1

When one works with Euler's equation in its vorticity formulation, one needs to work with the transport equation

$$(B\cdot \nabla)j=(j\cdot \nabla)B.$$

If one just uses cartesian coordinates this is (Einstein's notation assumed)

$$\left(B_\alpha\frac{\partial j_\beta}{\partial x_\alpha}-j_\alpha\frac{\partial B_\beta}{\partial x_\alpha}\right)e_{\beta}=0.$$

In a more geometrical formulation, this means that the Lie bracket of the fields, $[B,j]=0$ vanishes. Working with this I came across this apparent contradiction

On the one hand, the Lie bracket is a geometrical quantity, whose expression in coordinates does not depend on the choice of parametrization I take. So if I wanted to use cilindrical coordinates, the equation above should be identical but substituting $\alpha,\beta\in \{1,2,3\}$ for $\alpha,\beta\in \{\rho,\phi,z\}$.

On the other hand, if one checks the formula for the quantities $(A\cdot \nabla)B$ in books, one sees that the resulting $(A\cdot\nabla)B-(B\cdot\nabla)A$ cannot be written like that in cilindrical or spherical coordinates. In particular, there are some annoying factors $\frac{1}{\rho}$ messing up the neat expression I would like to obtain (for instance, in Wikipedia's page)

I am sure that there is some subtlety (maybe not too subtle) I am not seeing, but I don't know if it is because my premise is wrong (the equation is not $[j,B]=0$), or what.

Thank you in advance :)

DaniS
  • 13

1 Answers1

-1

When you understand 2D polar coordinates you understand cylindrical coordinates. Formulas (2) and (3) in this post show how components of vectors and the coordinate basis vector fields transform. In other words, one and the same vector field $\boldsymbol{V}$ can be written as $$\tag{A} \boldsymbol{V}=V^x\partial_x+V^y\partial_y=V^r\partial_r+V^\phi\partial_\phi $$ where (in slightly different notation than in the linked answer) \begin{align} \partial_r&=(\cos\phi)\,\partial_x+(\sin\varphi)\,\partial_y\,,\\[2mm] \partial_\varphi&=-r(\sin\phi)\,\partial_x+r(\cos\varphi)\,\partial_y\,,\\[2mm] V^r&=(\cos\varphi)V^x+(\sin\varphi)V^x\,,\\[2mm] V^\phi&=-\tfrac1r(\sin\phi)V^x+\tfrac1r(\cos\phi)V^x\,. \end{align}

  • I invite you to derive these equations from the chain rule.

  • You already see the "annoying factors" $r$ and $\frac1r$ lurking in these equations but note that the following will become particularly simple when they stay on the right hand sides rather than being incorporated into, say, $\frac1r\partial_\phi\,.$ This is just a normalization of the basis vector field but it has the drawback that the new basis $\{\partial_r,\frac1r\partial_\phi\}$ is not longer holonomic. This non holonomic form is used in many $\nabla$ formulas displayed in Wikipedia.

  • Now I show what is coordinate invariant when using the holonomic bases (aka "coordinate bases") $\{\partial_x,\partial_y\}$ and $\{\partial_r,\partial_\phi\}\,:$

  • With another vector field $$\tag{B} \boldsymbol{U}=U^x\partial_x+U^y\partial_y=U^r\partial_r+U^\phi\partial_\phi $$ we can look at the The Lie bracket \begin{align} [\boldsymbol{U},\boldsymbol{V}]&=\boldsymbol{U}\boldsymbol{V}-\boldsymbol{V}\boldsymbol{U}\, \end{align} and show that it is a tensor. First, it is obviously coordinate invariant since the two vector fields are. It remains to show that this commutator equals the difference of two directional derivatives by which it is a coordinate invariant first order differential operator, hence, a tensor. To do this let's write (A) and (B) in index notation $$ \boldsymbol{V}=V^a\partial_a=V^\mu\partial_\mu $$ where Latin indices denote the Cartesian coordinates and Greek indices the polar coordinates. Same for $\boldsymbol{U}\,.$ Then, by the product rule \begin{align} \boldsymbol{U}\boldsymbol{V}&= (U^a\partial_a)(V^b\partial_b)=(U^a\partial_aV^b)\,\partial_b+U^aV^b\partial_a\partial_b\,. \end{align} The first term is the directional derivative $\partial_{\boldsymbol{U}}\boldsymbol{V}$ and the second term a second order derivative that is symmetric in $a$ and $b$ (hence in $\boldsymbol{U}$ and $\boldsymbol{V}\,$). Since in the difference $\partial_{\boldsymbol{U}}\boldsymbol{V}- \partial_{\boldsymbol{V}}\boldsymbol{U}$ the second order derivatives cancel we obtain $$ [\boldsymbol{U},\boldsymbol{V}]=\partial_{\boldsymbol{U}}\boldsymbol{V}- \partial_{\boldsymbol{V}}\boldsymbol{U}\,. $$

  • Final remark: these facts hold in any coordinate system (not only in Cartesian and polar coordinates) as long as we use coordinate bases $\partial_a$ and $\partial_\mu\,.$

Kurt G.
  • 17,136