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January 12

quotient space of light cones

At point p in spacetime S, the light cone at p is the set of points x with d(p,x)=0 under the Minkowski metric. d(·,·)=0 is an equivalence relation on spacetime so that immediately makes me ask what the Quotient space (topology) S/d(·,·)=0 is supposed to mean physically, if anything. (Forgive the hacky notation). Of course there would also be such a relation in GR, and a quotient space.

Is this concept actually used in physics? Does it have a name? A bit of web search didn't find anything. Thanks. ~2026-23335-5 (talk) 07:02, 12 January 2026 (UTC)

It is not an equivalence relationship. An event C on the light cone of an event B which is on the light cone of an event A will normally not be on the light cone of A. This would explain why you did not find anything. It would only happen if the three events are colinear (on a single line (geodesic)). JRSpriggs (talk) 14:41, 12 January 2026 (UTC)

(edit conflict) Is d(°_⌣°)b really an equivalence relation? This requires d(p,q) = 0 ∧ d(q,r) = 0 → d(p,r) = 0 to hold universally. In fact, even in a Minkowski universe with just one spatial dimension, the pair of equations d(p,q) = 0 ∧ d(q,r) has a solution in q for any choice of p and q, so taking the equivalence closure contracts everything to the trivial one-point space.  ​‑‑Lambiam 15:04, 12 January 2026 (UTC)

Doh! Thanks. (From OP). ~2026-25905-6 (talk) 17:28, 13 January 2026 (UTC)

January 13

Bell experiment

The many Bell experiments demonstrate that entanglement connects particles across space. Is it known (either experimentally or predicted by theory) that they are also connected across time?

Thought experiment: Alice and Bob create a set of entangled particle pairs (a1,b1), (a2,b2), etc. Alice keeps the a_i's in her lab while Bob takes the b_i's on his relativistic starship for a quick run to Alpha Centauri and back, near the speed of light the whole way. So very little time has passed for him and the b_i's when he returns, while Alice and the a_i's are maybe 10 years older. Now they do a Bell experiment on those entangled pairs. What happens?

More realistically, I wonder if there's a version that can be done in particle accelerators, letting one set of particles spin around in a cyclotron while the other set is held stationary somehow. Or maybe some version of the relativity demonstrations where they fly atomic clocks in opposite directions on airplanes.

Thanks ~2026-25905-6 (talk) 17:33, 13 January 2026 (UTC)

According to the theory as it is not understood even by quantum physicists, Bob's foray to Alpha Centauri won't make a difference. Or, rather, we are lacking a theoretical model that would allow us to make a different prediction.  ​‑‑Lambiam 22:11, 13 January 2026 (UTC)

Yes, entanglement works just fine across time as well as space. This idea leads to things like the delayed-choice quantum eraser. In your version, where Alice and Bob make their measurements after they have returned to the same place, each particle can "know" how the other is being measured and what the results were. In that case, you could explain the results with a local hidden-variable theory. The usual Bell experiment requires a spacelike separation in order to make sure that any hidden variables have to be nonlocal. (The standard technical term is spooky action at a distance.) --Amble (talk) 22:58, 13 January 2026 (UTC)

January 14

On Board Diagnostics

Our excellent article could be improved with information on vulnerabilities and strategies to combat relay theft of up-market automobiles. Not something that concerns this old bike rider, but it seems to be a rising problem. Some scary (to push-button start vehicle owners) documentaries on Australia's ABC and SBS TV networks. Doug butler (talk) 12:33, 14 January 2026 (UTC)

We have an article on relay attacks. What is the link with on-board diagnostics?  ​‑‑Lambiam 15:03, 14 January 2026 (UTC)

Thank you for the link. Apparently similar attacks are made on motor vehicles. The article, for which I failed to find a link, mentioned vehicles such as Toyota LandCruisers being stolen and shipped in containers to countries that also drive on the left side of the road. Africa was mentioned.

The article mentioned tools available on the internet by which a car with pushbutton start can be opened electronically and then started (presumably by physical connection to the OBD socket but that fact wasn't mentioned). The article then morphed into relay attacks, which from the wikilink you supplied, is a quite different technique used by people with similar aims and morals.

