Jekyll2017-09-04T18:30:57+00:00http://rantonels.github.io/rantonels.github.iomusings of an apprentice holomancerA simple and rigorous proof of 26/10 dimensions in string theory2017-05-05T00:00:00+00:002017-05-05T00:00:00+00:00http://rantonels.github.io/critical-dimension<p>…does not exist.</p>
<p>At least, I couldn’t manage. The requirement of a specific and large number of spacetime dimensions (26 for the simpler bosonic string theory, 10 for the more sophisticated superstrings) is one of the most misunderstood aspects of the theory and definitely the essential ingredient for the majority of negative gut feelings towards it. It seems like there should be more effort from string peeps to explain in the simplest way possible the origin of these bizarre numbers to non-experts.</p>
<p>To be more specific, the real magical number is not the dimension \(D\) itself of spacetime, but \(D-2\), the number of dimensions <strong>transverse</strong> to the string, the ones it can oscillate <em>into</em> (minus one for the time dimension, and minus one for the dimension longitudinal to the string). In other words, the 1D string traces a 2D surface in spacetime, called a <em>worldsheet</em>. The magic number is the number of <em>remaining</em> directions available to the string, so \(D-2\).</p>
<div>
<figure>
<img class="picture" src="/images/woleshet.png" />
<figcaption>This picture has been badly drawn at least one thousand times, but I felt like badly drawing it yet one more time. </figcaption>
</figure>
</div>
<p>Thus the magic number is 24 (or 8 for the superstring). 24 is truly a mystical number. John Baez provides a <a href="http://math.ucr.edu/home/baez/numbers/24.pdf">fantastic account</a> of why 24 is his favourite number, hinting at scattered mathematical gems in which it makes unexpected appearances. Some of them look like absolutely nothing else than pure numerology:</p>
<script type="math/tex; mode=display">1^2 + 2^2 + 3^2 + \ldots + 23^2 + 24^2 = 70^2</script>
<p>and <a href="https://archive.lib.msu.edu/crcmath/math/math/c/c023.htm">this only works</a> for \(24\), except 0 and 1 of course. (If you like <strong>tough</strong> math puzzles, try proving this. I don’t). It is nothing else than incredible that this funny identity is related to unexpectedly complex and fascinating mathematics (monstrous moonshine), <em>and</em> string theory (which acts as the “glue” that brings together monstrous moonshine). Also relevant is the appearance of 24/2 in the crazy series</p>
<script type="math/tex; mode=display">1 + 2 + 3 + 4 + \ldots "=" -\frac{1}{12}</script>
<p>In fact, that is why in bosonic string theory \(D-2 = 24\). In superstrings, the equivalent madness is</p>
<script type="math/tex; mode=display">1 - 2 + 3 - 4 + \ldots "=" \frac{1}{4}</script>
<p>which gives \(D-2 = 8\). You should at this point be convinced that I’ve lost it completely. After all, these sound like random unrelated facts (and “facts”) and even though each one of them could be easily explained to a layman, I’m not really explaining what the relationship should be with the number of dimensions in string theory. I sound mad not because I say things that are wrong, but because these things are incoherent and disconnected. The problem is that the connective tissue is too technical, either mathematically, or physically, or stringly, for me to explain simply.</p>
<p>How to <em>convince</em> someone that \(D-2 = 24\) with the least effort possible? It’s much easier if they accept the crazy equation \(1+2+3+\ldots = -1/12\); but that’s clearly not satisfyingly rigorous. Even if it is possible to make sense of it, for example, with ζ-regularization, aka with heat-kernel regularization + analytic renormalization (and there is already a lot of material on this around) this would <em>still</em> be unsatisfactory, since there is no non-shady reason for which all of this manipulation should have anything to do with the physics. Is it possible to arrive to the correct results <em>without</em> this crazy equation; actually, without ever encountering any “irregularity” to regularize? Yes, sure. But exactly how <strong>brief</strong> and elementary can we be doing that?</p>
<p>I’ve found that proofs of \(D=26\) can be classified roughly as such:</p>
<ol>
<li><strong>Light-cone quantization</strong>
<ol>
<li>with crazy eqt. Requires a bit of knowledge about the polarizations of massive/massless vector bosons.</li>
<li>without crazy eqt. Requires essentially studying all of the string’s quantization / Virasoro algebra.</li>
</ol>
</li>
<li><strong>Conformal field theory</strong>
<ol>
<li>without crazy eqt. Requires knowing conformal field theories and conformal anomalies.</li>
</ol>
</li>
<li><strong>Modular invariance</strong>
<ol>
<li>with crazy eqt. Requires fairly elementary QM and math.</li>
<li>without crazy eqt. Requires elementary QM but an f-ton of basic, but tedious complex analysis.</li>
</ol>
</li>
</ol>
<p>Proof 1.1, I’ve done it <a href="https://www.reddit.com/r/askscience/comments/3y6jxc/how_does_that_divergent_sum_which_equals_112/cyb1n94/">here</a> a while ago. It’s pretty easy, but you need to trust the crazy equation. The other proof with the crazy equation, 3.1, is in <a href="http://math.ucr.edu/home/baez/numbers/24.pdf">Baez’s slides</a>.</p>
<p>The crazy-free proofs curiously form an unholy trinity. Each only requires knowledge from one vertex of a triangle spanning between</p>
<ol>
<li>String theory</li>
<li>Theoretical physics</li>
<li>Mathematics</li>
</ol>
<p>for a certain (surely biased) interpretation of these terms. Proof 1.2 is the most common in introductory string theory, since you’re already learning the necessary string physics anyway, and it’s mathematically not intimidating. Proof 2.1 is for a more advanced study and gives a very clear physical interpretation of the critical number of dimensions as it is directly related to the cancellation of a conformal anomaly. Proof 3.2 I’ve never seen before - it seems ideal for people that know some basic complex analysis (residue theorem is all we need) and some really basic quantum mechanics, and requires next-to-no knowledge about string physics.</p>
<p>I’ve tried really hard to squeeze an <strong>easy</strong> proof that is “rigorous” (doesn’t use the crazy equation), but I have failed. I have come to believe the three vertices of the triangle are to be understood as translation in different languages of the same conceptual core, and that core is irreducible. So if two of the links in the chain are brought to a minimum complexity, the third has to eat it all up.</p>
<p>All I could manage is the following <strong>realization of proof 3.2</strong>. This will bring the theoretical and string physics to the background and will concentrate on the mathematics, and will be as rigorous as possible on the mathematical side (while occasionally being slightly handwavy on the physics). It turns out quite longer than what you would call elementary, but I find it satisfying that it builds the number 24 piece by piece (as 2, times 3, times 4). Consider it a <strong>letter of apology</strong> for releasing \(1+2+3+\ldots=-1/12\) into the wild; if you care, here it is.</p>
<h1 id="the-proof">The proof</h1>
<p>We have already talked about the worldsheet, the surface the string traces in spacetime. On this surface, we can set up a coordinate system \((\sigma^1,\sigma^2)\). Clearly, it is sensible physically that while observables might be written in terms of these coordinates, they should be invariant under coordinate changes \((\sigma^1,\sigma^2) \rightarrow (\sigma^{1}\prime,\sigma^2 \prime)\). After all, the coordinates are arbitrary to begin with. A simple example of (local) coordinate change is a dilation/rescaling: \( \renewcommand{\Re}{\operatorname{Re}}
\renewcommand{\Im}{\operatorname{Im}}\)</p>
<script type="math/tex; mode=display">(\sigma^1, \sigma^2) \rightarrow (\lambda \sigma^1, \lambda\sigma^2)</script>
<p>so we understand that our theory’s observables must be <strong>invariant under such rescalings</strong>. This is easy to do classically, but becomes highly non-trivial when the thing becomes quantum-mechanical. This scale symmetry is part of the so-called worldsheet <em>conformal</em> symmetry of string theory.</p>
<p>Imagine a situation in which a string loops in time to trace a torus in spacetime. Surely, a lot of possible tori shapes are all equally valid choices. We want to compute the probability for this process to happen with a specific torus shape, or better the <em>quantum amplitude</em>, whose squared modulus is the probability. A basic point of quantum mechanics is that we can calculate the amplitude by summing \(\exp(i E_i t) \) over all possible states \(i\) each with energy \(E_i\), through the time passed \(t\). Thus</p>
<script type="math/tex; mode=display">Z = \sum_i e^{i E_i t}</script>
<p>is the total amplitude. \(Z\) should respect our desired symmetries.</p>
<p>An important point is that a composite system AB of two non-interacting subsystems A and B has \(Z_{AB} = Z_A Z_B\). Because of this nice property, we can first concentrate on computing the \(Z\) of a string oscillating in <strong>one</strong> single transverse dimension, and then take that to the power of \(D-2\) to make it wiggle in \(D-2\) transverse dimensions.</p>
<div>
<figure>
<img class="picture" src="/images/stringharmonics.jpg" />
<figcaption>Just like actual vibrating strings, string theory strings have harmonics which are integer multiples of the fundamental.</figcaption>
</figure>
</div>
<p>So a string is almost exactly like an actual guitar strings and has infinite oscillation modes, or harmonics, or overtones. <strong>Each one</strong> of them is a harmonic oscillator, which in the quantum version has energy levels</p>
<script type="math/tex; mode=display">E_n = \omega \left( \frac{1}{2} + n \right)</script>
<p>Right? So the \(Z\) of one quantum harmonic oscillator is</p>
<script type="math/tex; mode=display">Z_{QHO} = \sum_{n=0}^\infty e^{i E_n t} = e^{\frac{i}{2}\omega t} \sum_{n=0}^\infty e^{i n \omega t} = \frac{e^{\frac{i}{2} \omega t}}{1 - e^{i\omega t}}</script>
<p>The geometric series I just summed appears to have ratio \(|r| = 1\), which means it actually does not converge. Let’s rip off the bandaid right now: the \(Z\)s, defined naively as we did, <em>almost never</em> converge in a Lorentzian (i.e. spacetime) theory. Often we physicists say they are “oscillatory” because we sum a bunch of \(exp(ix)\) factors, and then handwave that they somehow magically cancel, but this is a <strong>cheap lie</strong> - we just mean they don’t converge. This is true not just in string theory but in all of quantum field theory, or standard quantum mechanics, or even the lowly quantum harmonic oscillator.</p>
<p>How can we move forward? The correct way to do it (and to word it) with is to make time a complex variable. By giving it an imaginary part, these \(Z\)s are made to converge. Not only: when you compute directly observable quantities from this, and then take \(t\) back to the real axis, you recover sensible answers (which match experiment when possible). So, this is not a trick; this prescription is our <strong>definition</strong> of what it means to have a quantum theory in spacetime.</p>
<p>tl;dr: don’t worry, assume \(t\) is complex and in the upper half-plane \(\Im t > 0 \).</p>
