Is this 'Einstein' transforming away the r=2m singularity.

I would like to know if my understanding here is correct.
It's an Einstein and Rosen paper.
Are Einstein and Rosen showing how to take away the singularity at r=2m. I think we now call this the coordinate singularity at r=2m.
My interest here is that it would be the first time I have come across Einstein himself mentioning that this particular singularity at r=2m can be transformed away.

http://www.nevis.columbia.edu/~zajc/acad/W3072/EinsteinWormhole.pdf

The PDF page 75. Section 2. The Schwarzschild Solution.

I get the impression their trying to construct particles and it puts me in mind of Penrose's twisters.
Never the less, ( I think) it shows that Einstein himself thinks the the r=2m singularity of the Schwarzschild solution to the EFE can be transformed away.

 

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http://www.mathpages.com/home/kmath375/kmath375.htm

Not sure if that offers any solution or answer to your question sweetpea, my main intention is to give this a bump in case its been missed and lost.

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r = 2m is a coordinate singularity in schwarzchild solution due to specific coordinates chosen by him...moreover coordinate singularities are not the part or character of the spacetime, they are present due to coordinates chosen, so can be gotten rid of by chosing different coordinates..r = 2m singularity can be removed by chosing Kruskal–Szekeres coordinates.....on the other hand r = 0 is a coordinate independent singularity can be termed as property of the spacetime...

Are you trying to say that Einstein thought r = 2m singualrity is real (non coordinate type) or coordinate type but he had no transformation at that time ?

 

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Thanks for that Paddo.
Remembering the reverence some give to every Einstein word. I find it interesting this is the man himself explaining away the singularity at r=2m, I think.

Remember according to some we are all POPSICLES (every frame) on the event horizon, what a blow if this shows Einstein himself telling us we can tranform away that particular singularity.

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Actually, he just rescaled \(r - \frac{2 Gm}{c^2} \to u^2, \quad dr \to 2 u \, du\)

So (in units where G=c=1), we have, \(r = u^2 + 2m\) and

\((1 - 2m/r)dt^2 - (1 - 2m/r)^{-1} dr^2 - r^2 (d \theta^2 + \sin^2 \theta \, d\phi^2 ) \\ = ( 1 - \frac{2m}{u^2 + 2m} ) dt^2 - ( 1 - \frac{2m}{u^2 + 2m} )^{-1} 4u^2 du^2 - (u^2 + 2m)^2 (d \theta^2 + \sin^2 \theta \, d\phi^2 ) \\ = \frac{u^2}{u^2 + 2m} dt^2 - 4 ( u^2 + 2m ) du^2 - (u^2 + 2m)^2 (d \theta^2 + \sin^2 \theta \, d\phi^2 )\)
by simple change of variables.

So the dr^2 term which used to be infinite at r=2m is now a finite du^2 term at u=0. But in both cases the dt^2 term is zero, which Einstein proposes is no big deal since it is possible to multiply both sides of the Einstein field equations by the determinant of the metric repeatedly until there is no formal division by zero.

So there still is a singularity in the equations of motion because the metric algebraically treats radial movement across the event horizon symmetrically. The raindrop metric does not, so the singularity corresponding to r=2m there simply does not exist.

 

Thank you rpenner for the clarification.

I note from Prof: Don Koks book there is also a way to show spacetime is “perfectly well behaved there “ that is at the event horizon, and this without using Kruskal -Szekeres coordinates.

From Prof: D. Koks book ‘Explorations in Mathematical-Physics: The Concepts behind an Elegant Language’.

I have in the past here ( as nimbus) quoted and linked to relevant passages in the book before. Thanks again

 

The interesting thing here is that in the coordinate time of a remote observer, nothing will ever actually reach the event horizon; by the shell theorem (at least for a nonrotating black hole) though nothing ever passes the event horizon, the mass and therefore the gravity remains as if it had. In the proper time of an observer falling toward the event horizon, they reach it with the acceleration the gravity environment dictates, but if they were able to "pull back at the last moment" then they would emerge into a universe trillions of years older than it is now.

The way to think of this is to say that the matter inside the Swarzchild radius at the moment the density becomes sufficient to form the event horizon is trapped in a space we can never access, and the matter outside but falling in is forever trapped in spacetime growing infinitesimally closer to the event horizon.

Since nothing ever actually crosses the event horizon, there is no formal singularity.

I don't know enough to say, but I think this means Einstein and Rosen were right. rpenner, can you confirm?

 

The singularity at r=2m is in the metric, not the curvature tensors, so it's not physical but involves a bad choice of coordinates. Like using polar coordinates at the origin.

