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February 7

Planck's law 1901 article entropy and integration

In his 1901 article, Planck go to the following equation:
$\vartheta =\nu f({\frac {U}{\nu }})$
and
${\frac {1}{\vartheta }}={\frac {dS}{dU}}$
to
${\frac {1}{\vartheta }}={\frac {1}{\nu }}f({\frac {U}{\nu }})$
And integrated (that is what he writes):
$S=f({\frac {U}{\nu }})$

For me, it's too short, not clear.
How can it go from ${\frac {1}{\nu f({\frac {U}{\nu }})}}$ to ${\frac {1}{\nu }}f({\frac {U}{\nu }})$?
And what operation describes the integration that follows?

Malypaet (talk) 13:50, 7 February 2024 (UTC)

I think he uses f as a general symbol to denote unknown functions. The form of the function may differ for the various quantities, but the important thing is that the argument is $U/\nu$ or $\vartheta /\nu$ in this combination. Not particularly good style, but he's Planck and we're nobody. --Wrongfilter (talk) 18:16, 7 February 2024 (UTC)

Same as Wrongfilter, but I went to the effort of typing this out, so I'm darn well gonna post it.

I don't think you can make the assumption that the two appearances of ${\textstyle f({\frac {U}{\nu }})}$ are the same function.

It's clearer if you keep the derivative on the left:

${\frac {dS}{dU}}={\frac {1}{\vartheta }}$

with

$\vartheta =\nu f({\frac {U}{\nu }})$

clearly leads to

${\frac {dS}{dU}}={\frac {1}{\nu }}f({\frac {U}{\nu }})$

as long as you don't require the two "f"s to be the same.

The same holds for the integration - after you integrate dS with respect to dU, you have another ${\textstyle f({\frac {U}{\nu }})}$, with no reason to believe they are the SAME function. PianoDan (talk) 18:29, 7 February 2024 (UTC)

It's easy when you know the result in advance to use magic functions to build a demonstration. I did notice that he used $f$ for different functions (at least five), but until those there was a logic that I understood and that's what I'm looking for here.

And please, don't say that we are nobody in front of someone who found an equation before its demonstration and who never wanted to tell how, by accident?

So far, I have managed to rewrite his article while remaining in the context of power and replacing the resonator with an element of surface $\lambda ^{2}$, I just stumble on these last two magic functions.

For example, if you rewrite this article from the beginning staying in power context, you get:

$\vartheta =\nu f_{7}({\frac {cU}{\nu }})$ where ${\frac {c}{\nu }}\equiv {\frac {1}{h\nu }}$ in time dimension. Just by logic.

Until 2019: $h={\frac {1}{c}}...$

I'll still try to find it myself.

I don't want to learn physics at Hogwarts School. Malypaet (talk) 23:19, 7 February 2024 (UTC)

It's not a question of "magic functions". Planck doesn't give a full expression for f here, because he is trying to establish a general property of ALL possible "f"s.

The point of that section of the paper was simply to demonstrate that the entropy is ONLY a function of ${\textstyle {\frac {U}{\nu }}}$, and NOT a function of the value of c. The derivation successfully demonstrates that fact, and that is its only purpose. PianoDan (talk) 00:17, 8 February 2024 (UTC)

@Wrongfilter My name is nobody, it's for the giant Cyclops Planck!

It's easier to understand when you know the result:

$U=f_{1}(\vartheta$)	$U=ln(\vartheta$)	$f1()=ln()$
$\vartheta =f_{2}(U)$	$\vartheta =e^{(U)}$	$f_{2}()=e^{()}$
${\frac {1}{\vartheta }}={\frac {1}{f_{2}(U)}}$	${\frac {1}{\vartheta }}={\frac {1}{e^{(U)}}}$
${\frac {1}{\vartheta }}=f_{3}{(U)}$	${\frac {1}{\vartheta }}={\frac {1}{e^{(U)}}}$	$f_{3}()={\frac {1}{e^{()}}}$

"c" is a constant and that's not the problem here. If we consider that the radiation circulates in the vacuum between the atoms of the cavity wall, at the microscopic level its value remains unchanged, while at the macroscopic level the speed of the radiation is given by $c_{m}={\frac {c}{n}}$ (n for refractive index), voilà.

Thank you all, our discussions allowed me to find the answer quickly.
Malypaet (talk) 13:59, 8 February 2024 (UTC)

Why isn't rain consistent around the world?

