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What's the physical meaning of $\vec A \times \vec \nabla$?

Gradient represents a field. What if we try we write "opposite" of that? I know that if direction of $\vec \nabla \times \vec A$ is positive z axis than $\vec A \times \vec \nabla$ will be negative z axis. But I wonder if it really gives a physical meaning. gradient represents a field. and divergence represent a field where all the pointing vector is diverging from the initial point. Curl is similar to field also (I don't have any idea how can I express my imagination for curl in sentences).

But what does it represent when we move gradient to back. $\vec{A} \cdot \vec \nabla$ and $\vec A \times \vec \nabla$

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    $\begingroup$ There is nothing wrong with that expression. You are free to write it. This quantity is a derivative operator and it is indeed used quite often in physics. $\endgroup$
    – Prahar
    Nov 15, 2021 at 10:11
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    $\begingroup$ It is ok to write it but usually $\vec \nabla \times \vec A \neq \vec A \times \vec \nabla $ $\endgroup$
    – Mauricio
    Nov 15, 2021 at 10:22
  • $\begingroup$ Possible duplicate: physics.stackexchange.com/q/531214/195139 $\endgroup$
    – Sandejo
    Nov 17, 2021 at 1:14
  • $\begingroup$ @Sandejo Yes! But I didn't find expected answer there.... $\endgroup$
    – Unknown
    Nov 17, 2021 at 5:40

2 Answers 2

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As written previously by @J.G, the following is the operator which could be applied to a function $$ \vec{A}\times \vec{\nabla} = e_{ijk}A_j \nabla_k,$$ while the following is a vector product of the del operator and a vector $$ \vec{\nabla} \times \vec{A} = e_{ijk} \nabla_j (A_k).$$ Normally they are not the same

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It's a vector-valued differential operator viz.$$(\vec{A}\times\vec{\nabla})\phi=\vec{A}\times\vec{\nabla}\phi,\,(\vec{A}\times\vec{\nabla})\cdot\vec{B}=\vec{A}\cdot\vec{\nabla}\times\vec{B},\,(\vec{A}\times\vec{\nabla})\times\vec{B}=(\color{blue}{\vec{\nabla}\vec{B}})^T\vec{A}-(\vec{\nabla}\cdot\vec{B})\vec{A}$$with the blue expression a matrix.

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  • $\begingroup$ I have edited my question.. $\endgroup$
    – Unknown
    Nov 16, 2021 at 17:03

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