The scalar Laplacian is defined as $\Delta A =\nabla\cdot\nabla A $. This makes conceptual sense to me as the divergence of the gradient… but I'm having trouble connecting this concept to a vector Laplacian because it introduces a double curl as $\Delta \mathbf{A}=\nabla(\nabla\cdot\mathbf{A}) – \nabla\times(\nabla\times \mathbf{A})$. I understand what curl is but I don't understand why it's introduced in the vector Laplacian.
[Math] Why does the vector Laplacian involve the double curl of the vector field
vector analysisVector Fields
Related Solutions
In geometric calculus literature (see, for example, Doran and Lasenby), such a function is called monogenic. Monogenic functions are generalizations of complex analytic (or holomorphic) functions. This condition is strictly stronger than being harmonic--all monogenic functions are harmonic, but not all harmonic functions are monogenic.
The term monogenic is not restricted to vector fields, as well; a scalar field with zero gradient would also be referred to as monogenic.
You can also consult this page by Gull, Lasenby, and Doran.
Edit: Phrased in the language of geometric calculus, we define the vector derivative of a vector field $u$ as $\nabla u$, given by
$$\nabla u = \nabla \cdot u + \nabla \wedge u$$
When $\nabla u = 0$, then $\nabla^2 u = \nabla \wedge (\nabla \cdot u) + \nabla \cdot (\nabla \wedge u) = 0$ as well, fulfilling the harmonic condition.
Yes, there's a more elegant way! It uses the language of differential forms, which has replaced the 19th-century language of gradients, divergences, and curls in modern geometry. You can appreciate the simplicity of this language even before learning how to read it:
For any 1-form $A$, $$\begin{align*} (\star d)(\star d)A & = (\star d \star)dA \\ \operatorname{curl} \operatorname{curl} A & = d^\dagger dA \end{align*}$$ Recalling that $\Delta = d d^\dagger + d^\dagger d$, we see that $$\begin{align*} \operatorname{curl} \operatorname{curl} A & = -d d^\dagger A + \Delta A \\ & = d(\star d \star)A + \Delta A \\ & = \operatorname{grad} \operatorname{div} A + \Delta A \end{align*}$$
This is the identity you wanted to prove, where $-\Delta$ is the vector Laplacian.
My favorite place to learn about differential forms is in Chapters 4 and 5 of Gauge Fields, Knots, and Gravity by John Baez and Javier Muniain.
Here's a rough glossary that should help you move between the language of differential forms and the old language of vector calculus. I'll start by telling you the various kinds of differential forms, and the basic operations on them.
- In $n$-dimensional space, there are $n+1$ kinds of differential forms, from 0-forms to $n$-forms. You can think of a $k$-form as a $k$-dimensional density. A 0-form is a function, and a 1-form is a row-vector field (in coordinate-free language, a dual-vector field).
- The exterior derivative is an operation $d$ that turns $k$-forms into $(k + 1)$-forms. As its name suggests, it generalizes the operation of differentiating a function.
- The Hodge star is an operation $\star$ that turns $k$-forms in to $(n - k)$-forms. It comes from the dot product between column vectors. In fact, the Hodge star encodes the same geometric information as the dot product: if you know one, you can reconstruct the other.
- The codifferential is an operation $d^\dagger$ that turns $k$-forms into $(k - 1)$-forms. In an odd-dimensional space, like ordinary 3-dimensional space, applying $d^\dagger$ to a $k$-form is the same as applying $(-1)^k \star d \star$. In an even-dimensional space, $d^\dagger$ always acts like $-\star d \star$.
If you keep in mind that a 0-form is a function and a 1-form is a row-vector field, all the familiar operations of vector calculus can be written in terms of the ones above.
- The gradient of a function $f$ is the 1-form $df$.
- The curl of a 1-form $A$ is the 1-form $\star dA$.
- The divergence of a 1-form $A$ is the function $\star d \star A$.
- The Laplacian of a function or 1-form $\omega$ is $-\Delta \omega$, where $\Delta = dd^\dagger + d^\dagger d$. The operator $\Delta$ is often called the Laplace-Beltrami operator.
With this glossary in hand, you should be able to follow the steps of the calculation above, which is mostly just translating back and forth between languages. The only tricky bit is getting the sign right when you rewrite $d^\dagger$ as $\pm \star d \star$: you have to figure out what kind of form $d^\dagger$ is being applied to.
Best Answer
The definition of Laplacian operator for either scalar or vector is almost the same. You can see it by noting the vector identity $$\nabla\times(\nabla\times A)=\nabla(\nabla\cdot A)-(\nabla\cdot\nabla)A$$ Plugging it into your definition produces still $$\Delta A=(\nabla\cdot\nabla)A$$