Outgoing FreeFall inertial coordinates in Schwarzchild Geom. (aside: GR inconsistent)

Zanket,

I am not sure that I even believe these results. So maybe you can look at them and see if they tell the truth. I am stunned.

The Schwarzchild Freefall inertial coordinates start here:

http://sciforums.com/showpost.php?p=1353875&postcount=3

http://www.modelofreality.org/Sect1_4.html

Instead of the inbound version (1.36) we will use the outbound version. We will have dR=0 (R=constant) for observers who are centered on outbound freefall frames. These frames are the ones that have an initial escape velocity with just barely enough energy to make it to infinity. We let

\(dR = dt - \frac{sqrt{r/2M}}{(1-2M/r)}dr\)


So

\(R= t - \left ( 2M ln \left| \frac{sqrt r - sqrt{ 2M}}{\sqrt r + sqrt{2M}} \right| + \frac{sqrt 2}{3 sqrt M} r^{3/2} + sqrt {8Mr} \right ) \)

If we let

\(dT = dt - \frac{sqrt{2M/r}}{(1-2M/r)}dr\)

Then

\(T= t - \left ( 2M ln \left| \frac{sqrt r - sqrt{ 2M}}{\sqrt r + sqrt{2M}} \right| + sqrt {8Mr} \right ) \)

Subtracting R from T (and noticing - signs):

\(T-R = sqrt{2/(9M)} r^{3/2} \)

And

\(r = (2M)^{1/3} \left (\frac{3}{2} (T - R) \right ) ^{2/3} \)


Finally, the Schwarchild metric becomes:

\(-d \tau^2 = - dT^2 + (4M/3)^{2/3} \frac{dR^2}{(T-R)^{2/3}} + (2M)^{2/3} \left(\frac{3}{2} (T-R) \right)^{4/3}d \Omega^2 \)

I have used mathematica to check this:

http://www.modelofreality.org/check.gif

Zanket,

I am stunned that this metric is well-behaved everywhere except at the singularity and T is everywhere timelike and R is everywhere spacelike! No coordinate singularity at the horizon!

Perhaps you are onto something. Perhaps you have a contradiction in your proof, but for the wrong reason!

Perhaps traditional Schwarzchild analysis is incorrect inside the horizon using accelerated Schwarzchild coordinates.

Don't forget that the Schwarzchild coordinates t,r,θ, and φ
are not inertial coordinates, they are accelerated coordinates. Anyway more on this later.

We can get a feel for this metric if we plot some radial light and stationary observer (R=constant) geodesics. Setting dτ=0 and dΩ=0 in the metric we get the radial light equation:

\( \frac{dT}{dR} = \pm \frac{(4M/3)^{1/3}}{(T-R)^{1/3}} = \pm sqrt{2M/r} \)

If we let M=3/4 for simplicity, and invert the equation we get:

\(\frac {dR}{dT} = \pm (T-R)^{1/3} \)

Using Mathematic to plot, then mirroring to get T back to the vertical axis, we see that the geodesics come right out of the horizon. The stationary observers (R=constant) come out within their own proper time, but before the beginning of Schwarzchild time. That is, they come right out, but t has gone to -∞ and back. An observer at infinity would never see them come out. This is in direct contradiction to regular Schwarzchild analysis inside the horizon, where r is the time coordinate and t the space coordinate, and things get weird because you really cannot plot r as a radius since it is time.

These inertial frames have spanned the horizon as you claim they must. I do not yet understand these equations. Perhaps you can find a mistake. Notice that the inbound light signal cannot even catch up to the accelerated Schwarzchild coordinates, very similar to how hyperbolic accelerated coordinates can outrun light.

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Andrew A. Gray

 

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I confess I'm struggling with this andrew. When you say t has gone to -∞ and back is there any opportunity to use c as a space/time conversion factor to yield a zero instead of an infinity?

 

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Andrew, can you clarify what the various quantities in your derivation are? I'm cunfused as to what the plot is illustrating. If you could summarize what T, R and the various colored lines represent, that'd be a big help.

 

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OK.

\(-d\tau^2 = -dT^2 + (4M/3)^{2/3} \frac{dr^2}{(T-R)^{2/3}} + (2M)^{2/3}(3/2(T-R))^{4/3} d \Omega^2 \)


This metric uses an inertial coordinate system of an observer who just barely has the radial escape velocity and will just barely "make it (freefall) to infinity".

Since dT has no coefficient in this metric, this means that if dR=0 and dΩ=0 (stationary in the coordinate system), that the coordinate time T and the proper time τ are the same. (This is called a stationary metric). The coordinate time and the proper time are the same for this observer.

R is the coordinate of successive observers who are "jettisoned" to infinity" one after another" with the escape velocity. Each observer has R=constant (dR=0). So for example, the first observer "jettisoned" could have R=0. The next, R=1. And so on. These observers could be thought of as "mapping" the R coordinate in space at any time T. The R coordinates are "freefalling" to infinity, one after another, similar to the ripples in the water after one drops a stone in.

