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July 29

Access to reference 9 in Petrichor

Garg, Anu (2007). The Dord, the Diglot, and an Avocado Or Two: The Hidden Lives and Strange Origins of Words. Penguin. p. 399. ISBN 9780452288614.

Do you have access to the page 399? I would like to find out what it says about the two researchers, Isabel Bear and Dick Thomas. (In ruwiki the same source is cited and it is written that Thomas was from UK, whereas the enwiki says they are both from Australia).

Thank you in advance for your help. Gryllida (talk, e-mail) 00:41, 29 July 2024 (UTC)

The ref says, "In 1964, two Australian researchers, I.J. Bear and R. G. Thomas..." but does not provide any further biographical informaion about either one. Our article links to enwiki articles about each of them: Isabel Joy Bear and Richard Grenfell Thomas. Bear's article makes a strong claim for her being Australian, even though she did work for a few years in the UK. Thoman's article does not have any hint of any national connection other than Australia. DMacks (talk) 01:02, 29 July 2024 (UTC)

Thanks, Bear's article says "In the 1950s Bear moved to the United Kingdom, where she worked at the Harwell Science and Innovation Campus. She moved to the University of Birmingham, where she worked as a postdoctoral researcher in the department of metallurgy. Whilst working in Birmingham Bear became interested in solid-state chemistry. Bear joined the Council for Scientific and Industrial Research (CSIRO) in 1953, [...]" -- does it mean she apparently worked in the UK between 1950 and 1953?

The Nature paper was in 1964. Gryllida (talk, e-mail) 04:05, 29 July 2024 (UTC)

Yes, Reference [3] after the first sentence in your quote says explicitly "

In the UK (1950-53)

During three years in the UK she was employed first as an Experimental Scientist in the Metallurgy Division of the Atomic Energy Research Establishment at Harwell, and later as a Research Assistant in the Metallurgy Department of Birmingham University." AlmostReadytoFly (talk) 15:45, 29 July 2024 (UTC)

The fastest Internet speed during rain?

So ChatGPT says the fastest Internet speed during rain is fiber topics, then cabled Internet, then mobile wireless Internet. I just want a 2nd opinion if anyone agrees or disagrees? Specifically, ChatGPT said:

• Most Affected: Mobile wireless internet is the most affected by rainy weather due to signal attenuation.
• Moderately Affected: Cabled internet can be affected if the infrastructure is old or damaged, but it is generally more resilient than mobile wireless internet.
• Least Affected: Fiber optic internet is the least affected by rain due to its well-protected, light-based transmission system.

Thanks. 66.99.15.162 (talk) 19:36, 29 July 2024 (UTC).

Fibre optic cable is always the fastest. The bot is right that wireless technology suffers from rain by signal attenuation, which, provided the protocol is designed to make use of good conditions, can slow down internet speed in rain. Some attenuation isn't too bad, as it reduces interference between nearby cellphone towers, without significantly affecting signal strength on short distances. Once the transmitter reaches maximum power, more rain reduces possible speed.

Wired technology is generally unaffected by rain, unless there's so much rain that it enters the cabinets housing routers etc. or causes landslides, ripping the cables apart. Maybe ChatGPT thinks (to the extend that machines can think) that copper cable networks tend to be older than fibre optic networks and therefore more susceptible to such water intrusion.

Copper networks are typically faster than wireless for the same reason as why speaking tubes are better than providing everybody with a megaphone: the more people shout over the same medium, the more confusion. PiusImpavidus (talk) 09:47, 30 July 2024 (UTC)

The speed of mobile wireless Internet access depends more on the generation of broadband cellular network technology deployed locally (2G, 3G, 4G, 5G) than on the weather conditions. If there is no cellular coverage, the only solution is satellite Internet access, which can stream at a high rate but has a high latency. In all cases (wireless, cable, fiber) the bandwidth may depend on the contract with the provider – often one can opt for a subscription with a higher rate at a higher cost. And in all cases the actual latency and streaming rate may be much lower than the promised one.  --Lambiam 09:57, 30 July 2024 (UTC)

Historic (pre-1800s) Wildfires in California

I read that historically, about 2-4 million hectares would burn a year in California, https://www.propublica.org/article/they-know-how-to-prevent-megafires-why-wont-anybody-listen and https://www.sciencedirect.com/science/article/abs/pii/S0378112707004379. A claim made in the sources was that smoke was a feature of the landscape, rather than an oddity as it is now. If the "extreme" modern season tends to be around 2 million ha with moderate air quality impact, what would be some rough estimates for the average pm2.5 levels across pre-1800s California in late summertime? Takedalullaby (talk) 21:01, 29 July 2024 (UTC)

Sounds like The Burning City by Larry Niven and Jerry Pournelle. They claim that the Los Angeles valley was originally called Iyáangẚ, "the valley of smoke". by the Tongva. Abductive (reasoning) 20:17, 30 July 2024 (UTC)

July 30

Rare quasi-cancer disease in children

Friends recently received a diagnosis for their toddler, and they told me what the doctor had called it, but now I can't remember its name. I know that it has a Wikipedia article, because I read the article when they first told me.

All I remember of the little guy's symptoms is that he frequently has joint pain at unexpected times. The doctor explained to them that the disease causes lesions in random places (including in the bones, if I remember rightly), and because these bulges occur in places where they shouldn't be, some interfere with ordinary movement and cause pain. The disease is treated with chemotherapy, and apparently there's some debate among the experts over whether it should be classified as a kind of cancer. I think the doctor gave a reasonably good prognosis for the disease with treatment and a dreadful prognosis without treatment. I don't remember if the Wikipedia article mentioned if the cause is known, or if it is, what causes it. It's not so rare that the exact number of diagnoses is known, but it's classified as rare (at least here in Australia) because it occurs only once per several thousand individuals. I've looked through Category:Syndromes with musculoskeletal abnormalities without finding it.

While this desk doesn't provide medical advice, remember that I'm not asking for diagnosis: I'm basically starting with a diagnosis and trying to work out the name. This is similar to the strep-infection question from Wikipedia:Kainaw's criterion. Nyttend (talk) 19:27, 30 July 2024 (UTC)

Histiocytosis? Ruslik_Zero 20:39, 30 July 2024 (UTC)

Ah, it's Langerhans cell histiocytosis. Now I remember my confusion when talking with the friends, since initially I thought they meant it was something pancreatic. Thanks for the pointer! Nyttend (talk) 21:35, 30 July 2024 (UTC)

July 31

Is there any simple necessary sufficient condition that conserves kinetic energy, without mentioning kinetic energy?

Note that the absence of external net force is a simple necessary sufficient condition that conserves momentum, without mentioning momentum (just as the absence of external torque is a simple necessary sufficient condition that conserves angular momentum, without mentioning angular momentum).

However, the absence of external net force is not a necessary condition (that conserves kinetic energy), because a given body's kinetic energy can be conserved even when external net force are exerted on that body, e.g. when it's in a circular orbit (in which case the space must have more than one dimension), or when the body elastically collides with a wall sharing a reference frame with an observer who measures the body's kinetic energy (in which case the space is allowed to have a single dimension).

The absence of external net force is not a sufficient condition (that conserves kinetic energy) either, as can be shown when two bodies inelastically collide with each other: The two body system's kinetic energy is not conserved [when the whole system is not seen at rest], although no external net force is exerted on that two body system.

HOTmag (talk) 07:11, 31 July 2024 (UTC)

In a system of two equal bodies circling each other around a common centre of mass, the net momentum of the system is constant, yet there are forces at play.  --Lambiam 09:18, 31 July 2024 (UTC)

The forces you are talking about, are internal ones, each of which is exerted on only a part of the two body system, on which no external force is exerted. Any way, for the sake of clarity, I've just added the word "external" to my first post. HOTmag (talk) 09:37, 31 July 2024 (UTC)

.

