There is only one True Singularity and It is at the Center

[pervect]

I don't know tensor calculus so this debate has gone way over my head. But I have one question: In any of these alternate co-ordinate systems, is there infinite time dilaton at the event horizon without infinite tidal force?

Yes. If you'll look back, even Pete has agreed that the tidal force is finite for an observer falling through the event horizon, and I believe that pretty much everyone else here agrees with me that the tidal force is finite even for an observer "hovering" at the event horizon.

It's possible to have a case where there is a horizon, with absolutely *no* tidal force whatsoever. This is the so-called "Rindler metric", associated with an accelerating observer.

If someone accelerates at a constant proper acceleration, he will observe a horizon that's located a distance c^2/a behind him known as the "Rindler horizon". This happens because photons emitted after a certain time can't reach him anymore, so his space-time becomes causally disconnected from that of the stationary observer.

See for instance the post by John Baez, keepr of the relativity FAQ, at

http://www.lns.cornell.edu/spr/2002-05/msg0041388.html

for some discussion of the Rindler Horizon. It's also discussed in "Gravitation", by MTW.

Note that while the _tidal forces_ are all zero with the Rindler metric, you can't stay stationary right at the horiozn point without an infinite acceleration. The same thing is true about the Scwarzschild horizon. This is discussed in "General Relativity", Wald, pg 152. I make this point in case you might be confusing a tidal force and an acceleration.

So horizons are associated with regions of space-time that become "disconnected" because photons can't pass between them. Horizons are global, not local. Horizons can occur without any tidal accelerations whatsoever, as the Rindler metric/ accelerating spaceship shows.

If you send a signal to an observer, and the photon never comes back, one could call this 'infinte time dilation", and that's one reason why infinite time dilations are associated with horizons. While this is one way of looking at it, its really a lot plainer to just say that the photon never arrives, and that the two space-time regions are causally disconnected, IMO.

 

[Ray Eston Smith Jr]

My question was whether that can be a finite tidal force and an infinite time dilation at a particular space-time position in one frame of reference.

Pete said that both the tidal force and time dilation are finite for an observer free-falling through the event horizon.

In the Rindler metric, my guess is that you would only observe infinite time dilation for that impossible observer hovering right at the horizon. Someone hovering a millimeter away would experience finite acceleration. That would mean an acceleration gradient approaching infinity. Because gravity and acceleration are equivalent that means tidal forces approaching infinity. (You don't need that impossible guy who's exactly on the horizon. I think you could look at a series of possible observers closer and closer to the horizon to get a series which would approach the limit of infinite acceleration, infinite tidal forces, and (I think) infinite time dilation.)

 

[pervect]

Pete said that both the tidal force and time dilation are finite for an observer free-falling through the event horizon.

Pete? You there?

In the Rindler metric, my guess is that you would only observe infinite time dilation for that impossible observer hovering right at the horizon. Someone hovering a millimeter away would experience finite acceleration. That would mean an acceleration gradient approaching infinity.

The acceleration needed does go as 1/x in the Rindler metric making the plane x=0 a singularity, and making the first part of your guess correct.

For comparison, the required acceleration for station-keeping for the Schwarzschild metric (black hole) is (Wald, p 152)

\frac{M}{r^2\sqrt{1-\frac{2M}{r}}}

The required acceleration to maintain station does go to infinity, and is less than infinity a short distance away from the horizon in both cases.

However, an observer near the horizon uses his own rods and clocks to measure the tidal force - using his own concepts of distance and time. The observer hovering near the black hole will have different coordinates that the Schwarzschild coordinates, the observer accelerating near the rindler horizon will also have his own set of local coordinates different from the other accelerating observer.

Using his own rods & clocks, the tidal force measured by any uniformly accelerating observer in flat-space time (this includes the observer at the Rindler horizon, and excludes the observer at the black hole where space-time is not flat at all) will be zero.

Of course, just like any uniformly observing observer, his frame size must be small enough to get this result. The frame size limit is distance << c^2/a. (Note that in a nice display of infinite regress, the observer accelerating near the rindler horizoin of the first accelerating observer has his own Rindler horizon!).

(You don't need that impossible guy who's exactly on the horizon. I think you could look at a series of possible observers closer and closer to the horizon to get a series which would approach the limit of infinite acceleration, infinite tidal forces, and (I think) infinite time dilation.)

Infinite acceleration, yes. Infinite tidal forces, no. Infinite time dilation, yes.

BTW, in practical terms, the usefullness of determining the tidal forces on an observer "hovering" right at the event horizon is probably minimal, since the observer is already undergoing an infinite non-tidal acceleration to hold station. It's an interesting mathetmatical fact, though. And it's definitely of interest in the more realistic case of an observer falling through the horizon. There it's important to know that the tidal forces are finite.

 

[pmb_phy]

Pete? You there?

Yes. I'm here.

