But first, some context. We know that Einstein’s theory of general relativity is incomplete. We know that because it cannot handle quantum properties. To complete General Relativity, we need a theory of quantum gravity. But progress in theory development has been slow and experimental evidence for quantum gravity is hard to come by because quantum fluctuations of space-time are so damn tiny. In my previous video I told you about the most promising ways of testing quantum gravity. Today I want to tell you about testing quantum gravity with black hole horizons in particular.

The effects of quantum gravity become large when space and time are strongly curved. This is the case towards the center of a black hole, but it is not the case at the horizon of a black hole. Most people get this wrong, so let me repeat this. The curvature of space is

*not*strong at the horizon of a black hole. It can, in fact, be arbitrarily weak. That’s because the curvature at the horizon is inversely proportional to the square of the black hole’s mass. This means the larger the black hole, the weaker the curvature at the horizon. It also means we have no reason to think that there are any quantum gravitational effects near the horizon of a black hole. It’s an almost flat and empty space.

Black holes do emit radiation by quantum effects. This is the Hawking radiation named after Stephen Hawking. But Hawking radiation comes from the quantum properties of matter. It is an effect of ordinary quantum mechanics and *not an effect of quantum gravity.

However, one can certainly speculate that maybe General Relativity does not correctly describe black hole horizons. So how would you do that? In General Relativity, the horizon is the boundary of a region that you can only get in but never get out. The horizon itself has no substance and indeed you would not notice crossing it. But quantum effects could change the situation. And that might be observable.

Just what you would observe has been studied by Niayesh Afshordi and his group at Perimeter Institute. They try to understand what happens if quantum effects turn the horizon into a physical obstacle that partly reflects gravitational waves. If that was so, the gravitational waves produced in a black hole merger would bounce back and forth between the horizon and the black hole’s photon sphere.

The photon sphere is a potential barrier at about one and a half times the radius of the horizon. The gravitational waves would slowly leak during each iteration rather than escape in one bang. And if that is what is really going on, then gravitational wave interferometers like LIGO should detect echoes of the original merger signal.

And here is the thing! Niayesh and his group did find an echo signal in the gravitational wave data. This signal is in the first event ever detected by LIGO in September 2015. The statistical significance of this echo was originally at 2.5 σ. This means roughly one-in-a-hundred times random fluctuations conspire to look like the observed echo. So, it’s not a great level of significance, at least not by physics standards. But it’s still 2.5σ better than nothing.

Some members of the LIGO collaboration then went and did their own analysis of the data. And they also found the echo, but at a somewhat smaller significance. There has since been some effort by several groups to extract a signal from the data with different techniques of analysis using different models for the exact type of echo signal. The signal could for example be dampened over time, or it’s frequency distribution could change. The reported false alarm rate of these findings ranges from 5% to 0.002%, the latter is a near discovery.

However, if you know anything about statistical analysis, then you know that trying out different methods of analysis and different models until you find something is not a good idea. Because if you try long enough, you will eventually find something. And in the case of black hole echoes, I suspect that most of the models that gave negative results never appeared in the literature. So the statistical significance may be misleading.

I also have to admit that as a theorist, I am not enthusiastic about black hole echoes because there are no compelling theoretical reasons to expect them. We know that quantum gravitational effects become important towards the center of the black hole. But that’s hidden deep inside the horizon and the gravitational waves we detect are not sensitive to what is going on there. That quantum gravitational effects are also relevant at the horizon is speculative and pure conjecture, and yet that’s what it takes to have black hole echoes.

But theoretical misgivings aside, we have never tested the properties of black hole horizons before, and on unexplored territory all stones should be turned. You find a summary of the current status of the search for black hole echoes in Afshordi’s most recent paper.

Minor grammar fix:

ReplyDelete“The statistical significance of this echo is was originally at 2.5 σ” should be “The statistical significance of this echo was originally at 2.5 σ”

Rick,

DeleteThanks, I have fixed that.

ReplyDelete:(

I do think that quantum signatures may exist in gravitational radiation. I have commented a couple of times on this here. Quantum hair on a stretched horizon may in a moments, ~ 10^{-24}sec or less, of coalescence stimulate the production of gravitons. These in a red shifted tortoise coordinate transform could result in BMS translations. These would be similar to echoes. I have tried to think of ways of trying to know how to empirically distinguish betweem them.

ReplyDeleteThe echo physics stems from the idea the horizon has a "hardness" of some sort. There are some ideas, a bit off the mainstream in black hole physics, that have the horizon replaced with a hardened surface of some kind. In a merger the gravitational wave mass-energy, which for a Schwarzschold Bh is 29% of the initial mass, will have a portion or amplitude of it scatter off the gravity field back into the merging BH. This is why the measurements have about 5% of the initial mass of 2 BHs converted to gravitational radiation. However, if the horizon is similar to a "glass" with an index of refraction that reflects gravitationa radiation there would then be this echo. Gravitational waves scatter back to the BH, then some amplitude reflects off this surface back out, where then again some scatter back in and so forth.

I tend to be somewhat skeptical of the idea the horizon has this "hardness." Even with firewall considerations this violation of the equivalence principle does not happen until the Page time when the BH is half it mass at about 7/8ths its Hawking duration. For stellar mass BHs with a duration of 10^{67}years this issue should not be showing up. The big bang was only 13.8x10^{10} uears ago.

There is an intuition for a hardened event horizon. Suppose you have a bound state of any type. This can be represented by the exchange of bosons, in Feynman diagram. The problem it is that, at some reference point, in order for the particles to keep their bind, the bosons will have to cross the horizon. If a human were to cross an event horizon, being a massive bound state of atoms, the parts that were right inside, would be disconnected from the outside.

