I If white holes were to exist in our Universe

[mad mathematician]

what astrophysicists and astronomers should observe in our universe to confirm their existence?

 

[javisot]

what astrophysicists and astronomers should observe in our universe to confirm their existence?

Standard answer 1- White holes do not exist, therefore there is nothing to observe

Answer 2- In LQG, according to Rovelli, the scenario could be different. But this doesn't correspond to the cosmology forum.

 

[mad mathematician]

Standard answer 1- White holes do not exist, therefore there is nothing to observe

Answer 2- In LQG, according to Rovelli, the scenario could be different. But this doesn't correspond to the cosmology thread.

that's the way people are doing science nowadays... LOL

 

[mad mathematician]

Strings don't exist, therefore string theory is refuted.

I thought it would take me more effort than that... cheers!

 

[Ibix]

The point about classical white holes is that they feature a singularity in your causal past. You literally cannot predict what would come out of one were one to exist and we know of no mechanism that would allow one to form, unlike black holes.

I don't know about Rovelli's speculation, but that question would belong in the Beyond the Standard Model forum.

 

[javisot]

that's the way people are doing science nowadays... LOL

Could you explain what you mean? According to consensus astronomy and astrophysics white holes don't exist, so your question doesn't make sense in that context.

You could ask about this in the context of LQG, but as I've already told you (and Ibix as well), that's not for this forum.

 

[DaveC426913]

Strings don't exist, therefore string theory is refuted.

String theory is not a theory, so that's sort of true.

That doesn't stop sciento-mathematicians from researching it, looking for evidence to vindicate it.

 

[mad mathematician]

The point about classical white holes is that they feature a singularity in your causal past. You literally cannot predict what would come out of one were one to exist and we know of no mechanism that would allow one to form, unlike black holes.

I don't know about Rovelli's speculation, but that question would belong in the Beyond the Standard Model forum.

So theoretically, GR doesn't even predict their existence.
Your answer is different than the other poster who claimed that they don't exist, therefore there's nothing to search for.
For me it's like in Dynamical Systems that we have sinks and sources, so I can conceive an existence of white holes as the sources.
I guess if we found another big bang in our own cosmos, then it would start another baby cosmos which will be sort of a white hole from our universe.
I guess one really should solve the QG conundrum... (I wish it was me, but I know my limits).

 

[javisot]

Your answer is different than the other poster who claimed that they don't exist, therefore there's nothing to search for.

I can't see the difference and you're wrong again, the white hole is part of the maximally extended Schwarzschild spacetime in Kruskal coordinates (it's part of GR). Its nonexistence is due to other reasons.

You are simplifying the answer I gave you, I gave you 2 answers, the second one is the one you would like to explore (in the beyond standar models forum)

 

[berkeman]

(Thread prefix changed A-->I)

 

[Ibix]

So theoretically, GR doesn't even predict their existence.

No, you can describe them. What you can't do is describe them forming, nor predict the consequences if you don't simply declare what those are (as is done in the maximally extended Schwarzschild spacetime).

For me it's like in Dynamical Systems that we have sinks and sources, so I can conceive an existence of white holes as the sources.

...but at the cost that you cannot predict the consequences. Perhaps unless you are considering one of the more speculaive models beyond the standard model.

However, I would note that you don't seem to be familiar with classical white holes. Grasping what LQG says about them may require a lot of groundwork.

 

[Drakkith]

that's the way people are doing science nowadays... LOL

Answers on an internet forum should not be taken as representative on how professional science is performed.

 

[PAllen]

The point about classical white holes is that they feature a singularity in your causal past. You literally cannot predict what would come out of one were one to exist and we know of no mechanism that would allow one to form, unlike black holes.

I don't know about Rovelli's speculation, but that question would belong in the Beyond the Standard Model forum.

Actually, the statement about causal past precludes formation. In the classical model, the WH singularity is in the past light cone of every event in spacetime.

 

[javisot]

I'd like to ask for an answer we can share in future discussions. A white hole is often described as the "time reversed" of a black hole, but what does it mean to be "the time reversed"?

