Wikipedia:Reference desk/Archives/Science/2022 April 4

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April 4

Weighing less than nothing

In Don Rosa's comic The Universal Solvent, the Ducks travel to the centre of the Earth in a vertical shaft. At one point, they notice gravity has started working in the opposite direction - it pulls them up instead of down. According to the story, this is because the majority of the Earth's mass is now above the Ducks instead of below them. Assuming travelling so far down were possible in the first place, is this really what would happen? JIP | Talk 11:20, 4 April 2022 (UTC)

Not really. At any point nearer to the top of the shaft than the centre of the Earth there would be more mass below. Think of the other hemisphere for a start. Only at the centre would gravity cease pulling down. Still, this is science fiction so interposing reality isn't always helpful! Martin of Sheffield (talk) 11:29, 4 April 2022 (UTC)

Thanks for the answer. It kind of figures. The further down they go, the less they weigh - but their weight stays positive the whole time, as the other hemisphere is still below them. I wonder how Rosa, who has a Bachelor's Degree in civil engineering, failed to spot this. JIP | Talk 11:43, 4 April 2022 (UTC)

Shell theorem says: "If the body is a spherically symmetric shell (i.e., a hollow ball), no net gravitational force is exerted by the shell on any object inside, regardless of the object's location within the shell." They would simply feel the gravity of the sphere closer to the centre while everything else would cancel out. I don't think the author actually thought the story was scientifically accurate. It's a Donald Duck story, not hard science fiction. PrimeHunter (talk) 11:56, 4 April 2022 (UTC)

It's actually quite a bit more complex than that once you factor in the changing density. See Gravity of Earth for the full, unexpurgated, gory details. Martin of Sheffield (talk) 12:05, 4 April 2022 (UTC)

Density variation as a function of depth is irrelevant to the shell theorem; are you saying density varies significantly in ways not spherically symmetric? —Tamfang (talk) 03:35, 6 April 2022 (UTC)

But the earth isn't a shell. Using shell theory a body 100 km down a hole experiences no gravitation attraction from the outermost 100 km of the earth's mass. The inner 6,361 km are still pulling down. Repeat the previous two sentences ad nauseum all the way down. What shell theory does tell us is that a problem like this is best handled by modelling the earth as an onion. As you pass through each layer you can discard it and only consider the layers below. Now consider what happens if our onion is composed of differing densities. The outermost layer might comprise only 1% of the mass, so the pull of gravity of the rest is 0.99g (I'm taking serious liberties with the "r²" term of the attraction). Now if the next layer actually held 10% of the onion's mass, then once that was passed through the pull would be 0.89g. Let's assume 5% for the next layer: 0.84% and so forth until we pass through the last layer at the heart of the onion and the pull disappears. I refer you back to Gravity of Earth for a more theoretical and less culinary explanation. Martin of Sheffield (talk) 08:12, 6 April 2022 (UTC)

I'm still mystified as to what you think you're contradicting. Whatever. —Tamfang (talk) 04:19, 10 April 2022 (UTC)

This statement: "Density variation as a function of depth is irrelevant to the shell theorem" Martin of Sheffield (talk) 10:45, 10 April 2022 (UTC)

Unless they mean they've already passed the center of the Earth within the shaft and was on the "opposite" side already, where "up" instead of "down" is pointing toward the center. GeorgiaDC (talk) 18:44, 4 April 2022 (UTC)

The shell theorem applies to shells (such as a Dyson sphere). However, Earth is a ball. 78.1.172.11 (talk) 21:11, 4 April 2022 (UTC)

Edit: Nvm I've read PrimeHunter's comment again and I see we don't actually disagree. 78.1.172.11 (talk) 21:20, 4 April 2022 (UTC)

• Assuming a spherically symmetrical, solid cow Earth, if you dug a giant whole through the dead center of the earth, and out of the other side, and then jumped in the hole, ignoring air-friction, you would be a harmonic oscillator, barely reaching the other side (your center of mass would clear the surface of the other side of the earth to the exact height your center of mass currently is on this side of the other), bouncing from one side to the other. This video from Science Asylum does a good job explaining the physics of the situation. --Jayron32 12:28, 4 April 2022 (UTC)

  If you instantly teleported out a hole and turned real physics back on what would happen? Sagittarian Milky Way (talk) 14:47, 4 April 2022 (UTC)

