Dr Jo Appears Confused by Quantum Theory

A writer I often admire, previously on Quora and more recently on Substack, is the New Zealand physician who goes by the pen name DrJo. His take-downs of anti-scientific nonsense (and sometimes actual fraud) are welcome and appear to be based on a good understanding of the relevant science and statistics, but getting closer to my own area of expertise, his take on quantum physics leaves me more doubtful.

Let’s first jump to the chase and look the concluding question in DrJo’s recent post on the subject:

“How can a space be both Hausdorff and non-local? What does ‘non-local’ even mean here? Can you explain the maths of this ‘non-local’ property, specifically the topology, without resorting to hand-waving?”

The main point is that DrJo appears here to be confusing the separability of points in space-time, just in the sense of how we can tell them apart (corresponding to his abuse of the word “Hausdorff”) and locality of a theory based on that space-time (which refers to some limit on how things happening at one point in space-time can affect those at another). Locality of a theory is a property that places some limit on how quickly actions at one place can affect the situation far away, but no such limit is required in order to tell events apart in either space or time. For example, I could still compute the distance from my home to the grocery store on the basis of how long it takes me to walk there – even if I had some way of instantly checking on the availability of a product and having them set it aside for me to pick up whenever I eventually arrive.

There is nothing wrong with feeling somewhat disoriented by the “spooky action at a distance” aspects of quantum theory – or with writing about that disorientation without having a technical background in the subject. But I do think that DrJo’s use of technical language for that purpose in his latest article is inappropriate – both because it is not correctly applied and because its effect is to create an aura of greater authority than is warranted.

Why do I say that? Well here goes.

DrJo’s post starts with what I consider a perfectly reasonable brief outline of the story behind the 2022 Nobel Prize in Physics (which was awarded to Alain Aspect, John F Clauser, and Anton Zeilinger for work on verifying the prediction of quantum mechanics that physical observations cannot be described by a theory in which the measurements correspond to permanent properties of classical particles which cannot somehow have instantaneous effects on one another). He follows this with a brief discussion of some aspects of the mathematical subject called topology which, while not wrong, strikes me as just introducing and throwing around technical terms for no useful purpose. And then comes the part about which I am actually “grumpy” but I hope not “inarticulate”:

Now here’s my problem. In terms of our current theories, spacetime is a Hausdorff differentiable manifold. Note the word ‘Hausdorff’ here. This assumption underpins both General Relativity and Quantum Mechanics.

But … if one of a pair of detectors interacts with one of a pair of entangled particles, and the other particle that is far removed instantaneously assumes a correlated state, how can this space possibly be Hausdorff? As has been pointed out ad nauseam by physicists like Einstein and Bell and Aspect and Clauser and Zeilinger, this is a non-local effect.

The points in the region of the second detector at an arbitrary distance from the first surely can’t be considered ‘Hausdorff’ (or ‘housed off’) if that distant interaction can influence them.

I guess the questions here are “How can a space be both Hausdorff and non-local? What does ‘non-local’ even mean here? Can you explain the maths of this ‘non-local’ property, specifically the topology, without resorting to hand-waving?”

Now, would a real physicist step up and explain.8 Please

OK. Let’s start with that first paragraph.

In terms of our current theories, spacetime is a Hausdorff differentiable manifold. Note the word ‘Hausdorff’ here. This assumption underpins both General Relativity and Quantum Mechanics.

The word ‘Hausdorff’ is not necessary here (as every differentiable manifold is Hausdorff) and appears to be just a way to appear impressive to the uninitiated by citing an esoteric-sounding, but not necessarily relevant, generalization when the actual spacetime frameworks of the physical theories have much stronger restrictions.

Next we have:

But … if one of a pair of detectors interacts with one of a pair of entangled particles, and the other particle that is far removed instantaneously assumes a correlated state, how can this space possibly be Hausdorff? As has been pointed out ad nauseam by physicists like Einstein and Bell and Aspect and Clauser and Zeilinger, this is a non-local effect.

