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

1. Aug 22, 2004

### 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.

2. Aug 22, 2004

### Janus

Staff Emeritus
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.

3. Aug 22, 2004

### 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).

4. Aug 22, 2004

### 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 Schwarszchild radius only if the observer was infinitely distant.

5. Aug 22, 2004

### 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.

6. Aug 22, 2004

### Staff: Mentor

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.

7. Aug 22, 2004

### pervect

Staff Emeritus
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.

8. Aug 22, 2004

### 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?

9. Aug 22, 2004

### pervect

Staff Emeritus
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).

10. Aug 22, 2004

### jcsd

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.

11. Aug 22, 2004

### pervect

Staff Emeritus
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.

12. Aug 22, 2004

### pervect

Staff Emeritus
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.

13. Aug 23, 2004

### pmb_phy

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

14. Aug 23, 2004

### 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).

15. Aug 23, 2004

Staff Emeritus
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.

16. Aug 23, 2004

### pmb_phy

Coordinates are used by observers to calculate physical quantities. Schwarzchild observers use Schwarzchild 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

17. Aug 23, 2004

### pervect

Staff Emeritus
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.

18. Aug 23, 2004

### chroot

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

- Warren

19. Aug 23, 2004

### pervect

Staff Emeritus
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.

20. Aug 23, 2004

### pmb_phy

I thought I mentioned that above??
I don't understand. What do you mean by this?
I wasn't speaking about a free-fall frame.

Pete