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November 26

Weight of neutron star material

From neutron star:

Neutron star material is remarkably dense: a normal-sized matchbox containing neutron-star material would have a weight of approximately 3 billion tonnes, the same weight as a 0.5-cubic-kilometer chunk of the Earth (a cube with edges of about 800 meters) from Earth's surface.

Neutron stars are extremely dense, to be sure, but what would that matchbox weigh on Earth? What is its mass? Tonnes are a mass measurement, not a weight measurement, so I don't know if this badly written passage means that it has a mass of approximately 3 billion tonnes or if it means that its weight on the neutron star is the same as the weight of a 3-billion-tonne object on Earth. First sounds reasonable because the star is very dense; second sounds reasonable because the star is very massive and has a tiny radius, so per gravitational constant, an extremely massive object with an extremely small radius of distance will cause any object on its surface to be extremely heavy. The statement has two citations, but I don't see the statement's information in either of them. Nyttend (talk) 09:52, 26 November 2025 (UTC)

It might be better to replace the word "weight" by "mass"; I assume the writer thought that lay readers were more familiar with weight than with mass. With weight expressed in mass units, the statement is actually true both on the neutron star and on earth; that would not be the case if the weight was properly given in Newtons, i.e. as a force. --Wrongfilter (talk) 11:15, 26 November 2025 (UTC)

Neutron Newtons? ←Baseball Bugs ^{What's up, Doc?} carrots→ 12:26, 26 November 2025 (UTC)

Not as tasty as the fig ones. --User:Khajidha (talk) (contributions) 13:39, 26 November 2025 (UTC)

Filling, though. :) ←Baseball Bugs ^{What's up, Doc?} carrots→ 14:03, 26 November 2025 (UTC)

Makes me feel weighted down and sluggish every time I eat them. --User:Khajidha (talk) (contributions) 14:22, 26 November 2025 (UTC)

What is the normal size of a matchbox, anyway? I would assume it varies by market. Better to use a cm3. —Tamfang (talk) 22:20, 26 November 2025 (UTC)

Maybe about the size of a Jimmy Neutron Matchbox car? — Preceding unsigned comment added by Baseball Bugs (talk • contribs) 00:28, 27 November 2025 (UTC)

I've left a related question at WP:RDH. Nyttend (talk) 04:06, 27 November 2025 (UTC)

November 29

FlUorine polymers

I'm looking for someplace to redirect perfluoroalkylether (PFAE) and perfluoropolyalkylether (PFPAE). Are these the same as perfluoropolyether (PFPE)? Or are there better targets? Should I just start stubs? -- Beland (talk) 02:39, 29 November 2025 (UTC)

PFAE ans PFPAE are mentioned in our article Krytox as being synonymous with PFPE. Industrial and commercial websites agree: [1], [2], [3], [4], [5].  ​‑‑Lambiam 15:24, 29 November 2025 (UTC)

Redirects added; thanks for the sourcing! -- Beland (talk) 17:23, 29 November 2025 (UTC)

November 30

Yellower Star

In night I see all stars of same white colour but one star is yellower than others. It twinkles like other stars but I am suspect it is some planet. ~2025-37195-37 (talk) 08:12, 30 November 2025 (UTC)

Where is it? Which constellation? There are many colored stars, but the colors are mostly not visible to the naked eye. See Stellar_classification#Harvard_spectral_classification for colors. However, there is a rule of thumb that if it twinkles, it is not a planet (because of the partial polarization of reflected light). Shantavira|^{feed me} 09:29, 30 November 2025 (UTC)

Polarisation has nothing to do with it — the reason why planets twinkle less than stars is that they have much larger angular sizes. The main planet in the evening sky at the moment is Saturn but it is not very conspicuous, as there are many stars of similar brightness. Later in the evening Jupiter rises but that doesn't strike me as particularly yellow, so without further information I guess the OPs question cannot be answered. --Wrongfilter (talk) 10:49, 30 November 2025 (UTC)

There are a bunch of apps for your phone that you can point at the sky and they'll tell you what you're looking at. Abductive (reasoning) 11:18, 30 November 2025 (UTC)

Also, next to not twinkling, none of the planets is clearly yellow.  ​‑‑Lambiam 13:27, 1 December 2025 (UTC)

If it's properly dark and you've got good eyesight, some bright stars are noticeably coloured. For example, Betelgeuse (the right shoulder of Orion, which you see on the left if you're on the Northern hemisphere) is noticeably red and Bellatrix (Orion's other shoulder) is blue. I suggest you use some planetarium program to find which star appears yellow to you. Stellarium is free and open source.

