 15 Earth g's from head to toe).

 

[JesseM]

$$\frac{2Gm}{r^3}$$

This is also a Newtonian equation--as I said, you can't assume Newtonian equations are still correct in GR, GR is a very different theory of gravity which only looks like Newtonian gravity in certain limits.

 

[George Jones]

This is also a Newtonian equation

Actually,

$$\frac{2Gm}{r^3}L,$$

where $L$ is the small radial spatial separation between two test masses, is both the Newtonian and general relativistic expression for tidal acceleration outside a spherically symmetric mass.

 

[George Jones]

I am just getting round to taking another look at the Kruskal extension. The last time I looked, I had many problems with it, at least the way it was presented, as it had quantities under square root signs that became negative. And that wasn't the only problem. It basically transforms only the r and t terms, and leaves the terms in theta and phi as is to get "Mixed Edington-Finkelstein coordinates".

I think have misinterpreted Kruskal-Szekeres coordinates.

However that does not allow construction of a 4d metric that I can recognize as supporting our physics. To be able to unequivocally say that a metric represents a universe like ours, it should be possible to get it in the form

a^2(t)(dx^2 + dy^2 + dz^2) - (cdt)^2

i.e. Minkowski with some tolerable rate of expansion or contraction

It's not supposed to look like a flat space Friedmann-Robertson-Walker spacetime.

 

[stevebd1]

Hi JesseM.

While I understand your argument that Newtonian equations cannot be assumed to be correct beyond the event horizon, I have seen stated a number of times 'While Newton's equation for gravity, g=Gm/r^2, is affected by coordinate acceleration, diverging at the event horizon, the Newtonian equation for tidal forces remains unchanged up to and possibly beyond the event horizon'. Another equation I've seen for calculating the gravity gradient at the event horizon is-

$$\frac{c^6\ dr}{4 (Gm)^2}$$

which looks like it's derived from the Newtonian equation and the Schwarzschild radius.

 

[JesseM]

Hi JesseM.

While I understand your argument that Newtonian equations cannot be assumed to be correct beyond the event horizon, I have seen stated a number of times 'While Newton's equation for gravity, g=Gm/r^2, is affected by coordinate acceleration, diverging at the event horizon, the Newtonian equation for tidal forces remains unchanged up to and possibly beyond the event horizon'.

My apologies, I jumped to conclusions there--since I knew the equation worked in Newtonian physics I thought you were just assuming it would work in GR (as PaulDent was doing in his gravitational acceleration calculation), but since you say you've seen it specifically stated that this equation works in GR (and George Jones backs this up) I'll take your word for it that it does.

 

[George Jones]

'the Newtonian equation for tidal forces remains unchanged up to and possibly beyond the event horizon'.

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.

 

[stevebd1]

I have heard r in relation to black holes being referred to as the 'reduced circumference' or the 'coordinate radius' but even then, they don't give the true distance to the singularity. One other definition I've seen in relation to Schwarzschild black holes is the free-float horizon to crunch distance which is expressed as $\tau_{max}\text{[metres]}=\pi M$ where M is the gravitational radius ($Gm/c^2$) which implies the 'true' distance to the singularity is marginally longer than coordinate length of 2M which establishes the event horizon.

 

[George Jones]

I have heard r in relation to black holes being referred to as the 'reduced circumference' or the 'coordinate radius'

Only outside the event horizon. Inside the event horizon, $r$ is a timelike coordinate.

but even then, they don't give the true distance to the singularity.

I don't think a "true distance" to the singularity is definable.

One other definition I've seen in relation to Schwarzschild black holes is the free-float horizon to crunch distance which is expressed as $\tau_{max}\text{[metres]}=\pi M$

I think that this is the proper time taken for an observer to fall from "rest" just above the event horizon to the singularity.

 

[stevebd1]

I had look in 'Exploring Black Holes' by Edwin Taylor & John Wheeler and they give 2 quantities in relation to the fall-in time from the event horizon. The one mentioned above which when divided by c is the maximum free-float horizon to crunch wristwatch time (basically from rest at the EH) and the other which is falling from rest at infinity. Assuming in that case that you're accelerating with space and space itself approaches c at the event horizon, your fall-in time from the event horizon would be reduced to -

$$\tau\text{[seconds]}=\frac{4M}{3c}=\frac{4Gm}{3c^3}\equiv\ 6.568\times10^{-6}\ \times\ \text{sol mass}$$

$$\tau\text{[metres]}=\frac{4}{3}M$$

where M is the gravitational radius.

I'm assuming this goes some way to demonstrating the timelike properties of r inside the event horizon.

For clarity, here's the maximum fall-in time (from rest at the EH)-

$$\tau_{max}\text{[seconds]}=\frac{\pi M}{c}=\frac{\pi Gm}{c^3}\ \equiv\ 1.548\times10^{-5}\ \times\ \text{sol mass}$$

$$\tau_{max}\text{[metres]}=\pi M$$

In both cases, the distance seems relative to how you cross the event horizon.

 

[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