**Does anyone care that a well-known instructor at Berkeley has no clue about modern physics?**

Dilaton, Bill, and Rehbock were among those who followed the newer exchanges under the comment by Richard Muller of Berkeley who has essentially stated that modern physics was a religion (link to Quora).

Some of the newer answers that have appeared there convey the old important points and some new ones – including the comments by David Simmons-Duffin of IAS Princeton (I am still proud for having sold him my furniture), Scott Stratton, and of course also Dilaton and Rehbock. The most important point is that it is simply not a scientific attitude to ignore all expectations – probabilistic statements – that have arisen from the scientific research. After all,

*all insights*that science is giving us only tell us that something is more likely or much more likely and something is less likely or much less likely. To dismiss the existing, albeit arguably "not 100% rigorous and loophole-free", arguments that the general postulates of quantum mechanics can't be deformed (and lots of other things that the researchers generally assume) means to place one's own prejudices above the scientific method.

I can imagine why Richard Muller is a popular instructor at Berkeley. At every school, there are lots of students who are not good at anything related to the thinking or science and Muller is making those feel better about themselves. It doesn't matter that you suck in mathematics and physics, he tells them: those who don't suck are just victims of hopes and a religion, anyway. The scientists who have actually discovered something and students who have understood something are no good – because everything may be totally different than they think, anyway.

Every lazy and mediocre scumbag must love such sermons.

But Muller's statements about any technical question – including elementary technical questions – seem to be rather flabbergasting. Bill has sent me a newer Muller comment that makes it very clear that he is dreaming about an event that will completely overthrow quantum mechanics (or overthrow "Copenhagen", which is clearly synonymous, but people similar to Mr Muller use the word "Copenhagen" because they believe that the Danish capital makes quantum mechanics sound derogatory or embarrassing).

Someone asked an elementary question about general relativity a few days ago:

Now, this is a rather standard question that the laymen and beginners often ask. I think that I was asking similar questions when I was beginning with general relativity some 25+ years ago. I have surely been repeatedly answering such questions, in the real life and on the Internet, e.g. in Why anything may ever fall into a black hole at all.What is the nature of space at event horizon and beyond that?

The way I imagine space is it gets more and more shrunk as we get closer and closer to anything with mass, which in this case is a black hole. And I know that gravity is nothing but a fall into the 3D pit created in space by a body with mass. But what is the nature of that pit at event horizon that nothing can escape up from there?

Before I discuss Mr Muller's unbelievable answer, let me try to offer mine. I will assume – as I almost always do – that the firewalls and similar things don't exist.

The event horizon is the lightlike hypersurface that separates the black hole interior from the black hole exterior. The black hole interior is composed of all the points in the spacetime that have the following property. If you draw any future-directed time-like trajectory starting at this point, none of them will ever get to the "asymptotic region" at infinity, outside and far away from the black hole.

This is what we mean when we say that no massive object can escape from the black hole, not even in principle. And the event horizon is the "last surface" from which you can no longer escape. From points an inch above the event horizon, it is possible to escape. From points one inch beneath the event horizon, it is not possible to escape.

But what does it look like near the event horizon?

When the black hole is large enough, a stellar mass black hole whose radius is comparable to miles or larger, the vicinity of the event horizon basically looks like a piece of a nearly flat 3+1-dimensional spacetime. Well, the Riemann curvature around those points is of order \(1/a^2\) where \(a\) is the black hole radius. But this curvature is "small" and doesn't change the intuition of the flat space "qualitatively". If you consider a box of the spacetime whose edge is much smaller than \(a\), for example a few meters (and the corresponding time one meter over \(c\)), it will be basically impossible for you to distinguish the region from a piece of the flat and empty Minkowski space, especially if you describe this region in the freely falling coordinates.

What do you see if you are near the event horizon? Where is it? Is it moving?

If you're near the event horizon and you allow gravity to attract you so that you are freely falling, the horizon looks like... It's a bit complicated – but not too complicated – situation so let me complete the statement as four paragraphs about an analogous situation and not a single sentence. ;-)

Imagine that you stand on the surface of Earth. Most TRF readers have surely done so in the past. At one moment, the event horizon is just a horizontal surface – imagine the floor of your apartment. And you are above the floor – meaning that you are outside the event horizon. We are creating an analogy in which the black hole is "inside the Earth".Note that the floor was moving "up" – it broke the time-reversal symmetry. The time-reversed situation in which the floor would be "shrinking" (moving down) would describe the "white hole" but basically due to the second law of thermodynamics (entropy never goes down), only black holes and not white holes may emerge in the real world.