DeleteConsider how long the horizon partitions a body. When freely falling in the horizon sweeps past at nearly the speed of light. Two particles at 10^{-13}cm will have a transient situation of 10^{-23} sec. So this is transient, but even for particles with a wide separation there is no measurable change. In that time a real boson from the interior particle can't reach the exterior one. However, if one is emitted the infalling particle will couple to it inside the BH as if nothing has happened. So the equivalence principle is upheld.

Deleteam not sure if matter falls so fast, notice that 10^-23s is a very long time for nuclear physics. Take a look at these tables: https://en.wikipedia.org/wiki/List_of_radioactive_nuclides_by_half-life#10%E2%88%9224_seconds_(yoctoseconds) https://en.wikipedia.org/wiki/Meson#Vector_mesons

DeleteThis is enough to disrupt nuclear binding. But, nevertheless, it's quite amusing to see that the order of magnitude is about the same.

I have a layperson's question. As I understand it, the horizon is the boundary from where no information can reach a stationary observer at infinity.

DeleteBut an infalling bound system is an accelerating system close to the horizon.

I am wondering whether the horizon seen from an infalling observer is closer to the center of the black hole than for an observer at infinity?

And I even ask myself whether an infalling observer will always see a horizon below herself?

See my entry on the geodesic deviation equation. For two masses separated by some distance there is a relative acceleration that increases as the two approach the BH. However, there is no blip or anything jerk at the horizon.

DeleteThe observer approaching a BH will witness the black nothingness ahead of their direction of fall. Once they cross the horizon this blackness persists. The event horizon is replaced with an apparent horizon. The apparent horizon is not an invariant as is the event horizon, but is frame dependent.

If you were falling into a black hole how could you tell when the event horizon turns into an apparent horizon? That would be difficult. The timing involved would require a timing of incredible precision.

The Hawking temperature of a black hole is T = ħc^3/8πGkM. The thermal partition function e^{-E/kT} is equivalent to the path integral measure e^{-Eτ/ħ} where we think of it → τ in a Euclideanization. Hence we have the time interval associated with the temperature τ = ħ/kT = 8πGM/c^3. This is just 4πr_s/c, which is the time it takes for a black hole not in a final state, say wildly perturbed, to return to equilibrium. For a solar mass BH this is about 10^{-4} seconds. It is also about the time any observer has inside a BH. For a BH of mass M we then have this time τ = 10^{-4}sec(M/M_sol) The large BH in the Milky Way center is 4×10^6 solar masses so the time it takes for this black hole to reach equilibrium is 600 seconds, or about 10 minutes. It is about the time you have before reaching the inner singularity. These multibillion solar mass BHs give you more time, and the last I checked there is now a 100 billion solar mass BH.

Why mention this? Will the mass of a BH is some integer number of Planck masses M = Nm_p. The Planck mass is m_p = √{ħc/G} or m_p = ℓ_p^{-1}√{ħ/c^2}. So, to determine when you cross the horizon means you have to have a timing capable of measuring N integrals of Planck time. Think of this as a sort of squeezed state where the uncertainty in time measurement has to be very small. To do this though you require a huge energy capacity to your clock, which is about equal to the mass of the black hole. This means in order to time just exactly when you cross the horizon you need a clock with as much mass-energy as the black hole you are entering. However, this is utterly perturbing to the black hole that any certainty of this sort is impossible.

This has a bearing on duplicating quantum states, or the no-cloning theorem. Suppose Bob teleports a state to Alice who crosses into a black hole. Bob then realizes if he goes into the black hole to meet up with Alice they would share the same state and voila the state is cloned without a classical signaling. However, his mission has a huge probability of failure. As Bob enters the BH he must find Alice ahead of the apparent horizon, but in order to do that he must be able to find the exact timing of the apparent horizon within the time interval associated with the temperature τ = ħ/kT = 8πGM/c^3. Bob can’t do this with any certainty. No-cloning theorem wins the day with black holes.

"The observer approaching a BH will witness the black nothingness ahead of their direction of fall."

DeleteThank you for this illuminating explanation.

I

tend to go back to the Rindler space model when I do not understand BH horizons. I now see it is also a good illustration in this case.

But this does not rule out the no-cloning theorem in principle. Scrmbling information when it crosses the horizon does the job in principle.

Delete“

ReplyDeleteThe curvature of space isnotstrong at the horizon of a black hole. It can, in fact, be arbitrarily weak.”I want to be sure I understand this. When you talk about the

curvatureof space/time, does that refer to the local rate-of-change in gravity? (change in escape velocity as a proxy?)I presume that at the event horizon, the local gravitational strength is very high (escape velocity approximately C) but the local rate at which escape velocity changes can be nearly zero.

Correct?

sean s.

Sean,

DeleteYes, the curvature is a local quantity. It is defined by the curvature tensor, though when I say it is not strong I mean that scalar contractions of the curvature tensor are smaller than the Planck scale.

The curvature is, loosely speaking, related to the tidal force. The tidal force is basically the gradient of the gravitional force. Your interpretation in terms of changes in the escape velocity sounds roughly correct, though this isn't terminology anyone in the field actually uses because it tends to bring up irrelevant questions like what is trying to escape how along exactly which path etc. Best,

Sabine

Sabine wrote:

Delete>Yes, the curvature is a local quantity. It is defined by the curvature tensor, though when I say it is not strong I mean that scalar contractions of the curvature tensor are smaller than the Planck scale.