The most obviously coloured object in the sky is Mars, but that currently appears so close to the Sun that it's practically invisible. PiusImpavidus (talk) 11:13, 30 November 2025 (UTC)

Since we don't know whether you are in the North or South hemisphere, it's impossible to tell what stars are currently visible to you at night. It should not be too difficult for you to find out which constellation the star in question is in. (My guess would be Capella.) {The poster formerly known as 87.81.230.195} ~2025-31359-08 (talk) 21:50, 30 November 2025 (UTC)

Globular map projection?

What would be the map projection of a true-perspective drawing of a globe, like these? I was thinking maybe just "globular projection", but that would seem too vague, since it can also describe things like the Nicolosi globular projection. ~2025-31275-58 (talk) 10:44, 30 November 2025 (UTC)

That should be a simple Orthographic map projection. --Wrongfilter (talk) 10:55, 30 November 2025 (UTC)

Audio recording of a Quagga

Please see Wikipedia:Reference desk/Humanities#Audio recording of a Quagga. Andy Mabbett (Pigsonthewing); Talk to Andy; Andy's edits 14:12, 30 November 2025 (UTC)

Function of ciguatoxins

I've just read that ciguatoxins produced by Gambierdiscus toxicus are largely harmless for larger species in food chain, even after biomagnification, and that they harm mostly warm-blooded animals, including humans. That looks odd because evolutionarily one would expect toxins to target direct threats which isn't the case here. That contrasts, for example, with capsaicin that evolved as a protection against pepper-eating animals. Is there an explanation? Brandmeister ^talk 22:06, 30 November 2025 (UTC)

G. toxicus may have evolved to produce ciguatoxins both for protection against being grazed by herbivorous fish, and also for some other metabolic reason, either as a necessary byproduct or for some function as yet undiscovered. The fish, meanwhile, may have then evolved immunity to the ciguatoxins (see Red Queen hypothesis). That a relative few of these fish may then be eaten by larger animals (including us) and those might also then be poisoned, is outside the evolutionary context of the dinoflagellate, so it's just our bad luck.

This paper discusses the general context of dinoflagellate toxin production, although it doesn't mention ciguatoxin specifically. {The poster formerly known as 87.81.230.195} ~2025-31359-08 (talk) 06:05, 1 December 2025 (UTC)

I also thought of immunity development, but, as I understand, this implies that herbivorous fish effectively won the evolutionary arms race which is postulated to be perennial. Maybe the toxin then continues to evolve to be more lethal? Brandmeister ^talk 20:52, 1 December 2025 (UTC)

Maybe, or maybe its other (hypothetical) metabolic role mitigates against its changing to do so, or maybe pollution and/or warming and/or other factor(s) is reducing the population of the immune grazing fish. Evolution is very hard to discern while it's going on. {The poster formerly known as 87.81.230.195} ~2025-31359-08 (talk) 23:57, 1 December 2025 (UTC)

Just to re-emphasize 08's earlier point: we call it a toxin because it's bad for us, but predation deterrence may have nothing to do with why that little beastie is making it; it may be used for something altogether unrelated or could be a by-product that just happens. For example, the liver of a polar bear is toxic to eat - you die of a vitamin A overdose - but the bear definitely didn't evolve that to stop pesky humans from eating it; from the bear's perspective, it's something that just happened without evolutionary pressure. Matt Deres (talk) 13:51, 2 December 2025 (UTC)

Planck's law for wavelength SI units question

I can't find an explanation for the SI unit "$Wsr^{-1}m^{-2}nm^{-1}$".
In some articles I find "$Wsr^{-1}m^{-3}$",
or "the power radiated per unit wavelength interval at wavelength 𝜆 by unit area of blackbody at temperature T",
or "The spectral radiance is defined as the radiant energy per unit of time emitted within a unit spectral bandwidth from a unit area of surface into a solid angle in a direction normal to the surface". I find also this text:"the power from wavelength 𝜆 to 𝛥𝜆", but after no trace of 𝛥𝜆 in the formulas ???
I assume that "$nm^{-1}$" is to match a bandwidth unit of 1 nanometer, but I don't know how to interpret it as there is no bandwidth parameter in the law.
So where can I find an official reference for the SI units of Planck's law ?
PS: In wikipedia I get the SI units but without explanation nor reference, and I was not lukky with Google ! Malypaet (talk) 23:24, 30 November 2025 (UTC)

1 nm = 1 × 10⁻⁹ m, so 1 nm⁻¹ = 1 × 10⁹ m⁻¹.

