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Do stringy electrons spin faster than light?

No. String theory's resolution of the old paradox is a sign of the hidden cleverness of string theory.

After he read some of my essays on the electron's spin, Tom W. Larkin asked an interesting question:

Does string theory resolve the paradox of (post-)classical physics that the electron, if imagined as a spinning ball of a very small radius, has to rotate faster than the speed of light for its spin to be \(\hbar/2\)?
One natural, fast, legitimate, but cheap reaction is to say: the electron isn't really a rotating ball. The spin may be carried even by a point-like particle, without any violations of relativity, as QED shows, so the paradox has never been there.

Of course that a string theorist is likely to answer in this way, too. Quantum field theory is a limit of string theory so any explanation that was OK within quantum field theory may be said to be correct within string theory, too. The paradox doesn't exist because the electron isn't a classical ball that gets the mass from the electrostatic self-interaction energy.

However, string theory does represent the electron (and other elementary particles) as some kind of an extended object which is qualitatively analogous to the rotating ball so some version of the "superluminal spinning" paradox may be said to reemerge in string theory. Does it cause inconsistencies within string theory?

It doesn't but the reasons are tricky and ingenious, if you allow me to lick Nature's buttocks a little bit.

The angular momentum (I will call it "spin") of a gyroscope etc. is equal to\[

\vec S = I\vec \omega

\] where \(\omega\) is the angular frequency and \(I\) is the moment of inertia. Up to constants of order one, the moment of inertia is equal to\[

I \sim mr^2

\] where \(r\) is some typical distance of the points of the object from the axis of rotation and \(m\) is the mass of the spinning object. That was some elementary classical mechanics, OK?

Now, if you assume that \(E=mc^2\) for the electron where \(m\) is the rest mass and that the full latent energy \(E\) is obtained as some electrostatic energy of the electron's charge \(e\) interacting with itself, via (up to numerical constants of order one)\[

E \sim \frac{e^2}{4\pi \epsilon_0 r},

\] then you may derive the typical distance \(r\) between the "pieces of the charge" of the electron i.e. the "classical electron radius" which is a few femtometers (that is \(10^{-15}\) meters).

If you know \(r\) and \(m\), you may calculate the moment of inertia \(I\sim mr^2\) as well as the angular frequency \(\omega\sim I/S\sim I/\hbar \). When you multiply this \(\omega\) by the radius \(r\) again, you get the velocity \(v\) of the points on the spinning surface of the electron. And it will be much higher than the speed of light!

I invite you to complete the steps above if you have never done so. You may evaluate all these things in the SI units, as if it were a basic school problem in mechanics. However, it's also nice to calculate it as an adult physicist, in the Planck units. In the Planck units, the electron mass is something like \(10^{-23}\). Similarly, setting the fine-structure constant to one for a while (we will have to return to this approximation), the electrostatic formula above implies that \(E\sim 1/r\) and the classical radius is therefore about \(10^{23}\) Planck lengths.

The moment of inertia is roughly \(mr^2\) so there are two factors of \(10^{23}\) "up" and one factor down, so it is again \(10^{23}\) in Planck units. The angular frequency is \(\omega\sim 1 / I\) where \(1\) means \(\hbar\) and it is again of order \(10^{-23}\), but if you multiply it by the radius \(10^{23}\) to get the speed, you get something of order one (the speed of light).

To quickly see (without real calculations) whether the speed on the surface is greater than the speed of light, we must be a bit more careful about the fine-structure constant. The mass was OK, \(10^{-23}\), but the electron radius is \(137\) times smaller than we have said because this reduced \(r\) has to cancel the \(\alpha=1/137\) that appeared in the numerator. The moment of inertia is \(137^2\) times smaller than we said, because it's \(mr^2\), and the angular velocity is therefore \(137^2\) times larger than we said because \(\hbar = I\omega\) has to be kept fixed. Even when multiplied by the \(137\) times smaller radius, we still get the velocity \(137\) times larger than we said.

The speed of the surface of the electron is therefore comparable to \(c/\alpha\sim 137 c\), up to numbers of order one that hopefully don't reduce \(137\) below one. You see that the speed of the classical electron is higher than the speed of light. A bummer.