.

Resolved

I've just thought about it, assuming that restmass does not change:

The conservation of velocity is a simple necessary sufficient condition that conserves momentum, without mentioning momentum.

The conservation of speed (i.e. of the absolute value of velocity) is a simple necessary sufficient condition that conserves kinetic energy, without mentioning kinetic energy. HOTmag (talk) 10:49, 31 July 2024 (UTC)

You write ‘The absence of external forces is not a sufficient condition ...’ This is debatable. I say that the kinetic energy of the constituents of the system doesn't necessarily contribute to the kinetic energy of the entire system. The energy associated with the movement of the inelastically colliding bodies relative to their common centre of mass is counted as internal energy of the system, not as kinetic energy. Take for example a bottle of warm gas in a circular orbit around Saturn. After a while, the gas cools down. The kinetic energy of the gas molecules decreases, but the kinetic energy of the bottle of gas remains the same; the thermal energy decreases. PiusImpavidus (talk) 18:08, 31 July 2024 (UTC)

Had the absence of external forces been a sufficient condition that conserves kinetic energy, then under that condition - the kinetic energy would've been conserved in all cases - hence in all reference frames and not only when the whole system is seen at rest. Hence, the absence of external forces can't be a sufficient condition that conserves kinetic energy, because when two bodies inelastically collide with each other while the kinetic energy is measured relative to any point that doesn't see the whole system at rest - then the system's kinetic energy does change after the collision - even though no external force is exerted on the system (because the system's momentum does not change). Any way, for the sake of clarity, I've just added this clarification to my paragraph you've quoted. HOTmag (talk) 20:54, 31 July 2024 (UTC)

We have two bodies, inelastically colliding. The kinetic energy of each of the bodies isn't conserved and external forces act on each of them. Namely, the force exerted by the other body. On the other hand, the kinetic energy of the system of two inelastically colliding bodies is conserved, as no external force acts on the system.

Consider the combined kinetic energy of two bodies with masses $m_{1}$ and $m_{2}$ and velocities ${\vec {v}}_{1}$ and ${\vec {v}}_{2}$:

$K={\frac {1}{2}}m_{1}{\vec {v}}_{1}^{2}+{\frac {1}{2}}m_{2}{\vec {v}}_{2}^{2}$

The velocity of the centre of mass is

${\vec {v}}_{cm}={\frac {m_{1}{\vec {v}}_{1}+m_{2}{\vec {v}}_{2}}{m_{1}+m_{2}}}$

Subtracting ${\vec {v}}_{cm}$ from ${\vec {v}}_{1}$ and ${\vec {v}}_{2}$ and rearranging some terms gives the combined kinetic energy of the two bodies in the centre-of-mass frame as

$K_{cm}={\frac {m_{1}m_{2}}{2\left(m_{1}+m_{2}\right)}}\left({\vec {v}}_{1}^{2}+{\vec {v}}_{2}^{2}-2{\vec {v}}_{1}\cdot {\vec {v}}_{2}\right)$

Now suppose we put the two masses into a black box and calculate the kinetic energy of the system in the box:

$K_{s}={\frac {m_{1}+m_{2}}{2}}{\vec {v}}_{cm}^{2}={\frac {1}{2\left(m_{1}+m_{2}\right)}}\left(m_{1}^{2}{\vec {v}}_{1}^{2}+m_{2}^{2}{\vec {v}}_{2}^{2}+2m_{1}m_{2}{\vec {v}}_{1}\cdot {\vec {v}}_{2}\right)$

Now add $K_{s}$ and $K_{cm}$ together and notice how the dot products cancel. Simple calculation shows that

$K=K_{s}+K_{cm}$

That was the maths, now the physics.

The two bodies each have their own kinetic energy. However, when we package the two bodies into a system, the system as a whole has a kinetic energy less than the kinetic energies of the constituent parts. The rest of the kinetic energy of the constituent parts isn't kinetic energy of the system, but internal energy of the system. And this internal energy is equal to the kinetic energy of the constituent parts in the centre-of-mass frame.

So, in your system of two inelastically colliding bodies, the kinetic energy of the bodies is indeed not conserved, but there's an external force acting on each of them. The kinetic energy of the system of two bodies (on which no external force acts) is conserved; it's the internal energy that decreases. PiusImpavidus (talk) 12:14, 1 August 2024 (UTC)

I agree to all of your maths, as well as to all of your claims about physics, except two sentences: "the kinetic energy of the system of two inelastically colliding bodies is conserved, as no external force acts on the system....The kinetic energy of the system of two bodies (on which no external force acts) is conserved".

The correct fact that "no external force acts on the system", is not sufficient for justifying your claim that the system's kinetic energy is conserved. As I've already pointed out in my previous response, the system's kinetic energy is only conserved when the whole system is seen at rest. For more details, see our article inelastic collision: "inelastic collisions do not conserve kinetic energy" [i.e. not in all reference frames].

Btw, for simplicity, let's assume that ${\vec {v}}={\vec {v}}_{1}=-{\vec {v}}_{2}$ and that $m=m_{1}=m_{2}$ (hence $K=K_{cm},$ and $K_{s}=0).$ HOTmag (talk) 14:23, 1 August 2024 (UTC)

The kinetic energy of the system $K_{s}$ only depends on the total mass of the system and $v_{cm}$. In the absence of external forces, neither of those change, so $K_{s}$ is conserved, in every reference frame. What the bodies do to each other is irrelevant. $K_{cm}$ and $K$, the difference between which is constant, are not conserved in an inelastic collision and the loss of energy is the same in every reference frame, including the centre-of-mass frame.

The nice thing I tried to show you is that the sum of the kinetic energies of the bodies can be broken into two parts, being the kinetic energy of the system of two bodies and the internal energy of the system, which can simply be added together. When you calculate the kinetic energy of a rock, you don't include the kinetic energies of all its vibrating atoms, right? Because that's the thermal energy of the rock. It's exactly the same here. PiusImpavidus (talk) 08:07, 2 August 2024 (UTC)

Thank you ever so much for your clarifications. So, I'm striking out the wrong thing in my first post (See above). HOTmag (talk) 11:12, 2 August 2024 (UTC)

Consider again a system of two equal bodies orbiting around a common centre of mass. Choose the coordinate system such that the common centre of mass of the two-body system is at rest. If one body's momentum equals $\mathbf {p}$ at some instant of time, that of the other at the same instant of time equals $-\mathbf {p} ,$ so the momentum of the system is $\mathbf {0} .$ Now apply external forces rotation-symmetrically to the objects so as to reverse their motion, making them circle again around their common centre of mass – which has not budged – but in the opposite sense. The symmetry guarantees that the momentum of the system remains $\mathbf {0}$ at all times, so the absence of external forces is not a necessary condition for conserving momentum.  --Lambiam 20:06, 31 July 2024 (UTC)

Your case does not involve an external force but rather involves an external torque. Any way, for the sake of clarity, I've just added this clarification to the first paragraph of my first post. HOTmag (talk) 20:54, 31 July 2024 (UTC)

...but equal and opposite applied forces like compression conserves momentum too. In other words, absence of external forces is a sufficient condition but it is not necessary. Modocc (talk) 21:35, 31 July 2024 (UTC)

By "equal", I guess you mean "having the same absolute value".

Anyway, when the external forces are equal [in their absolute value] and opposite, then the sum of those external forces is zero. Hence, saying that the external forces are equal [in their absolute value] and opposite, is like saying that there are no external forces. Hence, if you want to prove that the absence of external forces is not a necessary condition that conserves momentum, you will have to give an example in which the sum of the external forces is not zero. HOTmag (talk) 06:47, 1 August 2024 (UTC)

The fact that external forces can sum to zero does not imply they are absent... For example, without material internal resisting forces submarines and neutron stars implode. Modocc (talk) 08:18, 1 August 2024 (UTC)

All depends on what we mean by "absence of external forces". By "absence of external forces" I intend to include also all cases of external forces summing to zero. HOTmag (talk) 08:41, 1 August 2024 (UTC)

If you mean "absence of external net force" then why not write that instead? Also if F=0 then dP/dt=0 per Newton's second law, see Force. Modocc (talk) 12:09, 1 August 2024 (UTC)

Re. your first sentence: Ok, I'm adopting your current suggestion (See above in my first post).

