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

[Ray Eston Smith Jr]

As I understand it, a gravitational singularity is a point where gravitational force is infinite and consequently time stops (from the viewpoint of any outside observer), light from a source at the singularity would become infinitely red-shifted after traveling zero distance (in other words light couldn't leave the singularity), escape velocity is infinite, and tidal forces are infinite. I don't think any of these effects can occur anywhere by itself, without the other effects. You can't have stopped time without also having infinite tidal forces. If the speed of light "appears" to approach zero, that really means c is constant but time dilation is approaching infinity, which means tidal forces are approaching infinity, which means an object approaching that point would be torn apart (in any reference frame, although different reference frames would disagree on where that point is located on their coordinate axes.)

I have read in many places that the only physical singularity is at the center of a black hole. The singularity at the Schwarzschild radius is only a mathematical singularity due to the choice of coordinate system. In other words, in the Schwarzschild metric from a stationary frame, the physical singularity "appears" to be at the Schwarzschild radius, but it's actually at the center. In the stationary reference frame, the event horizon is like a goldfish bowl that makes the singularity/fish at the center look like it's on the outside of the bowl.

When a free-falling observer passes the Schwarzschild radius, he is not passing a physical singularity. So, he is not torn apart by infinite tidal forces. If he looks back, he does not see the universe blue-shift to infinity. He does see finite blue-shifting behind him as he approaches the singularity, but he will see no infinite blue-shift until he actually reaches the singularity. If, just short of the singularity (but well inside the Schwarzschild radius), he turns on his retro-rockets with thrust equal to the gravitational force at that point, then he will shift back to the stationary frame and he will see himself hovering just outside the Schwarzschild radiius instead of just outside the singularity. But, when he was free-falling toward the singularity, he observed a huge but finite blue-shifted time interval behind him. Now that he's back in the stationary frame, he still see that same amount of blue-shift behind him - he still sees that the universe behind him has aged enormously (but not infinitely). From the stationary frame, he expains all this as the effect of being very close the event horizon.

 

[Janus]

For one, it isn't the magnitude of gravitational force that causes time dilation, but the difference in gravitational potential

The formula for gravitational time dilation is

$$T = \frac{T_{o}}{\sqrt{1-\frac{2GM}{Rc^2}}}$$

The Schwarzschild radius is determined by

$$R = \frac{2GM}{c^2}$$

substituting this for R in the first equation:

$$T = \frac{T_{o}}{\sqrt{1-\frac{2GM}{\frac{2GM}{c^2}c^2}}}$$

$$T = \frac{T_{o}}{\sqrt{1-\frac{2GM}{2GM}}}$$

$$T = \frac{T_{o}}{\sqrt{1-1}}$$

$$T = \frac{T_{o}}{0}$$

And you get inifinite time dilation and red/blue shift at the Schwarzschild radius.

 

[jcsd]

Janus, IIRC infite time dialtion/shifting only occurs from the point of view of a remote observer (which is what your formula is for).

 

[Ray Eston Smith Jr]

Since the escape velocity at the Schwarzschild radiius is the speed of light, a distant free-falling observer would see infinite red-shifting of a light beam (i.e. he would see nothing) originating at the Schwarzschild radius only if the observer was infinitely distant.

 

[Ray Eston Smith Jr]

As jcsd says, the formulas apply to what is seen by a remote observer. But near a singularity any observer not actually in the singularity qualifies as remote. So the formulas describe the limits as one approaches a singularity. Within the singularity different formulas apply - the weird formulas which most people (not me) claim apply inside the Schwarzschild radius.

 

[russ_watters]

You seem to be mixing words again, Ray - if the singularity is a point, "within the singularity" doesn't really have any meaning. I think you mean within the Schwarzschild radius: within the event horizon.

 

[pervect]

Since the escape velocity at the Schwarzschild radiius is the speed of light, a distant free-falling observer would see infinite red-shifting of a light beam (i.e. he would see nothing) originating at the Schwarzschild radius only if the observer was infinitely distant.

What happens when light is emitted exactly at the event horizon of a non-rotating Scwarzschild black hole is that the light just sort of "hangs there" at R=2M.

So an observer at a finite distance does not see it. The only observer who can see it is one who passes through the event horizion.

 

[Ray Eston Smith Jr]

Janus - You gave the formula for gravitational time dilation. I'm assuming that's from a frame of reference at rest with respect to the gravitational source. But isn't there a time dilation associated with any acceleration? A free-falling observer would have an acceleraton relative to the gravitational source, therefore the time dilation he sees should be different than what the stationary observer sees. Therefore the point where he sees T approach infinity should be different. I'm guessing that point is at the singularity. Can you do the math for me please?

