Crossing the Event Horizon of a Black Hole

[Dmitry67]

And yet another example.
Look at our world from the SUPERLUMINAL rest frame.
You see the same - imaginary time etc.
But it does not affect how we observe things, right?

 

[Ich]

I found a weird metric with imaginary time and space:
ds² = -dx² + dt² +dy² + dz²
Obvoiusly such spacetimes can't exist, as you can't see your nose there.

 

[stevebd1]

No. They won't see anything at all. In order to "see" something light has to travel from somewhere outside to the retina and then the electrical impulses travel to your brain.

Inside the EH, all light cones are timelike and nothing (not even light) will travel backwards to hit the retinas.

I recommend you take a look at http://en.wikipedia.org/wiki/Gullstrand-Painlev%C3%A9_coordinates" which can be expressed in two forms, free-fall rain frame and global rain frame (r_s=2M)-

Free-fall rain frame-

c^2d\tau^2=c^2dt_r^2 - dr_r^2 - r^2\left(d\theta^2 + \sin^2\theta \, d\phi^2\right)

Global rain frame-

c^2d\tau^2=\left(1-\frac{r_s}{r}\right)c^2dt_r^2 - 2\sqrt{\frac{r_s}{r}}\ cdt_rdr - dr^2 - r^2\left(d\theta^2 + \sin^2\theta \, d\phi^2\right)

where

<br /> \begin{flalign}<br /> &amp;dt_r=dt-\beta\gamma^2dr\\[6mm]<br /> &amp;dr_r=\gamma^2dr-\beta dt<br /> \end{flalign}<br />

and

<br /> \begin{flalign} <br /> &amp;\gamma=\frac{1}{\sqrt{1-\beta^2}}\\[6mm]<br /> &amp;\beta=-\sqrt{\frac{r_s}{r}} <br /> \end{flalign}<br />

where \gamma is the Lorentz factor and \beta is the velocity of the rain frame relative to the shell frame.

The principle behind the form is-

\gamma=\frac{1}{\sqrt{1-(v/c)^2}}\equiv \frac{1}{\sqrt{1-(r_s/r)}}

Basically length contraction induced by velocity (which in turn is induced by curvature) balances out the length expansion induced by gravity. This is identical to 'Metric for the Rain Frame' shown on page B-13 of Exploring Black Holes by Taylor & Wheeler.

The free-fall rain frame proper time remains time-like all the way to the singularity, representing the frame of the in-falling observer while the global rain-frame proper time becomes negative at r<2M but with no geometric singularity at the event horizon (dr remains 1).

 

[JesseM]

1. All of the above coordinate transformations show that space or time separations below r=2M are imaginary.

Do you claim that the time separation of events along a timelike worldline is imaginary in Kruskal-Szkeres coordinates? If so I'm pretty certain your wrong, perhaps you'd care to address atyy's post #81 above. As atyy noted, it's true you'll get an imaginary value if you try to calculate the proper time along a spacelike curve, but this is always true in relativity, even in flat SR spacetime (and in KS coordinates it should be just as true outside the horizon as inside it).

 

[DiamondGeezer]

Do you claim that the time separation of events along a timelike worldline is imaginary in Kruskal-Szkeres coordinates? If so I'm pretty certain your wrong, perhaps you'd care to address atyy's post #81 above. As atyy noted, it's true you'll get an imaginary value if you try to calculate the proper time along a spacelike curve, but this is always true in relativity, even in flat SR spacetime (and in KS coordinates it should be just as true outside the horizon as inside it).

No I don't. If you read carefully you'll find I talk about the traditional spacetime separations between events. I know what the difference is between timelike and spacelike and its irrelevant to the arguments.

What I point out (tediously repeating myself I know) is that "inside" a black hole event horizon, spacelike and timelike separations are always imaginary. The various cited coordinate transformations end up with the same result, and curiously nobody seems to be perturbed by the mathematical gyrations which appear to explain what happens when a particle reaches the event horizon in order to allow it to reach the center of curvature.

All of this should tell you lots about the Schwarzschild Metric, but unfortunately nobody's paying any attention.

Nothing I have heard on this thread has shown any insight into the question of black holes or event horizons. Instead I read repeatedly the same arguments which are ipso facto mathematically and physically absurd when examined.

It's not arrogance on my part to say that the mathematical treatment which gives rise to the theory of black holes is badly flawed.