  If you returned air resistance into the picture, you would behave like a dampened harmonic oscillator, gradually slowing down and your return points would get further and further from either hole, until you came to a gentle rest floating at the exact center of the earth. --Jayron32 15:12, 4 April 2022 (UTC)

  Also, if you returned the Earth's actual gravitational irregularities, I doubt you'd ever make it to the center in one piece. You're going to be moving at terminal velocity pretty quickly, and given the irregularities in gravity on your long trip, you'll probably be pinging off the sides of your shaft rather violently. And, since part of the Earth are molten (liquid) rock and metal, your hole would very quickly fill in, and then exposed to the air above it, would cool down pretty quickly. You wouldn't make it past the Asthenosphere, which means you only would fall for a few kilometers before smacking into hot, partially solidified magma. --Jayron32 16:33, 4 April 2022 (UTC)

  Also, also, as I look at some information in Kola Superdeep Borehole, even within the crust, when you get down past about 10 km, the temperature is hot enough to boil water. You'd probably bake to death before you even hit the bottom. --Jayron32 16:36, 4 April 2022 (UTC)

  Also, also, also, consider that Caisson disease, aka "the Bends", is a real possibility at depths of that level, given the massive increase in air pressure you are likely to experience. --Jayron32 16:52, 4 April 2022 (UTC)

a decompression accident would be a problem on the way back up, oxygen toxicity due to its high partial pressure would get you on the way down. — Preceding unsigned comment added by 2A01:E34:EF5E:4640:1961:F229:5025:A78 (talk) 11:33, 5 April 2022 (UTC)

• I always thought that (assuming you didn't fill most of it with air compressed to specific gravities >>1) some spectacular implosion at at least c. the speed of sound of each Earth layer would happen, as most of the cylinder surface would have pressure of 1 to 3.6 million atmospheres. How much magma gas would come out of solution from this brief exposure to the hole's lower pressure, and how dramatic of an eruption that could cause I don't know. Would magma not reach the entrance because mantle is denser than continental crust or would it barely leak out or spurt 100 meters high or something stronger and less Hawaiian? How big or small would the earthquake be? How far from the hole would crack? And if it's wide enough for bouncing off the edge to not come first then how far could you fall before things get bad? Sagittarian Milky Way (talk) 23:32, 4 April 2022 (UTC)

Even worse than gravitational irregularities is the Coriolis force sending you off course. Unless you travel exactly Pole to Pole. PiusImpavidus (talk) 10:15, 5 April 2022 (UTC)

Ah yes, that too. So many harms from one hole. Sagittarian Milky Way (talk) 13:29, 5 April 2022 (UTC)

You've obviously met my first girlfriend. --Jayron32 10:55, 6 April 2022 (UTC)

And that's in a relatively cold section of the crust. In other parts of the world, e.g. the KTB boreholes in Bavaria, the temperature at that depth is so hot that the rock is liquid, and it's not physically possible to drill any further. 51.155.110.141 (talk) 20:36, 5 April 2022 (UTC)

The comic seems to say that their weight is decreasing (again because of the shell theorem) and that they will soon be weightless (at the center of the Earth), not that gravity is pulling them in the opposite direction. --Amble (talk) 19:18, 4 April 2022 (UTC)

Did the comic literally say they weigh "less than nothing"? --←Baseball Bugs ^{What's up, Doc?} carrots→ 22:19, 4 April 2022 (UTC)

It's often asserted that a person jumping into a hole more than 35 miles deep would not hit the bottom, but would float at that depth due to the density of the air there. But that assumes air is an ideal gas, as well as neglecting the increasing temperature and the shell theorem. It would be interesting to know the true person-floating depth if it exists. catslash (talk) 22:57, 5 April 2022 (UTC)

According to the Internet air at the temperature of 35 miles would have to be past the right edge of the chart (10,000 bar) to float a person. Sagittarian Milky Way (talk) 02:50, 6 April 2022 (UTC)

The effect of changing gravity would cancel out; what matters is the relative densities of the falling body and the medium. —Tamfang (talk) 04:25, 10 April 2022 (UTC)

Nitrogen triiodide; is the purple cloud toxic?

I asked this question at Talk:Nitrogen triiodide#Toxicity?, but I thought that it was worth reposting here.