<p>To get back on track, we had that one string in one dimension had infinite oscillation modes, which clearly have frequencies which are integer multiples of the fundamental. In fact, \(\omega = 1,2,3,\ldots\) and so the \(Z\) for our 1D string is the product</p>
<script type="math/tex; mode=display">Z_{1,L} = \prod_{k=1}^\infty \frac{e^{\frac{i}{2}k t}}{1- e^{ikt}} = e^{\frac{i}{2} (1+2+3+\ldots)t } \left(\prod_k 1- e^{ikt}\right)^{-1}</script>
<p>…and we fell off the deep end. The madness of \(1+2+3+\ldots\) catched up with us. If we were weak in spirit, we would succumb and substitute \(1+2+3+\ldots \rightarrow - 1/12\), and we would get the correct answer, skipping almost all of the math in this article. But we are not here for this, we are here to make that \(24\) come out <strong>without</strong> using witchcraft. Therefore, let’s proceed.</p>
<p>Why did that divergent sum appear in that exponent in the first place? If you trace our calculations back it comes from the zero-point energies of the QHOs, the \(E_0 = \frac{\omega}{2}\). However, zero-point energies are <strong>arbitrary</strong>, and that one was just a useful conventional choice. Ultimately there is an ambiguity in building a quantum HO from a classical HO called an <em>ordering</em> ambiguity, since you need to convert commuting variables \(p,q\) into non-commuting operators \(\hat p, \hat q\) and there is no preferential “quantization” of things like \(pq\). Is the quantization \(\hat p \hat q\)? Or is it \(\hat q \hat p\)? The difference is a constant, an arbitrariness in the zero-point energy. Thus our most conservative bet is that the ZPOs here actually sum to a finite, but unknown value:</p>
<script type="math/tex; mode=display">Z_{1,L}^{-1} = e^{-irt} \; \prod_k 1 - e^{ikt}</script>
<p>where \(r\) is an unknown real number. Note that the mad calculation would give the magic value \(r = 1/24\). So, for my next two tricks:</p>
<ul>
<li>I am going to show, using a symmetry, that \(1/r\) is the number of transverse dimensions.</li>
<li>I am going to show, using another symmetry, that \(r = 1/24\) anyway, even without \(1+2+3+\ldots = - 1/12\)</li>
</ul>
<p>To start talking about these symmetries, I need to relate my variable \(t\) to the shape of the torus. The relationship is that the torus is built by gluing same-colour edges in this parallelogram:</p>
<div>
<figure>
<img class="picture" src="/images/parallelocheap.png" />
<figcaption></figcaption>
</figure>
</div>
<p>where \(t\) is represented as a point in the complex plane. (The \(2\pi\) side is the consequence of a convention that made it so that the frequencies \(\omega = 1,2,3,\ldots\) were integers). Or to clean up the notation, with \(\tau = t/2\pi\) (and a simple rescaling, which I remind is a symmetry):</p>
<div>
<figure>
<img class="picture" src="/images/parallelo.png" />
<figcaption></figcaption>
</figure>
</div>
<p>Note that when \(\tau\) moves to the real axis the torus degenerates. That makes physical sense: can a string really loop back unto itself in real time? That’d be time travel. It can only happen in imaginary time or at least in the upper half-plane.</p>
<p>Defining \(q := e^{i 2\pi \tau}\), our \(Z_{1,L}\) becomes</p>
<script type="math/tex; mode=display">Z_{1,L}^{-1}(\tau) = q^r \; \prod_k (1-q^k)</script>
<h2 id="first-trick">First trick</h2>
<p>First of all, if we shift \(\tau \rightarrow \tau + 1\), we get different parallelograms:</p>
<div>
<figure>
<img class="picture" src="/images/modulart.png" />
<figcaption></figcaption>
</figure>
</div>
<p>but they make the same torus after gluing (check it!). If the shape does not actually change, then the physical quantities shouldn’t either.</p>
<p>However, under this transformation our amplitude does transform:</p>
<script type="math/tex; mode=display">Z_{1,L}^{-1} \rightarrow e^{2\pi i r} \, Z_{1,L}^{-1}</script>
<p>That makes no sense… until we remember that this is only for <strong>one</strong> dimension. For \(D-2\) transverse dimensions, the total amplitude is \(Z_{1,L}^{D-2} \), which will remain invariant if</p>
<script type="math/tex; mode=display">r(D-2) = 1</script>
<p>this equation defines the <em>critical dimension</em> of the string theory. If we will find an integer value of \(1/r\), then the theory will make sense only in \(D = 1/r + 2\) spacetime dimensions.</p>
<p>Now, proving that \(r = 1/24\) will not be as easy. Let’s try.</p>
<h2 id="second-trick">Second trick</h2>
<p>The other symmetry we will consider is to send \(\tau \rightarrow - 1/\tau\). The resulting tori are not identical but they are similar - they are rescaled versions of eachother. You can check that the original \(1\) side matches with the new \(-1/\tau\) side and the original \(\tau\) side matches with the new \(1\) side. As we said, rescalings of the worldsheet should be a symmetry.</p>
<p>Thus the total amplitude must be invariant under this; however we should not expect \(Z_{1,L}^{D-2}\) to necessarily also be. Huh? Wasn’t \(Z_{1,L}^{D-2}\) the total amplitude? No, I lied to protect you from the harsh truth. The truth (part of it) is that <strong>oscillation on a string always come in pairs</strong>. There are many possible characterizations of this dichotomy: left-movers and right-movers, holomorphic and antiholomorphic, sine and cosine…</p>
<p>What matters is that to our \(Z_{1,L}\) (the \(L\) is now understood to stand for “left”) there will also be a paired \(Z_{1,R}\) for the sister oscillations. Thankfully, I’ll just handwave that \(Z_{1,R} = Z_{1,L}^* \) so that the overall amplitude is the product \(|Z_{1,L}^{D-2}|^2\).</p>
<p>However, that is still not all: while we have accounted for oscillations of the string about a “reference” position, we also need to account for the <em>overall</em> movement of the centre of mass in space. We want the string to be back to its original point after time \(\tau\) if we want it to loop into a torus; but we cannot just <em>want</em> it, we need to include the probability amplitude for it to happen.</p>
<p>What is the probability amplitude for a quantum particle in 1D to stay in the same place after a certain amount of time? A hint is that for <strong>purely imaginary times</strong>, the Schroedinger equation is the <a href="https://en.wikipedia.org/wiki/Heat_equation">heat equation</a>. And an infinitely concentrated speck of heat evolves under the heat equation in 1D by <a href="https://en.wikipedia.org/wiki/Heat_kernel">spreading into a gaussian whose peak decreases as</a> \((\operatorname{time})^{-1/2} \). Thus, let me guess the amplitude for the centre of mass is something like this</p>
<script type="math/tex; mode=display">(\Im \tau)^{-1/2}</script>
<p>If so, putting everything back together the total amplitude should be something like</p>
<script type="math/tex; mode=display">Z \sim (\Im \tau)^{-\frac{D-2}{2}} \, | Z_{1,L} |^{2(D-2)}</script>
<p>Ok, now under \(\tau \rightarrow -1/\tau\),</p>
<script type="math/tex; mode=display">\Im \tau \rightarrow \frac{1}{ | \tau |^2} \Im \tau</script>
<p>therefore if \(Z_{1,L}\) were to transform in some way similar to this:</p>
<script type="math/tex; mode=display">Z_{1,L} \rightarrow \tau^{-1/2} Z_{1,L}</script>
<p>our total amplitude would be invariant, and our theory would make sense. I’m now set to prove that this can only happen if \(r=1/24\).</p>
<h3 id="sorcery">Sorcery</h3>
<p>Let’s do some preliminary rewriting.</p>
<script type="math/tex; mode=display">Z_{1,L}^{-1} (\tau) = q^r P(\tau)\,,\quad P(\tau) := \prod_{\ell=1}^\infty (1-q^\ell)</script>
<p>Then, a more comfortable presentation of the product \(P(\tau)\):</p>
<script type="math/tex; mode=display">- \log P(\tau) = - \sum_{\ell=1}^\infty \log(1-q^\ell) = \sum_{l,k = 1}^\infty \frac{1}{k} q^{k\ell} = \sum_{k = 1} \frac{1}{k} \frac{q^k}{1-q^k} = \sum_{k=1} \frac{1}{k} \frac{1}{q^{-k} -1}</script>
<p>I’ve used the Taylor expansion for \(-\log(1-x) \) and the geometric series. If you are not convinced, you can check more carefully these steps (including the sum swap) are sensible for \(\Im \tau > 0\).</p>
<p>Now:</p>
<p>The following psychedelic argument is due to Siegel (yeah, that Siegel). We start for no apparent reason from this function of a complex variable \(w\), with \(\tau\) as a parameter:</p>
<script type="math/tex; mode=display">f(w) = \cot w \cot \frac{w}{\tau}</script>
<p>and then, introducing another real parameter \(\nu\) (our finishing move will be to send this to infinity), we construct the combination</p>
<script type="math/tex; mode=display">g(w) = \frac{f(\nu w)}{w}</script>
<p>Let’s count the poles of \(g\). A quick inspection shows there is a set of simple poles at \(w = \pm \frac{\pi k}{\nu}\) and another at \(w = \pm \frac{\pi k \tau}{\nu} \), for \(k = 1, 2, 3, \ldots\); plus a triple pole at \(w = 0\). The residues are easily computed as follow:</p>
<script type="math/tex; mode=display">\operatorname{Res}_{\pm\frac{\pi k}{\nu}} g = \frac{1}{\pi k}\cot \left( \frac{\pi k}{\tau} \right)</script>
<script type="math/tex; mode=display">\operatorname{Res}_{\pm\frac{\pi k \tau}{\nu}} g = \frac{1}{\pi k}\cot \left( \pi k \tau \right)</script>
<script type="math/tex; mode=display">\operatorname{Res}_{0} g = - \frac{1}{3} (\tau + \tau^{-1})</script>
<p>If the residue at the triple point doesn’t seem obvious, recall that the Taylor expansion of the cotangent at zero starts \(\cot s \sim \frac{1}{s} - \frac{s}{3} \).</p>
<div>
<figure>
<img class="picture" src="/images/trighocplotcrop.png" />
<figcaption>Colour wheel plot of \(g(w)\), for \(\tau = i\), \(\nu = 1\). The white specks are the poles, and the order is how many times the colours repeat circling them.</figcaption>
</figure>
</div>
<p>Now, it should be clear the intention is to exploit \(g\) for an application of the residue theorem. We consider the path \(\gamma\) running ccw around the parallelogram with vertices \(1,\tau,-1,-\tau\).</p>
<p>The residue theorem says</p>
<script type="math/tex; mode=display">\frac{1}{2\pi i} \int_\gamma \frac{f(\nu w)}{w} d\omega = \sum_{p \in \operatorname{poles}} \operatorname{Res}_p g</script>
<p>The sum is actually only over those poles that are inside the parallelogram, but we’ll see in a little while that’s not something we should worry about.</p>
<p>Now, I would like to use this equation to prove the identity we actually need. I will only prove it, however, for \(\tau\) on the imaginary axis, because it’s simpler, but it will actually be true for all \(\Im \tau > 0\). Since both sides of the equation I’ll derive are holomorphic in \(\tau\) over the upper half-plane, them agreeing on a line is enough to prove they are always equal. Long story short: assume \(\tau\) is purely imaginary for now, but at the end we can just drop this assumption.</p>
<p>Let’s do the integral on the parallelogram, which is now a rhombus. \(g(w) = f(\nu w) w^{-1} \) doesn’t seem to be an easy function to guess an antiderivative for. So let’s take the \(\nu \rightarrow \infty\) limit. It’s not hard to see that \( f(\nu w) \) converges to the constant \((1,-1,1,-1)\) respectively on the four segments of \(\gamma\) (if you don’t see it, write \(\cot s\) using complex exponentials). Thus in the limit the integral is</p>
<script type="math/tex; mode=display">\left( \int_1^\tau - \int_\tau^{-1} + \int_{-1}^{-\tau} - \int_{-\tau}^{1} \right) \frac{dw}{w}</script>
<p>Easy! The integral of \(1/w\) is \(\log w\), so let me just plug that in and… aaagh! <strong>Branches</strong>! We need to choose a branch for the log, and make sure the path does not jump over the branch cut. We can use symmetry to rewrite</p>