The other metric,

\(-d\tau^2 = -dT^2 + (4M/3)^{2/3} \frac{dr^2}{(R-T)^{2/3}} + (2M)^{2/3}(3/2(R-T))^{4/3} d \Omega^2 \)

see

http://www.modelofreality.org/Sect1_4.html

is for ingoing freefalling observers "dropped from infinity". These observers are dropped from infinity "one after another" and crash into the singularity. This diagram is easier for me to understand:

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If we let M=3/4 for simplicity, and substitute in (1.42) r=0, we find that R-T=0 yields the singularity for both systems.

If we substitue r=2M=2(3/4), we get

T=R-1 (ingoing)

and

T=R+1 (outgoing)

for the horizon.

Light geodesics are given by equation (1.43) ingoing and similar for outgoing. (See above).
Remember: in a flat spacetime diagram, light geodesics are 45[sup]o[/sup] lines if we let c=1.

In the simple ingoing diagram above, if light is emitted from outside the horizon from the R=4.25 observer, the outgoing light simply makes it to infinity, and the ingoing light simply hits the singularity.

The "inward pointing" light emitted from the R=1.2 observer inside the horizon simply hits the singularity. The "outward pointing" light from observer R=1.2 struggles to get out but eventually is pulled to the singularity because of the horizon.

Both observers at R=1.2 and R=4.25 crash into the singularity after their inward pointing light beams crash.

Now the other outgoing diagram, I do not completely understand yet. Perhaps there is an error, but I cannot find one.

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The R=1.6 observer, above the horizon, releases an outward pointing light beam and it makes it to infinity (red, to the left from the green R=1.6 observer). However, (according to this diagram) since the Schwarzchild coordinates are accelerated coordinates, (and accelerated towards the horizon), the inward pointing light beam never makes it to the horizon in this system (red, to the right of the R=1.6 green observer). This is similar to the external Schwarzchild coordinates where nothing makes it to the horizon. And we know that accelerated coordinates can outrun light (see Gravitation, accelerated observers).

Now for the weirdness. The observer at R=.5, below the horizon, releases an outward pointing light beam as well as in inward pointing light beam. This is what I do not understand. This plot says that the outward pointing light beam just goes right out through the horizon, as well as the observer himself. However, we know from the external Schwarzchild coordinates that the t coordinate has come from -∞ before this is finished (see below). A Schwarzchild observer at infinity would never see this happen. Not only that, it appears that the two parts of the geodesic are split in that they both "go away" from -∞ at the horizon in the plot below. This feature does not appear in the R,T coordinates, and since T is timelike, the R=constant observers must go towards the future. This says that the weird internal Schwarzchild analysis may be incorrect.

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Not only that, since the Schwarzchild coordinates are accelerated coordinates and very weird inside the horizon, it is r that is the time coordinate and t is the space coordinate inside the horizon. So the analysis is difficult to say the least. We see from the above diagram that since r is the time coordinate below the horizon, then we conclude that r is diminishing only. Timelike coordinates only go one way. So maybe I am missing something. Maybe accelerated (noninertial) time coordinates do not have to go in one direction. I do not know yet. I do know that accelerated coordinates can have "negative energy".

Now the ingoing light beam emitted by observer R=.5 does not really go towards the singularity. The singularity is accelerating away from it. And the horizon starts to catch up to it from behind, but never does.

These plots say that nothing comes out of the horizon according to Schwarzchild time. These plots also say that using the proper time of these observers, that they just come on right through the horizon, no problem. According to these plots, Zanket was right! The horizon must be covered by inertial coordinates, and here they are!




Andrew A. Gray

 

Thanks for that, andrew. I'm going to have to read it again when I've got some quiet time, but can I say for now that:

The R=1.6 observer, above the horizon, releases an outward pointing light beam and it makes it to infinity (red, to the left from the green R=1.6 observer). However, (according to this diagram) since the Schwarzchild coordinates are accelerated coordinates, (and accelerated towards the horizon), the inward pointing light beam never makes it to the horizon...

says to me that there can be no "below the horizon". Time is going slower and slower, and at the horizon it stops. That's the end of it. The R=.5 observer below the horizon is starting from "beyond the end of time". It's not a real situation. That's why I was talking about flipping the infinite t to a zero c, and saying black holes are frozen stars with no central singularities. Time is the problem. I know it sounds odd, but T isn't fundamental. Check out my RELATIVITY+ essays and TIME EXPLAINED. I've been asking for feedback but was met with hostility and calls for the trashcan, and sadly they got moved to pseudoscience this week. But they should be of some value.

PS: I like to think I've been somewhat supportive of Zanket, but don't share his view that he scuppers relativity. I guess I seek to fix it.