Then

   1 W sr⁻¹ m⁻² nm⁻¹

= (1 W sr⁻¹ m⁻²) × (1 nm⁻¹)

= (1 W sr⁻¹ m⁻²) × (1 × 10⁹ m⁻¹)

= (1 × 10⁹ W sr⁻¹ m⁻²) × (1 m⁻¹)

= 1 × 10⁹ W sr⁻¹ m⁻² m⁻¹.

 ​‑‑Lambiam 13:24, 1 December 2025 (UTC)

So:

"1 $Wsr^{-1}m^{-3}$"

$=10^{-9}\cdot Wsr^{-1}m^{-2}nm^{-1}$

If you multiply the law by $\Delta \lambda$ (aka $\Delta \lambda =10^{-9}m$) as bandwidth unit of 1 nanometer, you get the same result without unit trick:

$=10^{-9}\cdot Wsr^{-1}m^{-2}$

But in radiance !

Therefore, I am always looking for an official reference by an organism like BIPM... Malypaet (talk) 16:22, 1 December 2025 (UTC)

In BIPM I found only the radiance, nothing about "spectral radiance".

The same in wikipedia List of physical quantities.

And I have no money to put in bying ISO/IEC 80000, vol 6 or 7, to check if it is there.

I have only found this in wikipedia:

The Spectral radiance in wavelength is the watt per steradian per square metre per metre (W·sr⁻¹·m⁻³)—commonly the watt per steradian per square metre per nanometre (W·sr⁻¹·m⁻²·nm⁻¹).

How should we interpret "commonly"?

Is this yet another strange, counterintuitive quantum quirk used to obtain a result that conforms to the law?

Because otherwise, with the law and standard SI units, we have a calculation that yields a result $10^{9}$ greater than the measurement !
Malypaet (talk) 21:46, 1 December 2025 (UTC)

The sentence in the article is confusing. There is no official SI unit for any physical quantity. You can read that the metre is "the" SI unit of length. But but but... the nanometre is also an SI unit, and it is also a unit of length. What is special about the metre as an SI unit is that is an SI base unit – but the choice of SI base units has been dictated more by convention and practicality than by any principled or theoretical considerations. Spectral radiance does not have its own base unit, but can be expressed in a so-called SI derived unit, being a product of powers of base units. Scientist and engineers have their own practical reasons for using other SI units than base units and derived units.

This underlies what is going on here. The situation for spectral radiance in wavelength is basically the same as for length. These two units for spectral radiance in wavelength are both SI units. The numerical values of the same quantity, expressed in either of these two units, differs by a factor of 10⁹. So why are there two units in use? For the same reason that the metre and the nanometre coexist as units of length. Scientists expressing the wavelength of the radiation under consideration in metres, which is more common for radio waves, will usually prefer W sr⁻¹ m⁻² m⁻¹ as the unit. Scientists expressing the wavelength of the radiation under consideration in nanometres, which is more common for visible light, will usually prefer W sr⁻¹ m⁻² nm⁻¹ as the unit.

The SI derived unit for radiance is W sr⁻¹ m⁻³, which is a more conventional way of denoting derived units, but of course in the end the same as that used by radio broadcasting engineers. The different way of denoting it betrays the background: the factor m⁻² is for the reciprocal ("per") of an area, and the m⁻¹ is for the reciprocal of a wavelength.  ​‑‑Lambiam 01:44, 2 December 2025 (UTC)

Thanks for these explanations, it's clear.

But if you keep the spectral radiance of "W sr⁻¹ m⁻³", the calculation differs to the measurement by a factor of 10⁹.

In the experiments I have read, the SI units tricks is a way to indirectly introduce the bandwidth in a bandwidth unit of one nanometer, that is a factor of 10^-9. Malypaet (talk) 10:11, 2 December 2025 (UTC)

The numerical values may be different, but combined with the respective units they denote the same magnitude of a physical quantity.

Consider the following problem. An object moves with a uniform speed of 50 km per hour. How long does it take to traverse a mile?

