mad mathematician
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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 observemad mathematician said:what astrophysicists and astronomers should observe in our universe to confirm their existence?
javisot said: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.
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.mad mathematician said:that's the way people are doing science nowadays... LOL
String theory is not a theory, so that's sort of true.mad mathematician said:Strings don't exist, therefore string theory is refuted.
So theoretically, GR doesn't even predict their existence.Ibix said: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.
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.mad mathematician said:Your answer is different than the other poster who claimed that they don't exist, therefore there's nothing to search for.
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).mad mathematician said:So theoretically, GR doesn't even predict their existence.
...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.mad mathematician said: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.
Answers on an internet forum should not be taken as representative on how professional science is performed.mad mathematician said:that's the way people are doing science nowadays... LOL
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.Ibix said: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.
Consider, for example, the Oppenheimer-Snyder 1939 idealized model of a spherically symmetric star collapsing to a Schwarzschild black hole.javisot said:What does it mean to say that a white hole is the "time reversed" of a black hole?
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)PeterDonis said: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.
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##.javisot said: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?
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 said: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 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 said: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.
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 said: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"?
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 said: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?
By 0 accretion, I mean that this black hole does not accrete. Its associated white hole also does not eject anything. Simply test particles.otennert said: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 said: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.
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 said: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:
Right, I mean that doesn't change the properties of the black/white hole.(no?)otennert said: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.
Why?javisot said:I find the last paragraph contradictory to the first two
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).PeterDonis said:Why?
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 said: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).
Ok, sorry, that's clear, thanks Ibix and Peter.Ibix said: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).
No it doesn't, correct.javisot said:Right, I mean that doesn't change the properties of the black/white hole.(no?)