LOG#019. Triangle mnemonics.

Today, we are going to learn some interesting mnemomic tricks using the celebrated Pythagorean Theorem from your young years at the school. We will be using some simple triangles to remember some of the wonderful formulae of Special Relativity. It is quite nice and surprising that basic euclidean trigonometry tools  are very helpful in the abstract realm of relativistic theories, a theory based on non-euclidean geometry! It is one of the great powers of Mathematics. Its amazing ability to model big things with simple pictures and equations. There is no other language like the mathematical language, and despite its terrible looking sometimes, it is awesome and beautiful most of the time. Of course, like it happens in Arts, mathematical beauty can be very subjective. So, you have to be trained in order to admire and love its great features.

The first triangle we are going to study is this one:

If we apply the Pythagorean Theorem to this triangle, we get:

X^2+(c\tau)^2=(ct)^2

This equation is indeed related to the square or “norm” of the spacetime vector (a.k.a., spacetime event)

\mathbb{X}\cdot \mathbb{X}=-c^2\tau^2=\mathbf{X}^2-c^2t^2

and thus, in complete agreement with the above triangle

\mathbf{X}^2+(c\tau)^2=(ct)^2

And it can be easily stated with words or concepts as:

\boxed{(\mbox{SPACE})^2+(\mbox{PROPER TIME})^2=(\mbox{TIME})^2}

You also get:

\sin \varphi= \dfrac{\tau}{t}

\cos \varphi=\dfrac{X}{ct}

\tan \varphi = \dfrac{X}{c\tau}=\dfrac{v}{c}=\beta

i.e.

\boxed{\tan \varphi=\beta}

Moreover,

\gamma^2=\dfrac{1}{1-\beta^2}

Elementary trigonometry yields:

1-\tan ^2\varphi=2-\sec^2\varphi=\dfrac{2\cos^2\varphi-1}{\cos^2\varphi}=\dfrac{\cos 2\varphi}{\cos^2\varphi}

\boxed{\gamma^2=\dfrac{\cos^2\varphi}{\cos 2\varphi}}

In the same way, we can draw other cool triangle:

Now, we can relate the Pythagorean theorem with the squared momenergy vector:

\mathbb{P}\cdot \mathbb{P}=-(mc)^2=\mathbf{P}^2-\dfrac{E}{c^2}

(mc^2)^2+(\mathbf{P}c)^2=E^2

or, in agreement with the last triangle:

(mc^2)^2+(Pc)^2=E^2

In words, it means simply that

\boxed{(\mbox{MASS})^2+(\mbox{MOMENTUM})^2=(\mbox{ENERGY})^2}

There, K(v) is the relativistic kinetic energy we have studied before in this blog. From this triangle, simple trigonometry provides:

\sin \phi = \dfrac{1}{\gamma}

\cos \phi=\dfrac{Pc}{E}

\tan \phi=\dfrac{mc}{P}

\dfrac{E}{mc^2}=\gamma=\dfrac{1}{\sin \phi}

\dfrac{K}{mc^2}=\gamma - 1=\dfrac{1-\sin \phi}{\sin \phi}

\dfrac{P}{mc}=\dfrac{1}{\tan \phi}

A more refined version of the above triangle is given by the following drawing ( the relativistic kinetic energy is now written as T instead of K):

In summary, triangles are cool!

Remark(I): The relationship (mc^2)^2+(\mathbf{P}c)^2=E^2=E^2(P,m) is also called dispersion relation in SR. Some theories beyond the Standard Model of Particle Physics and/or “extended” relativies can modify this equation. However, any known experiment seems to be consistent with SR and this dispersion relationship as far as I know.

Remark(II): Massless particles with m=0, i.e., particles with rest mass equal to zero, satisfy

\boxed{E=Pc}

They are called ultra-relativistic particles. This is the case of massless gauge bosons like photons, gluons and likely gravitons. It was thought that the neutrinos were massless too. In recents years, though, we have managed conclusive evidence that neutrinos are not massless and they have a very tiny mass. They are yet ultra-relativistic particles, since we can indeed use yet the equation E=Pc up to a great degree of precision, but they are ultimately MASSIVE particles, and, for them, the exact mass-energy equation reads

\boxed{E=\sqrt{(mc^2)^2+(\mathbf{P}c)^2}}

This equation is of course the energy-mass general rule for any massive particle like neutrinos, massive fermions, massive gauge bosons, the Higgs particle, and so on.

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