• Solution 1: Go all imperial. 50 km per hour ≈ 31.0686 mph. So it takes 1/31.0686 hours = 3600/31.0686 s ≈ 115.873 s.
• Solution 2: Go all metric. 1 mile = 1.609344 km.  So it takes 1.609344/50 hours = (1.609344/50) × 3600 s ≈ 115.873 s.

In the end, it does not depend on the units. Planck's law holds in imperial units just as much as it does in metric units.  ​‑‑Lambiam 16:06, 2 December 2025 (UTC)

Sorry, but with the bolometer of a spectroradiometer you get a power per square meter ( W m⁻² ).

This power depends not only on the wavelength, but also on the bandwidth. If you double the bandwidth, you approximately double the power, right ?

So taking an example with an online blackbody calculator (https://www.omnicalculator.com/physics/blackbody-radiation):

With a temperature at 1000k, for a wavelength of 550nm (and a bandwidth of 1nm ?) you get:

a spectral radiance = 10 307 W sr⁻¹ m⁻² m⁻¹

or

a spectral radiance = 0.000010307 W sr⁻¹ m⁻² nm⁻¹

The total radiance for all the spectrum = 18 049 W sr⁻¹ m⁻²

Therefore, one sees logicaly that the quantity associated with m⁻¹ or nm⁻¹, cannot be the wavelength, but the bandwidth.

Now if the bolometer gives 0.000010307 W m⁻², what about the spectral radiance ?

And what I have missed ? Malypaet (talk) 18:50, 2 December 2025 (UTC)

“If you double the bandwidth, you approximately double the power, right ?”

Yes, that's right.

“With a temperature at 1000k, for a wavelength of 550nm (and a bandwidth of 1nm ?) you get:

a spectral radiance = 10 307 W sr⁻¹ m⁻² m⁻¹”

The spectral radiance is the radiance within a particular band (centred around 550nm in this case), divided by the width of that band, in the limit where we take the bandwidth very narrow. If narrow enough, so that the spectrum has no features narrower than the chosen bandwidth, the resulting spectral radiance is independent of the bandwidth. In this theoretical case, we make the bandwidth infinitesimally narrow (which is quite a bit less than 1nm), although for practical measurements it's a bit more than that.

“Now if the bolometer gives 0.000010307 W m⁻², what about the spectral radiance ?”

What the bolometer gives you is the spectral radiance integrated over some bandwidth and integrated over some solid angle. Unless otherwise specified, it's safe to assume that this bandwidth is the full spectrum (that is, zero to infinity), or at least the part over which the bolometer is sensitive. The integral over the solid angle is, unless otherwise specified, taken over the full extend of the source.

Practice your calculus. PiusImpavidus (talk) 17:20, 3 December 2025 (UTC)

Thanks.

“If you double the bandwidth, you approximately double the power, right ?”

Yes, that's right.

...the resulting spectral radiance is independent of the bandwidth....

You are contradicting yourself.

So if you divide the bandwidth by 2, you approximately divide the power by 2, etc...theoretically down to zero power, and zero spectral radiance.

But Monochromatic radiation is idealized and does not exist in nature, even with lasers and their linewidth.

Therefore, logically, bandwidth should be included in the calculus of the blackbody spectral distribution.

Now, if you have a 1 nm bandpass filter in front of the bolometer, $\Delta \lambda =1nm$, I am currious to see how do you integrate this $\Delta \lambda$ into your calculus (not $d\lambda$), from the planck's law?

And with $\Delta \lambda =2nm$ ?

I know, for this last case, you divide the power by 2, to have a bandwidth unit of 1nm, and we return to the previous case.

And remember that here, energy is quantified and cannot be infinitesimal.
Malypaet (talk) 23:26, 3 December 2025 (UTC)

Say the price of eludium per microgram is 13 ₹. Then 7 mg of eludium costs 91,000 ₹, and 14 mg costs 182,000 ₹. Double the amount and you double the price. The price per microgram remains the same, it is independent of the actual weight of an amount of eludium purchased. Same for radiance; it is energy flux per area per bandwidth, so it is independent of the bandwidth.  ​‑‑Lambiam 09:01, 4 December 2025 (UTC)

Two days ago, you write per wavelength for $nm^{-}1$, now per bandwidth?