 

[pervect]

Janus, IIRC infite time dialtion/shifting only occurs from the point of view of a remote observer (which is what your formula is for).

You might want to consider what the null geodesic of the light ray emitted at the event horizion looks like. Then it's fairly easy to answer the question of which observers intersect this geodesic (see the light).

 

[jcsd]

You might want to consider what the null geodesic of the light ray emitted at the event horizion looks like. Then it's fairly easy to answer the question of which observers intersect this geodesic (see the light).

The worldline I suppose would, just be 'along' the event horizon, so I suppose what you are saying is that tie dialion will be infinite for any observer not falling into the hole.

 

[pervect]

The worldline I suppose would, just be 'along' the event horizon, so I suppose what you are saying is that tie dialion will be infinite for any observer not falling into the hole.

The way I look at it is that the first question to ask is if the photon ever arrives. If it never arrives, it doesn't really make sense to ask about its red shift, or the time dilation. Those question arise after one has confirmed that the photon does arrive.

 

[pervect]

You can't have stopped time without also having infinite tidal forces.

What makes you think this? And what do you mean by stopped time?

I'm guessing that your motivation may be that you're very interested in black holes and want to understand them better (it's hard to be sure at this point - you wind up making a lot of statements that sound definite rather than questioning/curious). If your motivation is curiosity, I'd like to recommend a book to you, "Black Holes: Einstein's Outrageous Legacy" by Kip Thorne. It's got a nice opening prologue where the protagonist visits several black holes of differing sizes, and reports on his observations.

 

[pmb_phy]

What makes you think this? And what do you mean by stopped time?

In Schwarzschild coordinates the tidal forces are infinite (i.e. at least one component of the Riemann tensor is infinite) a the event horizon and at the event horizon time has stopped, i.e. if we could see a clock there then we wouldn't see the hands move. But since we can't detect a clock or anything else at the event horizon then we can't detect a clock stopping nor can we detect an infinite tidal force. For a free-fall frame the tidal forces are finite and time does not stop at the event horizon. However a clock can't remain at rest at the event horizon either so...

Pete

 

[Ray Eston Smith Jr]

Thank you, Pete. It is very to difficult to get straight answers on the internet or from popular science books about the differences between the stationary frame and the free-fall frame. I think people tend to forget there are reference frame differences, so they mix together contradictory facts from the two different frames, or they deny a fact from one reference frame because it's not true in the other frame.

If tidal forces at the event horizon are infinite in the stationary frame but finite in the free-falling frame, doesn't that mean that the event horizon must be at different locations in the two frames. A falling astronaut can't be both ripped apart and not ripped apart at the same time in the same place.

If time does not stop at the event horizon as viewed by an outside observer in the free-falling frame, doesn't that mean that free-faller does not experience an infinite blue-shift as he crosses the event horizon?

What do you think of my theory that the stationary frame event horizon is at a different space-time location (I think at 0 distance and 2M time interval from the center) than the space-time location of the free-falling Schwarzschild radius (I think at 2M distance and 0 time interval from the singularity).

 

[selfAdjoint]

In Schwarzschild coordinates the tidal forces are infinite (i.e. at least one component of the Riemann tensor is infinite) a the event horizon and at the event horizon time has stopped, i.e. if we could see a clock there then we wouldn't see the hands move. But since we can't detect a clock or anything else at the event horizon then we can't detect a clock stopping nor can we detect an infinite tidal force. For a free-fall frame the tidal forces are finite and time does not stop at the event horizon. However a clock can't remain at rest at the event horizon either so...

Pete

But the Schwarzschild coordinates are just COORDINATES. Longitude becomes undefined at the Earth's poles, but the geometry does nothing strange there. Similarly the coordinate singularity at the event horizon can be eliminated by choosing a different coordinate system, which GR always allows us to do.

 

[pmb_phy]

But the Schwarzschild coordinates are just COORDINATES. Longitude becomes undefined at the Earth's poles, but the geometry does nothing strange there. Similarly the coordinate singularity at the event horizon can be eliminated by choosing a different coordinate system, which GR always allows us to do.

Coordinates are used by observers to calculate physical quantities. Schwarzschild observers use Schwarzschild coordinates. Therefore a Schwarzschild observer reckons that clocks closer to a BH run at the same rate as clocks measured by far-away observers. However this is not the case for all other observers, namely the observers who measure things locally.

Regarding choosing different coordinates - This is not the same as choosing different coordinates in Euclidean geometry where there are no physical consequences. Choosing different coordinates in GR has physical consequences. If, in SR, I measure a force then the magnitude of that force will depend on the coordinates I choose to evaluate that force. These different coordinate systems correspond to frames of reference moving relative to each other. In GR choosing different coordinates has a similar effect, but now its gravitational time dilation that is playing a role, not the time dilation related to speed.