I since found that one counter to relay attacks is a "sleeping battery" or "sleeping fob battery" replacement for the button battery in the owner's smart key, which activates with physical movement, then after maybe three minutes disconnects. Doug butler (talk) 19:26, 14 January 2026 (UTC)

This may be the scary article you read. The attack schematically depicted in our Relay attack article will still work with a sleeping fob awakened by motion. What will (literally) foil the attack is keeping the key in a pocket or bag lined with a conductive material, like a wire mesh or tinfoil, creating a Faraday shield.  ​‑‑Lambiam 21:28, 14 January 2026 (UTC)

The very first news article on Australia's channel 9 about the discovery of these stolen Landcruisers said something along the lines of "The criminals drilled into the vehicles and connected to the wiring system". This was never repeated in later bulletins. Is it possible to start these particular models by connecting to the OBD socket maybe?~2026-34591-5 (talk) 01:38, 17 January 2026 (UTC)

Water treatment and phosphorus

Doesn't modern water treament (example: [1]) remove phosphorus from sewage? If so why is phosphorus still banned for detergents for consumers? I have a reference question (talk) 16:47, 14 January 2026 (UTC)

The article phosphates in detergent has lots of references. Not all waste water is treated in modern equipment and many countries have found it simpler to ban these materials as other detergents are available. Mike Turnbull (talk) 17:25, 14 January 2026 (UTC)

Plus waste water that escapes treatment during storm surges (a hot topic in the UK now). Alansplodge (talk) 17:30, 14 January 2026 (UTC)

Centripetal acceleration in cyclotron/synchrotron

Particle accelerator descriptions often say the energy of the accelerated particles, i.e. basically how fast they are made to move. But of course they also undergo centripetal acceleration a as they go around the circular chamber. a depends on the radius as well as the particle energy, and that's not always so easy to find.

Does anyone know if any data is available for the highest a's available? Per the principle of equivalence (PE), this a can't be distinguished from a high gravity field. So is a potentially high enough to be interesting for studying quantum behaviour in high gravity? Does that amount to the combination of QM and GR? Since PE is a (classical) GR thing, could it be that it breaks down enough for this approach to not really work?

This is related to my earlier question about entanglement across time. Thanks. ~2026-30246-1 (talk) 20:45, 14 January 2026 (UTC)

If you know the maximal frequency $f$ and the radius $r$ of a cyclotron, the maximal magnitude centripetal acceleration equals

$a=(2\pi f)^{2}r.$

I don't readily see which aspect of quantum theory might be impacted by a very high acceleration. It seems to me that it will be very difficult to observe quantum properties experimentally while the system being observed is circling millions of cycles per second in a cyclotron.  ​‑‑Lambiam 22:52, 15 January 2026 (UTC)

But I don't know the frequency, and even if I did, that's for just one cyclotron. Now I'm trying to remember why I wanted the high gravity. The original question was about spinning half of an entangled pair in the cyclotron while holding the other half in an ion trap, then recovering the spinning part and doing a Bell experiment where the particles now had differing "ages" due to relativistic time dilation. ~2026-30246-1 (talk) 23:05, 15 January 2026 (UTC)

Possible figures for the MSU K500: f = 21.5 MHz and r = 0.66 m, resulting in a ≈ 12×10¹⁵ m/s². For the speed, we'd then have v/c ≈ 0.3, giving a Lorentz factor γ ≈ 1.05 .  ​‑‑Lambiam 08:28, 16 January 2026 (UTC)

January 17

Separating frozen burgers?

is there actually a way to do this without thawing the entire stack of burger patties out? A knife between them doesn't work and putting a cleaver between them and hitting it with a hammer doesn't separate them either. ~2026-35003-7 (talk) 00:27, 17 January 2026 (UTC)

Yes, by separating them before you freeze them in the first place, and preventing their freezing together by sliding a bit of plastic wrap or parchment paper between them. -- Avocado (talk) 03:48, 17 January 2026 (UTC)

