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Dead Stars Don’t Radiate

johncarlosbaez.wordpress.com

151–155 of 155 posts

Re: Dead Stars Don’t Radiate

#151
post #148

Earlier quoted context omitted.

Apart from my MSci in Physics... Perhaps you could post some links to the spacetime diagrams you are talking about?

The diagram on the Wikipedia page for Kruskal-Szekeres coordinates[1] does the job. There you see the trajectory of some infalling observer along with some future light cones[2] of points along that trajectory and the event horizon marked as the dashed line. The usual Schwarzschild r and t coordinates are also shown as the pale hyperbolas. Say the trajectory that's drawn on the diagram is the trajectory of your feet.…

Thanks! Hmm, I think we're talking about slightly different things but it's been too long since I studied it to put it in the right words :)

I completely agree that spacetime can be "flatish" for a large block hole, but the event horizon does still represent a boundary right?

Consider the edge case of crossing the event horizon itself at some speed I agree that your head will pass through the future light cone of your feet, and so could do somethign to affect your head (by emitting something falling slower than your head), but I'm not sure any light rays could follow that path.

Re: Dead Stars Don’t Radiate

#152
post #127
post #8

Earlier quoted context omitted.

You have to remember the "one particle in the pair fails to escape the event horizon" explanation is a simplification of the alleged reality, which is the scattering of particles (or fields) in the presence of an event horizon. As far as I know there is no intuitive, non-mathematical way to describe this accurately, so science communicators of all stripes tend to approximate it in ways that can mislead the audience.…

Not a physicist, but the more accurate “intuitive” explanation I read is that an accelerating observer sees thermal radiation in a vacuum. This is called the Unruh effect [0]. And since a black hole requires an accelerated observer to not be pulled in you will always have thermal radiation coming from the black hole UNLESS you are free falling into it. Physicists please correct me where I’m wrong! [0]: https://en.m.w…

> accelerating observer

Note that this is proper acceleration which is measurable locally with e.g. a weighted spring, contact with a piezoelectric scale, or any other sort of accelerometer apparatus.

> a black hole reuires an accelerated observer to not be pulled in

There are an infinite number of free-fall trajectories outside a black hole, most of which never go near the black hole in the first place. There are also infinite numbers of hyperbolic orbits which "graze" a black hole and an infinity of various elliptical and (quasi-)circular orbits.

An accelerometer on one of these trajectories will report zero proper acceleration. Yet none of these trajectories cross the horizon from the outside.

There are additionally free-fall trajectories which cross the black hole horizon from the outside. Who knows what happens not long after that: flat-space physics like the Standard Model of Particle Physics curved spacetime physics like General Relativity give conflicting answers.

Finally there are also an infinity of trajectories which are somewhere properly accelerated. Most of these won't cross the horizon, but some can: one can turn on one's rocket engines and carefully steer a course that crosses the horizon.

(There are multiple types of horizon; the interesting one is the apparent horizon which can be measured locally with various types of apparatus. An event horizon -- if there is one -- can only be determined with reference to the entire global spacetime and all its contents from infinite past to infinite future. There's a variety of other horizons too. Visser catalogued some of them in https://journals.aps.org/prd/abstract/10.1103/PhysRevD.90.12...>).

Note that trajectories in my paragraphs above correspond to everywhere non-spacelike curves of all sorts. Free-falling trajectories are geodesics (timelike or lightlike) and one can grind out the geodesic equation for the central mass in a Schwarzschild spacetime, for example.

Finally, proper acceleration is difficult to maintain for long, let along perpetually. (Although one can stand on the surface of a rocky planet and with an accelerometer measure over a very very very long term practically constant acceleration, and might for practical reasons want to assign that approximately constant acceleration to e.g. little "g"). So most curves where there is some proper acceleration also have some free-falling segments eventually.

One consequence is that a traveller who undergoes different accelerations (including none -- free-fall) along its path through spacetime will count different numbers of particles at different points along the traveller's sometimes-geodesic/sometimes-accelerated timelike curve.

A black hole doesn't really change this other than that the formation of some types of horizon induces an acceleration between freely-falling observers before the horizon forms and observers after. The later freely-falling observers see more particles than the earlier ones, and they bunch up not very far (but not preciesly on) the appropriate horizon.

The Unruh effect introduces an Unruh horizon attached to an observer undergoing proper acceleration. That observer sees more particles when it is accelerating than when it is/was freely-falling.