That's what I wrote; the power per nanometer of bandwidth remains approximately the same. However, in your analogy [ power in W ≡ price in ₹ ] and [ bandwidth in nm ≡ weight in micrograms ], the power is not independent of the bandwidth, just as the price is not independent of the eludium: 0 micrograms of eludium ≡ a price of 0 ₹.

However, the expression "per bandwidth" is inappropriate, because power is proportional to bandwidth (product) and not inversely proportional (divide) as with wavelength in Planck's law. I prefer the expression "within the bandwidth," which I found in an older article, but I don't know the antonym for "per."

I just noticed that if we multiply Planck's law by a bandwidth of 10⁻⁹m, while keeping the meter for all quantities, we find the same arithmetic value as with the simple Planck's law in W sr⁻¹ m⁻² nm⁻¹ and I am looking for an explanation, voilà. Malypaet (talk) 18:19, 4 December 2025 (UTC)

I think you are mistaken; I did not write "per wavelength" What I wrote was, "Scientists expressing the wavelength of the radiation under consideration in nanometres, which is more common for visible light, will usually prefer W sr⁻¹ m⁻² nm⁻¹ as the unit." That is something different. Of course, someone using nanometres for wavelengths will naturally express the width of a band from, say, 32.2 nm to 35.8 nm also in nanometres, and say that the bandwidth is 3.6 nm.

As I wrote before, the validity of Planck's law does not depend on the system of units used. I am not sure what you mean by multiplying a law by a bandwidth.  ​‑‑Lambiam 20:31, 4 December 2025 (UTC)

Sorry for my mistake.

"Same for radiance; it is energy flux per area per bandwidth, so it is independent of the bandwidth. ​‑‑Lambiam 09:01, 4 December".

Here bandwidth is strange, as there is no bandwidth parameter in Planck's law.

So, take the bandwidth parameter as $\Delta \lambda$ (1nm for example).

Now from Stefan-Boltzmann law#Derivation from Planck's law in radiance:

$L(_{T})={\frac {\sigma }{\pi }}T^{4}$

Then integrate this law for the summ of all bandwidth for wavelength 0 to $\infty$:

$L(_{T})=\int _{0}^{\infty }I({\lambda },T)\cdot d\lambda$

Where $I({\lambda },T)$ is the planck's law in wavelength.

Then replace $d\lambda$ by $\Delta \lambda$, and the integral by a summation of the mean wavelength ${\tilde {\lambda }}$ (in a bandwidth) from 0 to $\infty$ by step of $\Delta \lambda$, as it is done in blackbody experiment.

And here you are.

So, you have the bandwidth $\Delta \lambda$ as a product of planck's law, with the result in radiance, not spectral radiance. The ratio${\frac {\Delta \lambda }{\lambda }}$ is dimensionless, and you have no more wavelength dimension nor Temperature, while they are present in the law.

And you can check that you get the same arithmetic value for a bandwidth of 1nm ($10^{-9}m$ for $\Delta \lambda$), as planck's law in $Wsr^{-1}m^{-2}nm^{-1}$.
Malypaet (talk) 23:50, 4 December 2025 (UTC)

You keep referring to Planck's law as if it is a physical quantity ("multiply the law by $\Delta \lambda$", "$I(\lambda ,T)$ is the planck's law in wavelength"). This does not make sense. Planck's law is a veridical statement, confirmed by observation. You cannot multiply Planck by a physical quantity, and neither can you do this for the law of which he is the eponym.

I have no idea where you get this symbol $L(_{T})$ from. It does not occur in the section Stefan–Boltzmann law § Derivation from Planck's law in radiance or any other section in that article.

Perhaps (or perhaps not), this article "The Units of Spectral Radiance" may help to clear up some confusion.  ​‑‑Lambiam 11:04, 5 December 2025 (UTC)

Good diversion, yes, L stands for radiance or luminance in French.

But $L=f(T)=\sigma T^{4}$, so why not $L(T)$ as there is $L(\lambda )$ or $I(\lambda ,T)$?

I agree that Planck's law has been regularly verified. I just noticed the absence of a bandwidth parameter in the law, whereas experiments take it into account.

In any case, thank you for this constructive exchange. Malypaet (talk) 17:20, 5 December 2025 (UTC)

I do not understand why you bring in the Stefan-Boltzmann law, since it seems to have nothing to do with the original question.

