tag:blogger.com,1999:blog-88830342019-04-21T10:34:28.260+02:00atdotdeWhat comes to my mind. Physics, politics, computers, rants. All CC-BY-SA unless otherwise stated.Unknownnoreply@blogger.comBlogger198125tag:blogger.com,1999:blog-8883034.post-25692531746501944332019-03-29T12:02:00.000+01:002019-03-29T12:02:20.890+01:00Proving the Periodic TableThe year 2019 is the <a href="https://www.iypt2019.org/">International Year of the Periodic Table</a> celebrating the 150th anniversary of Mendeleev's discovery. This prompts me to report on something that I learned in recent years when co-teaching "Mathematical Quantum Mechanics" with mathematicians in particular with <a href="http://www.mathematik.uni-muenchen.de/~hkh/">Heinz Siedentop</a>: We know less about the mathematics of the periodic table) than I thought.<br /><br /><img alt="" class="mw-mmv-final-image jpg mw-mmv-dialog-is-open" crossorigin="anonymous" height="213" src="https://upload.wikimedia.org/wikipedia/commons/b/b2/D._Mendeleev%27s_Periodic_table_from_his_book.JPG" width="320" /><img alt="" class="mw-mmv-final-image jpg" crossorigin="anonymous" height="320" src="https://upload.wikimedia.org/wikipedia/commons/thumb/b/b3/Medeleeff_by_repin.jpg/1024px-Medeleeff_by_repin.jpg" width="256" /><br /><br />In high school chemistry you learned that the periodic table comes about because of the orbitals in atoms. There is Hundt's rule that tells you the order in which you have to fill the shells in and in them the orbitals (s, p, d, f, ...). Then, in your second semester in university, you learn to derive those using Sehr\"odinger's equation: You diagonalise the Hamiltonian of the hyrdrogen atom and find the shells in terms of the main quantum number $n$ and the orbitals in terms of the angular momentum quantum number $L$ as $L=0$ corresponds to s, $L=1$ to p and so on. And you fill the orbitals thanks to the Pauli excursion principle. So, this proves the story of the chemists.<br /><br />Except that it doesn't: This is only true for the hydrogen atom. But the Hamiltonian for an atom nuclear charge $Z$ and $N$ electrons (so we allow for ions) is (in convenient units)<br />$$ a^2+b^2=c^2$$<br /><br />$$ H = -\sum_{i=1}^N \Delta_i -\sum_{i=1}^N \frac{Z}{|x_i|} + \sum_{i\lt j}^N\frac{1}{|x_i-x_j|}.$$<br /><br />The story of the previous paragraph would be true if the last term, the Coulomb interaction between the electrons would not be there. In that case, there is no interaction between the electrons and we could solve a hydrogen type problem for each electron separately and then anti-symmetrise wave functions in the end in a Slater determinant to take into account their Fermionic nature. But of course, in the real world, the Coulomb interaction is there and it contributes like $N^2$ to the energy, so it is of the same order (for almost neutral atoms) like the $ZN$ of the electron-nucleon potential.<br /><br />The approximation of dropping the electron-electron Coulomb interaction is well known in condensed matter systems where there resulting theory is known as a "Fermi gas". There it gives you band structure (which is then used to explain how a transistor works)<br /><br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><img alt="" class="mw-mmv-final-image svg mw-mmv-dialog-is-open" crossorigin="anonymous" height="217" src="https://upload.wikimedia.org/wikipedia/commons/thumb/4/4b/NPN_Band_Diagram_Active.svg/2560px-NPN_Band_Diagram_Active.svg.png" style="margin-left: auto; margin-right: auto;" width="400" /></td></tr><tr><td class="tr-caption" style="text-align: center;">Band structure in a NPN-transistor</td></tr></tbody></table>Also in that case, you pretend there is only one electron in the world that feels the periodic electric potential created by the nuclei and all the other electrons which don't show up anymore in the wave function but only as charge density.<div><br /></div><div>For atoms you could try to make a similar story by taking the inner electrons into account by saying that the most important effect of the ee-Coulomb interaction is to shield the potential of the nucleus thereby making the effective $Z$ for the outer electrons smaller. This picture would of course be true if there were no correlations between the electrons and all the inner electrons are spherically symmetric in their distribution around the nucleus and much closer to the nucleus than the outer ones. But this sounds more like a day dream than a controlled approximation.</div><div><br /></div><div>In the condensed matter situation, the standing for the Fermi gas is much better as there you could invoke renormalisation group arguments as the conductivities you are interested in are long wave length compared to the lattice structure, so we are in the infra red limit and the Coulomb interaction is indeed an irrelevant term in more than one euclidean dimension (and yes, in 1D, the Fermi gas is not the whole story, there is the Luttinger liquid as well).</div><div><br /></div><div>But for atoms, I don't see how you would invoke such RG arguments.</div><div><br /></div><div>So what can you do (with regards to actually proving the periodic table)? In our class, we teach how Lieb and Simons showed that in the $N=Z\to \infty$ limit (which in some sense can also be viewed as the semi-classical limit when you bring in $\hbar$ again) that the ground state energy $E^Q$ of the Hamiltonian above is in fact approximated by the ground state energy $E^{TF}$ of the Thomas-Fermi model (the simplest of all density functional theories, where instead of the multi-particle wave function you only use the one-particle electronic density $\rho(x)$ and approximate the kinetic energy by a term like $\int \rho^{5/3}$ which is exact for the three fermi gas in empty space):</div><div><br /></div><div>$$E^Q(Z) = E^{TF}(Z) + O(Z^2)$$</div><div><br /></div><div>where by a simple scaling argument $E^{TF}(Z) \sim Z^{7/3}$. More recently, people have computed more terms in these asymptotic which goes in terms of $Z^{-1/3}$, the second term ($O(Z^{6/3})= O(Z^2)$ is known and people have put a lot of effort into $O(Z^{5/3})$ but it should be clear that this technology is still very very far from proving anything "periodic" which would be $O(Z^0)$. So don't hold your breath hoping to find the periodic table from this approach.</div><div><br /></div><div>On the other hand, chemistry of the periodic table (where the column is supposed to predict chemical properties of the atom expressed in terms of the orbitals of the "valence electrons") works best for small atoms. So, another sensible limit appears to be to keep $N$ small and fixed and only send $Z\to\infty$. Of course this is not really describing atoms but rather highly charged ions.</div><div><br /></div><div>The advantage of this approach is that in the above Hamiltonian, you can absorb the $Z$ of the electron-nucleon interaction into a rescaling of $x$ which then let's $Z$ reappear in front of the electron-electron term as $1/Z$. Then in this limit, one can try to treat the ugly unwanted ee-term perturbatively.</div><div><br /></div><div>Friesecke (from TUM) and collaborators have made impressive progress in this direction and in this limit they could confirm that for $N < 10$ the chemists' picture is actually correct (with some small corrections). There are <a href="https://www.macs.hw.ac.uk/~chris/icms/GeomAnal/friesecke.pdf">very nice slides</a> of a seminar talk by Friesecke on these results.</div><div><br /></div><div>Of course, as a practitioner, this will not surprise you (after all, chemistry works) but it is nice to know that mathematicians can actually prove things in this direction. But it there is still some way to go even 150 years after Mendeleev.</div>Unknownnoreply@blogger.com1tag:blogger.com,1999:blog-8883034.post-15845107713584675762019-03-16T10:43:00.000+01:002019-03-16T10:43:18.866+01:00Nebelkerze CDU-Vorschlag zu "keine Uploadfilter"Sorry, this one of the occasional posts about German politics and thus in German. This is my posting to a German speaking mailing lists discussing the upcoming EU copyright directive (must be stopped in current from!!! March 23rd international protest day) and now the CDU party <a href="https://www.handelsblatt.com/dpa/wirtschaft-handel-und-finanzen-roundup-cdu-nationale-umsetzung-der-urheberrechtsreform-ohne-uploadfilter/24110748.html">has proposed</a> how to implement it in German law, although so unspecific that all the problematic details are left out. Here is the post.<br /><br /><div class="" style="font-family: Helvetica; font-size: 12px;">Vielleicht bin ich zu doof, aber ich verstehe nicht, wo der genaue Fortschritt zu dem, was auf EU-Ebene diskutiert wird, sein soll. Ausser dass der CDU-Vorschlag so unkonkret ist, dass alle internen Widersprüche im Nebel verschwinden. Auch auf EU-Ebene sagen doch die Befuerworter, dass man viel lieber Lizenzen erwerben soll, als filtern. Das an sich ist nicht neu.</div><div class="" style="font-family: Helvetica; font-size: 12px;"><br class="" /></div><div class="" style="font-family: Helvetica; font-size: 12px;">Neu, zumindest in diesem Handelsblatt-Artikel, aber sonst habe ich das nirgends gefunden, ist die Erwähnung von Hashsummen („digitaler Fingerabdruck“) oder soll das eher sowas wie ein digitales Wasserzeichen sein? Das wäre eine echte Neuerung, würde das ganze Verfahren aber sofort im Keim ersticken, da damit nur die Originaldatei geschützt wäre (das waere ja auch trivial festzustellen), aber jede Form des abgeleiteten Werkes komplett durch die Maschen fallen würde und man durch eine Trivialänderung Werke „befreien“ könnte. Ansonsten sind wir wieder bei den zweifelhaften, auf heute noch nicht existierender KI-Technologie beruhenden Filtern.</div><div class="" style="font-family: Helvetica; font-size: 12px;"><br class="" /></div><div class="" style="font-family: Helvetica; font-size: 12px;">Das andere ist die Pauschallizenz. Ich müsste also nicht mehr mit allen Urhebern Verträge abschliessen, sondern nur noch mit der VG Internet. Da ist aber wieder die grosse Preisfrage, für wen die gelten soll. Intendiert sind natürlich wieder Youtube, Google und FB. Aber wie formuliert man das? Das ist ja auch der zentrale Stein des Anstoßes der EU-Direktive: Eine Pauschallizenz brauchen all, ausser sie sind nichtkommerziell (wer ist das schon), oder (jünger als drei Jahre und mit wenigen Benutzern und kleinem Umsatz) oder man ist Wikipedia oder man ist GitHub? Das waere wieder die „Internet ist wie Fernsehen - mit wenigen grossen Sendern und so - nur eben anders“-Sichtweise, wie sie von Leuten, die das Internet aus der Ferne betrachten so gerne propagiert wird. Weil sie eben alles andere praktisch platt macht. Was ist denn eben mit den Foren oder Fotohostern? Müssten die alle eine Pauschallizenz erwerben (die eben so hoch sein müsste, dass sie alle Film- und Musikrechte der ganzen Welt pauschal abdeckt)? Was verhindert, dass das am Ende ein „wer einen Dienst im Internet betreibt, der muss eben eine kostenpflichtige Internetlizenz erwerben, bevor er online gehen kann“-Gesetz wird, das bei jeder nichttrivialen Höhe der Lizenzgebühr das Ende jeder gras roots Innovation waere?</div><div class="" style="font-family: Helvetica; font-size: 12px;"><br class="" /></div><div class="" style="font-family: Helvetica; font-size: 12px;">Interessant waere natuerlich auch, wie die Einnahmen der VG Internet verteilt werden. Ein Schelm waere, wenn das nicht in großen Teilen zB bei Presseverlegern landen würde. Das waere doch dann endlich das „nehmt denjenigen, die im Internet Geld verdienen dieses weg und gebt es und, die nicht mehr so viel Geld verdienen“-Gesetz. Dann müsste die Lizenzgebühr am besten ein Prozentsatz des Umsatz sein, am besten also eine Internet-Steuer.</div><div class="" style="font-family: Helvetica; font-size: 12px;"><br class="" /></div><div class="" style="font-family: Helvetica; font-size: 12px;">Und ich fange nicht damit an, wozu das führt, wenn alle europäischen Länder so krass ihre eigene Umsetzungssuppe kochen.</div><div class="" style="font-family: Helvetica; font-size: 12px;"><br class="" /></div><div class="" style="font-family: Helvetica; font-size: 12px;">Alles in allem ein ziemlich gelungener Coup der CDU, der es schaffen kann, den Kritikern von Artikel 13 in der öffentlichen Meinung den Wind aus den Segeln zu nehmen, indem man es alles in eine inkonkrete Nebelwolke packt, wobei die ganzen problematischen Regelungen in den Details liegen dürften.</div>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-8883034.post-58525385107130592772019-03-06T15:06:00.004+01:002019-03-06T15:24:15.199+01:00Challenge: How to talk to a flat earther?Further down the rabbit hole, over lunch I finished watching <a href="https://www.netflix.com/title/81015076">"Behind the Curve"</a>, a Netflix documentary on people believing the earth is a flat disk. According to them, the north pole is in the center, while Antarctica is an ice wall at the boundary. Sun and moon are much closer and flying above this disk while the stars are on some huge dome like in a planetarium. NASA is a fake agency promoting the doctrine and airlines must be part of the conspiracy as they know that you cannot directly fly between continents on the southern hemisphere (really?).<br /><br />These people are happily using GPS for navigation but have a general mistrust in the science (and their teachers) of at least two centuries.<br /><br />Besides the obvious "I don't see curvature of the horizon" they are even conducting experiments to prove their point (fighting with laser beams not being as parallel over miles of distance as they had hoped for). So at least some of them might be open to empirical disprove.<br /><br />So here is my challenge: Which experiment would you conduct with them to convince them? Warning: Everything involving stuff disappearing at the horizon (ships sailing away, being able to see further from a tower) are complicated by non-trivial diffraction in the atmosphere which would very likely turn this observation inconclusive. The sun being at different declination (height) at different places might also be explained by being much closer and a Foucault pendulum might be too indirect to really convince them (plus it requires some non-elementary math to analyse).<br /><br />My personal solution is to point to the observation that the declination of Polaris (around which I hope they can agree the night sky rotates) is given my the geographical latitude: At the north pole it is right above you but is has to go down the more south you get. I cannot see how this could be reconciled with a dome projection.<br /><br />How would you approach this? The rules are that it must only involve observations available to everyone, no spaceflight, no extra high altitude planes. You are allowed to make use of the phone, cameras, you can travel (say by car or commercial flight but you cannot influence the flight route). It does not involve lots of money or higher math.<br /><br /><div class="separator" style="clear: both; text-align: center;"></div><div class="separator" style="clear: both; text-align: center;"><a href="https://avishekgyawali.com/wp-content/uploads/2017/12/flat-theory-720x500.