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# Are the directions of electric fields lines affected by Gravity?

1. Jul 27, 2012

### Edward Solomo

I recall that the path of light itself can be altered by gravity, then, being part of the electromagnetic force, then is it safe to assume that the paths of electric fields lines can also be warped?

I would imagine that the consequences would be enormous for electric flux in curved space.

For instance, the flux about a point charge is given by (1/4)(1/pi)(1/Eo)Q/r^2 by Gauss's law INTEGRAL[E * dA] take a sphere so the E field is tanget to the sphere and decreases at a uniform rate from the center, then you get Q/Eo = E(4pir^2), then solve for E.

However, if we're in a region of highly curved space, then the E field no longer retains its uniform rate of change from the center of a point charge. The flux lines would appear something like the picture attached below.

Eventually all of the flux lines would concentrate themselves at a singular point, as if the field of the point charge originated from two or more locations.

Or conversely, in a rapidly expanding area of space, which would be negatively curved, the E-field would then diminish even faster than 1/r^2, much like a dipole, which decreases at a rate of 1/r^3. In a region with a strong enough negative curve, the electric field about a point charge could reach zero over a finite distance (or possibly have reverse effects, an electric field under time reversal? Such as electrons attracting other electrons?)

Of course this sounds like a bunch of nonsense, but the Gaussian laws would suggest this given my current understanding, and thus my understanding must be severely flawed!

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2. Jul 27, 2012

### Reptillian

Yep, electric field lines are warped since the space itself is warped. One of the more interesting ideas I've seen is the idea that electrons themselves are a kind of microscopic wormhole. Electric field lines converge on the wormhole giving the sort of point particle behavior we've come to expect. Positrons would be the other end of the wormhole. :D The probelm I see with this view is that you still have a sort of electromagnetic field filling all of spacetime separate from gravity. Which I suppose isn't all that different from quantum field theory in which every fundamental particle has a unique field associated with it that fills the universe. Seems conceptually icky to me. Ideally, the electromagnetic field would itself be understood in terms of geometry.

3. Jul 27, 2012

### pervect

Staff Emeritus
Field lines work fine with General Relativity, and are a fine way to visualize electromagnetism. The tubes of the field lines are representations of anti-symmetric tensors, also known as a "differential forms", of rank 2, referred two as two-forms.

Elelctromagnetism turns out to be represented by a rank 2 anti-symmetric tensor, so it's a perfect match.

See for instance http://125.71.228.222/wlxt/ncourse/DCCYDCB/web/condition/9.pdf , "Teaching Electromagnetic Field Theory Using Differential Forms", IEEE TRANSACTIONS ON EDUCATION, VOL. 40, NO. 1, FEBRUARY 1997. This doesn't specifically talk about the usefulness of two forms with GR, but it does talk about using two forms to teach E&M.

Gravity turns out to be much harder - in general it takes a rank 4 tensor to represent it, and while it has a high degree of symmetry, it's not a differential form.

4. Jul 27, 2012

### Muphrid

It may be easier to talk about EM fields not as "differential forms" but in terms of bivectors. The EM field is a vector field in spacetime but a field of oriented planes. The most natural way to think of this iis to take the electric field lines you know and extend them in the time dimension to draw out sheets. This is the natural object that describes the EM field, and boosting these planes is what gives the relationship between electric and magnetic fields.

This also explains why gravity isn't usually said to have field lines. The Riemann tensor isn't a bivector field like the EM field; rather, it's a linear operator on bivectors.

5. Jul 27, 2012

### Q-reeus

Maybe an occasion to at last get some kind of resolution to what was proposed but never settled here: https://www.physicsforums.com/showthread.php?p=3946413
Seems to be from above entries a general agreement E field will be warped by gravitation. That makes good sense to me, yet how is that squared with the generally accepted as mathematically sound Reissner-Nordstrom charged black hole, where at large r the E field, owing to charge infalling at the infinitely redshifted BH EH (event horizon) is asymptotically close to the flat spacetime expression q/(4πε0r2)? Does it make any sense to say E field lines could in general be gravitationally bent somehow (as suggested by illustration in #1) yet with absolutely no change in field strength - which the mere existence of a RN BH surely implies? To put this concretely in a very simple and less extreme context, suppose we have a uniformly charged thin spherical mass shell. What does the marriage of EM with GR predict for the dependence of radial E field at large r on the mass M of that shell? Or suppose instead of the surface charge, a 'point' electric dipole is placed at the center of that mass shell. Will the field lines be altered at large r owing to M (assuming, apart from any purely gravitational effects, shell electromagnetic transparency - i.e. εr=1)?

6. Jul 27, 2012

### Staff: Mentor

Hi, Q-reeus, yes, we never really got closure on the questions raised in that other thread. I actually did work through a lot of the computations but never got to a point where I felt ready to post them. If I have time I'll try to go back and look again.

Regarding the specific point in your quote above, in a R-N BH the E field lines are not bent; they all stick straight out, radially, from the BH, all the way out to infinity. That has to be true by spherical symmetry. So that particular case may not be the best one to investigate whether and how gravity can bend EM field lines in a more general scenario that doesn't have that special symmetry.

7. Jul 27, 2012

### Q-reeus

Hi Peter. It would be good to see things resolved there (and/or here) if possible (big if!).
Agreed and I probably could have clarified slightly better that E field bending was not implied in as you say spherically symmetric RN case. However the point there was that imo field warpage of any kind is inconsistent with RN implied total absence of coupling between source charge gravitational potential and far field radial E field strength (radial was specified). Logically *static* EM field (no disputing EM radiation does couple to gravity) is either effected or not by gravity as a package - can't in general admit to directional warpage without logically field strength also being effected. And you know my arguments re finite RN E field.

8. Jul 27, 2012

### Staff: Mentor

I'm not sure I would describe such a coupling, if it is present in the spherically symmetric RN case, as "field warpage". But I think by "warpage" you really mean "some observable effect on the field", which seems more general to me and would cover the RN case.

9. Jul 28, 2012

### jartsa

Two identical charges side by side in a gravity field might repel each other in these alternative ways:
1: both experience some lift
2: both experience some down force
3: for some very odd reason the identical charges experience a different vertical force
4: the forces are opposite and horizontal

Well quite obviously number 4 is the only reasonable of these alternatives.

We know that two observers side by side in a gravity field must look upwards, if they want to see each other, but electric charges are felt as being in the direction where they really are.

10. Jul 28, 2012

### Q-reeus

Yes it just means here 'any departure from flat spacetime field configuration' - field pattern or field strength.

It may be worth reminding at this stage of a finding you made in another thread that isotropic Minkowski spacetime spatial metric components interior to a spherical mass shell are identical to asymptotic exterior Schwarzschild values at infinity. Consequently if one collapses together the plates of a charged parallel plate capacitor lying inside such a shell, it's coordinate determined travel distance is identical to that when done at infinity. Alternately simply discharge the plates at constant separation - the capacitor field effective volumes are identical inside and at infinity. Yet we know that the energy release when operation is performed within the shell is depressed wrt infinity by the redshift factor √-gtt = √(1-2GM/(rc2)).

And that imo inescapable fact presents a rather difficult dilemma for RN supporters (i.e. practically whole of GR community). I raised this scenario several times in the other thread linked to in #5, but worth raising it here again. Cuts right through all the mathematical elegance and sophistication of the standard picture. Which basically wants to have it's cake (RN metric with E field unaffected by infinitely depressed grav redshift) and eat it too (standard grav redshift applies to emitted radiation etc.).

Could splitting between an invariant 'active' charge |E| = qa/(4πε0r2) and potential effected 'passive' charge F = qpE, with qp = q√-gtt, allow just that? It certainly works for the collapsing capacitor plates discharge scenario energy-wise. And for the static plates discharge too if we identify gravitationally depressed electrostatic energy density with W = 1/2√-gttε0|E|2. Even works if applied to output of a dipole oscillator where field output is reduced solely owing to redshifted frequency - both input and output power drops as -gtt (radiation field strength drops proportional to frequency and thus as √-gtt), as required. Alas there is at least one fatal failing.

Suppose we separate two charged spheres apart via a dielectric rod, equal and opposite charges but with one difference - one sphere is more massive than the other. If RN is true, qa's are identical and hence magnitude of E field induced by one charged sphere on the other. However we had to have qp dependent on potential, so action and reaction are unequal, with a greater force exerted on the less massive sphere. Newton not happy.

A better approach imo that has no such inconsistency at least for our static scenario was suggested in #248 in the other thread linked in #5. Leave charge as just plain q but with coordinate values for vacuum permittivity and permeability given by: ε,μ = (1/√-gtt)(ε00). We now have that coordinate determined light speed c, EM energy and power densities, and Newton's 3rd law for electrostatics (and similarly for magnetostatics) all work out good. I do not claim this can be automatically extended to more complex situations involving general motion in non-static, non-symmetric spacetimes, but then it does not fall badly at the first hurdle either!

11. Jul 31, 2012

### Q-reeus

They're going for gold in London, but a different type of gold applies to this thread - as one golden oldie goes 'Silence is golden, golden...' Or another: 'The kiss of death, from Mr Gold Finger...' Gee, hate to think that's my label here. Anyway just in case someone wants to step out of the shadows, here's another thought following on from #10.

Say we have a steady current loop as magnetic dipole of moment m, arranged coaxial with and centred about an electric dipole p of the same energy (as determined by the mutual mm and pp forces of attraction/repulsion between other identical such dipoles). Two such mp pairs, but one pair with opposite relative orientation between m and p, will, provided separations are not too close to invalidate 'point dipole' approximation, experience no net interaction forces or torques. That's in flat spacetime. What about if these pairs are within the spherical mass shell of #10? Still flat spacetime there, but potential depressed by √-gtt. Locally, non-interaction between dipole pairs must still hold. Which is interesting, since coordinate value of drift speed of magnetic dipole circulating currents is depressed wrt infinity by √-gtt, and so therefore the magnitude of the magnetic dipole moments m - assuming as per RN metric implies, charge itself as a source of E field is invariant wrt potential.

