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Does electron's magnetic field look like that of a bar magnet?

Yes. It's just the laymen's whole way of thinking about it that is wrong.

Brian Bi asked a question on the Physics Stack Exchange,

What does the magnetic field of the (quantum-mechanical) electron look like?
He seems annoyed by the idea that "quantum spinning electrons" have dipole magnetic fields because the magnetic field of a magnetic dipole looks like this:

It's a nice, classical picture of the magnetic field. On the other hand, the spin \(\vec J={\mathbf S}\) of an electron (the arrow above a letter or the bold face represent three-dimensional vectors) is a bizarre quantum observable, an operator. So they can't be proportional, he thinks. Either the formula relating the spin and the magnetic moment\[

\mathbf{\vec\mu} = \frac{g_e \mu_b}{\hbar} \mathbf{S}

\] has to be wrong or we have to deny the formula for the magnetic field (which we may call the "equation under attack")\[


\] he believes. The idea of the laymen – including all the "interpreters" of quantum mechanics – is that in the microscopic world, the structure of all the equations, shapes of fields have to change but the reasoning stays the same. They think that the microscopic world is described by a "different classical theory".

However, the truth is just the opposite. The microscopic phenomena are described by the "same but quantum theory" as their macroscopic counterparts. Indeed, many equations of motion in quantum mechanical theories may be obtained from the classical ones by adding simple hats above everything (or understanding that they're there without writing them). But the hats often make a lot of deep, conceptual difference.

I remember pretty clearly that when I was 15 or so, I was slightly confused by the question about the "superpositions of the electron's magnetic fields" as well. So it may be a good idea to say a few words and equations.

The main message of this blog post is completely analogous to the 2012 text The electron is spinning, after all. In that blog post, I argued that it is totally correct to say that the electron is spinning. The electron's spin is the same "kind" of angular momentum we know from a gyroscope. It may be added to the angular momentum of the gyroscopes. When lots of equally spinning electrons are absorbed by a gyroscope, the gyroscope will start to spin, too. However, because the spin is just \(\hbar/2\), most of the "classical stereotypes" about the angular momentum (and everything else) and "shortcuts" to solve the problems become invalid.

But quantum mechanics, like classical physics, does contain the concept of the angular momentum and the electron's spin contributes to the angular momentum just like the gyroscopes! By definition (the most general definition), the angular momentum is the quantity conserved whenever the laws of physics are rotationally symmetric, as Emmy Noether taught us, and the electron's spin is no less angular-momentum-like than the gyroscope's angular momentum!

In this text, I am analogously saying that the magnetic field around the electron – which is a small magnet – exists, is the same "kind" of the magnetic field that big magnets possess, and in fact, it is given by the same formula as the magnetic field of bar magnets. However, it's all the other things a layman might expect about the behavior of the magnetic field that work differently due to the effects of quantum mechanics.

OK, let's start.

An electron is a tiny magnet with the magnetic moment that is proportional to the spin and that I repeat for you,\[

\mathbf{\vec \mu} = \frac{g_e \mu_b}{\hbar} \mathbf{S}

\] Both sides of the equation are three-dimensional vectors. The right-hand side is also an operator on the Hilbert space. The operator of the spin \({\mathbf S}\) is \(\hbar/2\) times the triplet of the Pauli matrices \(\vec\sigma\). The latter don't commute, have off-diagonal elements, and so on.

The left hand side seems to determine the nice pictures of the magnetic fields which is why it looks classical. But that's just an illusion. Because the right hand side is an operator on the Hilbert space, the left hand side has to be an operator, too. Note that it would be wrong to assume that only the expectation values of \(\mathbf S\) are proportional to the magnetic field. Expectation values are just statistical averages but the laws of physics – such as the formula for the electron's magnetic moment – have to constrain the observables in the individual experiments in some way, not just the statistical averages.

So the fundamental equations of motion (the Heisenberg equations of motion) and similar relationships simply have to be full-fledged operator equations.

It is very hard for a person who is not thinking quantum mechanically (yet) to imagine that the magnetic moment \(\vec \mu\) is an operator, too. If it is an operator, then so should be the magnetic field created by such a dipole\[


\] If we allowed the postulates of quantum mechanics for fields around magnets, the beginners are afraid, then we could easily end up with superpositions of different profiles of magnetic fields and that would be as bad as the complex superpositions of a dead cat and an alive cat.