The laws of evaporation physics should be consistent around the world, right? So how can some place rain a lot more than other places. And let's make this about large cities near lakes/oceans. In the U.S., southern California had to open a saltwater treatment plant in 2015 as they are sucking the Colorado River try, while being net to an ocean, which is true of Washington state next to an ocean but rains heavily. What is the missing variable?

Also, what is the lowest altitude for rainfall? Thanks. 170.76.231.162 (talk) 17:36, 7 February 2024 (UTC).

The surface of the Earth is not consistent, so why would we assume that the application of the "laws of evaporation physics" will result in a single, uniform result globally? Different latitudes have different temperature profiles (e.g. it's a lot warmer near the equator and a lot colder near the north and south poles). Geography itself has effects, such as rain shadow, where warm, moist air follows the breeze up a mountain range and cools as it goes up, resulting it it raining on the mountains, and dry air coming down the other side. --OuroborosCobra (talk) 19:16, 7 February 2024 (UTC)

Then there's oceanic currents, and the Earth's tilted axis, and its non-circular orbit around its star, and... Bazza (talk) 19:58, 7 February 2024 (UTC)

You are too close to the horse latitudes and the west side of continents exaggeration of the horse latitudes. This is why Afrique Nord to west India is dry and the rest of the Eurafrasian horse is wet. With a transition zone in between. The horse latitudes exist cause Earth has three pairs of convection cells (the average wind circuits of the atmosphere). The air sinks on average in the horse latitudes reducing rain. The southbound California current also reduces rain on that side by causing subsidence. While the northbound West Atlantic express boosts East Coast rain if anything. All mountains to the west of you reduce rain. So Mojave is dry, Arizona is dry. Also fossil fuels is making the driest latitude which is in Baja (not the end or Tijuana, somewhere in between) move closer to you and many other effects. Sagittarian Milky Way (talk) 23:09, 7 February 2024 (UTC)

Death Valley: 282 ft (86m) below sea level, 2.2 in (56mm) annual rainfall. DOR (ex-HK) (talk) 16:27, 13 February 2024 (UTC)

Lowest natural surface on Earth, Dead Sea c. minus 1,400 feet and drying up, has had rain. It is true that rain dries continuously in a desert though, sometimes evaporating completely before reaching the ground. Sagittarian Milky Way (talk) 17:20, 13 February 2024 (UTC)

Also the East Coast would have less rain if the Gulf of Mexico and Caribbean were full-blown land. Sagittarian Milky Way (talk) 17:36, 13 February 2024 (UTC)

laser cutting glass

This video[1] shows glass being cut by a laser. There appears to be some sort of "white smoke" during the process. I'm trying to understand what's going.

The melting point of Silicon dioxide is approximately 1713 °C, the boiling point is around 2,950 °C.

I'm guessing that it's a combination of the following factors:

1. The glass in the cut area is melted, and remains attached to the the two pieces. After the cut, the melted parts solidifies as a part of the two resultant pieces. The total mass of the two resultant pieces is the same as the original piece.

2. The glass in the cut area is melted and then vaporized, the glass vapour is removed by the fume extraction system. The total mass of the two resultant pieces is less than the original piece.

3. The glass in the cut area is melted, and micro particles of it (in either liquid or solid state) are carried away as a "white smoke", and micro particles of it are also splattered around the machining area. The total mass of the two resultant pieces is less than the original piece.

Which one of these factors apply in this case?

If it's a combination of two or more factors, which one is the dominant one? OptoFidelty (talk) 20:04, 7 February 2024 (UTC)

It's likely closest to #2 in your list. See laser ablation. It's not necessarily melting and then evaporating, however, it could be sublimating (direct from solid to gas). Yes, some amount of mass is being lost (or at least no longer useful to these pieces, as I suppose you could recover and reuse the smoke material for a nice atom economy with the fume extractor), although whether you say that the larger or smaller resulting pieces of glass is losing that mass (or some ratio between them) is basically up to where you designate the division between the pieces. In this case, they seem to want the smaller piece of glass that they cut out, so they would program the machine to cut it out such that the tolerances of the dimensions of the final piece are within required specifications. Rather than cutting "directly on the line," if you were, it would intentionally cut slightly outside of that line knowing the diameter of the "drill." The same would be done with a mechanical process. --OuroborosCobra (talk) 20:21, 7 February 2024 (UTC)

Thank you!