Youtube is full of videos showing Nitrogen triiodide exploding with a purple cloud. Our article on Nitrogen triiodide should say whether that purple cloud is toxic.

The first paragraph of the article says "...releasing a purple cloud of iodine vapor", but later the article says

"The dry material is a contact explosive, decomposing approximately as follows:

8 NI₃ · NH₃ → 5 N₂ + 6 NH₄I + 9 I₂"

So which is it? "Iodine vapor"? Or is the purple cloud a combination of Ammonia, Ammonium iodide, and Elemental Iodine (I₂)? And is the purple cloud an actual vapor or is it a cloud of solid (liquid?) Iodine particles? Iodine boils at 113.7 °C and Ammonium iodide boils at 551 °C so (based upon my extensive education (which consists of getting a C- in high-school chemistry)) it seems that only the Ammonia would be a vapor, while the purple stuff would be an aerosol.

Could someone who understands chemistry please write up a paragraph for that article with citations explaining what is in that purple cloud and whether it is toxic? 2600:1700:D0A0:21B0:5A9:3B45:A28C:DE50 (talk) 18:54, 4 April 2022 (UTC)

Iodine is a solid under standard conditions, but it sublimes under these conditions to produce some violet iodine vapour - see Iodine#Properties. Mikenorton (talk) 22:01, 4 April 2022 (UTC)

Ammonia is a colourless gas. I could find no mention of the colour of gaseous ammonium iodide, which suggests it is not remarkable. So the markedly violet colour of the cloud is rather likely purely due to the presence of iodine in the gas state.  --Lambiam 22:54, 4 April 2022 (UTC)

The pictures at [1] look a lot more like a smoke or a dust than a gas to me. Could it be mostly a fine powder of iodine crystals? Of course the crystals create the gas, so there would have to be be gas as well. Either way, could it be that all the article needs is a "for more information see Iodine#Toxicity" link? 2600:1700:D0A0:21B0:5A9:3B45:A28C:DE50 (talk) 02:03, 5 April 2022 (UTC)

If you watch the video, you can see that the cloud looks like vapour more than dust once it starts to disperse.  --Lambiam 09:35, 5 April 2022 (UTC)

What do we do when we see a chemical explosion and a cloud and we're not sure how hazardous it is?

Well, it turns out that this is very common!

I pull out my handy paper copy of the Department of Transportation Emergency Response Guidebook (2020 edition, available for free in PDF format and probably available for free at your local US fire department!)

This is a guidebook for first responders "during the initial phase of a transportation incident involving hazardous materials/dangerous goods". How often do we find a weird chemical with unknown safety parameters? Too often.

Page 1 of the book is the decision-making flow-chart. Follow along - the flow chart will direct you to Guide 111: "Unidentified Cargo" - and lists all the potential hazards. "Evacuation: Immediate precautionary measure - Isolate spill or leak area for at least 100 meters (330 feet) in all directions."

If you say the dry material is a "contact explosive": That'd be ... Guide 112 - "EVACUATION - Immediate precautionary measure. Isolate spill or leak area immediately for at least 500 meters (1/3 mile) in all directions. MAY EXPLODE AND THROW FRAGMENTS 1600 METERS (1 MILE) OR MORE IF FIRE REACHES CARGO."

Later when we have time for it, we can pull out the MSDS and figure out how to comply with the necessary local and national chemical handling guidelines. In the short term, don't play with chemicals if you aren't trained to deal with them and their hazards.

Meanwhile - if you think people who are skilled and trained in chemistry have the time to micro-analyze the safety concerns of every possible type of chemical they might encounter, and evaluate whether this particular purple cloud matches that particular textbook example, and then clearly cite the details on a free encyclopedia, you are grossly misinformed. Don't mess around with chemicals, don't mess around with explosives, and if you're not sure if it's hazardous, assume it is hazardous and don't mess around with it.

Nimur (talk) 23:18, 4 April 2022 (UTC)

Got it. You have chosen to criticize someone for daring to ask that a Wikipedia article on Nitrogen triiodide not give the reader two different answers for what it turns into when it explodes.

And we shouldn't have information such as Iodine#Toxicity, Phosphorus#Precautions, Organic mercury#Toxicity and safety, Thermite#Hazards, or any of the other thousands of places where Wikipedia talks about the hazards of a substance. Got it.

