Wikipedia:Reference desk/Archives/Science/2021 November 9

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November 9

Scientific fraud

This is about Elisabeth Bik. How does she know there's an intention to cheat rather than a printing error? I mean, in publishing scientific papers there are many links, humans and machines. At any step of the process something could go wrong and an image be swapped for another.

E.g. in my study of sociology I have computed the same correlation as someone else did in a book I had never heard of. That was very clear to my professor. But he could not prove intention to cheat, since he provided us with a 30 years old data file. He vowed to never again provide such old files to his students.

People sometimes stumble upon the same correlation, or the same words. That's why afterwards I have always scanned myself my own papers for plagiarism. Which not only keeps the trouble away, but also provides one with new bibliographical sources. tgeorgescu (talk) 02:24, 9 November 2021 (UTC)

Speaking of "how does one know whether something was a printing error?", see this article from 2010. --184.145.50.17 (talk) 06:25, 9 November 2021 (UTC)

What makes you say that Elisabeth Bik knows there is "an intention to cheat"? When she signals issues with published research papers, she calls them "concerns". Many involve reuses of identical images alleged to have been obtained from independent experiments. While such a reuse may occasionally result from an unfortunate accidental mix-up (not a "printing error"), when this happens again and again in the publications of an author, or five times in one publication, it becomes IMO very hard to ascribe this apparent pattern to something unintentional.  --Lambiam 09:52, 9 November 2021 (UTC)

From here: "Overall, 3.8% of published papers contained problematic figures, with at least half exhibiting features suggestive of deliberate manipulation." (emphasis mine). There's the hedging of "suggestive", but clearly an indication that an attempt has been made to identify the intentional ones. Matt Deres (talk) 20:56, 9 November 2021 (UTC)

• Parts of what Bik reports on her blog can be explained by mistakes. Others, not so much. Consider this recent example of images containing repeated elements - the explanation seems to be that they took photos from elsewhere without mentioning it and photoshopped over covered parts, which is clearly a breach of ethics. Tigraan^{Click here for my talk page ("private" contact)} 11:26, 10 November 2021 (UTC)

For an explicit example: "Please note that I am not suggesting here that moving this data point from the treatment group to the placebo group was done on purpose. Mistakes can happen, in particular if there is a rush to publish, and I do not know exactly what has happened here."^[1] This is IMO not "hedging" on the part of Bik, so as to avoid being caught in commitment, but a frank appraisal of (lack of) certainty.  --Lambiam 09:44, 11 November 2021 (UTC)

Definition of Volt

Volts = energy / charge. Each galvanic reaction can add a fixed amount of electrical energy to a closed circuit. Volts equal energy per chemical reaction divided by electric charge.

$Volts={\frac {EnergyPerChemicalReaction}{ElectricCharge}}$

Within a capacitor, the relation between work and charge is:

$W={\frac {1}{2}}VQ={\frac {1}{2}}CV^{2}$

Therefore:

$Volts=2{\frac {Energy}{Charge}}={\sqrt {2{\frac {Energy}{Capacitance}}}}$

A galvanic cell can charge a capacitor until capacitor volts equal nominal volts. Therefore, energy per charge entering the capacitor can differ from energy per charge within the capacitor. Please account for the difference in energy per charge. — Preceding unsigned comment added by Vze2wgsm1 (talk • contribs) 09:00, 9 November 2021 (UTC) Vze2wgsm1 (talk) 09:05, 9 November 2021 (UTC)

Would that be a homework question? Your first challenge is that when charging a capacitor, the voltage starts off from zero and then ramps up. So that you will have to integrate the (current going in) x (voltage) over time to work out the total energy put in. Now secondly your galvanic cell puts out a "fixed" voltage. So what will happen when you attach a fixed voltage to a zero voltage? It is basically the same as shorting out your battery. So where does the energy go in a short circuit? The same effect will happen when attaching an uncharged capacity to your cell. So user:Vze2wgsm1 where will the energy go? Graeme Bartlett (talk) 11:02, 9 November 2021 (UTC)

