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If I define the vector as $V_i=V^T_i+V^L_i$ and the transverse part is defined by $$V^T_i=\Big(\delta_{ij}-\frac{\partial_i\partial_j}{\partial^2}\Big)V_j$$ then is is obvious that $\nabla.V^T=0$ as well as $\nabla \times V^L=0$. What happened if I took the the divergence of the cross product of two different vectors with only transverse component? Are they zero too? For example is $\nabla.(A^T\times V^T)=0?$

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The identity you're looking for is $$ \nabla \cdot (\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot (\nabla \times \mathbf{A}) - \mathbf{A} \cdot (\nabla \times \mathbf{B}) $$ so indeed, if the curl of both factors in the cross product vanish, the divergence of the cross product vanishes, too.

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+1, but one should add that these identities are easier to identify in k-space, since then they are algebraic k identities rather than differential identities (although the two are obviously the same, psychologically, I find k-identifies slightly easier to internalize than cross-product identities). – Ron Maimon Sep 18 '12 at 13:23
Actually using this identity give me the term above! I would like to vanish it somehow! Using this identity again will just take me to the previous step. – aries0152 Sep 18 '12 at 13:53
I don't understand what you want, aries, or more likely, you don't understand something. You assumed $\nabla\times A=0$ and $\nabla\times B=0$, see the first line below your question's displayed equation where $A\to A^T$ and $B\to V^T$, so if those vanish, the combination in the identity vanishes, too, doesn't it? – Luboš Motl Sep 18 '12 at 14:46
If I understand well from the question, aries assumes that $\nabla\cdot A=0$ and $\nabla\cdot B=0$, and not that $\nabla\times A=0$ and $\nabla\times B=0$. – Cristi Stoica Sep 18 '12 at 15:09
Oh, I see, then it doesn't hold, obviously. – Luboš Motl Sep 18 '12 at 15:10

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