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="500" data-original-width="720" height="222" src="https://avishekgyawali.com/wp-content/uploads/2017/12/flat-theory-720x500.jpg" width="320" /></a></div><br />Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-8883034.post-38290574542983782102019-02-12T08:17:00.000+01:002019-02-12T08:20:47.524+01:00Bohmian Rapsody<h3>Visits to a Bohmian village</h3><br />Over all of my physics life, I have been under the local influence of some Gaul villages that have ideas about physics that are not 100% aligned with the main stream views: When I was a student in Hamburg, I was good friends with people working on algebraic quantum field theory. Of course there were opinions that they were the only people seriously working on QFT as they were proving theorems while others dealt with perturbative series only that are known to diverge and are thus obviously worthless. Funnily enough they were literally sitting above the HERA tunnel where electron proton collisions took place that were very well described by exactly those divergent series. Still, I learned a lot from these people and would say there are few that have thought more deeply about structural properties of quantum physics. These days, I use more and more of these things in my own teaching (in particular in our Mathematical Quantum Mechanics and Mathematical Statistical Physics classes as well as when thinking about foundations, see below) and even <a href="https://arxiv.org/abs/1803.04993">some other physicists start using their language</a>.<br /><br />Later, as a PhD student at the Albert Einstein Institute in Potsdam, there was an accumulation point of people from the Loop Quantum Gravity community with Thomas Thiemann and Renate Loll having long term positions and many others frequently visiting. As you probably know, a bit later, I decided (together with Giuseppe Policastro) to look into this more deeply resulting in a <a href="https://arxiv.org/abs/hep-th/0409182">series</a> <a href="https://arxiv.org/abs/hep-th/0610193">of</a> <a href="https://arxiv.org/abs/0912.3011">papers</a> there were well received at least amongst our peers and about which I am still a bit proud.<br /><br />Now, I have been in Munich for over ten years. And here at the LMU math department there is a group calling themselves the <a href="http://www.mathematik.uni-muenchen.de/~bohmmech/index.php">Workgroup Mathematical Foundations of Physics</a>. And let's be honest, I call them the Bohmians (and sometimes the Bohemians). And once more, most people believe that the Bohmian interpretation of quantum mechanics is just a fringe approach that is not worth wasting any time on. You will have already guessed it: I did so none the less. So here is a condensed report of what I learned and what I think should be the official opinion on this approach. This is an informal write up of a<a href="https://arxiv.org/abs/1902.03752"> notes paper</a> that I put on the arXiv today.<br /><br />Bohmians don't like about the usual (termed Copenhagen lacking a better word) approach to quantum mechanics that you are not allowed to talk about so many things and that the observer plays such a prominent role by determining via a measurement what aspect is real an what is not. They think this is far too subjective. So rather, they want quantum mechanics to be <i>about</i> particles that then are allowed to follow trajectories.<br /><br />"But we know this is impossible!" I hear you cry. So, let's see how this works. The key observation is that the Schrödinger equation for a Hamilton operator of the form kinetic term (possibly with magnetic field) plus potential term, has a conserved current<br /><br />$$j = \bar\psi\nabla\psi - (\nabla\bar\psi)\psi.$$<br /><br />So as your probability density is $\rho=\bar\psi\psi$, you can think of that being made up of particles moving with a velocity field<br /><br />$$v = j/\rho = 2\Im(\nabla \psi/\psi).$$<br /><br />What this buys you is that if you have a bunch of particles that is initially distributed like the probability density and follows the flow of the velocity field it will also later be distributed like $|\psi |^2$.<br /><br />What is important is that they keep the Schrödinger equation in tact. So everything that you can do with the original Schrödinger equation (i.e. everything) can be done in the Bohmian approach as well. If you set up your Hamiltonian to describe a double slit experiment, the Bohmian particles will flow nicely to the screen and arrange themselves in interference fringes (as the probability density does). So you will never come to a situation where any experimental outcome will differ from what the Copenhagen prescription predicts.<br /><br />The price you have to pay, however, is that you end up with a very non-local theory: The velocity field lives in configuration space, so the velocity of every particle depends on the position of all other particles in the universe. I would say, this is already a show stopper (given what we know about quantum field theory whose raison d'être is locality) but let's ignore this aesthetic concern.<br /><br />What got me into this business was the attempt to understand how the set-ups like Bell's inequality and GHZ and the like work out that are supposed to show that quantum mechanics cannot be classical (technically that the state space cannot be described as local probability densities). The problem with those is that they are often phrased in terms of spin degrees of freedom which have Hamiltonians that are not directly of the form above. You can use a Stern-Gerlach-type apparatus to translate the spin degree of freedom to a positional but at the price of a Hamiltonian that is not explicitly know let alone for which you can analytically solve the Schrödinger equation. So you don't see much.<br /><br />But from Reinhard Werner and collaborators I learned how to set up qubit-like algebras from positional observables of free particles (at different times, so get something non-commuting which you need to make use of entanglement as a specific quantum resource). So here is my favourite example:<br /><br />You start with two particles each following a free time evolution but confined to an interval. You set those up in a particular entangled state (stationary as it is an eigenstate of the Hamiltonian) built from the two lowest levels of the particle in the box. And then you observe for each particle if it is in the left or the right half of the interval.<br /><br />From symmetry considerations (details in my <a href="https://arxiv.org/abs/1902.03752">paper</a>) you can see that each particle is with the same probability on the left and the right. But they are anti-correlated when measured at the same time. But when measured at different times, the correlation oscillates like the cosine of the time difference.<br /><br />From the Bohmian perspective, for the static initial state, the velocity field vanishes everywhere, nothing moves. But in order to capture the time dependent correlations, as soon as one particle has been measured, the position of the second particle has to oscillate in the box (how the measurement works in detail is not specified in the Bohmian approach since it involves other degrees of freedom and remember, <a href="https://en.wikipedia.org/wiki/Dirk_Gently%27s_Holistic_Detective_Agency">everything depends on everything</a> but somehow it has to work since you want to produce the correlations that are predicted by the Copenhagen approach).<br /><br /><table align="center" cellpadding="0" cellspacing="0" class="tr-caption-container" style="margin-left: auto; margin-right: auto; text-align: center;"><tbody><tr><td style="text-align: center;"><a href="https://1.bp.blogspot.com/-lTT_scLBW_M/XGFFndrDvpI/AAAAAAAALrc/0kijm_ZZFU8W6XL7NA8nb-xcdHjpzRDbQCLcBGAs/s1600/bohmtrajectories.png" imageanchor="1" style="margin-left: auto; margin-right: auto;"><img border="0" data-original-height="338" data-original-width="557" height="194" src="https://1.bp.blogspot.com/-lTT_scLBW_M/XGFFndrDvpI/AAAAAAAALrc/0kijm_ZZFU8W6XL7NA8nb-xcdHjpzRDbQCLcBGAs/s320/bohmtrajectories.png" width="320" /></a></td></tr><tr><td class="tr-caption" style="text-align: center;">The trajectory of the second particle depending on its initial position</td></tr></tbody></table><br /><br />This is somehow the Bohmian version of the collapse of the wave function but they would never phrase it that way.<br /><br />And here is where it becomes problematic: If you could see the Bohmian particle moving you could decide if the other particle has been measured (it would oscillate) or not (it would stand still). No matter where the other particle is located. With this observation you could build a telephone that transmits information instantaneously, something that should not exist. So you have to conclude you must not be able to look at the second particle and see if it oscillates or not.<br /><br />Bohmians tell you you cannot because all you are supposed to observer about the particles are their positions (and not their velocity). And if you try to measure the velocity by measuring the position at two instants in time you don't because the first observation disturbs the particle so much that it invalidates the original state.<br /><br />As it turns out, you are not allowed to observe anything else about the particles than that they are distributed like $|\psi |^2$ because if you could, you could build a similar telephone (at least statistically) as I explain the in the paper (this fact is known in the Bohm literature but I found it nowhere so clearly demonstrated as in this two particle system).<br /><br />My conclusion is that the Bohm approach adds something (the particle positions) to the wave function but then in the end tells you you are not allowed to observe this or have any knowledge of this beyond what is already encoded in the wave function. It's like making up an invisible friend.<br /><br />PS: If you haven't seen "Bohemian Rhapsody", yet, you should, even if there are good reasons to <a href="https://www.attitude.co.uk/article/is-this-just-fantasy-5-problems-with-bohemian-rhapsodys-portrayal-of-freddie-mercurys-life-and-sexuality-1/20196/?fbclid=IwAR1ErvovvVwAhEgqYM3OmhQfchn7KaLOJMetYJ2zbsvYpacrgRDUX_Cps0M">criticise</a> the dramatisation of real events.Unknownnoreply@blogger.com1tag:blogger.com,1999:blog-8883034.post-44984355172575453682019-01-17T20:43:00.002+01:002019-01-17T20:43:52.504+01:00Has your password been leaked?Today, there was <a href="https://www.troyhunt.com/the-773-million-record-collection-1-data-reach/">news</a> about a huge database containing 773 million email address / password pairs became public. On <a href="https://haveibeenpwned.com/">Have I Been Pawned</a> you can check if any of your email addresses is in this database (or any similar one). I bet it is (mine are).<br /><br />These lists are very probably the source for the spam emails that have been around for a number of months where the spammer claims they broke into your account and tries to prove it by telling you your password. Hopefully, this is only a years old LinkedIn password that you have changed aeons ago.<br /><br />To make sure, you actually want to search not for your email but for your password. But of course, you don't want to tell anybody your password. To this end, I have written a small perl script that checks for your password without telling anybody by doing a calculation locally on your computer. You can find it on <a href="https://github.com/atdotde/HIBPPasswort/blob/master/hibp.pl">GitHub</a>.Unknownnoreply@blogger.com1tag:blogger.com,1999:blog-8883034.post-26395064813694019392018-10-26T17:11:00.000+02:002018-10-27T10:39:54.418+02:00Interfere and it didn't happenI am a bit late for the party, but also wanted to share my two cents on the paper <a href="https://www.nature.com/articles/s41467-018-05739-8">"Quantum theory cannot consistently describe the use of itself"</a> by Frauchiger and Renner. After sitting down and working out the math for myself, I found that the analysis in <a href="https://arxiv.org/abs/1810.07065">this paper</a> and the <a href="https://www.scottaaronson.com/blog/?p=3975">blogpost by Scot</a> (including many of the the 160+ comments, some by Renner) share a lot with what I am about to say but maybe I can still contribute a slight twist.<br /><br /><h4>Coleman on GHZS</h4><div>My background is the talk <a href="https://www.youtube.com/watch?v=EtyNMlXN-sw">"Quantum Mechanics In Your Face"</a> by Sidney Coleman which I consider as the best argument why quantum mechanics cannot be described by a local and realistic theory (from which I would conclude it is not realistic). In a nutshell, the argument goes like this: Consider the three qubit state state </div><div><br /></div><div>$$\Psi=\frac 1{\sqrt 2}(\uparrow\uparrow\uparrow-\downarrow\downarrow\downarrow)$$</div><div><br /></div><div>which is both an eigenstate of eigenvalue -1 for $\sigma_z\otimes\sigma_z\otimes\sigma_z$ and an eigenstate of eigenvalue +1 for $\sigma_x\otimes\sigma_x\otimes\sigma_z$ or any permutation. This means that, given that the individual outcomes of measuring a $\sigma$-matrix on a qubit is $\pm 1$, when measuring all in the z-direction there will be an odd number of -1 results but if two spins are measured in x-direction and one in z-direction there is an even number of -1's. </div><div><br /></div><div>The latter tells us that the outcome of one z-measurement is the product of the two x-measurements on the other two spins. But multiplying this for all three spins we get that in shorthand $ZZZ=(XXX)^2=+1$ in contradiction to the -1 eigenvalue for all z-measurments. </div><div><br /></div><div>The conclusion is (unless you assume some non-local conspiracy between the spins) that one has to take serious the fact that on a given spin I cannot measure both $\sigma_x$ and $\sigma_z$ and thus when actually measuring the latter I must not even assume that $X$ has some (although unknown) value $\pm 1$ as it leads to the contradiction. Stuff that I cannot measure does not have a value (that is also my understanding of what "not realistic" means).</div><div><br /></div><h4>Fruchtiger and Renner</h4><div>Now to the recent Nature paper. In short, they are dealing with two qubits (by which I only mean two state systems). The first is in a box L' (I will try to use the somewhat unfortunate nomenclature from the paper) and the second in in a box L (L stands for lab). For L, we use the usual z-basis of $\uparrow$ and $\downarrow$ as well as the x-basis $\leftarrow = \frac 1{\sqrt 2}(\downarrow - \uparrow)$ and $\rightarrow = \frac 1{\sqrt 2}(\downarrow + \uparrow)$ . Similarly, for L' we use the basis $h$ and $t$ (heads and tails as it refers to a coin) as well as $o = \frac 1{\sqrt 2}(h - t)$ and $f = \frac 1{\sqrt 2}(h+f)$. The two qubits are prepared in the state</div><div><br /></div><div>$$\Phi = \frac{h\otimes\downarrow + \sqrt 2 t\otimes \rightarrow}{\sqrt 3}$$.</div><div><br /></div><div>Clearly, a measurement of $t$ in box L' implies that box L has to contain the state $\rightarrow$. Call this observation A.</div><div><br /></div><div>Let's re-express $\rightarrow$ in the x-basis:</div><div><br /></div><div>$$\Phi =\frac {h\otimes \downarrow + t\otimes \downarrow + t\otimes\uparrow}{\sqrt 3}$$</div><div><br /></div><div>From which one concludes that an observer inside box L that measures $\uparrow$ concludes that the qubit in box L' is in state $t$. Call this observation B.</div><div><br /></div><div>Similarly, we can express the same state in the x-basis for L':</div><div><br /></div><div>$$\Phi = \frac{4 f\otimes \downarrow+ f\otimes \uparrow - o\otimes \uparrow}{\sqrt 3}$$</div><div><br /></div><div>From this once can conclude that measuring $o$ for the state of L' one can conclude that L is in the state $\uparrow$. Call this observation C.</div><div><br /></div><div>Using now C, B and A one is tempted to conclude that observing L' to be in state $o$ implies that L is in state $\rightarrow$. When we express the state in the $ht\leftarrow\rightarrow$-basis, however, we get</div><div><br /></div><div>$$\Phi = \frac{f\otimes\leftarrow+ 3f\otimes \rightarrow + o\otimes\leftarrow - o\otimes \rightarrow}{\sqrt{12}}.