Sabine, the obvious contraction of the Riemann tensor is the Ricci tensor, which is zero in empty space. And so, the contraction of the Ricci tensor is zero.

Is there another way to contract the Riemann tensor that you have in mind that does not give zero? Or are you taking into account quantum corrections a la Hawking?

Can you clarify?

All the best,

Dave

Dave,

DeleteYou take higher order contractions.

Sabine wrote:

Delete>You take higher order contractions.

The trace of the Ricci tensor is fully contracted, and I had thought that is the only non-zero scalar associated with the Riemann tensor due to the anti-symmetry in the first pair and second pair of indices.

Or by "higher order," do you mean that you multiply the Riemann tensor by itself and then start appropriate contractions? I can see how that might give a non-zero result -- I have not thought that through.

Thanks.

Dave

Dave,

Delete"do you mean that you multiply the Riemann tensor by itself and then start appropriate contractions"Yes, it's a standard procedure/not saying anything new here.

The question came up in a thread some years ago whether there are any non-trival solutions where the Rieman-tensor is non-vanishing but all the higher order contractions happen to be zero. I think the outcome was that none of us knew of a proof that this can't be, but that there doesn't seem to be any example for it either. In any case, it's not a worry you need to have for the solutions we usually deal with. (Though I'll admit it's an interesting question.)

Scalar contractions of the curvature tensor (and of its derivatives) is not enough. There are non flat spacetimes (including some with singularities) for which all scalar curvature contractions are zero.

Deletewhich?

DeleteH.-J. Schmidt, Why do all the curvature invariants of a gravitational wave vanish?,

DeleteG. Sardanashvily (Ed.) New frontiers in gravitation, Hadronic Press (1996) 337-344,

gr-qc/9404037; H.-J. Schmidt, Lectures on Mathematical Cosmology, gr-qc/0407095.

The Riemann curvature of importance in the Schwarzschild metric is R_{trtr} = GM/r^3c^2. The Schwarzschild radius if r_s = 2GM/c^2 which is the radius of the event horizon. The Riemann curvature at the horizon is then R_{trtr} = ½(r_s)^{-2}. This is clearly smaller for a larger Schwarzschild radius or equivalently a larger black hole. The geodesic deviation equation describes the relative acceleration between two masses separated by a vector U^ρ and moving along V^σ as

DeletedU^μ/ds = R^μ_{νρσ}V^νU^ρV^σ

The geodesic deviation at the event horizon is then finite. It does not diverge and the reason is because both masses are freely falling. If I were to try to fix one mass stationary and let the other mass fall there would be a divergence once one mass reaches the horizon. That would however take an infinite time, or for practical purposes a very long time, to happen. This in part answers more completely the question above posed to me about why an event horizon should not have some hardened element to it.

The relationship to Hawking radiation temperature is with the formula T = ħc^3/8πGkM, for ħ the Planck-Dirac constant of action and k the Boltzmann constant. I give a somewhat qualitative derivation of this here

https://physics.stackexchange.com/questions/265494/what-happen-if-hawking-radiation-is-not-found/265501#265501

Clearly the larger the mass of the black hole the less Hawking radiation there or the temperature is smaller. This temperature is in units of solar masses given by T = 6.17×10^{-8}K(M_sol/M). This means a standard stellar or multi-solar mass BH is around a billionth of a Kelvin from absolute zero.

stor,

DeleteIt seems to me that this is a consequence of considering infinitely extended plane waves (or rather a direct product with an essentially flat space). Is this relevant for a realistic situation?

I have not looked at Stor's references yet, but I strongly suspect the vanishing of contracted curvature is the same as saying the Ricci curvature tensor and scalar are zero for vacuum. Gravitational waves are type N vacuum solutions of the Einstein field equations.

DeleteSabine,

DeleteWhat is a realistic space-time, and why it cannot happen for the realistic space-times?

Lawrence,

No, that is not true. The Schwarzschild solution is a vacuum solution, hence the Ricci tensor and scalar are zero, but the Kretschmann scalar is not zero.

stor,

DeleteI did not say it cannot happen, I said I do not know of an example. A realistic space-time is one that actually describes processes occurring in nature. Try to figure out a way to create an infinitely extended plane wave and maybe you'll see why I don't think it's realistic.

(The Schwarzschild metric, for the record, is also not realistic.)

@Stor: The Riemann tensor is

DeleteR_{abcd} = C_{abcd} + Ricci tensor and scalar terms.

The Ricci stuff zero in vacuum and the Riemann tensor is then just the Weyl tensor. The Gauss's law derivation in GR for the Petrov types evaluate curvature in a closed surface not bounding matter. So the Kretschmann scalar R^{abcd}R_{abcd} is then determined by the Weyl tensor.

Both black holes, Petrov type D, and gravitational waves as Petrov type N are vacuum solutions.

Can I conclude correct that a blackhole can generate G-waves by shaking surrounding-space. But inner G-waves could not escape it due to the nature of the blackhole itself ? So we could not detect native G-waves from the blackhole ?

ReplyDeleteQuestion that may be somewhat tangential to the post topic: when two BHs merge and the GWR is detected by LIGO (etc), what's observed seems to closely match what's predicted, which includes a "ringdown". As I understand it, the model has some sort of gap which gets extrapolated over, around the actual merger itself (and formation of a new BH) (it may be that the gap is tiny).

ReplyDeleteIs this true? If so, what is the gap due to, perhaps a breakdown of known physics?