 

Andrew, thanks for the excellent analysis and the summary. I haven’t had much time for physics lately. Just a few comments for now, until I can further study your posts above:

Note that the other thread shows without using math that GR requires that the “horizon must be covered by inertial coordinates”. Not only does your analysis show it, but also it is implied by a more casual look at what GR predicts. The theory incorporates SR, so that SR’s relativity principle becomes the following, as stated by Thorne: “All small, inertial (freely falling) reference frames in our real, gravity-endowed Universe must be "created equal"; none can be preferred over any other in the eyes of the laws of physics.” If all inertial frames are equivalent in GR, then it must be true that Y (defined in the other thread) can pass outward through the horizon, because that is the case when Y covers all of X initially, and SR predicts that any inertial frame K can cover all of an inertial frame J that is wholly above a horizon. We know that GR also contradictorily predicts that Y cannot pass outward through the horizon (otherwise objects at rest in Y could do likewise), and that is how GR is self-inconsistent.

It’s interesting what you note about “accelerated coordinates can outrun light”. You’re talking about a Rindler horizon of SR. I have often wondered why GR does not predict the existence of what I call a “Rindler hole”. This would be an object from which light moving radially outward always increases in altitude, but the height of such light asymptotes to a specific altitude, so that the object is black (no emissions received) when viewed above that altitude, but there need be no central singularity. GR should predict such objects, since the equivalence principle establishes the equivalence between an accelerated observer (who can outrun light) and an observer fixed at a specific altitude. I think that GR’s omission of this prediction is either something that is predicted by GR’s equations but no one has noticed it, or is something that the equations do not predict because GR is flawed. (And we know it’s flawed in at least one way because it’s self-inconsistent.)

I suspect that what may have happened is this: Einstein created field equations that incorporate the self-inconsistency I describe above. Schwarzschild found the first solution to these equations, for Schwarzschild geometry. Physicists including Einstein looked at Schwarzschild’s solution, saw that the escape velocity can be c at r > 0, and interpreted from that the prediction of black holes (that interpretation required because otherwise they would have to accept that objects can move faster than c as directly measured, in which case SR would be violated), rather than seeing those predictions as a warning sign that the theory has a serious problem. Perhaps you are the first person to do the analysis above and see mathematically that GR is self-inconsistent in that the interpretation (the prediction of black holes) contradicts what the math predicts. Taylor and Wheeler give a clue on pg. 2-22 here that black holes are predicted by interpretation only; search for “interpretation” therein.

In a future thread I will show that escape velocity must be always less than c in a theory of gravity that is consistent with SR.

 

QUOTE:
. . . says to me that there can be no "below the horizon". Time is going slower and slower, and at the horizon it stops. That's the end of it. The R=.5 observer below the horizon is starting from "beyond the end of time". It's not a real situation. That's why I was talking about flipping the infinite t to a zero c, and saying black holes are frozen stars with no central singularities. Time is the problem. I know it sounds odd, but T isn't fundamental.


Farsight,


I wouldn't get too worked up over this. Remember, GR uses light to define time. So if light cannot travel somewhere, "it is outside the scope of time". But it is easy to imagine existence inside the horizon, outside the scope of Schwarzchild time.




Zanket,

I am not sure that I am ready to say GR is inconsistent. The Schwarzchild coordinates are accelerated coordinates. Accelerated coordinates are tricky, and if something comes up that seems weird in accelerated coordinates, then one must be very careful not to jump to conclusions.

Einsteins principles really are only valid for inertial frames.

Now, once again back to your proof:

QUOTE:
But GR predicts that nothing may pass outward through a horizon.


But now we know that this is probably not true. (It only seems true to Schwarzchild observers at infinity, who never can see something pass outward.)


QUOTE:
Then Y cannot extend below the horizon.


Again, probably not true. Y probably can extend below.


QUOTE:
Then X cannot be an inertial frame.


Yes, it probably can (IF analysis is correct). Then GR is probably not inconsistent. Interior Schwarzchild analysis is probably what is incorrect.


For the SR accelerated "horizon", see Gravitation:

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In the above figure 6.1, we see that zones II and III cannot send a light signal to the accelerated observer, so we have an x=t "black plane".

Zanket, this is actually a very good analogy for the Schwarzchild horizon. Here are the hyperbolic coordinates:

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Notice that the hyperbolic horizon is at

ξ[sup]1'[/sup] → -g[sup]-1[/sup]


It would be tempting to set up an X and Y inertial coordinate system near ξ[sup]1'[/sup] → -g[sup]-1[/sup] and using the same reasoning in your "proof" to conclude that Y cannot span the hyperbolic horizon.

However, since this is flat spacetime, we know this to be false.

