Demonstration of Electromagnetic Tensor antisymmetry

Question

I've already made a post about this topic here, but I realized that I didn't understand the explanation on that post. in Chapter 7 of Rindler's book on relativity, in section about electromagnetic field tensor, he states that

and introducing a factor 1/c for later convenience, we can ‘guess’ the tensor equation, $$ F_\mu= \frac{q}{c} E_{\mu \nu} U^\nu$$ thereby introducing the electromagnetic field tensor$$E_{\mu \nu}$$ We would surely want the force $F\mu$ to be rest-mass preserving, which, according to (6.44) and (7.15), requires $$F_\mu U^\mu = 0$$. So we need $$E_{\mu \nu} U^\mu U^\nu = 0$$ for all $ U^\mu$ , and hence the antisymmetry of the field tensor $$E_{\mu \nu}= −E_{\nu \mu}$$\

. . . I'm really confused about the correct way to show that the equation $E_{\mu \nu} U^\mu U^\nu = 0$ implies the fact that $E_{\mu\nu}$ is antisymmetric tensor. What is the correct demonstration of this implication?

OBS: i've saw some posts answering this kind of question with bilinear maps notation, instead of component notation. If possible, please make some demonstration using the index notation as in the post.

Answer 1

First decompose $E$ as the sum of its symmetric and antisymmetric parts: $$E_{ab} = E_{(ab)} + E_{[ab]}\:.$$ Now the idea is to prove that $$E_{(ab)} =0\tag{0}$$ so that $E_{ab} = E_{[ab]}$ is antisymmetric.

To this end observe that, in our hypothesis, $$0= E_{ab}U^aU^b = E_{(ab)}U^aU^b + E_{[ab]}U^aU^b\:,\tag{1}$$ where $$E_{[ab]}U^aU^b= E_{[ba]}U^bU^a= -E_{[ab]}U^bU^a = -E_{[ab]}U^aU^b =0\:.$$ Here (1) implies $$E_{(ab)}U^aU^b =0\:.\tag{2}$$ Writing $U=X+Y$, we have from (2) $$E_{(ab)}X^aX^b + E_{(ab)}Y^aY^b + 2E_{(ab)}X^aY^b=0\:.\tag{3}$$ where we have used $$E_{(ab)}X^aY^b= E_{(ab)}Y^aX^b$$ as a consequence of the symmetry of $E_{(ab)}$. Using again (2) in (3) for $U=X$ and $U=Y$, we obtain, for every choice of $X$ and $Y$, $$E_{(ab)}X^aY^b=0\:.$$ In other words, all matrix elements of the matrix of elements $E_{(ab)}$ vanish, so that $E_{(ab)}=0$ and (0) is true conluding the proof.

Answer 2

It's probably clearer to go backwards:

$$ E_{\mu\nu} = -E_{\nu \mu}\\ \Leftrightarrow E_{\mu\nu}U^\mu V^\nu = -E_{\nu\mu}U^\mu V^\nu \hspace{1em} \forall U, V\\ \Leftrightarrow E_{\mu\nu}U^\mu V^\nu = -E_{\mu\nu}U^\nu V^\mu \hspace{1em} \forall U, V \hspace{1em}\text{Relabel RHS} \mu \leftrightarrow \nu\\ \Leftrightarrow E_{\mu \nu} (U^\mu + V^\mu)(U^\nu + V^\nu) = 0 \hspace{1em} \forall U, V $$

Now we recognise that the addition of $V$ in the final equation doesn't actually change the condition, and without losing any generality we may take it to be zero. So, we establish $E_{\mu\nu}U^\mu U^\nu \Leftrightarrow E_{\mu\nu} = - E_{\nu\mu}$

The only property used here is linearity of each tensor.