More generally: what could extremely detailed (sensitivity, time resolution) GWR observations of BH mergers tell us about the limits of GR? Whether they are stellar mass BHs, or SMBHs (which LISA may be able to detect one day).

In the case of a binary neutron star merger, the result may be the formation of a BH. Again, could extremely sensitive GWR observations constrain GR in new ways?

I thought a hard horizon followed from some versions of string theory-- the fuzzball thing. Masses of hyperextended strings. And this, I thought, was supposed to solve the information loss problem. So detecting echoes might have less to do with quantum gravity and more to do with confirming a particular variant or set of variants of string theory. Which would be rather amazing.

ReplyDeleteYou said:

ReplyDelete> That’s because the curvature at the horizon is inversely proportional to the third power of the black hole’s mass.

Small correction, the curvature at the horizon is inversely proportional to the *second* power of the black hole's mass. In geometrized units, curvature is units of 1/(length)^2. At a radius r away from a black hole of mass M, you will see a Riemann curvature of typical size

|R| ~ GM/r^3.

When you get to horizon scale, take r ~ GM, so you get that the horizon curvature goes as

|R_h| ~ 1/(GM)^2.

Leo,

DeleteYes, that's right of course. Ugh. Sorry for the blunder.

There are no event horizon nor singularity in my conclusions. The Schwartschild solution can be impossible because of 'no rotation no energy' logic and Kerr-like black holes can lack double horizons by subtle modifications in GR when causal interactions throughout considered. I studied and found echos in 2015.

ReplyDeleteDisclaimer: I'm still little ignorant. :)

If the temperature of a black hole is less than the temperature of the Cosmic Microwave Background, why wouldn't black holes be gaining mass from the CMB in the current epoch?

ReplyDeleteSteve Bullfox asked:

Delete>If the temperature of a black hole is less than the temperature of the Cosmic Microwave Background, why wouldn't black holes be gaining mass from the CMB in the current epoch?

You're right, they do. Black-hole evaporation is a theoretical issue, not something happening to any (astronomical-sized) black holes.

But, the theoretical possibility raises various apparent paradoxes, and we hope that trying to resolve these paradoxes may help us better understand things such as quantum gravity.

ReplyDelete"The curvature of space is not strong at the horizon of a black hole. It can, in fact, be arbitrarily weak."To be picky, it can also be very strong, but for very small black holes (not what LIGO is detecting).

Interestingly, when people first thought about objects similar to black holes, about 250 years ago, they didn't think of keeping the mass constant and decreasing the radius and hence increasing the density, but rather keeping the density constant while increasing the radius and hence increasing the mass.

When calculating the radius at which the escape velocity is the speed of light, the Schwarzschild radius, GR and Newtonian theory (if one thinks of light as particles moving at the speed of light) give the same result. However, in GR, there is

no wayone can get out, while in Newtonian theory, while a ballistic particle can't escape, one could get out with a rocket, or a ladder, or whatever..That also means that in GR with a very massive black hole, tidal forces would be negligible at the horizon, nothing strange would be apparent, yet one could cross the point of no return without even noticing it until later.

A simple and rogue analogue. Imagine a river which gets gradually more narrow, which makes the water speed up. A few miles away there is a waterfall. You move on this river with a boat with a maximum speed of 5 miles/hr. As soon as you cross the line where the waterspeed is 5 miles/hr you are doomed to go down the waterfall, without noticing anything yet.

ReplyDelete27-JAN-2020

ReplyDeletehello -

re: the local idea of curvature:

If relativity says that no "object" is "in" space, but is rather,

a spatial extension (locally), then the LOCAL idea of FLAT, say in

the case of a photon inelastically scattering from a charged mass, seems always absurd. What is a flat spinor or gauge field in that situation?

More to the point perhaps, what is the relation at the point of such an interaction (collision point), between the ratio of incident energy to kinetic energy of recoiling scatterer (mass), and the (constant) scalar curvature of the manifold defined by the interaction? Could they be the same?

cheers,

mj horn

ReplyDelete"Interestingly, when people first thought about objects similar to black holes, about 250 years ago, they didn't think of keeping the mass constant and decreasing the radius and hence increasing the density, but rather keeping the density constant while increasing the radius and hence increasing the mass."It just occurred to me

whythey probably did this: it was probably easier to imagine adding more and more matter to make the object larger and larger than to imagine compressing matter to some extremely high density.Sabine said... "We know that Einstein’s theory of general relativity is incomplete. We know that because it cannot handle quantum properties."

ReplyDeleteDon't get me wrong even with very limited understanding relative too you I tend to think the same thing but the comment it begs the question; why not equally say either QM, or GR, or both are likely incomplete?

I guess it is because General Relativity only explains gravity while Quantum Theory explains "Everything Else".

DeleteIt is a bit of a shame though because GR is quite an elegant description of the fabric of the Universe while the Standard Model seems like a fairly arbitrary list of rules for handling all of the smaller stuff within it.

mj horn;

ReplyDeleteIs that really true? Does GR say that “no

objectis in space”? That seems wrong. What am I missing?sean s.

Gutentag Sabine, wie geht es dir? Im English 'Backreaction' sagt DURCHFALL. Entschuldigung mir, aber das ist recht! Und Albert Einstein (Avram Moos) ist kaput! Er war kein Wissenschaft ('God does not play dice'???). Ich mag Mileva Einstein (du kennst?). Auf wiedersehen, DN

ReplyDelete

ReplyDelete"why not equally say either QM, or GR, or both are likely incomplete?"Some people do. Roger Penrose, for example, thinks that QM is affected by gravitational effects.