Yes, Planck's law has no bandwidth parameter. It also has no area parameter. If $L$ is the energy flux per area per bandwidth, and $A$ is the area, then $L\cdot A$ is the energy flux for that area per bandwidth – assuming that $L$ is constant over the area. Likewise, if $\Delta \lambda$ is the bandwidth, then $L\cdot \Delta \lambda$ is the energy flux per area for the given bandwidth. At least, that is a good approximation when $\Delta \lambda$ is sufficiently small so that $L$ varies linearly over the range of wavelengths $[\lambda _{0}{-}{\tfrac {1}{2}}\Delta \lambda ,\lambda _{0}{+}{\tfrac {1}{2}}\Delta \lambda ].$ Otherwise you should use

$\int _{\lambda _{0}-{\tfrac {1}{2}}\Delta \lambda }^{\lambda _{0}+{\tfrac {1}{2}}\Delta \lambda }L(\lambda )\,d\lambda .$

 ​‑‑Lambiam 21:50, 5 December 2025 (UTC)

I just used the reverse example of going from Planck's law to Stefan–Boltzmann law, so one arrives at $I(\lambda ,T)\cdot \Delta \lambda$, with the SI units of radiance that gives a spectroradiometer.

Then you divide by $\Delta \lambda$ to get the spectral radiance of Planck's law, and, therefore,you can write "per bandwidth" (-^-).

With a bandwidth of 1nm, if you use the unit $nm^{-1}$ in spectral radiance, you cancel the division by $\Delta \lambda$ (per bandwidth) and get the same arythmetic value as the spectroradiometer.

Thats all. Malypaet (talk) 23:22, 5 December 2025 (UTC)

In quantum physics you can have two different definition for the same quantity:

"Spectral radiance is the radiant energy per time interval, per wavelength interval, per area, and per solid angle that is received by a detector oriented normal to the source of radiation. Spectral radiance is abbreviated L(λ), where λ is the center wavelength of the detector."

From: "https://www.biospherical.com/support/spectral-radiance/"

"wavelength interval"="bandwidth",and has the same physical dimension as wavelength, doesn't it ? Malypaet (talk) 21:47, 2 December 2025 (UTC)

An other one:

Remember the spectral radiometric quantities have units including per nanometer or per Hz (the nanometer or Hertz represents an interval not an absolute value).

https://sites.science.oregonstate.edu/chemistry/courses/ch660/Supplement%201.pdf
Malypaet (talk) 22:34, 2 December 2025 (UTC)

And a serious one:

https://www.nist.gov/system/files/documents/pml/div685/pub/sp250-1.pdf

Spectral radiance denoted $L_{\lambda }$ is defined as the radiant flux at a given point, direction, and wavelength per unit of projected area, solid angle, and wavelength interval.
Malypaet (talk) 22:50, 2 December 2025 (UTC)

The first and last definition are clearly equivalent. The Remember sentence is not a definition.  ​‑‑Lambiam 09:08, 4 December 2025 (UTC)

December 3

Medical-biology: interpersonal difference of in-door drowsiness

Something I searched around but didn't get a great answer for:

People can have very different capacity to tolerate reduced O₂ / increased CO₂ level, even with similar living experiences (e.g., always at sea-level) and physical fitness (e.g., stamina and strength). Why? What contributes to this difference?

Q2: How do we call this phenomenon? I've came across both hypoxemia and hypoxia, but they seem to describe only severe cases than mere drowsiness.

Thank you a lot. iris 3:04p, Edited 1:19p the next day to remove personal medical stuff. ~2025-38114-80 (talk) 07:04, 3 December 2025 (UTC)

It sounds like you need to pay your doctor a visit. ←Baseball Bugs ^{What's up, Doc?} carrots→ 08:48, 3 December 2025 (UTC)

I know. Sadly our outpatient departments here focus mainly on solving identifiable diseases, and the doctors' response is basically "go out once in a while".