But what's special about the event horizon is that it is a light-like (null) surface. Its proper volume is zero, much like the proper length of the light-like (null) trajectories is zero. If the floor were sitting at one point, its world volume would be a non-degenerate 2+1-dimensional manifold. But the event horizon is "null". The right way to achieve the "null" character of the horizon is to make the floor move by the speed of light.

So being close to a black hole event horizon is just like standing above the horizontal floor on Earth – a floor that is moving up by the speed of light. If you don't do anything – if you allow your body to fall freely – the floor will catch you and the black hole will devour you. You will find yourself "under the floor" which means "inside the black hole". When you're beneath the floor, you can never escape from the black hole again because the floor is still rushing away from you by the speed of light and you have no chance to catch up with it.

Alternatively, before the floor devours you, you may decide to start your jets and try to escape. But your jets have to run for a very long time for you to be safe. In the actual black hole geometry, it becomes a bit easier to become safe again when you're far enough from the event horizon again. But the "immediate" short-term situation is exactly the same as if you are above the floor moving up by the speed of light.

It's really essential that there is no "extreme curvature" or anything "extreme yet observable" that you could see or smell near the event horizon. The event horizon is just a null surface – a floor moving at the speed of light in the locally Minkowski coordinates – and there exists a nearly flat spacetime above it just like there exists a nearly flat spacetime beneath it. Locally, the "floor" looks like a fictitious surface. Nothing special is going on near the surface. There is no plastic substance over there. The precise location of the event horizon may only be determined globally – the event horizon is connected with the question "from which places you have at least some chance to escape from the black hole's grip in the future". To answer this question, you have to investigate what the whole future spacetime looks like. The event horizon isn't connected with any special "local" conditions. Every ball around an event horizon point looks just like a ball in the flat spacetime. You don't see any qualitative change of the scenery when your spaceship crosses the horizon in the real-world case.

Everyone who got a well-deserved A in any general relativity course will agree with me. The event horizon is not a special locus "locally". Locally it looks like any "fictitious" surface we imagine anywhere in space – which is moving by the speed of light. Too bad, most laymen are never explained this important point – in some sense, a very simple yet fundamental point. And although general relativity has been known exactly for 100 years, even many people who consider themselves "professional physicists" are confused about this simple point.

Now, on the contrary, most people are quickly exposed to the black hole solution in the Schwarzschild coordinates:\[

\eq{

ds^2 &= -\zav{ 1 - \frac{2M}{R} } dt^2 + \frac{dR^2}{1-2M/R} +\\

&+ R^2 (d\theta^2+\sin^2\theta\cdot d\phi^2)

}

\] Here, the event horizon is the surface defined as \(R=2M\). The geometry seems singular around that surface because \(g_{tt}\to 0\) and \(g_{RR}\to \infty\). However, these singular values of the metric are due to the awkward choice of coordinates (the Schwarzschild coordinates). There exist other coordinates that make it obvious that the spacetime is perfectly non-singular – very weakly curved – in the immediate vicinity of the event horizon.

When I say that the Schwarzschild coordinates are awkward, I don't want to say that they are awkward from all points of view. They are only awkward because they make the vicinity of the event horizon look singular even though it is not. But the Schwarzschild coordinates are very nice for other reasons. In particular, they make it obvious that the geometry is "static": it doesn't depend on the coordinate \(t\) at all.

All the coordinates that make the vicinity of the event horizon look as smooth as it is unavoidably violate this condition: they inevitably obscure the "static" character of the metric. Why? It's simple. Because in all these coordinates, there is a time coordinate \(\tau\) and it is always possible to ask "what is the value of \(\tau\) – the moment – at which a particular observer defined by some distances at the beginning crossed the event horizon?" Because in all the regular coordinates, it's possible for the observer to fall into a black hole, there has to be a preferred value of \(\tau\) which means that different values of \(\tau\) cannot be as good as others.