All charges from a galvanic cell have the same energy per charge. When a 2 volt galvanic cell is charging a capacitor with zero or 1 volt, then a conflict exists between energy per volt in and energy per volt stored. Vze2wgsm1 (talk) 13:31, 9 November 2021 (UTC)

The galvanic cell provides a fixed voltage, the capacitor starts at zero voltage. But we know that the sum of all voltages over a loop in the circuit must be zero, so there must be some voltage over some other component. The wires have some finite resistance and there's the internal resistance of the galvanic cell. Sending a current through that resistance may have something to do with it. PiusImpavidus (talk) 14:24, 9 November 2021 (UTC)

Could be superconductive conductors between the galvanic cell and the capacitor. The capacitor's plates could be superconductive.Vze2wgsm1 (talk) 15:05, 9 November 2021 (UTC)

The differences are always due to energy lost via heat. The second law of thermodynamics strikes again... --Jayron32 14:25, 9 November 2021 (UTC)

No. A capacitor's heat loss is via internal resistance. $P=V^{2}/R$ Vze2wgsm1 (talk) 14:58, 9 November 2021 (UTC)

Not only energy lost to heat. The difference between the voltage across the cell and the voltage across the capacitor includes other impedances such as inductance. If you try to eliminate dissipative effects using superconductors, effects like kinetic inductance can become important. There are still dissipative effects (such as interactions with nearby dielectric materials), but these can be made very small [2]. The circuit then becomes a high-Q LC oscillator. --Amble (talk) 17:46, 9 November 2021 (UTC)

Capacitor or other discharge into a resistor can remove both the charges and the energies associated with the charges.

During oscillation within an LC circuit, all the charges and the energy associated with the charges can exhaustively convert into magnetic energy. Then polarity reverses and magnetic energy can exhaustively convert into the charges and energy associated with the charges in the capacitor. Vze2wgsm1 (talk) 19:03, 9 November 2021 (UTC)

Even if the wires and the plates of the capacitor are superconductive, you still have the internal resistance of the galvanic cell, basically the resistance of the electrolyte, which isn't superconductive. And if you get the resistance low enough, inductance will get important. Now the energy delivered by the galvanic cell not stored in the capacitor when it gets to the same voltage as the cell will be stored in the magnetic field. Then the voltage on the capacitor will overshoot the voltage of the cell, using the energy from the magnetic field. As noted above, it's an LC circuit. PiusImpavidus (talk) 09:35, 10 November 2021 (UTC)

The problem is that a 2 V galvanic cell can supply a 1 volt capacitor w/o violating the conservation of energy law.

A 2 V galvanic cell with high electrolyte resistance can deliver 1 V to an external circuit if external circuit resistance equals electrolyte resistance. Normally, electrolyte resistance is low enough to be negligible.

$NominalVolts=ExternalCircuitVolts+ElectrolyteVolts$

When galvanic cells are in series, they all add their energy to the same charge. Supplied energy per charge can be much higher than the capacitor's energy per charge.

$Vtotal=V1+V2+V3+...$

Normally, the inductance of the short wires between the cell and the capacitor is too low to hold much magnetic energy.Vze2wgsm1 (talk) 12:24, 10 November 2021 (UTC)

You first want to neglect resistance, then you want to neglect inductance, then you say it doesn't add up. Indeed, but you cannot neglect these things. Remember, things are only negligible compared to something else.

