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tisdag 16 juli 2019

How to Understand that the Special Theory of Relativity is Unphysical

 Two space traveling Lorentz invariant twins both one year older than the other: Deep but unphysical.
Einstein's special theory of relativity is loaded with paradoxes, such as the twin paradox, ladder paradox, cooling paradox and Ehrenfest's paradox, expressing effects of time dilation and space contraction. In classical physics one paradox would be enough to kill a physics theory, but not so in modern physics with the special theory of relativity as corner stone, where the presence of paradoxes instead is taken as evidence that the theory is deep.

The paradoxes of special relativity express physical paradoxes such as traveling twins both ageing more slowly than the other. So even if special relativity is deep it cannot be a theory about true physics if it contains paradoxes, because true physics cannot be paradoxical that is contradictory.
Two twins cannot both physically age more slowly than the other.

The quickest way to understand that special relativity is unphysical, and as such can carry seemingly physical (but unphysical) paradoxes, is to recall Einstein's starting point as a description of so-called events recorded by space-time coordinates $(x,t)$ with $x$ a space coordinate and $t$ time coordinate. Einstein thus consider events to have no extension in space so that one single space coordinate $x$ is enough to describe its location in space.

Einstein's special theory concerns the description of events in two different (Euclidean) coordinate systems assumed to move with respect to each other with a certain constant velocity, so-called inertial systems.

In classical physics the connection between such systems is given by a Galilean coordinate transformation where space coordinates are transformed to match the difference in motion between the two coordinate systems while time coordinates remain the same.

In special relativity the connection is instead given by the Lorentz transformation where also time coordinates are transformed by mixing space into time. Special relativity thus boils down to the Lorentz transformation and all the paradoxes from physical point of view such as time dilation and space contraction, are consequences of the Lorentz transformation mixing space into time.

Einstein's catch is that Maxwell's equations for electromagnetics, like any wave equation, has the same analytical expression in all inertial systems under the Lorentz transformation (but not under the Galilean transformation).

Einstein then postulates that physical laws are laws which take the same form in all inertial systems connected by the Lorentz transformation, and thus declares that Maxwell's equations express a physical law.   In circular reasoning Einstein then argues that because a physical law expresses physics the Lorentz transformation expresses physics and therefore special relativity is a physical theory as a mathematical theory about physics.

Einstein's great contribution to modern physics is viewed to be his postulate that physical laws (must) have the same analytical expression in all inertial systems connected by Lorentz transformation, in other words must be Lorentz invariant.

Einstein's basic idea is thus that (true) physical laws (must) be Lorentz invariant. What we now first must understand is that this whole idea is absurd: Of course physical laws in their physical meaning must be independent of choice of coordinate system, but it is absurd to ask that they would literally have the same analytical expression. It is like claiming that a statement translated to different languages would not only have the same meaning but also the same notational expression letter by letter. This would mean that there was only one language, which is absurd. What is not absurd but rational is to expect that the same physical law will have different analytical expressions in different coordinate systems.

Next, we return to Einstein's starting point as a study of events without spatial extension, which we will see is the very reason Einstein can make the absurd claim that Maxwell's equations are Lorentz invariant. Now, solutions to Maxwell's equations represent physical waves and waves have extension in space. And now comes the catch: Maxwell's equations come along with initial conditions, which describe the initial configuration of a wave with extension in space at a certain initial time. And initial conditions are not Lorentz invariant because they mix space into time. Only Einsteinian events without extension in space can be claimed to be Lorentz invariant. A detailed account of the mathematics by  physics is given in Chapter 5 and 11 of Many-Minds Relativity.

Einstein's insistence on Lorentz invariance thus builds on the misconception that initial conditions with extension in space as physics, can be reduced to events without extension in space as unphysics. This is absurd and is the root to all the paradoxes of special relativity resulting from mixing space into time by Lorentz transformations without physics.

Einstein insisted on Lorentz invariance forgetting that wave equations have initial conditions with extension in space.  The result is a lot of modern physics formed to be Lorentz invariant which cannot be physics.

The basic trouble with modern physic preventing progress is generally viewed to be that quantum mechanics is incompatible with relativity theory in that Schrödinger's equations are not Lorentz invariant. The above analysis indicates that this is a ghost problem which should not be allowed to prevent progress. Asking for Lorentz invariance is unphysical. There is no incompatibility. There can be no incompatibility between physical theories because physics cannot be incompatible with itself. Twins cannot be incompatible.

Einstein confessed at several occasions that his knowledge of mathematics was superficial:
• I neglected mathematics...because my intuition was not strong enough to differentiate the fundamentally important from the dispensable erudition.
It is therefore not so strange that Einstein could be misled to give the Lorentz transformation a meaning which lacked physical reason. What is strange is that his delusion has come to represent the highest level of understanding of a modern physicist even that of Ed Witten as the smartest living physicist, by many viewed to be smarter than Einstein with an extraordinary deep understanding of mathematics...

PS The reason Einstein's unphysical events without extension in space and the associated Lorentz transformation have come to serve as a corners stone of modern physics, is that it fits with the modern physics core concept of elementary particle as an object without extension in space. But this concept is loaded with poison in the form of infinities and divergent integrals and more, and so modern physicists have been driven into a despair of string theory, in 11 space dimensions on a spatial scale 15 orders of magnitude smaller than estimated scale of a proton, all way beyond any scientific reason.

A better way out is to accept that there are no particles without extension in space, only waves with extension in 3d space, all according to Schrödinger, the inventor of quantum mechanics. Without particles the special theory of relativity has no physical meaning and scientific relevance. What has  Ed Witten to say about this revelation?

måndag 8 juli 2019

Was Einstein a Swindler?

What is the most compelling argument showing that Einstein's special theory of relativity as based on the Lorentz transformation connecting observations by different observers in different coordinate systems (inertial systems) moving with constant velocity with respect to each other, is unphysical and thus void of scientific content?

You find this argument in a previous post showing that although Maxwell's equations, as a form of wave equation, describing the propagation of electromagnetic waves including light, which is the central object of study in the special theory of relativity, takes the same mathematical form in different inertial systems, and thus appears to be invariant as requested by the special theory of relativity, initial conditions are not invariant and thus the whole point of relativity theory evaporates. We recall that an initial condition represents a configuration of an object extended in space, like a wave form, at a special instant of time.

The fact that this is not seen in presentations of relativity is a result of Einstein's restriction to consider events in space-time as isolated flashes at specific coordinates in space $x$ at specific instances of time $t$ described by space-time coordinates $(x,t)$. The unphysical aspect of such isolated flash-like events, is that they have no spatial extension and thus do not appear in the form of initial conditions for a wave equation. By restricting events to concern objects of no spatial extension, the non-invariance of initial conditions for a wave equations with a collapse of the basic idea of special relativity, can be hidden and success can be declared. This is what Einstein did, and the world was stunned!

But real objects/waves have extension in space, even flashes, and so their physics cannot be described by the special theory of relativity. The special theory of relativity is thus unphysical and as such is loaded with physical paradoxes, including the Ladder Paradox arising because a ladder has spatial extension.

Are you convinced by this argument? That the special theory is unphysical because it concerns physics without extension in space. If you are convinced, what is then your conclusion about the status of modern physics with the special theory of relativity declared as a corner stone? If the corner stone is unphysical, what about the building erected on the corner stone? So was Einstein correct when he described himself as a swindler? For more evidence see Many-Minds Relativity, in particular section 5.9 and Dr Faustus of Modern Physics. Or maybe you say that we must leave physics to physicists even if they are misled by a swindler?

PS1 The above argument shows that the idea of Lorentz invariance of Maxwell's equations is misconceived. The logical conclusion made in Many-Minds Relativity following an idea of Ebenezer Cunningham, is that formulation of Maxwell's equations requires specification of a Euclidean spatial coordinate system with the observer in normal case tied to its origin. Such a coordinate system acts like an aether for propagation of electromagnetic waves, and there are thus as many aethers as Euclidean coordinate systems. Einstein said that there is no aether, and then there can be no Maxwell equations and no electromagnetic waves and no light...

PS2 The logical conclusion from the Michelsen-Morley null result is that there are many aethers, as many as there are Euclidean coordinate systems and that physical laws in general take different forms in different coordinate systems, while expressing the same physical reality. Einstein's idea that true physical laws take the same formal mathematical form in all (inertial) coordinate systems represents a fundamental misconception of the meaning of a physical law. It is like claiming that a statement about a physical fact necessarily must have the same form in all languages, while it is clear to everyone with a rational mind that different languages express the same thing in different ways and not with the same words. Yes, Einstein was a swindler and led modern physics into a quagmire, but this is something modern physicists are unable to fathom. If you think this analysis does not capture reality ask your favourite physicist about the physics of special relativity and notice that you get no meaningful response.

måndag 1 juli 2019

$E=mc^2$: Definition or Physical Fact?

All these fifty years of conscious brooding have brought me no nearer to the answer to the question, 'What are light quanta (photons)?' Nowadays every Tom, Dick and Harry thinks he knows it, but he is mistaken. (Albert Einstein, 1954)

We continue exploring the meaning of the most famous equation of physics $E=mc^2$, which Einstein suggested in 1905 to be a consequence of his special theory of relativity and struggled through all his life to justify theoretically, however without success.

The equation $E=mc^2$ carries the same ambiguity as the basic postulate of Einstein's special theory of relativity, the constancy of the speed $c$ of light, for which it is never clear if it is only a definition true by logic, or a law of physics which may be true or false.

Is then $E=mc^2$ a definition or a physical law as a statement about a physical fact, which may be false?

We start with the following natural question: From where does the factor $c^2$ come, which attributes the energy $mc^2$ to mass of size $m$?

This question can be given an answer for a photon of frequency $\nu$, which can be observed e.g. through the photoelectric effect, to have the energy
• $E=h\nu$,
with a properly specified Planck's constant $h$. We can now, if we want, attribute the mass $m=\frac{h\nu}{c^2}$ to a photon, to get
• $E=h\nu =mc^2$
simply by definition.  We can do this because the physics of a photon is unclear.  We can supplement by naturally attributing the momentum $p=mc$ to a photon of mass $m$ and speed $c$ and so obtain $E=pc$ as an equivalent form of $E=mc^2$ (also discussed in the previous post).

In short, we can argue that for the fictitious concept of a photon (compare with the Einstein quote above), energy and mass indeed can be viewed to be, in the words of Einstein,  different manifestations of the same thing (namely energy).

By claiming that a radiating body loses the mass attributed to emitted photons (although the loss is too small to be measured), we can then give the relation $E=mc^2$ a general meaning beyond photons, still however essentially by definition with eventual physical meaning remaining to be explained.

That mass indeed can be converted to energy in nuclear fission and fusion processes, was a surprise to Einstein and cannot be seen as a consequence of his special theory of relativity, because it has no connection to nuclear physics. The  first quantum field theory Standard Model proof (with quarks and gluons) of $E=mc^2$ was presented only in 2008.

The general idea of a connection between mass and energy is as old as physics, with in particular the kinetic energy of a body of mass $m$ moving with speed $v$ being equal to $\frac{1}{2}mv^2$. Moreover, the factor $c^2$ was suggested prior to Einstein by Poincare and Hasenörhl preceded by Heaviside and Wien, among others. But it was Einstein who got the Prize.

Recall that in Maxwell's equations for electromagnetic waves in vacuum, with the electric field $\hat E$ (in Gaussian units) satisfying an equation of the form
• $\frac{1}{c^2}\frac{\partial^2\hat E}{\partial t^2}-\nabla^2\hat E=0$,
with the factor $\frac{1}{c^2}$ serving the same role as mass $m$ in a mechanical wave equation with connection to energy, thus supporting a relation of the form $m=\frac{E}{c^2}$, which was the original form suggested by Einstein, among others, as an expression for (fictitious) mass rather than (real) energy.

Recall that with the new meter standard as a certain fraction of a light second, the speed of light is by definition equal to exactly 1 (light second per second). For perspective see Many-Minds Relativity. So today, the constancy of the speed of light is a matter of definition. To Einstein in 1905 it was both a definition and a physical fact with truth secured by pure logic as the unique nature of Einstein's physics, which so impressed the world, although there has been many critical views from start which however have been muted since they did not fit the success story of modern physics.

Is it then true that a radiating body loses mass, even if the emitted energy comes from a source of internal heat energy as a form of kinetic vibrational energy measured by temperature (and not mass) according to a new analysis of blackbody radiation? I will seek to return with an answer.

