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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.

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    $\begingroup$ Hint: decompose $E$ as the sum of its symmetric and antisymmetric parts. $\endgroup$ – Valter Moretti Aug 5 '20 at 3:43
  • $\begingroup$ @ValterMoretti So if $E_{\mu \nu} U^\mu U^\nu = (E_{(\mu \nu)} + E_{[\mu \nu]} )U^\mu U^\nu = 0 , \ \forall U$. Commuting the dummies index, we have $E_{\nu \mu} U^\nu U^\mu = (E_{(\nu \mu)} + E_{[\nu \mu]}) U^\nu U^\mu = 0, \ \forall U$. Using the properties of Symmetric and Antisymmetric parts of $E$ and commuting $U^\mu$ and $U^\nu$, we get $ E _{\\nu \mu} U^\mu U^\nu = (E_{(\mu \nu)} +-E_{[\mu \nu]} U^\mu U^\nu = 0 , \ \forall U$. So, what I need to do now to complete the demonstration? $\endgroup$ – Lil'Gravity Aug 5 '20 at 21:24
  • $\begingroup$ Summing the first and the last identity you have that $2E_{(ab)}U^aU^b=0$. Here, assuming $U=X+Y$ and using the symmetry of indices, you find $2E_{(ab)}X^aY^b=0$. $\endgroup$ – Valter Moretti Aug 6 '20 at 6:26
  • $\begingroup$ Arbitrariness of $X$ and $Y$ implies that the symmetric part of $E$ must vanish (all its matrix elements are zero). Hence $E$ has only antisymmetric part: it is antisymmetric. $\endgroup$ – Valter Moretti Aug 6 '20 at 6:28
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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.

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  • $\begingroup$ Thanks for the Answer! $\endgroup$ – Lil'Gravity Aug 6 '20 at 11:20
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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.

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