The question over the horizon amounts to a question on how firmly quantum mechanics or quantum gravitation upholds the equivalence principle of general relativity. The Page time, about the time where a black hole has quantum emitted half of its initial quantum states or mass. A quanta of Hawking radiation is entangled with the black hole. At around half its mass quantum states are entangled with the black hole, but also with previously emitted states of Hawking radiation. The problem is that a bipartite entanglement can’t evolve into a tripartite entanglement by unitary evolution.

ReplyDeleteFor tripartite entanglements Kirwan polytopes are representations concurrences τ_i|ψ_{ABC}⟩ with

τ_1|ψ_{ABC}⟩ = 1/3(τ_{A|BC} + τ_{B|CA} + τ_{C|AB})

τ_2|ψ_{ABC}⟩ = 1/3(τ_{A|B} + τ_{B|C} + τ _{C|A}).

The first of these is a representation of entanglements with three states according to how each state is entangled with the other two in an entanglement. The second describes three estates in entanglements purely according to bipartite entanglements. The difference τ_1|ψ_{ABC}⟩ - 2τ_2|ψ_{ABC}⟩ = τ_3|ψ_{ABC}⟩ with

τ_3|ψ_{ABC}⟩ = τ_{A|BC} - τ_{A|B} - τ_{A|C}.

is an obstruction called the monogomy principle τ_{A|BC} ≥ τ_{A|B} + τ _{A|C}. This is a residue of entanglement called the 3-tangle that has no description in a 2-body system. τ_1|ψ_{ABC}⟩ ≤ τ_1(|GHZ⟩) and τ_2|ψ _{ABC}⟩ ≤ τ_2(|W⟩) are bounded above by the GHZ and W states respectively. This is an obstruction to the holographic conservation of information and quantum gravitation, which is often called a firewall. This can be extended to higher entanglements and Kirwan polytopes of higher dimension, though things becomes more difficult.

This is a topological obstruction, and straight forwards quantization of gravitation and the holographic principle face this problem. Unitary groups G_u do not describe this process, and it is only amenable to SLOCC transformations such as GL(2,C). Unitary groups have closed orbits or paths of evolution and moduli that are connected by Cauchy sequences and convergence conditions. GL(2,C) is noncompact and does not have these nice properties. The Hausdoff conditions for unitarity no longer apply. The relationship between unitarity and the equivalence principle is subtle and could be a cornerstone to the problem of quantum gravitation.

The firewall is a “fix” on this problem where the equivalence principle at the horizon is abandoned in favor of unitarity. When a black hole reaches the Page time, about 7/8ths the duration of a black hole, the horizon becomes a sort of singularity or end of physics. Anything that reaches this simply ceases to exist. This might in some ways happen, just as from a classical perspective there is some change of physics at a singularity. It is likely there is some sort of phase transition in physics.

Now, we can ask whether these echoes are a manifestation of the firewall. That is unlikely IMO. A stellar mass black hole has a duration of 10^{67}years. If echoes are due to some firewall issue we have two difficulties. Even a very young black hole will by statistics start of have quantum dynamics with 3-tangle obstruction problems. However, even a black hole formed shortly after the big bang will have a tiny number of its quantum bits in this “frustration.” Any black hole is at 1:10^{57} or less of its duration and so any issue of this sort is most likely a negligible perturbation. One of course might speculate that with something like Penrose’s CCC black holes might pass through this cycling of cosmologies. Of course then one must ask questions about the population of black holes. So the firewall is most likely not a source for the echoes of gravitational radiation from black hole merges.

28-JAN-2020

ReplyDeleteHi Sean -

I can't testify to the "truth" of it, but the source of my opinion that

it's truth is credible enough. . .

In his (1952) note to the fifteenth edition of Relativity (1961, Three Rivers Press), Einstein wrote:

"Physical objects are not in space, but are spatially extended. In

this way the concept of 'empty space' loses its meaning."

I can't pretend to understand it, but the more I think on it, the

more interesting the idea gets.

cheers,

mj horn

mj horn,

DeleteThe preceding sentence from Einstein was:

>“I wished to show that space-time is not necessarily something to which one can ascribe a separate existence, independently of the actual objects of physical reality."

It seems likely that he was just making the point that matter/energy (and pressure and the cosmological constant) shape the geometry of space-time: space-time is not something whose geometry is independent of matter. This is just what General Relativity is all about.

Or Einstein may have been alluding to his (unsuccessful) hope that General Relativity would fully implement Mach's principle (it does not).

In any case, as several of us scientists have pointed out here, verbal summaries of theories in physics, especially when simplified for a popular audience, are often misleading. The definitive statement of a theory in physics is the mathematical formulation. And the issue you are raising is not any issue at all in the actual math.

Dave

sean s. and mj horn:

DeleteEinstein has in the last 30 years of his life permanently redefined what space is and which properties it has. It looks like a very vague philosophical matter. I have worked through his 30 years of development and my impression is not that his understanding has settled. - Perhaps we should better follow the way of Lorentz, where space is nothing than emptiness without further properties. I think it can be shown that more is not needed. That would be a Copernican step.

antooneo wrote:

Delete>Einstein has in the last 30 years of his life permanently redefined what space is and which properties it has.

