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Verify the vector identity:

$\nabla \cdot (a \times b) = b \cdot \nabla \times a - a \cdot \nabla \times b $

Given that:

$a = (R_a, S_a, T_a)$, $b = (R_b, S_b, T_b)$ and $\nabla = (\frac{\partial} {\partial x},\frac{\partial} {\partial y},\frac{\partial} {\partial z} )$.

Where $R_i, S_i, T_i$ have continuous partial derivatives.

I attempted it on paper but cannot get it to work. Can anyone show me how this is done?

My working:

LHS:

$\nabla \cdot (a \times b) = \frac{\partial}{\partial x}(S_a T_b - S_b T_a) + \frac{\partial}{\partial y}(T_a R_b - R_a T_b) + \frac{\partial}{\partial z}(R_a S_b - R_b S_a) $

RHS:

$\nabla \times a = (\frac{\partial}{\partial y} T_a - \frac{\partial}{\partial z} S_a)\hat i - (\frac{\partial}{\partial x} T_a - \frac{\partial}{\partial z} R_a)\hat j + (\frac{\partial}{\partial x} S_a - \frac{\partial}{\partial y} R_a)\hat k$

$b \cdot (\nabla \times a) = R_b(\frac{\partial}{\partial y} T_a - \frac{\partial}{\partial z} S_a) - S_b(\frac{\partial}{\partial x} T_a - \frac{\partial}{\partial z} R_a) + T_b(\frac{\partial}{\partial x} S_a - \frac{\partial}{\partial y} R_a)$

Now I computed the other part of the RHS as well however I can't see how I would be able to maninipulate it into the LHS.

Anyone have any idea?

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  • $\begingroup$ Just compute it component-wise. It would be best if you were to show your work, so that we can identify your error. $\endgroup$ – Michael Burr Jan 28 '17 at 12:14
  • $\begingroup$ Note that the chain rule derivation cannot work for curl. $\endgroup$ – MathArt Mar 5 '19 at 13:08
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Use the Leibniz rule in the firts term:

$$ \frac{\partial}{\partial x}(S_aT_b - S_bT_a) = T_b\frac{\partial S_a}{\partial x} + S_a\frac{\partial T_b}{\partial x} - T_a\frac{\partial S_b}{\partial x} - S_b \frac{\partial Ta}{\partial x}. $$

Expanding the other two you should get the desired equality.

Another approach can be using index notation.

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