$$</div><div><br /></div><div>so with probability 1/12 we find both $o$ and $\leftarrow$. Again, we hit a contradiction.</div><div><br /></div><div>One is tempted to use the same way out as above in the three qubit case and say one should not argue about contrafactual measurements that are incompatible with measurements that were actually performed. But Frauchiger and Renner found a set-up which seems to avoid that.</div><div><br /></div><div>They have observers F and F' ("friends") inside the boxes that do the measurements in the $ht$ and $\uparrow\downarrow$ basis whereas later observers W and W' measure the state of the boxes including the observer F and F' in the $of$ and $\leftarrow\rightarrow$ basis. So, at each stage of A,B,C the corresponding measurement has actually taken place and is not contrafactual!</div><div><br /></div><h4>Interference and it did not happen</h4><div>I believe the way out is to realise that at least from a retrospective perspective, this analysis stretches the language and in particular the word "measurement" to the extreme. In order for W' to measure the state of L' in the $of$-basis, he has to interfere the contents including F' coherently such that there is no leftover of information from F''s measurement of $ht$ remaining. Thus, when W''s measurement is performed one should not really say that F''s measurement has in any real sense happened as no possible information is left over. So it is in any practical sense contrafactual.</div><div><br /></div><div>To see the alternative, consider a variant of the experiment where a tiny bit of information (maybe the position of one air molecule or the excitation of one of F''s neutrons) escapes the interference. Let's call the two possible states of that qubit of information $H$ and $T$ (not necessarily orthogonal) and consider instead the state where that neutron is also entangled with the first qubit:</div><div><br /></div><div>$$\tilde \Phi = \frac{h\otimes\downarrow\otimes H + \sqrt 2 t\otimes \rightarrow\otimes T}{\sqrt 3}$$.</div><div><br /></div><div>Then, the result of step C becomes</div><div><br /></div><div>$$\tilde\Phi = \frac{f\otimes \downarrow\otimes H+ o\otimes \downarrow\otimes H+f\otimes \downarrow\otimes T-o\otimes\downarrow\otimes T + f\otimes \uparrow\otimes T-o \otimes\uparrow\times T}{\sqrt 6}.$$</div><div><br /></div><div>We see that now there is a term containing $o\otimes\downarrow\otimes(H-T)$. Thus, as long as the two possible states of the air molecule/neuron are actually different, observation C is no longer valid and the whole contradiction goes away.</div><div><br /></div><div>This makes it clear that the whole argument relies of the fact that when W' is doing his measurement any remnant of the measurement by his friend F' is eliminated and thus one should view the measurement of F' as if it never happened. Measuring L' in the $of$-basis really erases the measurement of F' in the complementary $ht$-basis.</div>Unknownnoreply@blogger.com3tag:blogger.com,1999:blog-8883034.post-25246650707401177892018-10-17T16:37:00.000+02:002018-10-17T20:55:14.592+02:00Bavarian electoral systemLast Sunday, we had the election for the federal state of Bavaria. Since the electoral system is kind of odd (but not as odd as <a href="https://en.wikipedia.org/wiki/First-past-the-post_voting">first past the post</a>), I would like to analyse how some variations (assuming the actual distribution of votes) in the rule would have worked out. So, first, here is how actually, the seats are distributed: Each voter gets two ballots: On the first ballot, each party lists one candidate from the local constituency and you can select one. On the second ballot, you can vote for a party list (it's even more complicated because also there, you can select individual candidates to determine the position on the list but let's ignore that for today).<br /><br />Then in each constituency, the votes on ballot one are counted. The candidate with the most votes (like in first past the pole) gets elected for parliament directly (and is called a "direct candidate"). Then over all, the votes for each party <b>on both ballots</b> (this is where the system differs from the federal elections) are summed up. All votes for parties with less then 5% of the grand total of all votes are discarded (actually including their direct candidates but this is not of a partial concern). Let's call the rest the "reduced total". According to the fraction of each party in this reduced total the seats are distributed.<br /><br />Of course the first problem is that you can only distribute seats in integer multiples of 1. This is solved using the <a href="https://en.wikipedia.org/wiki/Largest_remainder_method">Hare-Niemeyer-method</a>: You first distribute the integer parts. This clearly leaves fewer seats open than the number of parties. Those you then give to the parties where the rounding error to the integer below was greatest. Check out the wikipedia page explaining how this can lead to a party losing seats when the total number of seats available is increased.<br /><br />Because this is what happens in the next step: Remember that we already allocated a number of seats to constituency winners in the first round. Those count towards the number of seats that each party is supposed to get in step two according to the fraction of votes. Now, it can happen, that a party has won more direct candidates than seats allocated in step two. If that happens, more seats are added to the total number of seats and distributed according to the rules of step two until each party has been allocated at least the number of seats as direct candidates. This happens in particular if one party is stronger than all the other ones leading to that party winning almost all direct candidates (as in Bavaria this happened to the CSU which won all direct candidates except five in Munich and one in Würzburg which were won by the Greens).<br /><br />A final complication is that Bavaria is split into seven electoral districts and the above procedure is for each district separately. So there are seven times rounding and adding seats procedures.<br /><br />Sunday's election resulted in the following distribution of seats:<br /><br />After the whole procedure, there are 205 seats distributed as follows<br /><br /><br /><ul><li>CSU 85 (41.5% of seats)</li><li>SPD 22 (10.7% of seats)</li><li>FW 27 (13.2% of seats)</li><li>GREENS 38 (18.5% of seats)</li><li>FDP 11 (5.4% of seats)</li><li>AFD 22 (10.7% of seats)</li></ul><div>You can find all the total of votes on <a href="https://www.landtagswahl2018.bayern.de/sitzeberechnung_wahlkreise.html">this page</a>.</div><div><br /></div><div>Now, for example one can calculate the distribution without districts throwing just everything in a single super-district. Then there are 208 seats distributed as</div><div><br /></div><div><ul><li>CSU 85 (40.8%)</li><li>SPD 22 (10.6%)</li><li>FW 26 (12.5%)</li><li>GREENS 40 (19.2%)</li><li>FDP 12 (5.8%)</li><li>AFD 23 (11.1%)</li></ul><div>You can see that in particular the CSU, the party with the biggest number of votes profits from doing the rounding 7 times rather than just once and the last three parties would benefit from giving up districts.</div></div><div><br /></div><div>But then there is actually an issue of negative weight of votes: The greens are particularly strong in Munich where they managed to win 5 direct seats. If instead those seats would have gone to the CSU (as elsewhere), the number of seats for Oberbayern, the district Munich belongs to would have had to be increased to accommodate those addition direct candidates for the CSU increasing the weight of Oberbayern compared to the other districts which would then be beneficial for the greens as they are particularly strong in Oberbayern: So if I give all the direct candidates to the CSU (without modifying the numbers of total votes), I get the follwing distribution:</div><div>221 seats</div><div><ul><li>CSU 91 (41.2%)</li><li>SPD 24 (10.9%)</li><li>FW 28 (12,6%)</li><li>GREENS 42 (19.0%)</li><li>FDP 12 (5.4%)</li><li>AFD 24 (10.9%)</li></ul><div>That is, there greens would have gotten a higher fraction of seats if they had won less constituencies. Voting for green candidates in Munich actually hurt the party as a whole!</div></div><div><br /></div><div>The effect is not so big that it actually changes majorities (CSU and FW are likely to form a coalition) but still, the constitutional court does not like (predictable) negative weight of votes. Let's see if somebody challenges this election and what that would lead to.</div><div><br /></div><div>The perl script I used to do this analysis is <a href="http://atdotde.de/~robert/wahl.html">here</a>.<br /><br /><b>Postscript:</b><br />The above analysis in the last point is not entirely fair as not to win a constituency means getting fewer votes which then are missing from the grand total. Taking this into account makes the effect smaller. In fact, subtracting the votes from the greens that they were leading by in the constituencies they won leads to an almost zero effect:<br /><br />Seats: 220<br /><span style="background-color: white; font-family: Menlo; font-size: 11px;"><ul><li>CSU <span class="Apple-tab-span" style="background-color: transparent; white-space: pre;"> </span>91 <span class="Apple-tab-span" style="background-color: transparent; white-space: pre;"> </span>41.4%</li><li>SPD <span class="Apple-tab-span" style="background-color: transparent; white-space: pre;"> </span>24 <span class="Apple-tab-span" style="background-color: transparent; white-space: pre;"> </span>10.9%</li><li><span style="background-color: transparent;">FW </span><span class="Apple-tab-span" style="background-color: transparent; white-space: pre;"> </span><span style="background-color: transparent;">28 </span><span class="Apple-tab-span" style="background-color: transparent; white-space: pre;"> </span><span style="background-color: transparent;">12.7%</span></li><li><span style="background-color: transparent;">GREENS </span><span class="Apple-tab-span" style="background-color: transparent; white-space: pre;"> </span><span style="background-color: transparent;">41 </span><span class="Apple-tab-span" style="background-color: transparent; white-space: pre;"> </span><span style="background-color: transparent;">18.6%</span></li><li><span style="background-color: transparent;">FDP </span><span class="Apple-tab-span" style="background-color: transparent; white-space: pre;"> </span><span style="background-color: transparent;">12 </span><span class="Apple-tab-span" style="background-color: transparent; white-space: pre;"> </span><span style="background-color: transparent;">5.4%</span></li><li><span style="background-color: transparent;">AFD </span><span class="Apple-tab-span" style="background-color: transparent; white-space: pre;"> </span><span style="background-color: transparent;">24 </span><span class="Apple-tab-span" style="background-color: transparent; white-space: pre;"> </span>10.9%</li></ul><div>Letting the greens win München Mitte (a newly created constituency that was supposed to act like a bad bank for the CSU taking up all central Munich more left leaning voters, do I hear somebody say "Gerrymandering"?) yields</div><div><br /></div><div>Seats: 217</div></span><div style="background-color: white; font-family: Menlo; font-size: 11px; font-stretch: normal; line-height: normal;"></div><ul><li>CSU <span class="Apple-tab-span" style="white-space: pre;"> </span>90 <span class="Apple-tab-span" style="white-space: pre;"> </span>41.5%</li><li>SPD <span class="Apple-tab-span" style="white-space: pre;"> </span>23 <span class="Apple-tab-span" style="white-space: pre;"> </span>10.6%</li><li>FW <span class="Apple-tab-span" style="white-space: pre;"> </span>28 <span class="Apple-tab-span" style="white-space: pre;"> </span>12.9%</li><li>GREENS <span class="Apple-tab-span" style="white-space: pre;"> </span>41 <span class="Apple-tab-span" style="white-space: pre;"> </span>18.9%</li><li>FDP <span class="Apple-tab-span" style="white-space: pre;"> </span>12 <span class="Apple-tab-span" style="white-space: pre;"> </span>5.5%</li><li>AFD <span class="Apple-tab-span" style="white-space: pre;"> </span>23 <span class="Apple-tab-span" style="white-space: pre;"> </span>10.6%</li></ul><div>Or letting them win all but Moosach and Würzbug-Stadt where the lead was the smallest:</div><div><br /></div><div>Seats: 210</div><br /><div style="background-color: white; font-family: Menlo; font-size: 11px; font-stretch: normal; line-height: normal;"></div><ul><li>CSU <span class="Apple-tab-span" style="white-space: pre;"> </span>87 <span class="Apple-tab-span" style="white-space: pre;"> </span>41.4%</li><li>SPD <span class="Apple-tab-span" style="white-space: pre;"> </span>22 <span class="Apple-tab-span" style="white-space: pre;"> </span>10.5%</li><li>FW <span class="Apple-tab-span" style="white-space: pre;"> </span>27 <span class="Apple-tab-span" style="white-space: pre;"> </span>12.9%</li><li>GREENS <span class="Apple-tab-span" style="white-space: pre;"> </span>40 <span class="Apple-tab-span" style="white-space: pre;"> </span>19.0%</li><li>FDP <span class="Apple-tab-span" style="white-space: pre;"> </span>11 <span class="Apple-tab-span" style="white-space: pre;"> </span>5.2%</li><li>AFD <span class="Apple-tab-span" style="white-space: pre;"> </span>23 <span class="Apple-tab-span" style="white-space: pre;"> </span>11.0%</li></ul><br /><div style="background-color: white; font-family: Menlo; font-size: 11px; font-stretch: normal; line-height: normal;"> </div><div><span style="font-variant-ligatures: no-common-ligatures;"><br /></span></div><div style="background-color: white; font-family: Menlo; font-size: 11px; font-stretch: normal; line-height: normal;"></div></div>Unknownnoreply@blogger.com1tag:blogger.com,1999:blog-8883034.post-1785472283772026702018-03-29T21:35:00.000+02:002018-03-29T21:35:21.326+02:00Machine Learning for Physics?!?Today was the last day of a nice <a href="https://indico.mpp.mpg.de/event/5578/overview">workshop</a> here at the Arnold Sommerfeld Center organised by Thomas Grimm and Sven Krippendorf on the use of Big Data and Machine Learning in string theory. While the former (at this workshop mainly in the form of developments following <a href="http://hep.itp.tuwien.ac.at/~kreuzer/CY/">Kreuzer/Skarke</a> and taking it further for F-theory constructions, orbifolds and the like) appears to be quite advanced as of today, the latter is still in its very early days. At best.<br /><br />I got the impression that for many physicists that have not yet spent too much time with this, deep learning and in particular deep neural networks are expected to be some kind of silver bullet that can answer all kinds of questions that humans have not been able to answer despite some effort. I think this hope is at best premature and looking at the (admittedly impressive) examples where it works (playing Go, classifying images, speech recognition, event filtering at LHC) these seem to be more like those problems where humans have at least a rough idea how to solve them (if it is not something that humans do everyday like understanding text) and also roughly how one would code it but that are too messy or vague to be treated by a traditional program.