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I have faced difficulties while trying to prove that

$$\nabla\times(\nabla\times\vec{A}) = \nabla(\nabla\cdot\vec{A})-\Delta\vec{A}$$

I don't have any clue how can I start to work with it. Any hint will be helpful.

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  • $\begingroup$ Maybe this question should be asked on Physics.SE? $\endgroup$ – TanMath Aug 1 '15 at 19:29
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    $\begingroup$ Firstly, this isn't a question about Maxwell's equations since this is a general fact about curls; secondly, you should have $\nabla\cdot \vec{A}$ in your second term, not $\nabla\times \vec{A}$. This is not a.. simple.. computation by any means. Have you ever proved any vector calculus identities? There is a very formulaic approach to these problems. $\endgroup$ – Cameron Williams Aug 1 '15 at 19:29
  • $\begingroup$ Have a look at this math.stackexchange.com/questions/1108020/… $\endgroup$ – David Quinn Aug 1 '15 at 19:41
  • $\begingroup$ This kind of vector identity proof is the perfect example for the method I describe(d) here. I'll post a detailed answer if you need more details. $\endgroup$ – wltrup Aug 1 '15 at 19:45
  • $\begingroup$ Possible duplicate of Proof for the curl of a curl of a vector field $\endgroup$ – Brahadeesh Oct 14 '18 at 8:46
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All you need to know is the definitions of (just taking $\mathbb{R}^3$ for simplicity) $$ \nabla=\left[\matrix{\frac{\partial}{\partial x}\\ \frac{\partial}{\partial y}\\ \frac{\partial}{\partial z}}\right]\quad\text{(a symbolic vector)},\qquad \vec A=\left[\matrix{A_1(x,y,z)\\A_2(x,y,z)\\A_3(x,y,z)}\right]\qquad\text{(a vector)} $$ the scalar product and the vector product of two vectors. Well, also $\Delta=\nabla\cdot\nabla$. Then just apply the products formally to the vector coordinates and keep the order so that the derivatives affect functions.

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  • $\begingroup$ i don't understand the phrase '...so that the derivatives affect functions' ?! $\endgroup$ – dsadfas Aug 2 '15 at 11:46
  • $\begingroup$ @dsadfas I mean that the coordinates of the vector $\nabla$ and the vector $A$ do not commute, it is important to keep the order right. For example, $\frac{\partial}{\partial x}f(x,y)$ (a function) is not the same as $f(x,y)\frac{\partial}{\partial x}$ (a derivative operator). $\endgroup$ – A.Γ. Aug 2 '15 at 13:02
  • $\begingroup$ thank you for your answer. i did it (finally) with the mention that I calculate the left side and than the right side to see if both are equal. I don't know how to calculate the right side and than to change the result to be like the left side. $\endgroup$ – dsadfas Aug 2 '15 at 13:31
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For some details missing from the derivation below, see my answer to this question.

First, we write the LHS in terms of its components, using the Kronecker delta and Levi-Civita symbols. (Note: I'll drop the vector arrow on $\vec{A}$ but it's a vector)

$$[\,\nabla \times (\nabla \times A)\,]_{i} = \epsilon_{ijk}\,\partial_{j}\,(\nabla \times A)_{k} = \epsilon_{ijk}\,\partial_{j}\,(\epsilon_{krs}\,\partial_{r}\,A_{s}) $$

The $\epsilon_{krs}$ are constants, so

$$[\,\nabla \times (\nabla \times A)\,]_{i} = \epsilon_{ijk}\,\epsilon_{krs}\,\partial_{j}\,\partial_{r}\,A_{s} $$

But

$$\epsilon_{ijk}\,\epsilon_{krs} = \delta_{ir}\,\delta_{js} - \delta_{is}\,\delta_{jr} $$

so

$$[\,\nabla \times (\nabla \times A)\,]_{i} = (\delta_{ir}\,\delta_{js} - \delta_{is}\,\delta_{jr})\, \partial_{j}\,\partial_{r}\,A_{s} = \delta_{ir}\,\delta_{js}\,\partial_{j}\,\partial_{r}\,A_{s} - \delta_{is}\,\delta_{jr}\,\partial_{j}\,\partial_{r}\,A_{s} $$

Simplifying,

$$[\,\nabla \times (\nabla \times A)\,]_{i} = \partial_{j}\,\partial_{i}\,A_{j} - \partial_{j}\,\partial_{j}\,A_{i} = \partial_{i}\,(\partial_{j}\,A_{j}) - (\partial_{j}\,\partial_{j})\,A_{i} = \partial_{i}\,(\nabla \cdot A) - \nabla^2A_i $$

