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I have been reading about differential forms from various sources. The usual vector-calculus operations of gradient, divergence, curl are nicely represented by the exterior derivative operator $d$ acting on 0-form, 1-form, and 2-forms respectively. For example, if $f$ is a 0-form (i.e. a scalar field) then its exterior derivative gives the gradient 1-form of $f$: $df=\partial_{x_1}f~dx_1+\partial_{x_2}f~dx_2+\partial_{x_3}f~dx_3$. Using the metric tensor for the $\{x_1,x_2,x_3\}$ coordinates I can convert this 1-form into a vector (which is what we engineers usually deal with). So far so good.

But in applications we often have to find the gradient of a vector field, in some convenient coordinate system, say (in my case) prolate spheroidal coordinates $(\xi,\eta,\phi)$. These are related to Cartesian coordinates $(x,y,z)$ by: $$x=d\sqrt{(\xi^2-1)(1-\eta^2)}\cos\phi\\ y=d\sqrt{(\xi^2-1)(1-\eta^2)}\sin\phi\\ z=d\xi\eta$$ in which $d>0$ is a constant. The range of spheroidal coordinates are: $\xi\geq 1,~-1\leq\eta\leq1,~0\leq\phi\leq2\pi$.

I have a vector field $\vec{u}=u_1(\xi,\eta,\phi)\vec{e}_\xi+u_2(\xi,\eta,\phi)\vec{e}_\eta+u_3(\xi,\eta,\phi)\vec{e}_\phi$, in which $\vec{e}$ are unit coordinate vectors. I need to find its gradient $\nabla\vec{u}$ (which is a second-order tensor) in spheroidal coordinates. How do I do that using differential forms? Using the metric tensor for the spheroidal coordinates, I can write $\vec{u}$ as a 1-form, but what next?

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  • $\begingroup$ How do you define the gradient of a vector field? Is it just the tensor $(\partial_i u_j)$? $\endgroup$
    – md2perpe
    Jul 25 '18 at 7:47
  • $\begingroup$ What kind of vector field do you have? Are you sure that it's proper to make it into a 1-form? That's not suitable for every vector field. $\endgroup$
    – md2perpe
    Jul 25 '18 at 8:02
  • $\begingroup$ @md2perpe In Cartesian coordinates gradient of a vector field is indeed $\partial_iu_j$. The vector field I am dealing with is a flow-velocity field induced by a moving spheroidal solid body. I don't want to make it into 1-form for its own sake. I want to find the gradient (in spheroidal coordinates) and I thought turning the vector into a 1-form may be required as an intermediate step (I am only guessing here). $\endgroup$
    – Deep
    Jul 25 '18 at 8:45
  • $\begingroup$ Is it for the viscosity term of the integral form of Navier-Stokes? $$\nu \oint_{\partial \Omega} \partial_i \vec u \, n^i \, dS$$ $\endgroup$
    – md2perpe
    Jul 25 '18 at 10:58
  • $\begingroup$ @md2perpe Not exactly. I am using something called reciprocal theorem to solve the problem, and one of the quantities that appears in it is the gradient of the velocity field. I can always write the vector field in Cartesian coordinates, take the gradient, and transform it to spheroidal coordinates (although I haven't given much thought to the details), but I was wondering if there is a more direct way to do it using differential forms, and without ever entering Cartesian coordinates. $\endgroup$
    – Deep
    Jul 25 '18 at 15:06
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If $\vec n$ is a vector field on $\mathbb R^3$, its "gradient" is actually its total covariant derivative. This makes sense on an arbitrary Riemannian manifold and is usually denoted by $\nabla \vec n$. If you want to compute it for $\mathbb R^3$ in different coordinates, you'll have to first compute the Christoffel symbols of the metric in those coordinates, and then $\nabla\vec n$ is the matrix-valued function whose components are given in any coordinate chart by $$ n^i {}_{;j} = \partial_j \xi^i + \sum_k \Gamma_{jk}^i \xi^k. $$ For details, check out any differential geometry book that treats Riemannian metrics. (You can try my Introduction to Riemannian Manifolds, but there are plenty of other good choices.)

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  • $\begingroup$ +1 I guess there is no straightforward way of doing it using just differential forms. $\endgroup$
    – Deep
    Jul 28 '18 at 4:23
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    $\begingroup$ @Deep: I would say the main reason why the total covariant derivative can't be expressed purely in terms of differential forms is that it's a 2-tensor that's not antisymmetric in its two indices. In the presence of a Riemannian metric, you can lower the upper index and treat it as a covariant tensor field, but it's a symmetric tensor field, not an antisymmetric one. $\endgroup$
    – Jack Lee
    Jul 30 '18 at 22:07

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