<br /><br />So, during some of the less entertaining talks I sat down and thought about problems where I would expect neural networks to perform badly. And then, if this approach fails even in simpler cases that are fully under control one should maybe curb the expectations for the more complex cases that one would love to have the answer for. In the case of the workshop that would be guessing some topological (discrete) data (that depends very discontinuously on the model parameters). Here a simple problem would be a 2-torus wrapped by two 1-branes. And the computer is supposed to compute the number of matter generations arising from open strings at the intersections, i.e. given two branes (in terms of their slope w.r.t. the cycles of the torus) how often do they intersect? Of course these numbers depend sensitively on the slope (as a real number) as for rational slopes [latex]p/q[/latex] and [latex]m/n[/latex] the intersection number is the absolute value of [latex]pn-qm[/latex]. My guess would be that this is almost impossible to get right for a neural network, let alone the much more complicated variants of this simple problem.<br /><br />Related but with the possibility for nicer pictures is the following: Can a neural network learn the shape of the Mandelbrot set? Let me remind those of you who cannot remember the 80ies anymore, for a complex number c you recursively apply the function<br />[latex]f_c(z)= z^2 +c[/latex]<br />starting from 0 and ask if this stays bounded (a quick check shows that once you are outside [latex]|z| < 2[/latex] you cannot avoid running to infinity). You color the point c in the complex plane according to the number of times you have to apply f_c to 0 to leave this circle. I decided to do this for complex numbers x+iy in the rectangle -0.74<x 0.35="" and="" as="" boundary="" contains="" corresponding="" like="" looks="" of="" other="" p="" picture="" set.="" some="" that="" the="" this:="" this="" y=""><div class="separator" style="clear: both; text-align: center;"><a href="https://3.bp.blogspot.com/-Cc-lEoFrjA8/Wr08gZizIjI/AAAAAAAALgU/NSktrpsD4fAajX2p_VxuuPxNd7A15IQkACLcBGAs/s1600/learnmandelbrot.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="501" data-original-width="501" height="320" src="https://3.bp.blogspot.com/-Cc-lEoFrjA8/Wr08gZizIjI/AAAAAAAALgU/NSktrpsD4fAajX2p_VxuuPxNd7A15IQkACLcBGAs/s320/learnmandelbrot.png" width="320" /></a></div><div class="" style="clear: both; text-align: left;">I have written a small mathematica program to compute this image. Built into mathematica is also a neural network: You can feed training data to the function Predict[], for me these were 1,000,000 points in this rectangle and the number of steps it takes to leave the 2-ball. Then mathematica thinks for about 24 hours and spits out a predictor function. Then you can plot this as well:</div><div class="separator" style="clear: both; text-align: left;"><br /></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-8HhkNAtvDo0/Wr09UHeMsrI/AAAAAAAALgc/TRnxq5VyItIJEomxyQuUF_dYDbgs0FO6ACLcBGAs/s1600/learnedmandelbrot.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="501" data-original-width="501" height="320" src="https://1.bp.blogspot.com/-8HhkNAtvDo0/Wr09UHeMsrI/AAAAAAAALgc/TRnxq5VyItIJEomxyQuUF_dYDbgs0FO6ACLcBGAs/s320/learnedmandelbrot.png" width="320" /></a></div><div class="separator" style="clear: both; text-align: left;">There is some similarity but clearly it has no idea about the fractal nature of the Mandelbrot set. If you really believe in magic powers of neural networks, you might even hope that once it learned the function for this rectangle one could extrapolate to outside this rectangle. Well, at least in this case, this hope is not justified: The neural network thinks the correct continuation looks like this:</div><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-uTaScAiWnro/Wr0_zViU4PI/AAAAAAAALgo/bZpyISjAFjAnSzLWk9y-yDcfOqLteDSHQCLcBGAs/s1600/learnmandelbrotextrapolated.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="501" data-original-width="501" height="320" src="https://2.bp.blogspot.com/-uTaScAiWnro/Wr0_zViU4PI/AAAAAAAALgo/bZpyISjAFjAnSzLWk9y-yDcfOqLteDSHQCLcBGAs/s320/learnmandelbrotextrapolated.png" width="320" /></a></div><div class="separator" style="clear: both; text-align: left;">Ehm. No.</div><div class="separator" style="clear: both; text-align: left;"><br /></div><div class="separator" style="clear: both; text-align: left;">All this of course with the caveat that I am no expert on neural networks and I did not attempt anything to tune the result. I only took the neural network function built into mathematica. Maybe, with a bit of coding and TensorFlow one can do much better. But on the other hand, this is a simple two dimensional problem. At least for traditional approaches this should be much simpler than the other much higher dimensional problems the physicists are really interested in.</div></x>Unknownnoreply@blogger.com3tag:blogger.com,1999:blog-8883034.post-61716206361556259252017-12-14T09:58:00.001+01:002017-12-14T09:58:36.040+01:00What are the odds?It's the time of year, you give out special problems in your classes. So this is mine for the blog. It is motivated by this picture of the home secretaries of the German federal states after their annual meeting as well as some recent discussions on Facebook:<br /><div class="separator" style="clear: both; text-align: center;"><a href="http://i.huffpost.com/gen/5625850/images/n-INNENMINISTERIUM-628x314.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="314" data-original-width="628" height="160" src="https://i.huffpost.com/gen/5625850/images/n-INNENMINISTERIUM-628x314.jpg" width="320" /></a></div>I would like to call it Summers' problem:<br /><br />Let's have two real random variables $M$ and $F$ that are drawn according to two probability distributions $\rho_{M/F}(x)$ (for starters you may both assume to be Gaussians but possibly with different mean and variance). Take $N$ draws from each and order the $2N$ results. What is the probability that the $k$ largest ones are all from $M$ rather than $F$? Express your results in terms of the $\rho_{M/F}(x)$. We are also interested in asymptotic results for $N$ large and $k$ fixed as well as $N$ and $k$ large but $k/N$ fixed.<br /><br />Last bonus question: How many of the people that say that they hire only based on merit and end up with an all male board realise that by this they say that women are not as good by quite a margin?Unknownnoreply@blogger.com2tag:blogger.com,1999:blog-8883034.post-29552792327471824022017-11-09T10:35:00.001+01:002017-11-09T10:35:22.989+01:00Why is there a supercontinent cycle?One of the most influential books of my early childhood was my "Kinderatlas"<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://images.booklooker.de/s/03249537_MTQxMzA=/Ravensburger-Kinderatlas.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="398" data-original-width="300" height="320" src="https://images.booklooker.de/s/03249537_MTQxMzA=/Ravensburger-Kinderatlas.jpg" width="241" /></a></div><div class="separator" style="clear: both; text-align: left;">There were many things to learn about the world (maps were actually only the last third of the book) and for example I blame <a href="https://thetheoreticaldiver.org/wordpress">my fascination for scuba diving</a> on this book. Also last year, when we visited the Mont-Doré in Auvergne and I had to explain how volcanos are formed to my kids to make them forget how many stairs were still ahead of them to the summit, I did that while mentally picturing the pages in that book about plate tectonics.</div><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-wz0JRcXGQos/WgQbY9hmgPI/AAAAAAAALZA/POM4JzAJyPYRYz7yswxn7vMtsoPhR3W_ACLcBGAs/s1600/IMG_0304.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="1600" data-original-width="1200" height="320" src="https://2.bp.blogspot.com/-wz0JRcXGQos/WgQbY9hmgPI/AAAAAAAALZA/POM4JzAJyPYRYz7yswxn7vMtsoPhR3W_ACLcBGAs/s320/IMG_0304.jpg" width="240" /></a></div><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://2.bp.blogspot.com/-hUniMIC9xac/WgQb1ZLJhYI/AAAAAAAALZE/xjSpHy8vFXQapyVTmOtMIpnhokFtRCN6wCLcBGAs/s1600/IMG_0310.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" data-original-height="348" data-original-width="1600" height="138" src="https://2.bp.blogspot.com/-hUniMIC9xac/WgQb1ZLJhYI/AAAAAAAALZE/xjSpHy8vFXQapyVTmOtMIpnhokFtRCN6wCLcBGAs/s640/IMG_0310.jpg" width="640" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: left;">But there is one thing I about tectonics that has been bothering me for a long time and I still haven't found a good explanation for (or at least an acknowledgement that there is something to explain): Since the days of Alfred Wegener we know that the jigsaw puzzle pieces of the continents fit in a way that geologists believe that some hundred million years ago they were all connected as a supercontinent Pangea.</div><a href="https://commons.wikimedia.org/wiki/File:Pangea_animation_03.gif#/media/File:Pangea_animation_03.gif"><img alt="Pangea animation 03.gif" src="https://upload.wikimedia.org/wikipedia/commons/8/8e/Pangea_animation_03.gif" /></a><br />By Original upload by <a class="extiw" href="https://en.wikipedia.org/wiki/User:Tbower" title="en:User:Tbower">en:User:Tbower</a> - <a class="external text" href="https://geomaps.wr.usgs.gov/parks/animate/" rel="nofollow">USGS animation A08</a>, Public Domain, <a href="https://commons.wikimedia.org/w/index.php?curid=583951">Link</a><br /><br />In fact, that was only the last in a series of supercontinents, that keep forming and breaking up in the "supercontinent cycle".<br /><a href="https://commons.wikimedia.org/wiki/File:Platetechsimple.png#/media/File:Platetechsimple.png"><img alt="Platetechsimple.png" height="640" src="https://upload.wikimedia.org/wikipedia/commons/thumb/2/2d/Platetechsimple.png/1200px-Platetechsimple.png" width="640" /></a><br />By <a class="new" href="https://commons.wikimedia.org/w/index.php?title=User:SimplisticReps&action=edit&redlink=1" title="User:SimplisticReps (page does not exist)">SimplisticReps</a> - <span class="int-own-work" lang="en">Own work</span>, <a href="https://creativecommons.org/licenses/by-sa/4.0" title="Creative Commons Attribution-Share Alike 4.0">CC BY-SA 4.0</a>, <a href="https://commons.wikimedia.org/w/index.php?curid=63509483">Link</a><br /><br />So here is the question: I am happy with the idea of several (say $N$) plates roughly containing a continent each that a floating around on the magma driven by all kinds of convection processes in the liquid part of the earth. They are moving around in a pattern that looks to me to be pretty chaotic (in the non-technical sense) and of course for random motion you would expect that from time to time two of those collide and then maybe stick for a while.<br /><br />Then it would be possible that also a third plate collides with the two but that would be a coincidence (like two random lines typically intersect but if you have three lines they would typically intersect in pairs but typically not in a triple intersection). But to form a supercontinent, you need all $N$ plates to miraculously collide at the same time. This order-$N$ process seems to be highly unlikely when random let alone the fact that it seems to repeat. So this motion cannot be random (yes, Sabine, this is a naturalness argument). This needs an explanation.<br /><br />So, why, every few hundred million years, do all the land masses of the earth assemble on side of the earth?<br /><br />One explanation could for example be that during those tines, the center of mass of the earth is not in the symmetry center so the water of the oceans flow to one side of the earth and reveals the seabed on the opposite side of the earth. Then you would have essentially one big island. But this seems not to be the case as the continents (those parts that are above sea-level) appear to be stable on much longer time scales. It is not that the seabed comes up on one side and the land on the other goes under water but the land masses actually move around to meet on one side.<br /><br />I have already asked this question whenever I ran into people with a geosciences education but it is still open (and I have to admit that in a non-zero number of cases I failed to even make the question clear that an $N$-body collision needs an explanation). But I am sure, you my readers know the answer or even better can come up with one.Unknownnoreply@blogger.com2tag:blogger.com,1999:blog-8883034.post-52566166367355955692017-06-16T16:55:00.000+02:002017-06-16T16:55:52.162+02:00I got this wrongIn <a href="https://atdotde.blogspot.de/2017/06/some-diy-ligo-data-analysis.html"><span id="goog_136133695"></span>yesterday's post<span id="goog_136133696"></span></a>, I totally screwed up when identifying the middle part of the spectrum as low frequency. It is not. Please ignore what I said or better take it as a warning what happens when you don't double check.<br /><br />Apologies to everybody that I stirred up!Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-8883034.post-54266413090808968632017-06-15T11:26:00.001+02:002017-06-16T16:47:20.122+02:00Some DIY LIGO data analysis<b>UPDATE: After some more thinking about this, I have very serious doubt about my previous conclusions. From looking at the power spectrum, I (wrongly) assumed that the middle part of the spectrum is the low frequency part (my original idea was, that the frequencies should be symmetric around zero but the periodicity of the Bloch cell bit me). So quite to the opposite, when taking into account the wrapping, this is the high frequency part (at almost the sample rate). So this is neither physics nor noise but the sample rate. For documentation, I do not delete the original post but leave it with this comment.</b><br /><br /><br />Recently, in the Arnold Sommerfeld Colloquium, we had Andrew Jackson of NBI talk about his take on the LIGO gravitational wave data, see <a href="http://www.physik.uni-muenchen.de/aus_der_fakultaet/kolloquien/asc_kolloquium/archiv_sose17/jackson/index.html">this announcement with link to a video recording</a>. He encouraged the audience to download the <a href="https://losc.ligo.org/events/GW150914/">freely available raw data</a> and play with it a little bit. This sounded like fun, so I had my go at it. Now, that <a href="https://arxiv.org/abs/1706.04191">his paper</a> is out, I would like to share what I did with you and ask for your comments.<br /><div><br /></div><div>I used mathematica for my experiments, so I guess the way to proceed is to guide you to<a href="http://atdotde.de/~robert/ligo%20writeup_moved.html"> an html export of my (admittedly cleaned up) notebook</a> (Source for your own experiments <a href="http://atdotde.de/~robert/ligo%20writeup_moved.nb">here</a>).</div><div><br /></div><div>The executive summary is that apparently, you can eliminate most of the "noise" at the interesting low frequency part by adding to the signal its time reversal casting some doubt about the stochasticity of this "noise".</div><div><br /></div><div class="separator" style="clear: both; text-align: center;"><a href="http://atdotde.de/~robert/HTMLFiles/ligo%20writeup_12.gif" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://atdotde.de/~robert/HTMLFiles/ligo%20writeup_12.gif" data-original-height="333" data-original-width="567" height="187" width="320" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: left;">I would love to hear what this is supposed to mean or what I am doing wrong, in particular from my friends in the gravitational wave community.