However, what the laymen still completely miss – especially because of the constant publication of dozens of increasingly wrong popular books about quantum mechanics – is that the complex superpositions of the different states of a cat are allowed to exist. Their existence in the Hilbert space is a general, universally valid postulate of quantum mechanics.

For cats, the superpositions only exist "in principle" because "in practice", we're not able to do measurements that depend on the relative phases and interference which is why the operator of the cat's "livelihood" may safely be assumed to be diagonal with respect to states we care about, and, consequently, it may be visualized like a classical observable in a regime where classical physics is a good approximation.

However, when it comes to the magnetic fields around the electrons, the superpositions (and other characteristically quantum phenomena and concepts) are absolutely vital for a proper understanding of the "character" of the magnetic field of an electron that is spinning up, for example.

Note that the electrons obviously have to have the magnetic field of a magnetic dipole in a sense, at least when it comes to the expectation values. The comparison of the electron's magnetic field and the magnetic field of a bar magnet may be thought of as an analogy. But it's actually much more than an analogy. The bar magnets' power to create the well-known field on the picture is literally composed of the electrons' power to produce a magnetic field. Ferromagnets' magnetic moment is nothing else than the sum of the magnetic moments of (some) electrons that just happen to spin in the same direction because their spins' aligned direction is energetically favored!

So if you have an electron spinning up, you may imagine that \(\vec\mu\) is a vector in the positive \(z\)-direction, and the formula for the expectation value of the magnetic field\[

\langle\mathbf{B}({\mathbf{r}})\rangle=\nabla\times{\langle\mathbf{ A }\rangle}=\frac{\mu_{0}}{4\pi}\left(\frac{3\mathbf{r}(\langle\mathbf{\vec\mu}\rangle\cdot\mathbf{r})}{r^{5}}-\frac{{\langle\mathbf{\vec\mu}\rangle}}{r^{3}}\right)

\] simply has to be right. In quantum mechanics, the expectation value may be written in terms of bra-vectors and ket-vectors so the equation above really means\[


\] I claim that a similar formula works for \(\ket\downarrow\) and \(\bra\downarrow\), too. Now, those things may look like a contradiction for the following reason:
The up- and down- spin states with respect to other axes different than \(z\), such as \(x\) and \(y\), are complex superpositions of the spin-up and spin-down states. But the magnetic fields we attribute to differently rotated bar magnets don't seem to be superpositions of each other, especially not complex superpositions.
Is there really a contradiction? Of course that there is no contradiction. Why?

Note that the last displayed equation relates some matrix elements with respect to the states \(\bra\uparrow\) and \(\ket\uparrow\). It's an equation for the vacuum expectation values. I have pointed out that in quantum mechanics, such laws wouldn't be enough. There should be a full-fledged operator equation from which the equation for the matrix elements may be extracted – by simply sandwiching the operator equation in between a bra vector and a ket vector.

So is there such a master operator equation? You bet. Why don't you just remove the bra vector and the ket vector? We are left with the equation we started with,\[


\] except that there are no bras, kets, langles, and rangles, and we must remember that \({\mathbf B}(\vec r)\) as well as \(\vec\mu\) have both arrows and hats above them (even though I generally omit all the hats because hats look like a luxurious decoration while being operators is what the observables want and are obliged to be by default – it's being classical that should be denoted by a cap or a wig). They are both three-dimensional vectors and operators on the Hilbert space. Also, you shouldn't forget that with respect to the state with one spin-up electron and the state with one spin-down electron, the operator \(\vec\mu\) is proportional to the Pauli matrices, with the coefficients that may be easily traced.

(The equation above is only OK when applied to one-electron states that contain nothing else. For more general states, we would have to discuss the spin of all particles in the system, and we would also have to add contributions from operators creating or annihilating physical photons because \(\mathbf B\) always contains them in the decomposition. These extra terms have vanishing vevs and matrix elements in the spin-up and spin-down states of one electron that contain no extra photons.)