Re. your second sentence: Of course, but what was wrong in my saying (ibid.) that "the absence of external [net] force is a simple necessary sufficient condition that conserves momentum, without mentioning momentum"? HOTmag (talk) 14:23, 1 August 2024 (UTC)

Nothing wrong, so I deleted the second part. Then in hindsight I deleted the first part because in many contexts it's understood, but not always. Then you restored the entire comment. Sigh. Modocc (talk) 15:02, 1 August 2024 (UTC)

Probably I had started responding to your response before you deleted it? Anyway, surprisingly, I didn't get any warning of "edit conflict" when I responded to your deleted response. HOTmag (talk) 15:56, 1 August 2024 (UTC)

Also, it is best practice to strike the original phrase(s) when modifying them so Lambiam's and my comments retain context. Modocc (talk) 15:54, 1 August 2024 (UTC)

Agree. Next time... HOTmag (talk) 15:57, 1 August 2024 (UTC)

${\vec {F}}\cdot {\vec {v}}=0$ PiusImpavidus (talk) 12:50, 1 August 2024 (UTC)

In my first post I gave two counter-examples:

1. On the one hand, when the body elastically collides with a wall sharing a reference frame with an observer who measures the body's kinetic energy, then the body's kinetic energy is conserved even though this case does not satisfy your condition. Hence it's not a necessary condition.

2. On the other hand, when two bodies inelastically collide with each other, the two body system's kinetic energy is not conserved [when the whole system is not seen at rest], even though this case does satisfy your condition. Hence it's not a sufficient condition. HOTmag (talk) 14:23, 1 August 2024 (UTC)

Both counter-examples fail:

1. While bouncing, ${\vec {v}}$ changes such that it is on average either 0 or parallel to the wall, whilst ${\vec {F}}$, the normal force, is perpendicular to the wall. Therefore, their dot product is zero, so the condition is satisfied.
2. I explained that one above. You have to make a distinction between the sum of the kinetic energies of the two bodies and the kinetic energy of the system of the two bodies. The former decreases in every reference frame, but the latter is conserved in every reference frame, so this doesn't prove that the condition is insufficient.

PiusImpavidus (talk) 08:39, 2 August 2024 (UTC)

So by ${\vec {v}}$ you meant the average velocity. I couldn't understand it before you made it clear. Instead of you, I would write: ${\vec {F}}\cdot ({\vec {v}}_{1}+{\vec {v}}_{2})=0.$

Anyways, your necessary sufficient condition could also be expressed verbally: "a zero external net force - or a zero average velocity - in every coordinate", right? HOTmag (talk) 11:12, 2 August 2024 (UTC)

By ${\vec {v}}$ I mean the instantaneous velocy, so the ${\vec {F}}\cdot {\vec {v}}$ is the instantaneous rate of change of the kinetic energy. If that's 0, the kinetic energy is always conserved.

The thing is, an instantaneous elastic collision is a strange thing. The force is infinite and acts during an infinitesimal time interval, the velocity is a step function changing exactly when the force in infinite. If we multiply the two, the result is zero except at the moment of collision, where it's undefined. So the maths don't work. And of course, no real collision is instantaneous. At the midpoint of any real collision, the kinetic energy (except the part associated with the motion parallel to the wall) has been converted to elastic energy, after the collision it has turned back into kinetic energy. So during a non-instantaneous elastic collision (BTW, no real collision can be fully elastic), kinetic energy isn't conserved at all times, but it is restored afterwards.

If you don't want to look into the details of the collision, you can say that it must be symmetrical, so that $\int {\vec {F}}\cdot {\vec {v}}\,dt=0$. If this integral isn't zero, the collision can't have been elastic. PiusImpavidus (talk) 08:15, 4 August 2024 (UTC)

1. Ok, so during the so-called "an elastic collision between a wall and a body", while the wall is exerting a non-zero external force on the body, the body's kinetic energy is not conserved, because it's converted to elastic energy. Agreed. But why does your condition have to mention the value V of the velocity, rather than only saying that "the kinetic energy is conserved if and only if no external force is exerted on the body along the axis of the body's motion", without having to mention the value V of the velocity? I guess that's because you want to include also the cases of so-called "elastic collisions" in which - the average external force is not zero (because the final momentum does not equal the initial momentum) - but the final kinetic energy does equal the initial kinetic energy, so the condition ${\vec {F}}=0$ (where F denotes the average external force exerted on the body along the axis of the body's motion) wouldn't be a sufficient condition, and that's why you had to strengthen it as: ${\vec {F}}\cdot {\vec {v}}=0$ where both F and V are average values, am I right?

2. I asked: "Is there any simple necessary sufficient condition that conserves kinetic energy, without mentioning kinetic energy?"

By asking that, I meant the following:

Is there any simple necessary sufficient condition, satisfying the following:

If a given system, carried (at the beginning of the process) an initial mass $m_{1}$ and an initial velocity ${\vec {v}}_{1}$ and an initial kinetic energy $E_{K1},$ and is carrying (right now) a final mass $m_{2}$ and a final velocity ${\vec {v}}_{2}$ and a final kinetic energy $E_{K2},$ then $E_{K1}=E_{K2}$ if and only if the condition holds.

Then I added, that the condition was not allowed to mention the kinetic energy.

Please note:

a). The kinetic energy can be defined as $\int {\vec {F}}\cdot d{\vec {s}}$ (when it's defined as zero if the momentum is zero), when ${\vec {F}}$ denotes the force and $d{\vec {s}}$ denotes the infinitesimal displacement, so I don't allow the condition to mention (at once) both ${\vec {F}}$ and $d{\vec {s}}$ (sorry for not making it clear before). Nor do I allow the condition to mention any set of properties (e.g. ${\vec {F}},{\vec {v}},dt$) that determines (at once) both ${\vec {F}}$ and $d{\vec {s}}$.

b). The kinetic energy is also determined once we are given any pair of the following three properties: mass-velocity-momentum, so I don't allow the condition to mention any such a pair either (sorry for not making it clear before). Nor do I allow the condition to mention any set of properties that determines any such a pair.

c). Your condition ${\vec {F}}=0$ is not necessary. Check: $m_{1}={\vec {v}}_{1}=m_{2}=1,{\vec {v}}_{2}=-1.$

d). Your condition ${\vec {F}}=0$ is not sufficient. Check: $m_{1}={\vec {v}}_{2}=1,m_{2}={\vec {v}}_{1}=2.$

HOTmag (talk) 19:28, 5 August 2024 (UTC)

Is there any simple necessary sufficient condition, that conserves a given system's total energy, without mentioning energy/mass?

Just as absence of external force is a simple necessary sufficient condition that conserves momentum, without mentioning momentum. HOTmag (talk) 10:57, 31 July 2024 (UTC)

The net work done by the system is equal to the net heat received. PiusImpavidus (talk) 17:45, 31 July 2024 (UTC)

It's not a necessary condition: Consider a harmonic oscillator, in which the total energy is conserved, even though the (changing) net work done by the system is not equal to the (zero) net heat received. HOTmag (talk) 21:09, 31 July 2024 (UTC)

What work? An isolated harmonic oscillator does no work on its surroundings. The parts forming the isolated harmonic oscillator do work on each other and their energies oscillate, but the whole system does no work and has constant energy.