You can also use a tool that is common for butchers: a bone saw like the one displayed here. Sawing frozen meat is AFAIK actually quite common, also when processing fish fillets frozen in blocks for instance, as a processing step to fish fingers. Regards, Grand-Duc (talk) 04:42, 17 January 2026 (UTC)

Put pieces of greaseproof paper between them before freezing. Alternatively heat the knife before cutting them apart, either in boiling water or over a flame. Shantavira|^{feed me} 09:21, 17 January 2026 (UTC)

Would a bolster (as in the brick chisel) and hammer work? Iapetus (talk) 09:54, 19 January 2026 (UTC)

Jewish space lasers, but only for kosher burgers. Clarityfiend (talk) 16:20, 20 January 2026 (UTC)

I'm wondering how you would put those thin separators in once they've been frozen? ←Baseball Bugs ^{What's up, Doc?} carrots→ 20:06, 20 January 2026 (UTC)

A bone saw will produce saw "dust", the quantity depending on the width of the blade, waste meat which could be sprinkled on the burger before grilling so not really waste. But in my experience, heating a knife in boiling water works just fine, possibly having to reheat it depending on the length of the parts to be separated. This is akin to spreading frozen butter portions on scones, dipping the knife into one's tea works wonders. — Preceding unsigned comment added by ~2025-30900-60 (talk) 17:30, 24 January 2026 (UTC)

Percentage of explained variation

In [2], the explained variation is the correlation coefficient or is it the correlation coefficient squared? This boils down to: is the variation explained by Abdi et. al. quite low or extremely low? tgeorgescu (talk) 23:36, 17 January 2026 (UTC)

What does the passage in the article say? The article is behind a paywall, and I'm not prepared to pay USD 56.00 to be able to answer the question. The coefficient of variation of a random variable is neither a correlation coefficient nor its square, and is normally something that is computed from experimental observations but not "explained". Did you mean "variance"? Variance is a square, but not the square of a correlation coefficient. The variance of a sum of independent random variables is equal to the sum of their variances, so parts of measured variance can be "explained" by attributing them to contributing independent factors.  ​‑‑Lambiam 08:15, 18 January 2026 (UTC)

@Lambiam: I mean the coefficient of determination. Is it applicable to meta-analysis? This is what I have learned during my Bachelor in sociology: R² says how much variance one has explained. tgeorgescu (talk) 21:33, 18 January 2026 (UTC)

Given a data set with variance $s^{2},$ where $s>0,$ you can scale it by dividing all values by $s.$ The variance of the scaled data set is then exactly equal to $1.$ As described in the section Coefficient of determination § As explained variance, the "explained variance" (and not "explained variation"), given a model, is then the variance of the values predicted by the model after likewise scaling by dividing all values by $s.$ If the model is perfect, the predicted values are the same as the observed values, so the explained variance equals the observed variance after scaling, which means it too is $1.$

Unfortunately, it is not possible to answer your specific questions given nothing more to go on than the abstract of the article.  ​‑‑Lambiam 21:38, 18 January 2026 (UTC)

It reports the correlation coefficient. In my judgment, the correlation coefficient squared is the explained variance. Is my judgment right or wrong? In my judgment, the correlation coefficient is weak, and if that gets squared, it becomes extremely weak. tgeorgescu (talk) 21:54, 18 January 2026 (UTC)

That is what is claimed in Coefficient of determination § As squared correlation coefficient, but when I do the calculations on a simple example, I get different results. The notations below are as in the Wikipedia article.

Take $\mathbf {y} =[90,110]^{\mathsf {T}}\!,~\mathbf {f} =[99,101]^{\mathsf {T}}\!.$ Then $SS_{\mathrm {res} }=9^{2}+9^{2}=182,$ while $SS_{\mathrm {tot} }=10^{2}+10^{2}=200.$ So

$R^{2}=1-{\frac {182}{200}}=0.19~.$

But there is a perfect linear relationship between $\mathbf {y}$ and $\mathbf {f} ,$ so $\rho ^{2}=1.$  ​‑‑Lambiam 22:55, 18 January 2026 (UTC)

January 18

Electrical "units"

A source in front of me says "All Station Masters who occupied a (railway-owned house) were supplied with ... 150 units of electricity per annum (a miniscule amount)". Does anyone have any idea what the unit was, e.g. kWh? That's what Google seems to think, but I'd like confirmation. The context is circa 1961-1964 in Kyabram, Victoria, and was apparently approximately equivalent to 32 gallons of kerosene for the same year, in the sense that staff were allocated one or the other, not both.