Y'know, antooneo, it does not look that way to almost all of us physicists who actually

agreewith Einstein! Most of us think that Special Relativity and General Relativity, both finished before Einstein was forty, are what redefined the nature of space and time, not something he did in later life.Considering that you

disagreewith Einstein (and that you seem not to be a physicist! -- what fields do you hold degrees in?), does it occur to you that perhaps you are not the best person to explain Einstein's actual views?Which reminds me: in the past you have kept pushing for three mutually contradictory propositions:

1) The empirical reality of the Sagnac effect proves special relativity is wrong.

2) Special relativity makes all the same predictions as your neo-Lorentzian relativity.

3) But neo-Lorentzian relativity is true.

Do you still adhere to all three of those propositions?

Because that sort of raises some questions about how you think.

And would you please tell everyone your real name, so that they can find your website and evaluate your theories for themselves? (Yes,

Iknow your name, but I am honoring your apparent desire not to have it posted here until you choose to reveal it.)30-JAN-2020

ReplyDeletePhysic Dave,

Calm down.

Try: SEC to AUX.

m.

mj horn,

DeleteChill out.

It's not my fault you are ignorant of physicis.

It's your fault.

Fix it.

Try: SEC to AUX. (I think you actually meant SCE to AUX.)

d

31-JAN-2020

ReplyDeleteDave -

Stop feeling guilty about my ignorance. I don't

recall blaming you.

But, you're right (apologies also to John Aaron):

"Signal Conditioning Electronics to Auxiliary."

Thankfully I was only seven, and didn't get the EECOM gig

that year.

We're all ignorant of physicis [sic].

And we're sick of it. That's why we're here, with heads on blog. Waiting.

The backreaction to ignorance is coming. Shhhh. Listen.

Sounds like a lahar of good ideas rumbling joyously

down steep slopes of hard work, faith in the truth you seek

(though, NOT guaranteed to coincide with what you might consider

desirable; see retailer for details) and the knowledge that

not all amateur earthlings are out to get you. Remember,

paranoia is just knowing all the facts.

Relax, it's all good.

Best, m.

Dave:

ReplyDeleteEinstein has repeatedly published about relativity until the end of his life. And the main reason for new publications was that he changed or adapted his concept of space and “new ether”. His last publication with adaptions was a new edition of “Mein Weltbild” in 1953. I refer you to the great book “Einstein and the ether” of Ludwik Kostro, a polish professor of relativity and follower of Einstein.

Sagnac is not a problem for Einstein in a static view. But a continuous transition from a curved beam to a straight one shows the logical conflict.

Regarding the alleged contradictory position: I can only recommend you to read carefully. I have said again and again: In normal tests (those ones to proof Einstein) the results of Einstein and Lorentz are the same. But the *method* is different. Einstein’s approach is based on principles and has to use the geometry of Minkowski and Riemann, whereas the “new-Lorentzian” way is a reductionistic one, which is based on independently known facts of physics, and it uses Euclidean geometry, which makes the treatment very simple.

About me: I do not find it sensible to say my name in an environment of contemptibility. But I can confirm that I am a physicist. I made my Ph.D. about particle physics at a particle accelerator. My adviser was the research director of that center. And the second reviewer was Herwig Schopper, later on the research director of CERN. When Schopper saw my thesis, he offered me a permanent employment which was exceptional at that time. But I did not take it because I had different plans.

antooneo wrote to me:|

Delete>But a continuous transition from a curved beam to a straight one shows the logical conflict.

(Sigh) As I pointed out to you long ago,

there is no continuous transition from the circle to the line.The whole idea of the Sagnac effect is that the rays meet up again at a place different from what one might expect (or with a phase different than one might expect):

they can meet up againonlybecause they are going in a circle!But if you change the circle to a straight line, the line is infinite in extent and cannot curl back on itself. The rays can then never meet up again. Therefore, no Sagnac effect.

Obvious to a child, even a young and not terribly bright child!

In any case, if the limit did exist, it would exist for both Einstein and Lorentz. If two functions, f and g are identical for all r, then their limit as r goes to infinity, if it exists, will also be the same.

You know that. You are just playing games.

But, why don't you tell everyone

your real nameso they can google your attempt at showing this??antooneo also wrote:

>But I can confirm that I am a physicist. I made my Ph.D. about particle physics at a particle accelerator.

Hmmmm..... Ph.D.s are usually given out by universities, not particle accelerators, at least here in the USA.

antooneo also wrote:

>About me: I do not find it sensible to say my name in an environment of contemptibility.

But

Ialready know who you are and I am your strongest critic. So, why are you afraid of other observers finding out? Is it just that if you reveal your name then I will be free to quote from your papers in detail and show how messed up your work is?Yeah, that's it. You do not want others here to see your work.

By the way, there are multiple errors in your write-up on the Sagnac effect, which I could easily point out, but talking about your write-up in detail would automatically involve revealing your name, and I am still respecting your desire not to do that.

In any case, my mathematical point that two functions f(r) and g(r) being equal for all finite r (true for Lorentzian vs. Einsteinian results for the Sagnac effect) must then have the same limit, if it exists, as r goes to infinity stands.

This is a well-known and rigorous result in mathematics, my friend. You are up against elementary math. And math wins.

Game. Set. Match.

Antooneo; re. “

ReplyDeletePerhaps we should better follow the way of Lorentz, where space is nothing than emptiness without further properties. I think it can be shown that more is not needed. That would be a Copernican step.”If you can actually show what you suggest, good for you! But I’m skeptical that you can. Best of luck on that.

Sincerely.And, please remember that the truth is “

NOT guaranteed to coincide with what you might consider desirable”.BTW, I’d like to take this opportunity to remind us all that

never in the history of Calming Down has telling someone “Calm down.” ever made them calm down. Or chill.And, no: paranoia is NOT “

just knowing all the facts.”sean s.