so

$$\nabla \times (\nabla \times A) = \nabla(\nabla \cdot A) - \nabla^2A $$

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  • $\begingroup$ There are two standard ways to write sums: using the symbol $\sum$ or Einstein's notation, where we contract upper and lower indices. You're not using either making this quite messy and confusing. $\endgroup$ – hjhjhj57 Aug 1 '15 at 22:43
  • $\begingroup$ The vector identity to be shown is valid independently of the coordinate system used, so one could choose any coordinate system to prove it with. If you choose a non-curvilinear coordinate system, then there is no difference between covariant and contravariant coordinates, in which case there is no distinction between upper and lower indices and one might as well write them all as lower indices. $\endgroup$ – wltrup Aug 1 '15 at 22:53
  • $\begingroup$ OFC, but that's just an abuse of notation. This may be trivial for the trained eye, but for someone new to the index notation it may be confusing. $\endgroup$ – hjhjhj57 Aug 1 '15 at 22:54
  • $\begingroup$ Perhaps, but that's a sin that we all commit aplenty around here. More importantly, considering that the OP is most likely at a point in his/her learning curve where he/she hasn't yet learned differential calculus in curvilinear coordinates, introducing this method - which may look already a bit obscure as it is - while also introducing a mystifying notation that will distract him/her from the goal at hand ("why are there upper and lower indices?") is probably not the most pedagogical approach. Let him/her learn and appreciate this with lower indices and then, later, learn the subtleties. $\endgroup$ – wltrup Aug 1 '15 at 23:01
  • $\begingroup$ @wltrup The line before the last one confuses me- why can you change the order of the derivatives? $\partial j \partial i A_j = \partial i (\partial j A_j)$ $\endgroup$ – Whyka Nov 11 '15 at 0:40
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Similar to @wltrup with notational changes, we have

$$\begin{align} \nabla \times \nabla \times \vec A&=(\partial_i \hat x_i)\times (\partial_j \hat x_j)\times (\hat x_k A_k)\\\\ &=\hat x_i\times(\hat x_j\times \hat x_k)\partial_i\partial_j(A_k) \tag 1\\\\ &=\left(\delta_{ik}\hat x_j-\delta_{ij}\hat x_k\right)\partial_i\partial_j(A_k)\tag 2\\\\ &=\hat x_j\partial_j\partial_iA_i-\hat x_k\partial^2_i(A_k) \tag 3\\\\ &=(\hat x_j \partial_j)(\partial_i A_i)-\partial^2_i(\hat x_kA_k) \tag 4\\\\ &=\nabla \nabla \cdot \vec A-\nabla^2\vec A \tag5 \end{align}$$

In going from $(1)$ to $(2)$ we made use of the vector triple product. Note that $\delta_{ij}$ is the Kronecker Delta with $\delta_{ij}=1$ for $i=j$ and $0$ otherwise.

In going from $(2)$ to $(3)$, we used the sifting property of the Kronecker Delta.

In going from $(3)$ to $(4)$, we rearranged terms.

In going from $(4)$ to $(5)$, we recognized the terms of the final result in terms of their tensor representations.

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    $\begingroup$ I believe you're missing a couple of hats in line 1. :) $\endgroup$ – wltrup Aug 1 '15 at 23:10
  • $\begingroup$ @wltrup Thanks! Nice catch. +1 for the comment. $\endgroup$ – Mark Viola Aug 1 '15 at 23:32
  • $\begingroup$ @dsadfas Please let me know how I can improve my answer. I really just want to give you the best answer I can. $\endgroup$ – Mark Viola Aug 6 '15 at 0:32
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I am coming to this party about three years too late, but the answers posted here, and some here as well: enter link description here, may leave readers with the false impression that the vector Laplacian acting on a vector field (appearing on the right-hand side) always takes the form $$\begin{align} \nabla^2 \vec{A} = \nabla^2(A_i)\,\hat{e}_i ~~~ {\text{(summation convention in force)}} \tag 1 \end{align}$$ in an arbitrary basis and coordinate system of $\mathbb{R}^3$, where $\nabla^2$ is the Laplacian operator in the coordinates associated with the basis. This is true only in a cartesian coordinate system and its orthonormal basis. For example, in the cylindrical coordinate system $\{\rho, \phi, z\}$, with orthogonal (but not orthonormal) basis $\{\hat{e}_\rho, \hat{e}_\phi. \hat{e}_z\}$, it might be supposed from these derivations that $$\begin{align} \nabla^2 \vec{A} = \nabla^2(A_\rho)\,\hat{e}_\rho + \nabla^2(A_\phi)\,\hat{e}_\phi + \nabla^2(A_z)\,\hat{e}_z, \end{align}$$ where $$\begin{align} \nabla^2(f) = \frac{1}{\rho}\frac{\partial}{\partial \rho}\left(\rho\frac{\partial f}{\partial \rho}\right) + \frac{1}{\rho^2}\frac{\partial^2 f}{\partial \phi^2} + \frac{\partial^2 f}{\partial z^2} \end{align}$$ is the Laplacian operator in cylindrical coordinates acting on a scalar function $f$. This is quite false, as I first learned long ago from the textbook ``Mechanics of Deformable Bodies'', by the master mathematical physicist Arnold Sommerfeld. In the answers to Problems I.3 and I.4 of the book, he lays bare the fallacy of this assumption. The first two components of $\nabla^2{\vec{A}}$ contain extra terms besides $\nabla^2(A_\rho)$ and $\nabla^2(A_\phi)$.

In fact, Sommerfeld clearly states in Chapter I, page 23, what seems to have been overlooked, or at any rate, not emphasized, in many of the answers posted here: the formula $\nabla \times (\nabla \times \vec{A}) = \nabla(\nabla\cdot \vec{A}) - \nabla^2\vec{A}$ is an identity in the sense of (1) being true, only in cartesian coordinates. For any other coordinate system and basis, the formula serves to define $\nabla^2\vec{A}$ by $$\begin{align} \nabla^2\vec{A} := \nabla(\nabla\cdot \vec{A}) - \nabla \times (\nabla \times \vec{A}), \end{align}$$ each term on the right-hand side being a well-defined operation in any coordinate system. So the upshot is that the formula can be shown to be true only in a cartesian coordinate system (as was done in the answers here); in any other coordinate system the formula instead defines the vector Laplacian acting on a vector field.

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