Please go away. Your answer was not helpful. 2600:1700:D0A0:21B0:5A9:3B45:A28C:DE50 (talk) 02:03, 5 April 2022 (UTC)

Telling an editor with over 16 years experience here to "go away" is not likely to get you anywhere. --←Baseball Bugs ^{What's up, Doc?} carrots→ 02:59, 5 April 2022 (UTC)

Let me apologize for my harsh tone.

My key point remains valid: even if we have reputable sources that describe the textbook-example chemistry reaction, there is no reasonable way to know if the particular instance in a particular demonstration is safe. The correct procedure - a procedure that I didn't invent, but cited from a reliable source!: assume the material is hazardous until other facts are known. Does this procedure apply in every context? Perhaps not. But, when we discuss hazards with anonymous participants on the internet - I feel that it is great to start from a position of zero assumptions, and aim for maximum safety. Please keep in mind that these discussions are archived and searchable - somebody in the future (who is not presently a participant) may find this discussion via a web-search; they might perform that search at some future time when they are seeing a strange and unidentified purple cloud and seeking information about it (...Imagine, under what circumstances, this Wikipedia archive might come up on some future human's screen - perhaps when a user types "is the purple cloud toxic" into a general-purpose internet search engine - or, speaks to some future voice-activated artificially-intelligent software-agent that was trained on a free database of information - and magically finds the answer on Wikipedia...! Ask yourself, profoundly, whether I am speaking in the hypothetical.) Other readers might have less experience than the person who asked the original question. This is just one motivating reason I write from such a cautious perspective: "assume it's a hazard" is fine advice, and probably more future-proof than a blanket assurance derived from one particular case.

It doesn't hurt to throw in a few "don't play with matches" every now and then - and I never mean to do so as an affront or insult to any particular individual - I only say these things for completeness.

Nimur (talk) 17:24, 5 April 2022 (UTC)

[2] says: "Although it is possible to make pure nitrogen triiodide (which is a red solid) by reacting boron nitride with iodine fluoride, the substance that is actually produced by the reaction I used to perform is more complex. It is an adduct of nitrogen triiodide and ammonia. An adduct is a not-quite compound. The two molecules, nitrogen triiodine and ammonia, are attracted to each other to form a new molecule that is represented by the usual NI3 and NH3 with a dot in between. Initially on formation there are five ammonias to each nitrogen triiodide, but as the substance dries it gives off ammonia to end up with a one to one adduct. What has formed in the solid are chains of nitrogen triiodide with ammonia molecules linking them."

I found the idea of an Adduct interesting. Is this worth covering in the article? 2600:1700:D0A0:21B0:5A9:3B45:A28C:DE50 (talk) 02:03, 5 April 2022 (UTC)

It's just iodine, from personal experience making the stuff without it being held in ammonia. Abductive (reasoning) 09:36, 5 April 2022 (UTC)

The ammonia and iodine gases that are released are toxic. The ammonia air level reported to be considered immediately dangerous to life or health is 300 ppm.^[3] The iodine air level reported to be considered immediately dangerous to life or health is 2 ppm.^[4][5]  --Lambiam 10:01, 5 April 2022 (UTC)

Special relativity : 3.4 Relativity without the second postulate

With only the principle of relativity , what about the CERN collider in which proton beams traveling in opposite directions collide at 99.9999991% of c? If you consider each beam as a frame of reference, for each, the other arrives at a speed close to 2c. And in case each beam is 50% of c you get a Lorentz factor = 1/0, so you can't apply the Lorentz transformation, only the Galilean transformation, right? In fact Einstein's relativity and the Lorentz transformation apply only to electromagnetic rays, not to inertial objects themselves with not invariant speed , but only to the reflection of electromagnetic rays on inertial objects! Didn't Einstein confuse what we see with what is? Malypaet (talk) 23:21, 4 April 2022 (UTC)

Hasn't the constancy of the speed of light been consistently experimentally confirmed?  --Lambiam 09:25, 5 April 2022 (UTC)

