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Sometimes I see this expression in my physics lectures where the professor writes:

$(\nabla \vec a)\vec b = (\vec b \nabla)\vec a$.

I wasn't sure about it, and I tried to explicitly calculate each expression (using the components) and as I expected, I got two different expressions which are not equal.

If I have to give an example, it would be the Tailor expansion of a magnetic field around a position vector $\vec r_0=0$:

$$\vec B(\vec r) = \frac 1 {0!}\vec B(\vec r)|_{\vec r= \vec r_0}(\vec r - \vec r_0)^0 + \frac 1 {1!} \nabla_\vec r \vec B(\vec r)|_{\vec r= \vec r_0}(\vec r - \vec r_0)^1 +...$$

Then by plugging in $\vec r_0=0$ I get:

$$\vec B(0) + \nabla_\vec r \vec B(\vec r)|_{\vec r= 0}\vec r $$

But somehow the 2nd term is expressed as:

$$\vec B(0) + (\vec r \nabla_\vec r) \vec B(\vec r)|_{\vec r= 0}$$

Can someone help me understand this symbolic ?

imbAF
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  • $$(\nabla a)\cdot\vec{b} = \sum_i (\partial_i a)b_i = \sum_i b_i\partial_i a = (\vec{b}\cdot\nabla)a$$ $a$ is not and cannot be a vector, but this equation can be repeated $3$ times for each component of a vector and everything works about by the linearity of both the dot product and derivative. – Ninad Munshi Dec 05 '21 at 12:54
  • refresh your page to see my updated comment. – Ninad Munshi Dec 05 '21 at 12:57
  • But in my example about the magnetic field, each term is certainly a vector – imbAF Dec 05 '21 at 12:57
  • Does my updated comment not answer your question? I'm pretty sure it does – Ninad Munshi Dec 05 '21 at 13:00

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