Mass-Energy Equivalence

The “mass” of any complex “body” or system of interacting component “bodies”, defined as its resistance to acceleration from rest, includes not just the masses of its components but also terms corresponding to the kinetic energy of their motions relative to one another and to the force fields acting between them.

The numerical factor relating the amount of energy corresponding to a unit of mass depends on the units of time and distance but it is exactly one if we use years for time and light-years for space or approximately one if we use feet for distance and nanoseconds for time. In terms of more normal units like metres and seconds the conversion factor is very large and one unit of mass corresponds to almost 17 powers of ten (ie 100,000,000,000,000,000 ) in units of energy.

Source: (1002) Alan Cooper’s answer to Can you give an intuitive description of what Einstein’s mass-energy equivalence equation means in one or two short sentences without describing the equation or the math? – Quora

Absolute Rest?

It is a common misunderstanding that what someone once jokingly called the “Big Bang” theory describes the universe as expanding from a single point.

What General Relativity actually suggests is just that in the distant past the universe was very dense (but still of infinite extent) and very hot, with everything flying apart so that since then it has been becoming less dense but not actually “expanding” in the sense of having a boundary that is moving outwards.

With regard to the second question about whether an object can “remain stationary”, in the absence of gravity (ie in Special Relativity) there is no way for an observer to identify whether or not it is moving in any absolute sense. But it is possible to identify whether an object is moving relative to any particular other object, or to the centre of mass of all the other material in its visible universe (so long as that visible universe contains just a finite total mass). And in General Relativity this can be done by checking for isotropy (sameness in all directions) of the microwave background radiation coming from the “Big Bang”.

Source: (1002) Alan Cooper’s answer to What is the rate of expansion of the universe from a single point? Can an object remain stationary in the universe? – Quora

Does Observing Change a Wave to a Particle?

The electron cloud is just a way of predicting what the observer will see on future observations of the electron. When an observation is made, the range of what might be possibly seen in future observations is reduced. So starting right from the moment an observation is made, the observer’s knowledge about the electron is described by a new wave which applies until the observer makes another observation. But the apparent wave behaviour does not cease. It just changes to that new wave, and the electron is never seen to behave like a classical particle. In fact, it is not actually ever either a classical particle with a definite position or a wave of some definite form, and the question of what it “is” may not have any real meaning. All we can talk about are the probabilities of various observations – which we know from experiments do not correspond to any classical picture of what is going on.

Source: (1001) Alan Cooper’s answer to What is it about the act of observing that changes an electron cloud into a single electron that behaves like a particle? – Quora

Do quantum fields have mass?

As so often happens, the answer to this question depends on what you mean by the words used.

Quantum fields are not things in the universe, but rather postulated quantities that are used in an attempt to provide descriptions of possible states of that universe. A state of the universe described in terms of such fields may or may not have mass, depending on the situation and on what you mean by “mass”.

And “mass” is also problematic. The parameter called “mass” in the equations defining a quantum field theory is related to the mass of a minimal (“single particle”) excitation of that field – and indeed this can sometimes be zero (as in the case of the Electromagnetic field for example). But a single photon does have energy and Viktor Toth tells us that Einstein said “E=mc^2” and so it must have mass. So what gives?

Well, when Einstein said “E=mc^2” (or perhaps a bit after the first time he said it) he understood that this was just in the particular situation of a system in its rest frame and a single photon does not have a rest frame!

For states involving many photons it may be possible to define a specific frame of reference by requiring some condition of balance between the observed frequencies (eg for two photons, the point from which they both appear to be departing in opposite directions with equal frequencies might perhaps be taken as the centre of mass). And in that case the force required to accelerate the entire system away from rest might be used to define a rest mass of the system that includes the kinetic energies of its components (in this case just the energies of the photons) as well as the masses of those components (which in the case of photons would be zero).

Source: (1001) Alan Cooper’s answer to Do quantum fields have mass? – Quora

Seeing the Curve

When swimming you can notice the effect over the width of a small lake:

But if you want to see a noticeable effect over just two meters then it depends on what you mean by “noticeable” (eg it would be a trivial high school exercise to calculate the largest radius at which a two meter tangent line ends at least one mm from the corresponding arc).

Source: (1001) Alan Cooper’s answer to How close to the Earth’s core would you need to be to see an ‘uneven’ water level?Not regarding temperature etc making it impossible, how deep would you have to dig to see a noticeable curvature in, lets say a 2m length of water? – Quora

Mass “Loss” in Chemical Reactions

In exothermic chemical reactions the combined masses of the resulting molecules is indeed slightly less than that of the inputs. But this does not make “the system’s mass appear to decrease” for two reasons. The superficial reason is that in chemical reactions the mass decrease due to energy loss from the system is far too small for us to detect, and so we did originally learn the very slightly wrong conclusion “that mass is conserved”. But the deeper reason is that, so long as the system is isolated (eg by including walls that reflect or absorb photons so that none can escape), then the system’s mass does not in fact decrease at all. This is because the system’s mass includes not just the masses of the individual molecules but also all of the energy in the system (including both photons and the thermal kinetic energy of relative motion of molecules). And if we do allow energy to escape, then the system’s mass is not conserved, as what we have learned more recently (though well over a hundred years ago now) is that what is conserved is not just the total mass of constituent particles but rather the total energy (where that includes a contribution from the masses of any particles within the system).

Source: (1001) Alan Cooper’s answer to In chemical reactions like CH₄ + 2O₂ → CO₂ + 2H₂O, we learn that mass is conserved.But since heat energy is released, E=mc² implies a tiny mass decrease. Heat is emitted as massless photons, so why does the system’s mass appear to decrease? – Quora

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