It follows that the operator \({\mathbf B}(\vec r)\) has to be a superposition of the Pauli matrices, too. In particular, with respect to the spin-up and spin-down basis of the one-electron (near the origin) Hilbert space, the operator \({\mathbf B}(\vec r)\) has lots of off-diagonal elements that you may easily quantify – they come from the Pauli matrices \(\sigma_x,\sigma_y\) i.e. from the corresponding components \(\mu_x,\mu_y\).\[

\bra\uparrow {\mathbf B} ({\mathbf r}) \ket\downarrow \neq 0.

\] If you want to know, it's the matrix \(\sigma_y\) that is pure imaginary so whenever there is a term in the expression for the magnetic field proportional to \(\mu_y\) in the matrix element of the magnetic field, the matrix element will acquire a pure imaginary contribution. The resulting complex numbers are multiplied by the complex amplitudes appearing in bra and ket vectors, if the latter are complex superpositions, and we finally obtain a real result of the magnetic field's vev in any state (because this operator is Hermitian) but all the relative phases matter.

The fact that the magnetic field \({\mathbf B}(\vec r)\), even when \(\vec r\) is macroscopically separated from the electron, like one meter, still behaves "quantum mechanically" in the sense that it has nonzero off-diagonal matrix elements is something that the laywoman implicitly assumes to be impossible, and even if you tell her that the assumption is wrong, she will actively fight against the unavoidable fact. (Note that I am politically correct so in order to increase the diversity and fairness, I use the word "she" for the idiot.) She believes that things that are "macroscopic" in the sense of being a meter away simply can't behave quantum mechanically.

But they always do. This in no way contradicts the fact that classical physics becomes a pretty good approximation for most phenomena at long distances. But classical physics only becomes a good approximation because if you calculate how strong effects resulting from quantum mechanics may be measured, they become much weaker (which often means much less likely!) as the distances become longer. Only if you calculate the final result, what is actually operationally measured by your apparatus, the quantum effects may become small. However, classical physics surely doesn't work at those distances because the quantum principles would disappear conceptually. The postulates of quantum mechanics never break down and they never cease to hold. In particular, if you want to see why quantum mechanics contains no contradictions, it's very likely that you won't be allowed to replace its whole parts by classical physics. It always hurts to do so. If something is quantum mechanical, everything that can be affected by it has to be quantum mechanical, too. Magnetic fields are produced by the quantum mechanical spin of the electron, so they have to be quantum mechanical operators, too!

In the previous paragraph, I said that the effects of quantum mechanics usually become weak and "hard to measure" at very long distances. Can we see it for the electron's magnetic field? Yes, of course.

What is totally silly is to imagine that the magnetic field around the electron is so strong that we should visualize it as a condensate of a macroscopic number of photons. If you interpret the field in terms of photons at all, the average number of photons in that field is very small, at most of order one. So you definitely need to consider all the effects of quantum mechanics analogous to the zero-point energy of the harmonic oscillator (well, not just "analogous to": the electromagnetic field literally has these zero-point energies, too). So for the calculation of the second powers of magnetic fields and their expectation values, the classical formulae will be wrong or "incomplete" due to the uncertainty principle. But at the linear level, the magnetic field of an electron (its expectation value) has the same functional dependence on \(\vec r\) as the magnetic field of a bar magnet.

You want to measure the magnetic field, right? You will surely be able to find a contradiction, to prove that the field can't be in the complex superpositions, won't you?

Of course, you won't. The electromagnetic field, like everything else in the Universe, is always allowed to be in any complex superposition of other two allowed states! What's probably wrong with the layman's or interpreter's intuition is that he still imagines that one may measure a physical system without affecting it.

But in quantum mechanics, it's never the case. Whenever you measure something, you affect the system and bring it to an eigenstate of the measured observable. Because of this change, the later measurement of any observable that doesn't commute with the first one will produce different results than if the first measurement won't have taken place! (Is that the right tense?)

It's extremely hard to measure very weak magnetic fields such as those produced by the magnet inside an electron. It won't be able to macroscopically move any "needle". Needles are just too heavy. You will need some very "light" probes if you want the probes to be visibly affected. If you think what it means, you will obviously be led to use another small particle like your "apparatus". The measurement of the electron's magnetic field will become equivalent to the measurement of its spin, whether you like it or not.

