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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?

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1 Answer 1

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The duel gets tenser and tenser.

$\epsilon^{xyab}\partial_a\partial_b=(-\epsilon^{xyba})\partial_a\partial_b=(-\epsilon^{xyba})\partial_b\partial_a=(-\epsilon^{xycd})\partial_c\partial_d=-\epsilon^{xyab}\partial_a\partial_b$

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

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

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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
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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
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