But… despite lots of badly written pop-sci nonsense, there is nothing in quantum theory which suggests that “the other particle that is far removed instantaneously assumes a correlated state”

And perhaps more importantly, wtf does he mean by “this space”? The discussion of experiments in which particles are separated usually requires our theory to involve observables whose possible values correspond to space-time coordinates relative to an observer, and the set of all possible such coordinate values can be assigned metrics in many ways – with the corresponding topologies all satisfying (among other things) the technical condition used to define what is called a Hausdorff space. So the question of whether or not there are surprising correlations between the spins of separated particles says nothing at all about whether or not the set of possible values of their positions can or cannot be endowed with a Hausdorff topology.

In fact no real physicist has ever claimed that the Bell-defying correlation of spins “is a non-local effect” but rather just that it would have to be a non-local effect IF the underlying physics was classical (with the spin or polarization values being permanent classical properties of the particles or photons in question).

Next we have:

The points in the region of the second detector at an arbitrary distance from the first surely can’t be considered ‘Hausdorff’ (or ‘housed off’) if that distant interaction can influence them.

But, leaving aside the fact that it’s spaces not points that are Hausdorff, there is absolutely no reason to think that the possibility of some instantaneous interaction causing influence from one point to another makes it impossible to distinguish those points in some other way.

Then at last we come to the final question – which I believe I have addressed above.

Source: Cracked up about Quantum Physics – Dr Jo

What optical physics principle makes the Earth’s curvature observable?

The fact that a sharp horizon exists at a distance which increases noticeably over a fairly small increase in altitude should be convincing evidence of the Earth’s curvature for anyone who thinks about it for a moment or two. And Domenico Barillari is right to emphasize that the only evidence for the Earth’s curvature that is visible in a static picture is the angle of depression of the horizon – which is NOT noticeable by eye (ie without using some scientific instrument) at any altitude that most of us will ever experience. But the only “optical physics principle” involved in either of those cases is the assumption that light travels in straight lines.

However, although we can deduce the Earth’s curvature, we do not actually see it in the form of a curved horizon. Pictures showing a horizon that curves down at the edges arise only when the camera’s line of sight is angled downwards and would look the same if the visible part of the Earth was a perfectly flat disc. And again, the only “optical physics principle” involved is the assumption that light travels in straight lines.

Source: (1001) Alan Cooper’s answer to When you see the Earth’s curvature from high altitude, what optical physics principle is at play that makes this phenomenon observable? – Quora

Inseparability of E&M

If two electron beams are travelling side-by-side, then, from the point of view of an observer relative to whom they are moving, the force between them has two components, one electrostatic and the other magnetic. But from the point of view of an observer who is travelling with them, the electrons in both beams are stationary so there is no current and the only force is electrostatic. The Lorentz transformation explains the difference in terms of a change of the distances between the charges in their direction of motion (and so a change of the charge densities in the beams); but regardless of that, what is seen as a sum of “electric” and “magnetic” forces in one frame is seen as just an “electric” field effect in the other, and so there is no inherent observer-independent distinction between the electric and magnetic components. The formulation in terms of a second order tensor allows us to express both the apparent electric and magnetic fields seen by an observer as aspects of a single Lorentz invariant tensor field; but while it describes how they are related as aspects of a single tensor field, I wouldn’t really say that it shows they are so related.

Source: (1000) Alan Cooper’s answer to How does the relativistic transformation of electric and magnetic fields show they are two aspects of a single electromagnetic field? – Quora

Mass to Energy easier than Energy to Mass?

The reason we say mass and energy are the same thing is because how we distinguish them is just a matter of bookkeeping – which depends on our point of view. But if you think of heat as a form of energy rather than of mass, then it is indeed easier to convert mass to energy than vice versa.

When a chemical reaction reorganizes matter into a state of lower potential energy, such as for example whenever we burn a fuel, the energy released typically gets converted to kinetic energy and quickly distributed among many particles of a larger system in the form of what we call heat (which is impossible to fully recover in the form of usable work). An observer looking in detail at the system will see each molecule as having a tiny bit less mass than its component atoms (by an amount corresponding to the binding energy) with the total of all these differences also corresponding to the increased kinetic (heat) energy of the molecules.

On the other hand, when we break apart a chemical bond then we must provide some amount of energy and the resulting masses of the components will add up to that much more than the mass of the molecule. But if we want to do so in a consistent way (such as in electrolysis of H2O or carbon capture from CO2) then we need to provide the energy in a very specific way that is not so trivially easy to set up as just exposing fuel to oxygen.

