Divergence of transposed gradient of vector quantity

Question

I would appreciate some help with this.

Is: $$ \nabla \cdot \left( \nabla \vec{v}\right)^T= \nabla \left( \nabla \cdot \vec{v}\right) $$

How can I show this? Is the gradient of a vector mathematically defined?

Best regards

Answer 1

My previous answer is incomplete, and it really illustrates why we need to take great care when using index notation. We first define the covariant derivative of a vector as well as the divergence of a matrix (in the standard basis): $$(\nabla \boldsymbol u)^i{}_j=\partial_ju^i \\ (\nabla\cdot\mathbf M)_j=\partial_kM^k{}_j$$ Writing these out explicitly, $$\nabla \boldsymbol{u} =\begin{bmatrix} \partial _{1} u^{1} & \partial _{2} u^{1} & \partial _{3} u^{1}\\ \partial _{1} u^{2} & \partial _{2} u^{2} & \partial _{3} u^{2}\\ \partial _{1} u^{3} & \partial _{2} u^{3} & \partial _{3} u^{3} \end{bmatrix} \\ \nabla\cdot\mathbf M=\begin{bmatrix} \begin{array}{l} \partial _{1} M^{1}{}_{1}\\ +\partial _{2} M^{2}{}_{1}\\ +\partial _{3} M^{3}{}_{1} \end{array} & \begin{array}{l} \partial _{1} M^{1}{}_{2}\\ +\partial _{2} M^{2}{}_{2}\\ +\partial _{3} M^{3}{}_{2} \end{array} & \begin{array}{l} \partial _{1} M^{1}{}_{3}\\ +\partial _{2} M^{2}{}_{3}\\ +\partial _{3} M^{3}{}_{3} \end{array} \end{bmatrix}$$ This means $$\nabla \cdotp \nabla \boldsymbol{u} =\begin{bmatrix} \begin{array}{l} \partial _{1} \partial _{1} u^{1}\\ +\partial _{2} \partial _{1} u^{2}\\ +\partial _{3} \partial _{1} u^{3} \end{array} & \begin{array}{l} \partial _{1} \partial _{2} u^{1}\\ +\partial _{2} \partial _{2} u^{2}\\ +\partial _{3} \partial _{2} u^{3} \end{array} & \begin{array}{l} \partial _{1} \partial _{3} u^{1}\\ +\partial _{2} \partial _{3} u^{2}\\ +\partial _{3} \partial _{3} u^{3} \end{array} \end{bmatrix}$$ If we rearrange the terms a bit, we see that $$\nabla \cdotp \nabla \boldsymbol{u} =\begin{bmatrix} \partial _{1}\begin{pmatrix} \partial _{1} u^{1}\\ +\partial _{2} u^{2}\\ +\partial _{3} u^{3} \end{pmatrix} & \partial _{2}\begin{pmatrix} \partial _{1} u^{1}\\ +\partial _{2} u^{2}\\ +\partial _{3} u^{3} \end{pmatrix} & \partial _{3}\begin{pmatrix} \partial _{1} u^{1}\\ +\partial _{2} u^{2}\\ +\partial _{3} u^{3} \end{pmatrix} \end{bmatrix}$$ Identifying $\nabla\cdot\boldsymbol u=\partial_1u^1+\partial_2u^2+\partial_3u^3$ this is $$ \begin{array}{l} \nabla \cdotp \nabla \boldsymbol{u} =\begin{bmatrix} \partial _{1}( \nabla \cdot \boldsymbol{u}) & \partial _{3}( \nabla \cdot \boldsymbol{u}) & \partial _{3}( \nabla \cdot \boldsymbol{u}) \end{bmatrix}\\ =\nabla ( \nabla \cdot \boldsymbol{u}) \end{array}$$ On the other hand if we consider $$( \nabla \boldsymbol{u})^{\intercal } =\begin{bmatrix} \partial _{1} u^{1} & \partial _{1} u^{2} & \partial _{1} u^{3}\\ \partial _{2} u^{1} & \partial _{2} u^{2} & \partial _{2} u^{3}\\ \partial _{3} u^{1} & \partial _{3} u^{2} & \partial _{3} u^{3} \end{bmatrix}$$ We see that $$\nabla \cdot ( \nabla \boldsymbol{u})^{\intercal } =\begin{bmatrix} \begin{array}{l} +\partial _{1} \partial _{1} u^{1}\\ +\partial _{2} \partial _{2} u^{1}\\ +\partial _{3} \partial _{3} u^{1} \end{array} & \begin{array}{l} +\partial _{1} \partial _{1} u^{2}\\ +\partial _{2} \partial _{2} u^{2}\\ +\partial _{3} \partial _{3} u^{2} \end{array} & \begin{array}{l} +\partial _{1} \partial _{1} u^{3}\\ +\partial _{2} \partial _{2} u^{3}\\ +\partial _{3} \partial _{3} u^{3} \end{array} \end{bmatrix}$$ We can see readily that this is $$ \begin{array}{l} \nabla \cdot ( \nabla \boldsymbol{u})^{\intercal } =\begin{bmatrix} \Delta \left( u^{1}\right) & \Delta \left( u^{2}\right) & \Delta \left( u^{3}\right) \end{bmatrix}\\ =( \Delta \boldsymbol{u})^{\intercal } \end{array}$$ It seems that Wikipedia has used the incorrect index order for the covariant derivative of a vector. They appear to think $$(\nabla\boldsymbol u)^i{}_j=\frac{\partial u^j}{\partial x^i}$$ When in fact it should be the other way around. To be ultra clear, we need to specify how the covariant derivative and divergence work. Letting ${}^p_q\mathbb{R}$ be the space of $(p,q)$ tensors over $\mathbb{R}$, then $$\nabla:{}^p_q\mathbb{R}\to{}^p_{q+1}\mathbb{R} \\ \operatorname{div}:{}^1_q\mathbb{R}\to {}_q\mathbb{R} \\ \Delta: {}^p_q\mathbb{R}\to {}^p_q\mathbb{R}$$ It is very important to distinguish between (column) vectors and covectors or row vectors.