Pete

 

[pervect]

In Schwarzschild coordinates the tidal forces are infinite (i.e. at least one component of the Riemann tensor is infinite) at the event horizon and at the event horizon time has stopped, i.e. if we could see a clock there then we wouldn't see the hands move. But since we can't detect a clock or anything else at the event horizon then we can't detect a clock stopping nor can we detect an infinite tidal force. For a free-fall frame the tidal forces are finite and time does not stop at the event horizon. However a clock can't remain at rest at the event horizon either so...

Pete

The actual tidal forces experienced by an observer are *not* infinite at the event horizion for an observer falling into a black hole.

Because the schwarzschild _coordinates_ are singular there, it's necessary to use the schwarzschild basis vectors (schwarzschild basis) to calculate the tidal forces (Reimann tensor).

See for instance MTW, pg 820-821. The tidal forces on the observer in the 'r' direction are -2M/r^3 (there are compressive terms of M/r^3 in the other spatial directions as well), and are independent of the r component of the velocity of the observer (they are the same for a stationary observer near the horizion, or for one falling into the black hole).

To calculate this yourself, if you use GRTensor II, use the schwb coordinates.

Black hole horizons are just one counterexample. The Rindler metric of an accelerating observer also has a horizon, and associated infinite red-shifts, without any infinite accelerations.

 

[chroot]

You could, of course, just use Eddington-Finkelstein coordinates, which have no coordinate singularities at the event horizon.

- Warren

 

[pervect]

Thank you, Pete. It is very to difficult to get straight answers on the internet or from popular science books about the differences between the stationary frame and the free-fall frame. I think people tend to forget there are reference frame differences, so they mix together contradictory facts from the two different frames, or they deny a fact from one reference frame because it's not true in the other frame.

It is also occasionaly difficult to get straight answers even on chat forums. Pete's usually pretty good, but everyone makes mistakes. Including, alas, myself. However, if you check out my previous reference

"Gravitation", Misner, Thorne, Wheeler - pg 820-821 - you'll see that the tidal forces are finite both for a stationary observer near the event horizon, and one falling into the black hole.

You might also want to see the quiz on the board about "Black Hole Basics", which also touches on this point.

 

[pmb_phy]

The actual tidal forces experienced by an observer are *not* infinite at the event horizion for an observer falling into a black hole.

I thought I mentioned that above??

Because the schwarzschild _coordinates_ are singular there, it's necessary to use the schwarzschild basis vectors (schwarzschild basis) to calculate the tidal forces (Reimann tensor).

I don't understand. What do you mean by this?

See for instance MTW, pg 820-821. The tidal forces on the observer in the 'r' direction are -2M/r^3 (there are compressive terms of M/r^3 in the other spatial directions as well), and are independent of the r component of the velocity of the observer (they are the same for a stationary observer near the horizion, or for one falling into the black hole).

I wasn't speaking about a free-fall frame.

Pete

 

[pervect]

You could, of course, just use Eddington-Finkelstein coordinates, which have no coordinate singularities at the event horizon.

- Warren

Yes - what I really meant to say was that one needed to use coordinates that were well-behaved at the horizon. I find the Schwarzschild _basis_ (schwb in GRTii) to be very convenient, and that's what MTW used as well for the discussion of the behavior of the Riemann at the horizion, but it's not the only possible choice of coordinates. MTW also uses infalling Eddignton-Finkelstein coordiantes in other parts of the text for infall, for instance.

 

[pervect]

I don't understand. What do you mean by this?

I wasn't speaking about a free-fall frame.

Pete

It's also true in the stationary frame. The Riemann in this particular case has the very interesting property that it's components are independent of any boost in the radial direction. So both infalling and stationary observers observe the same value for the components of the Riemann. In fact, the Riemann was worked out in MTW for the stationary observer, then the results were extended to the falling observer by observing the fact that the components of the Riemann were independent of a radial boost.

If you are a GRTii user, look at the schwb metric. I can type the formulas in from MTW if you don't have the text, but it's easier if you look at the text yourself if you have it. (I was under the impression you did?).

The trick is to use the the one-forms dt, dr, etc. as the basis, rather than the coordinates dr,dt. This means that one has an observer with unit basis one-forms dt, dr, etc. at the horizon. If the use of the dual basis is too confusing, you could always resort to Eddington-Finkelstein coordinates as chroot suggested (but the text reference I posted uses the coordinates I mentioned).

The important point is that the Riemann is well behaved at the horizon, for both a falling and a stationary observer.

 

[pmb_phy]

It's also true in the stationary frame.