The circuit can be described as a simple loop with a voltage source, a capacitor, a resistor and an inductor. When going around this loop, the voltages (over each component) must add up to zero. This means that, if the capacitor isn't charged to the same voltage as the voltage source, there must be a non-zero voltage over the resistor and/or the inductor. Physics just tells us that in this circuit resistance and inductance cannot both be negligible. The current or its time derivative will always get large enough to get enough voltage over the resistor or the inductor to get the sum of voltages to zero, and this tells you where the energy is going. PiusImpavidus (talk) 09:47, 11 November 2021 (UTC)

The definition of volt conflicts with the definition of electron volt. A capacitor can contain 1 volt and 1 electron volt of energy. A transformer can convert the 1 volt into a hundred volts. The hundred volts can only accelerate an external electron one electron volt. I think volts per electron volt could be a better definition of volt than energy per charge. Vze2wgsm1 (talk) 10:10, 12 November 2021 (UTC)

There's no conflict here. A capacitor that, when charged to 1 volt, holds 1 electronvolt of energy, must have a capacitance of 3.2·10^-19 farad, which is tiny. At this voltage, it holds a charge of 3.2·10^-19 coulomb or 2 electron charges, so there're 2 electrons on one side and 2 holes on the other. At such low charges you can no longer use Kirchhoff's circuit laws, as those assume charge is continuous.

A transformer can convert from 1V to 100V, but while doing so, it cuts the current (for continuous operation) or charge (for batch operation) by a factor of 100. If you start with a charge of 2 electron charges, that can't happen, so your transformer can't work. PiusImpavidus (talk) 11:38, 13 November 2021 (UTC)

An external electron can gain an electron-volt from a single volt. By selection of capacitance, a capacitor containing an electron-volt of energy can have any voltage value.

$Energy={\frac {1}{2}}CV^{2}$ — Preceding unsigned comment added by Vze2wgsm1 (talk • contribs) 11:00, 15 November 2021 (UTC) ````

Does the geoid have any concave "straight lines"? How long?

They must be so short that whether "straight line" means great circle/great ellipse/geodesic on a geoid or ellipsoid/loxodrome or intersection of one of the planes perpendicular to normal with the one of the roundoids, or coplanar with one of the centroids of one of the round surfaces or one of the points on the curving plumb bob line probably hardly affects the distance. Sagittarian Milky Way (talk) 16:18, 9 November 2021 (UTC)

Can you clarify with some context? I'm not sure what you are asking about. --OuroborosCobra (talk) 17:27, 9 November 2021 (UTC)

Is every "line segment" (a great circle is probably close enough at these tiny fractions of Earth's size) on the geoid convex? Or are there some concave ones? Probably around mountains, trenches and stuff. Extremely sightly concave.Sagittarian Milky Way (talk) 18:19, 9 November 2021 (UTC)

No, a geoid is not the surface of the earth, and so does not trace mountains and valleys and trenches. It is an idealized mathematical shape that approximates the shape the earth would take if its surface were entirely covered with water. It is related to, but slightly different than, the various Earth ellipsoids that also create mathematical approximations of the Earth's shape. What makes the geoid not an ellipsoid is that it takes into account such things as gravitational anomalies (areas of the earth where the surface gravity is higher or lower than average). However, these anomalies are NOT all that large, and their effect should not be over-estimated. As noted at Geoid, these deviations are no more than about +/- 100 meters, considering a great circle has a circumference of about 40,000,000 meters, or if you prefer a radius of 6,400,000 that means a vertical deviation of less than 100/6,400,000 or a little less than 15 parts per million. That is not enough to make even a noticeable dent in the surface. --Jayron32 19:16, 9 November 2021 (UTC)

I know all that, everyone knows valleys have concave lines. Does the geoid have them too? If the extremely smoothed out topography of the geoid can curve faster than the ~0.01 arc second/foot curvature of the Earth then you can have a concave line segment where the mean sea level bends the wrong way. Vertical deflection says plumb bobs "tilt" up to ~100 seconds of arc at asymmetric Himalayan mountains. But since big mountains are miles wide the mountain center of mass must be rather far from the point of max deflection (near the steeper slope?) so a geoid concavity may not be possible. Sagittarian Milky Way (talk) 21:04, 9 November 2021 (UTC)

I literally just explained that to you. Geoids don't follow mountains or valleys. They just don't. Even the Himalayas don't have enough of a gravitational attraction to alter the shape of the geoid great circle (or elipse, whatever) enough to alter the curvature enough for you to even notice. The most pronounced gravitational anomaly only has an effect of 15 parts per million. That will not "bend" the curve of ANY part of the geoid enough to even come close to "flat", never mind "concave". --Jayron32 12:08, 10 November 2021 (UTC)