To start with let us recall that the concept of mass $m$ connects to force $f$ and motion/acceleration $\frac{dv}{dt}$ with $v$ velocity through Newton's 2nd law $\, m\frac{dv}{dt}=f$, which can be used to define mass in terms of force and motion, as well as momentum $mv$ as the integral of force, and kinetic energy $m\frac{v^2}{2}$ as the integral of $fv$ as work. Force can be measured by a spring and motion by a meter stick and time, which defines mass threefold in terms of Newton's law, momentum and kinetic energy. The basic relation is Newton's 2nd law, while the integrals of $f$ and $fv$ are computed/collected in physical form as momentum and kinetic energy. Newton's 2nd law is Galilean invariant, while momentum and kinetic energy as integrals depend on initial velocity. Momentum and kinetic energy thus carries information about mass modulo initial velocity: If you travel at the same velocity as a cannon ball, its mass is hidden and you cannot detect it by being hit.

Defining mass by Newton's 2nd law in terms of force and motion/acceleration makes mass = inertial mass, from which equality of inertial mass and gravitational mass follows by definition, since gravitation appears as force. Einstein's Equivalence Principle as the basic assumption of the general theory of relativity is thus empty of physical content,  as the general nature of Einstein's physics jumping freely between definition and physical fact, as exposed in detail  Many-Minds Relativity. Take a look and get enlightened by understanding the confusion between definition and fact, which has corrupted modern physics into a mess of subjective epistemology instead of a science of objective ontology in the spirit of Enlightenment. This was understood by Einstein but he only gave cryptic evidence to this insight like in the above quote and
• If I would be a young man again and had to decide how to make my living, I would not try to become a scientist or scholar or teacher. I would rather choose to be a plumber or a peddler...

söndag 23 juni 2019

Einstein's 7 Erroneous Proofs of E=mc2

The book Einstein's Mistakes by Hans Ohanian gives a chronology of Einstein's many scientific mistakes including 7 erroneous proofs presented by Einstein of the crown jewel of his theory of relativity in the form of $E=mc^2$ stating proportionality between energy $E$ and mass $m$ with $c$ the speed of light. The need of a 7th proof indicates that proofs 1-6 are all incorrect and so it is not far-fetched to expect that also Einstein's 7th proof is incorrect.

To give perspective, let us recall the proof from Many-Minds Relativity chapter 14-15 of a related connection, this time between mass and momentum $P$ of the form $P=mc$. We recall that this relation can be seen as a consequence of a new Many-Minds form of Newton's 2nd Law stating the following connection between a velocity $v(t)$ and acceleration $\frac{dv}{dt}$ of a body of mass $m$ acted upon by a force $F=F(t)$ depending on time $t$:
• $\frac{m}{1+v}\frac{dv}{dt}=F$,
This form of Newton's 2nd Law results from measuring velocity of a moving object through Doppler shift $\frac{1}{1+v}$ of received signals from the object with the speed of light normalised to 1. It states that objects in approach/recession with respect to an observer, appear to be subject to an increase/decrease of mass connecting to acceleration. Here $v$ is negative in approach and positive in recession and with $v > -1$ in approach, but unlimited in recession allowing far away galaxies to recede faster than the speed of light as observed in large redshift.

Using that for  $\vert v\vert$ much smaller than 1, $\frac{1}{1+v}\approx 1-v$, Newton's 2nd Law takes the form
• $F\approx m\dot v -mv\dot v\approx (m+P)\dot v$
with $P=-mv$ momentum. This relation has the form of a classical Newton's 2nd Law with the mass $m$ augmented by $P$, which trades to a connection between momentum $P$ and mass $m$ of
the stated form $P=mc$ without normalisation to $c=1$.

We have thus given a proof of the relation $P=mc$, as an alternative the relation $E=mc^2$, which Einstein could not prove and maybe nobody else can.

PS1 In Einstein's special relativity also the recession speed is limited by the speed of light.  This is not what is observed, since galaxies outside the Hubble sphere at a distance of 4300 megaparsecs are by their redshift observed to recede faster than the speed of light. The apparent contradiction with Einstein's special theory of relativity is handled in the usual way: The special theory is correct but it does not apply to receding galaxies, for which instead the general theory of relativity must be used and the general theory is so complicated that contradictions is beyond human

PS2 The suspicion that $E=mc^2$ is just a matter of definition, which is true by defining mass in terms of force and acceleration through Newton’s 2nd Law (thus in terms of energy), and not a physical fact, which could be true or not, is growing stronger and stronger. Einstein is the master of ambiguity between definition and fact, with the constancy of the speed of light as a key example, which by physicists mislead by Einstein is viewed to be both a definition and a physical fact.

lördag 15 juni 2019

Demystifying the New SI Base Units.

In the previous post we observed that Planck's constant $h$ appears as a conversion factor connecting light of frequency $\nu$ with attributed energy $h\nu$ (in eV or Joule) through the photoelectric effect with the release of an electron from a surface exposed to light (of sufficient high frequency). The inner mechanics of the atoms delivering the electrons upon excitation by exposure to light does not enter into the discussion and so Planck's constant can be given a meaning in macroscopic physics, thus without quantum mechanics, as a trade between light and electron energy and then further to mechanical energy. Its role in quantum mechanics then appears as an after construction.

Let us now turn to Boltzmann's constant $k$ to see its connection to Planck's constant and macroscopic physics. Boltzmann's constant appears in Planck's universal law of blackbody radiation law of the form
• $E(\nu ,T) = W(a)\, kT\nu^2$,
• $W(a) = \frac{a}{\exp(a )-1}$ with $a = \frac{h\nu}{kT}$,
where $E(\nu ,T)$ is the (suitably normalised) intensity of radiation of frequency $\nu$ from a blackbody of temperature $T$ and $W(a)$ is a cut-off factor with $W(a)=1$ for small $a$ and
$W(a)$ small for medium to large $a$, expressing Wien's displacement law stating cut-off of high frequencies. We see that Planck’s constant only appears in the cut-off factor.

Experimental observation of $E(\nu ,T)$ makes it possible to determine $W(a)$ and thus $kT$ in terms of $h\nu$, from which Boltzmann's constant $k$ can be determined with respect to a chosen scale for temperatur $T$, or the other way around as in the new SI units by specifying by definition
• $k=1.380650\times 10^{-23}\, J/K$,
thus setting a new standard for Kelvin $K$ as measure of temperature. The connection between the energy measures $h\nu$ and $kT$ then shows to be
• $h\nu_{max} \approx 2.8214391\times kT$,
where $\nu = \nu_{max}$ gives maximum of the spectrum $E(\nu ,T)$.

Again, this can be done without having to invoke quantum mechanics in its standard form with $h$ as a "smallest quantum of action" as exposed in detail on Computational Blackbody Radiationwhich presents a derivation of Planck's law using deterministic continuum physics instead of as usual statistics of discrete quanta. In particular, the new derivation captures the universality of blackbody radiation beyond specific inner atomic mechanics.

The universality of Planck's law is expressed by the fact that an ideal blackbody can take the form of a set of oscillators without very specific inner structure. In particular different blackbodies with different inner structure can share the same temperature scale.

To sum up, both Planck's constant and Boltzmann's constant are specified by definition in the new SI units, from which the new units kilogram and Kelvin can be determined by macroscopic experiments without resort to quantum mechanics in its standard form.

Hopefully this helps to demystify both Planck's and Boltzmann's constant, and the new SI units.

onsdag 12 juni 2019

New Perspective on New Unit of Mass in terms of Planck's Constant

In the 2019 redefinition of the SI base units the kilogram as unit of mass is defined in terms of Planck's constant
• $h$ set by definition to exactly $6.62607015×10^{−34}$ Joule-second ($J\cdot s$),
where
• $Joule = Newton\times m = M\times\frac{m}{s^2}\times m=M\frac{m^2}{s^2}=Mc^2$
with $m$ meter, $s$ second, $M$ mass in kilogram and $c$ the speed of light.

This defines kilogram in terms of Planck's constant $h$, second $s$ and speed of light $c$ with meter $m$ defined in terms of $c$. The relation $E=Mc^2$, viewed as a profound discovery attributed to Einstein's relativity theory, then appears simply as a definition (of mass).

The connection to quantum mechanics comes by attributing a certain energy $h\nu$ to light of frequency $\nu$ through the law of the photoelectric effect
• $h\nu = eV_0 + \phi = eV_0 + h\nu_0$,
where $eV_0$ in electronVolts is the energy of a released electron with charge $e$ and $V_0$ a stopping potential in Volt, and $h\nu_0$ is the work to release an electron with $\nu_0$ a threshold frequency. This relation determines $h\approx 4.1357\times 10^{-15}$ in $eV\cdot s$, which fits with the new definition of $h$ in terms of $J\cdot s$ with the conversion $eV= 1.602176634×10^{−19} J$.

The photoelectric effect connects the macroscopic phenomena of light of different frequencies and stopping potential to the microscopic phenomenon of electron charge. In this connection there is nothing that says light of frequency $\nu$ is to be viewed as a stream of discrete photon particles of energy $h\nu$ and that Planck's constant $h$ has the physical meaning of a discrete smallest quantum of action.  Instead Planck's constant has the role of connecting light energy to electron potential energy ultimately to mechanical energy.

For a new continuum physics approach to blackbody radiation and the photoelectric effect with discrete quantum replaced by a threshold condition (as in the photoelectric effect), see Computational BlackBody Radiation.

The new definition of kilogram gives perspective on the very small size of Planck's constant $\sim 6.6\times 10^{-34}\, J\cdot s$ misleading to an idea of an absurdly small Planck length $\sim 1.6\times 10^{-35}\, m$ believed to have a physical meaning, in string theory in particular.  On the other hand, the length scale of atoms (and X-ray light of frequency about $3\times 10^{18}$) is about $10^{-10}\, m$ and that of a proton $10^{-15}\, m$, with the Planck scale 20 orders of magnitude smaller, way beyond any thinkable experimental exploration and thus meaning. Planck time  $\sim 5.3\times 10^{-44}\, s$ is even more absurd. No wonder that modern physics playing with Planck length and time is in a state of deep crisis, with the scientific madness come to full expression in the  Chronology of the Universe starting with the Planck Epoch before $10^{-43}\, s$ after Big Bang.

In Schrödingers equation $h$ multiplies the time derivative of the wave function, which means that
the atomic energy (potential + "kinetic" energy) of an eigenfunction of frequency $\nu$ is equal to $h\nu$, which comes to expression in the photoelectric effect.  Schrödinger's equation is a continuum model without any smallest quantum of action, only discrete eigenvalues representing different energies.

A reformulation of quantum mechanics in the form of a Schrödinger equation as a continuum model in real 3d space plus time without statistics, can be inspected at Real Quantum Mechanics.

To get an idea of the absurdly small Planck length scale $1.6\times 10^{-34}\, m$, one may compare with the estimated size of the observable Universe which is about $10^{27}\, m$ or $10^{33}\, \mu m$ with $\mu m$ mikrometer.

In short, Planck's constant $h$ converts light energy to mechanical energy through electron potential energy, and as such does not ask for a meaning as a "smallest quantum of action" in a mist of "quantisation" into absurdly small "quanta".

The new kilogram standard is specified to high precision using a Kibble balance, where gravitational force is balanced by an electromagnetic force (between two coils), which depends on Planck's constant. Mass is then derived by measuring the local gravitational constant.

lördag 25 maj 2019

Does Sabine Hossenfelder Exist?

Sabine Hossenfelder on BackReaction claims herself to be a modern physicist expressing truths about the state of modern physics:
• I do not know what it means for something to be “real” or “true.” You will have to consult a philosopher on that.
• If you want to claim that the Higgs-boson does not exist, you have to demonstrate that the theory which contains the mathematical structure called “Higgs-boson” does not fit the data. Whether or not Higgs-bosons ever arrive in a detector is totally irrelevant.
• Here is a homework assignment: Do you think that I exist? And what do you even mean by that?
• From November on, I will be unemployed, at least that is what it presently looks like: If you don't exist, can you be employed?
If this is the truth, no wonder that modern physics is in a state of crisis.

But Sabine's criticism of modern physics appears well founded and thus admirable. At the price of making employment difficult, which is even more admirable.  But she is not alone saying that modern  physics in a state of stalemate crisis without progress, as evidenced in the new book
The Universe Speaks in Numbers by Graham Farmelo summing up:
• ...the slow rate of progress of the string framework may presage a more sedate pace in fundamental physics that may persist for centuries to come.
Of course the pessimism of deep crisis can be turned around into its opposite, as expressed by the leading star of modern physics Arkani-Hamed:
• There has never been a better, more exciting time to be a theoretical physicist.
Concerning the crisis of modern physics it is commonly accepted that one reason is that the two basic building blocks, relativity theory and quantum mechanics, are contradictory/incompatile . If you dig a little deeper you will find that the underlying reason is that both theories are unphysical as exposed in detail as Many-Minds Relativity and Real Quantum Mechanics and also here.

Two theories which are physical cannot be contradictory, because physics which exists cannot be contradictory. But unphysical theories may well be contradictory, as ghosts can have contradictory qualities.

