- #1

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- Homework Statement
- Prove that Equ (1) becomes Equ (3) after the coordinate transformation given in Equ (2)

- Relevant Equations
- Equ 1:

$$\chi(\mathbf{k},\omega,\mathbf{R},T)=

\frac{\partial F}{\partial \omega} \frac{\partial G}{\partial T}

-\frac{\partial F}{\partial T} \frac{\partial G}{\partial \omega}

-\mathbf{\nabla_k}F\cdot\mathbf{\nabla_R}G

+\mathbf{\nabla_R}F\cdot\mathbf{\nabla_k}G \tag{1}

$$

here ##F,G## are any functions of ##(\mathbf{k},\omega,\mathbf{R},T)##.

Equ 2:

$$

\mathbf{P}=\mathbf{k}-\mathbf{A}(\mathbf{R},T) \tag{2a}

$$

$$

\Omega = \omega - U(\mathbf{R},T) \tag{2b}

$$

Equ 3:

$$

\begin{align}

\chi(\mathbf{k},\omega,\mathbf{R},T)= \chi(\mathbf{P},\Omega,\mathbf{R},T)

+\mathbf{E}\cdot

\left(

\frac{\partial F}{\partial \Omega} \mathbf{\nabla_P}G

-\mathbf{\nabla_P}F\frac{\partial G}{\partial \Omega}

\right)\nonumber \\

+\mathbf{B}\cdot \left(

\mathbf{\nabla_P}F \times \mathbf{\nabla_P}G

\right) \tag{3}

\end{align}

$$

here ##\mathbf{E}=-\mathbf{\nabla_R}U-\frac{\partial \mathbf{A}}{\partial T}## and ##\mathbf{B}=\mathbf{\nabla_R}\times\mathbf{A}##

**My Progress:**

I tried to perform the coordinate transformation by considering a general function ##f(\mathbf{k},\omega,\mathbf{R},T)## and see how its derivatives with respect all variable ##(\mathbf{k},\omega,\mathbf{R},T)## change:

$$

\frac{\partial}{\partial\omega} f = \frac{\partial\mathbf{P}}{\partial\omega}\cdot \mathbf{\nabla_P}f + \frac{\partial\Omega}{\partial\omega} \frac{\partial f}{\partial\Omega} + \frac{\partial\mathbf{R}}{\partial\omega}\cdot \mathbf{\nabla_R}f + \frac{\partial T}{\partial\omega} \frac{\partial f}{\partial T} \tag{4a}

$$

$$

\frac{\partial}{\partial T} f = \frac{\partial\mathbf{P}}{\partial T}\cdot \mathbf{\nabla_P}f + \frac{\partial\Omega}{\partial T} \frac{\partial f}{\partial\Omega} + \frac{\partial\mathbf{R}}{\partial T}\cdot \mathbf{\nabla_R}f + \frac{\partial T}{\partial T} \frac{\partial f}{\partial T} \tag{4b}

$$

$$

\mathbf{\nabla_k} f = \mathbf{\nabla_k} \mathbf{P} \star \mathbf{\nabla_P}f + \mathbf{\nabla_k}\Omega \frac{\partial f}{\partial\Omega} + \mathbf{\nabla_k}\mathbf{R} \star \mathbf{\nabla_R}f + \mathbf{\nabla_k} T \frac{\partial f}{\partial T} \tag{4c}

$$

$$

\mathbf{\nabla_R} f = \mathbf{\nabla_R} \mathbf{P} \star \mathbf{\nabla_P}f + \mathbf{\nabla_R}\Omega \frac{\partial f}{\partial\Omega} + \mathbf{\nabla_R}\mathbf{R} \star \mathbf{\nabla_R}f + \mathbf{\nabla_R} T \frac{\partial f}{\partial T} \tag{4d}

$$

here ##\star## indicates that I don't know what kind of operator (dot, cross) should be used. While I am very confident that 4a and 4b are correct, I am not sure about 4c and 4d. For example, in 4a I put dot operator between two "vector" quantities ##\frac{\partial\mathbf{P}}{\partial\omega}\cdot \mathbf{\nabla_P}f## because the RHS should be a scalar (just like LHS) and the only way to get scalar is to take dot product between vectors. I do the same in 4b.

In 4c, the LHS is a vector (##\mathbf{\nabla_k} f ##), so the RHS should also be a vector. But I have quantities which I don't how to solve. For example: ##\mathbf{\nabla_k} \mathbf{P}, \mathbf{\nabla_R} \mathbf{P}## and what should be ##\star## operator? a dot product, cross?

Using the expression given in (2), I can simplify these equations till:

$$

\frac{\partial}{\partial\omega} f = (0)\cdot \mathbf{\nabla_P}f + (1) \frac{\partial f}{\partial\Omega} + (0)\cdot \mathbf{\nabla_R}f + (0) \frac{\partial f}{\partial T} \tag{5a}

$$

$$

\frac{\partial}{\partial T} f = \left(-\frac{\partial\mathbf{A}}{\partial T}\right)\cdot \mathbf{\nabla_P}f +\left(- \frac{\partial U}{\partial T}\right) \frac{\partial f}{\partial\Omega} + (0)\cdot \mathbf{\nabla_R}f + (1) \frac{\partial f}{\partial T} \tag{5b}

$$

$$

\mathbf{\nabla_k} f = \mathbf{\nabla_k} \mathbf{P} \star \mathbf{\nabla_P}f + (0) \frac{\partial f}{\partial\Omega}+ \mathbf{\nabla_k}\mathbf{R} \star \mathbf{\nabla_R}f + (0) \frac{\partial f}{\partial T} \tag{5c}

$$

$$

\mathbf{\nabla_R} f = \mathbf{\nabla_R} \mathbf{P} \star \mathbf{\nabla_P}f - \mathbf{\nabla_R}U \frac{\partial f}{\partial\Omega} +\mathbf{\nabla_R}\mathbf{R} \star \mathbf{\nabla_R}f +(0) \frac{\partial f}{\partial T} \tag{5d}

$$

So, the relation between different derivatives is:

$$

\frac{\partial}{\partial\omega} = \frac{\partial }{\partial\Omega} \tag{6a}

$$

$$

\frac{\partial}{\partial T} = \left(-\frac{\partial\mathbf{A}}{\partial T}\right)\cdot \mathbf{\nabla_P} +\left(- \frac{\partial U}{\partial T}\right) \frac{\partial }{\partial\Omega} +\frac{\partial }{\partial T} \tag{6b}

$$

$$

\mathbf{\nabla_k} = \mathbf{\nabla_k} \mathbf{P} \star \mathbf{\nabla_P} + \mathbf{\nabla_k}\mathbf{R} \star \mathbf{\nabla_R}\tag{6c}

$$

$$

\mathbf{\nabla_R} = \mathbf{\nabla_R} \mathbf{P} \star \mathbf{\nabla_P} - \mathbf{\nabla_R}U \frac{\partial }{\partial\Omega} +\mathbf{\nabla_R}\mathbf{R} \star \mathbf{\nabla_R} \tag{6d}

$$

To proceed, I need to simplify expressions in 6c and 6d. But I don't know how to solve: ##\mathbf{\nabla_k} \mathbf{P} \star, \mathbf{\nabla_k}\mathbf{R} \star, \mathbf{\nabla_R} \mathbf{P} \star, \mathbf{\nabla_R}\mathbf{R} \star ##.

Someone please nudge me in right direction. Please, please. I have spent so much time writing this question.