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I How does the Vector Laplacian come about?

  1. Oct 29, 2017 #1
    So the Laplacian of a scalar is divergence of the gradient of a scalar field, and it comes out to the double derivative of the field in X, Y, and Z.

    My book says the Laplacian of a vector field is the double derivative of the X component of the field with respect to X, the double derivative of the Y component with respect to Y, and same with Z.

    I'm not sure how this comes about from the definition of the Scalar Laplacian.
     
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  3. Oct 29, 2017 #2
    The most general definition of the Laplacian is
    $$\nabla^{2}\mathbf{T}=\nabla\cdot\nabla\mathbf{T}$$
    where ##\mathbf{T}## is some tensor, which could be a scalar, vector, matrix, etc. The gradient of ##\mathbf{T}## in Cartesian coordinates is
    $$\nabla\mathbf{T}=\left[\nabla T_{x}\;\;\nabla T_{y}\;\;\nabla T_{z}\right]$$
    where
    $$\nabla T_{x}=\begin{bmatrix}\frac{\partial}{\partial x}T_{x} \\ \frac{\partial}{\partial y}T_{x} \\ \frac{\partial}{\partial z}T_{x} \end{bmatrix}$$
    and so on for ##T_{y}## and ##T_{z}##. The dot product ##\nabla\cdot\nabla\mathbf{T}## is just
    $$\nabla\cdot\nabla\mathbf{T}=\left[\nabla\cdot\nabla T_{x}\;\;\nabla\cdot\nabla T_{y}\;\;\nabla\cdot\nabla T_{z}\right]$$
    Thus if ##\mathbf{T}## is a scalar then the Laplacian returns a scalar, if ##\mathbf{T}## is a vector then the Laplacian returns a vector, and so on.
     
  4. Oct 30, 2017 #3
    How would you do ∇T if T is a vector? I thought gradient only acts on a scalar field.

    Just examining derivatives wrt X and components in X-hat direction to make it simpler.
    So gradient of scalar takes the derivative of field with respect to X, then tacks on the X-hat direction.

    So say vector T is only made of Tx in X direction.

    How would you do the gradient of Tx in X-hat direction? It seems like you would do derivative of Tx in the X-hat direction wrt to X, then tack on another X-hat direction, but that doesn't make any sense.
     
  5. Oct 30, 2017 #4
    I think you should view it within the calculus of differential forms. The Laplacian for any form (and hence also for vector fields, which are in one-to-one correspondence to 1-forms, and for functions, which are ##0##-forms) is per definition given by
    $$ \Delta = d \circ \delta + \delta \circ d \, : \Omega^p(M) \rightarrow \Omega^p(M)$$, where ##\delta## is the codifferential given by ##\delta = (-1)^a* d *## where I don't know the exact expression for ##a## at the moment, but it depends on the dimension of the underlying space and the signature of the metric. ##*## is the hodge dual, which maps ##p##-forms to ##(n-p)##-forms, where ##n## is the dimension of the underlying space.
    Applied to a function, ##\Delta## gives you the usual scalar Laplacian (if ##M=\mathbb{R}^n##), applied to one-forms and identifying one-forms and vectors gives you the vector-Laplacian (also if ##M = \mathbb{R}^n##).

    PS: This is in fact just the same NFuller wrote, but in a different representation, where you do not have to apply some nabla to a tensor. The advantage of the form-approach is, that it is valid for arbitrary (regular) coordinate systems and also for curved spacetimes.
     
    Last edited: Oct 30, 2017
  6. Oct 30, 2017 #5

    vanhees71

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    The vector Laplacian has to be handled with some care. In Cartesian components you simply have
    $$\Delta A^j=\partial_k \partial^k A^j=\delta^{ik} \partial_i \partial_k A^j.$$
    It's not as easy in general coordinates (even not in general orthogonal curvilinear coordinates). To define it unambiguously in a covariant way we note that in Cartesian coordinates
    $$\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\vec{\nabla} (\vec{\nabla} \cdot \vec{A})-\Delta \vec{A},$$
    and thus it's safe to define generally
    $$\Delta \vec{A}=\vec{\nabla} (\vec{\nabla} \cdot \vec{A}) - \vec{\nabla} \times (\vec{\nabla} \times \vec{A}),$$
    where the right-hand side has a well-defined covariant meaning.
     
  7. Oct 30, 2017 #6
    Sorry Metmann, that is way beyond me.
     
  8. Oct 30, 2017 #7
    Let's assume ##\mathbf{T}## has the form ##\mathbf{T}=T_{x}\hat{x}+T_{y}\hat{y}+T_{z}\hat{z}##. The gradient is
    $$\nabla\mathbf{T}=\begin{bmatrix}\frac{\partial}{\partial x}T_{x} & \frac{\partial}{\partial x}T_{y} & \frac{\partial}{\partial x}T_{z} \\ \frac{\partial}{\partial y}T_{x} & \frac{\partial}{\partial y}T_{y} & \frac{\partial}{\partial y}T_{z} \\ \frac{\partial}{\partial z}T_{x} & \frac{\partial}{\partial z}T_{y} & \frac{\partial}{\partial z}T_{z} \end{bmatrix}$$
    So just as the gradient of a scalar produces a vector, the gradient of a vector produces a matrix. Then the Laplacian is
    $$\nabla\cdot\nabla\mathbf{T}=\begin{bmatrix}\frac{\partial^{2}}{\partial x^{2}}T_{x} + \frac{\partial^{2}}{\partial y^{2}}T_{x} + \frac{\partial^{2}}{\partial z^{2}}T_{x} \\ \frac{\partial^{2}}{\partial x^{2}}T_{y} + \frac{\partial^{2}}{\partial y^{2}}T_{y} + \frac{\partial^{2}}{\partial z^{2}}T_{y} \\ \frac{\partial^{2}}{\partial x^{2}}T_{z} + \frac{\partial^{2}}{\partial y^{2}}T_{z} + \frac{\partial^{2}}{\partial z^{2}}T_{z} \end{bmatrix}$$
    Which is just a vector, as expected. As stated by Metmann, this is only valid in Cartesian coordinates and there are more general ways to describe this. However, this is probably the simplest to understand.
     