So the whole mystery about the "seemingly different" character of the elementary particles' magnetic fields on one hand and their spin on the other hand will go away. Of course that the magnetic fields will behave like the spin – quantum, non-commuting observables with lots of off-diagonal elements. After all, we really measure the spin. For example, you may scatter another particle off the electron and the predicted probabilities or cross sections will depend on the electron's spin, so you will gain some information. But by doing so, you affect the electron's spin, too. It may change. That's why the imperfection of the measurement of the magnetic field by "one scattering" can't really be improved by performing the scattering repeatedly. Each scattering changes the situation!

But that doesn't change the fact that the formula for the magnetic field of a magnetic dipole is still correct. As an example, consider orthohydrogen and parahydrogen. It's all about a subtlety of the \(H_2\) hydrogen molecule. Note that hydrogen gas in a bottle is composed of these diatomic molecules.

The molecules are composed of two hydrogen atoms. An ordinary, light hydrogen atom has a simple nucleus: a single proton. The proton has a spin and magnetic moment, too, although the magnetic moment is about 3 orders of magnitude smaller (the ratio is comparable to the inverse ratio of the proton and electron masses, i.e. 2,000, but it is not quite equal, due to the complicated quark composition of the proton).

In the orthohydrogen molecule, the spins (and therefore magnetic moments) of the two protons are aligned – pointing to the same direction. In the parahydrogen, they are antiparallel – pointing to opposite directions. The \(2\times 2 = 4\)-dimensional Hilbert space of the two protons' spins may be generated by a basis consisting of 3 basis vectors of the spin-one \((I=1\)) orthohydrogen; and 1 basis vector of the spin-zero (\(I=0\)) parahydrogen.

A point I want to make is that the usual "magnetic potential energy" of the two magnetic dipoles indeed does contribute to the slightly different energies of the orthohydrogen vs parahydrogen. The magnetic energy of a magnetic dipole in a magnetic field is \(-\vec\mu\cdot {\mathbf B}(\vec r)\) and you may indeed substitute the magnetic field \({\mathbf B}(\vec r)\) from the inverse power law formula I started with (this magnetic field is centered around the other particle with the magnetic dipole).

So everything is perfectly consistent. You must remember that quantum mechanics, its superpositions, nonzero commutators, inability to measure things without modifying their state, the probabilistic character of predictions, and so on, and so on, applies to everything, including electromagnetic fields. The Hilbert space of one electron in a given region (with some determined position-or-momentum information) is indeed exactly two-dimensional. The spin-up and spin-down states (with respect to any axis) may be chosen as a basis. The magnetic field around the electron is entangled with the spin of the electron itself (just like Schrödinger's cat and perhaps also Wigner's friend is entangled with the radioactive nucleus and the lethal hammer device) but its expectation value is still given by the usual formula from the bar magnet.

The complex superpositions of the "up" and "down" states of the electron produce the right expectation value of the magnetic field – like the field of a bar magnet rotated in a different direction – largely because of the off-diagonal elements of the operator \({\mathbf B}(\vec r)\). The off-diagonal elements are not needed for the calculation of the vev of the magnetic field in the state "up" or the vev in the state "down", but they are needed for the calculation of the vev in the states polarized with respect to all other axes.

Quantum mechanics always works, classical physics usually doesn't, and even if classical physics gives correct answers to some questions, however "obvious", they have to be verified using the formalism of quantum mechanics and the derivation of "obvious" things in quantum mechanics may often look nontrivial and require some algebra. That's in the special case when classical physics is essentially OK; I emphasize that it's usually not.

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reader Leo Vuyk said...

An alternative layman interpretation of the up and down spin states of the electron.
See perhaps also:
Calabi Yau shaped double Fermion spin states.

reader Uncle Al said...

Molecular hydrogen's 3:1 ortho:para spin states at ambient conditions are serious stuff for NASA. Enthalpy of vaporization of para-hydrogen is 904 J/mol at its 20 kelvin boiling point. Ortho to para net rotational energy there is 1091 J/mol, If you liquefy ambient hydrogen without spin equilibration (e.g., iron on charcoal), more than 45% of it will vaporize as the ortho-spin state decays over time. Those little circumflex top hats are important.

reader John McVirgo said...