Sean s.:

ReplyDelete“If you can actually show what you suggest, good for you! But I’m skeptical that you can. Best of luck on that. Sincerely.”

This is of course the question to be answered. You can find something in the internet if you search for the “origin of gravity”. But I can try to explain some essentials here.

It is well known and proven that the speed of light is reduced in a gravitational field; the stronger the field, the greater the reduction. (The formula is known.) That means that a light-like particle moving in a field underlies classical refraction. If this refraction is calculated for instance for photons passing a star, the resulting deflection is identical to Einstein’s calculation of a curved space around the star. But of course much simpler.

Next step: If this refraction is applied to the internal oscillation in a particle, this oscillatory motion undergoes the same deflection, and it lets the particle accelerate towards the stronger field. The according calculation yields the gravitational law of Newton.

I have used this approach to deduce the Schwarzschild solution. You may know that this is the application of Einstein’s field equations to a spherical gravitational field. This deduction (by the approach described above) is a simple exercise which a pupil at school can do. - The “normal” way is the use of Einstein’s field equations including Riemannian geometry. A horrible exercise in contrast! For the same result.

Summary: No need for specific properties of the space.

antooneo wrote:

Delete>It is well known and proven that the speed of light is reduced in a gravitational field; the stronger the field, the greater the reduction.

No, if measured by local clocks and local distance measurements, the speed of light is

notreduced in a gravitational field -- that is obvious from the principle of equivalence.I know that you have a quote from Einstein alluding to your point, but that is simply because Einstein is using the coordinate time measured by clocks far from the gravitating source.

Math matters. And the math says that the speed of light measured by local measuring devices is invariant.

And, no, your paper does not prove your point at all.

The key issue is your formula 2.1: you write that the speed of light depends on the velocity in a formula with an exponent that is either 1 or 1/2 depending on the direction the light is moving.

That strange formula, expressed in an eccentric manner, is in fact

derivedfrom the Schwarzschild solution: no one knew that formula before Schwarzschild found the Schwarzschild solution.It is really just a rather eccentric way of re-writing the Schwarzschild solution and is only true in the particular coordinates Schwarzschild chose to use.

So, what really happened is this:

I. Schwarzschild derived the Schwarzschild solution.

II. Someone noted that one (eccentric) way of viewing the Schwarzschild solution was to think in terms of a direction-dependent index of refraction for light, and derived the appropriate formula for this (eccentric) perspective.

III. You then just turned around and from that formula trivially worked backward to get the Schwarzschild solution!

You derived the Schwarzschild solutionfromthe Schwarzschild solution!Not exactly a great accomplishment.

I can similarly "prove" the Riemann hypothesis by assuming the Riemann hypothesis. I can even prove that the Flying Spaghetti Monster is real by just assuming the existence of the Flying Spaghetti Monster!

Idiotic.

To honestly derive the Schwarzschild solution, you would have to start with something that was not itself derived from the Schwarzschild solution.

The only reference you cite does not appear to provide the formula you use, the formula that was in fact itself derived from the Schwarzschild solution: in fact, that reference seems to have nothing to do with your paper at all (I am citing the version available on the arXiv so that anyone can see it).

Yes, I know you claim that "This equation was determined experimentally for the first time by the Shapiro experiment in 1970 for small changes in c." Except it wasn't. The Shapiro experiment merely helped to confirm the Schwarzschild solution, adding nothing significantly new to what was already known from Pound-Rebka and Eddington's observations. And, you certainly cannot "derive" your formula 2.1 from the Shapiro observations alone!

The Shapiro experiment merely ran a further check showing that the Schwarzschild solution was correct.

You disagree??? Then cite a reference that shows how to derive your formula 2.1 from the Shapiro observations alone, without assuming the Schwarzschild solution.

I think you will find it... challenging.

You did get your formula 2.1 from

somewhere. So why not tell us where you got it from so that everyone can see if it really was derived from the Shapiro time delay or, as I claim, from the Schwarzschild solution.Give us an honest citation, antooneo!

Of course, we have already proved that your claims about the Sagnac effect violate first-year calculus. Limits are limits.

antooneo vs. first-year calculus. Math wins.

Game. Set. Match.

Dave:

ReplyDeleteThere is of course a continuous transition possible from a circuit to a straight line. It means to increase the radius to infinity.

This is a standard exercise which we all have learned at school.

The specific situation at the Sagnac case is that the deviation of a) c measured in a fiber ring from b) the measurement in a straight piece according to Einstein, is a constant value. The value is always ‘v’ where ‘v’ is the surface speed of the ring. It does not depend on the radius. So, if the radius is increased to infinity, the discrepancy remains ‘v’. Also for the case of an infinite radius. But in this transition the geometrical and physical difference between the circuit and a straight line vanishes. That’s it! - And only one value can be true. Sagnac is true, otherwise the navigation systems using it could not work.

The conversation about the nature of space is interesting.

ReplyDeleteMy humble understanding is that space is only math and as descriptive geometry defined as emergency of physical matter interactions (propulsions by signaling) and general collapsing back together towards minimum (apparent attractive forces). In condensed body these fundamental phenomena are in balance.

Now, when massive particles construct proper time there is a sort of planckian delay of entanglement correlations and net acceleration of boosts is weaker than net 'acceleration' of collapsing - manifesting gravity.

This can be a useless simplification but it's really plausible to say that space itself is physically a meaningless concept.