<script type="math/tex; mode=display">2\left( \int_1^\tau + \int_1^{-\tau} \right) \frac{dw}{w}</script>
<p>and now that the path is all on the right, we can use the negative real axis as a branch cut and thus use the principal branch of the logarithm. Keeping track of everything at the end you get</p>
<script type="math/tex; mode=display">4 \log \left(\frac{\tau}{i} \right)</script>
<p>Nice. Now for the residues. If \(\nu \rightarrow \infty\), all the poles shrink and move closer to the origin; so in the limit they are all inside the rhombus and the sum is over all poles. Thus the sum is</p>
<script type="math/tex; mode=display">- \frac{1}{3} (\tau + \tau^{-1}) + 2 \sum_{k=1}^\infty \left( \frac{1}{\pi k} \cot(\frac{\pi k}{\tau}) + \frac{1}{\pi k} \cot(\pi k \tau) \right)</script>
<p>Note</p>
<script type="math/tex; mode=display">\cot s = \frac{ e^{is} + e^{-is} }{e^{is} - e^{-is} } = \frac{1 + e^{-2is}}{1 - e^{-2is}} = -1 + \frac{2}{1-e^{-2is}}</script>
<p>plug that in, and you’ll recognize that the sum of the residues becomes</p>
<script type="math/tex; mode=display">-\frac{1}{3} (\tau + \tau^{-1}) + \frac{4}\pi \sum_{k=1}^\infty \frac{1}{k} \left( \frac{1}{1-e^{-\frac{2\pi i k}{\tau}} } - \frac{1}{ 1- e^{-2 \pi i k \tau} } \right)</script>
<script type="math/tex; mode=display">=-\frac{1}{3} (\tau + \tau^{-1}) + \frac{2}{\pi} \left(\log P(-1/\tau) - \log P(\tau) \right)</script>
<p>Finally we’re starting to get back to Earth. This is starting to look like a theorem about our \(P(\tau)\) product. Now that we have LHS and RHS, let’s use the residue theorem.</p>
<script type="math/tex; mode=display">\frac{1}{2} \log\left(\frac{\tau}{i} \right) = - \frac{\pi i}{12} (\tau + \tau^{-1}) - \left(\log P(\tau) - \log(P(-1/\tau) \right)</script>
<p>There’s our magic number! Half of it at least. This still looks like gibberish though… let’s exponentiate for good measure:</p>
<script type="math/tex; mode=display">e^{\frac{2\pi i / \tau}{24}} P(-1/\tau) = \sqrt{\tau/i} \; e^{\frac{2 \pi i \tau}{24}} P(\tau)</script>
<p>This seems not much of a fact about \(P(\tau)\), but about the combination</p>
<script type="math/tex; mode=display">\eta(\tau) := e^{\frac{2\pi i \tau}{24}} P(\tau) = q^{\frac{1}{24}} \prod_{k=1}^\infty (1-q^k)</script>
<p>which we just proved transforms “nicely” under a \(\tau \rightarrow - 1/\tau\) transformation:</p>
<script type="math/tex; mode=display">\eta(-1/\tau) = \sqrt{\frac{\tau}{i}} \eta(\tau)</script>
<p>And that’s it! That’s what we were looking for. For \(Z_{1,L}^{-1}(\tau)\) to transform how we want, it has to be \(\eta(\tau)\) itself, and so \(r=1/24\), and so, finally</p>
<script type="math/tex; mode=display">D = 26</script>
<h1 id="conclusions">Conclusions</h1>
<p>Because of the philosophy behind this whole post, to only use mathematics as elementary as possible and to have it be self-contained, we sped by what is actually incredibly fascinating (and surely much more elegant) math. \(\eta(\tau)\) is, of course, the <a href="https://en.wikipedia.org/wiki/Dedekind_eta_function">Dedekind eta function</a>, and what we proved are its transformation properties under the modular group; more precisely that the modular discriminant \(\Delta(\tau) := (2\pi)^{12} \eta^{24}\) is a modular form of weight 12. String theory is tightly connected to this area of mathematics; in fact I hope that for those that already know this stuff this served as a sneak peek of what strings even have to do with modular forms. Anyways, I don’t think I would do justice to a subject I actually barely know the basics of, so I’ll shut up now.</p>
<p>An interesting question that pops up sometimes: if it’s so wrong, why does \(1+2+3+\ldots = -1/12\), or ζ-regularization / heat kernel / Abel summation or whatever you want to call it, give the same correct result as more rigorous paths, with one tenth of the effort?</p>
<p>I have <strong>no</strong> idea.</p>
<p>Instinctively I would babble something about “physically reasonable” or “analyticity” but frankly this is just pure madness. It’s definitely not a coincidence because the same trick works for superstrings too. In fact I’m not even sure what the exact relationship between the three different classes of proofs are, they look like completely different reasonings. In light-cone quantization, on a first reading you don’t even realize how the scale/conformal invariance even fits in - it’s very hidden. And if you happen to know some conformal field theory, go take a look at how that proof builds the 24. They don’t sound <em>exactly</em> like translations of the same thing, even though they <em>have</em> to be.</p>
<p>It’s fascinating how much information about the content of the theory the proofs of \(D = 26\) (or \(D=10\)) carry, and how much you already learn about strings just by trying to explain what is essentially the <strong>most basic</strong> fact about the theory. It is a limpid example of the fact that in string theory <em>everything fits</em>, nothing is just thrown there, everything is essential.</p>
<h1 id="addendum">Addendum</h1>
<p>There is perhaps a faster proof if you assume Euler’s pentagonal number theorem</p>
<script type="math/tex; mode=display">P(\tau) = \prod_{k=1}^\infty (1-q^n) = \sum_{k=-\infty}^{\infty} (-1)^k q^{k(3k-1)/2}</script>
<p>If you complete the square in the exponent you can rewrite</p>
<script type="math/tex; mode=display">P(\tau) = q^{-1/24} \sum_{k=-\infty}^{\infty} (-1)^k q^{\frac{3}{2}(k-\frac{1}{6})^2}</script>
<p>thus</p>
<script type="math/tex; mode=display">\eta(\tau) = q^{1/24} P(\tau) = \sum_{k=-\infty}^{\infty} (-1)^k q^{\frac{3}{2}(k-\frac{1}{6})^2}</script>
<p>With \(\eta\) in this form, it is <em>doable</em> to prove \(\eta(-1/\tau) = \sqrt{\tau/i} \eta(\tau) \) using Poisson resummation.</p>
<p>Still, invoking Euler’s theorem seems definitely like cheating, so I didn’t pursue this path.</p>
<h1 id="refs">Refs</h1>
<p>Baez, J. (2008). <a href="http://math.ucr.edu/home/baez/numbers/24.pdf">My favourite number: 24</a>. Talk at the University of Glasgow.</p>
<p>Siegel, C. (1954). A simple proof of η(-1/τ)=η(τ)√τ/i. Mathematika, 1(1), 4-4. doi:10.1112/S0025579300000462</p>…does not exist.Gravity, entropy, and life.2017-03-16T15:39:30+00:002017-03-16T15:39:30+00:00http://rantonels.github.io/gravity-entropy-and-life<p>Cosmology, or at least basic cosmology, models the world as perfectly homogeneous and isotropic. Everything in the study of the evolution of the Universe actually works out fine if this is at least true at all times on the largest scales, and observations confirm it pretty much is. But at smaller scales, our Universe does not appear like a featureless, homogeneous soup, but as an immense tree of fractal complexity and divisions and subdivisions. Superclusters, clusters, galaxies, stars, planets, mountains and craters. And then there's life on Earth, an incredible variety of desperate machines grasping for survival. And on the top of the tree (for what we know), human civilizations, our actions and thoughts and the information we produce and consume. All of this complexity is concentrated in minuscule oases of useful information lost in a huge, huge blackness filled only by what is essentially thermal noise.</p>
<p><!--more--></p>
<p>A pretty picture, but doesn't that sound... wrong? Is that supposed to happen? On the face of it, all of this complexity arises and continues to exist because of an intricate array of physical phenomena and interactions on which life piggybacks. The varied chemistry of carbon allows for the extraordinary machineries of biology, fluid mechanics drives the weather and the tectonic cycle, electromagnetism and the theory of conductivity is the foundation for all of electronics, computers, and the Internet. We could list thousands of examples, but you get the idea.</p>
<p>As any person who "fucking loves science" will tell you, we are made of stardust, in the sense that our existence is based on many natural phenomena in a mechanical Universe. You would assume that is a satisfying explanation: many complex elements and interactions, thus great complexity in the results. However, all of these interactions naturally push towards death and decay. Organic matter burns and turns into vapour and dissolves into the atmosphere. Atomic bombs explode. Light bounces around and loses coherence. Hard drives break. Information scrambles. It's just the second law of thermodynamics. And the "complexity" of the interactions never really matters - a triple pendulum or the entire standard model in a box both tend to the same result: thermal equilibrium, and nothing happening. In schools we teach two essential lessons of both life and science: that everything in existence is the product of a chaotic and uncaring Universe, and that anything worth something requires care and work to be preserved, and would otherwise disassemble in time. But we don't point out the apparent contradiction between the two for some reason. At least not from the physics, even though this is very much a physics issue.</p>
<p>Creationists (to the extent that they actually exist and aren't an elaborate practical joke by the US on the rest of the world) like to argue that it is impossible for life on Earth to have spontaneously arisen, because it would mean a transition from "disorder" to "order", or more precisely a spontaneous creation of new useful information, in violation of the second law of thermodynamics. This very abstract conceptualization of life as a decrease in entropy, as a transition from equilibrium to non-equilibrium, is not wrong and is actually the view Schrödinger tried to push in <a href="https://en.wikipedia.org/w/index.php?title=What_Is_Life%3F&_%28Schr%C3%B6dinger%29=" target="_blank">"What is Life?"</a>. The obvious mistake, as the fucking lover of fucking science will make very sure to point out, is forgetting about the massive thermonuclear reactor in the sky. The Earth is not a closed system and receives constant input of "useful" energy from the Sun. This "negative entropy" (more precisely free energy) drives the Earth system and keeps it out of equilibrium. Alternatively, the Earth still has to cool its core and there's a bit of free energy being provided to the crust from the temperature difference between the core and the coldness of empty space, and a few organisms can live on that. None of these things can last forever, but for now they work.</p>
<p>However, this begs the question. If all our free energy comes from the Sun, then... who made the Sun? What brought it all out of equilibrium? The solar system started out as a relatively featureless, less differentiated protostellar cloud of hydrogen and helium. It was <strong>dead</strong>. Now it's neatly organized in differentiated bodies, a nuclear furnace forging heavy elements, a bunch of random planets with different composition and varied moons, and one even has life and roads and <a href="http://www.matriciana.com/piatto_di_vera_amatriciana.jpg" target="_blank">bucatini all'amatriciana</a>. That <em>is</em> a potential violation of the second law. Hydrogen in a tank does not do that - it does not form tiny solar systems with tiny people, it fills the tank homogeneously and stays dead in all aspects. In fact, no substance does that. Essentially every system I can think of, when isolated in a small tank, will eventually "die" and decompose and reach thermal equilibrium. So why does a lot of hydrogen in space do the opposite?</p>