BTW, in practical terms, the usefullness of determining the tidal forces on an observer "hovering" right at the event horizon is probably minimal, since the observer is already undergoing an infinite non-tidal acceleration to hold station.

Remind me - An observer can't orbit a black hole just outside of the event hoirizon. The only orbits allowed for bodies with finite proper mass are outside the photon sphere. Do I remember correctly?

Pete

 

[pmb_phy]

Pete? You there?

Yes. I'm here.

Ray - I never said that. I said that a far away observer who is at rest outside the (non-rotating) black hole (known as a Schwarzschild observer) will reckon the tidal forces on an observer near the event horizon to be larger and larger as the observer approaches the event horizon.

By the way - The tidal forces on a body is a function of the velocity of the body.

BTW, in practical terms, the usefullness of determining the tidal forces on an observer "hovering" right at the event horizon is probably minimal, since the observer is already undergoing an infinite non-tidal acceleration to hold station.

Remind me - An observer can't orbit a black hole just outside of the event hoirizon. The only orbits allowed for bodies with finite proper mass are outside the photon sphere. Do I remember correctly?

Pete

 

[pmb_phy]

Pete? You there?

Yes. I'm here.

Ray - I never said that. I said that a far away observer who is at rest outside the (non-rotating) black hole (known as a Schwarzschild observer) will reckon the tidal forces on an observer near the event horizon to be larger and larger, approaching infinity, as the observer approaches the event horizon. This far away observer can never measure what happens at the event horizon. The quantity which reflects the strength of the tidal forces are the Riemann tensor. In Schwarzschild coordinates (those coordinates used by this far away, stationary observer - spherical spatial coordinates) some of the components are infinite at the event horizon.

By the way - The tidal forces on a body is a function of the velocity of the body.

What do I mean by "far away observer" and "reckons" is a bit more complicated. A Schwarzschild observer is an observer who records events using the Schwarzschild coordinates of an event. These are not observers located at the events themselves. By "reckon" I mean "determine" which can be done by using recording devices spread over the region of space to locally record events, but using clocks and rods which read off the Schwarzschild coordinates.. The recordings are then later analyzed by the Schwarzschild observer (hence the term "bookkeeper" coordinates which has been used too). It is also in this sense that a beam of light does not change frequency as it moves through the field of a non-rotating black hole. What is different is how local observers measure these frequencies and when compared with the frequencies measured locally by the far away observer (when the light gets far away) - they get different results.

The point is that one must be careful of who is measuring what.

BTW, in practical terms, the usefullness of determining the tidal forces on an observer "hovering" right at the event horizon is probably minimal, since the observer is already undergoing an infinite non-tidal acceleration to hold station.

Remind me - An observer can't orbit a black hole just outside of the event hoirizon. The only orbits allowed for bodies with finite proper mass are outside the photon sphere. Do I remember correctly?

Pete

 

[Ray Eston Smith Jr]

Post #13 in this thread from Pete:
"For a free-fall frame the tidal forces are finite and time does not stop at the event horizon."

Is that not equivalent to "Pete said that both the tidal force and time dilation are finite for an observer free-falling through the event horizon"?

If an observer can feel his acceleration, then he should be able to feel tidal forces if his feet are subject to different acceleration than his head.

 

[pmb_phy]

Post #13 in this thread from Pete:
"For a free-fall frame the tidal forces are finite and time does not stop at the event horizon."

Is that not equivalent to "Pete said that both the tidal force and time dilation are finite for an observer free-falling through the event horizon"?

Sorry. You're correct. That's the result of me having to read too fast. My appologies.

Pete

 

[Ray Eston Smith Jr]

Thanks, Pete. I'm particularly interested in the part about time not stopping at the event horizon (from the free-fall frame). Most popular accounts of falling into a black hole have the observer getting hit with the "blue sheet" as he passes the event horizon. But if the free-falling observer's time (as seen by another free-faller who's farther away) doesn't stop there, then I think he wouldn't see the "blue sheet" til he reaches the actual singularity.

Could someone please post some standard definitions to avoid confusion between reference frames, coordinate systems, spatial locations, local vs diistant, observer vs observed, mathematical vs actual singularity, time-component vs spatial components, real acceleration vs apparent accleration (if there is such a thing), etc? I try to use these terms carefully and accurately, but I might be garbling my messages by misusing terms.

 

[pmb_phy]

Thanks, Pete. I'm particularly interested in the part about time not stopping at the event horizon (from the free-fall frame).

You're welcome. When I said that time doesn't stop at the event horizon I say that with tounge in cheek. No clock can be a rest at the event horizon so it can't be there at two instants of time. The metric corresponding to a free-fall frame does not have a singularity at the event horizon.

Most popular accounts of falling into a black hole have the observer getting hit with the "blue sheet" as he passes the event horizon. But if the free-falling observer's time (as seen by another free-faller who's farther away) doesn't stop there, then I think he wouldn't see the "blue sheet" til he reaches the actual singularity.