By RN metric reckoning, the coordinate value for electric dipole moments p are unaffected by the surrounding mass shell (charge magnitude and displacement distance both indifferent to gtt). Locally then, within the shell, net interaction-free mp pairs, but by coordinate measure there is net interaction owing to dominance by the electric dipoles?! Just doesn't add-up imo. And this coordinate imbalance persists regardless of whether one applies the split into 'active' and 'passive' charge done in #10 or not. Oddly perhaps the force imbalance problem between charged spheres as per #10 cancels out when applied to an equivalent situation involving current loops of differing mass and therefore locally differing √-gtt factors. Small comfort.

Now try the suggested cure: depressed values for ε,μ = (1/√-gtt)(ε00). From the Wiki article here: http://en.wikipedia.org/wiki/Dipole, we have expressions for magnetic dipole:

and electric dipole:

At first sight the inverse relative locations of ε00 in the two expressions appears in conflict with having each directly modified by 1/√-gtt factor - seemingly implying an increased B field in conflict with an expected reduction in E field. Recall however that the magnetic moment m is that owing to circulating charges, which in coordinate measure not only circulate slower by √-gtt, but with reduced effective charge by the same factor. Hence m by itself is reduced by the factor -gtt, the field B then further modified by μ = 1/√-gttμ0 whereas electric moment p is unaffected in the above expression for E, it's reduced effective charge appearing outside of p itself - in the single modifier 1/√-gtt for ε0. Properly interpreted and applied then, there is imo a harmony here that works. That a well-known 94 yo solution to EFE's is directly under challenge seems to be a matter of sheer indifference here at PF. Or maybe Mr GF just generates too much perplexity/fear/loathing for comfort.

Last edited: Jul 31, 2012
12. Jul 31, 2012

### Staff: Mentor

Yes, I did make such a finding. But it doesn't mean what you think it means. The fact that the spatial metric coefficients are the same does not mean you can directly compare distances inside the shell to distances at infinity the way you are trying to. See further comments below.

However, your capacitor scenario is irrelevant to the questions you've raised about RN spacetime, because in the capacitor scenario, unless I'm misunderstanding something, the EM field is supposed to be zero everywhere except between the capacitor plates. So the EM fields of both capacitors are purely local, and a purely local analysis is all that can be applied to them.

You're not stating this precisely enough. The correct statement is that the *locally measured* capacitor plate separation is the same inside the shell and at infinity. See below.

Let's restate this scenario more precisely. We have two capacitors: C1 is at infinity, C2 is inside a spherical mass shell. The proper distance between the plates, measured locally, is D for both C1 and C2. Both capacitors are stipulated to start in the same state of charge, again as measured locally. That means that the same energy E, measured locally, will be released when we discharge each capacitor. So measured locally, the capacitors are identical.

It is true that an observer at infinity will see energy E coming from C1, but energy fE coming from C2, where f < 1 is the "redshift factor" inside the spherical mass shell. What accounts for the difference? Obviously that the energy was redshifted as it climbed out of the gravity well. It doesn't imply any local change in C2 as compared to C1; as we saw above, locally the two capacitors are identical. That means the energy redshifting has nothing to do with the local behavior of the EM field. (If you want, we can transport the energy from C2 to infinity by some means other than EM radiation.) The difference in observed energy received at infinity is purely due to the effects of the intervening spacetime; it has nothing to do with the local properties of C2 vs. C1.

13. Jul 31, 2012

### Staff: Mentor

I've only given your latest scenario a quick read-through, but you are making an awful lot of claims about what GR and the RN metric supposedly say, without backing them up with math. This has happened before. You also continue to insist on focusing on coordinate-dependent quantities instead of invariants, which has also happened before. And you insist on comparing coordinate-dependent quantities in different parts of spacetime (e.g., at infinity vs. inside the shell) and then insisting that these comparisons have some physical meaning. This has happened before as well. All of these are reasons why your proposed challenges are so often received with what you term "sheer indifference".

Furthermore, it's not clear to me exactly what you think you are challenging. Are you trying to claim that the R-N metric is not a solution of the EFE? That's ludicrous; it's easy to check, and thousands of physics undergraduates probably check it every year as a homework assignment. Are you trying to claim that the EFE is wrong? Good luck with that; within its domain of validity it has a lot of experimental support (to put it mildly). Are you trying to claim that the R-N solution, while mathematically correct, is somehow "unphysical"? Well, yes, it is, in at least two ways. One way, which is IMO not that big an issue, is that it assumes exact spherical symmetry; but that's easy to justify as an idealization which is often pretty well approximated by real objects. The other way, which is more of an issue, is that the interior of the R-N solution has a number of features that make it not very reasonable physically. But neither of those issues are relevant to your proposed challenge; you are also assuming perfect spherical symmetry, and your challenge can be applied purely in the exterior region of R-N spacetime. So I'm not sure exactly what you think your challenge is supposed to prove.

14. Jul 31, 2012

### Q-reeus

Sure it does - as obviously implied in #10 and explicitly stated in that other thread linked to in #5. We can, especially obviously for collapsing plates scenario, use ropes & pulleys, rods & bell cranks etc. and make a 1:1 correspondence that is entirely physically meaningful. Do you challenge that? How so exactly if you do?
Asserting my examples are irrelevant does not make them so - see my last comments.
A trivial true statement which ignores what's going on here. Go back and check context of that entire first main para in #10, esp. preceeding sentence to above. Actually, don't bother; here it is in full:
No reasonable excuse for taking it other than how it is clearly meant to be taken. And how it then subsequently leads on.
Another 'precise' and equally trivially true statement.
And via ropes & pulleys etc. we can tie that down to equivalent fields and forces acting 'down there'. As per earlier comments. And it's not just 'absolutes' - this time as per your opening comments in #13, think carefully about the example given in #11 - we have a number of ratios 'paradoxes' of RN making there. If it's all so easy to resolve within standard picture, offer your own full explanation/resolution please, up to your own standards of 'precision'. I maintain logic behind RN metric inevitably leads to paradox.
Please do not keep repeating that red herring - you aught to know perfectly well it is a false representation of what I have been saying all along. And I don't like going around in circles.
Well the intervening spacetime doesn't have any effect on charge as source of E if RN metric is true. Do you disagree? How so if so?
Having read your #13 best I don't respond to it because it's style and tone invites a slanging match we can do without. what would impress in a positive way wold be for you to apply that undoubted prowess with 'the math' to at the very least the simple scenario of #5, and actually commit to a definite prediction. Why was that not done quite some time ago in that other thread? Here I have given an increased choice of alternative configurations - all laughably simple for someone of your grasp of GR surely. The one time you tentatively presented what passed as a proper math solution, not for for charged shell but RN exterior E field was here:https://www.physicsforums.com/showpost.php?p=3969524&postcount=345
My somewhat delicate response was in #348, and you came back, very briefly, in #366 with what looked to be a quiet disowning of the original. Fair enough I suppose, but time to properly settle at minimum the charged shell case, wouldn't you say? Then we might come back to the highly relevant scenarios here and see it all in a new light.

So may I humbly suggest we put off vague accusations of 'imprecision' etc. Once the long awaited kosher GR solution for charged spherical mass shell materializes, discussions of objective substance will be possible. Of course I do not accept there is anything lacking in logical rigor to that already given here in #10 & #11, but I understand the imo underlying Sacred Cow mentality that is denied in word but adhered to in practice. Don't think I'm singling you out on that score - far from it. Must go. :zzz:

15. Jul 31, 2012

### Staff: Mentor

And any such linking mechanism will be affected by the curvature of spacetime in between, just as the energy being transmitted back upward is.

I know no such thing. I wasn't making a statement about what you were saying; I was making a statement about what GR says. Did you read the part where I said I was assuming that the EM field is zero everywhere except inside the capacitor plates?

If you'll remember, in one of those other threads I said I needed to go back and re-evaluate what I said about the charge integral being unchanging, since it didn't seem to be consistent with other "obvious" things about the R-N metric.

Because such things take time and effort. First, before even constructing a math model, I have to be sure I understand your scenario correctly. Your way of stating scenarios does not always make that easy. Please understand that I'm not saying that as a criticism, and I apologize for coming across as critical in my previous post. But it is a fact that I (and apparently others here on PF) sometimes find it difficult to understand your scenarios. That makes it difficult to model them in the math. Obviously *you* understand your own scenarios, but you can't model them in the math, so there has to be a translation step involved, and often it doesn't appear to work very well.

Second, please understand that I am participating in these discussions because I find them interesting and fun, but that doesn't mean they are at the top of my priority list. (I suspect this goes for others as well who have said things about trying to model these things in math.) I understand that you believe you have a knock-down refutation of GR, and I also understand that nobody here on PF has yet presented a counterargument that you consider valid. That doesn't change the fact that I, and probably others, are applying a heuristic that says that, when presented with a scenario like those you have proposed, which is claimed to refute GR, it is far more likely that the person who is proposing the scenario has made a mistake somewhere, than that it actually refutes GR. That doesn't mean we have actually found the mistake; nor is it a *proof* that there must be a mistake. It's just a heuristic judgment in order to set our priorities for how much time and effort we are willing to spend in trying to either verify your claims, or find your mistake.

