Calculation of the divergence of the advection of the perpendicular gradient of the potential times density using Clebsch coordinates

We would here like to calculate $$ \nabla\cdot\left(\mathbf{u}_E\cdot\nabla\left[n \nabla_\perp\phi \right]\right) $$ using cylindrical Clebsch coordinates, as this tensor identity has not been found in the literature.

NOTE: These are normalized equations. As $B$ is constant, we can choose $B_0$ so that the normalized $\tilde{B}=1$, thus, $B$ is excluded from these equations.

Also, we would like to compare this with $$ B\{\phi,\Omega^D\} $$

In [1]:
from IPython.display import display
from sympy import symbols, simplify, sympify, expand
from sympy import init_printing
from sympy import Eq, Function
from clebschVector import ClebschVec
from clebschVector import div, grad, gradPerp, advVec
from common import rho, theta, poisson
from common import displayVec


In [2]:
u_z = symbols('u_z', real = True)
# In reality this is a function, but as it serves only as a dummy it is here defined as a symbol
# This makes it easier to replace
f = symbols('f', real = True)
phi = Function('phi')(rho, theta)
n   = Function('n')(rho, theta)
# Symbols for printing
zeta, chi, epsilon = symbols('zeta, chi, epsilon')

Calculation of the $E\times B$ advection

We would now like to calculate

$$ \zeta = \nabla\cdot\left(\mathbf{u}_E \cdot\nabla\left[n\nabla_\perp\phi\right]\right) $$

We will do this by

  1. Calculate $n\nabla_\perp\phi$
  2. Define $\mathbf{u}_E$
    • By first calculating $\nabla_\perp\phi$
  3. Calculate $\mathbf{u}_E\cdot\nabla \left(n\nabla_\perp\phi\right)$
    • To check the different contributions we also
      • Calculate $\mathbf{u}_E\cdot\nabla f$
      • Calculate $\mathbf{a}\cdot \left(n\nabla_\perp\phi\right)$
  4. Take the divergence of the resulting vector
  5. Compare this with $B\{\phi,\Omega^D\}$

Calculation of $n\nabla_\perp\phi$

In [3]:
nGradPerpPhi = gradPerp(phi)*n

Defining $\mathbf{u}_E$

We have that $${u}_E = - \frac{\nabla_\perp\phi\times\mathbf{b}}{B}$$ Remember that we are working with normalized equations, so $B$ (which in reality is $\tilde{B}$) is equal to $1$.

NOTE: It migth appear that there is a discrepancy between having a coordinate system where $B$ is not constant where we have derived equation where $B$ is constant. This is because the cylindrical coordinate system is not a Clebsch system, but the metrics coinside. The Poisson bracket is the only place where $B$ is used explicitly, and care must be taken. The workaround is easy: Just multiply the Poisson bracket with $B$ to make it correct in cylindrical coordinates.

In [4]:
# The basis-vectors are contravariant => components are covariant
eTheta = ClebschVec(rho=0, theta=1, z=0, covariant=True)
eRho = ClebschVec(rho=1, theta=0, z=0, covariant=True)
B = eTheta^eRho
displayVec(B, 'B')
Blen = B.len()
display(Eq(symbols('B'), Blen))
b = B/(B.len())
displayVec(b, 'b')

NOTE: Basis vectors in $B$ are covariant, so components are contravariant

Calculation of $\nabla_\perp\phi$

In [5]:
gradPerpPhi = gradPerp(phi)

In [6]:
# Normalized B
BTilde = 1

# Defining u_E
ue = - ((gradPerpPhi^b)/BTilde)
displayVec(ue, 'u_E')

Calculation of $\mathbf{u}_E\cdot\nabla \left(n\nabla_\perp\phi\right)$

Calculation of $\mathbf{u}_E\cdot\nabla f$

In [7]:
ueDotGrad_f = ue*grad(f)

Calculation of $\mathbf{a}\cdot\nabla \left(n\nabla_\perp\phi\right)$

In [8]:
aRho,  aZ,  aTheta  = symbols('a^rho, a^z, a^theta')
a_Rho, a_Z, a_Theta = symbols('a_rho, a_z, a_theta')
aCov = ClebschVec(rho = a_Rho, z=a_Z, theta = a_Theta, covariant=True)
aCon = ClebschVec(rho = aRho,  z=aZ,  theta = aTheta,  covariant=False)

Using covariant vector

In [9]:
aCovDotNablaGradPhi = advVec(aCov, nGradPerpPhi)

Using contravariant vector

In [10]:
aConDotNablaGradPhi = advVec(aCon, nGradPerpPhi)

Calculation of full $\mathbf{u}_E\cdot\nabla \left(n\nabla_\perp\phi\right)$

In [11]:
ueDotGradnGradPerpPhi = advVec(ue, nGradPerpPhi)

In [12]:

Calculation of $\nabla\cdot\left(\mathbf{u}_E\cdot\nabla\left[n\nabla_\perp\phi\right]\right)$

In [13]:
div_ueDotGradnGradPerpPhi = div(ueDotGradnGradPerpPhi)
zetaFunc = div_ueDotGradnGradPerpPhi.doit().expand()
display(Eq(zeta, simplify(zetaFunc)))

Comparison with $B\{\phi,\Omega^D\}$

In cylindrical Clebsch coordinates, we have that $\mathbf{u}_E\cdot\nabla = \{\phi,\cdot\}$. However, we have normalized our equations so that $\tilde{B}=1$. As $B$ from the Clebsch system is not constant, we can achieve normalization by multiplying the Poisson bracket with the un-normalized $B$ (from the Clebsch system).

