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I have found numerous definitions for the divergence of a tensor which makes me confused as to trust which one to use.

In Itskov's Tensor Algebra and Tensor Analysis for Engineers, he begins with Gauss's theorem to define

\begin{equation} \text{div} ~\boldsymbol{S} = \lim_{V \to 0} \frac{1}{V} \int_{\partial V} \boldsymbol{S} ~\boldsymbol{n} ~da \end{equation}

which, resorting to some coordinates system, gives \begin{equation} \text{div} ~\boldsymbol{S} = \boldsymbol{S}_{,i} ~\boldsymbol{g}^i = S_{j}^{~~i} |_i ~\boldsymbol{g}^j \end{equation}

I actually like this definition because of its naturalness from beginning with Gauss's theorem. However, it requires choosing a basis. To define a coordinate-free divergence, I have come across multiple definitions:

One from this wiki article defines the divergence as

\begin{equation} (\boldsymbol{\nabla \cdot S}) \boldsymbol{\cdot a} = \boldsymbol{\nabla \cdot} ~(\boldsymbol{S ~a}) \end{equation}

where $\boldsymbol{a}$ is an arbitrary constant vector. This gives

\begin{equation} \boldsymbol{\nabla \cdot S} = S^{i}_{~~j} |_i ~\boldsymbol{g}^j \end{equation}

where the first index is contracted. Yet, another wiki article defines

\begin{equation} (\boldsymbol{\nabla}\cdot\boldsymbol{T})\cdot\mathbf{c} = \boldsymbol{\nabla}\cdot\left(\mathbf{c}\cdot\boldsymbol{T}^\textsf{T}\right) \end{equation}

to give the exact same result as the other wiki article. (Here I presume that Reddy's notation were used, where he uses dot product for denoting any product he can find! One problem with Reddy's notation is that I cannot figure out how he dot products a vector into a dyad, as in $\boldsymbol{e}_k \cdot \boldsymbol{e}_i\otimes\boldsymbol{e}_j$, so please do not advise me using his notation. This being said, I don't know what $\mathbf{c}\cdot\boldsymbol{T}^\textsf{T}$ means; is it $\mathbf{c}~\boldsymbol{T}^\textsf{T}$ where $\boldsymbol{T}^\textsf{T}$ is acting from the left on the vector $\mathbf{c}$? If so, I don't think this holds for a general curvilinear basis. I guess $\mathbf{c}~\boldsymbol{\cdot}\boldsymbol{T}^\textsf{T} = \boldsymbol{c}^\textsf{T}\boldsymbol{T}^\textsf{T}$ is more appropreate, but I don't reckon Reddy means this way.) This article also says that Itskov's result (contracting the second index) is actually true only for symmetric tensors, which Itskov never assumes.

Abeyratne's lecture notes (p. 64) uses this definition

\begin{equation} (\text{div} ~\boldsymbol{T})\cdot\mathbf{c} = \text{div} \left(\boldsymbol{T}^\textsf{T}\mathbf{c}\right) \end{equation}

where he claims that the second index gets contracted. I don't know whether $\text{div}$ and $\boldsymbol{\nabla \cdot}$ are different or the same.

Ogden's "Nonlinear Elastic deformations" puts it in a very nice way: that there are three possible contractions for the gradient of a 2nd rank tensor $\boldsymbol{\nabla}\otimes \boldsymbol{T}$, so defining the divergence is a matter of convention. He contracts the first index. But still, which one should one choose for a throughout consistency in his calculations. What is the definition of divergence?

Kelly's lecture notes were a little helpful, yet because of its different notations from other, I always get caught wondering if I am doing the right way. For example, he finds for the gradient of a tensor field that $\text{grad}~\boldsymbol{v} = (\boldsymbol{\nabla}\otimes\boldsymbol{v})^\textsf{T}$, but Ogden finds it with the transpose, and I believe they have used the same definitions to start with, namely the directional derivative. This will make much mess for me, as to define the divergence of the vector field whether as $\boldsymbol{\nabla\cdot v} = \text{tr} (\boldsymbol{\nabla}\otimes\boldsymbol{v})^\textsf{T}$ or as $\boldsymbol{\nabla\cdot v} = \text{tr} (\boldsymbol{\nabla}\otimes\boldsymbol{v})$.

Please help me organize my mind on the subject, and share with me your experience regarding the same notation conflictions and how you have overcome them.

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    $\begingroup$ See definition of tensor Laplacian for a definition of the divergence of a tensor field. In general, one has to specify the valence/type of the tensor field, and which slots to take the trace over. For vector fields, there's only one choice (also included in the general definition above), but for computation it's much better to use the Voss-Weyl formula, which I prove here. With this definition one can indeed prove Gauss' divergence theorem. $\endgroup$
    – peek-a-boo
    Jul 26, 2022 at 8:13
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    $\begingroup$ An equivalent definition of the divergence of a vector field on an oriented Riemannian manifold is as follows: the Riemannian metric and orientation give rise to a unique smooth volume form $\mu_g$. We then define for any smooth vector field $X$, it's divergence to be the unique smooth function such that it's Lie derivative satisfies $\mathscr{L}_X(\mu_g) =(\text{div }X) \mu_g$. This is very geometric since it says the divergence of a vector field is related to how volumes change(see the remarks at the end of my second link). $\endgroup$
    – peek-a-boo
    Jul 26, 2022 at 8:17
  • $\begingroup$ @peek-a-boo I need to spend some time to fully understand those links. But before that, let me ask a brief one: In Itskov's definition, divergence of a (1,1) tensor field results in a covector field; this gives rise to Gauss's theorem. The other definitions produce a vector fields. Are they incompatible with Gauss's theorem? $\endgroup$
    – Bjaam
    Jul 26, 2022 at 8:56
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    $\begingroup$ For the divergence theorem, you need to have a vector field $X$ (a $(1,0)$ tensor field). Then, its divergence is a smooth function $\text{div}(X)$ (i.e a $(0,0)$ tensor field). $\endgroup$
    – peek-a-boo
    Jul 26, 2022 at 8:57
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    $\begingroup$ The divergence of a $(1,1)$ tensor field is indeed a $(0,1)$ tensor field (i.e a covector field), but I don't see how he's using the divergence theorem on a $(1,1)$ tensor field. The only way I can even remotely make sense of this is to do everything in $\Bbb{R}^n$, but conceptually this is very unappealing. $\endgroup$
    – peek-a-boo
    Jul 26, 2022 at 9:10

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