Regarding, Sublimation (phase transition), what determines whether "solid->liquid->gas" happens or "solid->gas" happens? It it related to the speed of the heating?

For example:

1. Iodine is heated at room temperature. Only "solid->gas" occurs.

2. Silicon dioxide is heated slowly above its boiling point. Only "solid->liquid->gas" occurs.

But in the situation in the video, it's not clear whether it's "solid->liquid->gas" or "solid->gas", or a mix of both. Is there a way to calculate the behaviour if all the processing conditions are known? OptoFidelty (talk) 20:45, 7 February 2024 (UTC)

It depends upon both temperature and pressure. See phase diagram. Even for pure substances (and glass is usually intentionally not pure), calculating those values is non-trivial as you end up having to factor in a lot of things, such as calculating intermolecular forces. With a lot of effort, you might be able to do that with computational chemistry software packages, but one is better off figuring out the phase diagrams for given substances experimentally, such as through various forms of calorimetry. That said, phase diagrams for common substances have largely already been figured out experimentally, so you can usually look it up. I don't know how reliable it is, but here is one for silicon dioxide. However, that gets into another topic, which is "what is glass?" Look over at glass, and you will see that it isn't any single thing that can be just called "silicon dioxide;" that's just one type (and probably not the type in the video). What specifically that glass is made of, what crystal phase it is in, whether there are additives (such as in borosilicate glass, these are all going to have an impact on the conditions for melting, boiling, and sublimation. --OuroborosCobra (talk) 21:28, 7 February 2024 (UTC)

(edit conflict) For a one-component three-phase substance it depends on the pressure $P_{\text{tri}}$ of the triple point. A three-phase substance can only be in the liquid phase at pressures greater than $P_{\text{tri}}$. If it is heated at a lower pressure, it will go directly from solid to gas.  --Lambiam 21:35, 7 February 2024 (UTC)

True, but unless you already know the triple point pressure, you still can't just calculate it easily. --OuroborosCobra (talk) 23:12, 7 February 2024 (UTC)

Ordinary glass doesn't really have a crystalline phase, nor a melting point. At least, not one that can be practically demonstrated. It does have a glass transition. And most ordinary glasses are pretty liquid at 1000°C; no need to go all the way up to the melting point of silica. Which is good, otherwise glassmaking would've been impossible in antiquity.

The glass particles in the white smoke (which could also contain contaminants) must be fairly solid. If they had been fairly liquid, they would be glowing yellow.

Glass is also brittle. Quick localised heating can break off tiny particles without making them significantly liquid or heating them to incandescence. You should be able to tell when viewing the smoke particles under a microscope. PiusImpavidus (talk) 10:31, 8 February 2024 (UTC)

Couldn't the "smoke" also contain vapourized SiO₂ (boiling point 2,950 °C)?  --Lambiam 11:27, 9 February 2024 (UTC)

Classical Mechanics: I wonder where my mistake in the following calculation is:

0. (This setion was added later). Here, I'm only referring to theoretical Newtonian situations involving no potential energy (i.e. no chemical and nuclear particle bonds, nor physical force fields, nor deformation of solids).

1. Hence, a given system's kinetic energy is equal to the sum of the kinetic energies of the system's parts, regardless of whether the system consists of two bodies or two photons or whatever.

2. Let two bodies, having the same mass $m$ each, move by the same speed $v$ in opposite directions.

3. Hence, each body carries kinetic energy of ${\frac {1}{2}}mv^{2}.$

4. Hence, per #1, the whole system, consisting of both bodies, carries kinetic energy of ${\frac {1}{2}}mv^{2}+{\frac {1}{2}}mv^{2}=mv^{2}.$

5. On the other hand, per #2, the whole system is at rest, so it's kinetic energy must be zero.

I wonder how I can settle the contradiction between #4 and #5. I guess the mistake is in #1 or in #5, but I can't find it yet.

Since, besides the first sections 0-1, each of the other sections 2-5 is supposed to derive from a previous section, so can you precisely point at the section (of the above six) which is the first one you don't agree with? 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 20:31, 7 February 2024 (UTC)

What is your definition of a system being "at rest"? The concept applied to a collection of objects is usually understood to mean that each object in the collection is at rest. For a fluid it means that the particles of which it exists are all at rest. (See Hydrostatics.) A spinning disc such as a gyroscope is not at rest, even though for each infinitesimal part having some velocity v there is a corresponding infinitesimal equi-massive part having velocity −v.  --Lambiam 21:01, 7 February 2024 (UTC)

Don't you agree, that a given collection of two photons, that move in opposite directions, is a collection at rest? 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 21:42, 7 February 2024 (UTC)

You're confusing energy and momentum. Momentum is a vector quantity, so your two particles cancel each other out. The system has zero net momentum.