I understand that you think it *should* be at the top of all our priority lists (or at least higher than it is now), because you think you are right; you think you *have* found a refutation of GR. You even think it is obvious. But if you are really, really convinced that you have found a refutation of GR, then you should not be depending on us to provide the proof for you. *You* should learn the math yourself; *you* should learn the physics yourself. *You* should be able to write up your arguments and proofs in the standard language of the field. Yes, that's a lot of work; but if you're right, there's probably a Nobel Prize at the end of it. Of course part of the reason I am telling you this is that I think there's a very low probability that you are right; so I think that if you actually do all that work, you will end up just finding out for yourself where you have made mistakes in the scenarios you have proposed here. But I do think the effort is worthwhile quite independently of where you end up with regard to those particular scenarios. Of course that's another heuristic judgment of mine, with which you may disagree.

16. Jul 31, 2012

### pervect

Staff Emeritus
Gauss's law says that we can define the charge enclosed by a surface as the integral of the surface area multiplied by the electric field normal to the surface. There's a very similar theorem in two forms, when the integration process is carried out appropriately.

The easiest way to define the apropriate sense of integration is to use the idea of counting field lines. You can also envision it as a more traditional integral carried out by dividing the whole area into a large number of small pieces, by insisting that you use observers with unit diagonal metrics - i.e. Lorentzian observers - to carry out the integration of each piece.

Consider a space-like sphere surrounding a black hole. This can easily be visualized for any point outside the event horizon by a sphere of constant schwarzschild radius r. The surface area of said 3-sphere will be 4 pi r^2, by he definition of the Schwarzschild r coordinate.

The event horizon isn't space-like , but null. But you can't have a static observer at the event horizon anyway, so it makes no sense to ask what the E-field would be there.

So it is necessary and sufficient that (4/3 pi r^2) * E_normal = constant for charge to be conserved. Which is in fact the prediction of the RN metric when you convert the metric using the usual transformation rules to that of a locally Lorentzian observer.

Note that this is a different behavior than what a local measure of gravitation will give. The integral of local normal force * area (defined in the special way we did above) isn't constant for gravity.

Last edited: Aug 1, 2012
17. Jul 31, 2012

### Staff: Mentor

This should be "2-sphere" and surface area 4 pi r^2, correct? (Vs. "3-sphere" and 4/3)

This was my thought, too, in the other thread that Q-reeus linked to; it's why I later said that the calculation in the specific post he linked to didn't look right to me, because it didn't match other things. The calculation was:

$$Q = \frac{1}{4 \pi} \int{F_{ab} n^{a} u^{b} dS}$$

where $dS$ is the surface element of a 2-sphere, $n^{a}$ is the outward-pointing normal to the 2-sphere, and $u^{b}$ is the 4-velocity of a static observer--or, more relevant for this calculation, it's the 4-velocity of an observer whose worldline is orthogonal to the spacelike hypersurface in which the 2-sphere lies. Since none of the quantities in the integrand depend on the angular coordinates, the integral just gives a factor of $4 \pi r^2$ and we end up with

$$Q(r) = r^2 F_{ab} n^{a} u^{b}$$

I inserted $u^{b} = (1, 0, 0, 0)$, $n^{a} = (0, \sqrt{1 - 2M / r + Q^2 / r^2}, 0, 0)$, and $F_{10} = Q / r^{2}$ to obtain

$$Q(r) = Q \sqrt{1 - \frac{2M}{r} + \frac{Q^2}{r^2}}$$

which, as I noted in that old post, could be interpreted as the "charge at radius r" being "redshifted" relative to the "charge at infinity". However, by your argument, we should get $Q(r) = Q$ for all r, which would require $u^{b} = (1 / \sqrt{1 - 2M /r + Q^2 / r^2}, 0, 0, 0)$; i.e., the timelike vector would have to be normalized using the metric to be a unit vector. The latter seems more physically reasonable to me, but I haven't been able to convince myself definitely one way or the other, if I don't use the fact that I already know the answer ought to be constant, Q(r) = Q at all r.

18. Jul 31, 2012

### PAllen

Pervect did say:

" You can also envision it as a more traditional integral carried out by dividing the whole area into a large number of small pieces, by insisting that you use observers with unit diagonal metrics - i.e. Lorentzian observers - to carry out the integration of each piece."

This alone would rule out ub being (1,0,0,0) in RN coordinates. Also, I have never seen this symbol used for anything other than a timelike unit vector.

Last edited: Jul 31, 2012
19. Aug 1, 2012

### pervect

Staff Emeritus
Yes, I wanted the surface of the 3 sphere, which is a 2 sphere, and it's obviously 4 pi r^2

As far as your calculation goes, I haven't looked it over in great detail, but if u^b = (1,0,0,0) g_00 u^b u^b is not equal to 1, so it's not a unit length vector, because g_00 is not unity. At least if you're using the coordinates I think you are (Schwarzschild).

A four-velocity should have unit length, it's only (1,0,0,0) if g_00 = 1.

20. Aug 1, 2012

### Staff: Mentor

Ok, good.

Yes, I'm using Schwarzschild coordinates. I agree the u^b vector I wrote down is not a timelike unit vector (and so I shouldn't have called it a "4-velocity", as PAllen pointed out); however, it is the timelike Killing vector, correct? So the question is, which is appropriate for the formula I wrote down: the timelike unit vector (in which case the integral gives Q at any radius r), or the timelike Killing vector (in which case the integral gives Q(r), which includes a "redshift factor" at any finite r)?


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# Schumann's Resonance

The Schumann resonance (SR) is a set of spectrum peaks in the extremely low frequency (ELF) portion of the Earth's electromagnetic field spectrum. Schumann resonances are global electromagnetic resonances, excited by lightning discharges in the cavity formed by the Earth surface and the ionosphere.

## Description

This global electromagnetic resonance phenomenon is named after physicist Winfried Otto Schumann who predicted it mathematically in 1952. Schumann resonance occurs because the space between the surface of the Earth and the conductive ionosphere acts as a waveguide. The limited dimensions of the Earth cause this waveguide to act as a resonant cavity for electromagnetic waves in the ELF band. The cavity is naturally excited by energy from lightning strikes. Schumann resonances are observed in the power spectra of the natural electromagnetic background noise, as separate peaks at extremely low frequencies (ELF) around 8, 14, 20, 26 and 32 Hz.

The fundamental mode of the Schumann resonance is a standing wave in the Earth-ionosphere cavity with a wavelength equal to the circumference of the Earth. This lowest-frequency (and highest-intensity) mode of the Schumann resonance occurs at a frequency of approximately 7.8 Hz. The ninth overtone lies at approximately 60 Hz and thus the cavity is also driven by the North American power grid. Detectable overtones extend upwards into the kilohertz range.

Schumann resonances are used to track global lightning activity. Owing to the connection between lightning activity and the Earth's climate they can also be used to monitor global temperature variations and variations of upper water vapor. Extraterrestrial lightning may also be detected and studied with Schumann resonances. Schumann resonance has been used for research and monitoring of the lower ionosphere on Earth and was suggested for exploration of lower ionosphere parameters on celestial bodies. Schumann resonances can be used to track geomagnetic and ionospheric disturbances. More recently, Schumann resonances have been used for monitoring transient luminous events – sprites, elves, jets, and other upper atmospheric lightning. A new field of interest using Schumann resonances is related to short-term earthquake prediction. Schumann resonances have gone beyond the boundaries of physics, invading medicine, raising interest in artists and musicians, and gaining interest from fringe fields such as psychobiology.

## History

The first suggestion that an ionosphere existed, capable of trapping electromagnetic waves, was made by Heaviside and Kennelly in 1902 [1] [2]. It took another twenty years before Edward Appleton and Barnett in 1925 [3], were able to prove experimentally the existence of the ionosphere. However, even prior to this, the first documented observations of global electromagnetic resonances were made by Nikola Tesla in 1905 and formed the basis for his scheme for wireless energy transmission [4]. Although some of the most important mathematical tools for dealing with spherical waveguides were developed by Watson in 1918 [5], it was Winfried Otto Schumann who first studied the theoretical aspects of the global resonances of the earth-ionosphere waveguide system, known today as the Schumann resonances. In 1952-1954 Schumann, together with Köning, attempted to measure the resonant frequencies [6] [7] [8] [9] . However, it was not until measurements made by Balser and Wagner in 1960-1963 [10] [11] [12] [13] [14] that adequate analysis techniques were available to extract the resonance information from the background noise. Since then there has been an increasing interest in Schumann resonances in a wide variety of fields.

## Basic Theory

Lightning discharges are considered as the primary natural source of Schumann resonances. Lightning channels behave like a huge antenna which radiates electromagnetic energy as impulsive signals at frequencies below about 100 kHz [15]. These signals are very weak, but the earth-ionosphere waveguide behaves like a resonator at ELF frequencies and amplifies the spectral signals from lightning at the resonance frequencies [15].

In an ideal cavity, the resonant frequency of the $n$-th mode $f_{n}$ is determined by the Earth radius $a$ and the speed of light $c$ [6].

$f_{n} =\frac{c}{2 \pi a}\sqrt{n(n+1)}$

The real Earth-ionosphere waveguide is not a perfect electromagnetic cavity. Losses due to finite ionosphere electrical conductivity make the system resonate at lower frequencies than would be expected in an ideal case, and the observed peaks are wide. In addition there are a number of horizontal asymmetries – day-night transition, latitudinal changes in the Earth magnetic field, sudden ionospheric disturbances, polar cap absorption, etc. that complicate the Schumann resonance power spectra.