<p>The reason has to do with the unusual thermodynamic properties of gravity and gravitationally-dominated systems. Consider a gas of particles (which might be molecules, or stars, or galaxies!) interacting only gravitationally. Then the total internal energy of the system is the kinetic + gravitational potential energy:</p>
<p style="text-align:center;">$$ U = U_k + U_G $$</p>
<p style="text-align:left;">However, also recall that for the gravitational interaction the virial theorem implies $$ U_G = - 2 U_k $$ (at least as a time-average) so that you can simplify</p>
<p style="text-align:center;">$$ U = - U_k \,.$$</p>
<p style="text-align:left;">Now, there is a common myth that kinetic energy is proportional to the temperature - we don't need such a strong (and occasionally false) statement. We just need to know that it is an increasing function of temperature. More precisely, you could write \( U_k(T,P)\) or \( U_k(T,V)\) and it would be an increasing function of \( T \) in either case. This means that for a gravitational system, the heat capacity either at constant pressure or volume is <strong>negative</strong>:</p>
<p style="text-align:center;">$$ \frac{\partial U}{\partial T} |_P < 0 $$</p>
<p style="text-align:center;">$$ \frac{\partial U}{\partial T} |_V < 0 $$</p>
<p>That is, a gravitational system gets colder when you give it energy, and gets hotter when you take it away. Note this also holds if there are additional interactions other than gravitational</p>
<p style="text-align:center;">$$ U = U_k + U_G + U_\text{other} $$</p>
<p style="text-align:left;">provided the gravitational potential is large enough compared to the other forces.</p>
<p>This changes everything. Let me just review briefly how normal thermodynamic systems always push towards equilibrium. Imagine a system has some type of inhomogeneity in the form of a temperature difference between two subsystems C and H, and let's say C is colder and H is hotter. Heat flows spontaneously from the hotter to the colder system, and that is always true. As C receives heat, and like most systems has positive heat capacity, it gets a bit hotter. Conversely, H is donating heat and gets colder. The temperature difference is reduced and the subsystems walk back into equilibrium.</p>
<p>However, if the heat capacity is negative, this doesn't work. Heat still moves from H to C, but H gets hotter and C gets colder. All inhomogeneities are amplified and the system moves <em>away</em> from equilibrium. Thus gravitational systems possesses the essential ingredient for the creation of life. But still, the second law of thermodynamics seems to be broken and that is always a bad sign. The 2nd law is a very, very general statement; it could be simplified as just the idea that the information in an imprecise or coarse-grained description of reality can only degrade, not improve. There just aren't any exceptions to this.</p>
<p>How does gravity manage to reduce entropy? It is commonly claimed that the Universe after the Big Bang was in a "low-entropy" state, while what awaits it in the far-future (heat death?) is a "high-entropy" state, and this difference actually gives the arrow of time, and is the reason for which we define it such that the Big Bang is in the past and the heat death in the future. It is true that there is an entropy difference between these two extremities, and that the <em>total</em> entropy increases monotonically between these two values. But the specific entropy in the early Universe is not particularly low compared to what we experience here on Earth - the early Universe was a homogeneous state in thermal equilibrium. If you consider the particles that compose a human body, the entropy they have now is much, much lower than the same particles (to the extent that this makes sense) right after the recombination of hydrogen, simply because the first is a complex, ordered system, in fact a living being composed of literally tens of trillions of little working machines, capable of withholding information and performing calculations and reasoning, while the latter is just... noise. So there <em>is</em> a decrease of entropy. It is gravity that is able to create such lower and lower-entropy states through gravitational collapse.</p>
<p>If a gravitational system has a typical size \( R \), then its potential energy will go as \( U_G \sim - 1/R\) which means \( \frac{dU}{dR} > 0 \). Intuitive: if you give energy to a gravitationally-bound system, orbits get wider. Since we already know the heat capacity is negative, this means \( \frac{dT}{d R} < 0 \), and that the more it shrinks, the hotter it gets. Like a satellite in orbit: if you take orbital energy away from it, it grazes the atmosphere and starts burning up.</p>
<p>However, entropy decreases as the system shrinks - a normal behaviour for usual systems but unexpected considering the weird thing we saw with the temperature. Ignoring pressure for now, the first law implies</p>
<p style="text-align:center;">$$ dS = dU/T = \frac{1}{T} \left(\frac{dU}{dR}\right) dR \Rightarrow \frac{dS}{dR} > 0 $$</p>
<p style="text-align:left;">So gravitational collapse violates the second law of thermodynamics and should never happen. But we are forgetting about the bizzarre property of spontaneous divergence from equilibrium of gravitational systems that I introduced before: they can (and will in time) split into subsystems of ever-increasing temperature difference. We know which one is the one getting hotter and hotter: it's the part that collapses. The one getting colder and colder is "radiation": actual radiation (gravitational or electromagnetic) escaping the gravitating mass or just expelled matter that has become gravitationally unbound. "Radiation" can move in a very large space and is not constrained by gravity to satisfy the conclusion $$ -2U_k = U_G $$ of the virial theorem, and thus carries a very large entropy. So a mass <em>can</em> collapse, but it needs to sacrifice part of itself to store a very large entropy, so that the part that continues collapsing can have a decreased entropy while still satisfying the second law of thermodynamics as a whole. While this is not an unusual or unseen occurrence (e.g.: I can reduce the entropy of a deck of cards by sorting it, but the minimum computations necessary in my brain will increase entropy by a larger amount), gravity is special and unique among all physical phenomena in that it does this spontaneously.</p>
<p style="text-align:left;">Thus the early Universe collapses gravitationally expelling "radiation", and the process is repeated at various scales. We therefore find little pockets of very low entropy in a very large high-entropy expanse. (Black holes are an exception, but for the sake of simplicity let's put them aside). It explains the tree of increasing complexity and the giant cold expanse of sparse thermal garbage.</p>
<p style="text-align:left;">You can imagine life without the chemistry, nuclear physics, condensed matter physics we know. You can imagine intelligence made of plasma or computers made of dancing stars in clusters. If you have a powerful enough imagination, every cog in the architecture of life is replaceable. Except for one: gravity. Only gravity can reduce entropy and create complexity where it doesn't already exist. It is, ultimately, the origin of anything worthwhile in the Universe. So, if you really need to worship something, or need a mantra, or are stupid enough to want a tattoo of a physics formula, let it be this one:</p>
<p style="text-align:center;">$$ F = -\frac{GM_1 M_2}{r^2} $$</p>
<p style="text-align:left;">because it's the closest thing to a loving God physics is going to give us.</p>
<hr />
<p>Now, with this said, why the heat death? Well, the reason is that in all of this I meant gravity <em>gravity</em>, that is the long-range interaction between masses governed by Newton's law. Not all of general relativity and the range of phenomena it predicts. In particular in the presence of dark energy / cosmological constant, the acceleration of mass \( M_1\) due to mass \( M_2\) is corrected as</p>
<p style="text-align:center;">$$ \ddot {\vec r} = - \frac{GM_2}{r^2} + \frac{\Lambda c^2}{3} r $$</p>
<p>so, a repulsive force which grows linearly with distance. The thing with the virial theorem applied to an interaction potential that goes as \(r^2\) gives \(U_k = U_\Lambda\) and so this means a positive contribution to the heat capacity. If dark energy is dominant, and it will be in the not-so-far future, then it undoes the creative property of gravity. In a dark-energy dominated system, we move towards thermal equilibrium instead of away from it. The natural order according to which all things spontaneously degrade and die is recovered, and all things will spontaneously degrade and die.</p>
<p> </p>
<p> </p>{"login"=>"saselli", "email"=>"riccardo.antonelli@hotmail.it", "display_name"=>"rantonels", "first_name"=>"", "last_name"=>""}riccardo.antonelli@hotmail.itCosmology, or at least basic cosmology, models the world as perfectly homogeneous and isotropic. Everything in the study of the evolution of the Universe actually works out fine if this is at least true at all times on the largest scales, and observations confirm it pretty much is. But at smaller scales, our Universe does not appear like a featureless, homogeneous soup, but as an immense tree of fractal complexity and divisions and subdivisions. Superclusters, clusters, galaxies, stars, planets, mountains and craters. And then there's life on Earth, an incredible variety of desperate machines grasping for survival. And on the top of the tree (for what we know), human civilizations, our actions and thoughts and the information we produce and consume. All of this complexity is concentrated in minuscule oases of useful information lost in a huge, huge blackness filled only by what is essentially thermal noise.Is the Planck length the minimum possible length?2017-02-20T09:15:47+00:002017-02-20T09:15:47+00:00http://rantonels.github.io/is-the-planck-length-the-minimum-possible-length<p>Don't Betteridge away this question so quickly. It holds more surprises than I thought it did.</p>
<p>There is a <em>massive</em> amount of confusion about this in popular explanations. It traces back to what I believe is the essential issue with pop-physics: everything is fourth-hand, being passed on from layman to layman and degraded with each iteration. In this case I would guess a rough scheme of the story of this misunderstanding went a little bit like this:</p>
<ul>
<li>Scientists provide pop-sci with the notion that "the Planck length is the minimum meaningful distance", which is a simplified representation of a correct insight.</li>
<li>Pop-sci journalists and enthusiast pass the notion on, until it gets deformed into "the Planck length is like a pixel size for the Universe", which is incorrect.</li>
<li>Scientists notice the mistake and release a vaccine to clear out the misunderstanding: "the Planck length is not like a pixel size for the Universe. It is just the scale where quantum gravity becomes relevant". Which is definitely <em>correct</em>, but...</li>
<li>The pop-science cycle chews on this until the notion has been transformed into "the Planck length has never been a minimum distance, it is a misconception. It's just a scale where our current theories break down, and there is nothing to suggest you could not reach down to smaller scales". Which sounds reasonable, but it's not correct.</li>