</div><div class="separator" style="clear: both; text-align: center;"><br /></div><div><br /></div><div><br /></div>Unknownnoreply@blogger.com6tag:blogger.com,1999:blog-8883034.post-18980835894091586472017-06-08T11:48:00.002+02:002017-06-09T14:58:08.282+02:00Relativistic transformation of temperatureApparently, there is a long history of controversy going back to Einstein an Planck about the proper way to deal with temperature relativistically. And I admit, I don't know what exactly the modern ("correct") point of view is. So I would like to ask your opinion about a puzzle we came up during yesterday's after colloquium dinner with Erik Verlinde:<br /><div><br /></div><div>Imagine a long rail of a railroad track. It is uniformly heated to a temperature T and is in thermodynamic equilibrium (if you like a mathematical language: it is in a KMS state). On this railroad track travels Einstein's relativistic train at velocity v. From the perspective of the conductor, the track in front of the train is approaching the train with velocity v, so one might expect that the temperature T appears blue shifted while behind the train, the track is moving away with v and so the temperature appears red-shifted. </div><div><br /></div><div>Following this line of thought, one would conclude that the conductor thinks the rail has different temperatures in different places and thus is out of equilibrium. </div><div><br /></div><div>On the other hand, the question of equilibrium should be independent of the observer. So, is the assumption of the Doppler shift wrong? </div><div><br /></div><div>A few remarks: If you are worried that Doppler shifts should apply to radiation then you are free to assume that both in front and in the back, there are black bodies in thermal contact with the rail and thus exhibiting a photon gas at the same temperature as the rail.</div><div><br /></div><div>You could probably also make the case for the temperature transforming like the time component of a four vector (since it is essentially an energy). Then the transformed temperature would be independent of the sign of v. This you could for example argue for by assuming the temperature is so high that in your black body photon gas you also create electron-positron pairs which would be heavier due to their relativistic speed relative to the train and thus requiring more energy (and thus temperature) for their creation.</div><div><br /></div><div>A final remark is about an operational definition of temperature at relativistic speeds: It might be difficult to bring a relativistic thermometer in equilibrium with a system if there is a large relative velocity (when we define temperature as the criterium for two systems in contact to be in equilibrium). Or to operate a heat engine between he front part of the rail and the back while moving along at relativistic speed and then arguing about the efficiency (and defining the temperature that way).<br /><br /><b>Update one day later:</b><br />Thanks for all your comments. We also had some further discussions here and I would like to share my conclusions:<br /><br />1) It probably boils down to what exactly you mean when you say ("temperature"). Of course, you want that his at least in familiar situations agrees with what thermometers of this type or another measure. (In the original text I had hinted at two possible definitions that I learned about from a very interesting <a href="https://arxiv.org/abs/1212.2409">paper</a> by Buchholz and Solveen discussing the Unruh effect and what would actually be observed there: Either you define temperature that the property that characterises equilibrium states of systems such there is no heat exchange when you bring in contact two systems of the same temperature. This is for example close to what a mercury thermometer measures. Alternatively, you operate a perfect heat engine between two reservoirs and define your temperatures via<br />$$\eta = \frac{T_h - T_c}{T_h}.$$<br />This is for example hinted at in the Feynamn lectures on physics.<br /><br />One of the commentators suggested using the ratio of eigenvalues of the energy momentum tensor as definition of temperature. Even though this might give the usual thing for a perfect fluid I am not really convinced that this generalises in the right way.<br /><br />2) I would rather define the temperature as the parameter in the Gibbs (or rather KMS) state (it should only exist in equilibrium, anyway). So if your state is described by density matrix $\rho$, and it can be written as<br />$$\rho = \frac{e^{-\beta H}}{tr(e^{-\beta H})}$$<br />then $1/\beta$ is the temperature. Obviously, this requires the a priori knowledge of what the Hamiltonian is.<br /><br />For such states, under mild assumptions, you can prove nice things: Energy-entropy inequalities ("minimisation of free energy"), stability, return to equilibrium and most important here: passivity, i.e. the fact you cannot extract mechanical work out of this state in a cyclic process.<br /><br />2) I do not agree that it is out of the question to have a thermometer with a relative velocity in thermal equilibrium with a heat bath at rest. You could for example imagine a mirror fixed next to the track and in thermal equilibrium with the track. A second mirror is glued to the train (and again in thermal equilibrium, this time with a thermometer). Between the mirrors is is a photon gas (black body) that you could imagine equilibrating with the mirrors on both ends. The question is if that is the case.<br /><br />3) Maybe rails and trains a a bit too non-spherical cows, so lets better look at an infinitely extended free quantum gas (bosons or fermions, your pick). You put it in a thermal state at rest, i.e. up to normalisation, its density matrix is given by<br />$$\rho = e^{-\beta P^0}.$$<br />Here $P^0$ is the Poincaré generator of time translations.<br /><br />Now, the question above can be rephrased as: Is there a $\beta'$ such that also<br />$$\rho = e^{-\beta' (\cosh\alpha P^0 + \sinh \alpha P^1)}?$$<br />And to the question formulated this way, the answer is pretty clearly "No". A thermal state singles out a rest frame and that's it. It is not thermal in the moving frame and thus there is no temperature.<br /><br />It's also pretty easy to see this state is not passive (in the above sense): You could operate a windmill in the slipstream of particles coming more likely from the front than the back. So in particular, this state is not KMS (this argument I learned from Sven Bachmann).<br /><br />4) Another question would be about gravitational redshift: Let's take some curve space-time and for simplicity assume it has no horizons (for example, let the far field be Schwarzschild but in the center, far outside the Schwarzschild radius, you smooth it out. Like the space-time created by the sun). Make it static, so it contains a timeline Killing vector (otherwise no hope for a thermal state). Now prepare a scalar field in the thermal state with temperature T. Couple to it a harmonic oscillator via<br />$$ H_{int}(r) = a^\dagger a + \phi(t, r) (a^\dagger + a).$$<br />You could now compute a "local temperature" by computing the probability that the harmonic oscillator is in the first excited state. Then, how does this depend on $r$?</div>Unknownnoreply@blogger.com12tag:blogger.com,1999:blog-8883034.post-58424659282751179802016-12-09T14:48:00.001+01:002016-12-09T14:48:29.824+01:00Workshop on IoT Liability at 33c3After my recent <a href="https://atdotde.blogspot.de/2016/10/mandatory-liability-for-software-is.html">blog post</a> on the dangers of liability for manufacturers of devices in the times of IoT, I decided I will run a <a href="https://events.ccc.de/congress/2016/wiki/Session:Haftung_f%C3%BCr_Devices_und_Software_gestalten">workshop</a> at 33C3, the annual hacker conference of the Chaos Computer club. I am proud I could convince <a href="https://buermeyer.de/ulf/">Ulf Buermeyer</a> (well known judge, expert in constitutional law, hacker, podcaster) to host this workshop with me.<br /><br />The main motivation for me is that I hope that this will be a big issue in the coming year but it might still be early enough to influence policy before everybody commits herself to their favorite (snake oil) solution.<br /><br />I have started collecting and sorting ideas in a <a href="https://docs.google.com/document/d/1hTYnLYdzDjeZAs_45HFx_HfMaZI2FrYJTV9YaXCHwrw/edit?usp=sharing">Google document</a>. <a href="https://commons.wikimedia.org/wiki/File%3AInternet_of_things_signed_by_the_author.jpg" title="By Wilgengebroed on Flickr [CC BY 2.0 (http://creativecommons.org/licenses/by/2.0)], via Wikimedia Commons"><img alt="Internet of things signed by the author" src="https://upload.wikimedia.org/wikipedia/commons/thumb/0/01/Internet_of_things_signed_by_the_author.jpg/512px-Internet_of_things_signed_by_the_author.jpg" width="512" /></a>Unknownnoreply@blogger.com5tag:blogger.com,1999:blog-8883034.post-28507534590825136672016-11-26T17:19:00.002+01:002016-11-26T17:19:58.857+01:00Breaking News: Margarine even more toxic!One of the most popular posts of this blog (as far as resonance goes) was the one on <a href="https://atdotde.blogspot.de/2006/08/scaling-of-price-of-margarine.html">Scaling the Price of Margarine</a>. Today, I did the family weekend shopping and noticed I have to update the calculation:<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://1.bp.blogspot.com/-Hp34RKjWpPk/WDmxh3naCbI/AAAAAAAALUA/WrC0JAlclgM-_f6uc2WJ7ZHL33QAR50agCLcB/s1600/laetta%2B-%2B1.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="320" src="https://1.bp.blogspot.com/-Hp34RKjWpPk/WDmxh3naCbI/AAAAAAAALUA/WrC0JAlclgM-_f6uc2WJ7ZHL33QAR50agCLcB/s320/laetta%2B-%2B1.jpg" width="240" /></a></div>At our local Rewe branch, they offer the pound of Lätta for 88 cents while they ask 1.19Euro for half the pound. With the ansatz from the old post, this means the price for the actual margarine is now -9,78Euro/kg. This, by coincidence is approximately also the price you have to pay to get rid of waste oil.Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-8883034.post-30445981001602457042016-11-13T14:12:00.001+01:002016-11-13T14:12:31.254+01:00Theoretical diverBesides physics, another hobby of mine is scuba diving. For many reasons, unfortunately, I don't have much time anymore, to get in the water. As partial compensation, I started some time ago to contribute the <a href="https://subsurface-divelog.org/">Subsurface</a>, the open source dive log program. Partly related to that, I also like to theorize about diving. To put that in form, I now started another blog <a href="http://atdotde.de/theoreticaldiver/">The Theoretical Diver</a> to discuss aspects of diving that I have been thinking about.Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-8883034.post-78417764819897241552016-11-13T09:36:00.000+01:002016-11-13T11:52:37.483+01:00OpenAccess: Letter to the editor of Süddeutsche ZeitungIn yesterday's Süddeutsche Zeitung, there is an <a href="http://www.sueddeutsche.de/politik/kolumne-goldener-zugang-1.3244900">opinion piece</a> by historian <a href="https://de.wikipedia.org/wiki/Norbert_Frei">Norbert Frei</a> on the German government's OpenAccess initiative, which prompted me to write a letter to the editor (naturally in German). Here it is:<br /><br /><span style="font-family: "helvetica"; font-size: 12px;">Zum Meinungsbeitrag „Goldener Zugang“ von Norbert Frei in der SZ vom 12./13. November 2016:</span><br /><div style="font-family: Helvetica; font-size: 12px;"><br /></div><div style="font-family: Helvetica; font-size: 12px;">Herr Frei sorgt sich in seinem Beitrag, dass der Wissenschaft unter der Überschrift OpenAccess von außen ein Kulturwandel aufgezwungen werden soll. Er fürchtet, dass ihn die Naturwissenschaftler zusammen mit der Politik zwingen, seine Erkenntnisse nicht mehr in papiernen Büchern darlegen zu können, sondern alles nur noch zerstückelt in kleine Artikel-Happen in teure digitale Archive einzustellen, wo sie auf die Bitverrottung waren, da schon in kürzester Zeit das Fortschreiten von Hard- und Software dazu führen wird, dass die Datenformate unlesbar werden.</div><div style="font-family: Helvetica; font-size: 12px;"><br /></div><div style="font-family: Helvetica; font-size: 12px;">Als Gegenmodell führt er die Gutenberg-Bibel an, von der eine Mehrzahl der Exemplare die Jahrhunderte überdauert haben. Nun weiss ich nicht, wann Herr Frei das letzte Mal in seiner Gutenberg-Bibel geblättert hat, ich habe in meinem Leben nur ein einziges Mal vor einer gestanden: Diese lag in einer Vitrine der Bibliothek von Cambridge und war auf einer Seite aufgeschlagen, keine andere Seite war zugänglich. Dank praktischem OpenAccess ist es aber nicht nur den guten Christenmenschen möglich, eine Kopie zu Hause vorzuhalten. Viel mehr noch, die akademischen Theologen aus meinem Bekanntenkreis arbeiten selbstverständlich mit einer digitalen Version auf ihrem Laptop oder Smartphone, da diese dank Durchsuchbarkeit, Indizierung und Querverweisen in andere Werke für die Forschung viel zugänglicher ist.</div><div style="font-family: Helvetica; font-size: 12px;"><br /></div><div style="font-family: Helvetica; font-size: 12px;">Geschenkt, dass es bei der OpenAccess-Initiative eine Ausnahme für Monographien geben soll. Niemand will das Bücherschreiben verbieten. Es geht nur darum, dass, wer Drittmittel von der öffentlichen Hand erhalten will, nicht noch einmal die Hand dafür aufhalten soll, wenn sich dann die vor allem wissenschaftliche Öffentlichkeit über die Ergebnisse informieren will. Professorinnen und Professoren an deutschen Universitäten schreiben ihre wissenschaftlichen Veröffentlichungen nicht zu ihrem Privatvergnügen, es ist Teil ihrer Dienstaufgaben. Warum wollen sie die Früchte ihres bereits entlohnten Schaffens dann noch ein weiteres Mal den öffentlichen Bibliotheken verkaufen? </div><div style="font-family: Helvetica; font-size: 12px;"><br /></div><div style="font-family: Helvetica; font-size: 12px;">Ich kann mich noch gut an meinen Stolz erinnern, als ich das erste Mal meinen Namen gedruckt auf Papier sah, der das Titelblatt meiner ersten Veröffentlichung zierte. Jenseits davon ist es für mich als Wissenschaftler vor allem wichtig, dass das, was ich da herausfinde, von anderen wahrgenommen und weitergetrieben wird. Und das erreiche ich am besten, wenn es so wenig Hürden wie möglich gibt, dieses zu tun.</div><div style="font-family: Helvetica; font-size: 12px;"><br /></div><div style="font-family: Helvetica; font-size: 12px;">Ich selber bin theoretischer Hochenergiephysiker, selbstredend gibt es sehr unterschiedliche Fächerkulturen. In meinem Fach ist es seit den frühen Neunzigerjahren üblich, alle seine Veröffentlichungen - vom einseitigen Kommentar zu einem anderen Paper bis zu einem Review von vielen hundert Seiten - in arXiv.org, einem nichtkommerziellen Preprintarchiv einzustellen, wo es von allen Fachkolleginnen und -kollegen ab dem nächsten Morgen gefunden und in Gänze gelesen werden kann, selbst viele hervorragend Lehrbücher gibt es inzwischen dort. Diese globale Verbreitung neben einfachem Zugang (ich habe schon seit mehreren Jahren keinen papiernen Fachartikel in unserer Bibliothek mehr in einem Zeitschriftenband mehr nachschlagen müssen, ich finde alles auf meinem Computer) hat so viele Vorteile, das man gerne auf mögliche Tantiemen verzichtet, zumal diese für Zeitschriftenartikel noch nie existiert haben und, von wenigen Ausnahmen abgesehen, verschwinden gering gegenüber einem W3-Gehalt ausfallen und als Stundenlohn berechnet jeden Supermarktregaleinräumer sofort die Arbeit niederlegen ließen. Wir Naturwissenschaftler sind auf einem guten Weg, uns von parasitären Fachverlagen zu emanzipieren, die es traditionell schafften, jährlich den Bibliotheken Milliardenumsätze für unsere Arbeit abzupressen, wobei sie das Schreiben der Artikel, die Begutachtung, den Textsatz und die Auswahl unbezahlt an von der Öffentlichkeit bezahlte Wissenschaftlerinnen und Wissenschaftler delegiert haben und sie sich ausschliesslich ihre Gatekeeper Funktion bezahlen liessen. </div><div style="font-family: Helvetica; font-size: 12px;"><br /></div><div style="font-family: Helvetica; font-size: 12px;">Und da ich an Leserschaft interessiert bin, werde ich diesen Brief auch in mein Blog einstellen.