So the analogy goes like this:

Schwarzchild coordinates ↔ Hyperbolic coordinates
Schwarzchild horizon ↔ Hyperbolic horizon
Inertial frame can't span horizon? ↔ Inertial frame can't span horizon?
Schw'd coordinate problems behind horizon ↔ Hyper'c coordinate problems behind horizon
GR inconsistent? ↔ SR inconsistent?


We see from this analogy that the problem lies with the accelerated coordinates and not with GR or SR. And the problem is with the accelerated coordinates "behind the horizon". So it seems to me that GR and SR remain consistent.

So Zanket, please clarify why you still think that there is a GR contradiction "in light" of this new info.



Andrew A. Gray

 

OK. I’ll just hit some key points for now.

If we are right that something can pass outward through a horizon, then it follows unequivocally that GR is self-inconsistent. In the OP in the other thread the definition of a “horizon” is a “one-way surface surrounding a black hole, defined by the property that anything may pass inward through the horizon, but (in the non-quantum description) nothing, not even light, may pass outward.” So if we are right that something can pass outward through a horizon, then by definition it cannot be a horizon that that something can pass outward through; that is, “something can pass outward through a horizon” is a contradiction in terms. That “horizon” is not a horizon by definition, hence by definition there’s no black hole there, regardless what Schwarzschild observers do or do not observe. (It might be a Rindler hole, but it’s not a black hole.) Then GR is self-inconsistent by both predicting and precluding black holes.

Let’s assume hypothetically that GR’s equations do not predict black holes after all, and that physicists simply made a mistaken interpretation in predicting them. Can GR be self-consistent then? No. The Schwarzschild metric unmistakably predicts that the escape velocity is c at an r-coordinate r = 2M (geometric units). In GR, an object that falls freely from rest at infinity passes each altitude at the escape velocity there. Then if an object could pass outward through r = 2M, it could directly measure the velocity of an object that fell freely from rest at infinity to be >= c, which contradicts SR. GR cannot contradict SR and be self-consistent, because SR is incorporated into GR.

So there is no way for GR to be self-consistent if we are right that something can pass outward through a horizon. Either both your analysis and mine are wrong, or else GR is self-inconsistent. There are no other possibilities.

Keep in mind that self-inconsistency is concomitant with weirdness. If GR becomes self-inconsistent at and below its theorized horizon, then it should be weird there. Even if your analysis was wrong, my much simpler analysis is at much less risk of a mistake. It’s all of one short math-free paragraph, yet with all your experience you could not refute it. As I pointed out in the other thread, once it is shown that Y cannot cover all of X initially due to the horizon, then the relativity principle requires that GR be self-inconsistent.

Yes, your analysis shows that “Y probably can extend below” the horizon. So does mine in the OP in the other thread. Both of those conclusions are based on GR. But GR also contradicts our analyses; that’s why you said here that “we have just seen that it is impossible to set up an (approximate) inertial frame Y, centered on an escaping particle, that extends throughout X (straddles the horizon)”. You based that conclusion on GR too, and you haven’t refuted yourself just by doing a different analysis. Your conclusion in the other thread is still valid. So see that you have already implicitly agreed that GR is self-inconsistent. Do one analysis in GR, you get a prediction. Do a different analysis for the same scenario, you get a contradictory prediction. That’s because GR is self-inconsistent. It predicts contradictory behavior for the same scenario.

 

Andrew: Sorry, I haven't studied this thread like I hoped to. I printed it out ready for my train journey at the weekend, but it was truncated on the right and I had to give up.

I think it's key, andrew. "Outside the scope of time" means "outside the scope of motion" Or "outside the scope of space". It renders everything below the horizon hypothetical. I am reminded of the sixteen square metres of carpet needed to cover the spare bedroom. Coming home with a roll which is minus four metres in length and minus four metres in width says we've got to stop and look harder at what we're talking about. IMHO we're in negative carpet territory below the horizon.

Zanket: I take the view that a black hole is a frozen star with no central singularity, and you simply cannot straddle the horizon. It's a "brick wall" where there is no time left, no charge, and no motion. I see this as a fix for GR not a flaw, one that Einstein was working on in the fifties. Let me annotate your OP on the other thread starting from "Now the proof":

Let X be an inertial frame falling through the horizon of a black hole.

Not possible. You can't fall through the horizon. Time dilation goes infinite and it takes "beyond the end of time" to fall through it.

Since the spacetime throughout an inertial frame is flat, it must be possible to set up an inertial frame Y that extends throughout X and in which a free test particle, that is above the horizon and moves away from the black hole indefinitely, stays at rest. But GR predicts that nothing may pass outward through a horizon.

I think it ought to predict that nothing can pass inward through a horizon, and the interpretation that yields singularities is wrong.

Then Y cannot extend below the horizon (if only because otherwise a latticework of synchronized clocks, that stays at rest with respect to Y and is spread throughout Y, would be passing outward through the horizon), and so the spacetime cannot be flat throughout X. Then X cannot be an inertial frame. That is, an inertial frame cannot fall through the horizon of a black hole.