It is important not note that speeds do not add the same way as you think they do. Which is to say that if you have two objects moving relative to each other, the velocity of object 2 from the frame of reference of object 1 is not simply to add the velocities. It is approximately that at sufficiently low speeds (which is to say, that WELL within significant figures, you get the same answer whether you use the Newtonian calculation or the Einsteinian one). At speeds that close to the speed of light, you need to use the correct equation to calculate relative speeds, which is the Velocity-addition formula,

$v_{rel}={\sqrt {\frac {(\mathbf {v_{1}} -\mathbf {v_{2}} )^{2}-(\mathbf {v_{1}} \times \mathbf {v_{2}} )^{2}}{1-\mathbf {v_{1}} \cdot \mathbf {v_{2}} }}}.$

where c=1 (a common way to write these equations to simplify the numbers. This formula will simplify to simple addition of velocities for sufficiently small values of V. Your mistake is assuming that two particles colliding will have a combined velocity of 2c. They don't. They have a combined velocity of a very very very small amount less than c, using the Einsteinian velocity addition formula I cited above. Also, we know they do this because we did the experiments many many many times, and the results match the predictions of the theory.--Jayron32 12:14, 5 April 2022 (UTC)

There is an error in this formula. The correct one is

$v_{rel}={\frac {\sqrt {(\mathbf {v_{1}} -\mathbf {v_{2}} )^{2}-(\mathbf {v_{1}} \times \mathbf {v_{2}} )^{2}}}{1-\mathbf {v_{1}} \cdot \mathbf {v_{2}} }}.$

I corrected the error in the article as well. Ruslik_Zero 20:02, 6 April 2022 (UTC)

The Einsteinian calculation is based on the Lorentz transformation, itself based on the constancy of the speed of light, for the displacement of light. It therefore makes no sense to use this calculation for objects with non-constant speed. You explain how the addition of kinetic energies? Malypaet (talk) 12:33, 5 April 2022 (UTC)

I don't care if it doesn't make sense to you. Your ability to understand it doesn't matter here. The math derives naturally from the postulates, and is confirmed by experiment. There are formulations one could do that do not rely on the inability of massed particles from exceeding the constant speed of light, but those formulations are bonkers... It's the equivalent of the silly epicycle nonsense used to maintain a geocentric frame of reference in astronomy. The math works, but putting the Earth in orbit around the sun works better. In the same way, the constant speed of light, and the fact that objects with mass cannot exceed it works better than the alternative. It is the "heliocentric model" for relativity. --Jayron32 13:11, 5 April 2022 (UTC)

You haven't answered my question about adding kinetic energies. At the LHC, a proton at a speed close to c collides with an immobile target with an energy of 7Tev, it collides another proton in the opposite direction with an energy of 14Tev. So if in these two cases we have a relative speed close to c (and not 2c for the collision), how to explain the difference in energy of the impact? Malypaet (talk) 21:04, 5 April 2022 (UTC)

Imagine it getting heavier the closer it is to c so pushes have diminishing returns. Sagittarian Milky Way (talk) 02:42, 6 April 2022 (UTC)

"getting heavier the closer it is to c", but with which formula ? with newton's mechanics it's simple and here the predictions give the right result, the addition of the speeds gives the right value for the energy, right? Malypaet (talk) 12:18, 7 April 2022 (UTC)

See https://en.wikipedia.org/wiki/Mass_in_special_relativity#Relativistic_mass. Modocc (talk) 19:00, 7 April 2022 (UTC)

I do not dispute the constancy of the speed of light, but how one can extrapolate the Lorentz transformations applied to a displacement of a light ray between mirrors and based on the constancy of the speed of light, whatever the referential , therefore to an inertial object whose speed of movement is not invariable, whatever the frame of reference? A mirage in the desert doesn't show reality, does it? As according to Einstein's principle of relativity, there is no absolute frame of reference, with the colliders of Protons, we have very close to 2c for the shock of the protons, moreover it is reported there that during the shock the kinetic energies add up. So what do you have as a more serious explanation? Malypaet (talk) 12:19, 5 April 2022 (UTC)