## Measurements

Today Schumann resonances are recorded by many stations around the world. The electromagnetic sensors used to measure Schumann resonances consist of two horizontal antennas for receiving the magnetic field in the north-south and the east-west direction and one vertical antenna for observing the vertical electric field. Since Schumann resonance frequencies are extremely low, practical antennas would have to be hundreds of kilometers long. In addition, the Schumann resonance electric field is much smaller than the static electric field in the atmosphere and the Schumann resonance magnetic field is orders of magnitude smaller than the Earth magnetic field [16] . Therefore, special receivers and antennas are needed to measure Schumann resonances. The electric component is commonly measured with a ball antenna, suggested by Ogawa et al. in 1966 [17] , connected to a high-impedance amplifier. The magnetic field is measured with magnetic induction coils consisting of tens of thousands of turns around material with very high magnetic permeability.

## Applications

### Global lightning activity

From the very beginning of Schumann resonance studies, they were used to monitor global lightning activity by tracking changes in Schumann resonance field intensities. At any given time there are about 2000 thunderstorms around the globe [18]. Producing ~50 lightning events per second [19], these thunderstorms create the background Schumann resonance signal.

Determining the spatial lightning distribution from Schumann resonance records is a complex problem: in order to properly estimate the lightning intensity from Schumann resonance records it is necessary to account for the distance to lightning sources. The common approach is to make a preliminary assumption on the spatial lightning distribution, basing on the known properties of lightning climatology. An alternative approach is placing the receiver at the North or South Pole, which remain approximately equidistant from the main thunderstorm centers during the day [20]. A distinct method, not requiring preliminary assumptions on the lightning distribution [21] is based on the decomposition of the average background Schumann resonance spectra, utilizing ratios between the average electric and magnetic spectra and between their linear combinations.

#### Diurnal variations

The best documented and the most debated features of the Schumann resonance phenomenon are the diurnal variations of the background Schumann resonance power spectrum.

A characteristic Schumann resonance diurnal record reflects the known properties of global lightning activity. The vertical electric field, which is equally sensitive in all directions and therefore measures the global lightning, shows three dominant maxima, associated with the three “hot spots” of planetary lightning activity: 9 UT (Universal Time) peak, linked to the increased thunderstorm activity from south-east Asia; 14 UT peak associated with the peak in African lightning activity; and the 20 UT peak resulting for the increase in lightning activity in South America. The time and amplitude of the peaks vary throughout the year, reflecting the seasonal changes in lightning activity.

##### ”Chimney” ranking

In general, the African peak is the strongest, reflecting the major contribution of the African “chimney” to the global lightning activity. The ranking of the two other peaks – Asian and American is the subject of a vigorous dispute among Schumann resonance scientists. Experimental Schumann resonance data show a greater contribution from Asia than from South America. This contradicts optical satellite and climatological lightning data that show the South American thunderstorm center stronger than the Asian center [19]. The reason for such disparity remains unclear. Williams and Sátori [22] suggest that in order to obtain “correct” Asia-America chimney ranking, it is necessary to remove the influence of the day/night variations in the ionospheric conductivity (day-night asymmetry influence) from the Schumann resonance records. On the other hand, such “corrected” records presented in the work by Sátori et al. [23] show that even after the removal of the day-night asymmetry influence from Schumann resonance records, the Asian contribution remains greater than American. Similar results were obtained by Pechony et al. [24] who calculated Schumann resonance fields from satellite lightning data. Both simulations – those neglecting the day-night asymmetry, and those taking this asymmetry into account, showed same Asia-America chimney ranking. As for today, the reason for the “invert” ranking of Asia and America chimneys in Schumann resonance records remains unclear and the subject requires further, targeted research.

##### Influence of the day-night asymmetry

In the early literature the observed diurnal variations of Schumann resonance power were explained by the variations in the source-receiver (lightning-observer) geometry [10]. It was concluded that no particular systematic variations of the ionosphere (which serves as the upper waveguide boundary) are needed to explain these variations [25]. Subsequent theoretical studies supported the early estimations of the small influence of the ionosphere day-night asymmetry (difference between day-side and night-side ionosphere conductivity) on the observed variations in Schumann resonance field intensities [26].

The interest in the influence of the day-night asymmetry in the ionosphere conductivity on Schumann resonances gained a new strength in the 1990s, after publication of a work by Sentman and Fraser [27]. Sentman and Fraser developed a technique to separate the global and the local contributions to the observed field power variations using records obtained simultaneously at two stations. Sentman and Fraser interpreted the local contribution as ionosphere height variation. Their work convinced many scientists in the importance of the ionospheric day-night asymmetry and inspired numerous experimental studies. However recently it was shown that results obtained by Sentman and Fraser can be simulated with a uniform model (without taking into account ionosphere day-night variation) and therefore cannot be interpreted in terms of ionosphere height variation [28] .

Schumann resonance amplitude records show significant diurnal and seasonal variations which in general coincide in time with the times of the day-night transition (the terminator). This time-matching seem to support the suggestion of a significant influence of the day-night ionosphere asymmetry on Schumann resonance amplitudes. There are records showing almost clock-like accuracy of the diurnal amplitude changes [23]. On the other hand there are numerous days when Schumann Resonance amplitudes do not increase at sunrise or do not decrease at sunset. There are studies showing that the general behavior of Schumann resonance amplitude records can be recreated from diurnal and seasonal thunderstorm migration, without invoking ionospheric variations [26] [24]. Two recent independent theoretical studies have shown that the variations in Schumann resonance power related to the day-night transition are much smaller than those associated with the peaks of the global lightning activity, and therefore the global lightning activity plays a more important role in the variation of the Schumann resonance power [29] [24].

The relative importance of the day-night asymmetry in Schumann resonance amplitude records is still debated. Successful monitoring of global thunderstorm activity with Schumann resonances relies on the proper interpretation of experimental data. It is therefore vital to understand and correctly interpret the major features of Schumann resonance field power variations.

####  The “inverse problem”

One of the interesting problems in Schumann resonances studies is determining the lightning source characteristics (the “inverse problem”). Temporally resolving each individual flash is impossible,Template:Huh but there are intense ELF transient events, also named ‘‘Q bursts’’. Q-bursts are triggered by intense lightning strikes, associated with a large charge transfer and often high peak current [17]. Q-bursts can exceed the amplitude of the background signal level by a factor of 10 and appear with intervals of ~10sec [21], which allows to consider them as isolated events and determine the source lightning location. The source location is determined with either multi-station or single-station techniques. The multi-station techniques are more accurate, but require more complicated and expensive facilities.

###  Transient luminous events research

It is now believed that many of the Schumann resonances transients (Q bursts) are related to the transient luminous events (TLEs). In 1995 Boccippio et al. [30] suggested that sprites, the most common TLE, are produced by positive cloud-to-ground lightning occurring in the stratiform region of a thunderstorm system, and are accompanied by Q-burst in the Schumann resonances band. Recent observations [30] [31] reveal that occurrences of sprites and Q bursts are highly correlated and Schumann resonances data can possibly be used to estimate the global occurrence rate of sprites [32].

###  Climate change research

Global climate change is the subject of intense debate and concern. One of the important aspects in understanding global climate change is the development of tools and techniques that would allow continuous and long-term monitoring of processes affecting the global climate. Schumann resonances are one of the very few tools that can provide such global information reliably and cheaply.

####  Global temperature

Williams [1992] [33] suggested that global temperature may be monitored with the Schumann resonances. The link between Schumann resonance and temperature is lightning flash rate, which increases nonlinearly with temperature [33]. The nonlinearity of the lightning-to-temperature relation provides a natural amplifier of the temperature changes and makes Schumann resonance a sensitive “thermometer”. Moreover, the ice particles that are believed to participate in the electrification processes which result in a lightning discharge [34] have an important role in the radiative feedback effects that influence the atmosphere temperature. Schumann resonances may therefore help us to understand these feedback effects.

####  Upper tropospheric water vapor

Tropospheric water vapor is a key element of the Earth’s climate, which has direct effects as a greenhouse gas, as well as indirect effect through interaction with clouds, aerosols and tropospheric chemistry. Upper tropospheric water vapor (UTWV) has a much greater impact on the greenhouse effect than water vapor in the lower atmosphere [35], but whether this impact is a positive, or a negative feedback is still uncertain [36]. The main challenge in addressing this question is the difficulty in monitoring UTWV globally over long timescales. Continental deep-convective thunderstorms produce most of the lightning discharges on Earth. In addition, they transport large amount of water vapor into the upper troposphere, dominating the variations of global UTWV. Price [2000] [37] suggested that changes in the UTWV can be derived from records of Schumann Resonances.

###  Extraterrestrial lightning

Existence of Schumann resonances is conditioned primarily by two factors: 1) presence of a substantial ionosphere with electric conductivity increasing with height from low values near the surface (or a high-conductivity layer, in case of gaseous planets); 2) source of excitation of electromagnetic waves in the ELF range. In Solar System there are five candidates for Schumann resonance detection: Venus, Mars, Jupiter, Saturn and its moon Titan.

Modeling Schumann resonances on the planets and moons of the Solar System is complicated by the lack of knowledge of the waveguide parameters, and today there is no possibility to validate the results. Nevertheless, theoretical results aid to estimate the possibility of detecting Schumann resonances on a planet.

The strongest evidence for lightning on Venus comes from the impulsive electromagnetic waves detected by Venera 11 and 12 landers. Schumann resonances on Venus were studied by Nickolaenko and Rabinowicz [1982] [38] and Pechony and Price [2004] [39]. Both studies yielded very close results, indicating that Schumann resonances should be easily detectable on this planet, should any sensor survive long enough in the harsh Venusian environment.

On Mars lightning activity has not been detected, but charge separation and lightning strokes are considered possible in the Martian dust storms [40] [41] . Martian global resonances were modeled by Sukhorukov [1991] [42], Pechony and Price [2004] [39] and Molina-Cuberos et al. [2006] [43] . The results of the three studies are somewhat different, but it seems that at least the first two Schumann resonance modes should be detectable.