 

[Dmitry67]

DiamondGreezer, did you check my diagram?
Do you agree with it or not?

 

[DiamondGeezer]

DiamondGreezer, did you check my diagram?
Do you agree with it or not?

Yes.

No.

 

[Dmitry67]

Yes.

No.

Then put your version of events on
http://www.valdostamuseum.org/hamsmith/DFblackIn.gif

(Bob and Alice are freely falling into BH looking at each other)

 

[DrGreg]

I know what the difference is between timelike and spacelike and its irrelevant to the arguments.

What I point out (tediously repeating myself I know) is that "inside" a black hole event horizon, spacelike and timelike separations are always imaginary.

If you really did know the difference between timelike and spacelike then you would understand that "imaginary spacelike" means "real timelike" and "imaginary timelike" means "real spacelike".

It's not arrogance on my part to say that the mathematical treatment which gives rise to the theory of black holes is badly flawed.

Just because you don't understand something means you are right and the many, many thousands of relativists who have studied this, some of them eminent physicists and mathematicians, are all wrong?

 

[DiamondGeezer]

If you really did know the difference between timelike and spacelike then you would understand that "imaginary spacelike" means "real timelike" and "imaginary timelike" means "real spacelike".

No. It's irritating that I have to keep repeating myself to people who really should know better.

viz.,

(the square root of a negative quantity)\neq(the negative square root of a positive quantity)

Just because you don't understand something means you are right and the many, many thousands of relativists who have studied this, some of them eminent physicists and mathematicians, are all wrong?

If science was really decided by majority voting then relativity would have been rejected more than 100 years ago, when only Einstein understood it. You choose a really poor subject to make such a majoritarian fallacy.

It wasn't so long ago that the majority of experts knew that there were WMDs in Iraq - completely wrong of course.

I don't compare myself to Einstein - he was a genius. Nor do I criticize General Relativity in general. I don't claim to understanding of all of General Relativity.

But on this one small point, the majority on this particular issue are making a mathematical case which is false - and I believe I do have an answer but first there has to be an acknowledgment of a problem.

It does take a certain amount of arrogance to say that "I am right and everyone else is wrong" but there are so many examples of an outsider pointing out something that has somehow escaped the experts in the field that its no longer seen as weird.

I don't know of a single scientist who doesn't have a least 10 hypotheses that goes against what "the majority of experts in the field" believe to be true.

I am perfectly prepared to be wrong (and I may well be). But the argumentation on the issue of black holes and event horizons is mathematically weak and physically contrived - in my humble opinion.

 

[George Jones]

Suppose the components of the metric with respect to a particular coordinate system are given by

ds^2 = -dt^2 - dx^2 + dy^2 - dz^2.

What do the coordinates t, x, y, and z represent?

 

[DiamondGeezer]

Suppose the components of the metric with respect to a particular coordinate system are given by

ds^2 = -dt^2 - dx^2 + dy^2 - dz^2.

What do the coordinates t, x, y, and z represent?

George.

Please patronize someone else.

 

[hamster143]

DiamondGeezer, do you need the email of the president of physics?

 

[Ich]

Suppose the components of the metric with respect to a particular coordinate system are given by

ds^2 = -dt^2 - dx^2 + dy^2 - dz^2.

Hey, I claim the https://www.physicsforums.com/showthread.php?p=2497321#post2497321"for this example.

 

[DiamondGeezer]

DiamondGeezer, do you need the email of the president of physics?

No. I have seen the original cartoon though...

Unfortunately I'm not proposing some grandiose thought experiment which undermines relativity...I'm actually interested in the interpretations of the mathematics of black holes, which appear not to make sense from the known laws of mathematics.

Dull, but true.

ETA: Did you ever wonder how many PhDs in Phlogiston Theory there were when the atomic theory was first established? Did majorities work then, Greg?

 

[hamster143]

spacelike and timelike separations are always imaginary.

No. It's irritating that I have to keep repeating myself to people who really should know better.

viz.,

(the square root of a negative quantity)\neq(the negative square root of a positive quantity)

For any two infinitesimally separated events, metric gives us an interval, ds^2 = g_{ab} dx^a dx^b.

We DEFINE "spacelike separations" as those with ds^2 &gt; 0 and "timelike" as those with ds^2 &lt; 0 (or vice versa, depending on which textbook you use).