We define the vorticity-like field $\Omega^D$ to be $\Omega^D = \nabla\cdot\left(n\nabla_\perp\phi\right)$. In the Clebsch system this is written as

In [14]:
vortD = div(gradPerp(phi)*n)
display(Eq(symbols('Omega^D'), vortD.doit().expand()))

We now write $\chi = B\{\phi,\Omega^D\}$

In [15]:
poissonPhiVortD = Blen*poisson(phi, vortD)
chiFunc = poissonPhiVortD.doit().expand()
display(Eq(chi, chiFunc))

The difference $\epsilon$ between $\zeta = \nabla\cdot\left(\mathbf{u}_E\cdot\nabla\left[n\nabla_\perp\phi\right]\right)$ and $\chi = B\{\phi,\Omega^D\}$ is given by

$$\epsilon = \zeta - \chi$$

In [16]:
epsilonFunc = (zetaFunc - chiFunc).expand()
display(Eq(epsilon, epsilonFunc))

In fact we see that

\begin{align*} \epsilon - \left( -\frac{1}{\rho}[\partial_\rho\phi]\{n, \partial_\rho\phi\} -\frac{1}{\rho^3}[\partial_\theta\phi]\{n, \partial_\theta\phi\} +\frac{1}{\rho^4}[\partial_\theta n][\partial_\theta\phi]^2 \right) =\\ \epsilon - \left( \frac{1}{\rho}[\partial_\rho\phi]\{\partial_\rho\phi,n\} +\frac{1}{\rho^3}[\partial_\theta\phi]\{\partial_\theta\phi, n\} +\frac{1}{\rho^4}[\partial_\theta n][\partial_\theta\phi]^2 \right) = \end{align*}

In [17]:
epsMinusCorrection = epsilonFunc\
        (1/rho)*phi.diff(rho)*poisson(phi.diff(rho), n)\

What is more interesting is in fact that

\begin{align*} \epsilon - \xi = \epsilon - \frac{B}{2}\{\mathbf{u}_E\cdot\mathbf{u}_E, n\} \end{align*}

In [18]:
xi = (Blen/2)*poisson(ue*ue, n).doit()
epsMinusNewCorr = epsilonFunc - (Blen/2)*poisson(ue*ue, n).doit()


Where \begin{align*}



In [19]:

Note that the last term $\frac{1}{\rho^4}(\partial_\theta n)(\partial_\theta\phi)^2$ does not appear to come from the Poisson bracket. This is however the case, and comes from the part which contains $\frac{1}{2}\partial_\rho\left(\frac{1}{\rho}\partial_\theta \phi\right)^2 = \left(\frac{1}{\rho}\partial_\theta \phi\right)\partial_\rho\left(\frac{1}{\rho}\partial_\theta \phi\right)$ as $\partial_i (fg) = f \partial_i g + g \partial_i f$

To summarize, we have

\begin{align*} \zeta - (\chi + \xi) = \end{align*}

In [20]:
display((zetaFunc - (chiFunc + xi)).simplify())

Printing for comparison

In [21]:
S = expand(zetaFunc)

strS = str(S)

# phi rho derivatives
strS = strS.replace('Derivative(phi(rho, theta), rho)', 'phi_x')
strS = strS.replace('Derivative(phi(rho, theta), rho, rho)', 'phi_xx')
strS = strS.replace('Derivative(phi(rho, theta), rho, rho, rho)', 'phi_xxx')
# phi theta derivatives
strS = strS.replace('Derivative(phi(rho, theta), theta)', 'phi_z')
strS = strS.replace('Derivative(phi(rho, theta), theta, theta)', 'phi_zz')
strS = strS.replace('Derivative(phi(rho, theta), theta, theta, theta)', 'phi_zzz')
# phi mixed derivatives
strS = strS.replace('Derivative(phi(rho, theta), rho, theta)', 'phi_xz')
strS = strS.replace('Derivative(phi(rho, theta), rho, theta, theta)', 'phi_xzz')
strS = strS.replace('Derivative(phi(rho, theta), rho, rho, theta)', 'phi_xxz')
# Non-derivatives
strS = strS.replace('phi(rho, theta)', 'phi')

# n rho derivatives
strS = strS.replace('Derivative(n(rho, theta), rho)', 'n_x')
strS = strS.replace('Derivative(n(rho, theta), rho, rho)', 'n_xx')
# n theta derivatives
strS = strS.replace('Derivative(n(rho, theta), theta)', 'n_z')
strS = strS.replace('Derivative(n(rho, theta), theta, theta)', 'n_zz')
# n mixed derivatives
strS = strS.replace('Derivative(n(rho, theta), rho, theta)', 'n_xz')
# Non-derivatives
strS = strS.replace('n(rho, theta)', 'n')

newS = sympify(strS)
display(Eq(symbols('S_new'), expand(newS)))