joigus

In the case of wave-particle duality, there doesn't seem to be the words to subtitute them. Speaking from personal experience, it feels like people have given up on trying to understand this double nature in terms of old concepts, and prefer to say things like 'quantum behaviour,' 'quanta,' and the like. Another interesting case that can perhaps be subsumed here is when people realise that something that seemed very special is but a particular case of a more general phenomenon. In field theory, eg, the word 'charge' was reserved for 'electric charge,' while people who work in field theory today tend to call anything that's conserved 'Noether charge' or just 'charge.' For these people, energy, momentum or angular momentum components, etc, are just 'charges.' It tends to make life easier when the context is understood by everybody in the field. I'm not a language determinist, BTW. I think it's more of a two-way connection. Sometimes language affect our thinking, other times it's our thinking that ends up being affected by the way we think.

Oh, and the list goes on and on. There is no 'quality time' in scientific English, but in regular English, there is.

LOL. Another way of saying 'heavy.' There is also a curious effect that's more horizontal --discipline to discipline-- than temporal. I think your example of 'metal' has a horizontal component to it. Eg: Physicists use 'reversible' with a different meaning than chemists do. For a physicist, 'reversible' means the process is infinitely slow, and every intermediate state is one of equilibrium, while chemists may use it in that sense when discussing thermodynamics, but definitely often use it meaning that a chemical reaction is completely displaced in one direction. Biologists too.

Another interesting shift in nuances is: Gauge invariance -> Gauge symmetry -> Gauge ambiguity I've even read and heard pronounced the expression 'gauge junk.' That's probably because, while we think we understand very well what heat is about, in the case of gauge ambiguities it's not so clear. So you have to write down the equations to comply with an apparently beautiful and simple symmetry principle, and once you start your calculations you have to throw away half of it because it's junk? What's that about?

You definitely should. The very reason why there are gauge fields is better understood in terms of the Lagrangian formalism. It's a beautiful language to express every fundamental physical theory we know. It also makes hard problems look simple. The downside is perhaps that it's far less intuitive than thinking about forces of different kinds. Another downside --and a very big one, mind you-- is that systems with dissipation are not possible to describe by means of Lagrangians. That's because friction is an emergent behaviour.

There's so much one can do with words. As said above, a system stores energy, not heat. It can give it out, off or away by means of work or heat exchange. Words in science are as good as the mathematical concepts they help us remember, the precise experimental operations they represent, etc.

Neither do I. And I can assure you I won't. Absolutely. I would even go as far as to say that there are several assumptions that are unphysical in my picture. In my 'defence,' my attempt was not meant to illustrate how real wind behaves. A spatially-confined gust of wind that's hitting a wall obviously does not correspond to my simpleminded 'model' of particles coming from -infinity and bouncing off an infinite wall, back to -infinity, while keeping completely parallel to each other. It's just an illustration of how a continuous supply of particles hitting a wall at a certain velocity will transfer momentum per unit area, but you have to give up on F=ma, not because it's any the less true, but because that relation is not the one that's useful. What's useful is transfer of momentum per unit area per unit time. Unfortunately, my model cannot be made consistent with the continuity equation, which I now realise after having thought about it for a while --even if it gives the right result by tinkering with the numbers. I think Studiot is thinking of a slightly different setting than you are, with no ground to sustain the wind so as to produce eddies, as you have pictured. But maybe he will be willing to ellaborate on that. I do believe @Boltzmannbrain has probably understood the main point, without getting bogged down in finer details of zero-velocity areas, stationary currents circulating, and the like, of which I'm --partially, at least-- guilty. The take-home lesson is: A continuous supply of particles at speed v, and constant density d, hitting a wall, will exert a force on a wall that's more simply expressed as proportional to dv2. The calculation of the dimensionless coefficient being a matter of correctly applying fluid mechanics. What's going on is transfer of momentum. Shall we leave it at that? I am happy with that, at least, for the time being. But if you want to discuss more, I'll be happy to take a back sit and learn --or refresh my memory-- a little more.

There is the possibility that I made a mistake. Let's see what other people have to say. Now that I think about it, it's not so clear the the continuity equation is satisfied at the point of hitting the wall. \( 2v\rho \) mass per unit area per unit time is hitting the wall, while only \( v\rho \) mass per unit area per unit time is bouncing off. There seems to be a mismatch. Qualitatively, it's clearly all about transfer of momentum, which in this case is better looked upon in terms of mass flow and individual momentum transfer per particle. Getting the constants right is another matter. I'll probably have to think about it some more when I get the chance. Anyway, it's just fun thinking about these problems.

Not really. But you can make it one. I'm sure you've taught a teenager at some point in you life. If you want to share your recollections/conclusions on a new thread I, for one, would be interested.

No. Sometimes infinity appears in a mild form, that can be tamed. That's why we teach those instances to teenagers (eg limits.) Other times it's like you say. Still other times finite problems are every bit as difficult. And quantum superpositions are not superpositions of 'real' and 'imaginary.' What are you on about?

Just another WAG.

I'm OK with that. Every time mathematicians publish a proof, there's a time period during which the jury is still out. Another caveat on terminology though. Mathematically-minded people like to say 'QFT is a functor' --thereby 'functorial.' This is a concept in category theory. People also say 'action is a 'functional' --a function of function that produces a number. Those are very different concepts with very similar terms, so it's necessary to tread carefully.