If you quickly and naively think about the changes that string theory makes to this calculation, string theory makes the problem worse because the electron in string theory is smaller. The energy of the electron comes from very stiff strings inside, not from the electrostatic energy, so the extended string hiding in the electron isn't \(10^{-15}\) meters large but \(10^{-35}\) meters tiny or so, not far from the Planck length (the string length, about 100 or 1,000 times longer, would be a better estimate).

So the size of the electron has seemingly shrunk \(10^{20}\) times which means that the required velocity on the surface has to increase \(10^{20}\) times – hopelessly larger than the speed of light. Do the points on the string move with these excessive superluminal speeds?

The answer is, of course, No. String theory reduces the "classical radius" of the electron but it changes other things, too. Most importantly, it changes the relevant mass, too.

The trick is that the thing that is spinning isn't as light as the electron. It's as heavy as the Planck mass (again, more precisely, the string mass, the square root of the string tension). Why? Because the electron, like all massless and observably light particles, comes from the massless level of the string whose mass is constructed in a similar way as the massless level of the bosonic string theory which I pick as an example because of its simplicity. The massless open bosonic strings are given by\[

\ket\gamma = \alpha_{-1}^\mu \ket 0.

\] I called the one-string state a "photon". A similar relationship holds for massless closed strings (the graviton, the dilaton, and the \(B\)-field) but there are two alpha excitations (one left-moving and one right-moving) in front of the tachyonic ground state \(\ket 0\).

If we want to see how the string pretending to be a point-like particle is spinning, we may see that it's really the oscillator excitation \(\alpha_{-1}^\mu\) or the analogous excitations in the superstring case (that may include fermionic world sheet fields) that carries all the spin because it carries the \(\mu\) Lorentz vector index (or spinor indices, in the Green-Schwarz formalism for the superstrings). This \(\alpha\) oscillator literally corresponds to adding some relative motion to the individual points of the string, so that they move as a wave with one cycle (the subscript) around the string.

The tachyonic ground state \(\ket 0\) is not spinning. The tachyon is a scalar, after all. The squared mass of the tachyonic ground state (which is filtered out in superstring theory but may be still used as a useful starting point to construct the spectrum in the RNS superstring) is equal to \(-(D-2)/24\) times \(1/\alpha'\) for the open string – so it's exactly \(-1/\alpha'\) for \(D=26\) – because the term\[

\frac 12 \zav{ 1+2+3+4+5+\dots } = -\frac{1}{24}

\] is contributed by each of the \(D-2\) transverse spatial dimensions. I've discussed this semi-heuristic explanation of the zero-point energies in string theory many times (calculations that avoid all of this heuristics exist, too, but they prove that the semi-heuristic treatment involving the sum of positive integers is at least morally right whether people like it or not).

And the oscillator \(\alpha_{-1}^\mu\) is increasing the squared mass back to the massless level, by \(+1/\alpha'\). And it's the spinning part of the electron or other particles in string theory. So the relevant estimate for the mass of the "gyroscope" that we should use in string theory isn't the tiny electron mass but the string mass, \(1/\sqrt{\alpha'}\) or so.

The estimate for the speed of the "stringy surface" of the electron is easy to calculate now. In the string units \(\hbar=c=\alpha'=1\), the radius is one, the spin is one, the moment of inertia is one, the angular velocity is therefore also one, and so is the speed of the surface. This estimate is compatible with the assumption that the pieces of the strings never really move faster than light although they get close – and more accurate derivations within string theory may be shown to confirm this claim in some detail.

Note that no counterpart of the fine-structure constant such as the string coupling \(g_{\rm string}\) entered in our calculation based on string units. Everything was comparable to the string scale – which may differ from the Planck scale by a power of \(g_{\rm string}\) but the string scale really simplifies the calculation more than the Planck scale. For \(g_{\rm string}\sim 1\), you don't have to distinguish the string scale and the Planck scale.

The "string-scale" part of the electron in perturbative string theory is very heavy and therefore it's easy for it to produce the angular momentum of order \(\hbar=1\) even with velocities that don't breach the speed-of-light limit. And the tachyonic, negative contribution to the squared mass cancels most of the squared mass from the "positive excitation" and it makes the particle massless (or, when subleading effects are included, very light). This tachyonic part doesn't enter the calculations of the gyroscope.

