What is the definition of Curl in Mathematica?

Question

I have a usual mathematical background in vector and tensor calculus. I was trying to use the differential operators of Mathematica, namely Grad, Div and Curl. According to my knowledge, the definitions of Mathematica for Grad and Div coincides with those usually employed in tensor calculus, that is to say

\begin{align*} \text{grad}\mathbf{T}&:=\sum_{k=1}^{3}\frac{\partial\mathbf{T}}{\partial x_k}\otimes \mathbf{e}_k\\ \text{div}\mathbf{T}&:=\sum_{k=1}^{3}\frac{\partial\mathbf{T}}{\partial x_k}\cdot\mathbf{e}_k \\ \tag{1} \end{align*}

for any tensor $\mathbf{T}$ of rank $n\ge1$. $x_k$'s are Cartesian coordinates and $\mathbf{e}_i$'s are the standard basis for $\mathbb{R}^3$. $\otimes$ and $\cdot$ are the usual generalized outer and inner products which are also defined in Mathematica by Outer and Inner. The usual definition that I know from tensor calculus for the Curl is as follows \begin{align*} \text{curl}\mathbf{T}&:=\sum_{k=1}^{3}\mathbf{e}_k\times\frac{\partial\mathbf{T}}{\partial x_k}. \tag{2} \end{align*} However, it turns out that Mathematica's definition for curl is totally different. For example, it returns the Curl of a second order tensor as a scalar, while according to $(2)$ it should be a second order tensor.

I couldn't find a precise definition of Mathematica for Curl in the documents. I am wondering what this definition is. What is the motivation for this? and How it can be related to the definition given in $(2)$?

Below is a simple piece of code for you to observe the outputs of Mathematica when we apply the Grad, Div and Curl operators to scalar, vector and second order tensor fields. I would like to draw your attention to some observations. Curl of a scalar is returned as a second order tensor, which I don't understand why! Curl of a vector coincides with the usual definition of Curl used in vector calculus. Curl of second order tensor is returned as a scalar, which I don't understand again.

Var={Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]};
Sca=\[Phi][Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]];
Vec={Subscript[v, 1][Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]],Subscript[v, 2][Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]],Subscript[v, 3][Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]]};
Ten=Table[Subscript[T, i,j][Subscript[x, 1],Subscript[x, 2],Subscript[x, 3]],{i,1,3},{j,1,3}];
MatrixForm[Grad[Sca, Var]]
MatrixForm[Grad[Vec, Var]]
MatrixForm[Grad[Ten, Var]]
MatrixForm[Div[Sca, Var]]
MatrixForm[Div[Vec, Var]]
MatrixForm[Div[Ten, Var]]
MatrixForm[Curl[Sca, Var]]
MatrixForm[Curl[Vec, Var]]
MatrixForm[Curl[Ten, Var]]

I will be happy if someone can reproduce the following result for the curl of a second order tensor with Mathematica's Curl function.

\begin{align*} \text{curl}\mathbf{T}&:=\sum_{k=1}^{3}\mathbf{e}_k\times\frac{\partial\mathbf{T}}{\partial x_k}=\sum_{k=1}^{3}\mathbf{e}_k\times\frac{\partial}{\partial x_k}\left(\sum_{i=1}^{3}\sum_{j=1}^{3}T_{ij}\mathbf{e}_i\otimes\mathbf{e}_j\right)\\ &=\sum_{k=1}^{3}\sum_{i=1}^{3}\sum_{j=1}^{3}\frac{\partial T_{ij}}{\partial x_k}(\mathbf{e}_k\times\mathbf{e}_i)\otimes\mathbf{e}_j\\ &=\sum_{k=1}^{3}\sum_{i=1}^{3}\sum_{j=1}^{3}\sum_{m=1}^{3}\epsilon_{kim}\frac{\partial T_{ij}}{\partial x_k}\mathbf{e}_m\otimes\mathbf{e}_j \tag{3} \end{align*}

where $\epsilon_{kim}$ is the LeviCivitaTensor for $3$ dimensions. Consequently, we get

\begin{align*} \left(\text{curl}\mathbf{T}\right)_{mj}=\sum_{k=1}^{3}\sum_{i=1}^{3}\epsilon_{kim}\frac{\partial T_{ij}}{\partial x_k}. \tag{4} \end{align*}

Implementing $(4)$ in Mathematica, we obtain

CurlTen = Table[
   Sum[
    LeviCivitaTensor[3][[k, i, m]] 
     D[Subscript[T, i, j][Subscript[x, 1], Subscript[x, 2], Subscript[x, 3]], {Subscript[x, k]}], {k, 1, 3}, {i, 1, 3}],
   {m, 1, 3}, {j, 1, 3}];
MatrixForm[CurlTen]

Answer 1

The definition used (motivated by exterior calculus) is as follows:

Given a rectangular array $a$ of depth $n$, with dimensions $\{d, ..., d\}$ (so there are $n$ $d$'s) and a list $x = \{x_1, ..., x_d\}$ of variables, then

Curl[a, x] == (-1)^n (n+1) HodgeDual[Grad[a, x], d]

If $a$ has depth $n$, then Grad[a, x] has depth $n+1$, and therefore HodgeDual[Grad[a, x], d] has depth $d-(n+1)$. Clearly then we need $n < d$. Note that $n = 0$ is admitted, i.e. we can take the curl of a scalar function.

