Spinor Info

Last December, I wrote a blog entry in which I criticized one aspect of the LHC’s analysis of the scalar particle discovered earlier this year, which is believed to be the long sought-after Higgs boson.

The Higgs boson is a scalar. It is conceivable that the particle observed at the LHC is not the Higgs particle but an “impostor”, some composite of known (and perhaps unknown) particles that behaves like a scalar. Or, I should say, almost like a scalar, as the ground state of such composites would likely behave like a pseudoscalar. The difference is that whereas a scalar-valued field remains unchanged under a reflection, a pseudoscalar field changes sign.

This has specific consequences when the particle decays, apparent in the angles of the decay products’ trajectories.

Several such angles are measured, but the analysis used at the ATLAS detector of the LHC employs a method borrowed from machine learning research, called a Boosted Decision Tree algorithm, that synthesizes a single parameter that has maximum sensitivity to the parity of the observed particle. (The CMS detector’s analysis uses a similar approach.)

The result can be plotted against scalar vs. pseudoscalar model predictions. This plot, shown below, does not appear very convincing. The data points (which represent binned numbers of events) are all over the place with large errors. Out of a total of only 43 events (give or take), more than 25% are the expected background, only 30+ events represent an actual signal. And the scalar vs. pseudoscalar predictions are very similar.

This is why, when I saw that the analysis concluded that the scalar hypothesis is supported with a probability of over 97%, I felt rather skeptical. And I thought I knew the reason: I thought that the experimental error, i.e., the error bars in the plot above, was not properly accounted for in the analysis.

Indeed, if I calculate the normalized chi-square per degree of freedom, I get \(\chi^2_{J^P=0^+} = 0.247\) and \(\chi^2_{J^P=0^-} = 0.426\), respectively, for the two hypotheses. The difference is not very big.

Alas, my skepticism was misplaced. The folks at the LHC didn’t bother with chi-squares, instead they performed a likelihood analysis. The question they were asking was this: given the set of observations available, what are the likelihoods of the scalar and the pseudoscalar scenarios?

At the LHC, they used likelihood functions and distributions derived from the actual theory. However, I can do a poor man’s version myself by simply using the Gaussian normal distribution (or a nonsymmetric version of the same). Given a data point \(D_i\), a model value \(M_I\), and a standard deviation (error) \(\sigma_i\), the probability that the data point is at least as far from \(M_i\) as \(D_i\) is given by

\begin{align}
{\cal P}_i=2\left[1-\Psi\left(\frac{|D_i-M_i|}{\sigma_i}\right)\right],
\end{align}

where \(\Psi(x)\) is the cumulative normal distribution.

Now \({\cal P}_i\) also happens to be the likelihood of the model value \(M_i\) given the data point \(D_i\) and standard distribution \(\sigma_i\). If we assume that the data points and their errors are statistically indepdendent, the likelihood that all the data points happen to fall where they fell is given by

\begin{align}
{\cal L}=\prod\limits_{i=1}^N{\cal P}_i.
\end{align}

Taking the data from the ATLAS figure above, the value \(q\) of the commonly used log-likelihood ratio is

\begin{align}
q=\ln\frac{{\cal L}(J^P=0^+)}{{\cal L}(J^P=0^-)}=2.89.
\end{align}

(The LHC folks calculated 2.2, which is “close enough” for me given that I am using a naive Gaussian distribution.)

Furthermore, if I choose to believe that the only two viable hypothesis for the spin-parity of the observed particle are the scalar and pseudoscalar scenarios (e.g., if other experiments already convinced me that alternatives, such as intepreting the result as a spin-2 particle, can be completely excluded) I can normalize these two likelihoods and interpret them as probabilities. The probability of the scalar scenario is then \(e^{2.89}\simeq 18\) times larger than the probability of the pseudoscalar scenario. So if these probabilities add up to 100%, that means that the scalar scenario is favored with a probability of nearly 95%. Not exactly “slam dunk” but pretty darn convincing.

As to the validity of the method, there is, in fact, a theorem called the Neyman-Pearson lemma that states that the likelihood-ratio test is the most powerful test for this type of comparison of hypotheses.

But what about my earlier objection that the observational error was not properly accounted for? Well… it appears that it was, after all. In my “poor man’s” version of the analysis, the observational error was used to select the appropriate form of the normal distribution, through \(\sigma_i\). In the LHC’s analysis, I believe, the observational error found its was into the Monte-Carlo simulation that was used to develop a physically more accurate probability distribution function that was used for the same purpose.