It's very natural that string theory had to solve this puzzle – even though you could simply deny its existence – because string theory partly restored the assumptions that were used in the derivation of the nonsensical superluminal speed of spinning electron. You may see that string theory is a typical unifying theory that really wants to see all quantities of fundamental objects as being close to "one" in the Planck units. And if some quantity is much smaller than the natural Planck unit, e.g. if the electron is much lighter than the Planck mass, it's due to some cancellations that are known to occur almost everywhere in string theory.

But the fundamental parts of the explanations that matter – in this case, I mean the "positive-mass" part of the electron's gyroscope – universally work in the regime where "everything is of order one in some units". Whenever dimensional analysis is any useful in string theory, except for telling you that everything is comparable to the Planck/string scale, it's always in situations where some leading Planck-scale natural contributions "mostly cancel". Quantum field theorists like to think that any precise cancellation is "unnatural" but string theory offers us tons of totally sensible, justifiable, provable cancellations like that.

The cancellations resulting from supersymmetry represent a well-known example but string theory really does imply similar cancellations even in the absence of SUSY (or cancellations that don't seem to be consequences of SUSY), too.

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snail feedback (41) :

reader Anonymous said...

Dear Luboš,

is it possible to have You as "Vedoucí bakalářské práce"? As You might guess, I'm from the Czechia. I'm just curious.

reader Rehbock said...

"Licking Nature's Buttocks" LOL :-)

reader Shannon said...

Only Michelangelo could have painted this, right ? :-)

reader Luboš Motl said...

Dear Anon, nice, thanks. I may enjoy it and I would probably not be lazy for that but I think that because of paperwork issues, the default answer is No, it is not possible anytime soon.

reader John McVirgo said...

If my intuition serves me right, a"rigid" object rotating at the speed of light about its center of mass in its center of momentum frame isn't stable, since there isn't a Lorentz frame common to all its points moving at different velocities.

This is a rotational version of Bell's space-ship "paradox", right?

reader John Q said...


Have you commented on Max Tegmark's "everything is math/all mathematical structures exist" idea. I'd like to hear your take. Sorry that this is off-topic.

reader Gene Day said...

Obama is pretty much the same as Putin, doing what he feels is best for his country, consistent with the world view of his base. Our position is clearly wrong but having another man in that office would not make things better and it might make things infinitely worse. Imagine what might happen if we had someone rash (think John McCain) in the oval office. Caution is often a virtue.

reader Dan said...

I am a little bit confused, frankly even irritated, by this article.
Due to quantum mechanics, it is impossible even in principle to measure the speed of the surface of the classical stringy electron. I think you admit something equivalent, too. To me that's the end of the story. Specifically, I'd be perfectly happy if that imaginary surface moved at 10^20 c, or whatever, knowing I could never exploit this to violate causality in any way.
Why should I care about potential contradictions between unobservable quantities or their "resolutions"?
After teaching your readers so diligently and lucidly the rules of our quantum world, why in the world would you write an article about such naive classical non-observables?
No offense, but I prefer your quantum articles any day.

reader Rasmus Hammar said...

I think you are right in that it is the end of the story regarding speculations about the "speed of electrons surfaces", and that the uncertainty principle is enough, which it is, But nevertheless this article brought up many things more than that, whethertheless it was "necessary" or not. :). I like that you say "Why should I care about potential contradictions between unobservable quantities or their "resolutions"?" but shall refrain from answering due to my own uncertainty.

reader John Archer said...

It may have escaped your attention but England is NOT California. We are not you. We are not like you. We are our own people and want to stay that way. What you lot do there is your business. Delude yourselves if you wish, but don't try to sell your foolish crap here. And If you think indians and pakis here want to be English (whether we want them or not) you've got another think coming. They do NOT.

Incidentally, Mark Steyn doesn't share your view. OK, he's a Canadian but many of his Californian readers agree with him and not you. So you certainly don't speak for all Californians, let alone all Americans, many of whom I know to be extremely concerned about the 'diversity' being foisted on them, just as we are with what's happening here. Even Lincoln wasn't happy with it. So much for it being a quintessentially 'American' thing too. This muli-culti, multi-racial scam is a powder keg waiting to go off. It's only a matter of time.