In the Schwarzschild coordinates, we may consider surfaces "right outside the black hole" defined by \(R=2M+ \epsilon\) where \(\epsilon\) is a positive number. The laymen and beginners love to think that they are comfortably sitting at such a surface and that's why they are led to believe that all the "singularities" of the Schwarzschild geometry near the event horizon are observable.

But there is nothing "comfortable" about the surface \(R=2M+\epsilon\). In the terrestrial-floor analogy, this surface corresponds to the set of points at which a rocket stays a certain distance above the floor that is trying to devour everyone. The rocket has to be spending lots of fuel all the time. And if it wants to be sitting at the same value of \(R=2M+\epsilon\) forever, it must spend an infinite amount of fuel. Clearly, it's not possible. Any finite object that sits at \(R=2M+\epsilon\) will ultimately be devoured by the black hole. It runs out of the fuel and the floor just catches up with it or him (I won't say "her" because I am a Gentleman).

The only thing that changes when you cross the event horizon – when the floor catches up with you – is that you lose the chance (even the totally theoretical chance) to escape from the black hole interior. Whatever you do with your jets and rockets and whatever their technological power is, it is impossible to escape from the black hole interior. Such an escape is exactly as forbidden as the superluminal journeys are banned in special relativity. These statements are not just analogous; one is the special case of the generalization of the other. In any locally Minkowski coordinates, an "escaping trajectory" would have to be superluminal.

On the picture above – the "causal or Penrose diagram of the [evaporating] black hole" – the black hole interior is the purple triangle. Time goes "up" and massive objects must move along trajectories that are always "closer to vertical lines than the horizontal lines". The restriction \(v\lt c\) is translated to the condition that only angles less than 45° away from the vertical lines are allowed as trajectories of massive objects.

Once you are in the purple triangle – the black hole interior – all your future-directed (up-going) time-like (mostly vertical, at most 45° from the vertical lines) trajectories end in the horizontal, tooth-like "singularity" where you're crushed because this is a place where the spacetime curvature (and therefore the tidal force) finally goes to infinity (it was low and finite near the event horizon).

On the contrary, if you're in the green area of the spacetime, you are outside the black hole and you have a choice. You may either avoid the black hole – you will need to turn on your jets if you're too close to the event horizon. Or you may jump into the black hole – at least if this decision is taking place sufficiently before the moment when the black hole fully evaporates.

To determine the exact location of the event horizon, one actually has to know the "whole future of the spacetime". The location of the event horizon can't be directly determined from any easily observable phenomena in the past and the present. The event horizon is like a "moment when the Second World War became unavoidable". You don't know exactly when it was but there was arguably a moment when no alternative existed (well, except that the human history is never banning possible resolutions as "strictly" as the causal structure of the spacetime). But nothing special could have been observed at that moment.

In particular, no plastic foil may be sitting at the event horizon – because the event horizon is a lightlike surface which means that the plastic foil would be basically obliged to move at the speed of light which the massive objects aren't allowed to do. Also, you may stand on the surface of Earth because of the pressure from the solid matter that our planet is made of. But there can't be any "solid surface" inside the black hole (lower than the event horizon) because that surface would be a purely spacelike, 3+0-dimensional surface, and massive objects' trajectories can't be spacelike (a spacelike trajectory only occurs for the forbidden faster-than-light tachyons). So any idea that you may "stand on a permanently solid surface" at a fixed Schwarzschild \(R\) inside the black hole contradicts the strictest laws of physics – the most universal laws of relativity. There cannot be anything to safely stand on inside the black hole.

Note that the causal/Penrose diagram above obscures the "static" character of the black hole spacetime. It is the same geometry that was described by the Schwarzschild solution. In the Schwarzschild coordinates, the geometry didn't depend on \(t\) at all. The Penrose diagram surely depends on the coordinate \(y\) – which is a form of time. The picture is not infinitely stretched and blurred and homogenized in the vertical direction. But as I have said, this dependence of the spacetime geometry on the time coordinate is unavoidable if the coordinates describe the vicinity of the event horizon in a non-singular way.