The Special Theory of Relativity of Einstein is unphysical because the Lorentz transformation is not a transformation between physical coordinates, as strongly underlined by its inventor Lorentz, but misunderstood by the patent clerk Einstein believing that the transformed time is real and thus that time is relative. Quantum Mechanics is unphysical because its interpretation is statistical which makes it non-physical, because physics is not an insurance company. Here Einstein was right understanding that God does not play dice.

Concerning the crisis of modern physics, listen to
It is comforting to see that I am not alone in my criticism of relativity theory and quantum mechanics.

It is impossible to discuss these things with main stream physicists, since their common wisdom is that neither relativity theory nor quantum mechanics can be understood as rational science.

An example of the confusion is the hype about quantum computing with the sound criticism by Dyakonov in The Case Against Quantum Computing of course being dismissed by main stream physicist Lubos.

The confusion is exposed in an exhibition combining arts and science about quantum mechanics at Center for Contemporary Culture in Barcelona commented on at BackReaction. Here is an artistic expression of the quantum leap of an electron which infuriated Schrödinger:

In this context you are invited to a previous post on the true meaning of Planck's constant $h$ shows The text book view is that $h$ is a fundamental quantum of action connecting the energy
$E=h\nu$ to a particle/photon of light of frequency $\nu$ according the Planck-Einstein relation with light viewed as a stream of discrete particles/photons.

But it is not at all necessary to view light this way to understand the true meaning of Planck's constant, which is revealed through the way it is measured, that is through the photoelectric effect which simply connects light frequency to the energy unit of electronVolt.

Einstein gave a heuristic explanation of the photoelectric effect from an idea of light as a stream of particles. By the common Aristotle logical fallacy of confirming the assumption by observing the consequence, this has convinced modern physicists that light indeed consists of a stream of particles, which however is against all scientific rationale and a basic reason for the crisis of modern (particle) physics. Schrödinger understood that there are no particles. See posts on the photoelectric effect showing that it does not require Einstein's particle heuristics to be understood; wave mechanics serves much better!

torsdag 16 maj 2019

Why Does a Modern Physicist Buy the CO2 Global Warming Hysteria?

Sabine Hossenfelder on BackReaction claims herself to be a physicist, yet it seems as if she without
thinking buys the CO2 Global Warming Hysteria:
• Climate Change: There are no simple solutions.
• The Earth is warming. Human carbon-dioxide emissions are one of the major culprits. We have known this for a long time. But in the past two decades, evidence for global warming has become more noticeable on local levels, as with seasonal shifts, extreme weather events, declines in biodiversity and, depending on where you live, droughts. And it will get worse.
She does not acknowledge that there are many knowledgeable physicists who do not buy this story, because it has no support in known laws of physics, including Dyson, Happer, Singer and many more.

So how can it be that she jumps on the climate alarmist wagon? I have asked her, but she does want not to communicate with me, so I have to guess: Is it because of a monumental confusion concerning quantum mechanics, which makes it impossible for a modern physicist to grasp even basic physics of the climate system of the Earth, and what does that then say about modern physics?

PS To get perspective listen to Hearing of the House Committee on Natural Resources May 22.

tisdag 14 maj 2019

Yes, Quantum Mechanics Is Wrong

Sabine Hossenfelder on BackReaction has become a truth-teller about the state of modern physics, and now she tells the truth about quantum mechanics: It is wrong:
• Yes, I am sorry. But I have a message for you from the depth of abstract math: We know that quantum mechanics is wrong.
The same idea has been expressed by almost every notable physicist, including Roger Penrose:
• Physics Is Wrong, From String Theory to Quantum Mechanics.
This is a funny situation, unprecedented in the history of science, and it has been like that for now
more than 100 years.

A restart is obviously needed and my contribution is Real Quantum Mechanics. Take a look! And think! See also recent posts.

But Sabine Hossenfelder is a physicist and thus must seek to rescue the hopeless situation in one way or the other:
• And whatever your misgivings are about quantum mechanics, there is no denying that it is useful.
So quantum mechanics is wrong, but it is nevertheless useful? That does not make sense. A theory which is wrong cannot be truly useful, because from something which is wrong cannot come something which is right. Only if the theory is right in some sense, can it be useful. But no physicist has any clue to what is right about quantum mechanics, only what is wrong!

Sabine Hossenfelder resorts to the common trick of referring to yet a more incomprehensible theory in the form of Quantum Field Theory, when basic Quantum Theory is questioned. This was used by
Einstein when he increased the bet to general relativity when his special theory of relativity was questioned on most rational grounds. This

tisdag 7 maj 2019

Quantum Mechanics Still a Complete Mystery 2

Tim Maudlin introduces his new book Philosophy of Physics: Quantum Theory with the following credo:
• A physical theory should clearly and forthrightly address two fundamental questions: what there is, and what it does.
• The answer to the first question is provided by the ontology of the theory, and the answer to the second by its dynamics.
• The ontology should have a sharp mathematical description, and the dynamics should be implemented by precise equations describing how the ontology will, or might, evolve.
• The Copenhagen Interpretation, in contrast, does not. There is little agreement about just what this approach to quantum theory postulates to actually exist or how the dynamics can be unam- biguously formulated. Nowadays, the term is often used as short-hand for a general instrumentalism that treats the mathematical apparatus of the theory as merely a predictive device, uncommitted to any ontology or dynamics at all.
•  Sometimes, accepting the Copenhagen Interpretation is understood as the decision simply to use the quantum recipe without further question: Shut up and calculate. Such an attitude rejects the aspiration to provide a physical theory, as defined above, at all.
• Hence it is not even in the running for a description of the physical world and what it does. More specific criticisms could be raised against this legacy of Bohr, but our time is better spent presenting what is clear than decrying what is obscure.
The Copenhagen Interpretation is the text book interpretation of quantum mechanics, which Maudlin thus refutes as meaningless. I agree totally.

What does then Maudlin offer in the book instead of the Copenhagen Interpretation? Very little: pilot wave theory and many worlds theory, both failed attempts to give meaning to Schrödinger's wave function. The book thus gives yet another account of the mystery of quantum mechanics 100 years after its creation.

The whole problem comes from the multi-dimensionality of Schrödinger's equation asking for a statistical unphysical interpretation. How many new books will be written on the theme that physicists do not understand the quantum mechanics they preach?

But there is light in the tunnel: Real Quantum Mechanics.


text:
# Inequality with Two Minima

### Remark

$x\lt \min\{y,z\}\;$ exactly means that $x\;$ is less than both $y\;$ and $z,\;$ i.e., that $x\lt y\;$ and $x\lt z.\;$ It follows that the problem is equivalent to proving two independent inequalities:

(*)

$\displaystyle\sqrt{a-c}+\sqrt{b-c}\le\sqrt{\frac{ab}{c}}$

and

(**)

$\sqrt{a-c}+\sqrt{b-c}\le\sqrt{2(a+b-c)}.$

### Proof 1

Inequality (*) is equivalent to

$\displaystyle\sqrt{\frac{c}{b}\cdot\frac{a-c}{a}}+\sqrt{\frac{c}{a}\cdot\frac{b-c}{b}}\le 1.$

Indeed, the AM-GM inequality yields

\displaystyle\begin{align} \sqrt{\frac{c}{b}\cdot\frac{a-c}{a}}+\sqrt{\frac{c}{a}\cdot\frac{b-c}{b}} &\le \frac{1}{2}\left(\frac{c}{b}+\frac{a-c}{a}\right)+\frac{1}{2}\left(\frac{c}{a}+\frac{b-c}{b}\right)\\ &=\frac{1}{2}\left[\left(\frac{a-c}{a}+\frac{c}{a}\right)+\left(\frac{b-c}{b}+\frac{c}{b}\right)\right]\\ &=\frac{1}{2}(1+1)\\ &=1. \end{align}

Equality occurs if $\displaystyle\frac{c}{b}=\frac{a-c}{a}\;$ and $\displaystyle\frac{c}{a}=\frac{b-c}{b},\;$ both taking place if $\displaystyle c=\frac{ab}{a+b}.$

Inequality (**) is shown directly with the AM-QM inequality:

$\sqrt{a-c}+\sqrt{b-c}\le\sqrt{2[(a-c)+(b-c)]},$

with equality if $a-c=b-c,\;$ if $a=b.$

### Proof 2

For (*), observe that, by the Cauchy-Schwarz inequality

\displaystyle\begin{align} \sqrt{\frac{ab}{c}} &= \frac{\sqrt{ab}}{\sqrt{c}}\\ &= \frac{ \sqrt{[(a-c)+c][c+(b-c)]}}{\sqrt{c}}\\ &\ge\frac{\sqrt{\left[\sqrt{(a-c)c}+\sqrt{c(b-c)}\right]^2}}{\sqrt{c}}\\ &=\frac{\sqrt{(a-c)c}+\sqrt{c(b-c)}}{\sqrt{c}}\\ &=\sqrt{a-c}+\sqrt{b-c}. \end{align}

Inequality (**) is shown via Jensen's inequality for $f(x)=\sqrt{x}:$

$\displaystyle\frac{\sqrt{a-c}+\sqrt{b-c}}{2}\le\sqrt{\frac{(a-c)+(b-c)}{2}}.$

### Proof 3

Note that

$\displaystyle\sqrt{x}+\sqrt{y}\le\sqrt{x+y+\frac{xy}{z}+z}=\sqrt{\frac{(x+z)(y+z)}{z}}.$

With $x=a-c,\;y=b-c,\;$ and $z=c,\;$ we get (*).

For (**), by the Cauchy-Schwarz inequality,

$2(a+b-c)=[1+1][(a-c)+(b-c)]\ge \left(\sqrt{a-c}+\sqrt{b-c}\right)^2.$

### Proof 4

WLOG, $c\le a\le b\;$ such that $a=c+x,\;$ $b=c+x+y,\;$ $x+y\gt 0.$

For (*), we wish to prove $\sqrt{cx}+\sqrt{c(x+y)}\le\sqrt{(x+c)(c+x+y)},\;$ or equivalently (by squaring),

$cx+cx+cy+2c\sqrt{x(x+y)}\le cx+x^2+xy+c^2+cx+yx$

which simplifies to $c^2-2c{x(x+y)}+x(x+y)\ge 0,\;$ i.e., $(c-\sqrt{x(x+y)})^2\ge 0\;$ which is true. Equality is achieved when $c^2=x(x+y),\;$ i.e., when $c^2=(a-c)(b-c),\;$ or $\displaystyle c=\frac{ab}{a+b}.$

Observe now that (**) is equivalent to $\sqrt{x}+\sqrt{x+y}\le\sqrt{2(2x+y)}\;$ which, by squaring, reduces to $x+x+y+2\sqrt{x(x+y)}\le 4x+2y,\;$ i.e., $x+(x+y)-2\sqrt{x(x+y)}\ge 0,\;$ or, finally, $\left(\sqrt{x+y}-\sqrt{x}\right)^2\ge 0\;$ which is true. Equality when $y=0,\;$ i.e. when $a=b.$

### Proof 5

With $\displaystyle x=\frac{a}{c}\;$ and $\displaystyle y=\frac{b}{c},\;$ (*) becomes

$\sqrt{x-1}+\sqrt{y-1}\le\sqrt{xy}.$

Squaring gives $x-1+y-1+2\sqrt{(x-1)(y-1)}\le xy,\;$ and, subsequently, $2\sqrt{(x-1)(y-1)}\le (x-1)(y-1)+1,\;$ which is of the form $2X\le X^2+1,\;$ which is true.

(**) is shown as in Proof 1.

### Acknowledgment

Dorin Marghidanu has posted the problem and a solution (Proof 1) at the CutTheKnotMath facebook page, with additional solutions by Marian Dinca (Proof 2), by (Proof 3), by Imad Zak (Proof 4).

format: plain


text:
# KdV hierarchy

$\partial_t V + \partial_x^3 V = 6 V \partial_x V$

can be rewritten in the Lax Pair form

$\partial_t L = [L, P]$

where L is the second-order operator

L = − D2 + V

(D = d / dx) and P is the third-order antiselfadjoint operator

P = 4D3 + 3(DV + VD).

Note that P consists of the zeroth order and higher terms of the formal power series expansion of 4iL3 / 2).

One can replace P with other fractional powers of L. For instance, the zeroth order and higher terms of 4iL5 / 2 are

$P = 4D^5 + 5(D^3 V + V D^3) - 5/4 (D \partial^2_x V + \partial_x^2 V D) + 15/4 (D V^2 + V^2 D)$

and the Lax pair equation becomes

$\partial_t V + \partial_x^5 u = \partial_x (5 V_x^2 + 10 V V_xx + 10 V^3)$

with Hamiltonian

$H(V) = \int V_{xx}^2 - 5 V^2 V_xx - 5 V^4 dx.$

These flows all commute with each other and their Hamiltonians are conserved by all the flows simultaneously.

