The function exterior_diff( T, coord ) computes the exterior derivative of the (covariant) components of the anti-symmetric tensor T and returns them as a tensor_type of rank equal to $\mathrm{rank}\left(T\right)+1$.  The result is totally anti-symmetric and uses the antisymmetric indexing function (unless T is a scalar).

If T is not a scalar or vector, it should be indexed using the antisymmetric indexing function.  If it is not indexed using the antisymmetric indexing function and it is not a scalar or vector, then the routine will determine its anti-symmetric part and compare that with the tensor to see if the components are really totally anti-symmetric. If it is found not to be completely anti-symmetric, the routine exits with an error.

The result is computed first by finding the first partials of the tensor components and then antisymmetrizing them.

Simplification:  This routine uses the `tensor/lin_com/simp` and `tensor/partial_diff/simp` routines for simplification purposes.  The simplification routines are used internally by the partial_diff and antisymmetrize routines as they are called by exterior_diff.  By default, `tensor/lin_com/simp` and `tensor/partial_diff/simp` are initialized to the `tensor/simp` routine.  It is recommended that these routines be customized to suit the needs of the particular problem.

This command is part of the tensor package, so it can be used in the form exterior_diff(..) only after executing the command with(tensor). However, it can always be accessed through the long from of the command by using tensor[exterior_diff](..).

$\mathrm{with}\left(\mathrm{tensor}\right)\:$

$\mathrm{coord}≔\left[x\,y\,z\right]\:$

$\mathrm{tc}≔\mathrm{array}\left(\mathrm{antisymmetric}\,1..3\,1..3\right)\:$

$$\begin{gathered} \mathbf{for}i\mathbf{to}3\mathbf{do} \\ \mathbf{for}j\mathbf{from}i+1\mathbf{to}3\mathbf{do} \\ \mathrm{tc}\left[i,j\right]≔\mathrm{cat}\left('t'\,i\,j\right)\left(x\,y\,z\right) \\ \mathbf{end}\mathbf{do} \\ \mathbf{end}\mathbf{do}\; \end{gathered}$$$T≔\mathrm{create}\left(\left[-1\,-1\right]\,\mathrm{eval}\left(\mathrm{tc}\right)\right)$

$T≔table\left(\left[\mathrm{compts}=\left[\begin{array}{ccc}0& \mathrm{t12}\left(x\,y\,z\right)& \mathrm{t13}\left(x\,y\,z\right)\\ -\mathrm{t12}\left(x\,y\,z\right)& 0& \mathrm{t23}\left(x\,y\,z\right)\\ -\mathrm{t13}\left(x\,y\,z\right)& -\mathrm{t23}\left(x\,y\,z\right)& 0\end{array}\right]\,\mathrm{index\_char}=\left[\mathrm{-1}\,\mathrm{-1}\right]\right]\right)$

$\mathrm{ex\_der}≔\mathrm{exterior\_diff}\left(T\,\mathrm{coord}\right)$

$\mathrm{ex\_der}≔table\left(\left[\mathrm{compts}=array\left(\mathrm{antisymmetric}\,1..3\,1..3\,1..3\,\left[\left(1\,1\,1\right)=0\,\left(1\,1\,2\right)=0\,\left(1\,1\,3\right)=0\,\left(1\,2\,1\right)=0\,\left(1\,2\,2\right)=0\,\left(1\,2\,3\right)=\frac{\partial }{\partial z}\mathrm{t12}\left(x\,y\,z\right)-\frac{\partial }{\partial y}\mathrm{t13}\left(x\,y\,z\right)+\frac{\partial }{\partial x}\mathrm{t23}\left(x\,y\,z\right)\,\left(1\,3\,1\right)=0\,\left(1\,3\,2\right)=-\frac{\partial }{\partial z}\mathrm{t12}\left(x\,y\,z\right)+\frac{\partial }{\partial y}\mathrm{t13}\left(x\,y\,z\right)-\frac{\partial }{\partial x}\mathrm{t23}\left(x\,y\,z\right)\,\left(1\,3\,3\right)=0\,\left(2\,1\,1\right)=0\,\left(2\,1\,2\right)=0\,\left(2\,1\,3\right)=-\frac{\partial }{\partial z}\mathrm{t12}\left(x\,y\,z\right)+\frac{\partial }{\partial y}\mathrm{t13}\left(x\,y\,z\right)-\frac{\partial }{\partial x}\mathrm{t23}\left(x\,y\,z\right)\,\left(2\,2\,1\right)=0\,\left(2\,2\,2\right)=0\,\left(2\,2\,3\right)=0\,\left(2\,3\,1\right)=\frac{\partial }{\partial z}\mathrm{t12}\left(x\,y\,z\right)-\frac{\partial }{\partial y}\mathrm{t13}\left(x\,y\,z\right)+\frac{\partial }{\partial x}\mathrm{t23}\left(x\,y\,z\right)\,\left(2\,3\,2\right)=0\,\left(2\,3\,3\right)=0\,\left(3\,1\,1\right)=0\,\left(3\,1\,2\right)=\frac{\partial }{\partial z}\mathrm{t12}\left(x\,y\,z\right)-\frac{\partial }{\partial y}\mathrm{t13}\left(x\,y\,z\right)+\frac{\partial }{\partial x}\mathrm{t23}\left(x\,y\,z\right)\,\left(3\,1\,3\right)=0\,\left(3\,2\,1\right)=-\frac{\partial }{\partial z}\mathrm{t12}\left(x\,y\,z\right)+\frac{\partial }{\partial y}\mathrm{t13}\left(x\,y\,z\right)-\frac{\partial }{\partial x}\mathrm{t23}\left(x\,y\,z\right)\,\left(3\,2\,2\right)=0\,\left(3\,2\,3\right)=0\,\left(3\,3\,1\right)=0\,\left(3\,3\,2\right)=0\,\left(3\,3\,3\right)=0\right]\right)\,\mathrm{index\_char}=\left[\mathrm{-1}\,\mathrm{-1}\,\mathrm{-1}\right]\right]\right)$

$\mathrm{op}\left(1\,\mathrm{get\_compts}\left(\mathrm{ex\_der}\right)\right)$

$\mathrm{antisymmetric}$