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I'll add one more thing to my answer to make it as complete as possible. Though in most fields in mathematical physics we try to write everything in covariant form, because of the divergence in the Navier-Stokes equations, it is actually more convenient to write everything in contravariant form. Typically you will see the N-S equations written in the general coordinate-free form $$\rho\frac{\mathrm D\boldsymbol u}{\mathrm D t}=\nabla\cdot \boldsymbol \sigma+\rho\boldsymbol{a} \\ \text{along with the continuity equation} \\ \partial_t\rho+\nabla\cdot(\rho\boldsymbol u)=0$$ Both sides here have the units of force per unit volume. Let's expand this into coordinates. I will assume a general coordinate system to maintain the generality. $$\rho\left(\partial_tu^i+u^j\nabla_ju^i\right)=\nabla_j\sigma^{ij}+\rho a^i$$ The components of the Cauchy stress tensor $\boldsymbol \sigma$ are (see my question ) $$\sigma^{ij}=(-p+\lambda \varepsilon^k{}_k)g^{ij}+2\mu\varepsilon^{ij}$$ Or, in coordinate free form, $$\boldsymbol{\sigma}=(-p+\lambda \operatorname{tr}\boldsymbol\varepsilon)\mathbf{g}^{-1}+2\mu\boldsymbol{\varepsilon}$$ $\lambda$ is the bulk viscosity. You'll notice that this formula is quite a bit different to the one on Wikipedia, and that's because this is about where Wikipedia begins to go very badly wrong. Instead of the inverse metric $\mathbf{g}^{-1}$ they have the identity tensor $\mathbf{I}$. Again, while in Cartesian coordinates they are one and the same, the distinction is very important when working in other coordinate systems. Here $\boldsymbol\varepsilon$ is the rate-of-strain tensor, which we are defining contravariantly: $$\varepsilon^{ij}=\frac{1}{2}\left((\nabla u)^{ij}+(\nabla u)^{ji}\right)$$ Where $(\nabla u)^{ij}$ is defined using an index raising, $(\nabla u)^{ij}=g^{jk}(\nabla u)^i{}_k$. One can see that $\boldsymbol \varepsilon$ is actually the symmetric part of the tensor $[(\nabla u)^{ij}]$. It is a easy exercise to check that $$\operatorname{tr}\boldsymbol \varepsilon=\varepsilon^k{}_k=g_{kj}\varepsilon^{kj}=\operatorname{div}\boldsymbol u$$ And so in the case that $\boldsymbol u$ is incompressible, the Cauchy stress tensor reduces to $$\boldsymbol\sigma=-p\mathbf{g}^{-1}+2\mu\boldsymbol\varepsilon \\ \sigma^{ij}=-pg^{ij}+2\mu\varepsilon^{ij}$$ Hence in this incompressible case $$(\nabla\cdot \sigma)^i=\nabla_j\sigma^{ij}=-\nabla_jg^{ij}p+\mu\nabla_j\left((\nabla u)^{ij}+(\nabla u)^{ji}\right)$$ The first term, $\nabla_jg^{ij}p$ is just $\operatorname{grad}^i p$. The second term requires essentially the same computation from the previous part of this answer.

Answer 2

One should use the covariant derivative here. It is defined generally for higher order tensors. (One may also define the gradient as the sharp of the covariant derivative). Letting $\nabla$ represent the covariant derivative, $\nabla \boldsymbol v$ is a $(1,1)$ tensor, or matrix. In the standard basis it has components $$(\nabla \boldsymbol v)^i{}_j=\partial_jv^i$$ The divergence operator is also straightforward in these coordinates. To take the divergence of a vector, $$\nabla\cdot \boldsymbol v=\partial_kv^k$$ Similarly for a matrix, $$(\nabla\cdot \mathbf{M})_j=\partial_kM^k{}_j$$ Comparing the two expressions, $$(\nabla\cdot\nabla \boldsymbol v)_j=\partial_k(\partial_jv^k)\\ \text{and} \\ (\nabla(\nabla\cdot \boldsymbol v))_j=\partial_j(\partial_kv^k)$$ Since partial derivatives commute, the two expressions are equal.

To show this is true in any coordinate system, more formalism is needed.