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...

  • 1
    $\begingroup$ Why is $\displaystyle\frac{\partial A_z}{\partial z}=0$ necessarily? $\endgroup$ – Berci Nov 8 '12 at 13:15
  • $\begingroup$ @ 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 up $\endgroup$ – yCalleecharan Nov 8 '12 at 13:20

$A\cdot\nabla$ 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 $A$.

  • $\begingroup$ @ Harald Hanche-Olsen . Thanks. If my $A$ is a magnetic field vector such as the magnetic field intensity $H$, then your answer implies that there is no curvature in the field lines. 1 vote up. $\endgroup$ – yCalleecharan Nov 8 '12 at 12:08
  • $\begingroup$ @ Harald Hanche-Olsen. Can we say that $(A \cdot \nabla) A \neq 0$ if we have the field effects in more than one direction? My post showed that having only one component $H_z$ in that case leads to $(A \cdot \nabla) A = 0$. $\endgroup$ – yCalleecharan Nov 8 '12 at 12:15
  • $\begingroup$ @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. $\endgroup$ – Harald Hanche-Olsen Nov 8 '12 at 17:24

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