Like in the above discussion, you don't keep proper track of what's in your system and what's not. PiusImpavidus (talk) 12:25, 1 August 2024 (UTC)

Sorry for not clarifying myself. Anyway, I meant the following:

When a gravity pendulum is influenced by gravitation, the pendulum's total energy is conserved, even though the (changing) net work done by the gravitation on the pendulum is not equal to the (zero) net heat received by the pendulum's movement. HOTmag (talk) 14:23, 1 August 2024 (UTC)

There are two ways to look at this:

1. When the bob of the pendulum climbs, its kinetic energy is converted into potential energy, but still present in the bob. The total energy of the bob doesn't change and no work is done. When the bob descends, the potential energy is converted back into kinetic energy, but again no work is done. This is mathematically the easy way, but not entirely correct.
2. When the bob of the pendulum climbs, it performs work on the gravitational field. The bob loses energy, the field gains energy. When the bob descends, the field performs work on the bob. The field looses energy, the bob gains it. The energy of the bob isn't conserved, but an isolated bob without a gravitational field is no harmonic oscillator. The system of bob+string+gravitational field is a harmonic oscillator and energy in that is conserved. Unfortunately, this approach is mathematically very hard.

In physics, we store energy in fields all the time. PiusImpavidus (talk) 08:22, 2 August 2024 (UTC)

no work is done. What? Don't you agree that the work is equivalent to the change in kinetic energy? HOTmag (talk) 11:18, 2 August 2024 (UTC)

In physics, Work is the product of a force and a lasting, permanent displacement. Philvoids (talk) 18:43, 2 August 2024 (UTC) Underlining added for clarity.

If the force is denoted by ${\vec {F}}$ and the displacement is denoted by ${\vec {s}},$ then the work is defined as $W=\int {\vec {F}}\cdot d{\vec {s}},$ and it's equivalent to the change in the kinetic energy, isn't it? HOTmag (talk) 21:02, 3 August 2024 (UTC)

As I stated, it's not entirely correct, but it's mathematically the easy way. PiusImpavidus (talk) 08:17, 4 August 2024 (UTC)

The more correct alternative is stating that gravity does do work on the bob (approach 2), but then the bob can't have potential energy. PiusImpavidus (talk) 08:24, 4 August 2024 (UTC)

1. I'd asked: "Is there any simple necessary sufficient condition, that conserves a given system's total energy, without mentioning energy/mass?"

2. You'd answered: "The net work done by the system is equal to the net heat received".

I assumed your answer was entirely correct, so I presented a counter example, with a pendulum's bob being the system I'd asked about:

3. When the bob reaches its lowest point, the bob carries a positive kinetic energy. Agree?

4. When the bob reaches its highest point, the bob carries no kinetic energy. Agree?

5. The heat doesn't change. Agree?

6. Hence, the (changing) net work done by the gravitation on the bob is not equal to the (zero) net heat received by the bob's movement. Agree?

7. The bob's total kinetic energy is conserved. Agree?

8. There's a contradiction between: the combination of 1,2, and the combination of 6,7. Agree?

For every statement of the above eight, please indicate if you agree to it. HOTmag (talk) 19:13, 5 August 2024 (UTC)

To be pedantic, I'll assume the point of view where no gravitational potential energy is stored in the bob; it will be stored in the gravitational field instead.

1. Yes, that's what you asked.
2. Yes, that's what I answered.
3. Indeed, at the lowest point the bob has positive kinetic energy.
4. Indeed, at the highest point the bob has no kinetic energy. The energy has been stored in the potential energy of the gravitational field.
5. Heat doesn't change; it flows. But in this case, the heat flow is zero, so I agree.
6. Indeed, during each quarter swing (left–centre–right–centre–left), either positive or negative work is done by the gravitational field on the bob, but no heat ever flows.
7. No, the kinetic energy of the bob isn't conserved. It goes up and down all the time. The total energy of the oscillator (kinetic energy of the bob plus potential energy of the gravitational field) is conserved.
8. No, there's no contradiction. The answer at 2 was about the total energy of the oscillator, the statement at 7 is about the kinetic energy of the bob.

You forgot that the gravitational field is part of the oscillator too. It's as important as the bob. PiusImpavidus (talk) 08:45, 6 August 2024 (UTC)

Oh!!! sorry for the big mistake I made in question #7, when I drafted it, adding the redundant word "kinetic". I've just struck it out. Please answer again #7 and #8, and please explain to me what I still didn't understand: when the bob gets to its highest point, doesn't the bob carry a potential energy? If it doesn't, in your opinion, then could you back it with any authoritative source? From Wikipedia if possible. Thank you for your patience! HOTmag (talk) 12:26, 6 August 2024 (UTC)

There are two ways to view this. The lazy way says that the gravitational potential energy is stored in the bob, which is converted to kinetic energy when moving down. Many beginner's books tell this. But if gravity performs work on the bob, this point of view violates conservation of energy. One solution is to ignore work done by gravity, which is easy and what most physicists do most of the time. The other, and proper, way to look at this, explained above under 2, is that the potential energy is stored in the gravitational field, not in any object. Unfortunately, the maths get really hard if you try it that way. As you wondered about the work done by gravity, that's the point of view I'll use here – but I'll skip the maths. So:

7. The bob's energy isn't conserved. The energy moves back and forth between the kinetic energy of the bob and the potential energy of the gravitational field. The total energy of the oscillator, the sum of these, is conserved.

8. No contradiction. Field and bob do work on each other, but are both part of the system. The system performs no work on the outside world. The energy of the system is conserved. The energy of the bob isn't.

It's just a matter of proper accounting. PiusImpavidus (talk) 20:01, 6 August 2024 (UTC)

I'm pretty surprised to read now, that the attitude I've always been taught about, is only what "Many beginner's books tell".

Do you think Wikipedia may add some info about the new description you support?

Additionally and more important: I still don't understand what's wrong in the attitude you describe as the one presented in the beginner's books. You say that "if gravity performs work on the bob, this point of view violates conservation of energy". So first, please notice that also the new description you support violates conservation of energy, because when the bob gets to the highest point the bob has lost the whole kinetic energy the bob carried when it was at the lowest point. Second, according to the attitude you describe as the one presented in the beginner's books, gravity is not committed to the conservation of kinetic energy, but rather is only committed to the sum of the kinetic energy and potential energy, and this sum is always conserved under gravity, even when the potential energy is attributed to the bob rather than to the gravitational field. HOTmag (talk) 07:40, 7 August 2024 (UTC)

Many more advanced books don't mention the subject at all.

Suppose an elevated object caries gravitational potential energy. When we let it fall down, it looses this potential energy and gains kinetic energy. So the source of the kinetic energy was already in this object; no energy is transferred to the object when it falls. However, gravity provides a force pulling the object down en provides work to accelerate the object. So the energy is transferred to the object in the form of work by gravity and can't have been in the object beforehand. This is a contradiction. There are just two ways to solve it: either gravity does no work or the potential energy was stored somewhere else than in the object. As you asked about the work done by gravity, I was forced to take the latter approach.

First point: when the bob reaches its highest point, it has lost all its kinetic energy. Gravity has done negative work on the bob and the energy has been stored in the gravitational field. So energy has been conserved.

Second point: The sum of potential energy and kinetic energy is conserved, provided no work is performed on the system. But work is performed on the bob, by gravity. So if our system is just the bob, the potential energy can't be stored in the system, so it's not in the bob.

I never really thought about it for a long time after reading my first books on physics a long time ago, but it's actually pretty obvious. PiusImpavidus (talk) 09:41, 8 August 2024 (UTC)

Suppose an elevated object caries gravitational potential energy. When we let it fall down, it looses this potential energy and gains kinetic energy. So the source of the kinetic energy was already in this object.

What? According to the attitude you described as the one presented in the beginner's books, the source of the kinetic energy was not in the object, because every kinetic energy added to any object is given to the object by a force, being the gravity in our case. This is supposed to be obvious in classical mechanics, which really considers gravity to be a "force" .

no energy is transferred to the object when it falls.