I don't know what else to do than to refer you again to the article Velocity-addition formula, which does ALL of the math, in full three dimensions, in ALL of the gory details. If you start with the constancy of the speed of light, and that nothing with mass can exceed that speed, then you do the math, this is the results you get. Experiments confirm that math. If you, as an observer standing next to the particle accelerator measure the two particles velocity to each be at 99.9999991% of c, they do NOT collide with a combined speed of twice that value; they collide with a speed of much closer to but not over c. The measured energies from the collision confirm this predicted calculation. I don't know what else to say except that the math is in that article and experiments do confirm it. Einstein started with the postulates of invariant speed of light and that nothing with a mass could reach that speed, and this is the equation that falls out of that math. --Jayron32 12:28, 5 April 2022 (UTC)

All this became apparent 300 years before the formulae were developed to explain it. When the telescope was invented, astronomers knew the exact positions of Jupiter's satellites from occultation observations. Knowing their velocities and direction of travel they checked the observed times of the occultations against the times they would have been seen allowing for the effect the velocities would have had on the time taken for the light to arrive. They were astonished to find that they had no effect whatsoever - the perceived speed was always the same. 79.73.134.90 (talk) 14:17, 5 April 2022 (UTC)

Yes, Rømer's determination of the speed of light is a famous experiment. However, people still held on to old beliefs (people will often shoehorn experimental results that conflict with existing theories in bizarre ways, see the aforementioned epicycles), and it really wasn't until the negative results of the Michelson–Morley experiment that would lead special relativity, completely rewriting how we understand motion. Science is not a singular process, and the growth of scientific knowledge happens gradually, over centuries. After all, while we credit calculus to Newton and Leibniz, the basics of calculus actually date as far back as The Method of Mechanical Theorems by Aristotle, but were largely ignored as mathematical "tricks" until given formal form some two millenia later. Just like no one knew what Aristotle was on to, no one knew what Rømer was on to for centuries. --Jayron32 16:18, 5 April 2022 (UTC)

Far from affecting only electromagnetic rays (e.g. light) and not inertial objects as the OP supposes, Lorentz contraction affects any geometric quantity related to lengths, so from the perspective of a moving observer, areas and volumes will also appear to shrink along the direction of motion. See Lorentz_transformation#Physical_implications. A proton has mass and is in no way immunized from Lorentz transformation just by the OP describing it as "an inertial object whose speed of movement is not invariable, whatever the frame of reference". The double negative "not invariable" confuses me. Philvoids (talk) 17:19, 5 April 2022 (UTC)

When you write "will appear to shrink", it is by the different vision that the radar echo of the light on a volume in the 2 reference frames gives you, isn't it? But if you are blind, the light and its speed no longer affect the size of the volume and yet it still exists. Malypaet (talk) 20:40, 5 April 2022 (UTC)

• OP, the above repetitious appeal to authority is a bit depressing; its kind of like watching a novice chess player being hammered by other more experienced players. On the other hand, cows like myself do prefer chewing on cud, before digesting it. Anyway, you wrote "In fact Einstein's relativity and the Lorentz transformation apply only to electromagnetic rays, not to inertial objects...". That is not correct, reference frame transformations apply to all coordinates of space and time and therefore to all matter waves. With respect to relativity, your velocity addition of 2c that does not preserve c, a Galilean transformation, as you acknowledged, and is thus misapplied and not within the scope of the theory. --Modocc (talk) 18:05, 5 April 2022 (UTC)

  So you criticize my correction of the OP's misconceptions by then... correcting the OP's misconceptions... I'm not sure how to respond to that. I guess just say "I'm sorry, I was wrong" and move on? What modifications would you like me to make to my answers to make them less bad? --Jayron32 18:10, 5 April 2022 (UTC)

  Your corrections are fine. Modocc (talk) 19:02, 7 April 2022 (UTC)

It appears that a primary confusion of the OP is why the (seemingly empirical) constancy that applies to c also affects everything else. Another confusion might be why we so confidently postulate the invariance of c if it's empirically derived. GeorgiaDC (talk) 18:50, 5 April 2022 (UTC)

Very briefly, on both points, c can also be derived from the Maxwell equations and ends up depending only on the properties of vacuum, namely ε₀ and μ₀. Since there's no special treatment for any inertial (constant velocity) observer, the Maxwell equations and properties of vacuum should be the same; hence c should also be the same. Once c is fixed, you can conclude from spacetime diagrams that observers with different velocities end up with different time and space measurements even for the same space-time interval (i.e. the same starting and end events). GeorgiaDC (talk) 19:11, 5 April 2022 (UTC)