Even clever people make mistakes. Even large groups of very clever people sometimes succumb to groupthink. But when you bet against clever people, you are likely to lose. I thought I spotted an error in the analysis performed at the LHC, but all I really found were gaps in my own understanding. Oh well… live and learn.

Physics blog sites are abuzz about Eric Weinstein and his Amazing New Theory of Everything. For a moment, I actually confused him with Eric Weisstein, well known in physics and math circles as the founder of Mathworld, which, in the pre-Wikipedia days, was the Internet’s pre-eminent mathematics encyclopedia (only to be hijacked for a while by an unscrupulous CRC press). No, Weinstein is someone else: he is a mathematical physicist turned economist. In any case, he is no dummy, nor does he appear to be a crackpot. He is outside of academia, but, well, so am I, so who am I to complain?

So Weinstein gets invited to Oxford to give a public lecture, and he talks, for the first time, about ideas he has been working on for the past twenty years, about unifying physics.

This is greeted by a headline in The Guardian that reads, “Roll over Einstein: meet Weinstein“. Others follow suit, and soon physics news and blog sites far and wide discuss… what, exactly? Well, no-one really knows.

That is because Weinstein has not published anything yet. Not even a non-peer reviewed manuscript on arxiv.org. This is pointed out in one of the few sensibly skeptical blog posts, written by Jennifer Ouellette on Scientific American’s blog site. Ouellette actually quotes a tweet by Sean Carroll: “Pretty sure Einstein actually wrote research papers, not just gave interviews to newspapers.”

Ouellette goes on to quote Oxford cosmologist Andrew Pontzen, who observes that these “shenanigans” have “short-circuited science’s basic checks and balances”. I couldn’t agree more. This is true even if Weinstein turns out to be right in the end.

Which is conceivable, since Weinstein is no crackpot. But it is much more likely that his theory will join many others, including Garrett “surfer dude” Lisi’s aesthetically beautiful E8 theory, that just don’t have much to do with observable reality.

In a post a few days ago, I expressed my skeptical views concerning the interpretation of some of the recent Higgs results from CERN. I used a simple analogy, an example in which measuring the average height of the inhabitants in a set of buildings is used to determine which of them may house the Harlem Globetrotters.

However, I came to realize (thanks in part to some helpful criticism that my post received) that I left out one possibility. What if the buildings are too small? Or the ‘Trotters are just too, hmm, tired after a long party and end up in the wrong building? In that case, a measurement may look like this:

If we have an a priori reason to believe that, for whatever reason, the players are indeed spread out across several buildings, then we can indeed not expect to see a sharp peak at #4 (or whichever building is assigned to the Globetrotters); instead, we should see a broad excess, just what the CMS experiment is seeing when it measured the decay of the presumed Higgs boson into a τ⁺τ⁻ pair.

So is there an a priori reason for the data to be spread out like this? I believe there is. No instrument detects τ leptons directly, as their lifetime is too short. Instead, τ events are reconstructed from decay products, and all forms of τ decay involve at least one neutrino, which may carry away a significant portion of the lepton’s energy. So the final uncertainty in the total measured energy of the τ⁺τ⁻ pair can be quite significant.

In other words, many of the Globetrotters may indeed be sleeping in the wrong building.

Nonetheless, as my copy of the venerable 20-year old book, The Higgs Hunter’s Guide suggests, the decay into τ leptons can be a valuable means of confirmation. Which is perhaps why it is troubling that for now, the other major detector at the LHC, ATLAS, failed to see a similar broad excess of τ⁺τ⁻ events near the presumed Higgs mass.

A few weeks ago, I exchanged a number of e-mails with someone about the Lanczos tensor and the Weyl-Lanczos equation. One of the things I derived is worth recording here for posterity.

The Lanczos tensor is an interesting animal. It can be thought of as the source of the Weyl curvature tensor, the traceless part of the Riemann curvature tensor. The Weyl tensor, together with the Ricci tensor, fully determine the Riemann tensor, i.e., the intrinsic curvature of a spacetime. Crudely put, whereas the Ricci tensor tells you how the volume of, say, a cloud of dust changes in response to gravity, the Weyl tensor tells you how that cloud of dust is distorted in response to the same gravitational field. (For instance, consider a cloud of dust in empty space falling towards the Earth. In empty space, the Ricci tensor is zero, so the volume of the cloud does not change. But its shape becomes distorted and elongated in response to tidal forces. This is described by the Weyl tensor.