I agree, if by "no energy" you mean zero total energy, because in classical mechanics gravity is a conservative force, which actually conserves the sum of potential energy and kinetic energy. However, gravity does not conserve the kinetic energy, and a kinetic energy is transferred to the object when it falls.

First point. When the bob reaches its highest point, it has lost all its kinetic energy. Gravity has done negative work on the bob and the energy has been stored in the gravitational field. So energy has been conserved.

Correct. I apologize for my first point. I confused kinetic energy - which is actually not conserved, with total energy (being the sum of potential energy and kinetic energy in our case) - which is actually conserved.

Second point: The sum of potential energy and kinetic energy is conserved, provided no work is performed on the system.

What? According to the attitude you described as the one presented in the beginner's books, the sum of potential energy and kinetic energy is always conserved under any conservative force, while gravity is really considered to be a conservative force - in classical mechanics, regardless of the work done by that conservative force. According to this attitude, "Work" is equivalent to the change in kinetic energy if there is no change in the potential energy, and is also equivalent to the change in potentia energy if there is no change in the kinetic energy. HOTmag (talk) 11:51, 8 August 2024 (UTC)

First two paragraphs: yes, there's an inconsistency. That's what I was trying to explain. I'll consider it settled now.

Last paragraph: yes, gravity is a conservative force, so the sum of potential and kinetic energy is conserved. I was only acknowledging the possibility that other forces may be present. PiusImpavidus (talk) 16:05, 9 August 2024 (UTC)

I don't see any inconsistency. The attitude you described as the one presented in the beginner's books, does not agree to your claim that "the source of the kinetic energy was already in the object". Where is the inconsistency?

Nor does this attitude agree to your addition "provided no work is performed on the system", so again, where is the inconsistency? HOTmag (talk) 16:29, 9 August 2024 (UTC)

Physicists carefully define Energy a measurable property that is transferred in interactions and Mass the intrinsic quantity of matter in a body because these are fundamental to both our study and understanding of many Systems i.e. functional groupings of elements that act according to a set of rules. Posing yet another challenge to define a physical system "without mentioning those words" is just inviting responders into a handicapped word play that may provide the engaging debate that the OP is seeking to expose their own ideas but that is a misuse of the reference desk that should not continue. Underlining added for clarity. Philvoids (talk) 10:03, 2 August 2024 (UTC)

I didn't ask about defining something (without using specific words). I only asked about a necessary sufficient condition for the conservation of (total) energy (without mentioning energy/mass), which is a definitely different thing!

Just as the absence of external net force is a necessary sufficient condition for the conservation of momentum, without mentioning momentum. This is a definitely legitimate question. HOTmag (talk) 11:27, 2 August 2024 (UTC)

August 1

Imane Khelif

I'm posting this question in the most neutral and, I hope, respectful way possible. I do not intend to offend anyone or start a fight. What I'd like to know is the actual biological situation of Imane Khelif. Our article is not very clear on the subject. I understand that it is a delicate topic so some of the details may not be public. Does she have X-Y, X-X or some pathological variation of chromosomes? Does she have male or female genitalia? Does she have a specific medical condition (I am aware of the concept of intersex but my understanding is that it could be the outcome of different medical conditions)? Thank you! 79.35.53.87 (talk) 14:30, 1 August 2024 (UTC)

It appears that no details have been shared officially with the public. The IBA has declined to divulge any specifics, but its chairman declared on March 23 to TASS that, based on the results of DNA tests, "it was proved they had XY chromosomes". That is all we know.^[1]  --Lambiam 15:50, 1 August 2024 (UTC)

I believe there are over 50 different 'situations' that fall under the umbrella title of Intersex. We have no access to evidence suggesting which, if any, of these Imane Khelif may fall under, but the OP (and others) might find the article of overall interest.

When categorization for sporting purposes is (some might say as a matter of practicality) binary, but the human species is not, quandries are bound to arise, alas. {The poster formerly known as 87.81.230.195} 94.1.169.77 (talk) 01:16, 2 August 2024 (UTC)

One may also consider the gender of Schrödinger´s Cat. --2001:871:6A:1B71:292B:D775:91F7:34E5 (talk) 16:49, 3 August 2024 (UTC)

Schrödinger´s Cat's gender is not regarded as an unresolved quantum Wave function so how shall that help? Philvoids (talk) 22:12, 3 August 2024 (UTC)

August 3

Heat capacity, sand battery, check calculation

Could someone please check my work? Sand has a specific heat capacity of around 800 kJ/kg·K. I calculate that if you put 10 000 kWh (10 MWh) into 900 kg (0.6 cubic metres) of it, you raise its temperature by 50 °C. This video is talking about a massive ~40 cubic metre sand battery but it's capacity is only ~10 MWh? Either I'm wrong or the video is wrong, right? https://www.youtube.com/watch?v=KVqHYNE2QwE&t=377s 80.46.251.32 (talk) 01:05, 3 August 2024 (UTC)

According to the article Thermal energy storage, there is a prototype 8 MWh sand battery, built in Finland in 2022. The source given on this system says it is a "steel container, which is 4 m wide and 7 m high, [and it] is filled with 100 tonnes of builder’s sand", and is heated to 500 °C. There is a picture of the silo. So it seems that your calculation must be wrong. Abductive (reasoning) 06:09, 3 August 2024 (UTC)

The specific heat capacity of quartz sand is more like 800 J/kg·K, lower by a factor of 1000. The specific heat capacity is temperature-dependent. At 500 °C it is substantially higher, about 1230 J/kg·K.^[2]  --Lambiam 10:47, 3 August 2024 (UTC)

August 4

density of solid tritium

Has anyone ever actually measured this? I've only seen the density of the liquid discussed in the literature. Double sharp (talk) 07:08, 4 August 2024 (UTC)

see https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4922399/ where they measured the molar volume, effect of pressure and temperature on volume. Also see https://link.springer.com/article/10.1007/BF00683620 . Graeme Bartlett (talk) 11:01, 4 August 2024 (UTC)

Interferometer116.68.77.174 (talk) 17:48, 4 August 2024 (UTC)

My company got a new project - milk analyzer using FTIR spectroscopy. Now we are conducting a research about interferometer for this milk analyzer and got informations like beamsplitter - [material : KBr or other material that transmit and refract IR beam of 2.5 to 20um wavelength, 50/50 beam splitting coating, anti-reflective coating in mid IR region flatness-1 um for 10um lambda, smoothness-RMS roughness of less than 10nm, thickness- 1 to 5mm]; lens - [plano convex lens that should effectively transmit 2.5 to 20um range wavelength effectively]; Compensating Plate : [Material : Same as beamsplitter, Coating: anti-reflective in the mid-IR range, Anti-reflective coating : Effective in mid IR range, Flatness : 1um for 10um lambda, Smoothness : RMS roughness of less that 10nm, Thickness : match the optical path length difference introduced by your beam splitter]; Mirror: [Reflectance max from 2µm to 20µm, Flatness : 1um for 10um lambda]. Should I proceed with finding optics with these specifications? Do the provided details suffice to initiate the process? And also 2 mirrors are used in this interferometer. One is fixed and other is have a linear motion. When I asked the displacement range of moving mirror in chatgpt 1st time it replied as 0.625mm to 1.25mm. I again asked this question and it replied as 'in a Michelson interferometer, the moving mirror displacement is often around a few millimeters to achieve the desired spectral resolution. Specifically, a displacement range of 1 to 5 millimeters is common, though some high-resolution instruments may require larger displacements.'. I am confused. What is your opinion about this? 116.68.77.174 (talk) 17:48, 4 August 2024 (UTC)