Chinese goverment limited outpatient sessions to US$4-equivalant of fee, afterall. ~2025-38114-80 (talk) 10:59, 3 December 2025 (UTC)

Wikipedia is not allowed to diagnose medical problems. ←Baseball Bugs ^{What's up, Doc?} carrots→ 14:56, 3 December 2025 (UTC)

Concur with Bugs. This question started out as a request for information, but quickly shifted to asking us what was wrong with you. That's not something we should be doing, even if we might like to do so out of helpfulness. Your point about it getting worse really suggests that something is happening to you and should probably be reviewed by a qualified person that can examine you properly. Sorry. Matt Deres (talk) 15:37, 3 December 2025 (UTC)

Any options between visiting a doctor and asking random strangers on the Internet? As stated at the top of this page, we cannot give medical advice. At the same time, if everybody with some minor issue went to see their doctor, there would be such a queue that the people really ill would drop dead before reaching the surgery's doorstep. Unless there were more doctors, but that will inflate the cost of healthcare. Now people have told me that US healthcare is an order of magnitude more expensive than anywhere else in the world. PiusImpavidus (talk) 16:25, 3 December 2025 (UTC)

Birds flying North

In the morning beteen 6 AM and 7 AM, I look out my East-facing window while I lie in bed (in Rockville, Montgomery county Maryland). I see birds flying North (or sometimes NNE). This happens almost every day (that is when the weather is not so bad that they are grounded). It is rare to see a bird fly in any other direction. This seems odd to me, especially since it is late Autumn. Shouldn't they fly South? Why do they fly North specifically? I figure that this is a Biology question, so I am asking it here. JRSpriggs (talk) 23:59, 3 December 2025 (UTC)

While some Northern hemisphere bird species "fly south for the winter", many don't. You give no indication of what sort of birds: birds may move in the early morning from their roosts to feeding grounds some distance away; in the case of waterbirds or water-associated birds (like bald eagles) these may be lakes in the area (which I regularly see with geese where I live in Southern England, near some artificial lakes). {The poster formerly known as 87.81.230.1905} ~2025-31359-08 (talk) 03:04, 4 December 2025 (UTC)

Depending exactly where you are in Rockville, they could be headed for Lake Needwood or Lake Bernard Frank on some sort of daily commute. Do you never see them doing the reverse in the late afternoon? Mikenorton (talk) 10:15, 4 December 2025 (UTC)

Most likely daily commuters. Most migratory birds have reached their winter home by now.

The birds may not even take the same trajectory there and back again. For example, they may fly high in a tailwind and low in a headwind. Does the wind have a preferred direction in Maryland? On days with significant thermals, the boundary layer is thinner in early morning than late afternoon, so wind speed has a different effect on altitude during the morning commute than during the evening commute. Maybe their feeding area is visible from very far away, so they can fly in a straight line, but their sleeping area is harder to find, so they have to follow a string of landmarks. Maybe they take a detour to make use of thermals in the afternoon, which aren't present in early morning. Et cetera. PiusImpavidus (talk) 11:58, 4 December 2025 (UTC)

December 4

If an alien ship came towards us...

I have seen the controversy over the 3I/ATLAS comet, that Avi Loeb said it was an alien spaceship, but most evidence would suggest it isn't. But let's say, as an hypotethical case, that a real alien spacecraft was really flying towards us. Let's say, like alien versions of the Voyager 1 and Voyager 2. How closer should they get for us to see them and confirm their nature? And would it be possible to intercept and capture them, if they were confirmed to be technological? Cambalachero (talk) 03:32, 4 December 2025 (UTC)

Since the technological capabilities would be unknown, there's no way to know what the answer to that question would be. Potentially it might be possible to approach it, but what happens when or if it detects the approaching vehicle is unknown until it happens. ←Baseball Bugs ^{What's up, Doc?} carrots→ 03:52, 4 December 2025 (UTC)

Assuming the hypothetical, it is a reasonable further assumption that the extraneous intelligence behind the comet-like probe has the technology to disguise its nature that is so advanced that we, backward humans, stand no chance of confirming the hypothesis.  ​‑‑Lambiam 07:48, 4 December 2025 (UTC)

Whether we can confirm it's built by aliens depends on how hard the aliens tried to disguise it. If they simply put a 3 kilometre radio dish on the surface, it should be easy once we can resolve the object. And there are tricks to make images of objects we can't resolve.

Intercepting it would be hard. The delta-V required is just too much. The best chance would be to use a big rocket with extra upper stages to send a tiny probe, like a cubesat. But orbital insertion must be perfect, or it won't get within a few kilometres of the nucleus. Even then, the probe will have a relative velocity of tens of kilometres per second, allowing for just one snapshot and measuring the mass, if the probe isn't disabled first by a collision with a dust particle from the coma. Capture requires another big delta-V, so forget about it.