In the traditional case of $d=3$ and $n=1$ we have $d-(n+1)=1$ and that's why the curl of a vector is also a vector.

The HodgeDual operation starts by antisymmetrizing its first argument, and hence implicitly we really have something like

Curl[a, x] == (-1)^n (n+1) HodgeDual[Symmetrize[Grad[a, x], Antisymmetric[All]], d]

Finally, one more comment: The definition given assumes that we work with tensors given in components in a Cartesian coordinated and orthonormal basis. If that wasn't the case we would have to insert some additional metric factors. That's handled by the third argument of Curl for a variety of alternative coordinate systems.

Answer 2

I fail to find a reference for the definition used by Curl, but manage to figure out how the Curl is defined.

I'll use Einstein summation convention for simplicity. The defintion used by Curl is summarized as follows:

1. Curl of a scalar / zero-order tensor is defined as

$$\text{curl}\ \phi= \epsilon_{kim}\frac{\partial\phi}{\partial x_m}$$

2. Curl of a vector / first-order tensor is defined as

$$\text{curl}\ \mathbf{v}= \epsilon_{kim}\frac{\partial v_m}{\partial x_i}$$

3. Curl of a matrix / second-order tensor is defined as

$$\text{curl}\ \mathbf{T}=\color{red}{\frac{1}{2}} \epsilon_{kim}\frac{\partial T_{im}}{\partial x_k}$$

Check

ϵ = LeviCivitaTensor[3];

Table[Sum[ϵ[[k, i, m]] D[Sca, Var[[m]]], {m, 3}], {k, 3}, {i, 3}] == 
ϵ.D[Sca, {Var}] == 
TensorContract[ϵ\[TensorProduct]D[Sca, {Var}], {{3, 4}}] == 
Curl[Sca, Var]
(*True*)

Table[Sum[ϵ[[k, i, m]] D[Vec[[m]], Var[[i]]], {m, 3}, {i, 3}], {k, 3}] == 
TensorContract[ϵ\[TensorProduct]D[Vec, {Var}], {{3, 4}, {2, 5}}] == 
Curl[Vec, Var]
(*True*)

1/2 Sum[ϵ[[k, i, m]] D[Ten[[i, m]], Var[[k]]], {k, 3}, {i, 3}, {m, 3}] == 
1/2 TensorContract[ϵ\[TensorProduct]D[Ten, {Var}], {{2, 4}, {3, 5}, {1, 6}}] == 
Curl[Ten, Var]
(*True*)

The second question is simple:

Table[Sum[ϵ[[k, i, m]] D[Ten[[i, j]], Var[[k]]], {k, 3}, {i, 3}], {m, 3}, {j, 3}] == 
TensorContract[ϵ\[TensorProduct]D[Ten, {Var}], {{2, 4}, {1, 6}}] == 
(Curl[#, Var] & /@ Transpose@Ten // Transpose)
(* True *)

Answer 3

I suggest to modify your simple piece of code to make the output more informative. I define a function dim[] which is similar to the Dimensions[] function but is restricted to matrices and arrays. For example, try Dimensions[x+y] which returns {2} but dim[x+y] returns 0.

ClearAll[dim, id];
dim[x : (_List | _StructuredArray)] := Dimensions@x;
dim[x_Div] = $Failed;
dim[x_] = 0;
id[x_] := (Print[dim[x // Expand]]; x // MatrixForm);

Print["Grad"];
id@Grad[Sca, Var]
id@Grad[Vec, Var]
id@Grad[Ten, Var]
Print["Div"];
id@Div[Sca, Var]
id@Div[Vec, Var]
id@Div[Ten, Var]
Print["Curl"];
id@Curl[Sca, Var]
id@Curl[Vec, Var]
id@Curl[Ten, Var]

You wrote

I couldn't find a precise definition of Mathematica for Curl in the documents

The documentation for Curl[] is long and typical of of some of the more obscure Mathematica function documentation. The answer is buried in Properties & Relations where it states:

Curl[f, {x, y, z}] == (-1)^r (r + 1) HodgeDual[ Symmetrize[Grad[f, {x, y, z}], Antisymmetric[All]]]

but HodgeDual[] is even more obscure so I am not surprised that the precise definition is not given. It would be a good idea if it was available to users. So the answer to your first question is that it is still obscure. Even with a definition, one usually tries various examples and examines the results. Sometimes you have to use the function in not obvious ways. In any case, if a function does not do what you want, you can attempt to write your own function that does exactly what you want. You may decide that this is the better option in your situation.

It is good that you seem to have implemented your own version. I suggest an alternative version and a simple test

CurlTen == Transpose@Array[Curl[Ten[[All, #]], Var] &, 3]

which returns True. I hope this helps you in some small way even though I can't answer your first question.

Answer 4

I am limiting myself to fields defined in the three dimensional Euclidean point space. The curl of a tensor can be found in these simple steps:

1. Take the simple composition of the second-order tensor, T, with the LeviCivitaTensor[3]. This is effected by the command,

   Dot[T, LeviCivitaTensor[3]]
   

2. Take the transpose of the divergence of the above result using the command:

   curl[T_] := Transpose[Div[Dot[T, LeviCivitaTensor[3]], {x, y, z}]]
   

The result you get here is the second-order tensor that is the curl of the tensor T.