My opinion is that it is unsafe to rely on ChatGPT for any project-critical information. It can however be a useful source of links that you will have to trace and verify. In Wikipedia find Michelson interferometer and FTIR (Fourier-transform infrared spectroscopy). External links that I think may help are: Fourier Transform Infrared Spectroscopy as a Tool to Study Milk Composition, Standardization of milk mid-infrared spectra from a European dairy network, Visible and near-infrared spectroscopic analysis of raw milk for cow health monitoring: Reflectance or transmittance? and Human Milk Analysis Using Mid-Infrared Spectroscopy. Philvoids (talk) 23:32, 4 August 2024 (UTC)

IMO one cannot rely on ChatGPT for any factual information, critical or not. It may answer any factual question in an authoritative tone with something that sounds plausible but has no relation to reality.  --Lambiam 23:39, 5 August 2024 (UTC)

It sounds like ChatGPT is ready to run for public office. ←Baseball Bugs ^{What's up, Doc?} carrots→ 23:55, 5 August 2024 (UTC)

Villar A, Gorritxategi E, Aranzabe E, Fernández S, Otaduy D, Fernández LA (December 2012). "Low-cost visible-near infrared sensor for on-line monitoring of fat and fatty acids content during the manufacturing process of the milk". Food Chemistry. 135 (4): 2756–60. doi:10.1016/j.foodchem.2012.07.074. PMID 22980869. Philvoids (talk) 13:53, 6 August 2024 (UTC)

August 6

Is there any natural process (i.e. without human intervention), conserving the kinetic energy, without conserving the speed (i.e. the absolute value of velocity)?

By conserving the KE without conserving the speed, I mean the following:

Let a system, carry (at the initial moment) an initial mass $m_{1}$ and an initial velocity ${\vec {v}}_{1}$ and an initial kinetic energy $E_{K1},$ and carry (at the final moment) a final mass $m_{2}$ and a final velocity ${\vec {v}}_{2}$ and a final kinetic energy $E_{K2}.$

Is it natural (i.e. without human contact), that $E_{K1}=E_{K2},$ but $|{\vec {v}}_{1}|\neq |{\vec {v}}_{2}|$ [for every reference frame]? while the reference frame remains the same - during the whole process?

Mathematically, it's possible of course, but I wonder if it's also physically possible in any natural process.

Note, that I'm only asking about a change in speed, rather than about a change in velocity, because no question would arise, had I replaced speed by velocity, in which case there would be natural processes conserving the KE without conserving the velocity, e.g. when the system is any given body that elastically collides with a wall sharing a reference frame with an observer who measures the body's kinetic energy. HOTmag (talk) 07:10, 6 August 2024 (UTC)

This can only happen when the mass changes inversely proportional to the square of the speed. It can be done naturally through clever accounting. Suppose a rock changes speed and breaks into four equal parts; a natural process. We can choose to define our system such that three of the parts of the rock leave the system, reducing the mass to one quarter. We can also choose a frame of reference in which the speed of the remaining fragment doubles compared to the speed of the original rock. Then your condition is met. Obviously, this has no physical relevance. PiusImpavidus (talk) 09:04, 6 August 2024 (UTC)

Oh, I forgot to add a crucial condition: the reference frame must remain the same, during the whole process. Thanks to your exmaple, I've just added this crucial condition I'd forgotten (See above). Sorry for the confusion. HOTmag (talk) 12:35, 6 August 2024 (UTC)

I don't change the reference frame during the process; I just conveniently choose it beforehand. If both speed and mass change, I can always find a reference frame where the kinetic energy is conserved and I can also find one where it isn't. And as the physics cannot depend on the chosen reference frame (principle of relativity), this conservation of kinetic energy can't have physical relevance. PiusImpavidus (talk) 20:24, 6 August 2024 (UTC)

Ok, my question wants this process to be independent of the reference frame. I've just added this new condition [in brackets] to my original post. Again, I apologize for the confusion.

Let's put my original question this way: Some natural forces (e.g. the elastic force exerted by a spring), conserve the kinetic energy for the long term (with respect to the "initial moment" defined as such), for every reference frame. Is there any combination of natural forces - that conserves the kinetic energy (in the above sense), but at some moments - at which the kinetic energy turns out to have been conserved (with respect to the "initial moment" defined as such) under that force for every reference frame - the speed (i.e. the absolute value of velocity) does not? HOTmag (talk) 07:43, 7 August 2024 (UTC)

I take the rest frame of the system before the process happens. The speed of the system is 0, and so is the kinetic energy. Now the process happens and the speed changes. My reference frame is no longer the rest frame of the system, so the kinetic energy can no longer be 0. Which proves that I can always find a reference frame in which your condition is violated. This means that there is no process that satisfies your condition in every reference frame. PiusImpavidus (talk) 09:58, 8 August 2024 (UTC)

Your consideration can also be used to prove, that if the kinetic energy is conserved for every reference frame, then not only the speed - but also the velocity - is conserved for every reference frame, right? HOTmag (talk) 11:30, 8 August 2024 (UTC)

Yes, if the kinetic energy is conserved in every reference frame, the velocity must be conserved too. The mass too. In other words, not much can have happened. (Note: I've assumed we're dealing with inertial reference frames. I think you assumed that too.) PiusImpavidus (talk) 15:54, 9 August 2024 (UTC)

Thank you. HOTmag (talk) 17:34, 10 August 2024 (UTC)

Metastasis of glioblastoma

From Glioblastoma#Surgery:

GBM cells are widely infiltrative through the brain at diagnosis, and despite a "total resection" of all obvious tumor, most people with GBM later develop recurrent tumors either near the original site or at more distant locations within the brain.

Why is this cancer so widely infiltrative? Nyttend (talk) 22:03, 6 August 2024 (UTC)

The glial cells thought to be involved are astrocytes, which are like neurons in many ways except they don't carry electrical signals. They reach out along the same pathways as neurons and if they become cancerous are predisposed to reach into distant areas. Here is a review article. Combine that with the fact that the somatic immune system is barred from the brain, and they cannot be stopped or contained. Abductive (reasoning) 23:02, 6 August 2024 (UTC)

Abductive, could you add something to the article explaining this? Proper understanding of the review article demands a higher-than-I-possess understanding of human biology. Even the bits that I can understand are hard to interpret in context, e.g. I understand "Glioblastoma cells generally invade as single cells", but I don't know if it's at all relevant to the "why" question, and I don't want to go dumping content into the article and accidentally cause it to imply something not in the source. Nyttend (talk) 19:45, 7 August 2024 (UTC)

I gave it a shot. Neurobio is the most difficult biology field, along with immunobio. Abductive (reasoning) 20:15, 7 August 2024 (UTC)

It may not be especially infiltrative as compared to other tumors. The problem is that in case of brain the resection of tumors with sufficiently wide margins is impossible for obvious reasons. Ruslik_Zero 19:16, 7 August 2024 (UTC)

You might consider reading the review article I linked above. Abductive (reasoning) 19:47, 7 August 2024 (UTC)

Sound effects for Jeopardy dollar amounts heard on 1960s phones

I have this memory of using a telephone in the early 1960s and if I couldn't hear anything else, I could hear a sound (very faint) similar to the old sound effects of dollar amounts appearing on Jeopardy!. a sound effect used in the opening of Jeopardy! Masters. Any idea what I was hearing?— Vchimpanzee • talk • contributions • 22:10, 6 August 2024 (UTC)

• Old relays. Here is a Youtube video with the sound amplified. Abductive (reasoning) 23:08, 6 August 2024 (UTC)

That doesn't seem to be it. It was much more of an electronic or robotic sound. Maybe it was a change in the technology from the old relays to something new.— Vchimpanzee • talk • contributions • 15:55, 8 August 2024 (UTC)