</div><div><br /></div>Unknownnoreply@blogger.com4tag:blogger.com,1999:blog-8883034.post-73095928251685852732016-10-27T14:56:00.003+02:002016-10-27T15:12:54.729+02:00Daylight saving time about to end (end it shouldn't)Twice a year, around the last Sunday in March and the last Sunday in October, everybody (in particular newspaper journalists) take a few minutes off to rant about daylight savings time. So, for this first time, I want to join this tradition in writing.<br /><br />Until I had kids, I could not care less about the question of changing the time twice a year. But at least for our kids (and then secondary also for myself), I realize biorhythm is quite strong and at takes more than a week to adopt to the 1 hour jet lag (in particular in spring when it means getting out of bed "one hour earlier"). I still don't really care about cows that have to deliver their milk at different times since there is no intrinsic reason that the clock on the wall has to show a particular time when it is done and if it were really a problem, the farmers could do it at fixed UTC.<br /><br />So, obviously, it is a nuisance. So what are the benefit that justify it? Well, obviously, in summer the sun sets at a later hour and we get more sun when being outside in the summer. That sounds reasonable. But why restrict it to the summer?<br /><br />Which brings me to my point: If you ask me, I want to get rid of changing the offset to UTC twice a year and want to permanently adopt daylight saving time.<br /><br />But now I hear people cry that this is "unnatural", we have to have the traditional time at least in the winter when it does not matter as it's too cold to be outside (which only holds for people with defective clothing as we know). So how natural is CET (the time zone we set our clocks to in winter), let's take people living in Munich for an example?<br /><br />First of all: It is not solar time! CET is the "mean solar time" when you live at a longitude of 15 degrees east, which is (assuming the latitude) close to <a href="https://www.google.de/maps/@48.1454127,15,13z?hl=de">Neumarkt an der Ypps</a> somewhere in Austria not too far from Vienna. Munich is about 20 minutes behind. So, this time is artificial as well, and Berlin being closer to 15 degrees, it is probably Prussian.<br /><br />Also a common time zone for Germany was established only in the 1870s when the advent of railways and telegraphs make synchronization between different local times advantageous. So this "natural" time is not that old either.<br /><br />It is so new, that<a href="https://en.wikipedia.org/wiki/Christ_Church,_Oxford"> Christ Church college in Oxford</a> still refuses to fully bow to it: Their clock tower shows Greenwich time. And the cathedral services start according to solar time (about five minutes later) because they don't care about modern shenanigans. ("How many Oxford deans does it take to change a light bulb?" ---- "Change??!??"). Similarly, in Bristol, there is a famous clock with two minute hands.<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://upload.wikimedia.org/wikipedia/commons/thumb/2/2a/Exchangeclock.JPG/800px-Exchangeclock.JPG" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="238" src="https://upload.wikimedia.org/wikipedia/commons/thumb/2/2a/Exchangeclock.JPG/800px-Exchangeclock.JPG" width="320" /></a></div><br />Plus, even if you live in Neumarkt an der Ybbs, your sun dial does not always show the correct noon! Thanks to the tilt of the earth axis and the fact that the orbit of the earth is elliptic, this varies through the year by a number of minutes:<br /><div class="separator" style="clear: both; text-align: center;"><a href="https://upload.wikimedia.org/wikipedia/commons/thumb/7/7a/Equation_of_time.svg/512px-Equation_of_time.svg.png" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="274" src="https://upload.wikimedia.org/wikipedia/commons/thumb/7/7a/Equation_of_time.svg/512px-Equation_of_time.svg.png" width="320" /></a></div><br />So, "winter time" is in no way more natural than the other time zone. So we should be free to choose a time zone according to what is convenient. At least for me, noon is not the center of my waking hours (it's more 5,5 : 12). So, aligning those more with the sun seems to be a pretty good idea.<br /><br />PS: The title was a typo, but looking at it I prefer it the way it is...<br /><br />Unknownnoreply@blogger.com2tag:blogger.com,1999:blog-8883034.post-37901975775254653842016-10-24T14:55:00.003+02:002016-10-24T15:02:02.875+02:00Mandatory liability for software is a horrible ideaOver the last few days, a number of prominent web sites including Twitter, Snapchat and Github were effectively unreachable for an extended period of time. As became clear, the problem was that DynDNS, a provider of DNS services for these sites was under a number of very heavy DDoS (distributed denial of service) attack that were mainly coming from compromised internet of things devices, in particular web cams.<br /><br />Even though I do not see a lot of benefit from being able to change the color of my bedroom light via internet, I love the idea to have lots of cheap devices (I continue to have a lot of fun with C.H.I.P.s, full scale Linux computers with a number of ports for just 5USD, also for Subsurface, in particular those open opportunities for the mobile version), there are of course concerns how one can economically have a stable update cycle for those, in particular once they are build into black-box customer devices.<br /><br />Now, after some dust settled comes of course the question "Who is to blame?" and should be do anything about this. Of course, the manufacturer of the web cam made this possible through far from perfect firmware. Also, you could blame DynDNS for not being able to withstand the storms that from time to time sweep the internet (a pretty rough place after all) or the services like Twitter to have a single point of failure in DynDNS (but that might be hard to prevent given the nature of the DNS system).<br /><br />More than once I have now heard a <a href="http://www.taz.de/Kommentar-Hackerangriff/!5347856/">call for new laws</a> that would introduce a liability for the manufacturer of the web cam as they did not provide firmware updates in time that prevent these devices from being owned and then DDoSing around on the internet.<br /><br />This, I am convinced, would be a terrible idea: It would make many IT businesses totally uneconomic. Let's stick for example with the case at hand. What is the order of magnitude of damages that occurred to the big companies like Twitter? They probably lost ad revenue of about a weekend. Twitter recently made $6\cdot 10^8\$ $ per quarter, which averages to 6.5 million per day. Should the web cam manufacturer (or OEM or distributor) now owe Twitter 13 million dollars? I am sure that would cause immediate bankruptcy. Or just the risk that this could happen would prevent anybody from producing web cams or similar things in the future. As nobody can produce non-trivial software that is free of bugs. You should strive to weed out all known bugs and provide updates, of course, but should you be made responsible if you couldn't? Responsible in a financial sense?<br /><br />What was the damage cause by the <a href="https://en.wikipedia.org/wiki/Heartbleed">heart bleed bug</a>? I am sure this was much more expensive. Who should pay for this? OpenSSL? Everybody that links against OpenSSL? The person that committed the wrong patch? The person that missed it code review?<br /><br />Even if you don't call up these astronomic sums and have fixed fine (e.g. an unfixed vulnerability that gives root access to an attacker from the net costs 10000$) that would immediately stop all open source development. If you give away your software for free, do you really want to pay fines if not everything is perfect? I surely wouldn't.<br /><br />For that reason, the GPL has the clauses (and other open source licenses have similar ones) stating<br /><br /><blockquote class="tr_bq"> 11. BECAUSE THE PROGRAM IS LICENSED FREE OF CHARGE, THERE IS NO WARRANTY<br />FOR THE PROGRAM, TO THE EXTENT PERMITTED BY APPLICABLE LAW. EXCEPT WHEN<br />OTHERWISE STATED IN WRITING THE COPYRIGHT HOLDERS AND/OR OTHER PARTIES<br />PROVIDE THE PROGRAM "AS IS" WITHOUT WARRANTY OF ANY KIND, EITHER EXPRESSED<br />OR IMPLIED, INCLUDING, BUT NOT LIMITED TO, THE IMPLIED WARRANTIES OF<br />MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE. THE ENTIRE RISK AS<br />TO THE QUALITY AND PERFORMANCE OF THE PROGRAM IS WITH YOU. SHOULD THE<br />PROGRAM PROVE DEFECTIVE, YOU ASSUME THE COST OF ALL NECESSARY SERVICING,<br />REPAIR OR CORRECTION.<br /> 12. IN NO EVENT UNLESS REQUIRED BY APPLICABLE LAW OR AGREED TO IN WRITING<br />WILL ANY COPYRIGHT HOLDER, OR ANY OTHER PARTY WHO MAY MODIFY AND/OR<br />REDISTRIBUTE THE PROGRAM AS PERMITTED ABOVE, BE LIABLE TO YOU FOR DAMAGES,<br />INCLUDING ANY GENERAL, SPECIAL, INCIDENTAL OR CONSEQUENTIAL DAMAGES ARISING<br />OUT OF THE USE OR INABILITY TO USE THE PROGRAM (INCLUDING BUT NOT LIMITED<br />TO LOSS OF DATA OR DATA BEING RENDERED INACCURATE OR LOSSES SUSTAINED BY<br />YOU OR THIRD PARTIES OR A FAILURE OF THE PROGRAM TO OPERATE WITH ANY OTHER<br />PROGRAMS), EVEN IF SUCH HOLDER OR OTHER PARTY HAS BEEN ADVISED OF THE<br />POSSIBILITY OF SUCH DAMAGES.</blockquote><div>(capitalization in the original). Of course, there is "required by applicable law" but I cannot see people giving you software for free if you later make them pay fines.</div><div><br /></div><div>And for course, it is also almost impossible to make exceptions in the law for this. For example, a "non-commercial" exception does not help as even though you do not charge for open source software a lot of it is actually provided with some sort of commercial interest.</div><div><br /></div><div>Yes, I can understand the tendency to make creators of defective products that don't give a damn about an update path responsible for the stuff they ship out. And I have the greatest sympathy for consumer protection laws. But here, there collateral damage would be huge (we might well lose the whole open source universe every small software company except the few big one that can afford the herds of lawyers to defend against these fines).<br /><br />Note that I only argue for mandatory liability. It should of course always be a possibility that a provider of software/hardware give some sort of "fit for purpose" guarantee to its customers or a servicing contract where they promise to fix bugs (maybe so that the customer can fulfill their liabilities to their customers herself). But in most of the cases, the provider will charge for that. And the price might be higher than currently that for a light bulb with an IP address.</div><div><br /></div><div>The internet is a rough place. If you expose your service to it better make sure you can handle every combination of 0s and 1s that comes in from there or live with it. Don't blame the source of the bits (no matter how brain dead the people at the other end might be).</div><div><br /></div><div><br /></div>Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-8883034.post-86039624289600381512016-10-07T10:15:00.003+02:002016-10-07T12:07:57.756+02:00My two cents on this year's physics Nobel prizeThis year's <a href="http://www.nobelprize.org/nobel_prizes/physics/laureates/2016/press.html">Nobel prize</a> is given for quite abstract concepts. So the popular science outlets struggle in giving good explanations for what it is awarded for. I cannot add anything to this, but over at math overflow, mathematicians <a href="http://mathoverflow.net/questions/251470/topology-and-the-2016-nobel-prize-in-physics/251590#251590">asked for a mathematical explanation</a>. So here is my go of an outline for people familiar with topology but not so much physics:<br /><br />Let me try to give a brief explanation: All this is in the context of Fermi liquid theory, the idea that you can describe the low energy physics of these kinds of systems by pretending they are generated by free fermions in an external potential. So, all you need to do is to solve the single particle problem for the external potential and then fill up the energy levels from the bottom until you reach the total particle number (or actually the density). It is tempting (and conventional) to call these particles electrons, and I will do so here, but of course actual electrons are not free but interacting. This "Fermi Liquid" explanation is just and effective description for long wavelength (the IR end of the renormalization group flow) where it turns out, that at those scales the interactions play no role (they are "irrelevant operators" in the language of the renormalization group).<br /><br />The upshot is, we are dealing with free "electrons" and the previous paragraph was only essential if you want to connect to the physical world (but this is MATH overflow anyway).<br /><br />Since the external potential comes from a lattice (crystal) it is invariant under lattice translations. So Bloch theory tells you, you can restrict your attention as far as solving the Schrödinger equation to wave functions living in the unit cell of the lattice. But you need to allow for quasi-periodic boundary conditions, i.e. when you go once around the unit cell you are allowed to pick up a phase. In fact, there is one phase for each generator of the first homotopy group of the unit cell. Each choice of these phases corresponds to one choice of boundary conditions for the wave function and you can compute the eigenvalues of the Hamiltonian for these given boundary conditions (the unit cell is compact so we expect discrete eigenvalues, bounded from below).<br /><br />But these eigenvalues depend on the boundary conditions and you can think of the as a function of the phases. Each of the phases takes values in U(1) so the space of possible phases is a torus and you can think of the eigenvalues as functions on the torus. Actually, when going once around an irreducible cycle of the torus not all eigenvalues have to come back to themselves, you can end up with a permutation it this is not really a function but a section of a bundle but let's not worry too much about this as generally this "level crossing" does not happen in two dimensions and only at discrete points in 3D (this is Witten's argument with the 2x2 Hamiltonian above).<br /><br />The torus of possible phases is called the "Brioullin zone" (sp?) by physicists and its elements "inverse lattice vectors" (as you can think of the Brioullin zone as obtained from modding out the dual lattice of the lattice we started with).<br /><br />Now if your electron density is N electrons per unit cell of the lattice Fermi Liquid theory asks you to think of the lowest N energy levels as occupied. This is the "Fermi level" or more precisely the graph of the N-th eigenvalue over the Bioullin zone. This graph (views as a hyper-surface) can have non-trivial topology and the idea is that by doing small perturbations to the system (like changing the doping of the physical probe or changing the pressure or external magnetic field or whatever) stuff behaves continuously and thus the homotopy class cannot change and is thus robust (or "topological" as the physicist would say).