</ul>
<p>With this rant about popular physics out of the way, let me try to present some hopefully convincing evidence that the facts I claimed are correct are correct, and vice versa.</p>
<p><!--more--></p>
<p>First, let's clarify we're going to work in orders of magnitude. Factors of 2 or π do not matter, in fact essentially only the dependence on dimensionful constant will be of interest - in the end there is an overall fuzziness to it from our own ignorance about quantum gravity anyway. So, from the top to the bottom of this article, it's orders of magnitude.</p>
<p>Imagine trying to locate the position of <em>something</em>, an event, or an object, to a small precision \( \Delta X\). What is the most efficient way to do that? I mean: how do you perform this measurement while disturbing your <em>something</em> as little as possible? There is a lower bound given by relativistic quantum mechanics. We know all objects in quantum mechanics are ultimately de Broglie waves, and are bound by the Heisenberg uncertainty principle:</p>
<p style="text-align:center;">$$ \Delta X \Delta P \gtrsim \hbar $$</p>
<p>meaning the minimum momentum uncertainty that the probe we use to locate our position must be \( \Delta P \sim \hbar / \Delta X\). Now we throw in relativity. The energy of a relativistic particle is</p>
<p style="text-align:center;">\( E^2 = m^2 c^4 + P^2 c^2\)</p>
<p>which implies the minimum energy given a certain momentum is obtained for massless particles (or you can also use ultrarelativistic particles like \( m c^2 \ll Pc\). If you're shooting them at 1Gev, electrons look massless too). In that case, the absolute minimum energy kick we must necessarily give to locate a position to \( \Delta X\) is</p>
<p style="text-align:center;">$$ \Delta E \sim \frac{\hbar c}{\Delta X}$$</p>
<p style="text-align:left;">which is of course nothing else than the relationship between the energy and wavelength of a photon (minus or plus an order of magnitude, of course). So, let's use photons, since they're optimal in this sense, and if you want you can imagine using two opposite incident photons, so you don't give any linear momentum to your <em>something</em>, which would be pointlessly confusing.</p>
<p style="text-align:left;">So this is the famous energy-length scales relationship of relativistic QM, or better a bound: if you are at the energy scale \( E\), you cannot probe distances smaller than \( L \sim \frac{\hbar c}{E}\), or if you are probing distances \(L\), the energy cannot be lower than \( E \sim \frac{\hbar c}{L}\).</p>
<p style="text-align:left;">Now, gravity. General relativity generates a different energy-distance bound. If you have a certain energy \( E\) (or mass \( M = E/c^2\)) available, there is a minimum space you can squeeze it in: the most compact possible object is a black hole. The radius of the black hole is the Schwarzschild radius:</p>
<p style="text-align:center;">$$ r_S \sim \frac{GM}{c^2} = \frac{GE}{c^4}$$</p>
<p style="text-align:left;">Note the two constants \( c^2/G\) and \( c^4/G\), these are respectively the Planck linear mass density and Planck force. We'll talk about these later. For now, it's important to note that the GR bound has a linear, instead of inverse, relationship between space uncertainty and energy. It tells us you that to localize something to a position uncertainty \( \Delta X\), you can use <em>at most</em> the energy \( E \sim \frac{c^4}{G} \Delta X\), if you use any more, you create a black hole larger than the precision you want and your measurement is ruined.</p>
<p style="text-align:left;">Since the bounds we found are power laws, let's place them in a log-log plot where they are straight lines:</p>
<div>
<figure>
<img class="picture" src="/images/planck0.png" />
<figcaption>Blue is the contribution of relativistic quantum mechanics. Red is from general relativity. Sorry if you're daltonic.</figcaption>
</figure>
</div>
<p style="text-align:left;">One goes down, one goes up. When you combine the bounds, you get an <em>absolute</em> bound on the position uncertainty. Trusting these bounds, it seems as if there is a minimum length scale where the Compton and Schwarzschild lengths coincide. That's</p>
<p style="text-align:left;">$$ \frac{\hbar c}{E} = \frac{GE}{c^4} \Rightarrow E = \sqrt{\frac{\hbar c^5}{G}}$$</p>
<p style="text-align:left;">or</p>
<p style="text-align:left;">$$ \lambda_C = r_S = \sqrt{\frac{\hbar G}{c^3}} =: l_P$$</p>
<p style="text-align:left;">the infamous Planck length. So, within an order of magnitude, you shouldn't be able to probe shorter distances than \( l_P\). At \( l_P\), the energy we need to develop is the Planck energy \( E_P = \sqrt{\frac{\hbar c^5}G}\,,\) and the equivalent mass is the Planck mass \( M_P = E_P/c^2 = \sqrt{\frac{\hbar c}{G}}\,.\) Note that while the Planck length seems like a lower bound to all physics, the Planck mass or energy are boundaries between different areas of physics. \( M_P\) is the threshold between particle physics and gravity, being an upper bound on the mass of fundamental particles and a lower bound on the mass of a black hole. As a black hole evaporates down to the Planck mass from Hawking radiation, you would expect it to morph into particle(s) as it jumps this fence.</p>
<!--<p>[caption id="attachment_1110" align="alignnone" width="827"]<img class="alignnone size-full wp-image-1110" src="/assets/planckspectrum2.png" alt="planckspectrum" width="827" height="444" /> The discrete mass spectrum of particles should morph into the continuous spectrum of black hole masses as the Planck mass is overcome. Some kind of "liminal being" must interpolate in this unknown region between particle and black hole.[/caption]</p>--!>
<div>
<figure>
<img class="picture" src="/images/planckspectrum2.png" />
<figcaption>The discrete mass spectrum of particles should morph into the continuous spectrum of black hole masses as the Planck mass is overcome. Some kind of "liminal being" must interpolate in this unknown region between particle and black hole.</figcaption>
</figure>
</div>
<p style="text-align:left;">Actually, that's not really rigorous because the theories we used to derive the bounds are surely not valid up to the Planck scale. If we use relativistic quantum mechanics and raise the energy up to near the Planck energy, there is an unlimited number of gravitational corrections that join the party as the strength of gravity becomes larger and larger, and these corrections are "uncontrollable" and render the theory useless - the least confusing description of the non-renormalizability of gravity I can manage in one sentence. Conversely, if I start from classical general relativity and lower the mass down to near the Planck mass, higher loop (quantum) corrections appear and the same problem arises. Basically, both quantum field theory and general relativity should be replaced by a consistent theory of quantum gravity, which presents itself in full glory only when you scrape the Planck scale, and otherwise at larger length scales reconstructs quantum field theory and general relativity. In pictures, there is a "small" fuzzy patch we need to cut out from our diagram to account for our ignorance:</p>
<!--<p style="text-align:left;"><img class="alignnone size-full wp-image-1084" src="/assets/planck1.png" alt="planck1" width="1111" height="1111" /></p>--!>
<div>
<figure>
<img class="picture" src="/images/planck1.png" />
<figcaption></figcaption>
</figure>
</div>
<p style="text-align:left;">Now, none of us here is worthy of witnessing the face of God, but I think we all have some mild assumptions on what it surely does not look like. So while a lack of perfect knowledge of Planck-scale physics precludes us from announcing the Planck length is the minimum length scale (again, within an order of magnitude and certainly not as a "hard" limit) with absolute certainty, we cannot avoid wondering how bizarre things would need to be in quantum gravity for it not to be, and with a precise and scientific notion of bizarreness: the concept of naturalness. Our bound seems to suggest a bubbly or foamy spacetime, evaporating away as we zoom down to the Planck scale, where geometry itself is affected by quantum uncertainties. Note we found something like</p>
<p style="text-align:center;">$$ (\Delta x)^2 \gtrsim l_P^2$$</p>
<p style="text-align:left;">But we could as well swap \( \Delta x\) with \( \Delta t/c\), since we're definitely in the relativistic regime. In fact, instead of talking about Compton wavelength, we could have talked about Compton frequency/period, and we'd have obtained an equivalent-ish but much more eloquent</p>
<p style="text-align:center;">$$ \Delta x \Delta t \gtrsim l_P^2 /c$$</p>
<p style="text-align:left;">Huh. Time and space seem connected by a sort of uncertainty principle... a space-time uncertainty principle! And just like the familiar position-momentum HUP arises from a commutation relation \( [x,p] \sim \hbar\), so the possibility arises that space and time do not commute, and the commutator is of order the Planck area:</p>
<p style="text-align:center;">$$ [x,t] \sim l_P^2 / c$$</p>
<p style="text-align:left;">Or you could do this with two different space dimensions, instead of space and time. In any case, there is a suggestion that the geometry itself of spacetime is non-commutative, or that non-commutative geometry (for what little I understand of it at least) should play a key role in quantum gravity. Anyway, the space-time uncertainty principle is a weaker refinement of our bound and seems to suggest that rather than being impossible to squeeze \( \Delta x\) below the Planck length, it is indeed possible to probe smaller lengths provided you allow for a large time uncertainty. You could perhaps locate the position of an event as good as you want along a single axis if you are willing to allow a large indeterminacy in the time at which it happens, or in its position along other spatial axes.</p>
<!--<p>[caption id="attachment_1141" align="alignnone" width="827"]<img class="alignnone size-full wp-image-1141" src="/assets/planckarea.png" alt="planckarea" width="827" height="444" /> Uncertainty boxes saturating the space-time uncertainty principle. The Planck area acts as a sort of "minimum area" or quantum of area for spacetime, in a very similar way in which \( \hbar\) acts as the quantum of phase space volume in quantum mechanics.[/caption]</p>--!>
<div>
<figure>
<img class="picture" src="/images/planckarea.png" />
<figcaption>Uncertainty boxes saturating the space-time uncertainty principle. The Planck area acts as a sort of "minimum area" or quantum of area for spacetime, in a very similar way in which \( \hbar\) acts as the quantum of phase space volume in quantum mechanics</figcaption>
</figure>
</div>
<p style="text-align:left;">And now, we were talking about naturalness... for this fundamental space-time foaminess <em>not</em> to arise, so that we can simply move forward to lower length scales as if nothing happened, would require that all those infinite higher-loop quantum corrections all conspire to cancel this uncertainty to zero, while it would very naturally be around $$ l_P^2$$. It requires a huge degree of fine-tuning for no apparent reason. Instead, it reasonable to assume this doesn't happen - it's not a proof, but certainly rephrases the question in a way that makes it clear which possibility is the "standard" one and which one is absurd and exotic. And the only reason that for us at large length scales this foaminess is not visible is pure dimensional analysis!</p>