I recall those rapid electronic sound effects when Jeopardy revealed the dollar amounts (here is a Youtube clip [3] of them)... and on the telephone system they were called Touch-Tones. From Push-button telephone: "In 1963, the Bell System introduced to the public dual-tone multi-frequency (DTMF) technology under the name Touch-Tone, which was a trademark in the U.S. until 1984. The Touch-Tone system used push-button telephones. In the decades after 1963, rotary dials were gradually phased out on new telephone models in favor of keypads and the primary dialing method to the central office became touchtone dialing." In fact, DTMF touch-tones are still in use with landlines, but have become less noticeable with speed-dialing. In addition, in the U.S.landline phone use is way down, with 73% of households only having cell phones. [4] Modocc (talk) 17:55, 8 August 2024 (UTC)

I'm guessing hat I was hearing was what replaced relays on the newer systems that could handle touch tone telephones.— Vchimpanzee • talk • contributions • 21:21, 8 August 2024 (UTC)

Most likely, although in those early days I rarely used the telephone and when I did it was only to answer an incoming call. :-) Modocc (talk) 23:05, 8 August 2024 (UTC)

August 7

Density of Iridium and Osmium

Note A at Iridium says

"At room temperature and standard atmospheric pressure, iridium has been calculated to have a density of 22.65 g/cm3 (0.818 lb/cu in), 0.04 g/cm3 (0.0014 lb/cu in) higher than osmium measured the same way. Still, the experimental X-ray crystallography value is considered to be the most accurate, and as such iridium is considered to be the second densest element" with a reference to https://technology.matthey.com/content/journals/10.1595/003214089X3311416

I am curious: has anyone ever directly measured the density of a pure crystal of each by immersing them in water and experimentally confirmed these calculations? Or is the difference too small to measure using conventional methods?

Looking at Isotopes of iridium and Isotopes of osmium, would the answer be different if the pure crystals were made from the heaviest stable isotope? Are we even able to seperate the isotopes of iridium and osmium?

Note the "At room temperature and standard atmospheric pressure". See https://technology.matthey.com/content/journals/10.1595/147106714X682337 for some other temperatures and pressures.

Interesting but not reliable: https://www.answers.com/natural-sciences/What_is_densest_material_in_world The bit about plutonium surprised me.

--Guy Macon Alternate Account (talk) 15:07, 7 August 2024 (UTC)

Sure we can separate the isotopes. Isotopically pure ¹⁸⁴Os (the lightest stable-ish isotope) should be less dense than isotopically pure ¹⁹³Ir. Double sharp (talk) 12:39, 9 August 2024 (UTC)

August 8

Butch Wilmore

In the top image on [https://www.bbc.co.uk/news/articles/cg4yqepr469o], what's the blurring on Wilmore's chin? Any ideas? --Dweller (talk) Old fashioned is the new thing! 12:25, 8 August 2024 (UTC)

Looks like a bandage. It's visible in some Google Images pics. But I can't find an explanation. ←Baseball Bugs ^{What's up, Doc?} carrots→ 12:47, 8 August 2024 (UTC)

I only see it on pictures of him in a space suit, but it is present on a picture associated with a postponed launch (so it probably wasn't a shaving accident). I see that his chin is much closer to the collar of his space suit than is Sunita Williams's chin to her collar. I wonder if it was chafing. One size doesn't quite fit all? -- Verbarson  ^talk_edits 17:36, 8 August 2024 (UTC)

Possibly a protective anti-chafing patch.  --Lambiam 21:12, 8 August 2024 (UTC)

Seems likely. Thanks. --Dweller (talk) Old fashioned is the new thing! 15:14, 9 August 2024 (UTC)

August 9

Total daytime per latitude

Since the southern hemisphere's summer comes during the perihelion, I'm thinking it should be slightly shorter than the winter since the Earth moves slightly faster, so the same latitude on the northern hemisphere should have a little more total daytime over the course of the year. Now, I know that since any part of the Sun's disc above the horizon is counted as daytime, the day is "too long" by the time it takes the Sun to rise/set from nothing to half a disc being visible. In middle latitudes this takes longer than on the equator because the Sun's path is closer to horizontal. I'm wondering how are these two effect related in size. Do you have a shorter total daytime length over 1 year at 45°S than 40°N for example 78.2.180.96 (talk) 06:49, 9 August 2024 (UTC)

Here [5] are sunrise and sunset times for a selection of latitudes north and south of the equator, generally at 5° intervals. 2A02:C7B:21D:5400:7905:88D1:43B9:344C (talk) 09:29, 9 August 2024 (UTC)

August 10

Since gravity is formed by the stress–energy tensor, which apparently depends on the reference frame, does gravity depend on the reference frame?

HOTmag (talk) 17:38, 10 August 2024 (UTC)

Suppose we have a planet in a circular orbit around a star, using a reference frame in which the star is stationary and the rest of the universe doesn't spin around it. This is an ordinary, Keplerian orbit, as predicted by classical physics.

Now suppose a spaceship passes at high speed (0.8c or thereabouts) through this planetary system, in the plane of the planet's orbit. We switch to the reference frame of that spaceship. Special relativity tells us there's length contraction. The orbit of the planet turns into an ellipse with the star in the centre – not at one of the focal points. This is not a Keplerian orbit, so gravity must have done something funny.

Theoretically, it should be possible to apply the Lorentz transformation to the stress-energy tensor and solve the Einstein field equations for the orbit of the planet, which should give the same result as the Lorentz transformation of the Keplerian orbit, but I won't do the maths for you. PiusImpavidus (talk) 18:39, 10 August 2024 (UTC)

You have a wrong premise here, or at least one not aligned with the current way of looking at these things.

The stress–energy tensor does not in fact depend on the frame of reference. The components of the tensor depend on the frame of reference.

But the tensor is not its components. The tensor (field) is the underlying Platonic entity that can be viewed in any local coordinate system and its components extracted with respect to those coordinates, but the underlying thing remains the same.

This is the coordinate-free approach, which has been the preferred one since the mid-20th century. You don't necessarily have to agree with it; it's not the kind of thing that's subject to scientific confirmation or disconfirmation. But you probably ought to be aware that it's the dominant approach. --Trovatore (talk) 21:35, 10 August 2024 (UTC)

To be clear, this approach predicts the same outcomes for observations as approaches in which the tensor is identified with its (frame-of-reference dependent) representation. It is preferred because it gives one fewer headaches.  --Lambiam 01:41, 11 August 2024 (UTC)

August 11

A hypothetical male version of Jeanne Calment

Would a hypothetical male version of Jeanne Calment, if one will ever actually exist (including in the future), be(come) around 119 years old? (Since Calment herself was almost 122.5 when she died (122.45 years, to be more specific) and 122.45 - 3 = 119.45.) Based on the data here (List of the verified oldest people), it seems like the men are generally around three years younger than the women of the equivalent rank (for instance, comparing the 100th oldest man ever to the 100th oldest woman ever, or the 50th oldest man ever to the 50th oldest woman ever, or the 25th oldest man ever to the 25th oldest woman ever, et cetera). 172.59.128.60 (talk) 05:19, 11 August 2024 (UTC)

Aside from the fact that we don't do such guesswork here, there is doubt about the age Jeanne Calment actually reached. ←Baseball Bugs ^{What's up, Doc?} carrots→ 06:09, 11 August 2024 (UTC)

"there is doubt".  --Lambiam 13:58, 11 August 2024 (UTC)

Was Jeanne Calment the Oldest Person Who Ever Lived—or a Fraud?

