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In maxwell's eq there is del whose cross and dot products exist.

So what is del in cross vs dot product.

What's the difference when it's just a partial differential operator.

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2 Answers 2

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You're used to the definitions$$U\cdot V=U_iV_i,\,(U\times V)_i:=\epsilon_{ijk}U_jV_k.$$(I've used Einstein notation without worrying about index heights.) Similarly,$$\nabla\cdot V=\partial_iV_i,\,(\nabla\times V)_i:=\epsilon_{ijk}\partial_jV_k.$$Since derivatives don't commute with functions, the consequences are slightly more complicated than for "normal" vectors. For example, in$$\nabla\times(U\times V)=U(\nabla\cdot V)-V(\nabla\cdot U)+\color{red}{(V\cdot\nabla)U-(U\cdot\nabla)V},$$the red terms have no "vanilla" analog. On the other hand, derivatives commute with each other, so e.g. $\nabla\cdot\nabla\times V=0$.

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  • $\begingroup$ In terms of differenetial operators what would del cross vs dot be like? $\endgroup$
    – Mini kute
    Apr 17, 2022 at 21:22
  • $\begingroup$ @Minikute $\nabla\times$ needs to take a vector, which $\nabla\cdot$ can't return. But a third kind of del, the grad of a scalar, $(\nabla\phi)_i:=\partial_i\phi$, is a vector you can pass to curl, viz. $\nabla\times\nabla\phi=0$ (again, this is what you'd expect, because the derivatives commute). Edit: I just noticed your "vs", but I'll leave what I already wrote. Could you clarified what you meant by curl vs div? $\endgroup$
    – J.G.
    Apr 17, 2022 at 21:26
  • $\begingroup$ i don't know vector calculus $\endgroup$
    – Mini kute
    Apr 17, 2022 at 21:39
  • $\begingroup$ I want the formula in partial different form, this the expanded version of that kinky notation $\endgroup$
    – Mini kute
    Apr 17, 2022 at 21:42
  • $\begingroup$ I've given you that. For example, $(\nabla\times V)_1=\partial_2V_3-\partial_3V_2$. $\endgroup$
    – J.G.
    Apr 17, 2022 at 21:45
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They aren't real cross/dot products, it is a notational trick.

$\nabla × $ is called the curl.

$\nabla \cdot$ is called the divergence.

$\nabla \cdot \vec{F} = \frac{\partial F_{x}}{\partial x} \hat i + \frac{\partial F_{y}}{\partial y} \hat j + \frac{\partial F_{z}}{\partial z} \hat k$

$\nabla × \vec{F} = (\frac{\partial F_{z}}{\partial y} - \frac{\partial F_{y}}{\partial z})\hat i +(\frac{\partial F_{x}}{\partial z} - \frac{\partial F_{z}}{\partial x})\hat j +(\frac{\partial F_{y}}{\partial x} - \frac{\partial F_{x}}{\partial y})\hat k $

Curl and divergences computation is the same as a cross product or dot product, but instead multiplying the vectors, you differentiate the component.

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  • $\begingroup$ So how is it different is what i asked ofc it's a notation but that's not my question $\endgroup$
    – Mini kute
    Apr 17, 2022 at 21:19
  • $\begingroup$ Without notation what's the expanded form in partial differential operators. $\endgroup$
    – Mini kute
    Apr 17, 2022 at 21:25
  • $\begingroup$ Look up the definitions of curl and divergence, that will give more than any answer here could. instead of actually computing a cross product or dot product, where you would multiply the "vectors", you apply the differential operator $\endgroup$ Apr 17, 2022 at 21:33
  • $\begingroup$ How is this tied to vectors $\endgroup$
    – Mini kute
    Apr 17, 2022 at 21:38
  • $\begingroup$ Briefly: The curl is a measure of the rotation of a field at a point. And the divergence is a measure of how much the vector field flows out of a point. For the intuition behind the curl and divergence, look up "divergence theorem" or "stokes theorem" $\endgroup$ Apr 17, 2022 at 21:39

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