<p style="text-align:left;">Still, this picture is very alien compared to the naive "pixely" one. Distances shorter than the Planck scale don't make sense in the sense that the inherent quantumness of spacetime pushes back. It's like trying to squeeze water in your hand only to have it squish out from the sides.</p>
<p style="text-align:left;">But then, it does not end here. We <em>do</em> have an explicit example of a working theory of quantum gravity in which to verify these deductions more precisely: string theory. In string theory the space-time uncertainty relations arise from the weird property that strings have of stretching and becoming longer when you give them a lot of energy, more precisely</p>
<p style="text-align:center;">$$ \Delta x \sim \frac{l_S^2 }{c\hbar} \Delta E$$</p>
<p style="text-align:left;">where \(l_s\) is a constant known as the string length. (Now, the string and Planck lengths are not the same thing, but I'm going to intentionally confuse them since the difference is not essential for this reasoning). The name of the string length comes from the fact that when instead strings have <em>little</em> energy, they don't shrink down to zero size but rather settle on a minimum constant size:</p>
<p style="text-align:center;">$$ \Delta x \sim l_S$$</p>
<p style="text-align:left;">Now it should be easy to see that the high-energy "stretching" behaviour combined with $$ \Delta E \sim \frac{\hbar}{\Delta t}$$ implies the time-space uncertainty:</p>
<p style="text-align:center;">$$ \Delta x \Delta t \gtrsim l_S^2$$</p>
<p style="text-align:left;">and that the low-energy behaviour instead implies that the string length is the minimum sensible scale:</p>
<p style="text-align:center;">$$ \Delta x \gtrsim l_S$$</p>
<p style="text-align:left;">Ultimately, the foaminess is interpreted in (perturbative) string theory as due to the fact that strings are extended objects that cannot get smaller than \( \sim l_S\), and that a localization measurement could not be done with anything else but strings. Or rather, you could: you could use a D-brane (like a D0-brane, a D-particle) to perform a position measurement. This allows you to break the \( \Delta x \sim l_S\) barrier and measure smaller lengths, but you must still satisfy the space-time uncertainty principle and must therefore move your D-brane very slowly, and let the measurement take a proportionally long time.</p>
<p style="text-align:left;">This can be done much, much more precisely but I'm not bothering here of course. If you want to read more <a href="https://cds.cern.ch/record/434343/files/0004074.pdf">this</a> is great.</p>
<p style="text-align:left;">Also, as an aside: string theory provides the "liminal beings" interpolating between particle and black hole! They are... drum roll... strings. A little-excited string is just a particle, in fact a point particle when seen from our infrared, large scale perspective since \( l_S\) is very small. There is a discrete spectrum of excited states and those are the possible particles. All of them are essentially of size \( l_S\). As you give it more and more energy you approach the Planck energy and the number of possible states grows vertiginously. (As \( \exp( \text{constant} \cdot \frac{l_S}{\hbar c} \Delta E)\) actually). The string begins getting longer and longer; as the mass rises above \( M_P\) the spectrum of states becomes continuous and the string turns into a tangly, warm wooly membrane above a baby horizon, a black hole. This transition can be study pretty explictly.</p>
<p style="text-align:left;">Be it correct or not in terms of our Universe, the fact that string theory confirms, with a much more specific and quantitative presentation, our hand-wavy expectations about quantum gravity is notable. It is certainly a good sign.</p>
<h3 style="text-align:left;">Addendum: why no numeric values?</h3>
<p>Why haven't I mentioned anything about the actual values in SI units of the Planck length, mass...? Sure, the calculations are order-of-magnitude, but at least the orders of magnitude (e.g. \( l_P \sim 10^{-34} m\)) could be interesting to talk about, right? Well, the point is while I think the arguments above about what happens at the Planck scale are true, I don't necessarily believe the Planck scale must be where you would naively think it is based on the calculation above. Let me explain. We found a Planck length which is small if gravity is weak:</p>
<p style="text-align:center;">$$ l_P \sim G^{1/2}$$</p>
<p style="text-align:left;">However, the strength of gravity that we measure here might not be genuine, it might be a fake. This happens for example if there are extra compact dimensions which are large, that is \( \gg l_P\). In that case, when you are at small scales, smaller than the compact dimensions, you are in a higher-dimensional (say, (9+1)-dimensional) space with a certain strength of gravity \( G_{(10)}\) and a "ten-dimensional Planck length"</p>
<p style="text-align:center;">$$ {l_P}_{(10)} \sim G_{(10)}^{1/8}$$</p>
<p style="text-align:left;">(hey, why the eight root? Dimensional analysis!) which is the "true" actual Planck length where all the magic stuff happens. However, as you hop to larger length scales, larger than the compact dimensions, and return to our 4-dimensional Universe, gravity gets an extra weakening. The gravitational constant we measure (the "4-dimensional" big G) is the 10-dimensional one divided by the volume of the 6 extra dimensions</p>
<p style="text-align:center;">$$ G_{(4)} = \frac{G_{(10)}}{V_6} $$</p>
<p style="text-align:left;">So, our gravity is weaker than normal... which means our \( {l_P}_{(4)}\) calculated from \( G_{(4)}\) is an incorrect estimate, smaller than the true Planck length \( {l_P}_{(10)}\). Our argument has to be corrected because as dive below the scale of the extra dimensions, it becomes easier to create a black hole.</p>
<h2 style="text-align:left;">All Planck units</h2>
<p>The Planck length is a "minimum length", where the expression is to be understood in the sense I attempted to explain above. What about all the other Planck units? Not all of them are necessarily interpretable as bounds; I tried to compile them in a table here with my personal attempt at an interpretation of this type. It's a work in progress.</p>
<p>Planck units are all dimensionful quantities constructible from powers of \( G,c,\hbar\). I've decided to use units with \( k_B = 1\) because man, that's one stupid constant, and I'm not going to touch on anything electromagnetic. There is exactly one Planck quantity for a given unit, since the three fundamental constants are independent. Some Planck units only require a subset of the constants, and these are indeed relevant not to general quantum gravity but to some "subtheory" - it should become clear as you scroll down the table.</p>
<p>All of this make sense to within an order of magnitude, of course.</p>
<h2>Newtonian gravity (\( G\))</h2>
<p>\( G\), units \( M^{-1} L^3 s^{-2}\), gives the strength of gravity.</p>
<h2>Special relativity (\( c\))</h2>
<p>\( c\), units \( L T^{-1}\), Planck speed, maximum possible speed.</p>
<h2>Quantum mechanics (\( \hbar\))</h2>
<p>\( J_P = \hbar\), units \( M L^2 T^{-2}\), Planck angular momentum / Planck action, minimum possible angular momentum or action. Phase space volume is "granular" and cells have volume \( \hbar\).</p>
<h2>Quantum Newtonian gravity / non-relativistic quantum gravity (\( G, \hbar\))</h2>
<p>I found no combination of \( G,\hbar\) with a clear interpretation.</p>
<h2>Relativistic quantum mechanics (\( \hbar, c\))</h2>
<p>\( \hbar c\), units \( M L^3 T^{-2} = E L\), minimum energy per unit inverse length.</p>
<h2>General Relativity (\( G, c\))</h2>
<p>The bounds here are enforced by existence, and occasional inevitability, of black holes.</p>
<p>\( \lambda_P = \frac{c^2}{G}\), units \( M L^{-1}\), Planck linear mass density. Maximum possible linear mass density, and that of a black hole. I haven't done the calculation but I think cosmic strings with higher density will fragment into black holes.</p>
<p>\( F_P = \frac{c^4}{G}\), units \( M L T^{-2}\), Planck force, maximum possible force in a system at equilibrium and the force between two close black holes. Attempt to produce larger forces will result in enough energy to turn everything into a black hole. Also typical tension of a string.</p>
<p>\( W_P = \frac{c^5}{G}\), units \( M L^2 T^{-3}\), Planck power. Maximum power emitted by a black hole merger. Possibly maximum power in some sense I cannot be bothered to think about now.</p>
<h2>Quantum Gravity (\( G, c, \hbar\))</h2>
<p>All possible remaining units combinations can be made with these. The (soft) bounds here are enforced by the non-commutative nature of spacetime and the space-time uncertainty principle.</p>
<p>\( l_P = \sqrt\frac{G\hbar}{c^3}\), units \( L\), Planck length, minimum possible length in the sense described before.</p>
<p>\( t_P = \sqrt\frac{G\hbar}{c^5}\), units \( T\), Planck time, minimum possible time in the sense described before.</p>
<p>\( l_P^2 = \frac{G\hbar}{c^3}\), Planck area, whatever, you get these.</p>
<p>\( T_P = E_P = \sqrt\frac{\hbar c^5 }{G}\), units \( \Theta = E = M L^2 T^{-2}\), Planck temperature or Planck energy. Maximum possible temperature.</p>
<p>\( M_P = \sqrt\frac{\hbar G}{c^3}\), units \( M\), maximum mass of a particle and minimum mass of a black hole.</p>
<p>\( a_P = \sqrt\frac{c^7}{\hbar G}\), units \( L T^{-2}\), Planck acceleration, maximum possible acceleration. At this acceleration, a Rindler horizon appears a Planck length away and emits Planck-temperature Unruh radiation; no kind of structure or information can survive without being irreparably scrambled.</p>{"login"=>"saselli", "email"=>"riccardo.antonelli@hotmail.it", "display_name"=>"rantonels", "first_name"=>"", "last_name"=>""}riccardo.antonelli@hotmail.itDon't Betteridge away this question so quickly. It holds more surprises than I thought it did.Twin paradox for literal children2017-01-07T09:56:33+00:002017-01-07T09:56:33+00:00http://rantonels.github.io/twin-paradox-for-literal-children<p>When I was very young, I was presented with this challenging riddle:</p>
<div>
<figure>
<img class="picture" src="/images/space-svg.png" />
<figcaption>Recreated as accurately as possible according to my artistic abilities</figcaption>
</figure>
</div>
<p>Little boy must go to the house. There are two paths. Which path is shorter?</p>
<p>Don't worry, we won't need the actual answer to the riddle. We just need the much weaker insight that is also part of what this image is meant to teach to children: different paths between the same starting and ending point can have different <em>lengths.</em></p>
<p>If the boy carries a ruler, and he uses it to measures the length of the path as he follows it, he will find that the left path and the right path have different lengths. Or, unrolling a thread on the path, he will find that different amounts of thread are needed. (Guessing which path is actually shorter requires instead extremely advanced mathematical abilities and is left as an exercise for the intrepid reader, but as I said, we don't need it.)</p>
<p>Now, the obvious generalization.</p>
<p>Instead of using the page to represent space, we use it to represent spacetime:</p>