<br /><br />If we want to inquire about the quantum Hall effect, this picture is also useful: The Hall conductivity can be computed to leading order by linear response theory. This allows us to employ the Kubo formula to compute it as a certain two-point function or retarded Green's function. The relevant operators turn out to be related to the N-th level wave function and how it changes when we move around in the Brioullin zone: If we denote by u the coordinates of the Brioullin zone and by $\psi_u(x)$ the N-th eigenfunction for the boundary conditions implied by u, we can define a 1-form<br />$$ A = \sum_i \langle \psi_u|\partial_{u_i}|\psi_u\rangle\, du^i = \langle\psi_u|d_u|\psi\rangle.$$<br />This 1-form is actually the connection of a U(1) bundle and the expression the Kubo-formula asks us to compute turns out to be the first Chern number of that bundle (over the Brioullin zone).<br /><br />Again that, as in integer, cannot change upon small perturbations of the physical system and this is the explanation of the levels in the QHE.<br /><br />In modern applications, an important role is played by the (N-dimensional and thus finite dimensional) projector the subspace of Hilbert space spanned by the eigenfunctions corresponding to he N lowest eigenvalues, again fibered over the Brioullin zone. Then one can use K-theory (and KO-theory in fact) related to this projector to classify the possible classes of Fermi surfaces (these are the "topological phases of matter", as eventually, when the perturbation becomes too strong even the discrete invariants can jump which then physically corresponds to a phase transition).Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-8883034.post-88880790578901583592016-10-07T10:15:00.002+02:002016-10-07T10:42:17.891+02:00My two cents on this years physics Nobel prizeThis year's <a href="http://www.nobelprize.org/nobel_prizes/physics/laureates/2016/press.html">Nobel prize</a> is given for quite abstract concepts. So the popular science outlets struggle in giving good explanations for what it is awarded for. I cannot add anything to this, but over at math overflow, mathematicians <a href="http://mathoverflow.net/questions/251470/topology-and-the-2016-nobel-prize-in-physics/251590#251590">asked for a mathematical explanation</a>. So here is my go of an outline for people familiar with topology but not so much physics:<br /><br />Let me try to give a brief explanation: All this is in the context of Fermi liquid theory, the idea that you can describe the low energy physics of these kinds of systems by pretending they are generated by free fermions in an external potential. So, all you need to do is to solve the single particle problem for the external potential and then fill up the energy levels from the bottom until you reach the total particle number (or actually the density). It is tempting (and conventional) to call these particles electrons, and I will do so here, but of course actual electrons are not free but interacting. This "Fermi Liquid" explanation is just and effective description for long wavelength (the IR end of the renormalization group flow) where it turns out, that at those scales the interactions play no role (they are "irrelevant operators" in the language of the renormalization group).<br /><br />The upshot is, we are dealing with free "electrons" and the previous paragraph was only essential if you want to connect to the physical world (but this is MATH overflow anyway).<br /><br />Since the external potential comes from a lattice (crystal) it is invariant under lattice translations. So Bloch theory tells you, you can restrict your attention as far as solving the Schrödinger equation to wave functions living in the unit cell of the lattice. But you need to allow for quasi-periodic boundary conditions, i.e. when you go once around the unit cell you are allowed to pick up a phase. In fact, there is one phase for each generator of the first homotopy group of the unit cell. Each choice of these phases corresponds to one choice of boundary conditions for the wave function and you can compute the eigenvalues of the Hamiltonian for these given boundary conditions (the unit cell is compact so we expect discrete eigenvalues, bounded from below).<br /><br />But these eigenvalues depend on the boundary conditions and you can think of the as a function of the phases. Each of the phases takes values in U(1) so the space of possible phases is a torus and you can think of the eigenvalues as functions on the torus. Actually, when going once around an irreducible cycle of the torus not all eigenvalues have to come back to themselves, you can end up with a permutation it this is not really a function but a section of a bundle but let's not worry too much about this as generally this "level crossing" does not happen in two dimensions and only at discrete points in 3D (this is Witten's argument with the 2x2 Hamiltonian above).<br /><br />The torus of possible phases is called the "Brioullin zone" (sp?) by physicists and its elements "inverse lattice vectors" (as you can think of the Brioullin zone as obtained from modding out the dual lattice of the lattice we started with).<br /><br />Now if your electron density is N electrons per unit cell of the lattice Fermi Liquid theory asks you to think of the lowest N energy levels as occupied. This is the "Fermi level" or more precisely the graph of the N-th eigenvalue over the Bioullin zone. This graph (views as a hyper-surface) can have non-trivial topology and the idea is that by doing small perturbations to the system (like changing the doping of the physical probe or changing the pressure or external magnetic field or whatever) stuff behaves continuously and thus the homotopy class cannot change and is thus robust (or "topological" as the physicist would say).<br /><br />If we want to inquire about the quantum Hall effect, this picture is also useful: The Hall conductivity can be computed to leading order by linear response theory. This allows us to employ the Kubo formula to compute it as a certain two-point function or retarded Green's function. The relevant operators turn out to be related to the N-th level wave function and how it changes when we move around in the Brioullin zone: If we denote by u the coordinates of the Brioullin zone and by $\psi_u(x)$ the N-th eigenfunction for the boundary conditions implied by u, we can define a 1-form<br />$$ A = \sum_i \langle \psi_u|\partial_{u_i}|\psi_u\rangle\, du^i = \langle\psi_u|d_u|\psi\rangle.$$<br />This 1-form is actually the connection of a U(1) bundle and the expression the Kubo-formula asks us to compute turns out to be the first Chern number of that bundle (over the Brioullin zone).<br /><br />Again that, as in integer, cannot change upon small perturbations of the physical system and this is the explanation of the levels in the QHE.<br /><br />In modern applications, an important role is played by the (N-dimensional and thus finite dimensional) projector the subspace of Hilbert space spanned by the eigenfunctions corresponding to he N lowest eigenvalues, again fibered over the Brioullin zone. Then one can use K-theory (and KO-theory in fact) related to this projector to classify the possible classes of Fermi surfaces (these are the "topological phases of matter", as eventually, when the perturbation becomes too strong even the discrete invariants can jump which then physically corresponds to a phase transition).Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-8883034.post-39458583830332274852016-09-19T21:43:00.000+02:002016-09-19T21:43:06.273+02:00Brute forcing Crazy Game PuzzlesIn the 1980s, as a kid I loved my Crazy Turtles Puzzle ("Das verrückte Schildkrötenspiel"). For a number of variations, see <a href="http://www.geekyhobbies.com/the-crazy-game-puzzles-puzzled/">here</a> or <a href="http://www.penguin.com/static/packages/us/yr-microsites/crazygamesolution/index.php">here</a>.<br /><br />I had completely forgotten about those, but a few days ago, I saw a self-made reincarnation when staying at a friends' house:<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="https://4.bp.blogspot.com/-3EsTQ25UG2c/V-A8U8PBycI/AAAAAAAALK0/d7Wo4YqA95wxdatwCa6ZuR9_f69jA0ntgCLcB/s1600/IMG_0350%2B%25281%2529.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="320" src="https://4.bp.blogspot.com/-3EsTQ25UG2c/V-A8U8PBycI/AAAAAAAALK0/d7Wo4YqA95wxdatwCa6ZuR9_f69jA0ntgCLcB/s320/IMG_0350%2B%25281%2529.jpg" width="319" /></a></div><div class="separator" style="clear: both; text-align: center;"><br /></div><div class="separator" style="clear: both; text-align: left;"><br /></div>I tried a few minutes to solve it, unsuccessfully (in case it is not clear: you are supposed to arrange the nine tiles in a square such that they form color matching arrows wherever they meet).<br /><br />So I took the picture above with the plan to either try a bit more at home or write a program to solve it. Yesterday, I had about an hour and did the latter. I am a bit proud of the implementation I came up with and in particular the fact that I essentially came up with a correct program: It came up with the unique solution the first time I executed it. So, here I share it:<br /><br /><pre style="background-color: #001800; color: #55cc66;"><span style="color: #b96969;">#!</span><span style="color: #007997;">/usr/bin/perl</span><br /><br /><span style="color: #b96969;"># 1 rot 8</span><br /><span style="color: #b96969;"># 2 gelb 7</span><br /><span style="color: #b96969;"># 3 gruen 6</span><br /><span style="color: #b96969;"># 4 blau 5</span><br /><br />@karten <span style="color: #808030;">=</span> <span style="color: #808030;">(</span><span style="color: #778c77;">7151</span><span style="color: #808030;">,</span> <span style="color: #778c77;">6754</span><span style="color: #808030;">,</span> <span style="color: #778c77;">4382</span><span style="color: #808030;">,</span> <span style="color: #778c77;">2835</span><span style="color: #808030;">,</span> <span style="color: #778c77;">5216</span><span style="color: #808030;">,</span> <span style="color: #778c77;">2615</span><span style="color: #808030;">,</span> <span style="color: #778c77;">2348</span><span style="color: #808030;">,</span> <span style="color: #778c77;">8253</span><span style="color: #808030;">,</span> <span style="color: #778c77;">4786</span><span style="color: #808030;">)</span><span style="color: purple;">;</span><br /><br /><span style="color: #508050; font-weight: bold;">foreach</span> $karte<span style="color: #808030;">(</span><span style="color: #778c77;">0</span><span style="color: #808030;">..</span><span style="color: #778c77;">8</span><span style="color: #808030;">)</span> <span style="color: purple;">{</span><br /> $farbe<span style="color: #808030;">[</span>$karte<span style="color: #808030;">]</span> <span style="color: #808030;">=</span> <span style="color: #808030;">[</span><span style="color: #808030;">split</span><span style="color: #cc5555;"> </span><span style="color: maroon;">/</span><span style="color: maroon;">/</span><span style="color: #808030;">,</span>$karten<span style="color: #808030;">[</span>$karte<span style="color: #808030;">]</span><span style="color: #808030;">]</span><span style="color: purple;">;</span><br /><span style="color: purple;">}</span><br /><span style="color: #808030;">&</span>ausprobieren<span style="color: #808030;">(</span><span style="color: #778c77;">0</span><span style="color: #808030;">)</span><span style="color: purple;">;</span><br /><br /><span style="color: #508050; font-weight: bold;">sub</span> ausprobieren <span style="color: purple;">{</span><br /> <span style="color: #508050; font-weight: bold;">my</span> $pos <span style="color: #808030;">=</span> <span style="color: #400000;">shift</span><span style="color: purple;">;</span><br /><br /> <span style="color: #508050; font-weight: bold;">foreach</span> <span style="color: #508050; font-weight: bold;">my</span> $karte<span style="color: #808030;">(</span><span style="color: #778c77;">0</span><span style="color: #808030;">..</span><span style="color: #778c77;">8</span><span style="color: #808030;">)</span> <span style="color: purple;">{</span><br /> <span style="color: #508050; font-weight: bold;">next</span> <span style="color: #508050; font-weight: bold;">if</span> $benutzt<span style="color: #808030;">[</span>$karte<span style="color: #808030;">]</span><span style="color: purple;">;</span><br /> $benutzt<span style="color: #808030;">[</span>$karte<span style="color: #808030;">]</span> <span style="color: #808030;">=</span> <span style="color: #778c77;">1</span><span style="color: purple;">;</span><br /> <span style="color: #508050; font-weight: bold;">foreach</span> <span style="color: #508050; font-weight: bold;">my</span> $dreh<span style="color: #808030;">(</span><span style="color: #778c77;">0</span><span style="color: #808030;">..</span><span style="color: #778c77;">3</span><span style="color: #808030;">)</span> <span style="color: purple;">{</span><br /> <span style="color: #508050; font-weight: bold;">if</span> <span style="color: #808030;">(</span>$pos <span style="color: #808030;">%</span> <span style="color: #778c77;">3</span><span style="color: #808030;">)</span> <span style="color: purple;">{</span><br /> <span style="color: #b96969;"># Nicht linke Spalte</span><br /> $suche <span style="color: #808030;">=</span> <span style="color: #778c77;">9</span> <span style="color: #808030;">-</span> $farbe<span style="color: #808030;">[</span>$gelegt<span style="color: #808030;">[</span>$pos <span style="color: #808030;">-</span> <span style="color: #778c77;">1</span><span style="color: #808030;">]</span><span style="color: #808030;">]</span><span style="color: #808030;">-></span><span style="color: #808030;">[</span><span style="color: #808030;">(</span><span style="color: #778c77;">1</span> <span style="color: #808030;">-</span> $drehung<span style="color: #808030;">[</span>$gelegt<span style="color: #808030;">[</span>$pos <span style="color: #808030;">-</span> <span style="color: #778c77;">1</span><span style="color: #808030;">]</span><span style="color: #808030;">]</span><span style="color: #808030;">)</span> <span style="color: #808030;">%</span> <span style="color: #778c77;">4</span><span style="color: #808030;">]</span><span style="color: purple;">;</span><br /> <span style="color: #508050; font-weight: bold;">next</span> <span style="color: #508050; font-weight: bold;">if</span> $farbe<span style="color: #808030;">[</span>$karte<span style="color: #808030;">]</span><span style="color: #808030;">-></span><span style="color: #808030;">[</span><span style="color: #808030;">(</span><span style="color: #778c77;">3</span> <span style="color: #808030;">-</span> $dreh<span style="color: #808030;">)</span> <span style="color: #808030;">%</span> <span style="color: #778c77;">4</span><span style="color: #808030;">]</span> <span style="color: #808030;">!</span><span style="color: #808030;">=</span> $suche<span style="color: purple;">;</span><br /> <span style="color: purple;">}</span><br /> <span style="color: #508050; font-weight: bold;">if</span> <span style="color: #808030;">(</span>$pos <span style="color: #808030;">></span><span style="color: #808030;">=</span> <span style="color: #778c77;">3</span><span style="color: #808030;">)</span> <span style="color: purple;">{</span><br /> <span style="color: #b96969;"># Nicht oberste Zeile</span><br /> $suche <span style="color: #808030;">=</span> <span style="color: #778c77;">9</span> <span style="color: #808030;">-</span> $farbe<span style="color: #808030;">[</span>$gelegt<span style="color: #808030;">[</span>$pos <span style="color: #808030;">-</span> <span style="color: #778c77;">3</span><span style="color: #808030;">]</span><span style="color: #808030;">]</span><span style="color: #808030;">-></span><span style="color: #808030;">[</span><span style="color: #808030;">(</span><span style="color: #778c77;">2</span> <span style="color: #808030;">-</span> $drehung<span style="color: #808030;">[</span>$gelegt<span style="color: #808030;">[</span>$pos <span style="color: #808030;">-</span> <span style="color: #778c77;">3</span><span style="color: #808030;">]</span><span style="color: #808030;">]</span><span style="color: #808030;">)</span> <span style="color: #808030;">%</span> <span style="color: #778c77;">4</span><span style="color: #808030;">]</span><span style="color: purple;">;</span><br /> <span style="color: #508050; font-weight: bold;">next</span> <span style="color: #508050; font-weight: bold;">if</span> $farbe<span style="color: #808030;">[</span>$karte<span style="color: #808030;">]</span><span style="color: #808030;">-></span><span style="color: #808030;">[</span><span style="color: #808030;">(</span><span style="color: #778c77;">4</span> <span style="color: #808030;">-</span> $dreh<span style="color: #808030;">)</span> <span style="color: #808030;">%</span> <span style="color: #778c77;">4</span><span style="color: #808030;">]</span> <span style="color: #808030;">!