For the benefit of the other readers that might not know, I'll mention that Hans Ohanian wrote a related paper on:

What is Spin? AJP 54 (6), June 1986

"According to the prevailing belief, the spin of the electron or of some other particle is a mysterious internal angular momentum for which no concrete physical picture is available, and for which there is no classical analog. However, on the basis of an old calculation by Belinfante [Physica 6, 887 (1939)], it can be shown that the spin may be regarded as an angular momentum generated by a circulating flow of energy in the wave field of the electron. Likewise, the magnetic moment may be regarded as generated by a circulating flow of charge in the wave field. This provides an intuitively appealing picture and establishes that neither the spin nor the magnetic moment are "internal"--they are not associated with the internal structure of the electron, but rather with the structure of its wave field. Furthermore, a comparison between calculations of angular momentum in the Dirac and electromagnetic fields shows that the spin of the electron is entirely analogous to the angular momentum carried by a classical circularly polarized wave."

reader hroent said...

"... than if the first measurement had not taken place."

Or "hadn't" :)

reader Shannon said...

Ann, I do think that Wall Street and the City are pushing the politicians to provoke an Armageddon. A big war is their only solution to revive economic growth and profits... And no one cares who is right or wrong, just do it... Scary really.

reader Gene Day said...

To understand the US position vis-a-vis Russia you need to see the world from an American perspective, Shannon. We just do not have the experience of foreign invaders and have never felt seriously threatened in the security of our homes (despite 9/11). The last serious invasion of our homeland was more than 200 years ago and we were victorious even then. When Japan attacked Pearl Harbor in 1941, Hawaii was not a state but just a territory on a remote Pacific island. We have been living in a cocoon for our entire history.
The only real threat to our safety as a people has been Russia and the Soviet Union. No other power has ever had the capability to destroy us and the fact that they still do puts them in a unique position in the American psyche.
Of course our stance and the sanctions are terrible; we agree on that, but both of our parties support the sanctions because they are popular with the public. The political right is even more vociferous than the political left. The cause is purely emotional but recall that millions of American families built bomb shelters in their homes. Nothing like that has ever happened before.
You can ascribe the US stance as stupid, and it is, but it has a very deep emotional foundation. Fear trumps logic every time.

reader Luboš Motl said...

That's an insightful explanation, Gene.

reader sirernestbarker said...

Hi. The aversions you make here have been ricocheting around since the 60's. Man. Most assuredly hitherto as well, but I can only attest to those memes uttered by sly stupid hippies, the words passing through them like vessels, but spoken with unqualified certainty, as if passing along secret knowledge privvy only to those with exeptional intelligence and rare autonomy. Today the memeplex replicates its child memes mostly electronically and those who relay them sometimes refer to the masses as "sheeple".

reader Luboš Motl said...

First, the contradiction is indeed just "apparent" because the electron's spin doesn't come from a rotation of an actual ball.

The full theory, especially string theory, is of course completely consistent. You may also say that what is "spinning" in string theory when you have an electron is actually a much heavier object, with the mass comparable to the strring scale, and the corresponding speed needed to get the spin hbar out of this heavy object is indeed "just beneath" the speed of light. The total electron mass is much lighter because of some cancellations. So the electron contains positive and negative contributions to the mass/energy, both of them are comparable to the string scale, and they nearly cancel. If you imagine that only the positive-mass contribution is separated away from the axis, which is in some sense true, you indeed get the right value of the spin from subluminal motion.

reader sirernestbarker said...

I meant to add I suppose anything is possible but I think people who say things like this have not thought them through, they're just repeating things that they find attractive, which is experienced as a true statement. For instance how would a big war improve the economy or lead to a one world gov't?

reader anna v said...

You are describing how the hoi polloi react and are manipulated. I am talking of the puppet masters, if they exist or not.

reader Luboš Motl said...

Nope, the spin is the electron's intrinsic property, so surely no interaction with other physical particles is needed for that.

reader Shannon said...

Hi Gene, well this is not reassuring about the Americans. Russians, on the other hand, do know what war and hard times mean which is then probably why Putin wants to avoid conflict.

reader Leo Vuyk said...

Could we assume that the electron (and all fermions) is a rotating stringy Higgs-field information changer?
However, than we should make the Feynman diagrams more complex including the Higgs field.
see: Quantum Gravity and Electro Magnetic Forces by Dual Repulsive Vacuum Oscillation Spectra in FFF-Theory.

reader Shannon said...

Wow, you are one clever dude, sirernestbarker.

reader Tom said...