Anothersignalman (talk) 10:43, 18 January 2026 (UTC)

There is no doubt that, in Australia in the 1960s, the unit of electrical energy for purposes such as metering and pricing, was the kilowatt hour. (It remains so today.) Dolphin (t) 10:57, 18 January 2026 (UTC)

OK, do I need a separate source in the Wiki article where I quote the above line, or can I just substitute "kWh" in place of "units"? Anothersignalman (talk) 11:06, 18 January 2026 (UTC)

When quoting a source, it's best not to change the original wording. If it were me, I'd just make the word "units" a piped link to the article Kilowatt-hour – this meaning of 'unit' is near-ubiquitous in former British Empire territories. And incidentally, the correct spelling is "minuscule"; if the source really spells it with an 'i', you might want to add '[sic]' after it. {The poster formerly known as 87.81.230.195} ~2025-31359-08 (talk) 11:42, 18 January 2026 (UTC)

Thanks. Typo was mine, I've added the link and the text paraphrases anyway so I've substituted kWh in place of units. Anothersignalman (talk) 11:48, 18 January 2026 (UTC)

@Anothersignalman: You said that an apparent equivalent to "32 gallons of kerosene" was given. That offers you the opportunity to calculate or approximate whether the "units of electricity" were indeed kilowatt-hours, as the energy content of kerosene is known. One can assume that the energy was supplied primarily for heating and cooking purposes, so you've got to make a calculation of usable energy gained when burning that amount of fuel oil. If it is comparable to 150 kWh, then you're certain that "unit" is masquerading the real kWh, otherwise, you can look whatever other unit is more fitting. I can't do the math myself right now, sorry... Regards, Grand-Duc (talk) 13:51, 18 January 2026 (UTC)

All I know is that the staff members were given either the electricity or kerosene allowance. I'd be curious to know whether the energy quantity is similar, but it's not really all that important for the article I'm working on at the moment. Anothersignalman (talk) 14:14, 18 January 2026 (UTC)

One imperial gallon is about 1.2 US gallons, so these 32 gallons are about 38.4 imperialist US gallons. According to a page with the title "Residential Energy Cost Comparison", 7.4 US gallons correspond to 293 kWh. For 38.4 gallons, we then have the equivalent of about 1023 kWh. A different value is produced by this units converter, which equates 38.4 US gallons of kerosene with 1517 kWh. This is an order of magnitude off of 150 kWh.  ​‑‑Lambiam 22:04, 18 January 2026 (UTC)

Electricity and kerosine are not necessarily exact alternatives in usage, unless the kerosine was only used to run a generator where there was no metered electricity supply. Might 150 kWh and 32 gallons of kerosine have been equivalent in cost (allowing for a likely service charge on the electricity supply)? {The poster formerly known as 87.81.230.195} ~2025-31359-08 (talk) 22:49, 19 January 2026 (UTC)

In this case the kerosene was used in lamps, not in a generator, so I'm guessing the light output per gallon would be significantly less than an incandescent light bulb, and both of those are much lower than modern LED lighting. Anothersignalman (talk) 06:36, 20 January 2026 (UTC)

Fluorescent lamps are already much more efficient than incandescent lamps, almost as good as modern led lighting, and fluorescent tube lights were widely available in the 1960s. They were just not very widely used domestically.

Equivalent could be equivalent in energy content (when used for (ohmic) heating), electric output (burn the kerosene in a diesel engine powering a generator), light output (oil lamp versus incandescent or fluorescent) or cost (most interesting to the provider). My guess would be the last option, which I expect to be numerically close to the first option. PiusImpavidus (talk) 09:52, 20 January 2026 (UTC)