And if it is all as wonderful as you paint it in California, then why is immigration such a hot topic in the US. Oh wait, don't tell me — it's all those awful white racists that are stirring things up and nothing whatsoever to do with your government's failure (deliberate policy more like) to secure its borders?

"I would try to teach you the virtues of tolerance but you are obviously beyond hope."

Well, now that's very kind of you, and incredibly patronising. (BTW, would the lesson be at the point of a gun? Oh never mind.) And that would be your version of tolerance, would it, Cardinal Day? And if I deny you thrice, can I hope to be crowned Pope maybe, or whatever the procedure is?

You moral evangelicals are all the same. It doesn't matter where you are, not even in the land of the free, your impulse to authoritarianism is a threat to all around you. You're a bunch of religious fanatics.

As it happens I used to be very tolerant of diversoids, far more so than most others I can tell you. But reality intruded a long time ago with the explosion in numbers and now the ugly consequences. Truly ugly. All these changes happened in my lifetime and there was no consent to any of them. You do understand the importance of consent, don't you? No, I know you don't. At best you'll pay lip service to it, that's all. Indeed you will studiously ignore it in any reply you deign to make here. It's kind of inconvenient. So much for democracy.

Now, I would similarly try to teach you to pull your head out of your Darwin's loser's arse but fortunately I have no experience of your predicament so there's a risk that I might succeed only in having the rest of you fall in as well, leaving just your chocolate star on display, twinkling like that grin on that Cheshire Cat. Hey, that's not a bad idea. Someone should pin one of those twinklers on Obwanana. Colour matched and the wave of the future. Keeool!

I think you're barking mad.

reader John Archer said...

You're right, Shannon.

Note the smug self-righteousness of the evangelist on heat. Suffocating ain't it?

reader Luboš Motl said...

There isn't any Lorentz frame where points moving at the speed of light are rest - whether the points are rotating or not.

And there isn't any common inertial frame for all points moving at different velocities - there isn't even any Lorentz frame for any *single* point that is rotating (except for points on the axis).

Moreover, points of string never move strictly at the speed of light, just approach it. So I am not sure what the paradox could be and what you want to say.

However, what probably addresses what you're bothered about, whatever it is, is that special relativity doesn't allow any perfectly rigid objects. When you start to rotate a disk, the circumference contracts while the radius doesn't, so one violates circumference = 2*pi*r.

All disks in special relativity stretch or compress a little bit.

Note that this has no impact on strings which are not disks - they are 1-dimensional. One-dimensional lines never have any intrinsic curvature. And two-dimensional world sheets have curvature fully described by the scalar curvature which may be removed by a Weyl rotation so if we're only interested in the angles on the world sheet, enough to trace rotation, there is no curvature on the world sheet, too.

This special relativistic reasoning from the high school might be guesses to be unrelated to cutting-edge features of string theory but it's actually closely related. String theory's consistency in removing the infinities of quantized gravity *depends* on the strings' not having any intrinsic propagating curvature degrees of freedom i.e. on their immunity against the "Bell" (Dewan's-Beran's) spaceship paradox. A small piece of the world sheet always has the same tension, too.

reader Luboš Motl said...

Hi, see

reader Luboš Motl said...

Sorry, Dan, I can't see the subtleties what goes on in your head but what I see is simply wrong.

Why do you think that due to quantum mechanics, it's impossible to measure the speed of the rigid classical electron? Quantum mechanics doesn't think that individual quantities can't be *measured*. They can be measured perfectly. Velocity of a well-defined point, like a tooth on a rigid surface, may be measured with any accuracy. Quantum mechanics only restricts the accuracy with which two different non-commuting observables may be measured. But individually, velocities can be measured exactly as nicely as positions, energies, angular momenta, flavors, parity, or any other observable.

If the electron were composed of many points - and be sure that quantum mechanics allows objects to be composed of many pointlike particles, that's the quantum mechanical description of macroscopic objects - then one could talk about the velocity and speed of any such point. It would be given by an operator, it could be measured, it would be eigenvalues, and if the eigenvalue increased above the speed of light, it would be in conflict with relativity!

If one defines and measures velocity of a "rigid point" - with some matter - correctly, the velocity greater than the speed of light is always a violation of causality.

reader Luboš Motl said...

