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If we consider the vector $\left ( A \cdot \nabla \right) \: B$, we have in Cartesian coordinates

$\left ( A \cdot \nabla \right) \: B = \left ( A \cdot \nabla B_x \right ) e_x + \left ( A \cdot \nabla B_y \right ) e_y + \left ( A \cdot \nabla B_z \right ) e_z,$ which gives in full writing:

$\left ( A \cdot \nabla \right) \: B = \left (A_x \frac{\partial \: B_x}{\partial \: x} + A_y \frac{\partial \: B_x}{\partial \: y} + A_z \frac{\partial \: B_x}{\partial \: z} \right )e_x + \left (A_x \frac{\partial \: B_y}{\partial \: x} + A_y \frac{\partial \: B_y}{\partial \: y} + A_z \frac{\partial \: B_y}{\partial \: z} \right )e_y + \left (A_x \frac{\partial \: B_z}{\partial \: x} + A_y \frac{\partial \: B_z}{\partial \: y} + A_z \frac{\partial \: B_z}{\partial \: z} \right )e_z$

Now, if $B=A$ and $A=\left (0, \: 0, \: A_z \right )$ (i.e. $A_x=A_y=0$), we have from the above definition, only the following term along the unit vector $e_z$:

$\left ( A \cdot \nabla \right) \: A = A_z \frac{\partial \: A_z}{\partial \: z}$

But this is equal to zero since $\frac{\partial A_z}{\partial \: z}=0$, despite $A_z \neq 0$.

So we get $\left ( A \cdot \nabla \right) \: A =0$

I would like to know the physical interpretation of this result. What is the significance of this vector being zero in terms of field lines?

Thanks a lot...

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    @ Berci I really don't know. If only 1 field component $H_z$ exists and for $(A \cdot \nabla) A$ to be $0$, then $\frac{\partial A_z}{\partial z}$ should be zero. Is this right to think this way? 1 vote up2012-11-08

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$A\cdot\nabla$ is the derivative in the direction of $A$ (times the length of $A$), so $(A\cdot\nabla) A=0$ means that $A$ itself does not change when you move along $A$. Which in turn implies that the field lines are straight lines.

More generally, the field lines are straight lines if and only if $(A\cdot\nabla)A$ is a (not necessarily constant) multiple of $A$.

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    @yCalleecharan: Not sure I understand the question. But I suspect that any vector field on the form $A(x,y,z)=(f(z),g(z),0)$ is a counterexample, if I haven't misunderstood completely.2012-11-08