Sorry but I don't know what a blue sheet is.

Could someone please post some standard definitions to avoid confusion between reference frames, coordinate systems, spatial locations, local vs diistant, observer vs observed, mathematical vs actual singularity, time-component vs spatial components, real acceleration vs apparent accleration (if there is such a thing), etc? I try to use these terms carefully and accurately, but I might be garbling my messages by misusing terms.

That'd be a good idea. Sorry I have to sign off for several hours. I'm not supposed to be sitting too long. I'll try to think of a good way to describe/define each of those terms.

Pete

 

[pmb_phy]

reference frames - I now recall that I once tried to define those terms. As I recall nobody really agreed on what the term "frame of reference" means and I don't like arguing about what a word means. So I'll just give you one definition in the context of special relativity. This is by Ray D'Inverno from his text Introducing Einstein's Relativity, page 17

We shall refer to an observer's clock, ruler, and coordinate axes as a frame of reference.

Lots of people don't like that definition since it means that if you simply rotate your coordinate system, or translate it, then you have a new frame of reference contrary to what people have in mind by a frame of reference. But I think D'Inverno is correct in his definition since it works nicely in GR since rotating a coordinate system and changing to one which is in motion both coordinate transformations.


coordinate systems - A system to label points in a set.

spatial locations - Locations in space. Can't get more simpler than that. "Boston" is a spatial location (the center of Boston would be more precise since its a point). "Boston Marathon" is an event - it has a location and a time (really a time interval but who'es counting).

local vs diistant - Having a spatial location which is in some sense "near". A local measurement is a measurement that is done in the spatial location where the physical phenomena is being measured. Distant means in some sense "far". A distant measurement is a measurement that is done far away from the physical phenomena. E.g. If there is a clock in the g-field of a non-rotating BH at rest outside the event horizon and it emits light then a distant observer can measure the frequency of light, from his frame of reference, by waiting for the light to reach him and then measuring its frequency. The frequency, as measured by any single observer, does not change as the light moves through the gravitational field.

observer vs observed - Observer - that which is recording events or measuring physical phenomena. It can refer to a system of clocks and rods. Observed - that which is being measured.

mathematical vs actual singularity - Mathematical - A fluke in the coordinate system. The geometry is nice but the coordinate system isn't. Take a sphere as an example. There are no physical singularities, the surface is smooth. Coordinate systems can be chosen which behave badly even when the actual surface doesn't. I can choose coordinates anyway I'd like. I can choose a coordinate to the the reciprocal to the azimuthal coordinate. When the asimuthal coordinate (theta) may be zero the reciprocal, which does label points on the sphere in most places, does not behave well when the azimuth is zero. You then get 1/0. That is a mathematical singularity

time-component vs spatial components - Its easier to explain this in SR. In SR an event is a Lorentz 4-vector. It represents the spacetime displacement from an event which is chosen to be the origin. If I label an event by

\bold X = (ct, x, y, z) = (x^0, x^1, x^2, x^3)

then for obvious reasons x⁰ is called the time component of that 4-vector and x^k, k = 1,2,3 - are the spatial components of the 4-vector. The 4-velocity, U, is defined as

\bold U = \frac {d\bold X}{d\tau} = (U^0, U^1, U^2, U^3

The time component of this 4-vector is U⁰ and the spatial components are U^k. In general, using the labeling convention defined above, the time component of a 4-vector A is the quantity A⁰ and the spatial components are A^k.

real acceleration vs apparent accleration (if there is such a thing) - I don't know what the difference is. I do know that the are some people who like to say that if the 4-acceleration of a particle is zero then it "really" isn't accelerating and it it is non-zero then it "really" is accelerating. I don't think that is a very good way to look at it myself. But - Se la vie!

Pete

 

[pervect]

Yes. I'm here.
Remind me - An observer can't orbit a black hole just outside of the event hoirizon. The only orbits allowed for bodies with finite proper mass are outside the photon sphere. Do I remember correctly?

Pete

Yep. Exactly.

To satisfy the line-counter here on brief responses, let me put in another plug for the popularization "Black Holes & Time Warps" by Kip Thorne, co-author of "Gravitation". It popped into my mind beause Thorne actuallly mentions this very point in his virtual "exploartion" of several black holes - very good for a popular book, but then what does one expect from one of the co-authors of "Gravitation" :-)

I also like Thorne's writing style, YMMV.

 

[pmb_phy]

Yep. Exactly.

To satisfy the line-counter here on brief responses, let me put in another plug for the popularization "Black Holes & Time Warps" by Kip Thorne, co-author of "Gravitation".

I have that book.

Pete

 

[pervect]

I have that book.

Pete

Me too, I've been recommending it to Ray Eston Smith, and since my post was not accepted because it was too short otherwise, I thought I'd give it another plug