# Why is the Electromagnetic Field Strength Antisymmetric?

In the book "Relativity: Special, General and Cosmological", Rindler introduces the Electromagnetic Field Strength by

$$$$F_ \mu = \frac{q}{c} E_ {\mu \nu} U ^\nu$$$$

where $$F_ \mu$$ is a 4-force over a particle and $$U ^ \nu$$ is a 4-velocity of a particle. Rindler states if the 4-Force is rest mass preserving, i.e if the mass $$m$$ of the particle is constant, $$m = const$$, so the 4-force must satisfy

$$$$F_ \mu U^\mu = 0$$$$

then we have

$$$$E_ {\mu \nu} U ^\nu U^\mu = 0$$$$

for all $$U ^ \mu$$.

By this result, rindler states that the electromagnetic tensor must be antisymmetric, $$E_ {\mu \nu} = - E_ {\nu \mu}$$.

EDIT - My problem is: I know that assuming $$E _{\mu \nu}$$ is antisymmetric and $$U^\mu U^\nu$$ is symmetric implies $$E _{\mu \nu} U^\mu U^\nu = 0$$. But if I assume only the symmetry of $$U^\mu U^\nu$$ and that $$E _{\mu \nu}U^\mu U^\nu = 0$$ , how I obtain the result that $$E_{\mu \nu}$$ is antisymmetric, i.e, that $$E_{\mu \nu } = - E_{\nu \mu}$$?

It's a standard result for bilinear maps $$E$$ that $$E(u,u)=0\;\forall u\iff E(u,v) = -E(v,u)\;\forall u,v$$
The part that's slightly less obvious is going from left to right, though it's still just a one-liner: $$0 = E(u+v,u+v) = E(u,u) + E(u,v) + E(v,u) + E(v,v) = E(u,v) + E(v,u)$$
• this explains the relation in a antisymmetric tensor, but don't explain why the condition $E_{\mu \nu} U^\mu U^\nu = 0$ implies this antisymmetry Commented Jan 29, 2020 at 0:49
• @Lil'Gravity The bilinear map in question is $E(x,y)=E_{\mu\nu}x^\mu y^\nu$, so the condition $E_{\mu\nu}U^\mu U^\nu=0$ can be rewritten as $E(U,U)=0$. Commented Jan 29, 2020 at 2:21
• @Lil'Gravity From there we have that $E(U,V)=-E(V,U)$, which is rewritten as $E_{\mu\nu}U^\mu V^\nu=-E_{\mu\nu}V^\mu U^\nu$. Since both $\mu$ and $\nu$ are contracted, we can relabel them however we want, so in particular if we swap $\mu$ with $\nu$ on the right-hand side, we get $E_{\mu\nu}U^\mu V^\nu=-E_{\nu\mu}V^\nu U^\mu$. The components of $U$ and $V$, being scalars, commute, so we switch them and get $E_{\mu\nu}U^\mu V^\nu=-E_{\nu\mu}U^\mu V^\nu$. Since this is true for all $U$ and $V$, we can then say that $E_{\mu\nu}=-E_{\nu\mu}$. Commented Jan 29, 2020 at 2:26