Schwarzschild metric has a coordinate singularity at the event horizon. On the outside, separations (dt,0,0,0) are timelike and separations (0,dr,0,0) are spacelike. When you cross the event horizon, signs of g_{rr} and g_{tt} are reversed and those separations become spacelike and timelike, respectively.

I don't see a problem here.

 

[hamster143]

No, they both become imaginary as well.

The Kruskal-Szekeres tranformed coordinates are:

u=\left[\left(\frac{r}{2m^{*}}\right)-1\right]^{\frac{1}{2}}\exp\left(\frac{r}{4m^{*}}\right)\cosh\left(\frac{T}{4m^{*}}\right)

and

v=\left[\left(\frac{r}{2m^{*}}\right)-1\right]^{\frac{1}{2}}\exp\left(\frac{r}{4m^{*}}\right)sinh\left(\frac{T}{4m^{*}}\right)

which also become imaginary because of

\left[\left(\frac{r}{2m^{*}}\right)-1\right]^{\frac{1}{2}}

when r<2M

The correct statement would be that the transformation from Schwarzschild to Kruskal coordinates has a singularity at r=2M, for the same reason why the Schwarzschild metric has a singularity there. That is a problem with our initial (faulty) coordinate system, not with the underlying physics.

The Kruskal metric itself,

ds^{2} = \frac{32G^3M^3}{r}e^{-r/2GM}(-dT^2 + dR^2) + r^2 d\Omega^

is well-defined and has no singularities at the event horizon (T = R).

 

[DiamondGeezer]

The correct statement would be that the transformation from Schwarzschild to Kruskal coordinates has a singularity at r=2M, for the same reason why the Schwarzschild metric has a singularity there. That is a problem with our initial (faulty) coordinate system, not with the underlying physics.

The Kruskal metric itself,

ds^{2} = \frac{32G^3M^3}{r}e^{-r/2GM}(-dT^2 + dR^2) + r^2 d\Omega^

is well-defined and has no singularities at the event horizon (T = R).

The Schwarzschild Metric has no singularities at the event horizon either. Wherever did you get the idea that there was a singularity at r=2M ? Not from me.

The Kruskal Metric is defined in terms of two new coordinates U and V which ALSO go imaginary when r<2M. Same "faulty" coordinates?

 

[hamster143]

The Schwarzschild Metric has no singularities at the event horizon either. Wherever did you get the idea that there was a singularity at r=2M ? Not from me.

The Kruskal Metric is defined in terms of two new coordinates U and V which ALSO go imaginary when r<2M. Same "faulty" coordinates?

The Schwarzschild metric has a coordinate singularity at the event horizon: g_{tt} crosses through zero and g_{rr} diverges. The physical meaning of this singularity is that, at the event horizon, the lightcone becomes tangential to the event horizon (as shown in Dmitry's pictures).

There's no problem in the Kruskal Metric in terms of U and V, either:

ds^{2} = -\frac{32G^3M^3}{r}e^{-r/2GM}(dU dV) + r^2 d\Omega^2,

U V = \left(1-\frac{r}{2GM}\right)e^{r/2GM}.

Once again, U and V coordinates are good, but the transformation law from Schwarzschild metric has some kinks around the event horizon.

 

[DiamondGeezer]

For any two infinitesimally separated events, metric gives us an interval, ds^2 = g_{ab} dx^a dx^b.

We DEFINE "spacelike separations" as those with ds^2 &gt; 0 and "timelike" as those with ds^2 &lt; 0 (or vice versa, depending on which textbook you use).

Schwarzschild metric has a coordinate singularity at the event horizon. On the outside, separations (dt,0,0,0) are timelike and separations (0,dr,0,0) are spacelike. When you cross the event horizon, signs of g_{rr} and g_{tt} are reversed and those separations become spacelike and timelike, respectively.

I don't see a problem here.

Except that you have claimed that the "signs of g_{rr} and g_{tt} are reversed" when r<2M, which is wrong. Both separations spacelike or timelike become imaginary when r<2M - they do not become the other, they become something else entirely.

That is the mathematical fudge which I have been pointing out for pretty much every post in this thread.

I'm really tired of repeating myself over and over.

 

[hamster143]

Except that you have claimed that the "signs of g_{rr} and g_{tt} are reversed" when r<2M, which is wrong. Both separations spacelike or timelike become imaginary when r<2M - they do not become the other, they become something else entirely.