The KdV hierarchy are examples of higher order water wave models; a general formulation is

$\partial_t u + \partial_x^{2j+1} u = P(u, u_x , ..., \partial_x^{2j} u)$

where u is real-valued and P is a polynomial with no constant or linear terms; thus KdV and gKdV correspond to j=1, and the higher order equations in the hierarchy correspond to j=2,3,etc. LWP for these equations in high regularity Sobolev spaces is in KnPoVe1994, and independently by Cai (ref?); see also CrKpSr1992.The case j=2 was studied by Choi</span> (ref?).The non-scalar diagonal case was treated in KnSt1997; the periodic case was studied in [Bo-p3].Note in the periodic case it is possible to have ill-posedness for every regularity, for instance $\partial_t u + u_{xxx} = u^2 u_x^2$ is ill-posed in every Hs [Bo-p3]


text:
#### core, lib, modules: restructured source code tree

- new folder src/ to hold the source code for main project applications
- main.c is in src/
- all core files are subfolder are in src/core/
- modules are in src/modules/
- libs are in src/lib/
- application Makefiles are in src/
- application binary is built in src/ (src/kamailio)

Daniel-Constantin Mierla authored on 07/12/2016 11:03:51
Showing 1 changed files
 1 1 deleted file mode 100644 ... ... @@ -1,51 +0,0 @@ 1 -/* 2 - * Copyright (C) 2013 Crocodile RCS Ltd 3 - * 4 - * This file is part of Kamailio, a free SIP server. 5 - * 6 - * Kamailio is free software; you can redistribute it and/or modify 7 - * it under the terms of the GNU General Public License as published by 8 - * the Free Software Foundation; either version 2 of the License, or 9 - * (at your option) any later version 10 - * 11 - * Kamailio is distributed in the hope that it will be useful, 12 - * but WITHOUT ANY WARRANTY; without even the implied warranty of 13 - * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the 14 - * GNU General Public License for more details. 15 - * 16 - * You should have received a copy of the GNU General Public License  17 - * along with this program; if not, write to the Free Software  18 - * Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301 USA 19 - * 20 - */ 21 -/*! 22 - * \file 23 - * \brief Kamailio core :: STUN support 24 - * STUN support callback, used by the STUN module 25 - * \ingroup core 26 - * Module: \ref core 27 - */ 28 - 29 -#include "events.h" 30 -#include "ip_addr.h" 31 -#include "stun.h" 32 - 33 -int stun_process_msg(char* buf, unsigned int len, struct receive_info* ri) 34 -{ 35 - int ret; 36 - stun_event_info_t sev; 37 - 38 - ret = 0; 39 - LM_DBG("STUN Message: [[>>>\n%.*s<<<]]\n", len, buf); 40 - if(likely(sr_event_enabled(SREV_STUN_IN))) { 41 - memset(&sev, 0, sizeof(stun_event_info_t)); 42 - sev.buf = buf; 43 - sev.len = len; 44 - sev.rcv = ri; 45 - ret = sr_event_exec(SREV_STUN_IN, (void *) &sev); 46 - } else { 47 - LM_DBG("no callback registering for handling STUN -" 48 - " dropping!\n"); 49 - } 50 - return ret; 51 -}

#### Core Removed history, svn $id$ and doxygen udpates on the .c files

Olle E. Johansson authored on 03/01/2015 09:53:17
Showing 1 changed files
 ... ... @@ -1,6 +1,4 @@ 1 1  /* 2 - * $Id$ 3 - * 4 2  * Copyright (C) 2013 Crocodile RCS Ltd 5 3  * 6 4  * This file is part of Kamailio, a free SIP server. ... ... @@ -20,6 +18,13 @@ 20 18  * Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301 USA 21 19  * 22 20  */ 21 +/*! 22 + * \file 23 + * \brief Kamailio core :: STUN support 24 + * STUN support callback, used by the STUN module 25 + * \ingroup core 26 + * Module: \ref core 27 + */ 23 28   24 29  #include "events.h" 25 30  #include "ip_addr.h"

#### all: updated FSF address in GPL text

Anthony Messina authored on 04/07/2014 09:36:37 • Daniel-Constantin Mierla committed on 04/07/2014 09:37:36
Showing 1 changed files
 ... ... @@ -17,7 +17,7 @@ 17 17  * 18 18  * You should have received a copy of the GNU General Public License  19 19  * along with this program; if not, write to the Free Software  20 - * Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA 20 + * Foundation, Inc., 51 Franklin Street, Fifth Floor, Boston, MA 02110-1301 USA 21 21  * 22 22  */ 23 23

#### core: removed STUN from core

- Added hooks so that STUN messages can be passed to a module
- Removed STUN configuration file parameters

Peter Dunkley authored on 29/03/2013 21:47:57
Showing 1 changed files
 1 1 new file mode 100644 ... ... @@ -0,0 +1,46 @@ 1 +/* 2 + * $Id$ 3 + * 4 + * Copyright (C) 2013 Crocodile RCS Ltd 5 + * 6 + * This file is part of Kamailio, a free SIP server. 7 + * 8 + * Kamailio is free software; you can redistribute it and/or modify 9 + * it under the terms of the GNU General Public License as published by 10 + * the Free Software Foundation; either version 2 of the License, or 11 + * (at your option) any later version 12 + * 13 + * Kamailio is distributed in the hope that it will be useful, 14 + * but WITHOUT ANY WARRANTY; without even the implied warranty of 15 + * MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the 16 + * GNU General Public License for more details. 17 + * 18 + * You should have received a copy of the GNU General Public License  19 + * along with this program; if not, write to the Free Software  20 + * Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA 21 + * 22 + */ 23 + 24 +#include "events.h" 25 +#include "ip_addr.h" 26 +#include "stun.h" 27 + 28 +int stun_process_msg(char* buf, unsigned int len, struct receive_info* ri) 29 +{ 30 + int ret; 31 + stun_event_info_t sev; 32 + 33 + ret = 0; 34 + LM_DBG("STUN Message: [[>>>\n%.*s<<<]]\n", len, buf); 35 + if(likely(sr_event_enabled(SREV_STUN_IN))) { 36 + memset(&sev, 0, sizeof(stun_event_info_t)); 37 + sev.buf = buf; 38 + sev.len = len; 39 + sev.rcv = ri; 40 + ret = sr_event_exec(SREV_STUN_IN, (void *) &sev); 41 + } else { 42 + LM_DBG("no callback registering for handling STUN -" 43 + " dropping!\n"); 44 + } 45 + return ret; 46 +}


text:
# System of non-linear partial differential eqs from electrostatics

1. Dec 12, 2012

I have an electrostatics problem (shown here: https://www.physicsforums.com/showthread.php?t=654877) wich leads to the following system of differential equations:

$\frac{\partial E_z}{\partial z}=\frac{\rho}{\epsilon_0}$ (1)

$Z_i E_r \frac{\partial \rho}{\partial r}+(u_g+ Z_i E_z) \frac{\partial \rho}{\partial z} + \rho Z_i \frac{\partial E_z}{\partial z}=0$ (2)

Substituting eq. (1) into eq. (2):
$Z_i E_r \frac{\partial \rho}{\partial r}+(u_g+ Z_i E_z) \frac{\partial \rho}{\partial z} + \frac{\rho^2 Z_i}{\epsilon_0}=0$ (3)

Therefore I have a system of 2 equations (1 & 3) with 2 unknowns, the axial field $E_z$ and the charge density $\rho(z,r)$. The rest of the variables are known so they can be supposed as constants.

I'm not sure on how to solve it, I'm considering two options:
- derivate eq. (3) with respect to $z$ to substitute in eq. (1), but I don´t get rid of $E_z$ and the eq. (3) becomes more complicated.

- Solve by semi-implicit method, considering that $z=du_z/dt$, but since is an equation in partial derivatives I'm not sure on how to manage the term in $r$

I'm totally stuck on this, I'm asking for a direction of solving it, not for a solution, so any help would be grateful.

2. Dec 28, 2012

### haruspex

How about separation of variables? Of course, being nonlinear it's not going to lead to a family of solutions which can be summed, but maybe towards one solution. I tried Ez = y(r,z) = R(r)Z(z). Substituting for ρ I get the form
AR'Z'+BRZ"+CR2(ZZ"+Z'2) = 0
(where R' means differentiated wrt r, Z' wrt z)
Integrating wrt z:
AR'Z+BRZ'+CR2ZZ' = f(r)
which should yield to a second round of integration wrt z easily enough.
Integrating wrt r gets hard unless you can figure out f(r) from boundary conditions.
That's all I can think of.

3. Jan 4, 2013

Hi Haruspex, first of all thank you very much for your help.
Your suggestion of separating variables gave me the idea of instead of separating Ez, which is supposed to only depend on z, separate $\rho(r,z)=R(r)Z(z)$
Then, the Poisson eq:
$dE=\frac{RZ}{\epsilon_0}dz$
and the eq in $\rho$:
$AR'Z+BRZ'+CR^2Z^2=0$

Considering the Neumann boundary conditions in the axial axis, when r=0:
$\frac{\partial E_r}{\partial r}=0 (1); \frac{\partial \rho}{\partial r}=0 (2)$
while in the walls, r=R: $E_z=0 (3)$

Using #(2) into ρ eq: $BZ'+CR_0Z^2=0$
where R(0)=R0

Integrating now wrt z: $\int B\frac{Z'}{Z^2} dz + \int CR_0 dz= 0$
$R_0Z = \rho(0,z) = \frac{B}{zC}$

Substituting and integrating for Ez in the axial line (r=0):
$E_z=\frac{\rho}{\epsilon_0}dz=\frac{B}{\epsilon_0C}Ln(z)$

What do you think of this kind of solution? I'm not very sure about having an electrical field which depends on a logarithm of distance, because is negative for small values of z and is contiunously growing, while Ez should decrease as z increases (it moves away from origin).

Thanks for helping.

4. Jan 6, 2013

### haruspex

Whoa, I'm confused. In the OP you wrote ∂Ez/∂z = ρ/ε0.
Looks like you meant ∂E/∂z = ρ/ε0.
Is that right? If so, please clarify the relationship between E, Ez and Er. Do the subscripts denote partial derivatives? Components? Something else?

5. Jan 7, 2013

oops, sorry about the confussion, subscripts denote components of the electrical field:
$\vec{E}=E_r\vec{r}+E_z\vec{z}$, neglecting the variation in the azimuthal direction.
The Poisson equation shown is the result of operating in cylindrical components:
$\frac{1}{r}\frac{\partial (r E_r)}{\partial r}+\frac{\partial E_z}{\partial z}=\frac{\rho(r,z)}{\epsilon_0}$
for a known $E_r=\frac{U}{r Log(R_i/R_{tip})}$ and unknowns $E_z$ and $\rho(r,z)$

I hope the problem is better explained now.

6. Jan 7, 2013

### haruspex

It is, thanks, but I'm still stuck with an apparent contradiction. At different points in the thread you have written:
$\frac{\partial E_z}{\partial z}=\frac{\rho}{\epsilon_0}$ (1)
$\rho = \rho(r,z)$
Ez, which is supposed to only depend on z
Do you see my puzzlement?

7. Jan 8, 2013

Yes, now I see, since $\rho = \rho(r,z)$, $\frac{\partial E_z}{\partial z}=f(r,z)$.
Then I don´t know if it is a conceptual mistake from me, but I was thinking that since
$\vec{E}=E_r\vec{r}+E_z\vec{z}$,
each component was depending on each variable,
Er=E(r) and Ez=E(z),
and therefore the Poisson equation:
$\nabla \vec{E}=\frac{\rho(r,z)}{\epsilon_0}$
would descompose in its form:
$\frac{1}{r}\frac{\partial (r E_r(r))}{\partial r}+\frac{\partial E_z(z)}{\partial z}=\frac{\rho(r,z)}{\epsilon_0}$.
Since the term in Er is cancelled, the resulting equation would be:
$\frac{\partial E_z(z)}{\partial z}=\frac{\rho(r,z)}{\epsilon_0}$.
But you are right and I don´t know why I conclude this "paradox"
Wouldn´t it be possible separate the effects on the r and z directions or maybe the correct expression is
$\frac{\partial E_z(z)}{\partial z}=\frac{\rho(z)}{\epsilon_0}$?

8. Jan 8, 2013

### haruspex

Yes, I'm afraid that's wrong. It's quite normal for a component in one direction to depend on location in another.

9. Jan 9, 2013

oh, I see my error, the field is actually discomposed as $\vec{E}=E\vec{r}+E\vec{z}$, where I have named $E_r=E(r,z)\vec{r}$ and $E_z=E(r,z)\vec{z}$, but I think I'm getting lost with the nomenclature, because then I only know Er at the discharge point, when z=0: $E_r(r,0)=\frac{U}{rLog(R_i/R_{tip})}$ and therefore I can only use it as an initial condition? It would change completely my approach to the problem...


text:
# Difference between basis and subbasis in a topology?