This naive guessing gedanken has common tangent with the blog post topic; if the space definition would obey my simplificated rule, the black hole event horizon must be impossible or at least nothing inside it cannot be in form of matter energy due to planckian quantum effects. Hence my view turns to the concept of black star or gravastar or eco whatever, which really can echo.

Dave

ReplyDeleteThe mentioned speed of light is of course the coordinate speed, because the locally measured speed obscures the physical reality. We know that in a gravitational field clocks run more slowly and rulers contract. So it is a quite trivial consequence that the measurement of c always yields the nominal value. – This is a general situation in the view of Lorentzian relativity. LR therefore considers that the behavior of measurement tools has always to be taken into account.

The equation used for the reduction of speed follows originally from Einstein. Shapiro and others have confirmed this equation, so we can anyway use it as an experimentally proven relation. We find it also in the Schwarzschild metric but the originator was Einstein. How did Einstein come to it? Einstein has adapted his theory of gravity to the results of Newton (in the non-relativistic case), it was not originally deduced by him from basics. He found everything in his theory correct which fulfilled his expectations. Which is NOT a deduction in the sense of reductionism.

And you, Dave, confuse the following: The equation mentioned above is in the Schwarzschild metric and is NOT part of the Schwarzschild solution. Both are different issues. And it is ‘idiotic’ to confuse it.

You mention the use of this equation in the according site and ask me for a reference for it. If you are looking further down, to chapter 5, you will find a deduction of this relation. It is based on quantum mechanics (i.e. exchange particles). So it may also show a direction how to solve the problem of quantum gravity. And it is a true deduction, in contrast to Einstein’s reference to postulates.

Math wins, that is true. I have everything based on math.

antooneo wrote to me:

Delete>The equation [2.1] used for the reduction of speed follows originally from Einstein.

Really? Where?

Where did you get it from?

Obviously, it is not true in general -- for example, in a situation where you are dealing with gravitational waves rather than a central gravitating source.

It is specifically applicable only in the Schwarzschild case, and then only if you use the Schwarzschild radial coordinate (I, and other physicists, often choose other radial coordinates, and then your formula will not work).

Einstein did not know the Schwarzschild solution in 1915 when he finished working out General Relativity: if he had had the formula you quote in 1915, he

wouldhave had the Schwarzschild solution then, which he did not.Einstein did not have this formula when he worked out General Relativity.antoonero also wrote:

>We find it also in the Schwarzschild metric but the originator was Einstein. How did Einstein come to it? Einstein has adapted his theory of gravity to the results of Newton (in the non-relativistic case), it was not originally deduced by him from basics.

Again,

where did Einstein do this? I have never seen it in any of his papers from 1915 or earlier? Where did Einstein originate the formula you use?I honestly think you just make things up.But prove me wrong: show us where Einstein originated this formula (before Schwarzschild worked out the Schwarzschild solution in early 1916).

A simple question.

C'mon – you' really are just making this up, aren't you? Show us where Einstein worked out your formula 2.1.

This is really, really weird.

Einstein's gedanken experiment on equivalence principle was with space ship accelerating strongly. The light ray needs to bend when crossing the ship. The bending ray is longer than known distance. Still if you are in the ship you can measure only based on light and cannot get any difference. Also ouside the ship there is proper acceleration of measuring device or apparent acceleration of the ship.

ReplyDeleteWhen acceleration field is spherical there actually exists an outer inertial reference frame where the longer coordinate time can be measured.

That's the way Einstein might have resolved Sagnac effect as early as year 1912...1913.

Dave,

ReplyDeleteYou again mix up the Schwarzschild metric and the Schwarzschild solution. Please understand that both issues are not identical. The equation about the dependence of c from the gravitational field can be used in the spherical solution but is not restricted to it.

This equation can be found at Schwarzschild’s metric. However, Schwarzschild did not develop general relativity, Einstein did it. Schwarzschild only interpreted the GRT of Einstein and described applications, particularly this spherical solution, also black holes. Logically Einstein is the originator.

And I have also shown in the source you referred to that the equation can be deduced from quantum mechanics. So, neither Einstein nor Schwarzschild are really needed here.

Any practical use of Einstein’s theory depends on the use of the gravitational constant G, otherwise no quantitative results are possible. So the theory had to be adapted to it, whoever it did (Einstein or Schwarzschild). Neither Einstein nor Schwarzschild did take any measures to derive G from basic facts or performed own measurements.

Dear Sabine Hossenfelder,

ReplyDeleteyour claim "there are no compelling theoretical reasons to expect them".

I disagree. The reasons appear almost immediately if we accept the idea that the equivalence principle may fail in quantum gravity. In this case, almost automatically there appear some preferred coordinates.

What may force one to accept preferred coordinates is gravitational energy. As long as you have the Strong Equivalence Principle, you cannot localize gravitational energy. But energy is a key ingredient of quantum theory. The natural, straightforward way to localize it is a preferred time coordinate, so that translation invariance in the preferred time gives energy via the Noether theorem. But then the question appears, how will the preferred time look like near the horizon. If the preferred coordinates will give a stationary metric outside the star, it is almost obvious that there will appear some infinities near the horizon.

There are not many candidates for preferred coordinates in GR, if it is not the human wish to simplify particular computation but supposed to be a physical law, ruled by a physical equation. There is essentially nothing but harmonic coordinates. And harmonic time becomes infinite at the horizon.

Quite similar, one may expect another metric as a way to break the equivalence principle. Bimetric theories of gravity, in particular theories of massive gravity, have a similar effect, the horizon will be critical. An example of such a theory is Logunov's "relativistic theory of gravity".