<div>
<figure>
<img class="picture" src="/images/spime-svg1.png" />
<figcaption></figcaption>
</figure>
</div>
<p>Two twins, Blue and Red, are together at the same event A (it can be their 8th birthday at their home). They are also together at an event B, say Blue's graduation party again at home. But <em>between</em> A and B Blue and Red lead different lives. Blue stays home on Earth studying. Red embarks on a long trip through the Universe. They have taken different paths through spacetime.</p>
<p>And these paths have a <em>duration</em>. The time that Blue feels passing from A to B is the duration of <em>his</em> path, and the time that Red feels from A to B is the duration of <em>her</em> path. How do you measure duration more objectively? With an instrument called a clock. (Or a stopwatch, if you prefer).</p>
<p>The twins both start their clock at A. As they follow their own, different paths in spacetime the clocks "unroll the thread" and measure the durations of the paths until they meet again at B. They will indeed measure different values, and one of the twins will be younger, having experienced less time.</p>
<p>And that's it. There is no "warping" or "slowing" or "fastening" of time, just different paths which in general have different durations, just as there was no "stretching" or "squashing" of space when the little boy needed to get to the house. It was just different paths. (Nor is there any mysterious force affecting clocks or biological aging). And this is incredibly general: it works in special or general relativity in <em>any</em> spacetime.</p>
<p>Of course this does not tell you which twin is younger and by how much. It does not touch on the problem that is usually considered the paradoxical part (the question of the apparent symmetry etc etc). But often that is way too much for the layman. To many people, the fact that the twins have different ages when they meet again <em>is</em> the paradox!</p>
<p>The kind of relativistic gymnastics the average physicist (or knowledgeable mathematician) performs when asked this very simple question:</p>
<blockquote><p>How can one twin be younger than the other? How is that possible?</p></blockquote>
<p>is incredible. People pull out spacetime diagrams, light cones, square roots, "motion through time", vectors, four-vectors, invariant sets, acceleration, planes of simultaneity... probably because when you study relativity you meet all this stuff <em>before</em> the twin paradox is introduced, and mostly as an example, so one subconsciously assumes this stuff is necessary to understand the twins at the most basic level. It's really not, it's very simple.</p>
<p>I might even go a bit further into the opinion territory and just say that the way the whole thing is taught is backwards. Proper time is simplest, it is invariant, it is immediately observable with a clock that the observer carries, and it should be introduced first. Just like proper length and rulers. Everything else then follows, as you can map out spacetime with families of observers with clocks and rulers and construct coordinate systems, in which then you can study inertial and non inertial frames, Lorentz transformations, time dilation of any variety, relativistic beaming, whatever. Everything can be derived from proper time, because that is simply what the metric on spacetime is, and that's the actual fundamental entity in classical relativity. End of rant.</p>{"login"=>"saselli", "email"=>"riccardo.antonelli@hotmail.it", "display_name"=>"rantonels", "first_name"=>"", "last_name"=>""}riccardo.antonelli@hotmail.itWhen I was very young, I was presented with this challenging riddle:A geometric proof of Hawking radiation, through imaginary time2016-09-07T13:23:52+00:002016-09-07T13:23:52+00:00http://rantonels.github.io/a-geometric-proof-of-hawking-radiation-through-imaginary-time<p>Black holes are states at thermal equilibrium, at a positive temperature inversely proportional to the mass, and they therefore emit black body radiation at that temperature. Everybody knows that. But why? By far the most common explanation in divulgation is in terms of pair production of particles at the horizon, which is a shame because I personally find it very opaque. The most popular among physicists instead is the "standard" derivation based on diagonalizing the equations of motion for the fields in two different charts and the Bogoliubov transformations and blah blah blah. Technical. As reliable and as sexy as a brick.</p>
<p>It would be cool if there was a more elegant argument. Perhaps simpler, depending on your definition of simpler. Well, there is. Here it is:</p>
<p>Let's use natural units. \( \hbar = c = k_B = 1\).</p>
<p>(The inverse) <strong>temperature is periodicity in imaginary time</strong>. Consider an observable of a system which is a function of a duration of time, \( A(t)\). For example, if I give it a kick at time \(t'\), how big of a noise is it going to make at time \(t'+t\)? The answer to this question is an observable depending on the time interval \(t\). Ok, then more often than not these \(A(t)\) will be analytic functions and to some extent you will be able to extend them to function of <em>complex</em> \(t\). Then the statement is:</p>
<p>If all of these functions \( A(t)\) are periodic of period \(\beta\) in imaginary time, that is \(A(t + i \beta) = A(t)\), then the system is actually in thermal equilibrium at temperature \(T = 1/\beta\).</p>
<p>This is a very important and fundamental relationship. I can't really introduce it simply, the most elementary introduction I've seen yet is <a href="http://physics.stackexchange.com/questions/27416/what-is-a-simple-intuitive-way-to-see-the-relation-between-imaginary-time-perio">this</a>.</p>
<p>Now, assuming this is known, let's see the geometry part of this post. The Schwarzschild metric is:</p>
<p>$$ ds^2 = - (1-\frac{r_S}{r}) dt^2 + (1-\frac{r_S}{r})^{-1} dr^2 + r^2 d\Omega^2$$</p>
<p>but as we will see all that matters is what happens close to the horizon at \( r = r_S\). So we change variables \( r = r_S(1+\epsilon)\) and take \( \epsilon\) very small so we study the near-horizon geometry. (It's not actually <em>necessary</em> to take this limit - it just simplifies the following calculations without compromising the result). So</p>
<p>$$ds^2 = - \epsilon dt^2 + \frac{r_S^2}\epsilon d\epsilon^2 + r_S^2 d\Omega^2$$</p>
<p>Cool, now let's take a final change of variables to send this into a decent form, we try \( \frac{\rho^2}{4 r_S^2} = \epsilon\):</p>
<p>$$ ds^2 = - \frac{\rho^2}{4r_S^2} dt^2 + d \rho^2 + r_S^2 d\Omega^2$$</p>
<p>This is actually all well-known, it's just the <a href="https://en.wikipedia.org/wiki/Rindler_coordinates">Rindler metric</a>. Spacetime right outside a black hole is well approximated by uniformly accelerating spacetime, makes a lot of sense.</p>
<p>☆ But now, here's the actual quantum magic. Let's <strong>extend the metric to complex t</strong>. In particular we care about the imaginary axis, so \( t = i \tau\) with \( \tau\) real. So</p>
<p>$$ ds^2 = \frac{\rho^2}{4r_S^2} d \tau^2 + d\rho^2 + r_S d\Omega^2$$</p>
<p>Now forget about the \( \Omega\) part - that's not an issue. Look at the τ - ρ part. Looks familiar?</p>
<p>It's just the metric for the flat Euclidean plane, in polar coordinates: \( ds^2 = dr^2 + r^2 d\theta^2\). You just need to identify \( \theta = \tau/(2 r_S) \). Actually, however, the identification is not perfect because of the periodicity of \( \theta \), which is of course \( 2\pi \). If the periodicity is anything else than that, this geometry will have an angular defect or a conical singularity. For example, if the period is \( \pi \), then you can build this by taking a sheet of paper, marking a point along an edge, and gluing together the two pieces of the edge on the two sides of the marked point. You very clearly get a cone with a sharp tip. The tip has curvature, but we don't want curvature: curvature is proportional to stress-energy in general relativity, but black holes have no energy lying about at the horizon - they are empty.</p>
<div>
<figure>
<img class="picture" src="/images/icecream.jpg" />
<figcaption>Did the particle pair explanation bring you ice cream? Yeah, didn't think so.</figcaption>
</figure>
</div>
<p>The tip must go away and so the period must be \( 2\pi \). So the period of \( \tau \) is \( 4\pi r_S = 8\pi G M \). So <strong>black holes are states of temperature</strong> \( T = 1/(8\pi G M) \), as measured by observers whose time is \( t \), therefore those at infinity.</p>
<p>Simply <b>glorious</b>.</p>
<p>It's very close to being elementary, because except for the relationship between temperature and imaginary time everything is pretty much changes of variables. It's such a cool idea that the thermodynamics of the quantum black hole metric is encoded somewhat in its geometrical structure. And it's also very general: we did not need to setup any specific quantum field theory (= a quantum particle theory) on the black hole metric, as you do in the other derivations. In fact, we know whatever theory actually describes quantum gravity, it cannot be a QFT. So this kind of argument is very powerful.</p>
<p>On the other hand, this is a pretty informal argument. It's very creative, so to speak, but not particularly rigorous. There are a million points where you could nitpick. Therefore, this is only seen as a secondary reinforcement of the results of the more precise derivations. I won't comment further on all these million subtleties, because my intent here was only to introduce the argument in its deceiving beauty.</p>
<h3>Bonus round</h3>
<p>By the way, from the ☆ onwards this is also a proof of the Unruh effect - obvious because the latter and Hawking radiation are more or less the same thing. In the case of Unruh radiation the imaginary-time metric really is flat Euclidean space. In our case of a Schwarzschild black hole, this is only true in the near-horizon limit. If we take into account the whole exterior region, the imaginary-time metric is describable as and is in fact called "a cigar". The tip of the cigar is smooth (because we imposed it!) and of course when you zoom in becomes the flat plane. This doesn't really change the required periodicity of \( \tau\), which runs as an angular coordinate around the cigar.</p>
<h3>Refs</h3>
<p>I believe the first time the imaginary-time periodicity argument for Hawking radiation was appreciated is <a href="http://journals.aps.org/prd/abstract/10.1103/PhysRevD.15.2738">this 1977 Hawking + Gibbons paper</a>. Also please take a look at the last sentence of the abstract - pretty interesting.</p>{"login"=>"saselli", "email"=>"riccardo.antonelli@hotmail.it", "display_name"=>"rantonels", "first_name"=>"", "last_name"=>""}riccardo.antonelli@hotmail.itBlack holes are states at thermal equilibrium, at a positive temperature inversely proportional to the mass, and they therefore emit black body radiation at that temperature. Everybody knows that. But why? By far the most common explanation in divulgation is in terms of pair production of particles at the horizon, which is a shame because I personally find it very opaque. The most popular among physicists instead is the "standard" derivation based on diagonalizing the equations of motion for the fields in two different charts and the Bogoliubov transformations and blah blah blah. Technical. As reliable and as sexy as a brick.The Riemann zeta function, the primon gas, and supersymmetry.2016-05-30T14:20:33+00:002016-05-30T14:20:33+00:00http://rantonels.github.io/the-riemann-zeta-function-the-primon-gas-and-supersymmetry<p>The following is probably one of the weirdest unexpected bridges between abstract mathematics and theoretical physics I know of; the two weirdest features being that it's pretty simple to explain, and that the area of math it connects to is number theory. While the latter thing <em>can </em>occur occasionally, especially in the area of string theory, the former is basically close to impossible.</p>