I agree with this.

GR required this conclusion, yet the theory predicts otherwise, contradicting itself, so it is self-inconsistent.

Did it? Does it? If Einstein had come up with something in 1955 that cleared up your issue, would the theory we were left with still be called Relativity?

Note that:
• The definitions of flat spacetime and inertial frame in the supporting info account for the tidal force. The tidal force in X need not be nonexistent.
• X can be arbitrarily small and its duration can be arbitrarily short. Then the tidal force in X can be nonexistent in the limit.

It can't be nonexistent. If it was, you would have already met the horizon.

I think we're all agreed here that there's something wrong. I say it's a matter of interpretation and you say GR is inconsistent. But does this matter that much? After all, String Theory doesn't posit strings any more, but it's still going strong. So where do we go from here?

 

QUOTE:
"Outside the scope of time" . . . "renders everything below the horizon hypothetical."


Farsight,

Agreed!

However, in my other thread I show how stable holes can be pulsating since all matter probably has charge. This allows things beneath horizons to be periodically "exposed" to the outside world.



Zanket,

We are almost there. I choose this option at this point:

QUOTE:
. . .physicists simply made a mistaken interpretation. . .

. . .Can GR be self-consistent then? . . .

Then if an object could pass outward through r = 2M, it could directly measure the velocity of an object that fell freely from rest at infinity to be >= c, which contradicts SR. GR cannot contradict SR and be self-consistent, because SR is incorporated into GR.


Zanket, I at least want you to understand my SR analogy before you come to this conclusion. Apparently my previous statement here was not correct.

Zanket, cannot the same argument be made for the hyperbolic horizon? That is,

Then if an object could pass through the ξ[sup]1'[/sup]= -g[sup]-1[/sup] horizon, it could directly measure the velocity of an object to be ≥ c, which contradicts SR. SR cannot contradict itself and be self-consistent.


Apparently, both the Schwarzchild and the Hyperbolic "escape velocities" tend toward c near these horizons. Both of these are accelerated coordinates. But in the Hyperbolic case, the above statement is obviously false since we have a much better understanding of the underlying inertial coordinates. So, take note:

A particle infalling towards r=2M will have a sub-lightspeed velocity according to the outgoing R,T "jettisoned" freefall observer. Also, an infalling particle "in between the horizon and the singularity" will also always have a sub-lightspeed velocity.

So like I said, we are almost there. If by GR you mean Einstein's field equations PLUS the Schwarzchild traditional analysis, then we agree: GR is inconsistent.

But if GR simply means Einstein's field equations, and excludes incorrect application, then GR remains consistent.

How do you want to define?


Andrew A. Gray

 

Farsight, I’ll get back to you.

I think I do understand your SR analogy. I understand a Rindler horizon intuitively, and can imagine one surrounding an object to create what I call a Rindler hole. (Well, almost imagine, since there are issues.) Inside a Rindler hole, you could always move outward, but you still couldn’t pass outward through the Rindler horizon; your increasing r-coordinate could at best asymptote to the horizon. I understand why you are saying that your previous statement was incorrect: because your analysis here shows that an inertial frame can pass outward through r = 2M, in which case there are no black holes. BTW, the pages you showed from Gravitation are the source of the derivation of the relativistic rocket equations; see the section Below the rocket, something strange is happening... for the author’s description of a Rindler horizon.

Can you explain those last two statements in more of laymen’s terms? I don’t see how you have shown that. (Keep in mind that I’m not as mathematical as you.) If your analysis shows that, then it contradicts another longstanding analysis of GR, and not just an interpretation. Infalling free-fall coordinates here, derived from the Schwarzschild metric, show that an observer falling freely from rest at infinity reaches c at r = 2M; that could be directly measured, violating SR, if an object could remain stationary or be moving outward there. Your analysis of outgoing free-fall coordinates shows that an observer can move outward there. Then if you are right about that, GR is inconsistent with SR, hence self-inconsistent. Note that these are both inertial observers. The presence of a Rindler horizon would not deny this conclusion; that’s why I ignored your analogy before. Assuming that infalling free-fall coordinates are correct, the only way these two observers could not directly measure a velocity >= c for each other is if all observers at and below r = 2M must fall inward, which contradicts your analysis and is the interpretation that predicts black holes.

 

Zanket,

Consider three observers A, B and C near the Hyperbolic Horizon shown below.

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Observer A is sub lightspeed (45 degree lines are lightspeed world lines). He enters the "dark plane" and is gone. He can never again be seen by the hyperbolic frame. He is gone forever.

However, observer B is traveling at nearly the speed of light. He can make it quite a ways out to infinity. So as observers near the Hyperbolic Horizon, their escape velocity nears c.

