Curl from requiring invariance under inertial coordinate changes

[powerof]

While investigating about the curl I have found this interesting perspective:

http://mathoverflow.net/a/21908/69479

I lack the knowledge to do the derivation on my own so I would like to ask for your help. I am an undergraduate.

I do not understand what a "first order differential operator" looks like. Is it just a combination of first order partial derivatives, like $\sum c_i \frac{\partial}{\partial x_i}\vec{e}_i$ ($c_i$ are generic constants)? I do not know either how to check whether my operator commutes with translations and rotations in order to see that "it is a multiple of curl".

If this is submitted to the wrong section I apologize.

Thank you for your time.

 

[Twigg]

The first order differential operators aren't exactly what you put. The author of that post didn't mention that his D operator has to be a vector-valued first order differential operator on vector-valued arguments, so it's more like $D|_{(x,y,z)}[\vec{A}] = \sum_{i,j,k}^{3} c_{ijk}(x,y,z) \vec{e}_{i} \frac{\partial}{\partial x_{j}} A_{k}(x,y,z)$.

To test for invariance under translations and rotations, you have to first write how each affects the coordinates and basis vectors.

For translations: $(x,y,z) \to (x+\epsilon, y, z)$ (it doesn't matter which direction the displacement is in, because I haven't specified what direction x is)
$D|_{(x,y,z)} \to D|_{(x+\epsilon,y,z)}$ which implies that $c_{ijk}(x,y,z) = c_{ijk}(x+\epsilon,y,z)$ for all values of $\epsilon$, or in other words that $c_{ijk}$ is a constant for all x. Since you can repeat this for y and z, you know that $c_{ijk}$ have to be constant for all (x,y,z).

For rotations: The general formula for rotations about any axis is a pain to work with, but we can say that if D is invariant under all rotations then D is invariant under the subset of right-handed rotations through 90 degrees. (This is, in fact, equivalent to saying the D is invariant under the action of infinitesimal rotations, but conceptually simpler).

$Case 1: (x,y,z) \to (-y, x, z) = (x',y',z')$
$(\frac{\partial}{\partial x},\frac{\partial}{\partial y}, \frac{\partial}{\partial z}) \to ( - \frac{\partial}{\partial y}, \frac{\partial}{\partial x}, \frac{\partial}{\partial z})$
$(A_{x}, A_{y}, A_{z}) \to (-A_{y}, A_{x}, A_{z})$
$D[\vec{A}] = \sum_{i,j.k=1}^3 \vec{e}_{i} \frac{\partial}{\partial x_{j}} A_{k} = \sum_{i,j.k=1}^3 \vec{e}_{i} \frac{\partial}{\partial x'_{j}} A_{k}$
From this we obtain, $c_{311} \frac{\partial}{\partial x} A_{x} + c_{312} \frac{\partial}{\partial x}A_{y} + c_{321} \frac{\partial}{\partial y} A_{x} + c_{322} \frac{\partial}{\partial y} A_{y} = (-1)^{2} c_{311} \frac{\partial}{\partial y} A_{y} + (-1) c_{312} \frac{\partial}{\partial y}A_{x} + (-1) c_{321} \frac{\partial}{\partial x} A_{y} + c_{322} \frac{\partial}{\partial x} A_{x}$
And finally we get
$(c_{311} - c_{322}) (\frac{\partial}{\partial x}A_{x} - \frac{\partial}{\partial y}A_{y}) + (c_{312} + c_{321}) (\frac{\partial}{\partial x}A_{y}) + (c_{321} + c_{312}) (\frac{\partial}{\partial y}A_{x}) = 0$
Since each of the bracketed combinations of the partials of $\vec{A}$ are arbitrary smooth functions, the only way to solve this equation is if all the scalar coefficients (the parenthesized terms with the c's in them) are equal to 0. There are exactly 2 non-trivial normalized solutions for the coefficients:
Solution 1: $c_{311} = c_{322} = 1$, $c_{312} = c_{321} = 0$. (The sum of like partial derivatives of the x and y components of A, i.e. the 2D divergence of A restricted to the z-plane)
Solution 2: $c_{311} = c_{322} = 0$, $c_{312} = -c_{321} = 1$. (The difference of unlike partial derivatives of the x and y components of A, i.e. the z-component of the curl of A)

You don't really need to compute the conditions on $c_{ijk}$ for invariance under rotations in the x or y planes to see that only Solution 2 is invariant under general 3D rotations. Solution 1 is the ordinary 2D divergence for the values of A on points in the z-plane. It clearly won't be invariant under rotations about the x or y axes. You don't even need to repeat the derivation to see what solution 2 will look like for each set of rotations, you can just use the cyclic substitution property to see that $c_{ijk}$ has to be the Levi-Civita symbol (the coefficients of curl).

 

[powerof]

(This is, in fact, equivalent to saying the D is invariant under the action of infinitesimal rotations, but conceptually simpler).

Can you expand on this or give me some resources to study on my own? Thank you.

 

[Twigg]

Apologies for the late reply.

I only skimmed through this, but it seems to cover the essentials.
http://www.cmth.ph.ic.ac.uk/people/d.vvedensky/groups/Chapter7.pdf

I learned the mechanics of infinitesimal symmetries primarily from a book by Peter J. Olver, "Applications of Lie Groups to Partial Differential Equations." Here's a Springer link with all the information, but unfortunately it's not free. If you're an undergraduate, you should see if it's in your school library.
http://link.springer.com/book/10.1007/978-1-4684-0274-2

 

[Twigg]

For rotations: The general formula for rotations about any axis is a pain to work with, but we can say that if D is invariant under all rotations then D is invariant under the subset of right-handed rotations through 90 degrees. (This is, in fact, equivalent to saying the D is invariant under the action of infinitesimal rotations, but conceptually simpler).

If you were to go through the trouble of stating the full transformation for rotation about the z-axis, you'd get $(x',y',z') = (xcos\theta - ysin\theta, xsin\theta + ycos\theta, z)$. If you take a first-order Taylor series for sine and cosine with respect to the rotation angle theta, you'd get $(x',y',z') = (x - \theta y, y + \theta x, z)$, which is just $(x',y',z') = (x,y,z) + \theta * (-y, x, 0)$. We call the second term, the $(-y,x,0)$ an infinitesimal rotation of the point $(x,y,z)$, in the same sort of way that we think of a velocity as a tiny displacement in a tiny amount of time (ish). Notice that this infinitesimal rotation is effectively the same as the 90 degree rotation I introduced in my original proof, except for the z-component, which I canceled on both sides of the equation for the $c_{ijk}$'s anyways. Applying the infinitesimal rotation to D on A and applying D to the infinitesimal rotation on A and equating them is effectively the same as what I did above. If you want to get really fancy, you can state this same idea by saying that the commutator of D and the infinitesimal rotation is 0. This is the same as saying that D is a constant along integral curves of the infinitesimal rotation operator. (It's not obvious that the infinitesimal rotation is a partial differential operator, because my discussion here is semi-formal. See Olver chapter 1 for the general treatment. That chapter also explains well why the commutator tells you how one operator varies with respect to another.)