<p>A while back I was in a very ugly situation I needed to distract myself out of. I was struck by the fact that the Riemann zeta function looks a lot like a partition function from statistical mechanics. The Riemann zeta is:</p>
<p>$$ \zeta(\beta) = \sum_{n=1}^\infty n^{-\beta}$$</p>
<p>for \(Re(\beta) > 1\) of course, and then one extends analytically. The partition function of a system with energy levels \( E_n\) is given by</p>
<p>$$ Z(\beta) = \sum_n e^{-\beta E_n}$$</p>
<p>where \( \beta = \frac{1}{k_B T}\). Note that if you have a system with energy levels \( E_n = \log n\), its partition function will be precisely the Riemann zeta. It's curious how they even use the same letter.</p>
<h2>The bosonic primon gas</h2>
<p>After googling I found out such a system has indeed been investigated. Basically, imagine a single particle, let's call it a primon, had the available energy levels</p>
<p>$$ \log p_0, \log p_1, \ldots $$</p>
<p>where the \( p_i\) are the prime numbers.</p>
<p>Then imagine a system made by a bunch of these particles, a gas, and that they don't interact. Your system can have any number of primons, and each of these can be in any single particle state; moreover they are indistinguishable, and finally they are bosons so that you can place as many as you want in any single-particle state.</p>
<p>The system's state could then be described by just specifying how many primons you want on each level, so the occupation numbers \( a_0,a_1,a_2,\ldots\). The energy of this state is then</p>
<p>$$ E = a_0\log p_0 + a_1 \log p_1 + \ldots = \log \left( p_0^{a_0} p_1^{a_1} \ldots \right)$$</p>
<p>Of course, for the energy of the state to be finite, the total number of particles $$ N = a_0 + a_1 + \ldots$$ must be finite and so the \( a_i\) must be zero from a certain \(i\) onwards.</p>
<p>Note how a prime decomposition just popped up. The energy is \( \log n\) for some integer \( n = p_0^{a_0} p_1^{a_1} \ldots\) decomposed as powers of primes. By the prime factorization theorem, this decomposition exists for all integers and is unique. So the physics version of this theorem is that this system has the states \( |n\rangle\) for all the integers \( n\), with energy \(\log n\), occupation numbers given by the exponents in the prime decomposition, and most importantly no degeneracy: there is one and only one way to place primons on the energy levels to make a state of energy \( \log n\). Here's a super explicit example:</p>
<p>The state \(|40 \rangle\) has energy \(\log 40\). Since \(40 = 2^3 \cdot 5^1\), this state has \(3\) primons with energy \(\log 2\) and \(1\) primon with energy \(\log 5\). The occupation numbers are \((3,0,1,0,0,0,\ldots)\).</p>
<p>Cool. So the (grand canonical, actually) partition function of this thing is the sum over all possible states of the system of \(e^{-\beta E_n}\); but as we saw the states are indexed by all the positive integers, with energy \( \log n\), so actually</p>
<p>$$ Z(\beta) = \sum_n e^{-\beta E_n} = \sum_{n=1}^\infty n^{-\beta} = \zeta(\beta)$$</p>
<p>So yeah, the Riemann zeta is the partition function for the primon gas. But that's not all. This is still a gas of bosons, and normal statistical mechanics applies. In particular, we know the partition function must be</p>
<p>$$ Z(\beta) = \prod_i \left(1 - e^{-\beta \epsilon_i }\right)^{-1} $$</p>
<p>where \( \epsilon_i\) are the single-particle energies. Inserting our value for the energies we get:</p>
<p>$$ Z(\beta) = \prod_{p} (1 - p^{-\beta} )^{-1} $$</p>
<p>where the product is over prime \(p\)... we rediscovered Euler's celebrated product formula:</p>
<p>$$ \zeta(\beta) = \prod_p (1- p^{-\beta})^{-1}$$</p>
<p>but essentially only doing physics.</p>
<p>Now we could stop here and this would be fun all by itself. We could also muse on whether this allows for an alternative approach to the Riemann hypothesis. However, there's two main obstacles:</p>
<ul>
<li>the Riemann hypothesis is about \( \beta = \frac{1}{2} + it \); while purely imaginary \( \beta\)s would be immensely interesting physically (because of Wick rotation, relating very roughly temperature in the statistical mechanics system with time in the quantum mechanics equivalent) I wouldn't really know what to do with the shift by one-half.</li>
<li>the Riemann hypothesis is about zeroes. We have an easy interpretation of poles: for example the divergence at $$ \beta=1$$ in the partition function means the primon gas cannot actually get any hotter than that because it would require infinite energy - this an example of a Hagedorn temperature. But I wouldn't know what the zero of big zeta would mean physically.</li>
</ul>
<p>Sad. Maybe someone more informed can shed some light.</p>
<p>But that's not all we can squeeze out of this. Let me introduce you to:</p>
<h2>The fermionic primon gas</h2>
<p>So now we want all primons to be fermions. The only thing we need to change is that no two particles can share the same state because of Pauli exclusion, so occupation numbers must be \( 0\) or \(1\). This means that we cannot get the state \(|n\rangle\) for all integers \( n\), but only if \( n\) has a decomposition into primes where all the exponents aren't bigger than one. Equivalently, it has to be a product of distinct primes, aka a square-free number.</p>
<p>So the partition function is a sum over square-free numbers:</p>
<p>$$ Z(\beta) = \sum_{n \text{S.F}} n^{-\beta} = \sum_{n=1}^\infty |\mu(n)| n^{-\beta} $$</p>
<p>I was able to extend the sum to all integers by weighing with the absolute value of the Möbius function $$ \mu(n)$$ from number theory. This function is \( 0\) on non-square-free numbers, \( 1\) if your number is a product of an even number of primes, and \( -1\) otherwise. The Möbius function admits a crystalline interpretation as the operator giving the statistics of the state:</p>
<p>$$ \mu(n) = (-1)^{N} $$</p>
<p>that is, if there is an odd number of fermions, our state will be a fermion, even, it will be a boson. $$ \mu(n)$$ will tell us that. Anyways, the partition function can also be written using the known statmech result for a system of fermions:</p>
<p>$$ Z(\beta) = \prod_i (1+e^{-\beta \epsilon_i}) = \prod_p (1 + p^{-\beta}) = \prod_p \frac{1- p^{-2\beta}}{1-p^{-\beta}} = \frac{\zeta(\beta)}{\zeta(2\beta)} $$</p>
<p>So we found <em>another</em> famous number theory result</p>
<p>$$ \sum_n |\mu(n)| n^{-\beta} = \frac{\zeta(\beta)}{\zeta(2\beta)} $$</p>
<p>Now the final step for our trascendence beyond this material hyperplane is the</p>
<h2>Supersymmetric primon gas</h2>
<p><strong>Now</strong> we're talking. The two theories above describe respectively a bosonic and a fermionic particle species. A theory containing both would be supersymmetric and these particles would be superpartners. (Of course, we don't have to fiddle with the ultra-complex math of actual supersymmetry transformations in D-dimensional spacetime - because there's no spacetime here!). The supersymmetry would just be the symmetry sending bosons to fermions and viceversa.</p>
<p>This is nothing to be scared of. I'm just taking the two systems, the bosonic and the fermionic gas, and putting them together. Simple composite system. Nothing is interacting so everything factorizes. We just consider the total energy = energy of bosons + energy of fermions.</p>
<p>Now \( | n \rangle\) does not identify a single state. Calling the occupation numbers of bosons and fermions respectively \( a_i\) and \( b_i\), we still define the total \( n\) as</p>
<p>$$ n = p_0^{a_0 + b_0} \cdot p_1^{a_1 + b_1} \cdot \ldots $$</p>
<p>and the energy is indeed \( \log n\), but now giving \(n\) only fixes the total exponents \( a_i + b_i\) by prime decomposition; to fix the occupation number we also specify the total number built only with the \( b^i\):</p>
<p>$$ d = p_0^{b_0} \cdot p_1^{b_1} \cdot \ldots $$</p>
<p>so we can describe all states as \( |n,d\rangle\) where \(d\) is square-free. Note that \( d\) divides \(n\). So actually \(d\) must be a square-free divisor of n.</p>
<p>Let's forget about the partition function (which is just the product of the previous two, since this is just a composite system of the previous two systems). Let's talk expectation values. If you have an operator \( O\), you can compute the average value of the operator at a given temperature through:</p>
<p>$$ \langle O \rangle = \sum_{\text{states}} e^{-\beta E} O$$</p>
<p>and note the partition function is just the expectation value of 1. So we want to compute the expectation value of the operator $$ (-1)^{N_F}$$ counting fermions, defined above. This is</p>
<p>$$ \Delta = \langle (-1)^{N_F} \rangle = \sum_n \sum_{d|n} \mu(d) n^{-\beta}$$</p>
<p>I've swapped the operator with its representation as the Möbius function, and I should sum over square-free divisors of \( n\), but actually I can just sum over all divisors since \( \mu\) projects out the square-free ones.</p>
<p>Now the inner sum is</p>
<p>$$ \sum_{d|n} \mu(d) \;= 1 \text{ if }n>1,\; 0 \text{ if }n=1 $$</p>
<p>Why? Well there's a math reason and a physics reason. Math reason, I'll let you check it out in the Wiki article for the Möbius function. Physics reason, is supersymmetry. At any given energy, the set of states is supersymmetric, and there are as many fermionic (-1) states as there are bosonic (+1). So in the sum they cancel out to 0. The only exception is the ground state \( |1,0\rangle\).</p>
<p>So \( \Delta = 1\). But then, this is the expectation value computed in the supersymmetric theory; since this is a simple sum of the bosonic and fermionic theory, this should factorize:</p>
<p>$$ \Delta = \langle (-1)^{N_F} \rangle_B \langle (-1)^{N_F} \rangle_F $$</p>
<p>first one, the operator always takes the value 1, so the expectation value is the partition function in the bosonic case, good ol' \( \zeta(\beta)\). Second one, it's the sum \( \sum_{d=1}^\infty \mu(d) d^{-\beta}\). So here's the last number theory fact for today:</p>
<p>$$ \frac{1}{\zeta(\beta)} = \sum_{d=1}^\infty \mu(d) d^{-\beta}$$</p>
<p>It's not over yet: you can also prove the more substantial Möbius inversion formula, but I'm not going to talk about that. Read about it in this paper: <a href="http://projecteuclid.org/euclid.cmp/1104180135">http://projecteuclid.org/euclid.cmp/1104180135</a></p>
<p>End of transmission.</p>
<p> </p>{"login"=>"saselli", "email"=>"riccardo.antonelli@hotmail.it", "display_name"=>"rantonels", "first_name"=>"", "last_name"=>""}riccardo.antonelli@hotmail.itThe following is probably one of the weirdest unexpected bridges between abstract mathematics and theoretical physics I know of; the two weirdest features being that it's pretty simple to explain, and that the area of math it connects to is number theory. While the latter thing can occur occasionally, especially in the area of string theory, the former is basically close to impossible.