It doesn't literally follow the land but Hawaii looks like Hawaii flattened about 300:1, not enough to cancel convexity. It is semantics if 300:1 flattening is still following. But I have a better idea: a 10 millionth scale half density Earth of common rock would theoretically have a surface gravity of 1/20 millionth g, if you find ground 24-25 inches below a flat geoid (Salton Sea?) and put one there that should be good for 648000/π/20 million or 0.0103132403 seconds of deflection. At 0.001 radii from the side surface it'll be 648000/π/(20 million*1.001²)=0.0102926447 seconds or 0.0000205956 seconds less which would take at least ~0.636 millimeters of Earth curvature to overcome and over 0.639 at the pole, which is more than 0.001 radius distance from the rock ball. If you decrease the ball size the gravity decreases proportionally but so does the distance between the side surface and x% of side surface gravity. It might still be true that any concavity is either prevented by isostasy i.e. the thicker continental crust under mountains (2.83 vs denser for the rock the continental crust floats on) or unmeasurably small but it seems any dense ball in a field would do it (theoretically). ​Sagittarian Milky Way (talk) 21:09, 10 November 2021 (UTC)

The textbook Geodesy: The Concepts (Authors P. Vanícek, E.J. Krakiwsky, 2015), pg. 91, states "One must bear in mind that, in reality, the lows on the map are not concave. As stated earlier, the geoid is a convex surface[.]" --Amble (talk) 21:21, 9 November 2021 (UTC)

So still convex. Nice to know! Sagittarian Milky Way (talk) 23:42, 9 November 2021 (UTC)

The height variation of the geoid (±100 m) is too small to overcome the positive curvature of the reference ellipsoid. Ruslik_Zero 08:54, 10 November 2021 (UTC)

This also depends on the wobbliness of the height variation. Modelling it locally to a first approximation as $-A\cos(\varepsilon d)$, in which $A\approx 100\operatorname {m}$ and $d$ denotes the horizontal distance from a given point in some fixed direction, local convexity requires the value of $AR_{\operatorname {earth} }\varepsilon ^{2}$ not to exceed $1$.  --Lambiam 10:07, 10 November 2021 (UTC)

This can be rewritten as $A(\varepsilon )>1/(R_{\operatorname {earth} }\varepsilon ^{2})$. This means that the power in perturbations with frequency $\varepsilon$ has to decrease slower than $1/\varepsilon ^{2}$ for a concavity to arise. Ruslik_Zero 14:02, 10 November 2021 (UTC)

I think we should ask a clarification: "which geoid"? There has been much talk, here, about "the" geoid, as if there is exactly one geoid that is universally agreed. I do not stand alone in my nitpickery. Our very own Wikipedia articles cite this NOAA source: "What Is The Geoid?" - which provides "just a few examples of the difficulty in defining the geoid."

Putting aside questions of "uniqueness" of "the" geoid, there is a core definitional question. "The" geoid - in fact, every geoid (and there are many to choose from, with a sampling of useful geoids explained and documented at the National Geodetic Survey website - ... every geoid is just a mathematical model that describes the shape of the Earth. It is a representation of an abstract, geometrical surface. There are many different standards; and methods; and practical uses. For today's discussion, we could probably all agree that the most useful geoid is "the" geoid represented by one of the most common gravimetric anomaly implementations. It is a shape defined and standardized using (e.g.) the World Geodetic System; it is expressible mathematically in the form of a harmonic continuation (essentially, it is a practical extension of the pure spherical harmonics representation). For this discussion, let's grant that we only care about geoids that are expressed in this format; the values of each coefficient are not of particular interest: these values differ according to each particular geoid representation and its practical purpose, (e.g. a geoid designed to represent gravitational anomaly; a geoid designed to represent gravimetric deflection; and so on). Let's ignore any of the curved surfaces that are not expressed as some form of spherical harmonic series - even if they are the same surface, represented using different formats.