I was reading Topology from Munkres and got confused by the definition of a subbasis. What is/are the difference between basis and subbasis in a topology?

-

Bases and subbases "generate" a topology in different ways. Every open set is a union of basis elements. Every open set is a union of finite intersections of subbasis elements.

For this reason, we can take a smaller set as our subbasis, and that sometimes makes proving things about the topology easier. We get to use a smaller set for our proof, but we pay for it; with a subbasis we need to worry about finite intersections, whereas we did not have to worry about that in the case of a basis.

-

Consider $S=\{\{0,1\},\{0,2\}\}$. What is the topological space $T(S)$ generated by $S?$ By definition, $S$ will then be a subbasis of $T(S)$.

Well, we want to all requirements to hold true and find that $T(S) = \{\emptyset, \{0\}, \{0,1\}, \{0,2\}, \{0,1,2\}\}$ (check this!).

Is $S$ a basis? No, because you cannot write $\{0\}$ as a union of any elements in $S$.

So you see that subbasis and basis are two different notions, even for a very basic example.

A subbasis can be thought of, and is actually defined to be, the "smallest set that becomes my topological space if I complete it under the property of being a topological space, i.e. fulfiling the axioms of topological space".

The two terms are related nevertheless. Every basis is a subbasis, and in one of the equivalent definitions of subbasis you will find that you already get a basis from your subbasis.

-
Shouldn't a subbasis be able to produce every element of the topology by building the union of intersections of elements of the subbasis, where the intersections have to be finite? Something like: Let $(X, T)$ be the topological space and $S$ be a subbasis. Then: $\forall U \in T: U = \cup_{A = M_i \cap M_j \cap ... with M_{i,j} \in S}$ where all $A$ should be finite. How do you make the empty set this way? –  moose Feb 4 at 17:04
@moose You can make the empty set with the empty union of empty intersection. In other words the union of intersection of no sets. Kind of like how the vector space $\{0\}$ is generated by the basis $\varnothing$ since $0$ is an empty sum. –  Pratyush Sarkar May 24 at 23:21

The collection of sets $(-\infty,b)$ and $(a,\infty)$ for $a,b\in \mathbb{R}$ constitute a sub-basis for the standard topology on $\mathbb{R}$. The collection of sets $(a,b)$ for $a,b\in \mathbb{R}$ constitute a basis for the standard topology on $\mathbb{R}$. I suggest that you look at the definitions of "basis" and "sub-basis" and convince yourself that the claims that I have made are correct; this is probably the best way to answer your question.

The basic idea is that a basis is the collection of all finite intersections of sub-basis elements. The open sets in a topology are all possible unions of basis elements. So, the open sets in a topology are all possible unions of finite intersections of sub-basis elements.

If we ignore, momentarily, the fact that we are trying to generate a topology, a subbasis is any old collection of subsets of the space. Thus, any basis is a subbasis. However, a basis $\mathcal B$ must satisfy the criterion that if $U,V\in\mathcal B$ and $x$ is an arbitrary point in both $U$ and $V$, then there is some $W$ belonging to $\mathcal B$ such that $x\in W\subseteq U\cap V$.
To really understand what that means you have to look at examples. The family $\{X\}$ is a basis for the indiscrete topology on $X$. The family $\mathcal P(X)$, the powerset of $X$, is a basis for the discrete topology on $X$. If $X$ is large enough and you remove some subset $S\subset X$ from $\mathcal P(X)$, you can still have a basis if $S$ is not singleton but $\mathcal P(X)\setminus \{x\}$ is merely a subbasis when $x\in X$.
A more ambitious example is the definition of the topology on the product of two spaces $X$ and $Y$. The family $$\{U\times Y:U\subseteq X, U\text{ is open}\}\cup\{X\times V:V\subseteq Y,V\text{ is open}\}$$ is a subbasis but not necessarily a basis of the box topology on $X\times Y$.
Finally, suppose a family $\mathcal B$ is a basis. You can generate a topology $\tau_0$ using $\mathcal B$ as a subbasis and you can generate a topology $\tau_1$ using $\mathcal B$ as a basis. The topologies are actually the same, as you can see here. You just have to do more work to generate a topology from a subbasis $\mathcal S$, i.e. you have to include all finite intersections of members of $\mathcal S$ because it is not guaranteed that they are unions of members of $\mathcal S$.


text:
# B How to compute logarithm with pen and paper?

1. Jan 30, 2017

### late347

If you had something particularly nasty like

$log_{10}(9)$ = ???

I asked my teacher about how to begin approaching that kind of computation of logarithm.

He was not very interesyed in explaining the procedures of estimating the value of the log. Perhaps the procedure went beyond the scope of our math course at the time.

For sure, I think John Napier did not have a calculator which he could just push some buttons. Evidently Napier used some method to compute values which he made into logarithm tables so the logarithm tables could be used by other people to perform calculations.

2. Jan 30, 2017

### Staff: Mentor

Here's some history of how Napier did it:

http://www.maa.org/press/periodical...ar-function-john-napier-introduces-logarithms

Some folks use the series approximation to get a few logarithms and then add, subtract, and multiply them to get the logs of other numbers. As an example, given the log(2) and the log(3) then log(2)+log(3)=log(6) ...

You could use the appropriate natural log series as the starting point for computing the first natural logarithm and then go from there.

http://hyperphysics.phy-astr.gsu.edu/hbase/Math/lnseries.html

3. Jan 30, 2017

### late347

Thanks for pointing me in a direction, I will investigate the matter further from there, I think.

4. Jan 30, 2017

### Staff: Mentor

He made the logarithm tables so others didn't have to re-calculate it every time. A lot of Taylor series. Calculators are not magic - they only perform operations every human can do as well. They are just faster.

For a rough approximation:
$\log_{10}(9) = \frac{ln(10\cdot 0.9)}{ln(10)} = \frac{ln(10) + ln(0.9)}{ln(10)} = 1 + \frac{ln(0.9)}{ln(10)} \approx 1+ \frac{-0.1}{2.3} = 1-0.04 = 0.96$.

Close to the actual value of about 0.9542.

5. Jan 30, 2017

### StoneTemplePython

With respect to the natural log, you may consider using a Pade approximation -- this should lead to a much tighter approximation for a given number of terms in a taylor polynomial. The underlying continued fraction is quite intuitive -- though if you are looking for analytical properties, a linear combination of polynomials (i.e. a taylor polynomial) is a lot easier to use than nested fractions.

6. Jan 30, 2017

### FactChecker

Mathematicians of old would often hire "idiot savants" who could do amazing math calculations in their heads. They could generate tables. They can not describe how they do the calculations. Some talk about how different numbers taste, feel, or smell.

7. Jan 30, 2017

### Staff: Mentor

Do you have a reference for that? I'd be interested in reading about it.

I don't doubt its true since many great artists actually ran workshops with assistants doing much of the mundane work so perhaps mathematicians used their students in the same way.

8. Jan 31, 2017

### late347

We have only had the basics of sequences and series taught to us so far.

Are there any online resources for reading about how series and how they work? I sometimes browse around math stackexchange and saw that series can converge for example.

I suspect that these are taught somewhere in calculus based courses/books?

9. Jan 31, 2017

### Svein

No. Calculating numbers is in the realm of numerical mathematics, a very different discipline. By the way, I Googled "approximating logarithms" and got 20 million hits.

10. Jan 31, 2017

### FactChecker

I think I read that they used savants in Bell's "Men of Mathematics" long ago (30 years?). The part about the savants trying to describe their methods in terms of taste, smell, feel, etc. was something I gathered from several TV reports over a period of time.

11. Jan 31, 2017

### Staff: Mentor

Yes, I've seen those documentaries. There's one guy, Daniel Tammet who was able to recite PI to thousands of places and who was able to be fluent in Icelandic in a very short time. He also apparently has synesthesia where he sees numbers as colors

https://en.wikipedia.org/wiki/Daniel_Tammet


text:
D. Boxes And Balls
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

Ivan has n different boxes. The first of them contains some balls of n different colors.

Ivan wants to play a strange game. He wants to distribute the balls into boxes in such a way that for every i (1 ≤ i ≤ n) i-th box will contain all balls with color i.

In order to do this, Ivan will make some turns. Each turn he does the following:

1. Ivan chooses any non-empty box and takes all balls from this box;
2. Then Ivan chooses any k empty boxes (the box from the first step becomes empty, and Ivan is allowed to choose it), separates the balls he took on the previous step into k non-empty groups and puts each group into one of the boxes. He should put each group into a separate box. He can choose either k = 2 or k = 3.

The penalty of the turn is the number of balls Ivan takes from the box during the first step of the turn. And penalty of the game is the total penalty of turns made by Ivan until he distributes all balls to corresponding boxes.

Help Ivan to determine the minimum possible penalty of the game!

Input

The first line contains one integer number n (1 ≤ n ≤ 200000) — the number of boxes and colors.

The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 109), where ai is the number of balls with color i.

Output

Print one number — the minimum possible penalty of the game.

Examples
Input
31 2 3
Output
6
Input
42 3 4 5
Output
19
Note

In the first example you take all the balls from the first box, choose k = 3 and sort all colors to corresponding boxes. Penalty is 6.

In the second example you make two turns:

1. Take all the balls from the first box, choose k = 3, put balls of color 3 to the third box, of color 4 — to the fourth box and the rest put back into the first box. Penalty is 14;
2. Take all the balls from the first box, choose k = 2, put balls of color 1 to the first box, of color 2 — to the second box. Penalty is 5.

Total penalty is 19.


text:
# Bredon cohomology of a permutation action on $S^3$

I've seen a couple of similar questions asking to verify computations of Bredon cohomology here and here, so I will ask one such question myself.

Let $$\mathbb{Z}/2$$ act on $$S^3\subset \mathbb{C}^2$$ by restriction of a permutation action on $$\mathbb{C}^2.$$ I wanted to compute Bredon cohomology $$\mathcal{H}^*_{\mathbb{Z}/2}(S^3;\underline{\mathbb{Z}}).$$

I have a cell decomposition based on a decomposition of complex $$1$$-dimensional disk into $$3$$ cells: $$\mathbb{D}=D\sqcup T\sqcup *.$$ Here $$T\sqcup *=S^1=\partial \mathbb{D}$$ and $$D$$ is the interior of $$\mathbb{D}.$$ Then we have a decomposition of $$S^3=\mathbb{D}\times S^1 \cup S^1\times \mathbb{D}$$ into cells compatible with the $$\mathbb{Z}/2$$ action.

The fixed point set of an action is a circle given by $$\{z_1=z_2\}\cap S^3\subset \mathbb{C}^2.$$ Since the orbit category of $$\mathbb{Z}/2$$ consists of $$*$$ and $$\mathbb{Z}/2$$ there are the following equivariant chains: $$\begin{array}{|c|c|c|c|} \hline \operatorname{dim} &*& \mathbb{Z}/2 & \operatorname{cells corresponding to} \underline{C}_n(S^3)(\mathbb{Z}/2)\\ \hline 0 & \mathbb{Z} & \mathbb{Z} & * \times *\\ 1 & 0 & \mathbb{Z}\oplus\mathbb{Z},\quad \begin{pmatrix} 1 \\ 0\end{pmatrix}\xrightarrow{\overline{1}} \begin{pmatrix} 0 \\ 1\end{pmatrix} & T\times *,*\times T\\ 2 & 0 & \mathbb{Z}\oplus \mathbb{Z} \oplus \mathbb{Z},\quad \begin{pmatrix} 1 \\ 0\\0\end{pmatrix}\xrightarrow{\overline{1}}\begin{pmatrix} 0 \\ 1\\0\end{pmatrix};\;\begin{pmatrix} 0 \\ 0\\1\end{pmatrix}\xrightarrow{\overline{1}}\begin{pmatrix} 0 \\ 0\\-1\end{pmatrix} & D\times *, *\times D, T\times T\\ 3 & 0 & \mathbb{Z}\oplus \mathbb{Z},\quad \begin{pmatrix} 1 \\ 0\end{pmatrix}\xrightarrow{\overline{1}} \begin{pmatrix} 0 \\ 1\end{pmatrix} & D\times T, T\times D\\ \hline \end{array}$$

So it seems that the cochains valued in $$\underline{\mathbb{Z}}$$ are:

$$\begin{array}{|c|c|} \hline \operatorname{dim} & \\ \hline 0 & \mathbb{Z}\\ 1 & \mathbb{Z}\\ 2 & \mathbb{Z}\\ 3 & \mathbb{Z}\\ \hline \end{array}$$ Since $$(T\times T)^*=0$$ in cochains, we have $$\mathcal{H}^3_{\mathbb{Z}/2}(S^3;\underline{\mathbb{Z}})=\mathbb{Z}.$$ Differential $$d_1$$ is an isomorphism since $$\partial(D\times *)=T\times *.$$ It seems that $$\mathcal{H}^*_{\mathbb{Z}/2}(S^3;\underline{\mathbb{Z}})=H^*(S^3;\mathbb{Z}).$$

It is a bit odd to me that the quotient is a homological sphere. Sure, the group $$\mathcal{H}^3_{\mathbb{Z}/2}(S^3;\underline{\mathbb{Z}})=\mathbb{Z}$$ since orientation is preserved, but maybe I've missed some $$2$$-torsion in lower degrees?