However, if we measure the mass of the system as a whole (either gravitationally, or as inertia at starting from rest), then the result we get includes not just the masses of all its molecules but also their kinetic energies relative to its centre of mass. And that total (which includes all of the heat energy in the system) remains absolutely constant and never either increases or decreases.

Source: (858) Alan Cooper’s answer to Why is converting mass to energy easier than converting energy to mass if they are the same thing? – Quora

Gravitational Force and “Relativistic” Mass 

A Quoran asks: “I see questions about how mass increases with speed relative to an observer. Does the gravitational force between objects depend on the relative speed between them?

I think the answer to this question is basically yes (although the way General Relativity predicts trajectories is not usually expressed in terms of forces between objects).

Actually, the quantity normally identified as the “mass” of an object is a property of the object itself (that used to be called “rest mass”) which does not depend on the observer, though it is true that the apparent resistance to acceleration (which used to be called “inertial mass” or “relativistic mass”) does depend on the object’s speed of motion relative to the observer (and to the direction of the applied force relative to that motion). So if gravity were due to (rest) mass alone we might expect the answer to be no.

But in General Relativity the trajectory of a freely falling object is governed by an equation in which all kinds of energy (and momentum) contribute. So it it reasonable to expect that a relatively moving object has a stronger effect than one which is stationary relative to the observer. However this is not completely obvious as we need to rule out the possibility that the momentum contributions cancel out those of kinetic energy (like they do in the equation E^2-p^2=m^2 for example).

In order to really answer the question we need to restrict our attention to a situation in which the idea of an inter-particle force does arise as a good approximation. One such is the case of a relatively tiny mass in free fall around a larger one, in which case the Schwarzschild metric provides a good approximation. And in that situation there is an extra non-Newtonian term in the effective potential so that the centripetal acceleration is slightly stronger when the distance is smaller (which gives an extra “kick” at perigee and contributes to the precession of orbits). This stronger attraction could perhaps be interpreted as Newtonian attraction with an increased gravitational mass, and since the orbiting body is moving faster when closer to the central mass it may well look as if the central mass increases with the speed of the orbiting observer.

Perhaps it would also be possible to calculate the second derivative of the distance to the central mass in the coordinates of the observer and I would not be surprised to find that this is proportional to the combined total of mass and kinetic energy of that mass in those observer coordinates.

On the other hand, we should NOT expect the gravitational force between two objects to appear stronger from the point of view of a third observer passing by at high speed (since that would shorten the orbital period while time dilation should make it seem longer).

Source: (899) Alan Cooper’s answer to I see questions about how mass increases with speed relative to an observer. Does the gravitational force between objects depend on the relative speed between them? – Quora

Does mass increase with velocity? 

The claim of “mass increase with velocity” is based on using the word “mass” for something that is not a property of the object itself but rather depends also on the relative velocity of the observer.

Although this usage is not “wrong” and was fairly common in the early days, most physicists soon chose to use the word “mass” just for the invariant “rest mass” and followed the title of Einstein’s initial paper on the effect by just using the word “inertia” for the velocity-dependent quantity that used to be called the “inertial (or relativistic) mass”.

In order to prove that the observed “inertia” depends on the relative speed (and direction) of motion of the observer it is necessary first to define it.

One way is to define it as the ratio of applied force to observed acceleration, but this in turn requires a clear definition of force. For example you could perhaps look at the dynamics of a collision with an object of known mass at rest in the observer’s frame.

Or perhaps you could just read Einstein’s original paper on the subject.

Source: (915) Alan Cooper’s answer to Why does mass increase with velocity as an object approaches the speed of light? How can I prove it mathematically? – Quora

How do CO2 molecules heat their neighbours? 

By hitting them!

A CO2 molecule in the atmosphere, after having absorbed an infrared photon, is vibrating or rotating more vigorously than it was before. When such a molecule collides with a neighbouring molecule (most likely an N2 or an O2) some of that vibrational energy contributes to the speed or vibration or rotation of the one it hits.