Wow, thanks! That makes pretty good sense to me, especially the +/- contributions.

reader Luboš Motl said...

Tom, pleasure. Just to recall a topic of many blog posts here. The negative contribution to the (squared) mass of particles in string theory is proportional to 1+2+3+4+5+... = -1/12. So this is from zero-point energies that don't really carry any spin.

The spin comes from the positive contributions that balance this negative contribution, like (D-2)/24 in the case of bosonic string theory, to zero. In bosonic string theory, the massless level right above the tachyon is obtained as

alpha_-1^n | 0 >

Here, the vacuum is the one-string tachyon, and the excitation increases the squared mass by a positive amount and carries the spin at the same moment, because of the Lorentz index.

So this excitation (or similar excitations for other states) is really the "gyroscope" part of the particle, and "its" mass - only the positive contribution - should be used to calculate the spin.

At the end, the rough calculation is trivial dimensional analysis, everything is at the Planck scale. The "positive part of the electron" has the Planck mass, the latent energy, one-half of hbar of the spin, and the speed of rotating points on the string (rotating ball on the surface) is comparable to the speed of light. All these quantities are of order 1 in the Planck units, so everything works, of course. It's the small mass of the physical electron, and this is due to the cancellations with the "non-spinning" negative (zero-point-related) contribution to the squared mass.

Note that string theory really does it in a clever way. Of course, people don't even think about "rotating ball" models of electrons and related paradoxes from the history of physics when doing string theory. But the point is that whenever the paradoxes had at least some point, string theory has some secret loophole that resolves it.

Needless to say, a man-made theory that someone would just randomly "make up", like loop quantum gravity, can never "solve" any problem in this way, there is no reason why it should. At most, one may prepare a contrived theory that is supposed to answer a contrived puzzle - the amount of awkwardness that one puts in is at least equal to the amount of awkwardness that was removed by finding an explanation, so nothing really gets better if the theory sucks, and all non-QFT theories of the "interior" of elementary particles except for string theory suck in this sense, of course.

reader mr. critic said...

Gene Day, you're right about one thing and it's not any news. America knows war only from the tv. But could you tell us why do you think the US stance is stupid? What are you, Americans, going to lose from it? You think Russia is not a threat to you anymore? You think you know anything about the Russian mentality and reality?

You're a typical western man degraded by the high level of civilization. You think the world is full of fairies now and your enemy deserves mercy. Be assured that in response you won't receive any... America had played a major role in demolishing communism (no thanks to people like you) by the use of financial pressure. It's doing the same thing now against a country with no real economy, but nonetheless armed with strong delusions, which treats the West as weak.

Russia knows only one dimension of greatness - the military one. Only there and in the communist countries of Asia you can witness megalomaniacal war parades every year. Putin always wanted to be the strongest among the strong (his own words). He follows the philosophy of the jungle. And he views the West with its humanistic values and individual freedoms as weak. You think such power is no threat to any civilisation? It's certainly a rival in the power competition and why people like you choose to defend the other side and not their own country is one of the nature's miracles.

reader MikeNov said...

As long as the EU pushes, Russia will cave. It is not that they believe they cannot win, but for Russia what is primary is maintaining its energy profits. The worst outcome for Russia is if Europe gets so serious about sanctions that they ignore the Greens and develop shale. Has Russia threatened to withhold Gazprom anywhere except Ukraine, which is simply being told to pay the full price.

reader MikeNov said...

It's not clear that Russia is the bad guys vs the EU. EU is headed towards becoming a Muslim state, particularly UK and France.

reader John Archer said...

Yes, we need to be rid of them, and all the other asians, blacks and whatnot. They were brought in, or let in, with no mandate to do so. The complete antithesis of democracy, undermining the very demos on which it is built. I do not consent to wogs, blacks or any other alien element having any say in how I am governed in my own country.

No one wants pakis living anywhere near them. Same goes for the rest.

So, how do you suggest we go about restoring the status quo ante (pre 1948)? Assisted repatriation, just kick them out, or what?

reader Gene Day said...

You are putting so many words into my mouth that I scarcely know how to respond. My view of the world is not at all what you think so get off your high horse or please shut the hell up.
“Stupid” means ill thought, not necessarily completely wrong, and there is no better word to describe your insulting diatribe.
Putin is a rational man doing what any good Russian leader would do. That does not imply that he and his country are not dangerous. That inference is absurd and you should be ashamed of yourself.
TRF is generally an intelligent forum and it does not need your preaching.

reader Tom said...