• I think you're correct. $\underline{\Bbb Z}$ coefficients have the property that they compute the integral cohomology of the orbit space $S^3 / (\Bbb Z/2)$, and by a slightly different cell decomposition I think this space is homotopy equivalent to $S^3$. – Tyler Lawson Aug 16 '20 at 15:26

Your final answer is correct, but the cell structure you're using isn't a $$G$$-CW structure: $$T\times T$$ can't be used as a cell in this way.

I would approach it like this: The action of $$G = {\mathbb Z}/2$$ on $$\mathbb{C}\times\mathbb{C}$$ can be written as the representation $$\mathbb{C}\oplus\mathbb{C}^\sigma$$, where $$G$$ acts trivially on $$\mathbb{C}$$ and by negation on $$\mathbb{C}^\sigma$$. The sphere $$S(\mathbb{C}\oplus\mathbb{C}^\sigma)$$ is also the one-point compactification $$S^{1+2\lambda}$$, where $$\lambda$$ denotes the real line with $$G$$ acting by negation. This has a $$G$$-CW structure with

1. one $$G$$-fixed 0-cell,
2. one $$G$$-fixed 1-cell,
3. one $$G$$-free 2-cell, and
4. one $$G$$-free 3-cell,

so that the skeleta are $$*$$, $$S^1$$, $$S^{1+\lambda}$$, and $$S^{1+2\lambda}$$. From here you can work out that the $$\underline{\mathbb{Z}}$$-cochain complex is $$\mathbb{Z} \xrightarrow{0} \mathbb{Z} \xrightarrow{1} \mathbb{Z} \xrightarrow{0} \mathbb{Z}.$$

A way to check that the answer is correct is to write $$H_G^n(S^{1+2\lambda}) \cong \tilde H_G^n(S^0) \oplus \tilde H_G^n(S^{1+2\lambda}) \cong \tilde H_G^n(S^0)\oplus \tilde H_G^{n-1-2\lambda}(S^0)$$ and then use the known calculation of the $$RO(G)$$-graded cohomology of a point (originally due to Stong (unpublished), since published in various places).

• It is not a G-CW structure because fixed points are not a G-CW subspace? – Grisha Taroyan Aug 17 '20 at 8:15
• All the cells in a $G$-CW complex have to have the form $G/H\times D^n$ where $G$ acts trivially on $D^n$. $T\times T$ doesn't have that form, it looks like the disc of a nontrivial representation of $G$. – Steve Costenoble Aug 17 '20 at 10:23
• Is there a way to work with "cells" with nontrivial action of the stabilizer? I'm now aware that any cell with nontrivial action can be subdivided into "good" G-cells, but is there a way to avoid this step? – Grisha Taroyan Aug 17 '20 at 12:28
• Not easily, at least not if you want to calculate the integer-graded part of the cohomology. Any filtration gives rise to a spectral sequence that could, theoretically, be used for computation, but that's not going to be straightforward. There is a notion of $G$-CW($V$) complexes, using cells of the form $G/H\times D(V)$, but that calculates the cohomology in grading $V$. I think Stefan Waner first noticed this, Gaunce Lewis published an exposition, and then Stefan and I published a book using a very generalized version. – Steve Costenoble Aug 17 '20 at 14:47
• One other situation where you can use arbitrary cells: At least for $G = \mathbb{Z}/p$, if you have a space built out of cells of the form $G/H\times D(V)$ where $V$ is even-dimensional, but can vary, then, with some additional assumptions, you can conclude that the $RO(G)$-graded cohomology is a free module over the cohomology of a point, though the generators may not be where you expect them to be. I think this was first proved by Ferland and Lewis. – Steve Costenoble Aug 17 '20 at 17:21


text:
Lemma 8.5.4. Let $\mathcal{C}$ be a site. Let $\mathcal{S}_1$, $\mathcal{S}_2$ be categories over $\mathcal{C}$. Suppose that $\mathcal{S}_1$ and $\mathcal{S}_2$ are equivalent as categories over $\mathcal{C}$. Then $\mathcal{S}_1$ is a stack in groupoids over $\mathcal{C}$ if and only if $\mathcal{S}_2$ is a stack in groupoids over $\mathcal{C}$.

Proof. Follows by combining Lemmas 8.5.2 and 8.4.4. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
# The canonical open set which is equal to a set with the BP modulo meager sets is regular open (Kechris)

A set $$U$$ in a topological space $$X$$ is called regular open iff $$U=(\overline{U})^°$$.

Exercise $$(8.30)$$ (Kechris, "Classical Descriptive Set Theory") Prove that $$U(A)=\bigcup \{U\,\text{open}\mid U\Vdash A\}$$ is regular open. Moreover, if $$X$$ is a Baire space and $$A$$ has the BP (Baire property), then $$U(A)$$ is the unique regular open set $$U$$ with $$A=^*U$$.

Actually, the first part of this question has already been asked, but the answer doesn't seem to work (at least for me): indeed, what the hint allows to prove is that $$A$$ is comeager in $$(\overline{U(A)})^°$$, but how does the emptyness of $$(\overline{U(A)})^°\setminus U(A)$$ follow?

For the second part, assume that $$A=^*V$$ for some open regular open set in $$X$$. I don't see how the assumption should help.

NOTE: I decided to ask for it once again because the user Brian M. Scott seems to be no more active on this site .

The aforementioned hint is that $$U(A)\Vdash A$$. As you indicate, this allows to show $$(\overline{U(A)})^°\Vdash A$$.
By the hint and its definition, $$U(A)$$ it is the largest open set $$U$$ such that $$U\Vdash A$$. On the other hand, $$U(A)\subseteq (\overline{U(A)})^°$$. Hence $$U(A) = (\overline{U(A)})^°$$ and therefore it is regular open. (I fail to see which emptiness should be relevant here.)
For the second part, assume $$A$$ the BP and $$A=^*V$$ with $$V$$ regular open. Then $$V\Vdash A$$ and hence $$V\subseteq U(A)$$.
From $$A=^*V$$ we also conclude $$A\setminus V$$ is meager, and hence $$U(A)\setminus A \cup A\setminus V \supseteq U(A) \setminus V$$ is meager and has empty interior. This shows that $$U(A)\setminus V \subseteq \overline{V}$$, and therefore $$U(A) \subseteq (\overline{V})^° = V$$. We have both inclusions.
• Thank you for your plain answer. You are right, you fail to see the emptyness because the above expression is imprecise: I meant that $(\overline{U})^°\setminus U$ should be empty. I apologize for this. Thank you once again. Apr 6, 2019 at 15:48


text:
# Deriving the UMVUE for Rayleigh scale parameter

Let $$X_1,...,X_n$$ be iid with the pdf given by $$f(x|\theta)=2\theta^{-1}xe^{-x^2/\theta}$$ for $$x>0$$. My task is to find the UMVUE for $$\theta$$, and I’m given the following hint: “$$U(X)=\sum_{i=1}^nX_i^2$$ is sufficient for $$\theta$$, and $$U/\theta$$ is a multiple of $$\chi_{2n}^2$$”.

It’s my understanding that to find the UMVUE, you need to derive $$E[T|U=u]$$ where $$T$$ is an unbiased estimator and $$U$$ is a complete sufficient statistic. However, I’m not sure how to make sense of the given hint. Where does the Chi-squared come from?

Joint density of $$X_1,X_2,\ldots,X_n$$ is

\begin{align} f_{\theta}(x_1,x_2,\ldots,x_n)&=\prod_{i=1}^n f(x_i\mid\theta) \\&=\left(\frac{2}{\theta}\right)^n \left(\prod_{i=1}^n x_i\right) \exp\left(-\frac{1}{\theta}\sum_{i=1}^n x_i^2\right)\mathbf1_{x_1,\ldots,x_n>0}\quad,\,\theta>0 \end{align}

This pdf is a member of the one-parameter exponential family.

So it follows that a complete sufficient statistic for $$\theta$$ is indeed

$$U(X_1,X_2,\ldots,X_n)=\sum_{i=1}^n X_i^2$$

Yes it is true that the UMVUE of $$\theta$$ if it exists is given by $$E(T\mid U)$$ where $$T$$ is any unbiased estimator of $$\theta$$. This is what the Lehmann-Scheffe theorem says. As a corollary, it also says that any unbiased estimator of $$\theta$$ based on a complete sufficient statistic has to be the UMVUE of $$\theta$$. Here this corollary comes in handy.

To make sense of the hint given, find the distribution of $$Y=X^2$$ where $$X$$ has the Rayleigh pdf you are given.

Via change of variables, the pdf of $$Y$$ is

\begin{align} f_Y(y)&=f(\sqrt y\mid\theta)\left|\frac{dx}{dy}\right|\mathbf1_{y>0} \\&=\frac{1}{\theta}e^{-y/\theta}\mathbf1_{y>0}\quad,\,\theta>0 \end{align}

In other words, $$X_i^2$$ are i.i.d Exponential with mean $$\theta$$ for each $$i=1,\ldots,n$$.

Or, $$\frac{2}{\theta}X_i^2\stackrel{\text{ i.i.d }}\sim\text{Exp with mean }2\equiv \chi^2_2$$

Thus implying $$\frac{2}{\theta}\sum_{i=1}^n X_i^2=\frac{2U}{\theta} \sim \chi^2_{2n}$$

So,

\begin{align} E_{\theta}\left(\frac{2U}{\theta}\right)=2n\implies E_{\theta}\left(\frac{U}{n}\right)=\theta \end{align}

Hence the UMVUE of $$\theta$$ is $$\boxed{\frac{U}{n}=\frac{1}{n}\sum_{i=1}^n X_i^2}$$

However, we did not require finding the distribution of $$X_i^2$$ since it is easy to show directly that $$E_{\theta}(U)=\sum_{i=1}^n \underbrace{E_{\theta}(X_i^2)}_{\theta}=n\theta$$

• Read it this time. Thanks for the response. – David Nov 19 '18 at 16:02
• If I were to also find the UMVUE estimator of $\theta^{-1}$, how could I alter this process? – David Nov 20 '18 at 23:11
• @DavidS The procedure is the same, except you would have to find an unbiased estimator of $\theta^{-1}$ based on the complete sufficient statistic $U$. It is reasonable to start with $E(1/U)$ and see whether it results in the form $k/\theta$. Then $1/(kU)$ is your UMVUE. – StubbornAtom Nov 21 '18 at 5:27


text:
• 应用
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# P2896 [USACO08FEB]一起吃饭Eating Together

• 414通过
• 1.1K提交
• 题目提供者 FarmerJohn2
• 评测方式 云端评测
• 标签 动态规划,动规,dp 递推 队列 USACO 2008
• 难度 普及/提高-
• 时空限制 1000ms / 128MB
• 提示：收藏到任务计划后，可在首页查看。

## 题目描述

The cows are so very silly about their dinner partners. They have organized themselves into three groups (conveniently numbered 1, 2, and 3) that insist upon dining together. The trouble starts when they line up at the barn to enter the feeding area.

Each cow i carries with her a small card upon which is engraved Di (1 ≤ Di ≤ 3) indicating her dining group membership. The entire set of N (1 ≤ N ≤ 30,000) cows has lined up for dinner but it's easy for anyone to see that they are not grouped by their dinner-partner cards.

FJ's job is not so difficult. He just walks down the line of cows changing their dinner partner assignment by marking out the old number and writing in a new one. By doing so, he creates groups of cows like 111222333 or 333222111 where the cows' dining groups are sorted in either ascending or descending order by their dinner cards.

FJ is just as lazy as the next fellow. He's curious: what is the absolute mminimum number of cards he must change to create a proper grouping of dining partners? He must only change card numbers and must not rearrange the cows standing in line.