I think that the popular books are usually doing a very bad job in explaining the observations in the black hole spacetime. But those people who learn general relativity – including all the good undergraduate students who took a general relativity course – do understand these basic things. Unfortunately, Mr Richard Muller is not among those. What did

*he*answer?

When you say space is more and more "shrunk", I assume what you mean is that a lot of space is squeezed into the region near the black hole.But this is just wrong. There are no trajectories "right above" the event horizon at all whose proper length would be infinite. There is no "infinite amount of volume" being concentrated above the event horizon. The event horizon is just a floor beneath you – and the proper distance between you and the floor (measured at an arbitrary "time slice") is finite. In this sense, the volume is unavoidably finite, too.

In the Schwarzschild coordinates, we see that \(g_{RR}\to \infty\). But \(\sqrt{g_{RR}}\) is the relevant coefficient by which the coordinate changes \(dR\) have to be multiplied to get the proper length. And if you integrate \(\int dR / \sqrt{1-2M/R}\) over an interval \(2M\lt R \lt 2M+\epsilon\), you get a finite answer because the integral \(\int dx / \sqrt{x}\) converges near \(x=0\) and the problems are equivalent. The indefinite integral is \(2\sqrt{x}\) which is finite both for \(x=0\) and \(x=+\epsilon'\). The quantity \(g_{RR}\) goes to infinity near the horizon but its integrals are finite.

So the proper distance from the event horizon is finite in the \(t={\rm const}\) as defined by the Schwarzschild coordinates. The proper distance from the horizon measured in differently sliced surfaces is

*even lower*because as you should know, the "straight" line in the Minkowski spacetime is the "longest one". The lines more similar to the mixed, light-like ones have a lower proper length.

Instead of having explained anything that is correct, Mr Muller has "vindicated" the elementary misconception of the layperson who has asked the question. Clearly, none of these two people understand the geometry near the event horizon.

Indeed, if you measure distance by the time it takes light to cover it, then the distance to the event horizon is infinite -- assuming you are using the standard coordinate system that is fixed at infinity.This sentence is flawed in a funny way. It is true that in the Schwarzschild coordinates, the light rays starting outside the black hole never quite reach the event horizon – they need \(\Delta t = \infty\) to get there, an infinite amount of the Schwarzschild coordinate time. But it is completely wrong to introduce this coordinate-dependent statement in the way that Mr Muller has offered above, especially because of the word "measure".

The point – and it is really an essential point of GR that Einstein has always identified as one of the keys to GR – is that we never "directly measure" coordinates in general relativity. Coordinates just don't have a direct physical meaning. That's the "paradigm shift" we have to undergo if we want to go from Newton's (or special relativistic) spacetime geometry to GR. We would need to fill the whole spacetime with a 4D skeleton of sticks and clocks that are informing us about the coordinates at each point. But the spacetime doesn't come equipped by such a skeleton. Instead, we measure times and distances by rulers and clocks and they always measure the "proper distances" and "proper times" of some sort.

But all the proper distances from the horizon – and the proper time that any finite object needs to fall into the black hole (beneath the event horizon) – are always finite. So Mr Muller's claim is just totally misguided. He is effectively trying to sell a singular character of inappropriate (Schwarzschild) coordinates as a true physical effect related to measurements. But it is not one. There is nothing infinite that one may measure in the vicinity of the event horizon and by his misunderstanding of this crucial fact, Mr Muller shows that he is as ignorant about general relativity as the author of the original question.

If you use an accelerating coordinate system, falling into the black hole, then the distance is finite.Again, totally wrong words. It's the Schwarzschild coordinate system that should be called "accelerating". The non-accelerating, natural coordinate system is unavoidably one associated with an observer who freely falls into the black hole. In general relativity, it's the world lines of freely falling observers that are straight – they are geodesics – and it's the straightest possible lines that should also be called non-accelerating.

At any rate, the proper distances are finite in all time slices, as I have explained, so Muller's sentence would be wrong even if you redefined the word "accelerating".

Gravity is not exactly a fall into a 3D pit. It is a fall into a 4D space-time pit. The time part of that is important. The typical diagram that shows curvature of space alone is somewhat misleading, although it helps the reader start to imagine curvature in space.Let me assume that this paragraph basically meant that Mr Muller has already seen the Penrose diagram which pictures the whole spacetime. But the previous wrong assertions unambiguously imply that he has misunderstood what the diagram actually implies about the "finiteness of things" near the event horizon.