Observers like C who are "behind the horizon", will never catch up to it.

Notice, however that there is really nothing there at the horizon. The whole "dark plane" phenomena comes from the fact that the hyperbolic frame is accelerating. "Spanning the horizon" with inertial coordinates is no problem. This is flat spacetime.

QUOTE:
"A particle infalling towards r=2M will have a sub-lightspeed velocity according to the outgoing R,T "jettisoned" freefall observer. Also, an infalling particle "in between the horizon and the singularity" will also always have a sub-lightspeed velocity."


Well, as I previous noted: The outgoing freefall metric is well-behaved everywhere and T is everywhere timelike and R is everywhere spacelike! No coordinate singularity at the horizon!

What does this mean? Well, a well behaved 3+1 metric will always have sublight particle goedesics by construction. Particle velocities approach c asymtotically only in metrics with coordinate singularities. To see this, consider a our well-behaved 3+1 metric:

\(-d \tau^2 = g_{oo} dT^2 + g_{11}dR^2 + g_{22}d \theta^2 + g_{33}d \phi^2 \)

where g[sub]oo[/sub] is inherently negative.
For light, dτ=0, so if we divide through by g[sub]oo[/sub] dT[sup]2[/sup] for light we have:

\(1 = \frac{g_{11}dR^2}{|g_{oo}|dT^2} + \frac{g_{22}d\theta^2}{|g_{oo}|dT^2} + \frac{g_{33}d\phi^2}{|g_{oo}|dT^2} \)
or
\( 1 = V_R^2 + V_\theta^2 + V_\phi^2 \)

For particle geodesics, dτ≠0, and similarly we get:

\( 1 - \frac{d\tau^2}{|g_{oo}|dT^2} = V_R^2 + V_\theta^2 + V_\phi^2 \)

So all particle geodesics are clearly sub-lightspeed by design.


Here are the outgoing freefall frames again:

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Notice that the horizon is moving at c in these coordinates, but light traveling in the same direction never catches up or leaves the horizon behind.

Edit: This next statement is incorrect.
The singularity is also traveling at c in these coordinates. Nothing ever catches up to it in these inertial frames.
Edit: The singularity is spacelike, as light is not at 45[sup]o[/sup] everywhere in this diagram. AAG 5-13-07

It is extremely similar to the hyperbolic horizon. It also seems that there is nothing really there at the Schwarzchild horizon either. The underlying inertial frames are well behaved.

Since the Schwarzchild coordinates are being accelerated away, a horizon forms.

One last thing. The Schwarzchild velocity is defined as the velocity that an observer at fixed r would measure. But this velocity would be impossible to measure since the coordinate r=2M is moving at c relative to an inertial frame. (A coordinate can move at c with no problem.) So measuring the Schwarzchild velocity fixed at the r=2M coordinate would be impossible since it is moving at c.

So again, I see no contradictions except that the Schwarzchild analysis was done incorrectly.


Andrew A. Gray

 

I’m confused by your post as a whole. It seems like you are contradicting yourself. Do you still agree that something (a material object or light) can pass outward through r=2M?

The Schwarzschild observer (a hovering observer) is at a fixed r, right? Then the hyperbolic horizon (or the similar one you are proposing) is also at a fixed r, right?

Are you saying that you think the interpretation of the Schwarzschild metric that predicts black holes is incorrect? That is, do you think black holes are precluded by Einstein’s field equations? Or what?

 

Zanket,

If the previous analysis is correct (and I still do not see any mistakes),
then yes, I believe that light and particles can escape the horizon, in their own proper time.
The external observer, however would never be able to see it happen,
so in this sense, it is more like a "frozen hole", like farsight says.

QUOTE:
"The Schwarzschild observer (a hovering observer) is at a fixed r, right?
Then the hyperbolic horizon (or the similar one you are proposing) is also at a fixed r, right?"


Technically it is at a fixed ξ[sup]1'[/sup]. The "horizon" is at fixed ξ[sup]1'[/sup] = -g[sup]-1[/sup].

Also, for the hyperbolic frame, notice that for path A in the above diagram, that time ξ[sup]0'[/sup]
goes to infinity before observer A gets to the "horizon". So in a sense,
according to the accelerated observer, it actually is a "frozen plane", nothing able to get past
according to ξ clocks.


QUOTE:
"Are you saying that you think the interpretation of the Schwarzschild metric that predicts black holes is incorrect?
That is, do you think black holes are precluded by Einstein’s field equations? Or what?"


Well, according to the external observer "at infinity", nothing comes out of the horizon,
as it is frozen in time. He never sees anything go into or out of the horizon. Ever.
So in that sense, it is a "black hole".

However, just like a falling observer can fall through the horizon and "see" below it in his own proper time,
an observer can be jettisoned outward through the horizon and see the outside within his own proper time.
At least that is what the equations say.