FJ的奶牛们在吃晚饭时很傻。他们把自己组织成三组（方便编号为1, 2和3），坚持一起用餐。当他们在谷仓排队进入喂食区时，麻烦就开始了。

每头奶牛都随身带着一张小卡片，小卡片上刻的是Di（1≤Di≤3）表示她属于哪一组。所有的N（1≤N≤30000）头奶牛排队吃饭，但他们并不能按卡片上的分组站好。

FJ的工作并不是那么难。他只是沿着牛的路线走下去，把旧的号码标出来，换上一个新的。通过这样做，他创造了一群奶牛，比如111222333或333222111，奶牛的就餐组按他们的晚餐卡片按升序或降序排列。

FJ就像任何人一样懒惰。他很好奇：怎样他才能进行适当的分组，使得他只要修改最少次数的数字？由于奶牛们已经很长时间没有吃到饭了，所以“哞哞”的声音到处都是，FJ只好更换卡号，而不能重新排列已经排好队的奶牛。

## 输入输出格式

输入格式：

* Line 1: A single integer: N

* Lines 2..N+1: Line i describes the i-th cow's current dining group with a single integer: Di

• 第1行：一个整数：n

• 第2~n+1行：第i-1行描述第i个奶牛目前分组。

输出格式：

* Line 1: A single integer representing the minimum number of changes that must be made so that the final sequence of cows is sorted in either ascending or descending order

一个整数，表示必须做出的最小变化数，以便以升序或降序排序最终序列。

## 输入输出样例

输入样例#1： 复制
5
1
3
2
1
1

输出样例#1： 复制
1

## 说明

感谢@一思千年 提供翻译

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text:
## Adobe Photoshop 2022 (Version 23.1) Crack Full Version

The King is dead, long live the King! At least for a little while. Michigan’s market price of an ounce of gold (and its inverse, i.e., the interest rate) can be estimated from the perspective of the yield curve. We first built a time-series of the 10-year treasury yield from 1986 to today. The data is available from FRED. (click to enlarge) The first thing that we notice is that, at first blush, the yield curve is in a inverted wedge, i.e. the two-year and 10-year yield are both higher than the yield of longer-term bonds. This is a precursor to a recession. We can then fit a curve to the data to find the level of the yield curve (i.e., the spot where the curve intersects the 100, 200, and 300 bps levels), as well as their slopes. (click to enlarge) The slope of the yield curve is given by where b and m are constant coefficients and A is the area of the yield curve. The yield curve, including the slope and the zero point, is then integrated to find the market price of the Treasury, which we can convert to the spot price of gold using the relation (click to enlarge) Gold prices are then plotted against the yield of the 10-year UST. The slope of the line in the figure above is about 1.5%, very close to the prevailing trend in the price of gold. This is an indicator of the current cycle of asset price inflation. There is also a significant gap between the market price of gold and the market value of the present government debt outstanding, that is the IOU. (click to enlarge) There are two potential interpretations of this gap. On the one hand, we can assume that the gap is due to inflation because the value of IOUs is inflated due to the fact that they are denominated in nominal, rather than real, dollars (i.e., the fraction of nominal IOUs that is in gold). This is the same argument used to support the view that the risk of default on the IOU is decreasing as nominal bonds become cheaper. On the other hand, we can assume that it is due to finance channeling. The first point is that, in this scenario, the real-dollar value of government bonds is increasing, which tends to

## What’s New in the?

Q: decreasing a function without finding its minimum Given $f:\mathbb{R}\to\mathbb{R}$ that satisfies $f(x)>0$ and $\lim_{x\to-\infty}f(x)$ is finite. Can I conclude that there is an $x_0$ such that $f(x)>0$ for all $x\geq x_0$? A: How about $$f(x) = \begin{cases} 1 & x\in[1,2)\\ \frac1{x^2} & x\in[-1,-1/2)\\ 0 & \text{otherwise} \end{cases}$$ I cannot see what you want here. EDIT: I think I finally know what you want to achieve. Basically I want to show that there is a $x_0>-\infty$ such that $f(x)>0$ for all $x\ge x_0$. There is a well known theorem which can be found in any good book on analysis. It says that if a function $f$ is positive and non decreasing on an interval $I$ then it can be extended to an entire function. As we know there is no limit point on $\mathbb R$ so $f$ does not have a point of discontinuity so by the Riemann-Lebesgue lemma, the function $f$ can be extended to the whole $\mathbb R$. Therefore there is a $x_0>-\infty$ such that $f(x)>0$ for $x\ge x_0$. Description Description Bed with a view and river views along the River Liffey. Experience the tranquil surroundings of this modern apartment located in the exclusive riverside development of North Wall Quay, off the Phoenix Park. This apartment offers the best in value and space with living/dining areas and modern kitchen/diner, ensuite bathroom and balcony or roof terrace. The apartment is light and bright with picturesque views and is conveniently located a short stroll from Temple Bar and the Square. This location offers excellent public transport links to the city centre. What is Gumtree? Gumtree.com.au is Australia’s leading

## System Requirements:

Minimum: OS: Windows 7/8/8.1/10 Processor: Intel Core i3, i5, i7 Memory: 4GB RAM Graphics: NVIDIA GeForce GTX 560 DirectX: Version 11 Network: Broadband Internet connection Recommended: Memory: 8GB RAM Graphics: NVIDIA GeForce GTX 770 DirectX: Version


text:
Table 3:
GP surrogate model parameters. Results from the combination of $n$-way ANOVA analysis without interactions and the posthoc Tukey–Kramer's comparison tests of the effects of four parameters on the mean of $RDEμ/rsucc$. The table shows the values of the parameters sorted according to the corresponding regressed $RDEμ/rsucc$ (shown left) for the respective combinations of dimension and half of the optimization run. Values with the mean response according to Tukey–Kramer significantly higher than the respective lowest mean response are marked with $↓$. Stars after the values sign the statistical significance of rejecting the ANOVA test hypotheses that the effects of the respective parameters on the mean responses of $RDEμ/rsucc$ are negligible. Significant results of Tukey–Kramer's comparisons between the values of the respective parameters are marked in the gray lines of each block (1 $=$ best value, 2 $=$ second etc.). Stars $★$/$★★$/$★★★$ sign the $p$-value lower than 0.05/0.01/0.001 for the respective tests.
dimPart of RunTrainset Selection$rmaxA$$Nmax$Covariance Function
2-D 0.40 $k$-NN (TSS2) 0.41 0.40 $20·D$ 0.40 $KMate´rnν=5/2$
0.41 population (TSS4) 0.41 0.41 $15·D$ 0.41 $KMate´rnν=3/2$
0.41 clustering (TSS3)   0.41 $10·D$ 0.41 $KSE$
0.41 recent (TSS1)
2-D ii 0.42 $k$-NN (TSS2) 0.42 2$★★$ 0.42 $20·D$ 0.42 $KSE$
0.42 recent (TSS1) 0.42 4 $↓$ 0.42 $15·D$ 0.42 $KMate´rnν=5/2$
0.42 clustering (TSS3)   0.42 $10·D$ 0.42 $KMate´rnν=3/2$
0.42 population (TSS4)
1<$★★$
3-D 0.38 recent (TSS1) 0.38 0.38 $15·D$ 0.38 $KMate´rnν=3/2$
0.38 $k$-NN (TSS2) 0.39 0.38 $10·D$ 0.38 $KMate´rnν=5/2$
0.38 clustering (TSS3)   0.38 $20·D$ 0.39 $KSE$
0.39 population (TSS4)
3-D ii 0.38 recent (TSS1)$★★★$ 0.39 0.39 $15·D$ 0.39 $KMate´rnν=3/2$
0.39 $k$-NN (TSS2)$★★★$ 0.39 0.39 $10·D$ 0.39 $KMate´rnν=5/2$
0.39 clustering (TSS3) $↓$   0.39 $20·D$ 0.40 $KSE$
0.40 population (TSS4) $↓$
1<$★$3, 1<$★★★$4, 2<$★$
5-D 0.34 $k$-NN (TSS2)$★★★$ 0.34 2$★★★$ 0.34 $10·D$ 0.34 $KMate´rnν=3/2$$★★★$
0.34 recent (TSS1)$★★★$ 0.35 4 $↓$ 0.34 $15·D$ 0.34 $KMate´rnν=5/2$$★★★$
0.34 clustering (TSS3)$★★★$   0.34 $20·D$ 0.35 $KSE$$↓$
0.35 population (TSS4) $↓$
1<$★★★$4, 2<$★$4, 3<$★$1<$★★★$ 1<$★★★$3, 2<$★★★$
5-D ii 0.35 $k$-NN (TSS2)$★★★$ 0.35 2$★★$ 0.35 $15·D$ 0.35 $KMate´rnν=5/2$$★★★$
0.35 recent (TSS1)$★★★$ 0.36 4 $↓$ 0.35 $10·D$ 0.35 $KMate´rnν=3/2$$★★★$
0.35 clustering (TSS3)$★★★$   0.35 $20·D$ 0.36 $KSE$$↓$
0.37 population (TSS4) $↓$
1<$★★★$4, 2<$★★★$4, 3<$★★$1<$★★$ 1<$★★$3, 2<$★★$
10-D 0.34 $k$-NN (TSS2)$★★★$ 0.33 4$★★★$ 0.34 $20·D$ 0.33 $KMate´rnν=5/2$$★★★$
0.34 recent (TSS1)$★★★$ 0.35 2 $↓$ 0.34 $15·D$ 0.34 $KSE$$↓$
0.34 clustering (TSS3) $↓$   0.34 $10·D$ 0.35 $KMate´rnν=3/2$$↓$
0.35 population (TSS4) $↓$
1<$★$3, 1<$★★★$4, 2<$★★$1<$★★★$ 1<$★★$2, 1<$★★★$
10-D ii 0.32 $k$-NN (TSS2)$★★★$ 0.32 4$★★★$ 0.33 $20·D$ 0.33 $KMate´rnν=5/2$$★★★$
0.33 recent (TSS1) $↓$ 0.34 2 $↓$ 0.33 $15·D$ 0.34 $KSE$$↓$
0.34 clustering (TSS3) $↓$   0.34 $10·D$ 0.34 $KMate´rnν=3/2$$↓$
0.34 population (TSS4) $↓$
1<$★★$2, 1<$★★★$3, 1<$★★★$1<$★★★$ 1<$★★★$2, 1<$★★★$
20-D 0.35 $k$-NN (TSS2)$★★★$ 0.34 4$★★★$ 0.35 $20·D$$★★★$ 0.34 $KMate´rnν=5/2$$★★★$
0.36 population (TSS4) $↓$ 0.37 2 $↓$ 0.35 $15·D$$★★★$ 0.34 $KMate´rnν=3/2$$↓$
0.36 clustering (TSS3) $↓$   0.36 $10·D$$↓$ 0.39 $KSE$$↓$
0.36 recent (TSS1) $↓$
1<$★$2, 1<$★★★$3, 1<$★★★$1<$★★★$1<$★★★$3, 2<$★★$1<$★$2, 1<$★★★$3,
2<$★★★$
20-D ii 0.34 $k$-NN (TSS2)$★★★$ 0.33 4$★★★$ 0.34 $20·D$$★★★$ 0.32 $KMate´rnν=5/2$$★★★$
0.35 recent (TSS1) $↓$ 0.36 2 $↓$ 0.35 $15·D$$★★★$ 0.34 $KMate´rnν=3/2$$↓$
0.35 population (TSS4) $↓$   0.36 $10·D$$↓$ 0.38 $KSE$$↓$
0.36 clustering (TSS3) $↓$
1<$★$2, 1<$★★★$3, 1<$★★★$4, 2<$★$1<$★★★$1<$★★★$3, 2<$★★★$1<$★★★$2, 1<$★★★$3,
2<$★★★$
dimPart of RunTrainset Selection$rmaxA$$Nmax$Covariance Function
2-D 0.40 $k$-NN (TSS2) 0.41 0.40 $20·D$ 0.40 $KMate´rnν=5/2$
0.41 population (TSS4) 0.41 0.41 $15·D$ 0.41 $KMate´rnν=3/2$
0.41 clustering (TSS3)   0.41 $10·D$ 0.41 $KSE$
0.41 recent (TSS1)
2-D ii 0.42 $k$-NN (TSS2) 0.42 2$★★$ 0.42 $20·D$ 0.42 $KSE$
0.42 recent (TSS1) 0.42 4 $↓$ 0.42 $15·D$ 0.42 $KMate´rnν=5/2$
0.42 clustering (TSS3)   0.42 $10·D$ 0.42 $KMate´rnν=3/2$
0.42 population (TSS4)
1<$★★$
3-D 0.38 recent (TSS1) 0.38 0.38 $15·D$ 0.38 $KMate´rnν=3/2$
0.38 $k$-NN (TSS2) 0.39 0.38 $10·D$ 0.38 $KMate´rnν=5/2$
0.38 clustering (TSS3)   0.38 $20·D$ 0.39 $KSE$
0.39 population (TSS4)
3-D ii 0.38 recent (TSS1)$★★★$ 0.39 0.39 $15·D$ 0.39 $KMate´rnν=3/2$
0.39 $k$-NN (TSS2)$★★★$ 0.39 0.39 $10·D$ 0.39 $KMate´rnν=5/2$
0.39 clustering (TSS3) $↓$   0.39 $20·D$ 0.40 $KSE$
0.40 population (TSS4) $↓$
1<$★$3, 1<$★★★$4, 2<$★$
5-D 0.34 $k$-NN (TSS2)$★★★$ 0.34 2$★★★$ 0.34 $10·D$ 0.34 $KMate´rnν=3/2$$★★★$
0.34 recent (TSS1)$★★★$ 0.35 4 $↓$ 0.34 $15·D$ 0.34 $KMate´rnν=5/2$$★★★$
0.34 clustering (TSS3)$★★★$   0.34 $20·D$ 0.35 $KSE$$↓$
0.35 population (TSS4) $↓$
1<$★★★$4, 2<$★$4, 3<$★$1<$★★★$ 1<$★★★$3, 2<$★★★$
5-D ii 0.35 $k$-NN (TSS2)$★★★$ 0.35 2$★★$ 0.35 $15·D$ 0.35 $KMate´rnν=5/2$$★★★$
0.35 recent (TSS1)$★★★$ 0.36 4 $↓$ 0.35 $10·D$ 0.35 $KMate´rnν=3/2$$★★★$
0.35 clustering (TSS3)$★★★$   0.35 $20·D$ 0.36 $KSE$$↓$
0.37 population (TSS4) $↓$
1<$★★★$4, 2<$★★★$4, 3<$★★$1<$★★$ 1<$★★$3, 2<$★★$
10-D 0.34 $k$-NN (TSS2)$★★★$ 0.33 4$★★★$ 0.34 $20·D$ 0.33 $KMate´rnν=5/2$$★★★$
0.34 recent (TSS1)$★★★$ 0.35 2 $↓$ 0.34 $15·D$ 0.34 $KSE$$↓$
0.34 clustering (TSS3) $↓$   0.34 $10·D$ 0.35 $KMate´rnν=3/2$$↓$
0.35 population (TSS4) $↓$
1<$★$3, 1<$★★★$4, 2<$★★$1<$★★★$ 1<$★★$2, 1<$★★★$
10-D ii 0.32 $k$-NN (TSS2)$★★★$ 0.32 4$★★★$ 0.33 $20·D$ 0.33 $KMate´rnν=5/2$$★★★$
0.33 recent (TSS1) $↓$ 0.34 2 $↓$ 0.33 $15·D$ 0.34 $KSE$$↓$
0.34 clustering (TSS3) $↓$   0.34 $10·D$ 0.34 $KMate´rnν=3/2$$↓$
0.34 population (TSS4) $↓$
1<$★★$2, 1<$★★★$3, 1<$★★★$1<$★★★$ 1<$★★★$2, 1<$★★★$
20-D 0.35 $k$-NN (TSS2)$★★★$ 0.34 4$★★★$ 0.35 $20·D$$★★★$ 0.34 $KMate´rnν=5/2$$★★★$
0.36 population (TSS4) $↓$ 0.37 2 $↓$ 0.35 $15·D$$★★★$ 0.34 $KMate´rnν=3/2$$↓$
0.36 clustering (TSS3) $↓$   0.36 $10·D$$↓$ 0.39 $KSE$$↓$
0.36 recent (TSS1) $↓$
1<$★$2, 1<$★★★$3, 1<$★★★$1<$★★★$1<$★★★$3, 2<$★★$1<$★$2, 1<$★★★$3,
2<$★★★$
20-D ii 0.34 $k$-NN (TSS2)$★★★$ 0.33 4$★★★$ 0.34 $20·D$$★★★$ 0.32 $KMate´rnν=5/2$$★★★$
0.35 recent (TSS1) $↓$ 0.36 2 $↓$ 0.35 $15·D$$★★★$ 0.34 $KMate´rnν=3/2$$↓$
0.35 population (TSS4) $↓$   0.36 $10·D$$↓$ 0.38 $KSE$$↓$
0.36 clustering (TSS3) $↓$
1<$★$2, 1<$★★★$3, 1<$★★★$4, 2<$★$1<$★★★$1<$★★★$3, 2<$★★★$1<$★★★$2, 1<$★★★$3,
2<$★★★$
Close Modal