It was long suggested that lightning dischargers take place on Titan [44] , but recent data from Cassini-Huygens seems to indicate that there is no lightning activity on this largest satellite of Saturn. Due to the recent interest in Titan, associated with the Cassini-Huygens mission, its ionosphere is perhaps the most thoroughly modeled today. Schumann resonances on Titan received more attention than on other celestial bodies. Schumann resonances on Titan were studied by Besser et al. [2002] [45] , Morente et al. [2003] [46] , Molina-Cuberos et al. [2004] [47] , Nickolaenko et al. [2003] [48] and Pechony and Price [2004] [39] . It appears that only the first Schumann resonance mode might be detectable on Titan.

Jupiter is the only planet where lightning activity is well established. Existence of lightning activity on this planet was predicted by Bar-Nun [1975] [49] and it is now supported by data from Galileo, Voyagers 1 and 2, Pioneers 10 and 11 and Cassini. Saturn is also expected to have intensive lightning activity, but the three visiting spacecrafts – Pioneer 11 in 1979, Voyager 1 in 1980 and Voyager 2 in 1981, failed to provide any convincing evidence. Even the strong storm monitored on Saturn by the Cassini spacecraft produced no visible lightning flashes. Little is known about the electrical parameters of Jupiter and Saturn interior. Even the question of what should serve as the lower waveguide boundary is a non-trivial one in case of the gaseous planets. There seem to be no works dedicated to Schumann resonances on Saturn. Up to date there was only one attempt to model Schumann resonances on Jupiter [50]. Should someone find a way to hang Schumann resonance sensor in the Jupiter’s atmosphere, the Schumann resonances there should be easily detectable.

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Thread: Complete the square for trigonometry

1. Complete the square for trigonometry

Hi
I am working on a long problem and the solution i am getting differs from the answer in the textbook.
It boils down to completing the square. The method i use works in the first half of the problem but the book gets a different answer in the 2nd half. Basically for the coordinates of the centre of a circle my x coordinate is correct, but my y coordinate is incorrect.
Can somebody complete the square on the following and give me coordinates for the centre of the circle. My answer is (x=0.5, y=-0.5). The book gives the y as -1.

The expression is: 4x^2+4y^2-4x+8y-11=0.

My solution has : y^2+2y becomes y^2+y+0.25 (b=0.5)= (y+0.5)^2-0.25
The book has: y^2+2y becomes (y+1)^2-1

Help appreciated
John

2. Re: Complete the square for trigonometry

Originally Posted by celtic1234
Hi
I am working on a long problem and the solution i am getting differs from the answer in the textbook.
It boils down to completing the square. The method i use works in the first half of the problem but the book gets a different answer in the 2nd half. Basically for the coordinates of the centre of a circle my x coordinate is correct, but my y coordinate is incorrect.
Can somebody complete the square on the following and give me coordinates for the centre of the circle. My answer is (x=0.5, y=-0.5). The book gives the y as -1.

The expression is: 4x^2+4y^2-4x+8y-11=0.

My solution has : y^2+2y becomes y^2+y+0.25 (b=0.5)= (y+0.5)^2-0.25 Mr F says: Expand this out. Do you get y^2 + 2y? The book is correct. Go back and review how to complete the square. Or perhaps, by expanding the book's answer, you will see why it is has to be correct ....

The book has: y^2+2y becomes (y+1)^2-1

Help appreciated
John
..

3. Re: Complete the square for trigonometry

thanks will do

4. Re: Complete the square for trigonometry

Thank you so much-i'm an idiot!!!. Thanks again for your help-found my mistake
John


text:
Trigonometry (11th Edition) Clone

Published by Pearson

Chapter 7 - Applications of Trigonometry and Vectors - Summary Exercises on Applications of Trigonometry and Vectors - Page 345: 10

Answer

There are two possible triangles. The angles of one triangle are as follows: $A = 28^{\circ}10', B = 36^{\circ}10',$ and $C = 115^{\circ}40'$ The lengths of the sides are as follows: $a = 21.2~yd, b = 26.5~yd,$ and $c = 40.5~yd$ The angles of the other possible triangle are as follows: $A = 28^{\circ}10', B = 143^{\circ}50',$ and $C = 8^{\circ}$ The lengths of the sides are as follows: $a = 21.2~yd, b = 26.5~yd,$ and $c = 6.3~yd$

Work Step by Step

Let $a = 21.2~yd$ Let $b = 26.5~yd$ Let angle $A = 28^{\circ}10'$ Let angle $B$ be the angle that subtends the side $b$. We can use the law of sines to find $B$: $\frac{a}{sin~A} = \frac{b}{sin~B}$ $sin~B = \frac{b~sin~A}{a}$ $sin~B = \frac{(26.5)~sin~28^{\circ}10'}{21.2}$ $sin~B = 0.59$ $B = arcsin(0.59)$ $B = 36^{\circ}10'$ We can find angle C: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-28^{\circ}10'-36^{\circ}10'$ $C = 115^{\circ}40'$ We can use the law of sines to find $c$: $\frac{c}{sin~C} = \frac{a}{sin~A}$ $c = \frac{a~sin~C}{sin~A}$ $c = \frac{(21.2~yd)~sin~115^{\circ}40'}{sin~28^{\circ}10'}$ $c = 40.5~yd$ Note that there is another possible value for the angle $B$: $B = 180^{\circ}- 36^{\circ}10' = 143^{\circ}50'$ We can find angle C: $A+B+C = 180^{\circ}$ $C = 180^{\circ}-A-B$ $C = 180^{\circ}-28^{\circ}10'-143^{\circ}50'$ $C = 8^{\circ}$ We can use the law of sines to find $c$: $\frac{c}{sin~C} = \frac{a}{sin~A}$ $c = \frac{a~sin~C}{sin~A}$ $c = \frac{(21.2~yd)~sin~8^{\circ}}{sin~28^{\circ}10'}$ $c = 6.3~yd$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.


text:
+0

# Please help solve

0
75
4

A man on the 14th floor of a building sees a bucket (dropped by a window washer) pass his window and notes that it hits the ground 1.6 s later. Assuming a floor is 13 ft high (and neglecting air friction), from which floor was the bucket dropped?

Guest Jul 22, 2017
Sort:

### 3+0 Answers

#1
+650
+1

The unknown starting height of the bucket’s fall is (h). The bucket falls (16t^2) feet in (t) seconds.  The man observes the bucket passing floor 14 at unknown time (t), and it continues to fall for (13ft*14) = 182ft in 1.6s

$$\begin{array}{|lll|} \hline &(h-16t^2) - (h-16(t+1.6)^2) &=& 182ft&\\ &-16t^2+16 (t+1.6)^2&=&182\\ &51.2t+40.96&=&182\\ &51.2t&=&141.04\\ &t&=&2.7546875\\ \text { }\\ &16(2.7546875)^2 &=& 121.4 \text{ft above observation floor. }\\ &\dfrac{121.4}{13}& =& 9.3 \text{ Floors above observation floor. }\\ &14+9 &=& 23 \text { The bucket fell from the }23^{rd} \text {floor }\\ \hline \end{array}$$

------

Lancelot Link says the best place to watch an opera is from the rafters or a chandelier, but you need a good pair of opera glasses to read their lips. It helps to know Italian, too.

GingerAle  Jul 23, 2017
edited by Melody  Jul 23, 2017
#2
0

Non-sequitur

/ˈnɒn ˈsɛkwɪtə/

noun1.

a statement having little or no relevance to what preceded it.

Guest Jul 23, 2017
#3
+650
+1

Well, Mr. BB I am impressed that you know a Latin phrase.  Are you fluent in pîg Latin?  I bet you are!  I know a Latin phrase that instantly brings you to anyone’s mind.  Persona non grata

Like usual, and seemingly always, Mr. BB, you are wrong. Lancelot Link’s metaphorical comment definitely relates by a logical sequence to the kinematic equation it follows. I can explain this, but (again) we are back to wasting time and annoying the pîg.

In this in this instance, I will try, because I love Lancelot, and truly appreciate his humor and wisdom, and the unique way he blends them to convey knowledge and information.  I don’t expect you to understand this or acknowledge it if you do. But I’m sure there a few on here who will understand and appreciate Lancelot’s teaching methods.

Lancelot attached this comment to the end of a systematic series of related equations expressed in the form of hypotheticals to demonstrate macro logic from micro observations.  The hypothetical depicted above is not likely to be a casual observational experience in the real world. For this experiment to have any accuracy, one needs careful observation and high-precision timing devices.  In a related hypothetical, the observer notes the travel time from the top to the bottom of a window and derives the starting height of the object. Again, no one in the real world can casually measure this objects travel time in tenths of seconds.

Lancelot’s opera metaphor extends to general observations. When one has a macro view, a tool is needed for the micro view. Here, this tool is opera glasses.   To understand the (Italian) opera one needs to understand the Italian language.   This is also true for physics. One needs to understand the language of physics: how gravity, velocity, distance, and time relate to each other.

Plus, the thought of a chimp watching an opera, while hanging from a rafter or sitting in a chandelier, is just plain funny to most humans and we genetically enhanced chimps.

----

One final note:  The phrases Non sequitur and Non- sequitur are different.  When used as a noun, it should not have a hyphen. Using a hyphen makes it a phrasal adjective. This is a major change in the meaning.  If you wish to point out potential esoteric communication errors, you really should have higher standards.

GingerAle  Jul 24, 2017

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text:
# Does the Banach algebra $L^1(\mathbb{R})$ have zero divisors?