 

If you were falling into a black hole - would the event horizon move away from you as you approached it?

 

No. If you are falling into a BH, from your perspective nothing unusual happens and you cross the EH to your eventual doom. [ignoring tidal gravity effects]
From my point of view watching you falling from a safe distance, as you approach the EH, I will see you gradually red shifted further and further along the spectrum, until fading from view, but I will never see you crossing the EH.
Here is an excellent site about all types of BH's
http://casa.colorado.edu/~ajsh/home.html

 

You appear to be saying that the event horizon moves away from you (or vanishes) as you approach it - which would be "Yes" not "No".

Edit- a choice of coordinate system made by the infalling observer would seem to be singularity free (at least event horizon free) . Compare and contrast this with Farsight's analysis.

 

No, I'm not saying that. From your frame approaching the EH, you cross it and fall to the Singularity region at the center.
From my remote frame standing off at a safe distance, you approach the EH and you are gradually redshifted beyond what my eyes can see. So I get my infra red telescope and then see you maintaining that approach but again being totally red shifted until beyond the range of my "scope. All the while time is slowing down, but I will never see you as stopped or crossing the EH.

With Farsight's analysis, he seems to make up his own pop science story and always has.

 

By using equation 3a in the paper, instead of equation 3, he was able to get rid of the division by zero that came up from using equation 1 (The Principal of Equivalence). There were also problems with gravity turning negative, but it also removes this problem as well. They actually mention that it would not be possible if it went negative.

 

This is a really contradictory statement. In the first sentence here, you say the speed of light is zero. In the second sentence, you say it is still the speed of light. Neither observer will see the light just hover in mid air. It is no wonder why Farsight is so confused about this.

Time would stop for an observer falling into the event horizon. For that observer, there would have just not been enough time for the light from him to reach anything outside of the black hole. The lifetime of the universe would pass before his eyes in an instant, so everything else in the universe just wouldn't be there anymore by the time the light left him. Objects would end up getting sucked into the black hole before the light even left the black hole to reach them, because time would have almost stopped for him while all the time it took for everything else to get sucked in passed by normally.

 

Apologies: wrong thread.

 

And you are a teacher? The poor kids!
http://casa.colorado.edu/~ajsh/home.html
http://jila.colorado.edu/~ajsh/insidebh/intro.html

 

http://casa.colorado.edu/~ajsh/singularity.html#r=1

Photons do not orbit in circles at the horizon, just skimming the surface. The place where photons orbit in circles is the photon sphere, at 1.5 Schwarzschild radii. Photons emitted at the horizon fall in; except that if a photon is emitted exactly vertically outward exactly at the horizon, then it will hover at the horizon, not moving at all.

You do NOT see all the future history of the world played out. Once inside the horizon, you are doomed to hit the singularity in a finite time, and you witness only a finite (in practice rather short) time pass in the outside Universe.

In order to watch the history of the Universe unfold, you would have to remain outside the horizon, the Schwarzschild surface. One way to watch all the history of the Universe would be to stay just above the horizon, firing your rockets like crazy just to stay put. The Universe would then appear not only speeded up, but also highly blueshifted (probably roasting you in gamma rays), and concentrated in a tiny piece of the sky just above you.

 

Not yet, but they would be better off getting their information from someone that doesn't get all their facts from news articles written by journalist. The blue dot there is a picture of an entire galaxy. The red dots are another galaxy that is behind it. This nonsense is starting to spill over into another thread now, because you posted it on the wrong thread.

 

Again, you contradict yourself. You say time is sped up, but you fail to realize that time would approach zero at the horizon. Someone cannot notice their own time dilation, so to them it would appear as though they fell right in. But, if they looked up they would be nailed by every other star in the galaxy, and then they would collide with every other SMBH in every other galaxy before they even hit the singularity from time outside of the black hole appearing to be sped up. It would be sped up so much outside, they wouldn't be hit by the light you gave off (traveling the speed of light) until they got sucked inside. They would become a boiling stew of subatomic particles, and then they would be spewed out of a white hole on the other side.

 

All Frames of References are as valid as each other: A SR Postulate:

The only nonsense or playing games is that performed by yourself and mentioned by the two others in the other thread.
Like I said, and like I referenced and supported: A photon emitted directly radially away from just this side of the EH will appear to hover there forever from a local point of view: From a distant frame of reference, the photon will be redshifted to infinity.
Farsight has been preaching his nonsense for ages, and all have told him he is totally wrong, just as you are totally wrong in appearing to agree with him.
The speed of light is a constant: It does not vary in a vacuum> The coordinate speed of light is just that light has further to travel in curved spacetime.