So the statement, "nothing can ever get out of a black hole" seems technically incorrect to me.

It would be interesting to get the geodesic equations of motion on the other side of the planar "horizon"
in terms of the ξ' coordinates, and see if ξ[sup]0'[/sup] acts like t does inside the Schwarzchild horizon.
I'll bet it does. From the metric 6.18 (see above), ξ[sup]0'[/sup] becomes spacelike on the other side of the horizon,
and there appears to be no time coordinate.


Andrew A. Gray

 

OK, I understand that we’re assuming that for now.

Do you agree that infalling free-fall coordinates (here) show that an object that falls freely from rest at infinity passes r=2M at c? If so, then if an object can pass outward through r=2M as you say, how do you explain that these two objects could not pass right by each other and directly measure a velocity (for each other) >= c?

My understanding is that, given infalling free-fall coordinates, GR would violate SR unless everything at r <= 2M falls (which led to the interpretation that predicts black holes). Your analysis says otherwise; then it shows that GR is self-inconsistent.

That seems illogical to me. If an object can pass outward through r=2M, then it can reach an external observer who hovers just above r=2M. How could such external observer never see it happen when the object could even touch that observer?

Or are you saying that the external observer noninertially accelerates and moves outward? I thought by “external observer” you meant an observer hovering at a fixed r > 2M.

The universe the observer jettisons into is not in a different dimension where external observers are missing. What prevents the jettisoned observer who is at, say, r=2.1M, from touching an external observer who hovers there?

And if so, then there are no black holes, since by definition nothing can pass outward through the horizon of a black hole. It may look black to an external observer, but it need not be a singularity; the surface can be at a fixed r > 0. If we are right, then GR does not allow black holes.

Do you think that GR’s math is okay, but it was just a bad interpretation that led to the prediction of black holes?

 

You have your event horizon on the wrong side of the singularity.
You have diagrammed a white hole, not a black hole.

 

First to Pete,

Well, unless I have made a mistake in the math, then if we consider this outgoing coordinate equation again:

\(r=(2M)^{1/3} \left ( \frac{3}{2} (T-R) \right )^{3/2}\)

and we let r=2M and we let M=3/4 for simplicity, then we get

\( \frac 3 2 = (\frac 3 2 )^{1/3} \left ( \frac 3 2 (T-R) \right )^{3/2} \)

or

\( 1 = T- R \)

This line, representing the "horizon", is a 45[sup]o[/sup] line passing thru T=1, R=0.

If we let r=0, then we get

\( 0 = T-R \)

This is a 45[sup]o[/sup] line passing thru the (R,T) origin.

Hence, it seems like the r=2M "horizon" is on the "top side" of the r=0 line.

And if we used the "bottom side" of the r=0 line , then it seems like there would be no positive "increasing time T observer" that could escape r=0. But this we know is incorrect since observers jettisoned from just outside the horizon can make it to infinity for sure.

So it seems to me that the graph is correct, at least if the previous math is correct.


Next, to Zanket,

"Do you agree that infalling free-fall coordinates show that an object that falls freely from rest at infinity passes r=2M at c?"


No. Now I understand that velocities in accelerated coordinates are not proper velocities. Look at this diagram again:

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The ξ[sup]1'[/sup]= -g[sup]-1[/sup] coordinate is moving at c as measured by an inertial frame. That means that any inertial observer that crosses ξ[sup]1'[/sup]= -g[sup]-1[/sup] will "see" the "horizon" moving at lightspeed. Very similar "paradoxes" apply here as well if one gets caught up in the accelerated-coordinates-velocity trap. For example, the "hyperbolic velocity" of an observer fixed at x[sup]1[/sup]=1 would then be c as it passed the "planar horizon". But this is flat spacetime.

The same is true for the r=2M coordinate. It is moving at lightspeed relative to any inertial observer. So no inertial oberver ever travels at c according to any other inertial observer. The accelerated-coordinate Schwarzchild velocity of an object may reach c, but this is not a proper velocity, just like the hyperbolic velocity is not really proper. And it is much easier to see the truth in flat spacetime. The Laws of Physics do not really apply to accelerated frames.

It would be interesting to plot the outgoing jettisoned geodesics in the infalling coordinates. Perhaps I will do this later if I get the time.

By "external observer", I meant one that is far away from the hole. To this observer, all events near the horizon seem "frozen".

However, the accelerated observer, accelerating just outside the "horizon" at fixed r can see something escape from the hole, it would seem to me. (if previous equations are correct.)

QUOTE:
"And if so, then there are no black holes, since by definition nothing can pass outward through the horizon of a black hole. It may look black to an external observer, but it need not be a singularity; the surface can be at a fixed r > 0. If we are right, then GR does not allow black holes."


