Can I stick my arm in past the event horizon?

[rootone]

By definition, an object inside the EH cannot by any means get outside of it,
To do that in principle would require a greater than infinite amount of force to be applied to the object, and that of course is nonsensical.

 

[Stephanus]

By definition, an object inside the EH cannot by any means get outside of it,
To do that in principle would require a greater than infinite amount of force to be applied to the object, and that of course is nonsensical.

Sorry, I still can picture or understand that.
Force is not identical with speed, altough closely related. The more force you have the faster you move, right?
So, on Earth a 1 ton rocket must have 9800 Newton force just to stay still.
What about the table that I write above.
The Schwarzshild radius for a 1000 trillion solar mass black hole is 300 thousands light year.
And the force needed for a 1 tons rocket just to to stay still at EH is 30 Newton?
Is that right?

 

[rootone]

An object outside of the event horizon experiences the gravity of the black hole.
If there are no other forces being applied to the object this will result in the object accelerating towards the BH.
However we could in principle attach an extremely powerful engine to the object and apply a force opposite to gravity, so the object can escape, (though probably not in great shape).
The closer we get to the EH, the more powerful the engine has to be to provide enough force to counter the gravity.
At the EH itself the amount of force needed goes to infinity.

 

[pervect]

No, it wouldn't. Look at it in a local inertial frame in which the observer hovering just above the horizon is momentarily at rest at the instant his arm reaches out. One limiting case would be the end of the arm freely falling through the horizon, starting at that instant--i.e., the worldline of the end of the arm is just a straight vertical line in a spacetime diagram, while the observer's body follows a hyperbola. In this limiting case, the observer would be unable to pull his arm back after a fairly short length of his own proper time, but the arm would be moving very fast relative to him by that time, so it would, in his new local inertial frame (at the instant where he can no longer pull his arm back), be greatly length contracted.

I was considering the case where the arm was born-rigid, at least in some sort of limit. I can see that I never actually explicitly said that - my bad. I'd agree that if you let the hand free-fall, it will reach the event horizon in a finite amount of (proper) time. But when you let the hand free-fall, while the body is holding station (a static observer), the arm must be stretching.

Part of the solution to the problem is realizing that the arm eventually must stretch, but I wanted to delay the onset of the stretching as much as is physically possible.

The advantage of having a born-rigid arm is that it corresponds naturally to the ruler definition of "distance". We need some operational way of defining what we mean by distance. The radar distance to the event horizon is obviously infinite, but the distance as measured by a ruler - specifically a born-rigid ruler - is finite. At least in the applicable limit, to be precise one would need to say that the limit of the ruler distance as one approaches the event horizon exists and is finite, unlike the radar distance. Of course, one can't actually have the end of the ruler actually reach the event horizon, the only class of objects that can hold station at the event horizon are massless objects such as light.The whole issue of dynamics is a bit of an afterthought, and not all that compatible with the notion of absolute born-rigidity. I don't see any practical problems (other than requiring a proper acceleration of 10^22 gravities, that is) with reaching within 1 micron of the horizon. The time dilation would be a modest million:1 between your body and your idealized pointlike hand, certainly significant but you still wouldn't notice the relativistic effects much if you kept the coordinate velocity of your "hand" below a meter/second.

 

[PeterDonis]

So, for a black hole 1000 trilion times solar mass (still below the mass of the observable universe, right).
Schwarzshild radius is 300 thousands ly. But the gravity acceleration in EH is 3cm/s?
Or am I doing the wrong calculation?

You are doing the wrong calculation. You are evidently using the Newtonian formula, $a = G M / R^2$, to compute this. That formula is not correct; the correct GR formula is

$$
a = \frac{GM}{R^2 \sqrt{1 - \frac{2GM}{c^2 R}}}
$$

According to this formula, $a$ increases without bound as $R \rightarrow 2GM/c^2$, i.e., as the horizon is approached.

What is the force of the rocket must have to stay at EH perpendicular to the singularity?

It is impossible for a rocket, or any object with nonzero rest mass, to remain static at the horizon. Only massless objects, which move at the speed of light, can stay static at the horizon.

 

[PeterDonis]

We need some operational way of defining what we mean by distance.

One obvious way of doing this if you want to consider events on both sides of the horizon is to set up a local inertial frame that covers events on both sides, and using "ruler distance" in that frame, i.e., operationally defining "distance" as that measured by rulers at rest in this local inertial frame.

One such frame is easily defined: it's the one in which the observer's body is following a Rindler worldline--a hyperbola of the form $x^2 - t^2 = 1 / a^2$, where $a$ is the observer's proper acceleration. This hyperbola obviously never crosses the horizon (which will just be the line $t = x$ in this frame). The end of the arm, at least in the simplest idealized case that allows a reasonable dynamics other than free fall, will follow a hyperbola with a smaller proper acceleration but the same x intercept; its equation will be of the form $(x - k)^2 - t^2 = 1 / b^2$, where $b < a$ is the arm's proper acceleration and $k = (1 / b) - (1 / a)$. This hyperbola will intersect the horizon at a coordinate time given by the equation $(t - k)^2 = t^2 + 1/b^2$ (just substitute $x = t$ into the equation for the second hyperbola), and the ruler distance will be the difference between this value (which will also be the $x$ value of that event) and the $x$ value of the first hyperbola (the one for the observer's body) at the same $t$. I'll leave the details as an exercise for the reader.

