How do you take this multivector derivative that has a nested variable?

I want to compute the multivector derivative $\partial_\psi$ of $\Omega_B\cdot \mathcal{I}(\Omega_B)$ where $\mathcal{I}$ is a function that is linear in bivectors, $\mathcal{I}(\Omega_B) = \int d^3x \rho(x) x\wedge (x\cdot \Omega_B)$. $\psi$ is an even multivector in 3D space (so it contains scalar and bivector terms). $\Omega_B$ is a function of $\psi$, $\Omega_B\equiv -\psi^\dagger \dot{\psi} + \dot{\psi}^\dagger \psi$. $\Omega_B$ is a bivector because the second term is the negative reverse of the first term. The overdot signifies time derivative and the dagger signifies reverse. This is for a Lagrangian method so we are treating $\dot{\psi}$ and $\dot{\psi}^\dagger$ as constants. I use $\hat{\partial}$ instead of $\partial$ when a specific term is being operated on. This question relates to the geometric algebra Lagrangian for rigid body motion.

What I have so far: $$ \begin{align} \partial_\psi \Omega_B\cdot (\Omega_B) &= \partial_\psi \langle \Omega_B \mathcal{I}(\Omega_B) \rangle \\ &= \hat{\partial}_\psi\langle\hat{ \Omega}_B \mathcal{I}((\Omega_B)\rangle + \hat{\partial}_\psi\langle \Omega_B \hat{\mathcal{I}(}(\Omega_B) \rangle \\ \end{align} $$

The first term is: $$ \begin{align} \hat{\partial}_\psi\langle\hat{ \Omega}_B \mathcal{I}(\Omega_B)\rangle &= \hat{\partial}_\psi \langle -\hat{\psi}^\dagger \dot{\psi}\mathcal{I}(\Omega_B) + \dot{\psi}^\dagger\hat{\psi} \mathcal{I}(\Omega_B) \rangle \\ &= \hat{\partial}_\psi \langle -\hat{\psi}^\dagger \dot{\psi}\mathcal{I}(\Omega_B) + \hat{\psi} \mathcal{I}(\Omega_B) \dot{\psi}^\dagger\rangle \\ &= -(\dot{\psi}\mathcal{I}(\Omega_B))^\dagger + \mathcal{I}(\Omega_B)\dot{\psi}^\dagger \\ &= 2\mathcal{I}(\Omega_B)\dot{\psi}^\dagger \end{align} $$ where I used $\Omega_B^\dagger = -\Omega_B$.

How is the $\hat{\partial}_\psi\langle \Omega_B \hat{\mathcal{I}}(\Omega_B) \rangle$ term calculated? This is also supposed to end up as $2\mathcal{I}(\Omega_B)\dot{\psi}^\dagger$.

Solution 1:

The second term is:

$$ \begin{align} \hat{\partial}_\psi\langle \Omega_B \hat{\mathcal{I}}(\Omega_B) \rangle &= \hat{\partial}_\psi \langle \hat{\mathcal{I}}(\Omega_B) \Omega_B \rangle \\ &= \int d^3x\rho \sum_J a^J \langle x\wedge (x\cdot ((a_J*\partial_\psi)(-\psi^\dagger\dot{\psi} + \dot{\psi}^\dagger\psi)))\Omega_B\rangle \\ &= \int d^3x\rho \sum_J a^J \langle x\wedge (x\cdot (-a_{J+}^\dagger\dot{\psi} + \dot{\psi}^\dagger a_{J+}))\Omega_B\rangle \\ &= 2\int d^3x\rho \sum_J a^J \langle x\wedge (x\cdot (\dot{\psi}^\dagger a_{J+})) \Omega_B\rangle \\ &= 2\int d^3x\rho \sum_J a^J \left\{\langle xx\cdot(\dot{\psi}^\dagger a_{J+})\Omega_B\rangle - \underbrace{\langle x\cdot(x\cdot(\dot{\psi}^\dagger a_{J+}))\Omega_B \rangle}_{=0}\right\} \\ &= 2\int d^3x\rho \sum_J a^J \frac{1}{2}\langle x\left(x\dot{\psi}^\dagger a_{J+} - \dot{\psi}^\dagger a_{J+}x\right)\Omega_B\rangle \\ &= 2\int d^3x\rho \sum_J a^J \frac{1}{2}\langle a_{J+}\Omega_B x^2\dot{\psi}^\dagger - a_{J+}x\Omega_Bx\dot{\psi}^\dagger \rangle \\ &= 2\int d^3x\rho \sum_J a^J \frac{1}{2}\langle a_{J+}x\left( x\Omega_B - \Omega_B x)\right) \dot{\psi}^\dagger\rangle \\ &= 2 \sum_J a^J a_{J+} * \left(\int d^3x \rho x\wedge (x\cdot \Omega_B)\dot{\psi}^\dagger \right)\\ &= 2\mathcal{I}(\Omega_B)\dot{\psi}^\dagger \end{align} $$