text:
# Find the mean slope of $f(x) = 2x^3-6x^2-90x+6$ on the interval $[-5,8]$

I need to find the mean slope of $$f(x) = 2x^3-6x^2-90x+6$$ on the interval $$[-5,8]$$. I am getting $$\frac{518}{13}$$ but this is wrong. Here is my work.

Using the mean value theorem $$f'(c)=\frac{f(b) - f(a)}{b - a}$$

$$f(b) = 2(-5)^3-6(-5)^2-90(-5)+6 = -250-150+450+6 = 56$$

$$f(a) = 2(8)^3-6(8)^2-90(8)+6 = 1024-384-66 = 574$$

$$\frac{56-574}{-5-8}=\frac{518}{13}$$

• The mean slope is just the slope of the secant line. – coreyman317 Mar 31 at 20:57

Your error is a simple mistake in computing $$f(8)$$. To be precise, you write that $$2(8)^3-6(8)^2-90(8)+6 = 1024-384-66,$$ where indeed $$2\times8^3=1024$$ and $$6\times8^2=384$$. You seem to have computed $$9\times8$$ instead of $$90\times8$$.
• Yes, but you compute $9\times8$ instead of $90\times8$ (I have updated my answer to point this out more clearly) – Servaes Mar 31 at 21:02


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# All Questions

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### Why do higher pitches appear to be louder?

It may just be in a few cases, but in the case of a flute, a higher pitch appears to come with a perceived higher volume. Is this simply because you need to put more energy into the flute to get a ...


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## Reblog: A Mathematical Trauma

Feature photo (above) by Jimmie via flickr.

My 8-year-old daughter’s first encounter with improper fractions was a bit more intense than she knew how to handle.

I hope you enjoy this “Throw-back Thursday” blast from the Let’s Play Math! blog archives:

Photo (right) by Old Shoe Woman via Flickr.

Nearing the end of Miquon Blue today, my youngest daughter encountered fractions greater than one. She collapsed on the floor of my bedroom in tears.

The worksheet started innocently enough:

$\frac{1}{2} \times 8=\left[ \quad \right]$

Don’t miss any of “Let’s Play Math!”:  Subscribe in a reader, or get updates by Email.

## More Than One Way to Solve It

Photo by Eirik Newth via flickr.

In a lazy, I-don’t-want-to-do-school mood, Princess Kitten was ready to stop after three math problems. We had gotten two of them correct, but the last one was counting the ways to paint a cube in black and white, and we forgot to count the solid-color options.

For my perfectionist daughter, one mistake was excuse enough to quit. She leaned her head against me as we sat together on the couch and said, “We’re done. Done, done, done.” If she could, she would have started purring — one of the most manipulative noises known to humankind. I’m a soft touch. Who can work on math when there’s a kitten to cuddle?

Still, I managed to squeeze in one more puzzle. I picked up my whiteboard marker and started writing:

DONE
DOEN
DNOE
DENO
DNEO
ONED
ODNE

## The (Mathematical) Trouble with Pizza

Photo by George Parrilla via flickr.

Kitten complained that some math programs keep repeating the same kind of problems over and over, with bigger numbers: “They don’t get any harder, they just get longer. It’s boring!”

So we pulled out the Counting lessons in Competition Math for Middle School. [Highly recommended book!] Kitten doesn’t like to compete, but she enjoys learning new ideas, and Batterson’s book gives her plenty of those, well organized and clearly explained.

Today’s topic was the Fundamental Counting Principle. It was review, easy-peasy. The problems were too simple, until…

Pizzas at Mario’s come in three sizes, and you have your choice of 10 toppings to add to the pizza. You may order a pizza with any number of toppings (up to 10), including zero. How many choices of pizza are there at Mario’s?

[The book said 9 toppings, but I was skimming/paraphrasing aloud and misread.]

• Can you figure out the answer?

## Quotable: What to Do When You’re Stuck

When a kid is feeling bad about being stuck with a problem, or just very anxious, I sometimes ask him to make as many mistakes as he can, and as outrageous as he can. Laughter happens (which is valuable by itself, and not only for the mood — deep breathing brings oxygen to the brain). Then the kid starts making mistakes. In the process, features of the problem become much clearer, and in many cases a way to a solution presents itself.

## Does It Work?

While I was collecting entries for the Math Teachers at Play #35 blog carnival, I ran across this post by Dave Lanovaz:

Don’t miss any of “Let’s Play Math!”:  Subscribe in a reader, or get updates by Email.

## If It Ain’t Repeated Addition, What Is It?

[Photo by SuperFantastic.]

Keith Devlin’s latest article, It Ain’t No Repeated Addition, brought me up short. I have used the “multiplication is repeated addition” formula many times in the past — for instance, in explaining order of operations. But according to Devlin:

Multiplication simply is not repeated addition, and telling young pupils it is inevitably leads to problems when they subsequently learn that it is not.

I found myself arguing with the article as I read it. (Does anybody else do that?) If multiplication is not repeated addition, then what in the world is it?

## Diagnosis: Math Workbook Syndrome

Photo by otisarchives3.

I discovered a case of MWS (Math Workbook Syndrome) one afternoon, as I was playing Multiplication War with a pair of 4th grade boys. They did fine with the small numbers and knew many of the math facts by heart, but they consistently tried to count out the times-9 problems on their fingers. Most of the time, they lost track of what they were counting and gave wildly wrong answers.

## How to Teach Math to a Struggling Student

photo by MC Quinn via flickr (CC BY 2.0)

Paraphrased from a homeschool math discussion forum:

Help! My daughter struggles with arithmetic. I guess she is like me: just not a math person. She is an outstanding reader. When we do word problems, she usually has no trouble. She’s a whiz at strategy games and beats her dad at chess every time. But numbers — yikes! When we play Yahtzee, she gets lost trying to add up her score. The simple basics of adding and subtracting confuse her.

Since I find math difficult myself, it’s hard for me to know what she needs. What’s missing to make it click for her? She used to think math was fun and tested well above grade level, but I listened to some well-meaning advice and totally changed the way we were schooling. I switched from using workbooks and games to using Saxon math, and she got extremely frustrated. Now she hates math.


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 Tr. Mat. Inst. Steklova, 2002, Volume 236, Pages 218–225 (Mi tm292)

Doubling Operations and Monodromy of Generalized Knizhnik–Zamolodchikov Equations

V. P. Leksin

Kolomna State Pedagogical Institute

Abstract: The Knizhnik–Zamolodchikov (KZ) equations associated to root systems of $A$ and $B$ types are considered whose coefficients belong to the algebras of chord diagrams and symmetric chord diagrams, respectively. Explicit formulas for the monodromy of these equations are derived with the use of doubling operations in the algebras of chord diagrams. Reductions of the KZ equations with matrix coefficients to Schlesinger deformations are considered. The simplest solutions to the Schlesinger equations under such reductions are found.

Full text: PDF file (192 kB)
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English version:
Proceedings of the Steklov Institute of Mathematics, 2002, 236, 206–213

Bibliographic databases:

UDC: 517.538+514

Citation: V. P. Leksin, “Doubling Operations and Monodromy of Generalized Knizhnik–Zamolodchikov Equations”, Differential equations and dynamical systems, Collected papers. Dedicated to the 80th anniversary of academician Evgenii Frolovich Mishchenko, Tr. Mat. Inst. Steklova, 236, Nauka, MAIK «Nauka/Inteperiodika», M., 2002, 218–225; Proc. Steklov Inst. Math., 236 (2002), 206–213

Citation in format AMSBIB
\Bibitem{Lek02} \by V.~P.~Leksin \paper Doubling Operations and Monodromy of Generalized Knizhnik--Zamolodchikov Equations \inbook Differential equations and dynamical systems \bookinfo Collected papers. Dedicated to the 80th anniversary of academician Evgenii Frolovich Mishchenko \serial Tr. Mat. Inst. Steklova \yr 2002 \vol 236 \pages 218--225 \publ Nauka, MAIK «Nauka/Inteperiodika» \publaddr M. \mathnet{http://mi.mathnet.ru/tm292} \mathscinet{http://www.ams.org/mathscinet-getitem?mr=1931022} \zmath{https://zbmath.org/?q=an:1020.32008} \transl \jour Proc. Steklov Inst. Math. \yr 2002 \vol 236 \pages 206--213

• http://mi.mathnet.ru/eng/tm292
• http://mi.mathnet.ru/eng/tm/v236/p218

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This publication is cited in the following articles:
1. Golubeva V.A., “On the Riemann-Hilbert correspondence for generalized Knizhnik-Zamolodchikov equations for different root systems”, Differential Equations and Quantum Groups - ANDREY A. BOLIBRUKH MEMORIAL VOLUME, Irma Lectures in Mathematics and Theoretical Physics, 9, 2007, 189–207
•  Number of views: This page: 118 Full text: 41 References: 20


text:
What are the limitations of using GET & POST to communicate with a web service?
By in on Mar 03 2011
• Mar, 2011 3

It is less secured than SOAP.

It prevents the user from passing structures and objects as arguments

It prevents the user from passing ByRef arguments

• 0