In parallel, you don't necessarily need constancy of c to begin with. Here are some derivations (https://iopscience.iop.org/article/10.1088/0143-0807/24/3/312/meta https://aapt.scitation.org/doi/abs/10.1119/1.10490 http://www2.physics.umd.edu/~yakovenk/teaching/Lorentz.pdf) that start off with the properties of spacetime (homogeneity, isotropy, etc) and/or the group property of applying transformations. The result would not be tied to the "speed of light" or speed of anything per se. Anyhow, the transformation function eventually ends up depending on a parameter k (or k²). If it's infinite, we live in a universe with Galilean transformation; if finite, we have Lorentz. GeorgiaDC (talk) 19:23, 5 April 2022 (UTC)

What I understand with your answers, it is that nothing explains how one associates a specific formula with the characteristics of the light for objects not having these characteristics. You always use equations from the characteristics of light (Lorentz transformation) to justify their suitability for other objects. Your explanations look like a self-justifying loop. In colliders, the predictions of Newton's mechanics give the right energy result, even at speeds close to c (or c2?), this would tend to dissociate the applications of the laws, Lorentz transformation for light and Galilean transformation for inertial objects, right? Malypaet (talk) 12:29, 7 April 2022 (UTC)

AFAIK, Newtonian calculations do not give correct values. Also, Lorentz transformations are not used just "for light". They are used because of the speed limit set by light, not as a means of measuring light. Lorentz transformations are necessary because the limit of the speed of light introduce an asymptote into the geometry of objects moving in spacetime. That means relative calculations one would do in Euclidean geometry, things like the pythagorean theorem and the basic trigonometric functions, don't work in the same way; so the Galilean transformations, which use Euclidean math, don't work. You need to use Hyperbolic functions. Lorentz transformations are just what you get when you do hyperbolic functions in 4D Minkowski spacetime. But, and I cannot stress this enough, your misconception is that Lorentz transformations were derived to deal with light alone. They were not. They were invented to deal with the speed limit of light, which is to say that objects with mass can approach but not equal the speed of light. The do depend on the speed of light, because the speed of light defines the asymptote that controls the hyperbolic functions in question, so that's the connection there. --Jayron32 12:40, 7 April 2022 (UTC)

"the limit of the speed of light introduces an asymptote in the geometry of objects moving through spacetime", ok, I understand very well that using light reflection as a measuring instrument to describe this geometry in a space-time, this is the basis of special relativity stated by Einstein in chapter 2. But I insist, because if instead of this reflection, to measure this geometry in space-time it is these objects which directly emit regular bursts of light (chapter 2, burst of light reflected by mirror, replaced by object at speed v, burst of light in A, then B and again in A). In this last exercise of thought, you see that in the different frames of reference, the speed of light does not intervene, the clocks are synchronous, we remain in the transformations of Galileo. Moreover, the speed limit observed in accelerators is not surprising, since the means of propulsion by electric or magnetic fields are components of light, therefore logically with the same limitations, hence the colliders to exceed this limit, in all relativity. There is some logic there, right? Malypaet (talk) 22:10, 7 April 2022 (UTC)

Chapter 2 of what book? Unless we're all reading the same book, we can't possibly know what diagrams you are looking at or what text you are reading? --Jayron32 11:40, 8 April 2022 (UTC)

Translation:"On the Electrodynamics of Moving Bodies" by A Einstein.[6] Malypaet (talk) 21:18, 8 April 2022 (UTC)

Correct: constancy of c applies to light. Incorrect: constancy of c does not affect non-light inertial objects (I assume you mean objects with non-zero rest mass). Briefly, because c is fixed, you would notice that an observer (say B) with a different velocity will have a different definition of simultaneity from your own when manipulating light. Now, if principle of relativity still holds, then as far as B is concerned, they're at rest and you're moving. That means they can't tell, and that whatever definition of simultaneity derived from manipulating light must match the results from manipulating even massive objects! Unless of course principle of relativity is broken, but that is not and should not be the case for multitude of other reasons. The alternative derivations linked above (they might be a bit dense but I do recommend that you try to understand them) do not start from the speed of light in particular. As long as there's some evidence whether transformations are Lorentz xor Galilean, we would know whether k² is finite or infinite respectively. GeorgiaDC (talk) 20:42, 11 April 2022 (UTC)

References