Assume that the functions $f,g: \mathbb{R}\rightarrow \mathbb{R}$ are integrable and equal to zero on $(-\infty,0)$, (i.e $f,g \in L^+$). Then by Titchmarsh's theorem:

$f*g$ is zero almost everywhere iff $f$ or $g$ is zero almost everywhere.

Hence the Banach subalgebra $L^+$ of $L^1(\mathbb{R})$ has no zero divisors.

Is the same true for arbitrary integrable $f$ and $g$ on $\mathbb{R}$ ?

-
 See also this related MO thread – t.b. Dec 11 '11 at 23:04

To make Yemon Choi's idea more explicit:

The bump function

$$b(x):=\cases{\exp{-1\over 1-x^2} & \bigl(|x|<1\bigr) \cr 0 & \bigl(|x|\geq 1\bigr)\cr}$$

is $C^\infty$ and has compact support $[-1,1]$; whence it is in the Schwartz space ${\cal S}$ and is the inverse transform of its Fourier transform $\hat b\in {\cal S}$. The same is true for the functions $f_1(x):=b(x-1)$ and $f_2(x):=b(x+1)$ whose product is identically zero. But up to a constant factor this product is equal to $\hat f_1\ *\ \hat f_2$.

-
 Thanks a lot for answer. – L.T Dec 13 '11 at 13:22

The MO thread that t.b. links to doesn't seem to quite answer the original question, which if I've understood it correctly is:

do there exist $f,g\in L^1({\mathbb R})$, neither of which is a.e. zero, such that $f*g$ is a.e. zero?

I am pretty sure that the answer is yes, because the answer to the analogous question for $\ell^1({\mathbb Z})$ is yes. To see this, work on the circle and consider the simplest function $h_1$ which is piecewise linear on $[-\pi,\pi]$, is zero on the intervals $[-\pi,-\delta]$ and $[\delta,\pi]$ for some fixed $\delta$, and which is $1$ at $0$. (Triangular spike.) Then viewed as a continuous function on ${\mathbb R}/2\pi{\mathbb Z}={\mathbb T}$, the Fourier series of $h_1$ is absolutely summable. Now let $h_2(x) = h_1(x+\pi)$, clearly $h_1\cdot h_2=0$, which means that (FS of $h_1$) $*$ (FS of $h_2$) $=0$.

Now your question takes place on the line, not the circle, but my feeling is that the same construction should work. That is, let $h_1$ and $h_2$ be the same functions as before, but regard them as functions on ${\mathbb R}$. Then both should be the Fourier transforms of integrable functions $f_1,f_2\in L^1{\mathbb R})$, and once again $f_1*f_2=0$ since $h_1\cdot h_2=0$.

(Actually, thinking about it a bit more, I "know" that the answer to your question must be yes, because it is equivalent to the question "does the Fourier algebra $A({\mathbb R})$ contain non-trivial zero divisors?", and it's known that $A({\mathbb R})$ is a regular Banach algebra of functions, meaning that one can find functions which are zero on any prescribed closed subset $F$ and 1 on any prescribed compact set disjoint from $F$; pointwise product in $A({\mathbb R})$ corresponds to convolution in $L^1({\mathbb R})$.)

-
 Thanks it is helpful. – L.T Dec 12 '11 at 7:58


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text: Localized molecular orbitals ... each with electron density equally distributed among the two O-H bonds. 3. This video contains a discussion and a few examples demonstrating the concept of localized vs delocalized ... Chemical Bonding Difference ... bonds when they ... they become "delocalised". Are electrons in non-polar covalent molecules localised or delocalised amd mobile? Best Answer: a localized pi bond is shared between two atoms in a molecule example CH2=CHCH2CH2CH3. Models of chemical bonding ... of the resulting compounds as well as the energies and other properties of the bonds ... energy can be found by difference. What makes the difference between ionic and covalent bonds? The more delocalised electrons the stronger the bond 3. Resonance form I contains 2 localized bonds and 1 delocalized bond. ... What's the difference between sigma/pi bonds and single ... What is a delocalized chemical bond? The answer is that the Pi bonds do not exist as localized bonds that stay See also chemical bond. Models of chemical bonding What holds ... the lattice energy can be found by difference. Again, notice that in step 1 the arrow originates with an unshared electron pair from oxygen and moves towards the positive charge on nitrogen. What is the difference between delocalized and localized bonds? Electrons in SIGMA bonds remain localized between two atoms, ... they become "delocalised". How can you know a localized and delocalized chemical bond? what is the difference between the localised and delocalised electrons , please explain with the help of an example The movement of mobile electrons in chemical ... -bonds between C and N to the carbon atom in ... to a similar system where electrons are localized. The "delocalized" label also applies to bonds and charges. Chapter 9 Theories of Chemical Bonding 9-3 9-3 A covalent bond is the result of the overlap of orbitals on adjacent atoms. ... you can break bonds by hitting and they easily reform . How can you know a localized and delocalized chemical bond? A new $$\pi$$ bond forms between nitrogen and oxygen. Covalent bonds form between non-metal atoms. Ionic Bond: The bond it particularly is created b/w 2 or extra ions. So there cannot be alternating single-double bonds in the ring like Our Lewis did structure showed. A pi bond occurs when two sets of electron orbitals, involved in the bond, overlap. A delocalized bond is a bond that appears in some resonance forms, but not others. ... Chemical bonds are the force of attraction between two atoms. difference between delocalised and localised electrons ... Delocalised electrons keep on moving or are unstable as in metals ... Delocalised over a number of bonds. Otherwise we would end up with a nitrogen with 5 bonds, which is impossible, even if only momentarily. what are localised and delocalised types of bonding explain the difference between them in details 7ylkjepp-Chemistry - TopperLearning.com Covalent bond: this bond is formed after mutual What is a delocalized electron? ... Delocalised electrons are spread across more than one atom. Delocalized bonds and delocalized charges. The crystal chemistry and chemical bonding are discussed. Covalent substances. Analyses of the electronic structures of CaNiGe 2, SrNiGe 2, and SrNiSn 2, with the help of the electron localization function (ELF), indicate the coexistence of localized covalent and delocalized bonding between the metal atoms involved. ... with the chemical symbol Si. Difference between Delocalization is the phenomena by which electrons or bonds are transient in nature. delocalized bonding ... the smaller the energy difference between pi orbitals . The pi bond is only shared by the 1st and 2nd Carbon.


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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeApr 8th 2013

The Idea-section at quasi-Hopf algebra had been confused and wrong. I have removed it and written a new one.

• CommentRowNumber2.
• CommentAuthorzskoda
• CommentTimeApr 8th 2013
• (edited Apr 8th 2013)

Why removing it ? You removed the important and more general viewpoint how it came. The case you describe now in the idea section is just a very special case. Drinfel’d worked in much larger generality of rational CFTs than the more special work of Pasquier Roche Dijkgraaf about at the same time (or a bit later). What was wrong ?

To repeat, Drinfel’d considered the monoidal categories in rational CFTs. Often one had a braided monoidal category structure which could be considered as representations of (possibly weak) quasitriangular Hopf algebras. However, possibly changing the associativity coherences which is allowed in this framework allows for more general twists of Hopf algebras where the axioms involve elements coming from Yoneda when twisting the coherences. Those are quasiHopf algebras. Of course, this procedure can be done more generally also in the case not involving quasitriangular structure, but this is beyond the case of Drinfel’d and beyond the more special case of Dijkgraaf et al.

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeApr 8th 2013

The previous version started out by mixing up quasi-triangular Hopf with quasi-Hopf. Next it claimed that the representations of a Hopf algebra constitute a braided monoidal category, which is wrong. Then it said that “the Hilbert space is a representation of a Hopf algebra”, which I can’t make sense of. All that should stay removed.

But I guess what you want to see re-installed is the sentence

This framework allows for natural class of weakenings by changing associativity coherence, what amounts to the twist by a nonabelian bialgebra 3-cocycle and yields the notion of a quasi-Hopf algebra as introduced in…

I am now putting that back in.

• CommentRowNumber4.
• CommentAuthorzskoda
• CommentTimeApr 8th 2013
• (edited Apr 8th 2013)

I edited 1 above – it did not mix quasitriangular – this was the case in original motivation for quasiHopf algebras, what includes your case! I think this also needs reinsertion in some form.

“the Hilbert space is a representation of a Hopf algebra”, which I can’t make sense of.

The Hilbert space of the theory is a representation – it belongs to the braided monoidal category mentioned. For example for gauged Wess-Zumino model for su(n) the detailed structure of this representation is investigated in a series of works of Hadjiivanov, Furlan, Stanev and Todorov, and my undergraduate diploma was studying some aspects of this.

• CommentRowNumber5.
• CommentAuthorUrs
• CommentTimeApr 8th 2013
• (edited Apr 8th 2013)

Yeah, I guess the quasi-triangular Hopf/quasi-Hopf mixup I introduced when putting cross-links too hastily.

I have now edited quasi-Hopf algebra. Check if you agree.

What do you mean by “I edied 1 above”? You edited where?

• CommentRowNumber6.
• CommentAuthorzskoda
• CommentTimeApr 8th 2013
• (edited Apr 8th 2013)

I edited my 1st nForum comment above i.e. 2 (you respond so quickly, that you posted 3 and 5 when I was still editing additions to 2 and 4).

• CommentRowNumber7.
• CommentAuthorzskoda
• CommentTimeApr 8th 2013

I edited now the entry, do you agree now ?

• CommentRowNumber8.
• CommentAuthorUrs
• CommentTimeApr 8th 2013

Thanks! Good. I have slightly edited a bit more (mostly putting in more hyperlinks.)

