## 28 August 2014

### RIP, Jacques Friedel (1921-2014)

Jacques Friedel passed away yesterday. He was among the pioneers of the study of defects in solids and of the electronic structure of matter, one of the founders of my lab and my scientific great-grandfather.

## 25 August 2014

### The arbitrariness of words

(via phys.org) A statistical analysis of English terms, recently published in Phil. Trans. R. Soc. B (free preprint on the corresponding author's site), finds systematic relations between sound and meaning, refuting a pure arbitrariness of the linguistic sign.

Of course, such relations have been sought for —and found— ever since Plato (last year, I reviewed on this blog Genette's Mimologics, a great exploration of the topic.) The novelty is the quantitative aspect of the analysis: the authors define phonetic and semantic distances between pairs of terms and then measure the correlation of these distances, which is higher than expected by pure chance. Unfortunately, they give no intuitive illustration for the amplitude of the effect, expressed as an $$r$$-factor. So, finally, how systematic is the English language?

Another interesting result is that more systematic words are acquired earlier. The authors speculate that systematicity helps language development in its early stages but might hinder it later, when (the vocabulary being larger) it can lead to confusion.

## 24 August 2014

### Moments of inertia of triangular prisms

Now that we have determined the moments of inertia of regular and truncated equilateral triangles, it is time to calculate them for the corresponding right prisms. These bodies, with mass density $$\rho$$, can be seen as stacks of infinitesimally thin triangles of thickness $$\text{d}h$$ and surface density $$\text{d} \mu = \rho \text{d}h$$ (we preserve the notations from the previous posts and introduce the height of the stack, $$H$$. The inertia moments of the prisms are denoted by $$P$$, instead of $$I$$.)
The centers of mass of these sheets are all situated on $$z$$, so the total moment of inertia about this axis is simply the sum of the individual ones. We must simply replace $$\mu$$ by $$\rho H$$ in (1) and (3):
$\begin{array}{ll} P_z(L,H) &= \rho H L^4 \frac{\sqrt{3}}{48}\\ P^{\text{tr}}_z(L,a,H) &= \rho H \frac{\sqrt{3}}{48} [L^4 - 3 a^4 - 12 a^2 (L-a)^2] \end{array}\tag{5}$
The derivation is slightly more complicated for axis $$y$$, since we need to account for the variable distance between it and the centers of mass of the sheets (using, of course, the overworked parallel axis theorem!) Fortunately, we only need the integral $$\displaystyle \int_{-H/2}^{H/2} \text{d}h \, h^2 = \frac{H^3}{12}$$ to get:
$\begin{array}{ll} P_y(L,H) &= \rho H L^4 \frac{\sqrt{3}}{96} \left [ 1 + 2 \left ( \frac{H}{L} \right )^2 \right ] \\ P^{\text{tr}}_y(L,a,H) &= \rho H \frac{\sqrt{3}}{96} \left [ L^4 - 3 a^4 - 12 a^2 (L-a)^2 + 2 H^2 (L^2 - 3 a^2) \right ]\\ &= \rho H L^4 \frac{\sqrt{3}}{96} \left [ 1 + 2 \left ( \frac{H}{L} \right )^2 - 12 x^2 \left ( (1-x)^2 + \frac{x^2}{4} + \frac{x^2}{2} \left ( \frac{H}{L} \right )^2 \right ) \right ] \end{array}\tag{6}$where $$x = a /L$$.

## 23 August 2014

### Moment of inertia of a clipped triangle

After calculating the moment of inertia for an equilateral triangle, let us consider the same shape, but with clipped corners, as in the drawing below:

We will preserve the notations of the previous post, adding the superscript "tr" for the truncated shape: $$I_{z}^{\text{tr}} (L,a)$$ is the moment about the $$z$$ axis of the equilateral triangle with side $$L$$, clipped by $$a$$ at each corner (with $$a \leq L/2$$). We will also use the same strategy, writing the moments of the complete shape as a combination of its four fragments:
$I_{z}(L) = I_{z}^{\text{tr}} (L,a) + 3[I_{z}(a)+m(a)d^2] ,$
where $$d=(L-a)/\sqrt{3} .$$ Using the results obtained for the full triangle immediately yields:
$I_{z}^{\text{tr}} (L,a) = \frac{\sqrt{3}}{48} \mu \left [ L^4 - 3a^4 - 12 a^2 (L-a)^2\right ] \tag{3}$
Similarly, from:
$I_{y}(L) = I_{y}^{\text{tr}} (L,a)+I_{y}(a) + 2[I_{y}(a)+m(a)(L-a)^2/4]$
we get:
$I_{y}^{\text{tr}} (L,a) = \frac{\sqrt{3}}{96} \mu \left [ L^4 - 3a^4 - 12 a^2 (L-a)^2\right ] = I_{z}^{\text{tr}} (L,a)/2 \tag{4}$
The clipped shape preserves the threefold symmetry of the original one, so the same conclusion as to the in-plane isotropy of the inertia tensor holds. Also, $$I_{z} = 2 I_{y}$$ in both cases; I'm sure there is some elegant way to explain this, but I can't find it.
A quick check of results (3) and (4) is that $$I_{y,z}^{\text{tr}} (2L,L) = I_{y,z}(L) .$$ In this case, one retrieves the situation shown in the illustration to the previous post.

## 22 August 2014

(via Slashdot) Infoworld gives some career advice to young programmers. I'm surprised at how much sense these short observations also make for beginning scientists. OK, maybe point 7 is not that relevant, and point 4 should be rephrased as Do not re-invent the wheel. Otherwise, they are spot-on.

## 21 August 2014

### Mass moment of inertia of an equilateral triangle

As in previous posts, I would like to determine the moments of inertia of a solid body, this time an equilateral triangular prism. I will start in this post by a (very thin) equilateral triangle. The challenge is getting the result in the simplest way, making the most of the symmetry elements and taking advantage of the parallel axis theorem.

### Around the $$z$$ axis

The $$z$$ axis goes through the center of mass of the triangle of interest (gray central area of side $$L$$ in the illustration above) and is perpendicular to its plane. We denote the corresponding moment by $$I_z(L)$$. The moment of the large triangle, with side $$2L$$, is $$I_z(2L)$$. We can relate these two parameters in two ways:
• For a given shape and surface mass density, the moment of inertia scales as the size to the fourth power, on dimensional grounds. Thus, $$I_z(2L) = 16 I_z(L)$$.
• The large triangle can also be described as the rigid assembly of the small central triangle and the three adjacent ones. The parallel axis theorem yields:
$I_z(2L) = I_z(L) + 3 [I_z(L) + m(L) d^2]$
where $$m(L)= \mu L^2 \sqrt{3}/4$$ is the mass of the small triangle, with $$\mu$$ the surface mass density, and $$d=L/\sqrt{3}$$ is the distance between the centers of mass of the side triangles and the $$z$$ axis.

Combining these two expressions for $$I_z(2L)$$ immediately yields:
$I_z(L) = \mu L^4 \frac{\sqrt{3}}{48} \tag{1}$

### Around the $$y$$ axis

The $$y$$ axis is contained in the plane of the triangle and goes through its center of mass and one vertex. Using the same strategy as above, we get:
$\left\{ \begin{array}{ll} I_y(2L) &= 16 I_y(L)\\ I_y(2L) &= 2 I_y(L) + 2 [I_y(L) + m(L) (L/2)^2] \end{array} \right.$
where on the right-hand side of the second equality the first term corresponds to the central and top triangles (both their centers of mass are on axis $$y$$) and the second one to the side triangles, whose centers are shifted by $$L/2$$. Finally:
$I_y(L) = \mu L^4 \frac{\sqrt{3}}{96} = \frac{I_z(L)}{2} \tag{2}$

How about the $$x$$ axis? To answer this question, we start by noting that there are three equivalent directions within the plane of the triangle: $$y$$ and the axes (say, $$y'$$ and $$y''$$) going through the other two vertices: $$I_y(L) = I_{y'}(L) = I_{y''}(L)$$. This third-order symmetry in a two-dimensional space means that the inertia tensor is in fact isotropic in the plane of the triangle, with the same value $$I_{\bot}(L) = I_{y}(L) = I_{x}(L)$$ for any axis in this plane. The inertia tensor is then:
$\mathrm{I} = \left ( \begin{matrix} I_{\bot}(L) & 0 & 0 \\ 0 & I_{\bot}(L) & 0 \\ 0 & 0 & I_z(L) \end{matrix} \right )$
This isotropy of a tensorial property for a system that does not in fact have full rotational symmetry is a very useful result (albeit somewhat counterintuitive). To give only one example from a completely different area of physics: a cubic crystal cannot be birefringent!