Because the Ricci tensor is absent, the Weyl tensor fully describes gravitational fields in empty space. In a sense, the Weyl tensor is analogous to the electromagnetic field tensor that fully describes electromagnetic fields in empty space. The electromagnetic field tensor is sourced by the four-dimensional electromagnetic vector potential (meaning that the electromagnetic field tensor can be expressed using partial derivatives of the electromagnetic vector potential.) The Weyl tensor has a source in exactly the same sense, in the form of the Lanczos tensor.

The electromagnetic field does not uniquely determine the electromagnetic vector potential. This is basically how integrals vs. derivatives work. For instance, the derivative of the function \(y=x^2\) is given by \(y’=2x\). But the inverse operation is not unambiguous: \(\int 2x~ dx=x^2+C\) where \(C\) is an arbitrary integration constant. This is a recognition of the fact that the derivative of any function in the form \(y=x^2+C\) is \(y’=2x\) regardless of the value of \(C\); so knowing only the derivative \(y’\) does not fully determine the original function \(y\).

In the case of electromagnetism, this freedom to choose the electromagnetic vector potential is referred to as the gauge freedom. The same gauge freedom exists for the Lanczos tensor.

Solutions for the Lanczos tensor for the simplest case of the Schwarzschild metric are provided in Wikipedia. A common characteristic of these solutions is that they yield a quantity that “blows up” at the event horizon. This runs contrary to accepted wisdom, namely that the event horizon is not in any way special; a freely falling space traveler would never know that he is crossing it.

But as it turns out, thanks to the gauge freedom of the Lanczos tensor, it is easy to construct a solution (an infinite family of solutions, as a matter of fact) that do not behave like this at the horizon.

Well, it was a fun thing to compute anyway.

I have received a surprising number of comments to my recent post on the gravitational potential, including a criticism: namely that what I am saying is nonsense, that in fact it is well known (there is actually a resolution by the International Astronomical Union to this effect) that in the vicinity of the Earth, the gravitational potential is well approximated using the Earth’s multipole plus tidal contributions, and that the potential, therefore, is determined primarily by the Earth itself, the Sun only playing a minor role, contrary to what I was blabbering about.

But this is precisely the view of gravity that I was arguing against. As they say, a picture is worth a thousand words, so let me try to demonstrate it with pictures, starting with this one:

It is a crude one-dimensional depiction of the Earth’s gravity well (between the two vertical black lines) embedded in the much deeper gravity well (centered) of the Sun. In other words, what I depicted is the sum of two gravitational potentials:

$$U=-\frac{GM}{R}-\frac{Gm}{r}.$$

Let me now zoom into the area marked by the vertical lines for a better view:

It looks like a perfectly ordinary gravitational potential well, except that it is slightly lopsided.

So what if I ignored the Sun’s potential altogether? In other words, what if I considered the potential given by

$$U=-\frac{Gm}{r}+C$$

instead, where \(C\) is just some normalization constant to ensure that I am comparing apples to apples here? This is what I get:

The green curve approximates the red curve fairly well deep inside the potential well but fails further out.

But wait a cotton-picking minute. When I say “approximate”, what does that tell you? Why, we approximate curves with Taylor series, don’t we, at least when we can. The Sun’s gravitational potential, \(-GM/R\), near the vicinity of the Earth located at \(R=R_0\), would be given by the approximation

$$-\frac{GM}{R}=-\frac{GM}{R_0}+\frac{GM}{R_0^2}(R-R_0)-\frac{GM}{R_0^3}(R-R_0)^2+{\cal O}\left(\frac{GM}{R_0^4}[R-R_0]^3\right).$$

And in this oversimplified one-dimensional case, \(r=R-R_0\) so I might as well write

$$-\frac{GM}{R}=-\frac{GM}{R_0}+\frac{GM}{R_0^2}r-\frac{GM}{R_0^3}r^2+{\cal O}\left(\frac{GM}{R_0^4}r^3\right).$$

(In the three-dimensional case, the math gets messier but the principle remains the same.)