Of course, this definition of "distance" won't be the same as the "ruler distance" in the momentarily comoving inertial frame of either the observer's body or the end of the arm. The former is unusable in this case anyway, since all of the lines of simultaneity in the momentarily comoving inertial frames of the observer's body intersect the horizon at the spacetime origin of the above inertial frame; so they can't be used to describe events anywhere else on the horizon. The latter would work, and could be calculated similarly to the above; just boost by the velocity $v$ of the arm end in the above local inertial frame at the instant the arm end crosses the horizon. This is equivalent to switching to a different local inertial frame, one whose "time axis" is the worldline of a freely falling observer who is momentarily comoving with the arm's end at the instant it crosses the horizon, and using "ruler distance" in that frame.

 

[Stephanus]

You are doing the wrong calculation. You are evidently using the Newtonian formula, $a = G M / R^2$, to compute this. That formula is not correct; the correct GR formula is

$$
a = \frac{GM}{R^2 \sqrt{1 - \frac{2GM}{c^2 R}}}
$$

You again PeterDonis
Thanks for your invaluable help for me all this time.
So, if let R_s = Schwarzshild Radius.
The formula could be written as
$$
a = \frac{GM}{R^2 \sqrt{1 - \frac{R_s}{R}}}
$$
So while R approaching R_s everything changes? Like v approaching c in Lorentz formula?

This question if off topic, but I want to ask it anyway.
So, actually every Newton formula has to be rewritten to be included by relativity formula?
for example
$V_1 + V_2 = (V_1 + V_2) / (1 + \frac{V_1 V_2}{c^2})$.
And if I'm not mistaken this formula was derived somewhere in 1700s and in 1915 it changed.
And this is final? No more change in the future?

 

[wabbit]

Yes there is another change ! De Sitter relativity includes also the effect of a maximum length related to the cosmological constant, and provides the generalization to de Sitter spacetime (empty with a cosmological constant) of special relativity in Minkowski spacetime. I've been looking for an easy treatment of this so I could understand it but no luck so far : )

 

[PeterDonis]

if let Rs = Schwarzshild Radius.
The formula could be written as

a=GMR21−RsR−−−−−−√​

a = \frac{GM}{R^2 \sqrt{1 - \frac{R_s}{R}}}
So while R approaching Rs everything changes?

Yes.

So, actually every Newton formula has to be rewritten to be included by relativity formula?
for example
$V_1 + V_2 = (V_1 + V_2) / (1 + \frac{V_1 V_2}{c^2})$.

This is a formula from SR, not GR, but in general, yes, you should expect that Newtonian formulas will not be exactly right in relativistic contexts.

And this is final? No more change in the future?

We have no way of knowing that for sure. But the current relativistic formulas are the best ones we have.

 

[pervect]

It's simpler than all that, really. If we consider the metric

-(1+gz)dt^2 + dx^2 + dy^2 + dz^2

We know that $\partial / \partial t$ is a timelike Killing vector, because none of the metric coefficients depend on t. We also know that any killing flow generates a born-rigid flow

http://en.wikipedia.org/w/index.php?title=Killing_vector_field&oldid=627685256

Killing fields are the infinitesimal generators of isometries; that is, flows generated by Killing fields are continuous isometries of the manifold. More simply, the flow generates a symmetry, in the sense that moving each point on an object the same distance in the direction of the Killing vector field will not distort distances on the object.

In other words, distances don't change as we move from the past to the future along the timelike Killing flow, which is just the property that we want for an object to be Born rigid. We want a set of points that maintain a constant distance from our reference observer and also each other. We further presumed our reference observe is a static observer hovering at a constant height above the black hole.

Thus for a static observer, the time-like Killing flow generates a notion of distance which is equivalent to the "ruler distance".

The metric is especially simple in this case, because the spatial coefficients are all unity, so the coordinates represent spatial distances directly.

The remaining argument is just that for a sufficiently large black hole, the Riemann curvature tensor is small (i.e. the tidal forces, properly defined, are are not important), so that the above metric (which has a zero Riemann curvature) is a good approximation to the more complex black hole metric. If we really wanted to go through the trouble, we could compute the actual time-like Killing flow of the Schwarzschild metric and use it to determine the distances for a static observer without making any such approximations.

Because a time-like Killing flow exists, we can take advantage of it to generate a Born rigid flow. Since the flow is non-rotating, I don't believe we can require that a Born-rigid flow must be generated by a time-like Killing flow, but we can certainly take advantage of the existence of a time-like Killing flow to generate a Born rigid flow.

If we ask for the notion of "distance" that an infalling observer might have, I don't believe we can find an appropriate Killing flow :(. This doesn't mean that a Born rigid flow does not exist, but it does means that we need a different technique to find it.

 

[PeterDonis]

If we ask for the notion of "distance" that an infalling observer might have, I don't believe we can find an appropriate Killing flow :(.

You're right; there isn't. This is easy to see, since all of the Killing vector fields in Schwarzschild spacetime are known, and the only one with timelike integral curves is $\partial / \partial t$ (in Schwarzschild coordinates), and only outside the horizon.

However, as I noted in a previous post, we can cover the "arm extending" experiment with a local inertial frame, so we can just use rulers at rest in that frame as our standard of distance.