• CommentRowNumber9.
• CommentAuthorzskoda
• CommentTimeSep 18th 2019

More references, including 2019 monograph by Bulacu et al.

• CommentRowNumber10.
• CommentAuthorzskoda
• CommentTimeOct 3rd 2019

I wrote the definition of a quasibialgebra in much greater detail and the connection to MacLane’s pentagon.

• CommentRowNumber11.
• CommentAuthorzskoda
• CommentTimeOct 3rd 2019

Added a section on twisting of quasibialgebras by $2$-cochains.

• CommentRowNumber12.
• CommentAuthorzskoda
• CommentTimeOct 3rd 2019

It would be interesting to find a higher-categorical explanation of the nonabelian cocycles and cohomology for bialgebras as defined by Drinfeld and Majid. It is a nonabelian cohomology different from one for groups and it specialized to Abelian cohomology of groups and Lie algebras in particular cases. There are nonabelian 3-cocycles which are related to other nonabelian 3-cocycles in examples like quasi-Poisson structures, nonabelian group 3-cohomology, Pasquier-Roche-Dijkgraaf-Witten model etc. Recent nonassociative geometries coming from fluxes in M-theory and nongeometric string backgrounds as studied in these terms more recently by Luest, Szabo and others are also stated via quasi-Hopf algebras and Drinfeld-Majid 3-cocycles/associators.

1. Fixed a typo about the left and right counit corollaries of (iii). This now accurately reflects (3.1.11) from “Quasi-Hopf Algebras: A Categorical Approach” by Bulacu, et al.

Sean Sanford

2. Just fixing minor typos

Olivia Borghi

3. Just fixing minor typos

Olivia Borghi

4. The “suitable notion of antipode” is actually pretty complicated so I wanted to illuminate that.

Olivia Borghi

• CommentRowNumber17.
• CommentAuthorUrs
• CommentTimeJan 23rd 2021

Thanks!

I have just reinstantiated the term “a suitable kind of antipode” with the formula, to orient the reader. Okay?


text:
# Reynolds number and inertial force

The Reynolds number is defined as the ratio of ´inertial´ forces to viscous forces.

$$Re = \frac{\text{Inertial Forces}}{\text{Viscous Forces}}$$

Now, viscous forces make sense to me. They are frictional shear forces that come about due to the relative motion of the different layers in a flowing fluid, resulting in different amount of friction, hence, different viscosity values.

However, I am not really sure how to think about the 'inertial force'. This, to me, is somewhat of a dynamic effect since large Re numbers indicate turbulence in most cases, where there is a lot of motion, vortices and eddies. But what exactly is the inertial force and how can it be explained physically?

-

Inertial force, as the name implies is the force due to the momentum of the fluid. This is usually expressed in the momentum equation by the term $(\rho v)v$. So, the denser a fluid and is, and the higher its velocity, the more momentum (inertia) it has. As in classical mechanics, a force that can counteract or counterbalance this inertial force is the force of friction (shear stress).In the case of fluid flow, this is represented by Newtons law, $\tau_x = \mu \frac{dv}{dy}$. This is only dependent on the viscosity and gradient of velocity. Then, $Re = \frac{\rho v L}{\mu}$, is a measure of which force dominates for a particular flow.

The inertial forces are what gives rise to the dynamic pressure. Another way to look at the Reynolds Number is by the ratio of dynamic pressure $\rho u^2$ and shearing stress $μ v/ L$ and can be expressed as $$Re =\frac{\rho u^2} {μ v/ L} = \frac{ u L} {\nu}$$

At very high Reynolds numbers, the motion of the fluid causes eddies to form and give rise to the phenomena of turbulence.

-
I think your answer can be improved by putting more emphasis on the dynamic pressure to explain it in the context of forces. –  Bernhard Oct 8 '13 at 20:40
Thanks for you suggestion. –  mcodesmart Oct 8 '13 at 20:54
I'm not sure if I'm correct, but in the last latex equation, shouldn't it be vL/(v kinematic viscosity) rather than uL/v? Or could you explain why its written as rho.v.L/mu before? –  midnightBlue Nov 25 '14 at 20:43

The ratio of the inertial force to the viscous forces of the fluid is known as the Reynolds number. Now the viscous forces are the forces due to to the friction between the the layers of any real fluid. In fluid mechanics we take the fluid as in the continuum condition, which means fluid particles are very closely packed so necessarily there is friction between layers of fluid. The inertial forces are the forces which are due to due the particles of fluid resisting any change in momentum.

-

Newton’s second law can be written as m x a = f1 + f2 + … + fn. The term m x a having dimensions of force and being proportional to the mass of the body, which is a measure of its inertia, is often named inertia force. So, according to the previous equation, the inertia force is just the resultant force acting on the body under analysis. In Fluid Mechanics it is advantageous to use mass per unit volume of the body (fluid in this case), that is its density, so that Newton’s law (or rather, the Navier-Stokes equation) is written with the terms having dimensions of force per unit volume of fluid. When fluids flow, different types of forces act on the fluid. These are represented in the previous equation by f1, f2, … , fn. Suppose viscous forces are represented by f2. Back to the original question, the Reynolds number (Re) associated with the fluid flow would be, in this case, m x a / f2. So, in fluid flow, Re is a measure of the ratio between the resultant force (or inertia force) and viscous force acting on the fluid. Notice that the viscous force is part of the inertia force. In other words, Re is the ratio between the resultant force acting on the fluid and one of its components. Our intuition regarding effects of inertia forces is quite good because in daily life our muscles overcame inertia of static bodies all the time. Inertia of moving bodies are also easily perceptible when our velocity (for instance in a moving car) changes magnitude or direction, that is, when we are accelerated relative to the ground (notice that a is a factor in the inertia force). Effects of viscous forces are much more subtle and require specific experiments.

-
You can use latex commands to improve your answer. –  Ali Jun 13 '14 at 3:30


text:
# Thread: Comparison of ODE solvers

1. ## [SOLVED] Comparison of ODE solvers

Hi,

I was reading about Runge-Kutta methods and wanted to do a quick comparison of Euler's method, fourth order Runge-Kutta and MATLAB's ODE45.

I used the following differential equation:
yprime = @(t,y) vanderpoldemo(t,y,1);

I was expecting the Runge-Kutta to be closer to ODE45 than Euler, but for some reason it is not. As can be seen from the plots, the Runge-Kutta solution is further away form ODE45 (which I'm assuming is the most correct one) than Euler's.

By changing the constant term in Runge-Kutta from (1/6) to about (1/3.8) I get quite close to ODE45. Here's a copy-paste of Runge-Kutta from MATLAB so you know which constant I'm talking about;

y(:,i+1) = y(:,i)+(1/6)*h*(k1+2*k2+2*k3+k4);

Should I just assume that my implementation of Runge-Kutta is wrong? I can't find any errors.. Here's the implementation:

Code:
function [t,y] = runge_kutta_4order(yprime, tspan, y0, h)

t0 = tspan(1); tfinal = tspan(end);

% set up the t values at which we will approximate the solution
t = [t0:h:tfinal]';

% include tfinal even if h does not evenly divide tfinal-t0
if t(end)~=tfinal, t=[t tfinal]; end

m = length(t);
y = [y0 zeros(length(y0), length(t)-1)];

for i=1:(m-1)
k1 = feval(yprime,t(i),y(:,i));
k2 = feval(yprime,t(i)+0.5*h, y(:,i)+0.5*h.*k1);
k3 = feval(yprime,t(i)+0.5*h, y(:,i)*0.5*h.*k2);
k4 = feval(yprime,t(i)+h, y(:,i)+h.*k3);

y(:,i+1) = y(:,i)+(1/6)*h*(k1+2*k2+2*k3+k4);
end

2. Hi there,

ODE45 is, essentially, an RK4 method, and so you should see almost no error between the results that ode45 give, and the results that RK4 give.

I just ran my own code, with my own rk4 method, and the from rk4 and ode45 are identical, almost no difference whatsoever.

With that said, I can't see any particular problem with your implementation.

So, what I'll do is give you my code, and allow you to spot any errors yourself.

It could be that your stepsize is too large? What stepsize are you using?

Here is my code. I haven't commented it, but hopefully it is self-explanatory.

Code:
clear
clc
StartTime = 0;
EndTime = 20;
Stepsize = 0.01;
x = [2,2];
xdot = [0,0];
tout=zeros(1,(EndTime/Stepsize)+1);
xout=zeros((EndTime/Stepsize)+1, length(x));
xdout = xout;
i=0;

for time = StartTime:Stepsize:EndTime,

i          = i+1  ;
tout(i)    = time ;
xout(i,:)  = x    ;
xdout(i,:) = xdot ;

xdot = vanderpoldemo(time,x,1);
x = rk4('vanderpoldemo',Stepsize,time,x);
end

[TOUT,YOUT] = ode45(@vanderpoldemo,[StartTime EndTime],[2;2]);

figure(1)
hold on
plot(tout,xout(:,1),'r--')
plot(TOUT,YOUT(:,1),'b--')
title('Comparison of ODE Solvers')
xlabel('time (s)')
ylabel('Function value')
legend('Runge-Kutta 4','ODE45')

figure(2)
hold on
plot(tout,xout(:,2),'r--')
plot(TOUT,YOUT(:,2),'b--')
title('Comparison of ODE Solvers')
xlabel('time (s)')
ylabel('Function value')
legend('Runge-Kutta 4','ODE45')
Code:
function x = rk4(model,h,t,x)
k1 = feval(model,t,x,1)        ;
k2 = feval(model,t+0.5*h,x+h*k1'/2,1) ;
k3 = feval(model,t+0.5*h,x+h*k2'/2,1) ;
k4 = feval(model,t+h,x+h*k3',1)   ;
x = x + h*(k1' + 2*k2' + 2*k3' + k4')/6    ;
end
Code:
function dydt = vanderpoldemo(t,y,Mu)
Mu = 1;
dydt = [y(2); Mu*(1-y(1)^2)*y(2)-y(1)];
Let me know how you get on.

