Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

I cannot seem to prove that the derivative of the duel tensor = 0.

$$ \frac{1}{2}\partial_{\alpha}\epsilon^{\alpha \beta \gamma \delta} F_{\gamma \delta} = 0. $$

Writing this out I get (for some fixed $\alpha$ and $\beta$),

$$ \partial_{\alpha} (\partial_{\gamma}A_{\delta} - \partial_{\delta}A_{\gamma}). $$

From here I get stuck.

Any ideas?

share|improve this question

1 Answer 1

up vote 1 down vote accepted

The duel gets tenser and tenser.


$\Longrightarrow\ \ \epsilon^{xyab}\partial_a\partial_b=0$

More abstractly, if $A^{ab}=-A^{ba}$ and $S_{ab}=S_{ba}$, then $A^{ab}S_{ab}=0$.

share|improve this answer
So basically an antisymmetric tensor multiplied to a symmetric one is zero (although the derivatives are not tensors). Is this the principle at work? –  Shinobii Mar 24 '13 at 19:45
They are not tensors, but two partial derivatives always commute with each other, therefore they are symmetric in their indices and when summed over with an antisymmetric tensor, this gives zero. –  Santiago Casas Mar 24 '13 at 19:52
Makes sense, thanks to both of you for your insight! –  Shinobii Mar 24 '13 at 19:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.