The atmosphere doesn’t continue heating up though, because in the equilibrium situation the molecules in the atmosphere collectively emit radiation at the same rate as they absorb it. But they do this equally in all directions, so what they are absorbing from below gets re-directed and only half of it ends up going outwards (with the other half going back down to warm the Earth’s surface).

Source: (901) Alan Cooper’s answer to How does a CO2 molecule in the atmosphere, after having absorbed an infrared photon, transmit that energy to neighbour air molecules to heat up the atmosphere? What’s the mechanism? – Quora

Why F=ma?

Because of Quora’s habit of applying answers to unrelated questions I need to clarify that the question I am responding to here is “Why is Newton’s second law written in the form F = ma, since a more accurate form based on the formula F = dp/dt should be F = m dv/dt + v dm/dt = ma + v dm/dt, since according to the theory of relativity, mass also changes, as does velocity?

Well first, Newton’s second law is written in the form F = ma because that is essentially the way he wrote it and if it wasn’t written that way it wouldn’t be Newton’s law. The form F=dp/dt (which is more general but not more “accurate”) was known to him but he didn’t need to use it because he was referring to the special basic case of an object of a fixed mass (as opposed to something like a rocket expelling massive exhaust) and he chose to take just the simpler special case as his starting point and deduce the more general as a consequence.

He did not write his laws in Lorentz covariant form because he was unaware of almost everything about electromagnetism and radiation, and so had no reason to expect that the correct laws of physics were not actually perfectly Galilean covariant. (But in any case, as described in more detail in other answers, both the Lorentz covariant laws of special relativity and those of General Relativity can be expressed in a form which includes an equation of the form F=ma for appropriate definitions of F,m, and a).

But your claim that “according to the theory of relativity, mass also changes, as does velocity” is based on a concept of “relativistic mass” which is not a well defined property of an object and has long been abandoned as misleading and not useful.

ce: (960) Alan Cooper’s answer to Why is Newton’s second law written in the form F = ma, since a more accurate form based on the formula F = dp/dt should be F = m dv/dt + v dm/dt = ma + v dm/dt, since according to the theory of relativity, mass also changes, as does velocity? – Quora

If energy (E) is equal to mass (m), then why is E = mc^2?

It is not true that “energy (E) is equal to mass (m)”. What is true is that the mass (m) of a system is equal to the special case of the time component of its 4-momentum in the “rest” frame (where the net spatial component of momentum is zero), and that the units we typically use to measure the time component as “energy” (E) are chosen in such a way that the extra “kinetic” energy as seen by a moving observer is given approximately by (1/2)mv^2 rather than (1/2)m(v/c)^2 with units of space and time chosen in such a way that the limiting speed c is very large (and so that speeds we actually experience do not have to be associated with very tiny numbers).

Source: (952) Alan Cooper’s answer to If energy (E) is equal to mass (m), then why did Einstein add the multiplier c^2? Recall the equation E = mc^2. – Quora

Magnets Doing Work

Why do physicists say that a magnet can’t do work? They don’t!!!

What they do say is that a magnetic field does no work on a classical spinless point particle (and that a homogeneous magnetic field does no work on a particle with spin either). But this in no way precludes the field of a magnetic dipole doing work on another object with a magnetic moment – as we observe every day in electric motors, and when we see the attraction of oppositely oriented bar magnets (and the lifting of anything from iron filings to scrap ferromagnetic metal by virtue of the magnetization induced in them by a strong magnetic field).

The question of where the energy to do this work comes from is probably most simply answered by saying that it was put into the magnet by the work necessary to get its atomic spins into alignment, and then resides in the external magnetic field (which, when the magnet has done work, gets reduced or cancelled out by the opposite fields from the objects which have been attracted – but can be restored by pulling them apart again). What is completely wrong is the claim made in some answers (including the one with the highest number of upvotes!) that the source of energy has anything to do with the process of moving the magnet and holding it above the items it is lifting.

Of course, since magnetic fields (like all the other force fields we are familiar with) are conservative, there is no net work done in a cyclical process unless we have an external source of energy (like an electric current supply). But this is true of all physical systems and not any special property of magnets.

Source: (1002) Alan Cooper’s answer to Why do physicists say that a magnet can’t do work? Hold a magnet above a pile of iron powder, it lifts them. Work is done against the G field! How to explain this discrepancy? – Quora