Rasmus, you are wrong as well. The uncertainty principle in no way prevents us from measuring the velocity (only) totally accurately, and it in no way allows the measured values of the velocity to exceed the speed of light.

reader Dan said...

Thanks for your reply, Lubos.

You (correctly) estimated the size of the string representing the electron to be at most the string scale. In string theory distances shorter than the Planck scale (and, perturbatively, even the string scale) cannot be probed because there exist no probes that could distinguish them.

reader Holger said...



\frac 12 \zav{ 1+2+3+4+5+\dots } = -\frac{1}{24}

seems obviously wrong. But you won't type that accidentally, it must be an 'insider' - would you give me a hint?

reader Casper said...

I'm glad to see that Lubos is getting around to the most useful particles at last.

reader mr. critic said...

Funny, being a fascist asshole I've never proposed annihilation of large masses of humans, while you Motl - a renown democrat, are constantly using this kind of "jokes" and I see a notable correlation with your support of all things militant, like the separatists who are actually just unemployed, former war mercenaries dying of boredom. "degraded" was a poor choise of wording, sorry. It's an offense as much as it would be if I say that he listens to too much Rachmaninov and no country music at all. I simply don't understand the pacifism in this case. Russia is not a villain in the obvious sense, but its actions can be characterized as rational only in medieval context.

What you say about Russia is true, but you miss the result. I don't know if Reagan made all the good job consciously, but the arms race was one of the aspect of the competition that ruined the Soviets. Blinded by their zealotry the russians spent all their fortune on weapons and ultimately failed. It will happen again if Russia goes in this popular among its people direction, because it lacks the economic power to sustain its delusion of grandeur for long. I would really like too see this scenario in its end and the faces of all people that clap for Putin and his rationality. Maybe you think that if everyone treats Russia like a very good neighbour Putin will start to play with little puppies? And no, Russia will not respond to financial sanctions with nukes. The idea is amusingly childish. But such conversation is potentially endless, so it's pointless. Believe what you like. You don't have the luxury of the math instruments on this issues, so you can't prove any absolute truth.

And please, don't tell me which battle is old, because I'm fighting it here everyday. Yes, the tumors which Moscow is officially sponsoring on our territory, called political parties, made of the lowest of scum you can imagine. But you can't even imagine it because you're living at a very safe distance from the russian sphere of influence, which receives so much praise here, far away from its grasp.


Gene Day, ok, I was too quick on the draw, excuse me. But let me make another inference. The sanctions are ill thought, but they make a lot of damage to the Russian elite's affluence - the empty threats they're making reveals only panic. So if the sanctions were completely right they would make no damage at all? Then why call them sanctions and not a friendly hug or something? That's what Putin wants - a weak West, which only contemplate intellectually while he plays his game of dominance. And the main question is not if he's rational or not, but whether his behaviour is in our favor. It's certainly not in mine.

reader Erwin said...

omg Holger, if you don't know this I wonder how you follow Lubos' blog!

reader Rasmus Hammar said...

Thank you for the answer Lubos! You helped me remember some things that apparently super late night drinks can remove. Sorry also to Dilaton for taking away some of the joy the article gave you. Hope it wasn't too horrible :) All the best, Rasmus.

reader PlatoHagel said...

Lubos......the phrase is "kick some ass." K and and L being right next to each other can be as easy as hitting the wrong key.

reader Dan said...

Hi Dilaton, there is no (perturbative) probe smaller than the string scale which would be needed to probe inner structure of the electron. I never used the words uncertainty principle. Quantum is key here because in a classical theory any evolution implied by the equations of motion is observable in principle. In a quantum theory it suffices to show that the would-be source of inconsistency is unobservable.

reader bob sykes said...

Nice to see the infamous sum of integers pop up again.

reader Dilaton said...

Ok, I was probably a bit too harsh at least to you ;-)

The thing is that due to the public bullying of theoretical physics officially called "science journalism" among other things, there are only few places left in the internet these days, where such topics can seriously and peacefully be discussed. This makes me probably a bit overly sensitive ... :-/

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reader MarkusM said...

Maybe this helps

reader Luboš Motl said...

Note the word "perturbative". There exist probes able to see things at shorter scales, like D0-branes, which is why this comment has nothing to do with the essence of the problem. It is still possible in principle to see things related to the inner structure which is why the contradiction shouldn't be ignored in this way.

