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Let $\vec{F}(x, y, z)$ be a vector-valued function describing a vector-field. Then the rotation and divergence of the field are:

$\nabla \times \vec{F} = \text{curl}(\vec{F}) = \color{red}{(\frac{\partial{F_3}}{\partial{y}} - \frac{\partial{F_2}}{\partial{z}}, \frac{\partial{F_1}}{\partial{z}} - \frac{\partial{F_3}}{\partial{x}}, \frac{\partial{F_2}}{\partial{x}} - \frac{\partial{F_1}}{\partial{y}})}$

$\nabla \cdot \vec{F} = \text{div}(\vec{F}) = \color{blue}{\frac{\partial{F_1}}{\partial{x}} + \frac{\partial{F_2}}{\partial{y}} + \frac{\partial{F_3}}{\partial{z}}} $

if the elements involved in the above relations are placed in a matrix, one gets:

$A= \begin{bmatrix} \color{blue}{\frac{\partial{F_1}}{\partial{x}}} & \color{red}{\frac{\partial{F_1}}{\partial{y}}} & \color{red}{\frac{\partial{F_1}}{\partial{z}}} \\ \color{red}{\frac{\partial{F_2}}{\partial{x}}} & \color{blue}{\frac{\partial{F_2}}{\partial{y}}} & \color{red}{\frac{\partial{F_2}}{\partial{z}}} \\ \color{red}{\frac{\partial{F_3}}{\partial{x}}} & \color{red}{\frac{\partial{F_3}}{\partial{y}}} & \color{blue}{\frac{\partial{F_3}}{\partial{z}}} \end{bmatrix}$

  1. Isn't the above matrix the gradient, $\text{grad}$, of $\vec{F}$?

  2. Isn't this how a Jacobian looks like?

Additionally, this matrix could be represented as the sum of a diagonal and antisymmetric matrices, then, the elements of the $\text{curl}$ are elements of an antisymmetric matrix and those of $\text{div}$ are the trace of a diagonal matrix.

  1. If valid, what are all these three matrices called?

  2. Could it be said that $\text{curl}(\vec{F})$ measures how "unsymmetrical" is the matrix A?

Rainb
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Ziezi
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    The gradient is not a matrix, it is always a vector. So it is always the Jacobian. Every row of the Jacobian is the Gradient of a component (from the function) If you split it into two matrices,those are the jacobian when you split the vector field in two vector fields, one with a potential, and one with a vector potential. – Rainb Jul 25 '20 at 09:47
  • I just wanted to say that I thought this was a nice observation (and also that your presentation with a colored Jacobian was nice).

    I get the sense from your question that -- when you wrote it five years ago! -- you might have been just learning vector calculus. I could imagine not getting a response being a bit demoralizing. But I think it was quite insightful of you. I've worked with Jacobians on a daily basis for nearly a decade (albeit not in a context that motivates this kind of question) and hadn't noticed this before today.

    – colah May 29 '23 at 02:04

1 Answers1

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  1. Isn't the above matrix the gradient, grad, of ⃗ ?

  2. Isn't this how a Jacobian looks like?

Yes! The matrix you've written down is the Jacobian of the function describing the vector field (by mapping coordinates to vectors).

this matrix could be represented as the sum of a diagonal and antisymmetric matrices, then, the elements of the curl are elements of an antisymmetric matrix and those of div are the trace of a diagonal matrix.

If valid, what are all these three matrices called?

Could it be said that curl(⃗ ) measures how "unsymmetrical" is the matrix A?

Yes! In fact, decomposing the Jacobian into symmetric and anti-symmetric parts (the "strain-rate tensor" and "spin tensor" respectively) is an important idea in fluid dynamics.

David Roberts has a blog post which is likely more helpful than my explanation here (and which I'm partially drawing from). But to briefly expand on the basic ideas, let's discuss the relationship to divergence and curl piece by piece:

Divergence - The divergence is indeed the sum of the diagonal elements of the Jacobian. In other words, it's the trace. Roughly, you can think of the trace as meaning something like "how much vectors tend to be scaled by a matrix". (A nice way to see this is to observe that the trace is the sum of the eigenvalues, which describe how much a matrix scales vectors in a given direction.) Thus, for the Jacobian, it means something like "how much is the vector field locally being exaggerated in the same direction if we infinitesimally move" -- exactly what we intuitively think of as being the divergence.

Curl - Turning to the curl, as you observed, the curl only interacts with the antisymmetric part of the matrix. In the special case of 3x3 matrices, it happens to be the case that all antisymmetric matrices correspond to the cross-product with some vector. So the curl fully descries the anti-symmetric portion of the matrix! This isn't true in general (and this might be seen as related to why the basic notion of curl isn't defined in higher dimensions).

Shear - Note that divergence doesn't fully describe the symmetric part of the Jacobian. If you want a complete matrix decomposition, you'll need to also use the "shear rate tensor" (the symmetric Jacobian with the identity subtracted out to set the trace to zero). That is to say, divergence and curl don't fully describe a vector fields local structure! (This article discusses the pros and cons of not teaching shear.)

colah
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