Some of the radiation is massless or has so little mass that it flies out to infinity. Around some types of horizon near a central mass, we'd call that Hawking radiation (as opposed to e.g. Unruh radiation, or radiation associated with e.g. a cosmic horizon in an expanding universe). And as Baez notes, over lonnnnnnnnnng times one would want to take into consideration how the central mass evolves as these particles fly away. Over much shorter durations (fuel is limited!) one would want to take into consideration how Unruh radiation and proper acceleration co-evolve.

The idea in the paper (which Baez dismantles in the linked blog entry at the top) is that no type of horizon is necessary for an apparently particle-free vacuum to look like a particle-rich patch of spacetime. That's a pretty wild claim. Why don't electrons evaporate on their own, in that case?

Re: Dead Stars Don’t Radiate

#153
post #148

Earlier quoted context omitted.

The diagram on the Wikipedia page for Kruskal-Szekeres coordinates[1] does the job. There you see the trajectory of some infalling observer along with some future light cones[2] of points along that trajectory and the event horizon marked as the dashed line. The usual Schwarzschild r and t coordinates are also shown as the pale hyperbolas. Say the trajectory that's drawn on the diagram is the trajectory of your feet.…

Thanks! Hmm, I think we're talking about slightly different things but it's been too long since I studied it to put it in the right words :) I completely agree that spacetime can be "flatish" for a large block hole, but the event horizon does still represent a boundary right? Consider the edge case of crossing the event horizon itself at some speed I agree that your head will pass through the future light cone of you…

Ok I drilled down a bit and looks like you are right, although I'm still not sure I've built up a clear understanding! (https://physics.stackexchange.com/questions/187917/thought-e...). In fact that question (series of onions) is exactly how I visualised it...

I'm not quite sure from that discussion why an event horizon is equivalent to a body moving outwards at the speed of light but it does make some sense. GR is always fun!

I still don't have a good idea of the "slow moving crossing the event horizon" case" but I'll read around it some more

Maybe the difference is between "free fall through an event horizon" vs "hover" (as much as is possible) at an event horizon

Re: Dead Stars Don’t Radiate

#154
post #7
post #5

Without a gravity well whose escape velocity exceeds c, how are they supposing hawking radiation happens in this scenario? Both virtual particles-antiparticles survive (and promptly disappear because one didn't just cross an event horizon).

That one's a big white lie of how Hawking radiation works. It's not even an approximation, just a far-fetched metaphor that Hawking made up, presumably to satisfy science journalists.

Hawking 1974, "Black hole explosions?" (Nature paywall) https://www.nature.com/articles/248030a0> (sci-hub antipaywall) https://sci-hub.se/https://www.nature.com/articles/248030a0> ("... the time dependence of the metric during the collapse will cause a certain amount of mixing of positive and negative frequencies [...] Part of this wave will be scattered by the curvature of the static Schwarzschild solution outside the black hole ... Another part of the wave will propagate backwards into the star") was certainly not for science journalists and is not at all "far-fetched". The only missing piece here is possibly discussion of his particular approach to second-quantization of the matter fields and how one might interpret that; Hawking's peers fifty years ago did not need assistance there.

This was followed in detail by Hawking 1975 (open access via Project Euclid) https://projecteuclid.org/journals/communications-in-mathema...> where the most relevant bit surrounds eqns (2.27-28) and there is nothing wrong with talking about signed "contributions to the probability flux into the collapsing body" (as opposed to identifying those contributions as particles).

Hawking himself wrote simpler pop-sci explanations in a number of places, including his book, where a reader would not be expected to understand how canonicalizations of wave functions in other than position space and directly-measured particles differ. However none of those works is the equivalent of the brief "explosions?" letter and its follow-on.

Re: Dead Stars Don’t Radiate

#155
post #150

Earlier quoted context omitted.

Some "infinities" of singularity are at the center sure, but all the maximal Relativistic effects are at the EH surface. It's even proven that the entropy (informational content roughly) is equal to the EH area divided by the number of planc-length square areas, as the amount of quantum arrangements of information that are allowed "inside". That is a HUGE hint everything's remaining on the surface. For example, when…

I think you misunderstand what Holographic principle says. Even if all information is encoded in two-dimensional surface that forms the Bekenstein bound, that does not mean that anything changes when you cross the area. It only means that radiation which black holes emit can't emit information of the matter it has absorbed.

If Event Horizons can come in any number of dimensions (with the conventional things we call Black Holes having 2D of space-like degrees of freedom on the EH) that leads to the obvious conjecture that our 3D universe might itself be a 'manifold' (projection) from a higher dimensional 4D space (i.e. we live on a 3D EH). That 4D space might even itself be a projection from a 5D one, and so on. This is extremely similar to Holographic Principle, but very definitely a more generalized theory.
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