</span><span style="color: #808030;">=</span> $suche<span style="color: purple;">;</span><br /> <span style="color: purple;">}</span><br /><br /> $benutzt<span style="color: #808030;">[</span>$karte<span style="color: #808030;">]</span> <span style="color: #808030;">=</span> <span style="color: #778c77;">1</span><span style="color: purple;">;</span><br /> $gelegt<span style="color: #808030;">[</span>$pos<span style="color: #808030;">]</span> <span style="color: #808030;">=</span> $karte<span style="color: purple;">;</span><br /> $drehung<span style="color: #808030;">[</span>$karte<span style="color: #808030;">]</span> <span style="color: #808030;">=</span> $dreh<span style="color: purple;">;</span><br /> <span style="color: #b96969;">#print @gelegt[0..$pos]," ",@drehung[0..$pos]," ", 9 - $farbe[$gelegt[$pos - 1]]->[(1 - $drehung[$gelegt[$pos - 1]]) % 4],"\n";</span><br /> <br /> <span style="color: #508050; font-weight: bold;">if</span> <span style="color: #808030;">(</span>$pos <span style="color: #808030;">=</span><span style="color: #808030;">=</span> <span style="color: #778c77;">8</span><span style="color: #808030;">)</span> <span style="color: purple;">{</span><br /> <span style="color: #508050; font-weight: bold;">print</span> <span style="color: #cc5555;">"Fertig!</span><span style="color: #aa3333; font-weight: bold;">\n</span><span style="color: #cc5555;">"</span><span style="color: purple;">;</span><br /> <span style="color: #508050; font-weight: bold;">for</span> $l<span style="color: #808030;">(</span><span style="color: #778c77;">0</span><span style="color: #808030;">..</span><span style="color: #778c77;">8</span><span style="color: #808030;">)</span> <span style="color: purple;">{</span><br /> <span style="color: #508050; font-weight: bold;">print</span> <span style="color: #cc5555;">"</span><span style="color: #cc5555;">$gelegt</span><span style="color: #808030;">[</span><span style="color: #cc5555;">$l</span><span style="color: #808030;">]</span><span style="color: #cc5555;"> </span><span style="color: #cc5555;">$drehung</span><span style="color: #808030;">[</span><span style="color: #cc5555;">$gelegt</span><span style="color: #808030;">[</span><span style="color: #cc5555;">$l</span><span style="color: #808030;">]</span><span style="color: #808030;">]</span><span style="color: #aa3333; font-weight: bold;">\n</span><span style="color: #cc5555;">"</span><span style="color: purple;">;</span><br /> <span style="color: purple;">}</span><br /> <span style="color: purple;">}</span> <span style="color: #508050; font-weight: bold;">else</span> <span style="color: purple;">{</span><br /> <span style="color: #808030;">&</span>ausprobieren<span style="color: #808030;">(</span>$pos <span style="color: #808030;">+</span> <span style="color: #778c77;">1</span><span style="color: #808030;">)</span><span style="color: purple;">;</span><br /> <span style="color: purple;">}</span><br /> <span style="color: purple;">}</span><br /> $benutzt<span style="color: #808030;">[</span>$karte<span style="color: #808030;">]</span> <span style="color: #808030;">=</span> <span style="color: #778c77;">0</span><span style="color: purple;">;</span><br /> <span style="color: purple;">}</span><br /><span style="color: purple;">}</span></pre><br />Sorry for variable names in German, but the idea should be clear. Regarding the implementation: red, yellow, green and blue backs of arrows get numbers 1,2,3,4 respectively and pointy sides of arrows 8,7,6,5 (so matching combinations sum to 9).<br /><br />It implements depth first tree search where tile positions (numbered 0 to 8) are tried left to write top to bottom. So tile $n$ shares a vertical edge with tile $n-1$ unless it's number is 0 mod 3 (leftist column) and it shares a horizontal edge with tile $n-3$ unless $n$ is less than 3, which means it is in the first row.<br /><br />It tries rotating tiles by 0 to 3 times 90 degrees clock-wise, so finding which arrow to match with a neighboring tile can also be computed with mod 4 arithmetic.Unknownnoreply@blogger.com1tag:blogger.com,1999:blog-8883034.post-39113532392132865222016-06-20T22:12:00.001+02:002016-06-20T22:12:33.408+02:00Restoring deleted /etc from TimeMachineYesterday, I managed to empty the /etc directory on my macbook (don't ask how I did it. I was working on <a href="http://subsurface-divelog.org/">subsurface</a> and had written a perl script to move system files around that had to be run with sudo. And I was still debugging...).<br /><br />Anyway, once I realized what the problem was I did some googling but did not find the answer. So here, as a service to fellow humans googling for help is how to fix this.<br /><br />The problem is that in /etc all kinds of system configuration files are stored and without it the system does not know anymore how to do a lot of things. For example it contains /etc/passwd which contains a list of all users, their home directories and similar things. Or /etc/shadow which contains (hashed) passwords or, and this was most relevant in my case, /etc/sudoers which contains a list of users who are allowed to run commands with <a href="https://xkcd.com/149/">sudo</a>, i.e. execute commands with administrator privileges (in the GUI this shows as as a modal dialog asking you to type in your password to proceed).<br /><br />In my case, all was gone. But, luckily enough, I had a time machine backup. So I could go 30 minutes back in time and restore the directory contents.<br /><br />The problem was that after restoring it, it ended up as a symlink to /private/etc and user helling wasn't allowed to access its contents. And I could not sudo the access since the system could not determine I am allowed to sudo since it could not read /etc/sudoers.<br /><br />I tried a couple of things including a reboot (as a last resort I figured I could always boot in target disk mode and somehow fix the directory) but it remained in /private/etc and I could not access it.<br /><br />Finally I found the solution (so here it is): I could look at the folder in Finder (it had a red no entry sign on it meaning that I could not open it). But I could right click and select Information and there I could open the lock by tying in my password (no idea why that worked) and give myself read (and for that matter write) permissions and then everything was fine again.Unknownnoreply@blogger.com0tag:blogger.com,1999:blog-8883034.post-24954675567200031522016-05-24T12:03:00.001+02:002016-05-25T09:25:38.082+02:00Holographic operator ordering?Believe it or not, at the end of this week I will speak at <a href="https://www-m5.ma.tum.de/Allgemeines/LQPWorkshop">a workshop on algebraic and constructive quantum field theory</a>. And (I don't know which of these two facts is more surprising) I will advocate holography.<br /><br />More specifically, I will argue that it seems that holography can be a successful approach to formulate effective low energy theories (similar to other methods like perturbation theory of weakly coupled quasi particles or minimal models). And I will present this as a challenge to the community at the workshop to show that the correlators computed with holographic methods indeed encode a QFT (according to your favorite set of rules, e.g. Whiteman or Osterwalder-Schrader). My [kudos to an anonymous reader for pointing out a typo] guess would be that this has a non-zero chance of being a possible approach to the construction of (new) models in that sense or alternatively to show that the axioms are violated (which would be even more interesting for holography).<br /><br />In any case, I am currently preparing my slides (I will not be able to post those as I have stolen far too many pictures from the interwebs including the holographic doctor from Star Trek Voyager) and came up with the following question:<br /><br /><blockquote class="tr_bq">In a QFT, the order of insertions in a correlator matters (unless we fix an ordering like time ordering). How is that represented on the bulk side?</blockquote><br />Does anybody have any insight about this?<br /><br /><div class="separator" style="clear: both; text-align: center;"><a href="http://www.giantfreakinrobot.com/wp-content/uploads/2014/06/helpmeobiwankenobi.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" src="http://www.giantfreakinrobot.com/wp-content/uploads/2014/06/helpmeobiwankenobi.jpg" height="136" width="320" /></a></div>Unknownnoreply@blogger.com3tag:blogger.com,1999:blog-8883034.post-82510413377926125202016-04-21T14:55:00.000+02:002016-04-21T14:55:23.079+02:00The Quantum in Quantum ComputingI am sure, by now, all of you have seen Canada's prime minister <a href="https://www.youtube.com/watch?v=rRmv4uD2RQ4">"explain" quantum computers</a> at Perimeter. It's really great that politicians care about these things and he managed to say what is the standard explanation for the speed up of quantum computers compared to their classical cousins: It is because you can have superpositions of initial states and therefore "perform many operations in parallel".<br /><br />Except of course, that this is bullshit. This is not the reason for the speed up, you can do the same with a classical computer, at least with a probabilistic one: You can also as step one perform a random process (throw a coin, turn a Roulette wheel, whatever) to determine the initial state you start your computer with. Then looking at it from the outside, the state of the classical computer is mixed and the further time evolution also "does all the computations in parallel". That just follows from the formalism of (classical) statistical mechanics.<br /><br />Of course, that does not help much since the outcome is likely also probabilistic. But it has the same parallelism. And as the state space of a qubit is all of a Bloch sphere, the state space of a classical bit (allowing mixed states) is also an interval allowing a continuum of intermediate states.<br /><br />The difference between quantum and classical is elsewhere. And it has to do with non-commuting operators (as those are essential for quantum properties) and those allow for entanglement.<br /><br />To be more specific, let us consider one of the most famous quantum algorithms, <a href="https://en.wikipedia.org/wiki/Grover%27s_algorithm">Grover's database lookup</a>, There the problem (at least in its original form) is to figure out which of $N$ possible "boxes" contains the hidden coin. Classically, you cannot do better than opening one after the other (or possibly in a random pattern), which takes $O(N)$ steps (on average).<br /><br />For the quantum version, you first have to say how to encode the problem. The lore is, that you start with an $N$-dimensional Hilbert space with a basis $|1\rangle\cdots|N\rangle$. The secret is that one of these basis vectors is picked. Let's call it $|\omega\rangle$ and it is given to you in terms of a projection operator $P=|\omega\rangle\langle\omega|$.<br /><br />Furthermore, you have at your disposal a way to create the flat superposition $|s\rangle = \frac1{\sqrt N}\sum_{i=1}^N |i\rangle$ and a number operator $K$ that act like $K|k\rangle= k|k\rangle$, i.e. is diagonal in the above basis and is able to distinguish the basis elements in terms of its eigenvalues.<br /><br />Then, what you are supposed to do is the following: You form two unitary operators $U_\omega = 1 - 2P$ (this multiplies $|\omega\rangle$ by -1 while being the identity on the orthogonal subspace, i.e. is a reflection on the plane orthogonal to $|\omega\rangle$) and $U_s = 2|s\rangle\langle s| - 1$ which reflects the vectors orthogonal to $|s\rangle$.<br /><br />It is not hard to see that both $U_s$ and $U_\omega$ map the two dimensional place spanned by $|s\rangle$ and $|\omega\rangle$ into itself. They are both reflections and thus their product is a rotation by twice the angle between the two planes which is given in terms of the scalar product $\langle s|\omega\rangle =1/\sqrt{N}$ as $\phi =\sin^{-1}\langle s|\omega\rangle$.<br /><br />But obviously, using a rotation by $\cos^{-1}\langle s|\omega\rangle$, one can rotate $|s\rangle$ onto $\omega$. So all we have to do is to apply the product $(U_sU\omega)^k$ where $k$ is the ratio between these two angles which is $O(\sqrt{N})$. (No need to worry that this is not an integer, the error is $O(1/N)$ and has no influence). Then you have turned your initial state $|s\rangle$ into $|omega\rangle$ and by measuring the observable $K$ above you know which box contained the coin.<br /><br />Since this took only $O(\sqrt{N})$ steps this is a quadratic speed up compared to the classical case.<br /><br />So how did we get this? As I said, it's not the superposition. Classically we could prepare the probabilistic state that opens each box with probability $1/N$. But we have to expect that we have to do that $O(N)$ times, so this is essential as fast as systematically opening one box after the other.<br /><br />To have a better unified classical-quantum language, let us say that we have a state space spanned by $N$ pure states $1,\ldots,N$. What we can do in the quantum case is to turn an initial state which had probability $1/N$ to be in each of these pure states into one that is deterministically in the sought after state.<br /><br />Classically, this is impossible since no time evolution can turn a mixed state into a pure state. One way to see this is that the entropy of the probabilistic state is $\log(N)$ while it is 0 for the sought after state. If you like classically, we only have the observables given by C*-algebra generated by $K$, i.e. we can only observe which box we are dealing with. Both $P$ and $U_\omega$ are also in this classical algebra (they are diagonal in the special basis) and the strict classical analogue would be that we are given a rank one projector in that algebra and we have to figure out which one.<br /><br />But quantum mechanically, we have more, we also have $U_s$ which does not commute with $K$ and is thus not in the classical algebra. The trick really is that in this bigger quantum algebra generated by both $K$ and $U_s$, we can form a pure state that becomes the probabilistic state when restricted to the classical algebra. And as a pure state, we can come up with a time evolution that turns it into the pure state $|\omega\rangle$.<br /><br />So, this is really where the non-commutativity and thus the quantumness comes in. And we shouldn't really expect Trudeau to be able to explain this in a two sentence statement.<br /><br />PS: The actual speed up in the end comes of course from the fact that probabilities are amplitudes squared and the normalization in $|s\rangle$ is $1/\sqrt{N}$ which makes the angle to be rotated by proportional to $1/\sqrt{N}$. Unknownnoreply@blogger.com0