Today we will recall Euclid's proof of Pythagoras' Theorem, using the diagram below. The hypotenuse of the central triangle is fixed at length $1$, and edge lengths of $s$, $t$ and $x$ are marked for later use.

First, observe that the triangles $BCF$ and $ACK$ are congruent, since they each have a side of length $1$, a side of length $t$, and equal angles between these sides. Now consider the rectangle $BEFP$. Clearly, one half of its area is given by the triangle $BPF$. We can chop off the triangle $BPC$ to get back to the triangle $BCF$, which must therefore have one half of the area of the rectangle $CGEF$ (since $BPC$ and $GBC$ are congruent). Similarly, the triangle $ACK$ has one half the area of the square $BCJK$. The congruence of $BCF$ and $ACK$ then forces the blue areas $CGEF$ and $BCJK$ to have equal areas.

The only line in the diagram that was not required in the proof is the red line. It marks the point $Z$ on the blue square where the ratio of the distances $BZ$ to $BJ$ is given by $(1 - x)$, as shown. Pythagoras' theorem tells us that $(1 - x) = s^2$ and $x = t^2$. The angles that characterise the full diagram are now $\phi$ and $\epsilon$. They are related by the rule

$\cos^{2} \phi = \tan \epsilon$

Without the diagram, this looks like a rather odd rule for a pair of angles to obey. The reason it is interesting is the coincidence that it works when $\epsilon$ is the Cabibbo angle and $(\pi /2 - \phi)$ is the Weinberg angle.

6 years ago

And the length of the red line is now given by a simple ratio of the form $\cos \theta_{C} / \cos \theta_{W}$ (I'm probably out by some $\pi /2$ phase).

ReplyDeleteOops ... need to swap $C$ and $W$ ...

ReplyDelete... and of course, I am talking about a specific energy in the running of $\theta_{W}$. I assume that the reader can figure that out.

ReplyDeleteI think it's worth mentioning here

ReplyDeletepi/4 - atan(e) = Wienberg angle

e = (4pi*alpha)^.5 = 0.30282212

(the natural unit of charge in Heaviside-Lorentz units)

(see http://arxiv.org/abs/hep-ph/0609131)

Thanks, Dave! That looks really cool.

ReplyDeleteAnother intriguing coincidence: $\sin (\theta_W + \theta_C)$ is very close to $2/3$. In fact, if we fix it at $2/3$, and use Euclid's rule to relate the angles, we can solve ... to obtain $\tan \theta_C = 0.23163$ for the CKM parameter.

ReplyDelete