Naively, it seems to mean that the same solution (the maximally extended Schwarzschild solution) describes a single object from two different perspectives. Peter points out that this isn't correct, and so I'd like an answer I can share: What does it mean to say that a white hole is the "time reversed" of a black hole?

 

[PeterDonis]

What does it mean to say that a white hole is the "time reversed" of a black hole?

Consider, for example, the Oppenheimer-Snyder 1939 idealized model of a spherically symmetric star collapsing to a Schwarzschild black hole.

The time reverse of that would be a Schwarzschild white hole expanding into a spherically symmetric star.

Note that neither of these models uses the full maximal extension of Schwarzschild spacetime, so neither one has both a black hole and a white hole region.

 

[javisot]

Consider, for example, the Oppenheimer-Snyder 1939 idealized model of a spherically symmetric star collapsing to a Schwarzschild black hole.

The time reverse of that would be a Schwarzschild white hole expanding into a spherically symmetric star.

Note that neither of these models uses the full maximal extension of Schwarzschild spacetime, so neither one has both a black hole and a white hole region.

The first two paragraphs seem descriptive, simple, and accurate. I'll use them directly in future discussions to "speak the same language." (I find the last paragraph contradictory to the first two, but I'm sure it's just me who misunderstood)

 

[otennert]

I'd like to ask for an answer we can share in future discussions. A white hole is often described as the "time reversed" of a black hole, but what does it mean to be "the time reversed"?

Naively, it seems to mean that the same solution (the maximally extended Schwarzschild solution) describes a single object from two different perspectives. Peter points out that this isn't correct, and so I'd like an answer I can share: What does it mean to say that a white hole is the "time reversed" of a black hole?

Let's start with what I think everybody agrees with here: the Schwarzschild spacetime is static in the asymptotic region "I". Hence there is a time-reversal symmetry $t\mapsto -t$.

The Kruskal spacetime has the same symmetry $t\mapsto -t$, although it must not be called static in the regions beyond the horizons. But if you turn the Kruskal diagram upside down, everything looks the same. That's pictorial of course, so the transformation $t\mapsto -t$ in Schwarzschild coordinates translates into a respective transformation in Kruskal coordinates. Hence the white hole (often marked region "III") is a "time-reversed" version of the black hole (normally region "II").

In Eddington-Finkelstein coordinates, things look as follows: the advanced EF coordinates cover regions "I" (asymptotic region) and "II" (inside black hole). The retarded EF coordinates cover regions "I" and "III" (inside white hole). (Alternatively region "Ib" instead of "I" -- the second asymptotic region -- can be taken for the sake of the argument). Formally, the line elements in retarded and in advanced EF coordinates look like the time-reversed versions of each other, but the respective EF time coordinate $\bar{t}$ itself is not the same in both versions. That's what is meant by the "time-reversal" symmetry, put in quotes.

A test particle which somehow has made it over the white hole horizon into the asymptotic region at an infinite negative value for the Schwarzschild time coordinate $t$ is observed to be redshifted, and as it moves further away from the white hole horizon, the redshift effect gets less and less. As the white hole contains an past spacelike singularity, such test particles then may appear from all directions on the celestial sphere. That was my original explanation.

All is much clearer when looking at the respective spacetime diagrams, there are lots of them online.

 

[javisot]

Let's start with what I think everybody agrees with here: the Schwarzschild spacetime is static in the asymptotic region "I". Hence there is a time-reversal symmetry $t\mapsto -t$.

The Kruskal spacetime has the same symmetry $t\mapsto -t$, although it must not be called static in the regions beyond the horizons. But if you turn the Kruskal diagram upside down, everything looks the same. That's pictorial of course, so the transformation $t\mapsto -t$ in Schwarzschild coordinates translates into a respective transformation in Kruskal coordinates. Hence the white hole (often marked region "III") is a "time-reversed" version of the black hole (normally region "II").

In Eddington-Finkelstein coordinates, things look as follows: the advanced EF coordinates cover regions "I" (asymptotic region) and "II" (inside black hole). The retarded EF coordinates cover regions "I" and "III". (Alternatively region "Ib" instead of "I" -- a second asymptotic region can be taken for the sake of the argument). Formally, the line elements in retarded and in advanced EF coordinates look like the time-reversed versions of each other, but the respective EF time coordinate $\bar{t}$ itself is not the same in both versions. That's what is meant by the time-reversal symmetry.