A better option might be radar observation. This would have been almost possible for the Arecibo Telescope. PiusImpavidus (talk) 13:20, 4 December 2025 (UTC)

Loeb didn't say it was an alien spacecraft, he speculated that it could be, and calculations on which he partly based this have been shown to be erroneous, as 3I/ATLAS#Alien spacecraft speculation details. Loeb has a history of this sort of thing.

Actual alien spacecraft would (I suggest) be most likely to betray their nature by alterations in velocity and/or trajectory impossible to ascribe to any natural cause (rather than slight ones which natural phenomena can explain), and/or by a reflectance spectrum completely unlike a natural body. {The poster formerly known as 87.81.230.195} ~2025-31359-08 (talk) 21:11, 4 December 2025 (UTC)

You ask about an alien version of the Voyager probes with their 1.85 meter radius radio mirrors. Consider flat disks of this radius, with 100% albedo, and ideal conditions (disk is perpendicular to the direction of the Sun, and opposite to the Sun as seen from the observer). The disk at about 1 AU from the Sun, receives a fraction of ${\frac {1.85^{2}\pi }{4\pi (1.510^{11})^{2}}}$ of the Sun's radiation. It reflects it to one side, and with Lambert's cosine law that should give a factor of 4 if I remember correctly, so the luminosity as seen from 1 AU distance should be ${\frac {1.85^{2}}{(1.510^{11})^{2}}}$ times that of the Sun. Taking minus the base-10 logarithm, multiplying it by 2.5, and adding it to the Sun's apparent magnitude gives us an apparent magnitude of 27.7 or so.

The Vera C. Rubin Observatory has a limiting magnitude of 24.5 for single images. As this telescope makes regular surveys, it would be likely to discover the alien Voyager if close enough. We need to increase the brightness by 3.2 magnitudes (or a factor of $10^{3.2/2.5}\approx 19$) which can be done by decreasing the distance from 1 AU to $\approx {\frac {1}{\sqrt {19}}}{\text{ AU}}\approx 0.228$ AU.

In order to really discover that it's moving through the Solar system with some hyperbolic excess velocity, we'd need a few images. But we'd still know almost nothing about the object. If we want to resolve it optically, perhaps to see the magnetometer boom and the disk of the radio reflector, say resolve 0.5 meters or better (I don't think seeing an elongated shape alone is enough), with e.g. the Hubble Space Telescope, the resolution should be at the order of magnitude of $2\cdot 10^{-7}$ (see angular resolution), so it would need to be as close as 2500 km (from the Hubble Space Telescope). Highly unlikely, unfortunately.

What other way would there be to tell that it is special? Its RTGs would long be cold, so not from excess heat. Its radio would long be silent. If we illuminate it with microwave radiation (radar), it would have to be probably as close to see the structure, at least if we do it with a local array of radio telescopes. Larger combined radio telescopes have been used as for the Event Horizon Telescope, but as it took longer for them to image our Galaxy's black hole than M87's larger one, due to fluctuations at shorter timescales for our smaller black hole, it's probably not ready for imaging something which is in all likelihood rotating comparatively fast.... though let's assume it is (if the alien Voyager is not tumbling, the regularity of the rotation can help). Assume a radar wavelength of 2 millimeters, and an effective telescope size of something like 10000 km, to give a resolution of something like $2\cdot 10^{-10}$, so we could resolve it at 2.5 million km instead of 2500 or so. All to be taken more as order of magnitude estimates.

So I think the best possibility is having the luck of the object missing Earth only by a few million km and a really good worldwide-receiver radar system being used.

Icek~enwiki (talk) 21:37, 4 December 2025 (UTC)

I would expect the best detail would come from planetary radar, where the signal is transmitted from Earth, bounces off the target, and is received again on Earth (either at the same site or a different one). The ability to modulate the signal gives a big advantage. In particular, the combination of time-of-flight and doppler allows details on the target to be resolved in a way that is not limited by distance or array size (assuming enough signal-to-noise). Unfortunately, planetary radar will not be possible during 3I/Atlas's pass through the inner solar system [6], but perhaps it could work for another interstellar visitor. --Amble (talk) 17:25, 5 December 2025 (UTC)

December 6

energy mass

Is it right that every solid matter in this universe including liquid and gas came from energy and in beginning there was no matter? ~2025-38835-24 (talk) 04:18, 6 December 2025 (UTC)