What is also true? In Schwarzschild coordinates R⁰₁₀₁ is

$$R^0_{101} = \frac {r_s}{r^3}\frac{1}{1-r_s/r}$$

Now you tell me - what is the value of this component when r = r_s? In fact, the Riemann was worked out in MTW for the stationary observer, ...

Page and equation number please?

Pete

 

[pervect]

What is also true? In Schwarzschild coordinates
Page and equation number please?
Pete

pg 821, eq) 31.4a) and 31.4b)

$$R_{\hat t \har r \hat t \hat r} = \frac{-2M}{r^3}, etc...$$

This is the Scwarzschild _basis_, not the Scwarzschild _coordinates_.

It's in the section $31.2 titled "The Non-singularity of the gravitational radius"

 

[pmb_phy]

pg 821, eq) 31.4a) and 31.4b)

$$R_{\hat t \har r \hat t \hat r} = \frac{-2M}{r^3}, etc...$$

This is the Scwarzschild _basis_, not the Scwarzschild _coordinates_.

It's in the section $31.2 titled "The Non-singularity of the gravitational radius"

I was speaking about Schwazschild coordinates and nothing else. As I already agreed, tidal forces as measured by a freely falling observer are finite. That is what you just posted, i.e. the tidal force for a freely falling observer. That is what those equations mean

I would agree that the tidal force in Schwazschild coordinates are not very meaningful or useful at the event horizon since nothing can remain at rest there. I was merely pointing out that the Riemann tensor becomes infinite at the event horizon when expressed in Schwazschild coordinates. This is similar to the metric tensor becomming infinite there. I did not state or mean that there was something about the spacetime geometry which is singular there. The metric tensor becomming infinite at the event horizon has a physical consequence in that as a light emmiter gets closer to the event horizon the frequency of the light as seen by a faraway obsever becomes zero.

But for the life of me I can't see where you get "stationary frame" from the equations that you posted above. What is it you're referring to when you use the term "stationary frame"? Do you mean stationary with respect to the free-fall frame?

pete

 

[pervect]

Also, to anticipate some possible objections

R_a^bcd

is finite and very similar to the text's value for R_abcd so when we "feed" R two finite one-forms t_b and t_d for time and one finite one-form d_c for the displacement, we get out a finite tidal acceleration one-form via the geodesic deviation eq

R_a^bcd t_b d_c t_d

All of the components of the Riemann above are one of the following:

0, +/- M/r^3, or +/- 2M/r^3

 

[pervect]

I was speaking about Schwazschild coordinates and nothing else.

Well, to repeat myself a bit, the Schwarzschild coordinates are singular at the horizon, but that doesn't mean that the Riemann itself is singular at the horizon.

As I already agreed, tidal forces as measured by a freely falling observer are finite.

So far so good.

That is what you just posted, i.e. the tidal force for a freely falling observer. That is what those equations mean

No, it is not what those equations mean. Did you actually read the text? Do you have it? I realize it can sometimes take time to find such things, assuming you own the text. If you don't own it, I can type sections of it in, but it's more work. I'll do some of that now:

To calculate those curvature coordinates, proceed in two steps. (1) Calculate the components, not in the traveler's frame, but in the "static" orthonormal frame

eq 31.4a)

Note that this is a *static* frame.

Did you read what I said about the components of the tensor being independent of the radial velocity?

MTW also makes a number of remarks about this.

But for the life of me I can't see where you get "stationary frame" from the equations that you posted above. What is it you're referring to when you use the term "stationary frame"? Do you mean stationary with respect to the free-fall frame?

pete

I mean eq 31.4a) when I say "stationary frame", which MTW calls a "static frame".

Also, as chroot pointed out, one can get the Riemann tensor by using Eddington-Finklestein coordinates, if the use of the dual basis is too confusing. Note that the Riemann tensor is finite in these non-singular coordinates.

See attached for the worksheet where I did just this:

 

[pervect]

I don't think the attachment 'took'? I don't know what I'm doing wrong for sure, but I'm guessing that my file type of .rtf is not on the "approved list" of attachment types. I can generate latex, but it has a number of "Mapleisms", so I doubt if it would parse correctly if I just cut and pasted it into a response.

Also, I finally figured out/noticed the private message system here, and sent pmb one of my first PM's on this point.

 

[Ray Eston Smith Jr]

You might as well say the time dilation at the event horizon is just a matter of coordinates. There is a real infinite time dilation and real infinite tidal force at the fixed-frame event horizon. The question is - where is the event horizon? I believe it has to be at the same spatial location as the central singularity - the only point where time dilation and tidal forces are infinite. But, in the free-fall frame, the time dilation and tidal forces are finite at the Schwarzschild-radius spatial distance from the center - so that must be a different location than the fixed-frame event horizon.

 

[Ray Eston Smith Jr]

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?

 

[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