And we need a second question: "what do we mean with the word convex?"

In the purview of a pure, mathematical definition - convexity and concavity have very precise mathematical meanings. Specifically, there is the definition of a convex function. Now, this is really important - when we talk about "convexity" and "concavity", we need to be really careful. For a simple shape, mathematical function convexity is intuitively identical to the visualized image of convexity. From this simplification, we get such concepts as the convex hull (which also has a precise, mathematical definition). But as we get to more sophisticated shapes, the precise mathematical meaning of the word "convex" begins to drift from our elementary intuition about "convexity."

Why does this matter? Because most of our geoid models are truncated harmonic series. If you don't immediately know why that matters, ... please accept by argument from authority that it matters a lot. A perfect representation of the "abstract" perfect geoid would require an infinite continuation of that series. Truncating that series introduces error (small but non-zero). This truncation error can change whether the surface is, or is not, convex.

Ruslik is 100% on the right track, in the effort to define the concavity or convexity, with reference to the local curvature. The catch is: upon which surface are we checking convexity? Upon the geoid or upon a truncated spherical harmonic representation of that geoid? Because that small difference matters a lot. A spherical harmonic representation is a linear combination of convex basis functions. That is not an accident; it is an on-purpose. This mathematical representation cannot be non-convex. (It is no accident that it is a convex series). This specific representation of the shape allows mathematically-inclined individuals to use standard and conventional methods of optimization to do work relating to the geoid model. The trouble falls in two categories:

1. Is this numerical representation the best representation? Truncated spherical harmonic series are not the only way to represent the ideal geoid: we can actually use different mathematical expressions to describe (and to approximate) the same shape, and depending on which expressions we choose, that shape can be represented exactly- or approximately-. We do not have to approximate this shape using a convex series of basis functions; we can write a mathematical expression of the same surface, using a different formulation, which can be locally non-convex.
2. Is this surface the best surface? Truncated spherical harmonic series are not a perfect representation of the ideal geoid - we truncated an infinite series of harmonics. We presume that the ideal shape is well-approximated using only a finite number of terms. The approximation that we use to represent the shape is always convex; is same true about the exact shape? This is independent of any concerns about how we might measure exactly; I only describe how we might define a surface: we can define a non-convex surface, and then approximate it with a truncated harmonic series, and then we can rigorously compute the approximation-error.

Now, back to why this makes any difference at all: convexity is a property of a function that can have an effect over an infinitesimal extent. And this isn't only a theoretical concern: it is a practical concern - especially if one wishes to use conventional numerical methods to work with a curve or shape! When we truncate an infinite series because the last infinite-number-of-coefficients are very small, we are pretending that we don't care about such things. That is exactly the type of fallacy that leads to profound numerical errors in practice!

Wrapping up, the take-aways are: (a) we need to be really careful about how we are describing our terms. If we use our terminology interchangeably - substituting "geoid" vs. "approximation of the geoid"; "function convexity" vs. "convex hull" vs. "shape that sort of curves inward a bit"; then we are abusing mathematical terminology in a manner that leads us to a wrong answer. And if we are ignoring "bits" that are "very small," we are committing a really serious mathematical error. Even in elementary calculus, we learn that working with infinite series leads to strange consequences: we're literally discussing the sums of infinite numbers of infinitesimally-small bits. That sum can converge to zero, or it can converge to a finite number, or it can converge to infinity. The difference is in the details.