When ds^2 changes sign, that means that spacelike separations become timelike and timelike separations become spacelike. That's the definition of "spacelike" and "timelike". No mathematical fudge involved, just an imperfect coordinate system.

In Schwarzschild case, we built our coordinate system on the presumption that the whole system has time translation symmetry (there's a timelike Killing field at every point of spacetime). Only it turns out that we're wrong, and the Killing field sufficiently near the center is really spacelike. The hypersurface where Killing field goes from timelike to spacelike is called "event horizon" and our original coordinate system (which we built upon a faulty assumption) suffers a bunch of coordinate singularities there. Thus the motivation to write other coordinate systems, such as Kruskal, which aren't as ill-behaved on the surface of the event horizon.

 

[DiamondGeezer]

For any two infinitesimally separated events, metric gives us an interval, ds^2 = g_{ab} dx^a dx^b.

We DEFINE "spacelike separations" as those with ds^2 &gt; 0 and "timelike" as those with ds^2 &lt; 0 (or vice versa, depending on which textbook you use).

Schwarzschild metric has a coordinate singularity at the event horizon. On the outside, separations (dt,0,0,0) are timelike and separations (0,dr,0,0) are spacelike. When you cross the event horizon, signs of g_{rr} and g_{tt} are reversed and those separations become spacelike and timelike, respectively.

I don't see a problem here.

The Schwarzschild metric has a singularity at the event horizon: g_{tt} crosses through zero and g_{rr} diverges.

There's no problem in the Kruskal Metric in terms of U and V, either:

ds^{2} = -\frac{32G^3M^3}{r}e^{-r/2GM}(dU dV) + r^2 d\Omega^2,

U V = \left(1-\frac{r}{2GM}\right)e^{r/2GM}.

Sweet Jeebus do I have to repeat myself again?

U V = \left(1-\frac{r}{2GM}\right)e^{r/2GM} becomes negative when r<2GM, which means that for an infalling particle (d\Omega^2 =0) then the dU dV also becomes negative WHICH MEANS THAT THE SPACETIME SEPARATION BECOMES IMAGINARY BECAUSE ITS THE SQUARE ROOT OF A NEGATIVE NUMBER.

End of story. End of discussion.

 

[DiamondGeezer]

When ds^2 changes sign, that means that spacelike separations become timelike and timelike separations become spacelike. That's the definition of "spacelike" and "timelike". No mathematical fudge involved, just an imperfect coordinate system.

In Schwarzschild case, we built our coordinate system on the presumption that the whole system has time translation symmetry (there's a timelike Killing field at every point of spacetime). Only it turns out that we're wrong, and the Killing field sufficiently near the center is really spacelike. The hypersurface where Killing field goes from timelike to spacelike is called "event horizon" and our original coordinate system (which we built upon a faulty assumption) suffers a bunch of coordinate singularities there. Thus the motivation to write other coordinate systems, such as Kruskal, which aren't as ill-behaved on the surface of the event horizon.

At the event horizon (and below) there's no such thing as a spatial separation. All separations are (imaginary) and solely timelike - whatever that means.

 

[hamster143]

Sweet Jeebus do I have to repeat myself again?

U V = \left(1-\frac{r}{2GM}\right)e^{r/2GM} becomes negative when r<2GM, which means that for an infalling particle (d\Omega^2 =0) then the dU dV also becomes negative WHICH MEANS THAT THE SPACETIME SEPARATION BECOMES IMAGINARY BECAUSE ITS THE SQUARE ROOT OF A NEGATIVE NUMBER.

End of story. End of discussion.

... no. UV can become negative, that's not a problem. \frac{32G^3M^3}{r}e^{-r/2GM} remains positive. dU dV does not change sign at r=2M.

 

[Ich]

Except that you have claimed that the "signs of g_{rr} and g_{tt} are reversed" when r<2M, which is wrong. Both separations spacelike or timelike become imaginary when r<2M - they do not become the other, they become something else entirely.

Ah.
g_{tt}=1-\frac{2M}{r}
does not become negative for r<2M, but imaginary?

It seems this identifies your problem: You don't know what g_tt is, and so you wrongly claim a difference between an imaginary space coordinate and a time coordinate.

 

[hamster143]

At the event horizon (and below) there's no such thing as a spatial separation. All separations are (imaginary) and solely timelike - whatever that means.

When you're falling into a black hole, different parts of your body remain spatially separated all the way to the singularity.