Some researchers have cast doubt on the record of the celebrated supercentenarian. By Lauren Collins The New Yorker 2020/02/17 [6]. This article is behind a paywall, hence I have not read it yet. Anyway, I've always wondered whether researchers were able to rule out familiar fraud in which the daughter at some point assumes their mom's identity. With an age difference of 23 years that would make her "only" 99 when she died. According to her article some family photos were deliberately burned which is a red flag that may or might not have been dealt with adequately. In addition, I expect that the growing popularity of genetic genealogy will be able to shed further light on her claim, either confirming or disproving it. hModocc (talk) 15:06, 11 August 2024 (UTC)

I suspect that she really was 122.45 years old when she died and that the people who argue otherwise are merely spewing a Russian conspiracy theory. However, maybe this question will eventually be settled with DNA testing. I don't know. Though if so, then I would expect the conspiracy theory proponents to be humiliated. 172.59.128.60 (talk) 21:13, 11 August 2024 (UTC)

I live in Hawaii, where the number of centenarians is staggering, to the point where you will inevitably meet one at some point. For me, the question is why aren't there more supercentenarians here. Something seems to happen between 103-106 or so, but I don't know what it is. Maybe poor eyesight leads to more falls, and premature death. Viriditas (talk) 23:16, 11 August 2024 (UTC)

It is really a matter of how you define the notion of a (hypothetical) "male version" of a female person. Life expectancy depends on many factors other than sex – where one lives, family, education, lifestyle. Would your hypothetical male version of Jeanne Calment also have led a leisurely lifestyle within the upper society of Arles, pursuing hobbies such as fencing, cycling, tennis, swimming, rollerskating, playing the piano, and making music with friends? If the defining characteristic is narrowed to dying at an age with the same percentage in actuarial tables for the separate genders, these tables are time-dependent and population-dependent. I don't know how easy it will be to find such tables for France around 1997. And they will not extend to the age of 122 anyway, so one would need to replace them by a plausible mathematical model that fits the available data.  --Lambiam 13:46, 11 August 2024 (UTC)

Yes, it's not obvious what "a hypothetical male version of Jeanne Calment" means precisely.

There's lifespan data for people with ~50% and ~25% samples of her genetic make-up, an example of the many factors at play.

Father - 93

Mother - 86

First son Antoine - 4

First daughter Marie - 1

Second son Francois - 97

Second daughter Jeanne - 122

Jeanne's daughter Yvonne - 36

Jeanne's grandson Frederic - 36

Sean.hoyland (talk) 15:07, 11 August 2024 (UTC)

Frederic died in a car accident, so he didn't live to his full potential. Or was it in a motorcycle accident? Either way, my point here still stands. 172.59.128.60 (talk) 21:14, 11 August 2024 (UTC)

Well, it's entirely possible, given Murphy's Law, that the gene interaction network presumably partially responsible for Jeanne's lifespan, increases the likelihood of death by traffic accident in males. Sean.hoyland (talk) 13:50, 12 August 2024 (UTC)

They answer is yes, but you're not going to like the implications: A "study followed 81 castrated men and found that their lifespan was on average 14.6 years longer than non-castrated males. A study in Italy in 2014 found similar results, with castrated men living on average 13.5 years longer than non-castrated males." Viriditas (talk) 23:01, 11 August 2024 (UTC)

But male supercentenarians generally aren't castrated, are they?

Male hormones may limit life expectancy. As to why these supercentenarian men defy the odds, I cannot say. Viriditas (talk) 23:08, 11 August 2024 (UTC)

We have an article, profoundly depressing, on forced sterilisation. This article does not seem to mention any effects on life expectancy. However, multiple websites state significant increases on life expectancy (20% - 60%) in neutered dogs and cats. --Cookatoo.ergo.ZooM (talk) 06:32, 12 August 2024 (UTC)

Russia having very few validated supercentenarians

Is the reason for Russia having very few validated supercentenarians in part due to the fact that both Communism and Nazism caused an extraordinarily massive number of premature deaths in Russia? (Ditto for Ukraine and Belarus?) 172.59.128.60 (talk) 23:03, 11 August 2024 (UTC)

Vodka. Abductive (reasoning) 00:54, 12 August 2024 (UTC)

The 2021 population pyramid of the Russian Federation shows a steep drop when going from the 70–74 bracket (people born in or after 1947) to the 75–79 bracket (people born in or before 1946), and an even steeper drop in the transition from the 80–84 to the 85–89 bracket. The Russian pyramid has a much thinner spire than that of the UK, which shows more gradual transitions. The strong relative lack of supercentenarians will continue for another 35 years. A connection with the extreme hardship of WWII appears plausible.  --Lambiam 12:01, 12 August 2024 (UTC)

It's also noticeable that male-female ratio in the older age groups is much lower in the Russian pyramid than in the British, even for the 65–69 and 70–74 groups. AndrewWTaylor (talk) 13:51, 12 August 2024 (UTC)

August 12

Moffat and ozone

Who was "Dr Moffat of Hawarden", who wrote papers about ozone, invented an ozonometer (see for example [7]), and after whom "Moffat's Ozone papers" ([8]) appear to be named? What more can we know about him? Was he Thomas Moffat MD FRAS ([9])?

How do the papers work? Andy Mabbett (Pigsonthewing); Talk to Andy; Andy's edits 14:44, 12 August 2024 (UTC)

Seems very likely it was Thomas Moffat, since the PMC2439974 article mentions ozone 13 times and was written in 1856, soon after ozone had been characterised for the first time by Christian Friedrich Schönbein in 1840. The Google books reference was also written in 1856, by David T. Ansted. Mike Turnbull (talk) 15:31, 12 August 2024 (UTC)

The full name is given as Thomas Barbour Moffat ([10], p.17), an obituary is here, and genealogical data here. --Wrongfilter (talk) 15:41, 12 August 2024 (UTC)

Thank you, both. I have compiled data from the above sources into Thomas Barbour Moffat (Q128923496). There is also an obituary in the BMJ, which describes him as "the author of many papers on geology, meteorology, sanitation etc."; though I can only find one other (On Medical Meteorology (Q58709971)).

Thomas Barbour Moffatt, of the distinguished Moffatt clan of Sundaywell, Dumfriesshire. What's the origin of the name "Sundaywell", sometimes written "Sunday Well"? 91.234.214.10 (talk) 17:41, 12 August 2024 (UTC)

According to the Dumfriesshire OS Name Books, 1848-1858 "Sundaywell - [Situation] At E. [East] end of Sundaywell Moor. A good Spring, the water of which is deepened by a Stone dam round it, seemingly very old, There is a tradition, that, at Some remote time, there were great numbers of people Baptised here, The farm takes the name from the well." and "Sundaywell - A large Farm House with extensive outbuildings and garden the property of trustees of the late Alexander Moffatt Occupied by John Edgar". Mikenorton (talk) 21:27, 12 August 2024 (UTC)

How did Carl Linnaeus know about the species he described?

I was reading about the Pied kingfisher, whose article states that it was first described by Linnaeus in 1758. I assumed he must have visited "Persia and Egypt" in order to catalogue the species, as that's where he says it lives—but I don't see reference to any such expedition on the catalogue's article. Were their earlier catalogues that Linnaeus drew from? How did he know about this bird? ꧁Zanahary꧂ 16:19, 12 August 2024 (UTC)

To quote from this webpage from Berkeley, "Linnaeus continued to revise his Systema Naturae, which grew from a slim pamphlet to a multivolume work, as his concepts were modified and as more and more plant and animal specimens were sent to him from every corner of the globe". Mikenorton (talk) 20:51, 12 August 2024 (UTC)

photons

Suppose the Sun were part of a binary: identical twins, separation several million miles. When the stars were 'side-by-side' in our line of sight, the radiation received on earth would be twice its present value. But what would happen when one star was eclipsing the other? Would photons from the distant star pass unscathed through the nearer one? If not what would the received intensity be? What spectral changes if any would be observed compared to the familiar G2 spectrum? Renshaw 2 (talk) 16:37, 12 August 2024 (UTC)

Subjecting our planet to double the usual solar radiation is likely to leave no one alive to make the interesting observation of whether a star is transparent. Philvoids (talk) 17:11, 12 August 2024 (UTC)