# Stationary below the event horizon?

1. Jan 4, 2010

### yuiop

Kevin Brown http://www.mathpages.com/rr/s6-04/6-04.htm gives the acceleration of particle in the Schwarzschild metric as measured in terms of the proper time of the particle as:

$$\frac{d^2 r}{d\tau^2} = -\frac{m}{r^2}$$

Does this not cause a problem for those that assert Schwarzschild coordinates are invalid below the event horizon because it impossible for a particle to remain stationary below the event horizon? This is the acceleration in terms of proper time which is often taken to be the final abitrator of truth in determing what is really happening in GR. This equation is independant of the velocity of the particle and is clearly not infinite for values of 0<r<=2m and K. Brown derives the equation assuming the velocity is zero. Further more the coordinate acceleration for a stationary particle is not infinite at or below the event horizon either, except for the case r=0. Where does the noton that a particle can not remain stationary (spatially) below the event horizon come from?

2. Jan 4, 2010

### bcrowell

Staff Emeritus
Why do you expect the acceleration to be infinite? If $d^2r/d\tau^2$ was infinite after crossing the event horizon, then zero proper time would elapse from crossing the event horizon to hitting the singularity. But we know that doesn't happen -- the observer has a finite amount of proper time before he hits the singularity.

3. Jan 4, 2010

### yuiop

Sure that is the case for a freefalling observer, but it is often stated or implied that if the observer were to use thrust below the horizon to prevent himself falling into the singularity that the acceleration required would be infinite and arriving at the singularity is inevitable.

It is very easy to make calculations using the Schwarzschild metric by setting dr=0, but it is often stated that dr=0 is impossible below the event horizon. Why is that?

4. Jan 4, 2010

### pervect

Staff Emeritus
the coordinate 'r' becomes a time coordinate below the event horion. It's second derivative with respect to proper time exists, but doesn't have the physical significance you ascribe to it.

Note that the coordinate-independent proper acceleration needed to hold an object stationary at some Schwarzchild coordinate r is, in geometric units:

$$\frac{m}{r^2 \sqrt{1-\frac{2m}{r}}}$$

This is what an accelerometer would measure, for instance. And this becomes imaginary if r becomes too small.

refs: see for instance http://en.wikipedia.org/w/index.php?title=Komar_mass&oldid=323885426 or the appropriate section of Wald.

5. Jan 4, 2010

### bcrowell

Staff Emeritus
It sounds like you're thinking in terms of Newton's laws, so $F_{total}=F_1+F_2=ma=ma_1+ma_2$. I don't think that works here. By the equivalence principle, the acceleration of gravity can actually have any value you like, depending on what local coordinate system you pick. Gravitational force just isn't a useful concept in general relativity, except maybe as a semi-Newtonian approximation.

6. Jan 4, 2010

### yuiop

If we do not stick to concept that a distance coordinate remains a distance coordinate and a time coordinate remains a time coordinate then we could come up with all sorts of arbitary rubbish like rephrasing your equation as:

$$\frac{m}{t^2 \sqrt{1-\frac{2m}{t}}}$$

and equally the radius of a black hole could not be stated in terms of distance but rather has to be stated in terms of time. Does this not cause difficulties for such things as black hole thermodynamics that claim the entropy of a black hole is proportional to its surface area if spatial measurements have no meaning as far as a black hole is concerned?

I agree that your interpretation of your equation is correct, but why does your equation differ from that given by mathpages?
K. brown clearly states:

$$\frac{d^2 r}{d\tau^2} = -\frac{m}{r^2}$$..."is a measure of the acceleration of a static test particle at the radial parameter r"

He is also using proper time tau which is the time measured by a clock attached to the particle and so all observers should agree on what that clock measures.

7. Jan 4, 2010

### yuiop

Well, I was specifically referring to acceleration rather than force.

Secondly, the argument that "the acceleration of gravity can actually have any value you like, depending on what local coordinate system you pick" can be turned on its head and used against anyone that claims that acceleration below the event horizon is infinite, because I can reply "yes, but only in your coordinates". The question of whether a particle can remain stationary below the event horizon (ie avoid the singularity) or even exit back out of the black hole should have a coordinate independent answer.

8. Jan 4, 2010

### Phrak

I believe the event horizon itself is coordinate dependent, isn't it?

BTW, I'm not referring to the defined surface of a Schwarzchild solution, but the surface of coordinate singularities as percieved by an outside observer.

9. Jan 4, 2010

### yuiop

That might well be, but the event horizon should have some coordinate independent properties such as an observer below the event horizon can not send information to an observer that remains above the event horizon.

10. Jan 4, 2010

### yuiop

OK, I concede that your equation probably should differ from that given by mathpages as the acceleration given by K. Brown is the acceleration measured by an observer that does not feel any force because they are still in freefall despite being instantaneously stationary at the apogeee, while your equation is the acceleration of an observer that actually does feel a force. Is there a reasonably simple derivation of your equation?

11. Jan 4, 2010

### Phrak

Being only slightly less ignorant after my previous post, it appears true that we can refer to both an absolute event horizon and an apparent horizon dependent upon an observer's state of motion with respect to a distant observer.

But, in any case my objections to the observer dependent event horizon are irrelevant, as belatedly, I see you must be referring to the absolute, Schwarzschild horizon.

12. Jan 5, 2010

### pervect

Staff Emeritus
The amount of time I can devote to deconfusing you is rather limited, but the point is very simple.

If you consdier two points in the schwarzschild space-time

(t,r,theta,phi) and (t,r+dr,theta,phi)

by using the Schwarzschild metric, you can compute the Lorentz interval between these two points and find that it is just

ds^2 = dr^2 / (1-r/rs)

Note that this changes sign as r passes through the critical radius rs, the schwarzschild radius. The interval is spacelike for r > rs, and timelike for r < rs.

It's much easier to say that r is a time coordiante, and IMO easier for the lay person to understand, though it's not quite as precise as the more detailed mathematical statement above.

If you don't get it, that's too bad, your tone doesn't really want to make me spend much more time on this - even if I had it to spare....

13. Jan 5, 2010

### yuiop

Sorry about the tone. I must have been in a pedantic mood when I made that post. I guess I find the the physical implications of r becomes a time coordinate more confusing than helpful. I do understand the formalism though.

I have figured out why your equation for acceleration differs from the one given by mathpages. Your equation is in terms of proper distance and proper time while the equation given by mathpages is a mixture of Schwarzschild coordinate distance and proper time. Probably should have spotted that sooner.

Your comments are not wrong and you raise an interesting aspect. For example if one observer was to fall towards a black hole followed by a second observer some time later, the second observer would never lose sight of the first observer and would never be aware of when the first observer crossed the event horizon and posssibly would not be aware of when he himself crossed the event horizon.