3. Thank you so much. After looking at your code I noticed that I had a * instead of a + at one place in my implementation of rk4. I must be going blind.
Again, thanks!


text:
# How to deal with exercises with no solutions given?

Probably most people will acknowledge the importance of doing exercises when reading a mathematical textbook. Here I am talking about a textbook of similar level as those ones listed in GTM. However, there is usually no solution avalible for those exercises in most mathematical books. When one tries to do exercises without solutions, it seems to me that two problems may occur:

1. One cannot easily verify if his/her answer is correct. For example, if an exercise asks the reader to find all groups of order $52$, without a solution how can one make sure that s/he has listed all the possible answers without mistake? Even for those "show that" questions, it is still possible that the reader gives a false proof and contains some (possibly non-obvious) errors.

2. For those questions one just cannot solve after working on it for a long time, how can s/he find out the solution? I know that one can ask the question here or to a teacher or a fellow student, etc.; but is there any better way?

For myself, I find much more motivated to do those exercises with solutions, then compare my answers with the suggested ones, then continue to do more. But for those without solutions I can hardly even motivate myself to attempt them, for the two reasons mentioned above. Anyone has any suggestions on what we should do with those exercises in a textbook?

Any comments and suggestions will be much appreciated.

• What is the GTM? – Harald Hanche-Olsen Mar 16 '13 at 10:35
• @HaraldHanche-Olsen Based on the OP's reference to groups of order 52, I think it is "Graduate Texts in Mathematics", a series of books by Springer. – A.P. Mar 16 '13 at 10:41

For exercises with solutions, there is a terrible bias : you work much less to find the solution.

The thing is that you don't really need to find the solution of an exercise, a good exercise should let you think a lot of the profound nature of the problem.

The exercises without solutions are the best to learn, by far.

And after some days, if you still don't find the solution and you are frustrated, ask a fellow. It encourages discussion which is really good for learning.

Your problem $\#1$ is a fake problem, because at the end when you write a proof, you know if there is something that is not perfectly clear in your proof. Once you've learnt how to write a perfect detailed proof, there is no space for mistakes.

• I agree completely. As an additional solution for #2, note that you can always post a question on SE. I am aware also of compilation of answers for some textbooks that do not normally have answers (for instance, Hartshorne's Algebraic Geometry). – Jakub Konieczny Mar 16 '13 at 13:12
• I think this answer contains several oversimplifications. First: "The thing is that you don't really need to find the solution of an exercise". But sometimes you really do: many texts leave steps of theorems as exercises. If you can't do the exercise, what do you do, stop reading the text? "Once you've learnt how to write a perfect detailed proof, there is no space for mistakes." It sounds to me like you are saying that sufficiently mature mathematicians never write faulty proofs. This is a nice idea but it is unfortunately false: have you ever refereed a research paper? Or written one? – Pete L. Clark Jul 2 '13 at 11:08
• @JakubKonieczny, may I know where I can find the "compilation of answers for some textbooks that do not normally have answers"? – Zuriel Sep 20 '16 at 1:41
• Thanks for the answer! May I know if it is normal for the author(s) of the book have all the solutions and will be willing to share if we ask for it? – Zuriel Dec 27 '16 at 23:54
• @JakubKonieczny Without solutions I make too less exercises and get in time stress, because I get stuck on one for too long and can't ask too much to teachers. In addition, on SE most of my questions have not been answered. For projects it is normal that not every problem can be solved, but there is a difference in math subjects, where learning maths by making much exercises is of highest priority, and projects, where doing experience with research has priority. – Rocco van Vreumingen Apr 6 at 19:32

I think that it would be nice to have a separate book with full workout solutions. Remember that sometime there is more than one choice to get to the right answer.However the person that is making use of this book will have the option to choose the best way he understands the most and that he or she can easily remember. I think that every one should be mature enough not to cheat on himself,and use such a book as a tool to check his abilities.

• Thanks! However, is such a book always available for all subjects? I know that for more elementary subjects such as calculus or linear algebra this should not be an issue. But how about for algebraic topology or knot theory? – Zuriel Mar 1 '18 at 8:23


text:
HOSTING A TOTAL OF 318 FORMULAS WITH CALCULATORS

## Two Stage Dividend Discount Model

The two-stage model has an unstable initial growth rate, and can be either positive or negative. This initial phase lasts for a specified time and is followed by stable growth which lasts forever. The problem with this model is that the growth rate from the initial phase is assumed to change to the stable growth rate overnight.

## $\frac{d\left[1+{g}_{1}\right]}{k-{g}_{1}}\left\{1-\left[\frac{1+{g}_{1}}{1+k}{\right]}^{n}\right\}+\frac{d\left[1+{g}_{1}{\right]}^{n}\left[1+{g}_{2}\right]}{\frac{k-{g}_{2}}{\left[1+k{\right]}^{n}}}$

Here,g1= Growth rate during high growth phase.g2= Growth in constant growth phase after n. n = Length of high growth phase.Assume g1<>k and g2< k,d=dividend for the next year.

ENTER THE VARIABLES TO BE USED IN THE FORMULA

Similar formulas which you may find interesting.


text:
## Flow equivalence of G-sfts

Publikation: Bidrag til tidsskriftTidsskriftartikelForskningfagfællebedømt

In this paper, a $G$-shift of finite type ($G$-SFT) is a shift of finite type together with a free continuous shift-commuting action by a finite group $G$. We reduce the classification of $G$-SFTs up to equivariant flow equivalence to an algebraic classification of a class of poset-blocked matrices over the integral group ring of $G$. For a special case of two irreducible components with $G=\mathbb{Z}_2$, we compute explicit complete invariants. We relate our matrix structures to the Adler-Kitchens-Marcus group actions approach. We give examples of $G$-SFT applications, including a new connection to involutions of cellular automata
Originalsprog Engelsk Transactions of the American Mathematical Society 373 4 2591-2657 0002-9947 https://doi.org/10.1090/tran/7981 Udgivet - 2020

ID: 238589818


text:
rogers2013consensus
Consensus time and conformity in the adaptive voter model
Tim Rogers and Thilo Gross
Phys. Rev. E 88, 030102, 2013
The adaptive voter model is a paradigmatic model in the study of opinion formation. Here we propose an extension for this model, in which conflicts are resolved by obtaining another opinion, and analytically study the time required for consensus to emerge. Our results shed light on the rich phenomenology of both the original and extended adaptive voter models, including a dynamical phase transition in the scaling behavior of the mean time to consensus.
Figure 1: Dynamical phase transition in the scaling behavior of the mean time to reach consensus TC as a function of network size $$N$$. At the critical value $$\rho = 1/2$$ the growth of TC is linear, whilst being exponential for ρ above the critical point, and logarithmic below. In each case circles give the average over 100 simulation runs with $$k = 10$$ and $$\phi = 0.1$$, while the solid lines show the result of the theory developed in the main text – see equation (7).


text:
# latex table to continue on the same page

I need to generate a latex table using data from a csv file. The table has only two columns with about 500 rows. I would like to make best use of the space and continue the table on the same page, before moving onto the next page. Any directions please?

Thanks, Sam

• Maybe print on a roll of papyrus :o) You won't be limited. Seriously, how do you want a single sheet of paper to contain 500 lines? – Bernard Jun 7 '16 at 12:28
• Question is not so clear. Do you want directions on how to make a basic table? use csv for your table? Or just to make an extra, extra long page for your table to be placed? If you have made any concrete attempts thus far, please share your code in your answer using the code block tools to highlight them in gray. No judgment here as to why you might want to do this ;) – A Feldman Jun 7 '16 at 17:22

There are examples of how to split a long narrow table into two (Automatic way to break narrow tables into two columns?), but I don't know if that would work so well in your case.

Depending on exactly what you need in terms of layout, you could possibly use a simple itemize list in a multicols environment, in combination with e.g. datatool.

For example:

\documentclass{article}
\usepackage{datatool}
\usepackage{multicol}
\setlength{\columnseprule}{0.4pt}
\begin{document}
\begin{multicols}{4}
\begin{itemize}
\end{itemize}
\end{multicols}
\end{document}

where 500.txt goes like

a,b
0,0
1,1
2,4
3,9
4,16
5,25

and so on for 500 lines.


text:
#### Sentence Examples

• It possesses the five vowels a, i, u, e, o, both short and long, and one pure diphthong, au.
• From Corinth, an ancient inscription written 1 30vvrpoc 66v has recently been discovered, which shows that though Cleonae for B wrote E {, like the Corinthian ?j, and, as at Corinth, wrote for a vowel sound, the vowel thus represented was not short and long e and n) as at Corinth, but Il only, as in Xp g A, (X p i i a 1 Here 'a represents and the spurious diphthong is represented by a, as in (dycv, Doric infinitive -= a form which shows that c has at Cleonae the more modern form I as distinguished from the Corinthian Regarding three other questions controversy still rages.
• Thus the long i in ride, wine, &c., has become the diphthong ai, and the name of the symbol I is itself so pronounced.
• Latin e was identified with a native diphthong ei, and becomes zvy, as in rhwyf from remus.
• A hard-and-fast rule of pronunciation is that only vowel or diphthong sounds, or the letters" m," n," ng," k," t "and" p "are permissible at the end of words, and hence the final letter of all words ending in anything else is simply suppressed or is pronounced as though it were a letter naturally producing one or other of those sounds.