## My first physics paper

I was looking through a pile of old papers and I discovered a copy of my very first physics paper which I published in my 4th year in college - quite a long time ago. I remember I was attending a standard electromagnetism class and the teacher said: to obtain the invariants of the electromagnetic filed, compute the characteristic polynomial:

$$det(F -I_4 x) = 0$$

where $$F$$ is electromagnetic field tensor. The coefficients are the invariants.

I went home and double checked the math and surprise: the equation implied:

$$x^4 - (B^2 +E^2 ) x^2 + {(E\cdot B)}^2 = 0$$

So the next lecture I confronted the professor and showed that you get $$B^2 + E^2$$ instead of $$B^2 - E^2$$ and his statement was wrong. To my surprise he said that he knew it was wrong, but it was close to the answer and it must have a kernel of truth but he did not know how to fix it. Now in college I never liked functional analysis (and only recently I developed the right intuition in the area) but I was always good at algebra and I could come up with the answer to any algebra problem at first sight. So I said to the teacher: I know how to fix it: just make the electric field imaginary. The professor than said: OK, write it up and if you can do it we'll write a paper.

This was not that easy -it took me an afternoon- but I worked it at home starting backwards. Suppose you have a $$4 \times 4$$ diagonal matrix $$J$$ with the diagonal elements $$(1,1,1,i)$$ Then if you multiply the electromagnetic tensor $$F$$ with $$J$$ on the left and on the right: $$J F J$$ then you are in business with $$det(J F J -I_4 x) = 0$$. But why does this work? What is happening is a transition from a pseudo-orthogonal group to an orthogonal group. Here is how:

The electromagnetic tensor changes under a Lorentz boost $$\Lambda$$ like this:

$$F^{'} = \Lambda F {\Lambda}^t$$

where

$${\Lambda}^t Y \Lambda = Y$$

with $$Y$$ the diagonal matrix $$(1,1,1,-1)$$

Now if we sandwich $$F$$ with $$J$$ we get:

$$J F^{'} J = J \Lambda F {\Lambda}^t J = (J \Lambda J^{-1}) (J F J) {(J \Lambda J^{-1})}^t$$

But what about $$\Lambda$$ ? Here is the fireworks: $$Y = J J$$

So let us compute $${(J \Lambda J^{-1})}^t (J \Lambda J^{-1})$$:

$${(J \Lambda J^{-1})}^t J \Lambda J^{-1} = J^{-1} {\Lambda}^t J J \Lambda J^{-1} = J^{-1} {\Lambda}^t Y \Lambda J^{-1} =$$

$$= J^{-1} Y J^{-1} = J^{-1} J J J^{-1} = 1$$

Bingo: we manage to get an orthogonal transformation from a pseudo-orthogonal one. This is reminiscent of Dirac's trick because we basically take the square root of $$Y$$ and we get two $$J$$ matrix instead. This is how people used to write imaginary $$ict$$ time and preserve the usual orthogonal rotations. And for orthogonal group there is an elementary linear algebra theorem which states that the only invariants of a similarity transformation (which is how the electromagnetic tensor changes under a Lorentz transformation in the new complexified orthogonal transformation) are the coefficients of the characteristic polynomial.

So lo and behold, my first physics paper appeared in the Romanian Journal of Physics, Volume 38, Number 9, pages 873-875 in 1993. The teacher's name was Andrei Ludu and he was also doing seminars on quantum groups. At that time I had no clue on Lie algebras, let alone on quantum groups and I could not find any motivation or intuition to a bunch of very long and way too abstract things.

But guess what? Quantum groups are actually Hopf algebras and they have very interesting physics applications. I'll talk about it next time.