 

[DrGreg]

The Kruskal Metric is defined in terms of two new coordinates U and V which ALSO go imaginary when r<2M. Same "faulty" coordinates?

George already dealt with this in post #54.

The equations which relate Kruskal-Szekeres coordinates to outside-the-horizon Schwarzschild coordinates are different to the equations which relate Kruskal-Szekeres coordinates to inside-the-horizon Schwarzschild coordinates. (In case this is what you have been referring to when you say people are claiming that \sqrt{-x}=-\sqrt{x}, look more closely: the sinh and cosh are swapped over too.)

Note that it isn't the Kruskal-Szekeres coordinates that change; it's the Schwarzschild coordinates. There isn't a single Schwarzschild coordinate system; there are two disjoint systems, one strictly outside the horizon and a completely different one strictly inside. They just happen to share the same metric equation. There are no Schwarzschild coordinates defined actually on the horizon itself; although the horizon is represented by the limit r \rightarrow 2M, there is no finite value of t for any event on the horizon.

If you're calculating the proper time along the worldline of a particle falling through the horizon using Schwarzschild coordinates, you have no choice but to split the integral into two parts, because there are two different disjoint coordinate systems. To do it in a single integral you have to use a coordinate system, such as Kruskal-Szekeres, that is defined along the whole worldline.

I still think you'd find this easier to understand if you'd look at the Rindler example I keep on mentioning. Exactly the same issue arises there; the equations relating (T,R) to Minkowski coordinates (t,x) are different on either side of the horizon and the (T,R) coordinates are undefined on the Rindler horizon itself.

 

[Ich]

If you're calculating the proper time along the worldline of a particle falling through the horizon using Schwarzschild coordinates, you have no choice but to split the integral into two parts, because there are two different disjoint coordinate systems.

I have done this and, unless my memory fails me completely, the coordinate singularity vanishes before you come to evaluate the integral. No need to split the integral.

 

[DiamondGeezer]

When you're falling into a black hole, different parts of your body remain spatially separated all the way to the singularity.

Another "proof by assertion".

If there is no space to be measured, there cannot be "spacial separation". Also, there are no geodesics to the "singularity" because at the event horizon, all line integrals are normal to the spatial direction.

 

[DiamondGeezer]

George already dealt with this in post #54.

The equations which relate Kruskal-Szekeres coordinates to outside-the-horizon Schwarzschild coordinates are different to the equations which relate Kruskal-Szekeres coordinates to inside-the-horizon Schwarzschild coordinates. (In case this is what you have been referring to when you say people are claiming that \sqrt{-x}=-\sqrt{x}, look more closely: the sinh and cosh are swapped over too.)

Note that it isn't the Kruskal-Szekeres coordinates that change; it's the Schwarzschild coordinates. There isn't a single Schwarzschild coordinate system; there are two disjoint systems, one strictly outside the horizon and a completely different one strictly inside. They just happen to share the same metric equation. There are no Schwarzschild coordinates defined actually on the horizon itself; although the horizon is represented by the limit r \rightarrow 2M, there is no finite value of t for any event on the horizon.

Ah, but DrGreg if there are two "disjoint" systems then there cannot be a continuous worldline that connects the infalling particle from the Universe to the singularity.

If you're calculating the proper time along the worldline of a particle falling through the horizon using Schwarzschild coordinates, you have no choice but to split the integral into two parts, because there are two different disjoint coordinate systems. To do it in a single integral you have to use a coordinate system, such as Kruskal-Szekeres, that is defined along the whole worldline.

But Kruskal-Szekeres has the same problem with the inner and outer integrals as I've pointed out. What K-S does is define the two integrals where one takes the place of the other when going through zero at the event horizon, producing a continuous function.

But as I've pointed out, that is no more a solution than the original Schwarzschild system, since the spatial and time separations becomes imaginary anyway.

If you like, the various coordinate transformations all produce entirely timelike paths at the EH.

In the real universe outside of the event horizon, when ds^2<0 then spacetime separations are timelike. But below the event horizon, even the timelike separations are imaginary.

There is another coordinate transformation of the Schwarzschild Metric that prevents any infalling particle from reaching the event horizon entirely, let alone heading for the supposed singularity within. In that scenario, the prevention of particles reaching the EH is caused by quantum theory.

The bad news is, is that in that scenario, black holes do not exist. The good news is, is that the different transformation is testable and falsifiable from the others.