  9. Oct 31, 2017 #8
    Thanks. How do we know this, though? Is it just definition?
     
  10. Oct 31, 2017 #9
    That's just how it's defined. Although this is technically an abuse of notation, I like to think of the gradient as a vector
    $$\nabla=\begin{bmatrix}\frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{bmatrix}$$
    such that the gradient of a vector is the outer product
    $$\nabla(\mathbf{T}^{T})=\begin{bmatrix}\frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{bmatrix}\begin{bmatrix}T_{x} & T_{y} & T_{z}\end{bmatrix}=\begin{bmatrix}\frac{\partial}{\partial x}T_{x} & \frac{\partial}{\partial x}T_{y} & \frac{\partial}{\partial x}T_{z} \\ \frac{\partial}{\partial y}T_{x} & \frac{\partial}{\partial y}T_{y} & \frac{\partial}{\partial y}T_{z} \\ \frac{\partial}{\partial z}T_{x} & \frac{\partial}{\partial z}T_{y} & \frac{\partial}{\partial z}T_{z} \end{bmatrix}$$
     
  11. Oct 31, 2017 #10
    I see, so the definition of the Laplacian of a Scalar is different than the definition of the Laplacian of a Vector, but they use the same notation?
     
  12. Oct 31, 2017 #11
    The definition also works with a scalar variable such as ##\mathbf{T}=\phi##
    $$\nabla\phi=\begin{bmatrix}\frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{bmatrix}\begin{bmatrix}\phi\end{bmatrix}=\begin{bmatrix}\frac{\partial\phi}{\partial x} \\ \frac{\partial\phi}{\partial y} \\ \frac{\partial\phi}{\partial z} \end{bmatrix}$$
    This produces a vector where the components are just the spatial derivatives of the scalar function ##\phi##, just as one would expect.
     
  13. Oct 31, 2017 #12
    The definition is in fact the same, the notation of usual vector analysis is just not good enough to display this fact ;)

    I admit that my post might have been a bit over top, but consider the following (all valid only for ##\mathbb{R}^n##): You may know the differential of a function: ##df = \sum_i \frac{\partial f}{\partial x^i}dx^i##. This is nothing but the gradient on scalar fields, when identifiying ## a_i dx^i \leftrightarrow (a^0,\ldots,a^n)^T ## with ##a_i = a^i##.

    Now, if you have a vectorfield ##(X^0,\ldots,X^n) ## and identify it with ## X= \sum_i X_i dx^i ## with ##X^i=X_i##, than ## dX = \sum_{i,j} \frac{\partial X^i}{\partial x^j}dx^j\wedge dx^i ##, which is nothing but a generalization of the differential on functions. The "inverse" of ##d## is ##\delta## given for vectors as ##\delta (\sum_i X_i dx^i) = \sum_i \frac{\partial X_i}{\partial x^i}##, which is nothing but the divergence of ##\vec{X}##. Then ## \Delta f = (d + \delta)^2f =\delta(df) + d(\delta f) = \sum_i \frac{\partial^2 f}{\partial (x^i)^2} ##, where ##\delta f = 0##. There is also a formula for the codifferential for objects of the type ##a_{ij}dx^j \wedge dx^i## (derived from the general definition given in my previous post). Then you can define the Laplacian for vectors exactly the same way as for functions, giving you ##\Delta X = \sum_{ij} \frac{\partial^2 X_i}{\partial (x^j)^2}dx^i ## which is nothing but the Vector-Laplacian, i.e. scalar Laplacian in each component.

    In fact you get: ##d(\delta(X)) = \nabla(\nabla \cdot \vec{X}) = grad(div(\vec{x}))## and ##\delta (d(X)) = -\nabla \times (\nabla \times \vec{X}) =- rot(rot(\vec{X}))##, which is what vanhees71 gave.

    The same way (with the same definition) you get a Laplacian for all tensors.
     
    Last edited: Oct 31, 2017
  14. Nov 1, 2017 #13

    vanhees71

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    I always prefer the index calculus (aka Ricci calculus), from which it becomes clear how the various components "naturally" transform under rotations (for 3D "classical" vector analysis in Euclidean space). The gradient is a co-vector, i.e., it has lower indices ##\nabla_j=\partial_j## wrt. to a Cartesian (co-)basis. Only because in Euclidean space wrt. a orthonormal (Cartesian) basis the metric reads ##g_{ij}=\delta_{ij}## the gradient looks like vector components (and this holds true of course only wrt. Cartesian bases).
     
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