Kinetic energy is a scalar quantity, and so direction doesn't matter when you add it up. PianoDan (talk) 22:24, 7 February 2024 (UTC)

Since, besides the first sections 0-1, each of the other sections 2-5 is supposed to derive from a previous section, so can you precisely point at the section (of the above six) which is the first one you don't agree with? 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 07:44, 8 February 2024 (UTC)

1-4 only apply to the system's parts because the KE of any massive object is entirely dependent on its speed. To clarify, consider another example: A) A newspaper is at rest on someone's doorstep: v=0, m₀v=0, KE=0 and it has rest mass m₀. OK, and... B) Its molecules vibrate, thus: v_i>0, m_iv_i>0 and KE_i>0 for every ith molecule. Descriptions A and B differ in granularity, and because mass and energy are equivalent the sum of its internal energies E_i divided by the speed of light squared equals its rest mass m₀ even though it is still at rest on the doorstep (v=0, m₀v=0, KE=0), thus 5 is correct. Also the classical formula is an approximation and does not hold for photons. Clarifying 1 to say "...a given system's [internal] kinetic energy..." would distinguish it from the kinetic energy of 5. Modocc (talk) 00:54, 8 February 2024 (UTC)

you have discovered temperature. Greglocock (talk) 02:40, 8 February 2024 (UTC)

Since, besides the first sections 0-1, each of the other sections 2-5 is supposed to derive from a previous section, so can you precisely point at the section (of the above six) which is the first one you don't agree with? 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 07:57, 8 February 2024 (UTC)

More precisely, they've discovered internal energy (there's a connection to temperature, of course). Energy is a book-keeping device. Depending on the level of description of a system, energy can be booked to different accounts. The kinetic energy of the individual particles contributes to the internal energy (and total energy) of the two-particle system; in the rest frame of the two-particle system, it does not contribute to the kinetic energy. In a many-particle system one can distinguish between the mean motion of the particles (which make up a gas flow, for instance) and the random motion (which determines the temperature of the flowing gas). --Wrongfilter (talk) 07:01, 8 February 2024 (UTC)

Since, besides the first sections 0-1, each of the other sections 2-5 is supposed to derive from a previous section, so can you precisely point at the section (of the above six) which is the first one you don't agree with? 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 07:44, 8 February 2024 (UTC)

Number 4, which should be formulated as "the whole system ... has a total (or internal) energy of ..." (assuming there's no interaction between the particles). --Wrongfilter (talk) 07:53, 8 February 2024 (UTC)

Thank you. Are you sure, the section - which is the first one you don't agree with, is #4 - rather than #1? Note #4 directly derives from #1, so if you agree with #1 you apparently have to agree with #4. 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 07:57, 8 February 2024 (UTC)

Yes, I missed that. So in #1 it should be "The systems total energy is the sum of the kinetic energies of its constituent particles". This is true if there are no interactions between the particles, and if there are no further internal degrees of freedom apart from the translational motion of the particles (translate as appropriate for photons, where the term "kinetic energy" is also not quite right). --Wrongfilter (talk) 08:14, 8 February 2024 (UTC)

Thank you. I've just added section #0 above, which excludes potential energies. Could you point now at the section which is the first one you don't agree with? 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 08:43, 8 February 2024 (UTC)

I've already said all I have to say. --Wrongfilter (talk) 08:46, 8 February 2024 (UTC)

New question: Do you agree that #1 directly derives from #0? 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 08:50, 8 February 2024 (UTC)

I've told you what I think is wrong with #1, this is independent of your #0. Please don't make these threads endless. --Wrongfilter (talk) 08:55, 8 February 2024 (UTC)

I only wanted to be sure I understood well (which sometimes needs further clarifications). So, as I understand now (due to your last clarification in your last response), internal energy is still different from kinetic energy, despite #0. Thank you. 2A06:C701:746A:1600:7422:E3A5:9217:C109 (talk) 09:02, 8 February 2024 (UTC)