So when I used a constant previously, its value would have been \(C=-GM/R_0\) and this would be just the zeroeth order term in the Taylor series expansion of the Sun’s potential. What if I include more terms and write:

$$U\simeq-\frac{Gm}{r}-\frac{GM}{R_0}+\frac{GM}{R_0^2}r-\frac{GM}{R_0^3}r^2?$$

When I plot this, here is what I get:

The blue curve now does a much better job approximating the red one. (Incidentally, note that if I differentiate by \(r\) to obtain the acceleration, I get: \(a=-dU/dr=-Gm/r^2-GM/R_0^2+2GMr/R_0^3\), which is the sum of the terrestrial acceleration, the solar acceleration that determines the Earth’s orbit around the Sun, and the usual tidal term. So this is another way to derive the tidal term. But, I digress.)

The improvement can also be seen if I plot the relative error of the green vs. blue curves:

So far so good. But the blue curve still fails miserably further outside. Let me zoom back out to the scale of the original plot:

Oops.

So while it is true that in the vicinity of the Earth, the tidal potential is a useful approximation, it is not the “real thing”. And when we perform a physical experiment that involves, e.g., a distant spacecraft or astronomical objects, the tidal potential must not be used. Such experiments, for instance tests measuring gravitational time dilation or the gravitational frequency shift of an electromagnetic signal are readily realizable nowadays with precision equipment.

But it just occurred to me that even at the pure Newtonian level, the value of the potential \(U\) plays an observable role: it determines the escape velocity. A projectile escapes to infinity if its overall energy (kinetic plus potential) is greater than zero: \(mv^2/2 + mU>0\). In other words, the escape velocity \(v\) is determined by the formula

$$v>\sqrt{-2U}.$$

The escape velocity works both ways; it also tells you the velocity to which an object accelerates as it falls from infinity. So suppose you let lose a rock somewhere in deep space far from the Sun and it falls towards the Earth. Its velocity at impact will be 43.6 km/s… without the Sun’s influence, its impact velocity would have been only 11.2 km/s.

So using somewhat more poetic language, the relationship of us, surface dwellers, to distant parts of the universe, is determined primarily not by the gravity of the planet on which we stand, but by the gravitational field of our Sun… or maybe our galaxy… or maybe the supercluster of which our galaxy is a member.

As I said in my preceding post… gravity is weird.

The following gnuplot code, which I am recording here for posterity, was used to produce the plots in this post:

set terminal gif size 320,240
unset border
unset xtics
unset ytics
set xrange [-5:5]
set yrange [-5:0]
set output 'pot0.gif'
set arrow from 0.5,-5 to 0.5,0 nohead lc rgb 'black' lw 0.1
set arrow from 1.5,-5 to 1.5,0 nohead lc rgb 'black' lw 0.1
plot -1/abs(x)-0.1/abs(x-1) lw 3 notitle
unset arrow
set xrange [0.5:1.5]
set output 'pot1.gif'
plot -1/abs(x)-0.1/abs(x-1) lw 3 notitle
set output 'pot2.gif'
plot -1/abs(x)-0.1/abs(x-1) lw 3 notitle,-0.1/abs(x-1)-1 lw 3 notitle
set output 'pot3.gif'
plot -1/abs(x)-0.1/abs(x-1) lw 3 notitle,-0.1/abs(x-1)-1 lw 3 notitle,-0.1/abs(x-1)-1+(x-1)-(x-1)**2 lw 3 notitle
set xrange [-5:5]
set output 'pot4.gif'
set arrow from 0.5,-5 to 0.5,0 nohead lc rgb 'black' lw 0.1
set arrow from 1.5,-5 to 1.5,0 nohead lc rgb 'black' lw 0.1
replot
unset arrow
set output 'potdiff.gif'
set xrange [0.5:1.5]
set yrange [*:*]
plot 0 notitle,\
((-1/abs(x)-0.1/abs(x-1))-(-0.1/abs(x-1)-1))/(-1/abs(x)-0.1/abs(x-1)) lw 3 notitle,\
((-1/abs(x)-0.1/abs(x-1))-(-0.1/abs(x-1)-1+(x-1)-(x-1)**2))/(-1/abs(x)-0.1/abs(x-1)) lw 3 notitle

Update (September 6, 2013): The analysis in this blog entry is invalid. See my September 6, 2013 blog entry on this topic for an explanation and update.

It has been a while since I last wrote about a pure physics topic in this blog.

A big open question these days is whether or not the particle purportedly discovered by the Large Hadron Collider is indeed the Higgs boson.

One thing about the Higgs boson is that it is a spin-0 scalar particle: this means, essentially, that the Higgs is identical to its mirror image. This distinguishes the Higgs from pseudoscalar particles that “flip” when viewed in a mirror.