Agreed. The holes would not be black. They would look frozen and red shifted to the observer at infinity. The singularity concept seems to have come from the timelike r-coordinate, below r=2M. Since r can decrease, and since timelike variables are claimed to be uni-directional, then it was claimed that r must always march onward towards zero, no matter what. But r is an accelerated coordinate. I'm not sure that a timelike accelerated coordinate needs to be unidirectional. Heck, the ξ time coordinate becomes spacelike on the other side of the horizon, and there is no timelike coordinate. So treating accelerated coordinates like proper inertial coordinates was probably the source of the misinterpretation. If the outgoing math is correct, no singularities must form.


QUOTE:
Do you think that GR’s math is okay, but it was just a bad interpretation that led to the prediction of black holes?


Yes. Probably so.



Andrew A. Gray

 

But your graph can't be correct - according to your graph, it is impossible for anything to cross the event horizon from the outside.

I haven't attempted to confirm your maths - I may not have the ability - but your logic doesn't ring true.

I'm not happy with that graph for other reasons as well - for starters, the horizon is only moving at c for freefall observers at the horizon. It is moving faster than c for observers inside the horizon, and slower than c for observers outside the horizon...

I think that the horizon needs to be on the right hand side of the singularity in your diagram, and geodesics on both sides of the horizon should curve away from it.

 

Pete,

QUOTE:
But your graph can't be correct - according to your graph, it is impossible for anything to cross the event horizon from the outside.

This may not be a problem. The same is true for the Schwarzchild observer for inbound light:

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According to regular Schwarzchild analysis, the Schwarzchild time t also goes to infinity before light (or anything else) enters the horizon. So it is "impossible for anything to cross the event horizon from the outside" according to "us", as well. And since the jettisoned observer becomes very similar to the Schwarzchild observer when far away, it only makes sense that this would be the same for him as well.

Don't forget, at the horizon, the outgoing inertial observer has V[sub]Schwarz[/sub]=c, while the ingoing observer has V[sub]Schwarz[/sub]=-c. (improper accelerating coordinate velocities).
Thus, the outgoing inertial observer has a outward velocity of c relative to the r=2M Schwarzchild accelerating coordinate near the horizon. It would only make sense that sending a light signal back towards the horizon would be troublesome for it to get there. (Very similar to the trouble the ingoing observer has sending light back out of the horizon!)

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QUOTE:
I'm not happy with that graph for other reasons as well - for starters, the horizon is only moving at c for freefall observers at the horizon. It is moving faster than c for observers inside the horizon, and slower than c for observers outside the horizon...


Pete, I don't understand this.
We have for light at the horizon:

\(\frac{dR}{dT} = (T-R)^{1/3} = 1\)

At the horizon, T-R=1, and we have a 45[sup]o[/sup] line for light.

Inside the horizon, say at T-R=½, we have:

\(\frac{dR}{dT} = (T-R)^{1/3} = (.5)^{1/3}= .79\)

so we have we have a 52[sup]o[/sup] line for light.


Outside the horizon, say at T-R=2, we have:

\(\frac{dR}{dT} = (T-R)^{1/3} = (2)^{1/3}= 1.26\)

so we have we have a 38[sup]o[/sup] line for light.

Is this what you are referring to?


QHOTE:
I think that the horizon needs to be on the right hand side of the singularity in your diagram, and geodesics on both sides of the horizon should curve away from it.


Perhaps it is possible to put the horizon on the right side of the singularity. Remember:

\(T-R=\sqrt{2/(9M)} r^{3/2}\)

This is the same as:

\(T-R=\pm \sqrt{2r^3/(9M) }\)

Where the ± comes from the redundancy of the square root. So we actually get:

\(r = \pm (2M)^{1/3} \left ( \frac{3}{2}(T-R) \right )^{2/3} \)


So choose the minus sign and let r=2M=2(¾). Then we get for the horizon:

\(T-R = -1\)

And finally,

\(T= R-1\)

This puts the horizon on the right side of the singularity:

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However, we know that constant R lines make it out to infinity, especially if they start outside the horizon. So this means that if we pick the minus sign, that time T runs in the negative direction, and it appears that everything is still the same. Hmm.




Andrew A. Gray

 

OK, my understanding is that you think r=2M (technically, ξ[sup]1'[/sup]= -g[sup]-1[/sup]) is a hyperbolic horizon for this observer.

Given that, I can understand why you think the things you say are true. But I still see a contradiction in your logic. You say:

But then nothing prevents that something from reaching any r, including the one where the external observer sits. Which contradicts you; that is, you are contradicting yourself.

If something can pass outward through r=2M, then it can subsequently escape to infinity, according to GR. Before it reaches infinity it will pass the external observer, who can see it. Then there cannot be a hyperbolic horizon below the external observer.

I understand a hyperbolic horizon in SR. But it seems that any theory of gravity that incorporates SR and predicts that such a horizon is below any hovering observer (an accelerating observer who is fixed at some r) is self-inconsistent.