The actual correct "simple QM" or "simple QFT" explanation why it is not a problem is that the angular momentum has two terms, one from the motion (orbital angular momentum) and one that may be assigned even to point-like particles (the spin).

But string theory really eliminates this division - all properties of elementary particles come from motion of the strings, motion of one type or another. In this sense, the composition of a particle from a string is analogous to the composition of an atom from electrons and protons, and all the properties like the angular momentum *do* arise from the actual motion of points on the strings (analogously to electrons etc.) in a (super)space. The split to "orbital and spin" is eliminated as string theory unifies them.

That's why the order-of-magnitude estimates are not allowed to go wrong, and they don't go wrong.

More generally, even in perturbative string theory where the probes are "fuzzy", one is still able to make order-of-magnitude estimates of most quantities, and of course that most properties related to free (non-interacting) strings scale like a power of the string scale. The coefficient is sometimes divergent and cutoff-dependent etc. and requires extra discussions, but the dimensional analysis is *always* possible. It is just completely wrong to completely deny the legitimacy of this technique or the legitimacy of the statement that the estimate for the speed of particular points (on the string) shouldn't be parametrically above the speed of light.

reader Luboš Motl said...

I don't dream about Putin who plays with (even smaller) puppies. I think that it is *your* personal defect to dream about controlling other people - in this case even leaders of powers - at this intimate level.

Putin likes to play with whatever animals he likes. If he plays with the tiger, I don't really feel threatened and I know very well that you are not threatened, either. His pleasure while playing with tigers is actually correlated with the fact that he is considered the right guy for this job by his nation.

So if someone wants to hurt Putin or any other Russian or human beings just because he likes to play with tigers or *anything analogous*, then he is an evil person whose acts are really unfair. and that was the last droplet that made me finally ban you. You have been a huge pain in the ass.

BTW country music is OK but you surely don't believe that it is on par or more refined than classical Russian music, do you?

reader Edwin Steiner said...

See also this nice article by Terry Tao on this kind of "impossible" sums:

reader Tom said...

I give Obama credit for not being a warmonger, like McCain, but no way that man places much weight on what is “best for his country”.

reader John Archer said...

I wouldn't be so sure.

Maybe he is doing his best for his country. It's just that he has his limitations, the biggest one being that he currently doesn't happen to be its president — he's too busy fucking up America right now.

reader Holger said...

Arrrgh, thanks a lot, I expected some kind of dirty tricks like that! After this much of smoothing I need a smoothy now ...

reader Peter F. said...

It is something absolutely amazing about what you are telling us It seems Nature allows some anthropic aspects of itself to "reasonate" in tune with how it plucks its strings at and below the Planck-scale ♬ while it is peering at its Platonic Partiture of physically realizable possibilities.:-)

reader Kimmo Rouvari said...

Thirst for both realism and locality? :-) Be my quest!

I'll update the post (that trivial part) in couple of days.

reader Dan said...

Sorry, but the D0-brane argument is a ruse - they also have size of order the string scale. In fact, there is no stable particle in the spectrum that is smaller.
I do agree with the arguments about scaling and dimensional analysis. Also your logic is relevant for larger states, so it's no wonder it works in the limit of elementary particles like electrons.
Still, you can't ever measure the speed of "an electron's surface", even in principle.
Best, Dan

reader Luboš Motl said...

It's not true, Dan. D0-branes, when slow, do probe parametrically substringy distances. What they obey is one of the "new" uncertainty-principle-like inequalities,

delta x * delta t > alpha'

They may probe "x" much shorter than the string scale if they sit there and move very slowly and can't distinguish time with an error that is parametrically longer than the string scale.

I can't measure the "speed of electron's surface" in string theory because there's no well-defined surface but in perturbative string theory, the speed of a point on the string is a well-defined observable that, along with the momentum, all other observables on the one-string Hilbert space are actually built from!

reader Dan said...

Agreed, and thank you for this open and honest discussion. Big TRF fan since 2004 here, btw.

reader Luboš Motl said...

Thanks - and what a long-term loyalty. ;-)

reader Linda Serena said...

Hi Gene, that really reminds me that the USA, despite having always been a country of immigration, has absolutely no experience and no idea what non integrative Islamist mass immigration means. You will find out.

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