I think I got a reasoning that recovers the 1/2 factor that @sethoflagos posited from Bernouilli's equation and, by now, I'm totally convinced is right. In the general situation --but still idealised, as we are assuming an infinite wall, and a constant velocity field pointing right in the image, we'd have something like, An ensemble of particles that's equivalent as to momentum --and K.E.-- transfer is, This would give a velocity field and a density that are consistent with the continuity equation iff 1/2 the molecules are going right, and the other 1/2 is going left --but with half the velocity. Otherwise, we wouldn't be getting an overall velocity field going right with velocity v, as 2v-v=v, while v-v=0. So the density \( \rho \) must --on average-- be made up of 1/2 the molecules going right --and eventually hitting the wall--, and the other 1/2 going left --and therefore never hitting it again. Although we still have \( \rho\triangle V=\rho Sv\triangle t \), this gives us twice the molecules that are going to hit the wall within time \( \triangle t \). Therefore, I should have written, \[ \triangle p=\frac{1}{2}v\triangle m \] which gives \( \frac{1}{2}\rho v^{2} \) for the pressure. In the more realistic case, it is certainly true that molecules moving wildly in all kinds of oblique directions would contribute to the --constant-- density, but this is made up for in the simplistic model by assuming that they're not there, and the average leftward motion is totally carried by particles moving perpendicularly to the wall. Note for @Boltzmannbrain on aspects already mention by other members: Newton's equations of motion are valid thoughout, of course. But there are cases in which thinking in terms of momentum transfer is more useful. This is, I think, one of them. Please tell me whether you agree/disagree with this --admittedly-- drastically simplified model.

Here. I didn't remember very well. Mathematicians seem to have proved it's unsolvable, on account of incompleteness theorems: https://www.eurekalert.org/news-releases/490733

Sorry I can't comment much on that. I know next to nothing about Polyakov loops. From what I gather, those are Wilson loops applied to a QFT upon which you've performed a Wick rotation. I suppose they give you topological invariants of the corresponding thermal theory?

I think @studiot knows much more about projective geometry than I do. Also other members who are mathematicians, or much more educated in higher mathematics. The mass gap in YM is considered to be the toughest problem around concerning physics, BTW. Just saying...

That was I --speaking a la Gandalf. By the way, I still think @swansont's argument is the ticket here. @Lorentz Jr is right though. Any realistic situation will be riddled with turbulence. Still, you can concoct a non-turbulent system for which the energy/momenta analysis is equivalent. Crudely speaking, that's because you can picture all the molecules hitting the wall at speed 2v, and all of them bouncing off perpendicularly (to the wall) at speed -v, the wall absorbing the deficit of kinetic energy, and the flow of particles still being a stationary flow which on average has velocity v towards the wall. Momentum would be conserved, and the continuity equation wouldn't suffer in the least.

Mmmm. Sorry, I didn't see your factor 1/2. I don't think conservation of (kinetic) energy will give you the right answer here. There's no reason to assume collisions are elastic. OTOH, momentum is always conserved at a macroscopic level. That's why I think Swansont's suggestion is the most secure foundation for this kind of reasoning. Does that make sense?

Agreed. You can get Seth's 'hydrodynamic' formula, from Swansont's suggestion. Consider some amount of time \( \triangle t \). With @studiot's suggestion of 1 second, the maths is slightly simplified. This defines a reference layer of wind. \[ \triangle l=v\triangle t \] \[ \triangle V=Sv\triangle t \] Every \( \triangle t \) seconds the corresponding layer of wind will transfer momentum to the wall by an amount, \[ \triangle p=v\triangle m \] where, \[ \triangle m=\rho\triangle V=\rho Sv\triangle t \] And the resulting pressure would be, \[ P=\frac{F}{S}=\frac{v\triangle m}{S\triangle t}=\frac{\rho Sv^{2}\triangle t}{S\triangle t}=\rho v^{2} \] As Studiot pointed out, momentum parallel to the walls is canceled out. While the air molecules bouncing back cannot transfer momentum back to the wind, so to speak, as we're assuming the gust of wind to be constant --it has reached some static regime. I forgot to mention my \( \rho \) is @studiot's d --the constant density.

I see no way in which anybody could disagree with it. And that's not good, I think.

The video certainly makes it looks as if it's the nuclei that are working as forced oscillators. Looking forward to your comments.

OK. So what I see here is: (1) Indeed Dr Lincoln says 'the photon absorbs the atom' at about 5.34. But I think that's just a slip of the tongue. (2) He doesn't mention the amplitudes, nor does he mention the frequencies involved. A forced oscillator responds with an amplitude that depends on the forcing oscillation frequency, the friction coefficient, and the natural frequency of the oscillator. But the frequency that survives, after the transient has died out, is the same as the frequency of the driving force. He omits those aspects, I think, for the sake of simplifying the explanation. He's mainly concerned with what he perceives as ongoing confusion due to flawed arguments on YT and the like. If you treated every single atom in the crystal as dipole, you could in principle calculate, by solving Newton's equation, how much it accelerates. You would need a formula to calculate the additional radiation, which is not easy at all. Of course, none of these details are dealt with in the explanation. And I understand why they aren't.

I think you're still confused as concerns 'fields' in algebra vs 'fields' in physics. Very different things. It's like 'dog' in 'hot dog' vs 'dog' in 'I'm walking my dog.'

What is a centripetal velocity? You mean angular velocity? Of what? Please, pick a scenario and stay with it for as long as you can. Reaction of a plane, Euler forces, now atmospheres... These are very different things. What's next? I don't see a circular argument. More like a Brownian motion.

Excellent idea. Thank you.