Nimur (talk) 15:29, 10 November 2021 (UTC)

An obvious way of defining whether a surface enclosing a space is convex is to define this as the enclosed space being convex, being the 3D analogon of the definition of a closed convex curve. So then an ellipsoid is convex but a torus is not. Then a geoid represented by a truncated spherical harmonic series is not necessarily non-convex – just think of truncating immediately after the $Y_{0{,}0}$ term, in which case the geoid is a sphere.  --Lambiam 18:42, 10 November 2021 (UTC)

I was barking up the wrong tree, but a geoid represented by a truncated spherical harmonic series is also not necessarily convex.  --Lambiam 09:15, 11 November 2021 (UTC)

Regarding "convexity is a property of a function that can have an effect over an infinitesimal extent" leads one pretty quickly down the coastline paradox issue and the problem of the fractal nature of problems like this. I intentionally avoided that discussion, as it introduces a level of detail to the discussion that is unlikely to help the OP. But you've now added that detail. I doubt the OP will take any of this into consideration, or really anything you wrote, but it is the problem with trying to answer questions like this both sufficiently and accurately. --Jayron32 16:52, 10 November 2021 (UTC)

I did care if a lead BB or dense crystal at sea level could do it, though the order of magnitude of the longest such concave line on Earth would be more interesting. And now I know that for math convention convenience geoid models are defined in a way that can't represent concave local curvature (yes I know the difference between Venn diagram indentations and concave local curvature), even if the best-fit curve for mean sea level that one would choose if one were omniscient and thorough might have tiny zones of local curvature "bending the wrong way". Sagittarian Milky Way (talk) 21:45, 10 November 2021 (UTC)

Okay and now I read the rest and find out that conventional geoid models can represent concave shapes (but probably don't as the book says they're convex) Sagittarian Milky Way (talk) 21:50, 10 November 2021 (UTC)

Re. "A spherical harmonic representation is a linear combination of convex basis functions. [...] This mathematical representation cannot be non-convex.": Where are you getting this from? The individual spherical harmonics are not even positive-valued in every direction; much less does each individual harmonic represent a convex shape. Even if they did, the conclusion doesn't follow, because the coefficients can be negative. --Amble (talk) 19:06, 10 November 2021 (UTC)

I apologize for failure to cite a source; "I got it from memory," which is a pretty weak citation. I think the formal name for this result is the Funk-Hecke theorem. The proof that (certain) trigonometric polynomials are convex is Homework Problem 3.19 in the horrible CVX book.

Here is a Papers in Analysis, R.T. Seeley (1966) presenting "a concise and elementary exposition of spherical harmonics, including the Funk-Hecke theorem."

Nimur (talk) 19:21, 10 November 2021 (UTC)

Thanks. That exercise is for convex basis functions, with additional constraints on the coefficients (non-negative and ordered), none of which apply here. The spherical harmonic expansion used for the geoid can easily represent non-convex shapes. The familiar (non-convex) shapes of electron orbitals in atoms are based on a very similar expansion. --Amble (talk) 19:39, 10 November 2021 (UTC)

For those looking to build the wiki and turn Funk–Hecke theorem blue, the eponyms are Paul Funk and Erich Hecke. DMacks (talk) 04:34, 11 November 2021 (UTC)

Amble, I believe you are correct, insofar as "it is possible" to construct a geoid using non-convex basis functions. I spoke (wrote?) too broadly; my statements are not universally true. Even though I concede that point, my statements are correct in some instances: for practical purposes, some geoid models are intentionally built using only convex basis functions. If we're in disagreement about whether this specific constraint applies, all I can say is that I did open my earlier commentary with remarks to the effect - "there is not only one geoid." Among the many, many, many models, some are intentionally built to be convex; those are the ones that have been most useful to me, and in that respect, my bold statements re: "cannot be non-convex" ought not be generalized to every other model. Details, details, details... but, I concede that detail to Amble, who is correct. It is possible to use non-convex basis functions (including some spherical harmonic representations) to model a non-convex shape, and to call it a geoid, by relaxing those specific constraints.

Honestly, at this phase of my mathematical prowess, so I'm not in a position to prove- or disprove theorems - it would take a lot of time and effort, and I'm "out of currency" with respect to methods of analysis. But, as Jayron already wrote, we've probably long lost the original author of the original question in our details... but, maybe some of our other readers and contributors got something out of the endeavor! We found an actual red-link, which (to me) is a rare occurrence in this decade!

Nimur (talk) 11:27, 11 November 2021 (UTC)