One reason why nothing can escape from the event horizon is that, measured with falling light, it is infinitely far away.No, one can't say it in this way. All future-directed trajectories of light (null trajectories) starting at a point in the black hole interior end in the singularity. There is

*none*that goes outside the black hole which is why we can't talk about "its" length at all. It just doesn't exist. One can't say whether this null trajectory going from the interior outside is "finitely far away" or "infinitely far away" if it doesn't exist at all!

Light returning from the event horizon would also take infinite time, even to get out a little part of the way.This is another very dangerous coordinate-dependent statement. It may be right or wrong depending on the choice of the coordinates. But it's surely

*conceptually wrong*for Mr Muller to make tons of these coordinate-dependent claims and pretend that they are operationally meaningful or objectively physical.

From a general relativity point of view (ignoring quantum effects) a falling object never quite reaches the event horizon.This is 100% wrong and I have already explained why. There is absolutely no quantum mechanics needed to explain that it's wrong. Most objects sitting or flying near the event horizon will be devoured by the black hole – and they will see that it takes a finite time on their clocks. It's just the time needed for the horizontal floor beneath you – which is moving up by the speed of light – to catch up with you. If the floor is a foot beneath you, it takes a nanosecond for you to be inside the black hole.

The problem with Mr Muller and most laymen who are exposed to the Schwarzschild solution is that he and they don't actually understand the geometry. They see "something singular near the horizon" so they immediately assume that physical phenomena are singular near the horizon. But they are not. Also, they see that one needs a large \(\Delta t\) in the Schwarzschild coordinates to get close to the horizon and they assume that \(t\) is the "right time" to settle all questions about time in the presence of the black hole.

But the Schwarzschild coordinate \(t\) is (and similarly, \(R\) is) among the most misleading coordinates if one wants to understand the behavior of time and space in the vicinity of the event horizon. As one approaches the event horizon, the Schwarzschild coordinates \(t,R\) become "arbitrarily far" from any realistic coordinates linked to clocks and rulers used by any observer that may actually exist. Mr Muller clearly doesn't understand this important point.

This also implies that it takes an infinite time for a black hole to form.This sentence is absolutely wrong, too.

No, it also takes a perfectly finite time for a black hole to form. When a star that is a bit larger than the Sun – a few light seconds – loses the pressure that was protecting it from the collapse (I am talking about a perfectly realistic situation), it takes a few seconds for each point on the surface of the now-collapsing star to find itself inside the black hole.

Just look at the causal diagram again.

The yellow line is the world line of a point on the surface of the star. It's somewhere in the spacetime and at some early moment, it crosses the event horizon and the observer on the stellar surface is already doomed, in the purple black hole interior, unable to escape again.

There

*are*some points in the black hole interior and locally, the event horizon – the boundary of the interior – does look like the horizontal (spherical) floor expanding by the speed of light. So if the star's radius was a few light second, it simply takes a few seconds for this floor to expand sufficiently to devour the surface of the star!

Mr Muller clearly got his completely wrong answer by using the Schwarzschild time \(t\) and interpreting it incorrectly as the time measured by an observer which it's not. But he had to make several additional mistakes at the same moment. The spacetime where a black hole arises from a star is clearly not "static" – something is changing in that spacetime, regardless of the choice of coordinates – so there can't be any static coordinates and there is nothing that one could automatically call the "Schwarzschild coordinates" for this spacetime. It's because the solution should be static in the Schwarzschild coordinates – a defining condition – but the stellar collapse simply cannot be static.

That means (ignoring quantum effects) that no black hole can really form in finite time. Most theorists ignore this, in part because a partially-formed black hole looks and behaves very much like a full (primordial) black hole.Black holes always form in finite time when the star loses the required pressure to resist the collapse. And there can't be any "partially formed black hole". Most theorists (trained in relativity) "ignore" the idea that black holes cannot form in the finite time (and the idea that the formation of a black hole may be "partial") because most theorists (trained in relativity) know that only completely deluded beginners may believe the misconception that any such process needs an infinite time.

Sorry, Mr Muller. You got an "F" from general relativity. And you suck at many other levels, too.

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