So then, one way to distinguish the Higgs from other possibilities, including so-called pseudoscalar resonances, is by establishing that the observed particle indeed behaves either like a scalar or like a pseudoscalar.

Easier said than done. The differences in behavior are subtle. But it can be done, by measuring the angular distribution of decay products. And this analysis was indeed performed using the presently available data collected by the LHC.

Without further ado, here is one view of the data, taken from a November 14, 2012 presentation by Alexey Drozdetskiy:

The solid red line corresponds to a scalar particle (denoted by 0+); the dotted red line to a pseudoscalar (0−). The data points represent the number of events. The horizontal axis represents a “Matrix Element Likelihood Analysis” value, which is constructed using a formula similar to this one (see arXiv:1208.4018 by Bolognesi et al.):

$${\cal D}_{\rm bkg}=\left[1+\frac{{\cal P}_{\rm bkg}(m_{4\ell};m_1,m_2,\Omega)}{{\cal P}_{\rm sig}(m_{4\ell};m_1,m_2,\Omega)}\right]^{-1},$$

where the \({\cal P}\)-s represent probabilities associated with the background and the signal.

So far so good. The data are obviously noisy. And there are not that many data points: only 10, representing 16 events (give or take, as the vertical error bars are quite significant).

There is another way to visualize these values: namely by plotting them against the relative likelihood that the observed particle is 0+ or 0−:

In this fine plot, the two Gaussian curves correspond to Monte-Carlo simulations of the scalar and pseudoscalar scenarios. The position of the green arrow is somehow representative of the 10 data points shown in the preceding plot. The horizontal axis in this case is the logarithm of a likelihood ratio.

On the surface of it, this seems to indicate that the observed particle is indeed a scalar, just like the Higgs. So far so good, but what bothers me is that this second plot does not indicate uncertainties in the data. Yet, judging by the sizable vertical error bars in the first plot, the uncertainties are significant.

However, to relate the uncertainties in the first plot, one has to be able to relate the likelihood ratio on this plot to the MELA value on the preceding plot. Such a relationship indeed exists, given by the formula

$${\cal L}_k=\exp(-n_{\rm sig}-n_{\rm bkg})\prod_i\left(n_{\rm sig}\times{\cal P}^k_{\rm sig}(x_i;\alpha;\beta)+n_{\rm bkg}\times{\cal P}_{\rm bkg}(x_i;\beta)\right).$$

The problem with this formula, from my naive perspective, is that in order to replicate it, I would need to know not only the number of candidate signal events but also the number of background events, and also the associated probability distributions and values for \(\alpha\) and \(\beta\). I just don’t have all the information necessary to reconstruct this relationship numerically.

But perhaps I don’t have to. There is a rather naive thing one can do: and that would be simply calculating the weighted average of the data points in the first plot. When I do this, I get a value of 0.57. Lo and behold, it has roughly the same relationship to the solid red Gaussian in that plot as the green arrow to the 0+ Gaussian in the second.

Going by the assumption that my naive shortcut actually works reasonably well, I can take the next step. I can calculate a \(1\sigma\) error on the weighted average, which yields \(0.57^{+0.24}_{-0.23}\). When I (admittedly very crudely) try the transcribe this uncertainty to the second plot, I get something like this:

Yes, the error is this significant. So while the position of the green arrow is in tantalizing agreement with what one would expect from a Higgs particle, the error bar says that we cannot draw any definitive conclusions just yet.

But wait, it gets even weirder. Going back to the first plot, notice the two data points on the right. What if these are outliers? If I remove them from the analysis, I get something completely different: namely, the value of \(0.43^{+0.26}_{-0.21}\). Which is this:

So without the outliers, the data actually favor the pseudoscalar scenario!

I have to emphasize: what I did here is rather naive. The weighted average may not accurately represent the position of the green arrow at all. The coincidence in position could be a complete accident. In which case the horizontal error bar yielded by my analysis is completely bogus as well.

I also attempted to check how much more data would be needed to reduce the size of these error bars sufficiently for a true \(1\sigma\) result: about 2-4 times the number of events collected to date. So perhaps what I did is not complete nonsense after all, because this is what knowledgeable people are saying: when the LHC collected at least twice the amount of data it already has, we may know with reasonable certainty if the observed particle is a scalar or a pseudoscalar.

Until then, I hope I did not make a complete fool of myself with this naive analysis. Still, this is what blogs are for; I am allowed to say foolish things here.