Lubos, that is too cool (along with your new post). I especially like [But the point is that whenever the paradoxes had at least some point, string theory has some secret loophole that resolves it.] You’ve bumped up my drive to understand string theory, but I’ve still got a long ways to go. (Have to admit that I’m bogging down in the second volume of Messiah - that book takes a real effort. By the way, is the intro to QFT in Messiah good enough to allow one to skip to a good intro to String Theory?)

reader Gene Day said...

Putin will do what is best for Russia and for his own political position within Russia’s politics. That’s what all politicians do. Putin will avoid conflict unless he can profit from it. There is no reason to demonize him or his country. He does not eat little children for breakfast.

reader Luboš Motl said...

Dear Mr Critic, the communist bloc collapsed 25 years ago, maybe you could finally take this fact into account if you don't want to increasingly look like a rotten fossil.

The provocations against Russia are even more stupid *because* Russia considers the military power more important than other forms of power. The harassment against Russia - which has been largely innocent in all the events surrounding Ukraine - has already led to increased arms races, spectacular approval rate of Putin well above 80%, and astronomical economical losses in many nations.

This is exactly what a wise America wouldn't want - and for example, Reagan didn't want it, either.

Your comment that Gene is "degraded" is unacceptable, and so is you suggestion that America "should have no mercy" as if America were just standing above a villain Russia with a knife in its hand, dictating conditions. The situation looks nothing like that.

At least, you should realize that you have no power to decide whether you will have "mercy" on this blog. You write one more offensive comment similar to your latest one and you're banned. Deal. I won't have mercy with assholes like you.

reader Gene Day said...

Wow! That sets an entirely new standard of xenophobia.

Another solution comes to mind:
YOU could leave!

reader Luboš Motl said...

Right, John, I have already heard these tough words of yours. Could you please avoid repeating them, especially in off-topic contexts?

reader Tom said...

Well said, Gene. You left a “should” out of your second sentence, however. I don’t think it’s arguable that Obama is too worried, or in fact gives a shit, about what is best for the USA. So there is no telling how far this Western-Russian confrontation might ramp up.

reader John Archer said...

Yes, OK, Luboš, but it was an off-topic response to what I suppose was an equally off-topic statement. Still, your blog, your rules.

Nevertheless, I'd like to respond to Gene below.

reader John Archer said...

"Wow! That sets an entirely new standard of xenophobia."

You say that as if there were something wrong with it. On the contrary, it's perfectly normal and very understandable: Croatia, Bosnia, Czechoslovakia, Uganda, Ukraine ... and the French weren't too happy about the historical English territories in France, but then we weren't too happy about the Normans either. Even much closer to home: Scotland, and we're the SAME PEOPLE. No, it is your view that is abnormal. Indeed it's pathological in the extreme.

Ask those sub-continentals, say, if they'd be happy to host mass migration from sub-Saharan Africa. I know for fact they really, really don't like Africans. Maybe the Chinese would be happy to play host to them? What do you think?

A friend of mine was a Brahmin, and a well-educated one at that. He was adamant that blacks were subhuman (yes, subhuman!) and that his view was commonly shared back in his home country. He wasn't the only one of them I knew to express that view either. I guess he, and they, felt they had their feet comfortably under the table and every right to be here to happily confide such a view. [NO WAY! They have no rights here.] But they did complain that whitey was prejudiced against them. Haha! Talk about blind, eh.

Do you have any views on that? For that matter, any Indians out there want to contradict him/them?

"Another solution comes to mind: YOU could leave!"

How wonderfully magnanimous of you! You must have worked extra hard to come up with that one — squatters' rights trump owners'; you could build a civilisation on that, couldn't you?

Fucking brilliant in fact, but I have a better one: you put them all up at your place, then we'll see how much you like a veritable south-asian invasion. Your environment will look wonderfully hideous, and totally fucking ALIEN. You'll love it. Smell that curry 24/7. Then you can leave, or hang around and continue pumping out that 'diversity is wonderful' shit. But I guess you'll pump it out wherever you are.

You and your kind will be responsible for the coming race wars if clearer heads don't prevail in the meantime.