However, physics in the asymptotic region "I" of spacetime containing either the white-hole or the black hole region is the same.

I interpret this as, and correct me if I'm wrong, that the accretion of the Schwarzschild black hole is 0 (without adding quantum effects, only test particles with no effect on geometry), therefore the "ejection" of its time-inverse is 0, which allows the symmetry needed to describe the same object in two ways in the same solution.

 

[otennert]

I interpret this as, and correct me if I'm wrong, that the accretion of the Schwarzschild black hole is 0 (without adding quantum effects, only test particles with no effect on geometry), therefore the "ejection" of its time-inverse is 0, which allows the symmetry needed to describe the same object in two ways in the same solution.

I am not quite sure if I understand your question correctly. Of course, the whole area here is "no quantum" whatsoever, only classical GR. By accretion you mean a particle crossing the Schwarzschild horizon? What quantity do you refer to by "0"?

 

[javisot]

I am not quite sure if I understand your question correctly. Of course, the whole area here is "no quantum" whatsoever, only classical GR. By accretion you mean a particle crossing the Schwarzschild horizon? What quantity do you refer to by "0"?

Are we talking about a Schwarzschild black hole, that is, a vacuum solution to Einstein's equations, in which we manually place test particles that do not affect the geometry of spacetime, right?

 

[otennert]

Are we talking about a Schwarzschild black hole, that is, a vacuum solution to Einstein's equations, in which we manually place test particles that do not affect the geometry of spacetime, right?

Correct, otherwise they wouldn't be test particles, but would rather enter the material stress energy tensor in the rhs of the Einstein equations.

 

[javisot]

Correct, otherwise they wouldn't be test particles, but would rather enter the material stress energy tensor in the rhs of the Einstein equations.

By 0 accretion, I mean that this black hole does not accrete. Its associated white hole also does not eject anything. Simply test particles.

That's why I understood that:

I interpret this as, and correct me if I'm wrong, that the accretion of the Schwarzschild black hole is 0 (without adding quantum effects, only test particles with no effect on geometry), therefore the "ejection" of its time-inverse is 0, which allows the symmetry needed to describe the same object in two ways in the same solution.

 

[otennert]

By 0 accretion, I mean that this black hole does not accrete. Its associated white hole also does not eject anything. Simply test particles.

That's why I understood that:

No, that's wrong then. A test particle does of course get "swallowed up" (forgive the colloquialism), and a white hole does get "ejected". A test particle by definition just does not show backreaction.

 

[javisot]

No, that's wrong then. A test particle does of course get "swallowed up" (forgive the colloquialism), and a white hole does get "ejected". A test particle by definition just does not show backreaction.

Right, I mean that doesn't change the properties of the black/white hole.(no?)

 

[PeterDonis]

I find the last paragraph contradictory to the first two

Why?

 

[javisot]

Why?

The first two paragraphs seem perfect, I take them. The last paragraph, in relation to the first two, seems to say that the inverse description of the Oppenheimer-Snyder collapse can be called a "white hole" (this is fine), but that doesn't mean there is a white hole (this confuses me).

 

[Ibix]

The last paragraph, in relation to the first two, seems to say that the inverse description of the Oppenheimer-Snyder collapse can be called a "white hole" (this is fine), but that doesn't mean there is a white hole (this confuses me).

No, it says that the standard one is a black hole, the time-reverse is a white hole, but neither spacetime contains both a black and white hole (unlike the maximally extended Schwarzschild spacetime, for example).

 

[javisot]

No, it says that the standard one is a black hole, the time-reverse is a white hole, but neither spacetime contains both a black and white hole (unlike the maximally extended Schwarzschild spacetime, for example).

Ok, sorry, that's clear, thanks Ibix and Peter.

 

[otennert]

Right, I mean that doesn't change the properties of the black/white hole.(no?)

No it doesn't, correct.

 

[PeterDonis]

Thread closed for moderation.

 

[PeterDonis]

After moderator review, the thread will remain closed. An off topic subthread has been deleted. Thanks to all who participated.