In summary, the conversation discusses the process of coordinate transformation using a general function and its derivatives with respect to various variables. The equations 4a, 4b, 4c, and 4d are given, but the correct operator (dot or cross) is unclear. The simplified equations 5a, 5b, 5c, and 5d are shown, and the relationship between derivatives is derived in equations 6a, 6b, 6c, and 6d. However, the expression in 6c and 6d cannot be simplified without knowing the correct operator for ##\mathbf{\nabla_k} \mathbf{P} \star, \mathbf
  • #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.
 
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  • #2


First of all, great job on making progress and trying to understand coordinate transformations. It can be a challenging concept to grasp, but it seems like you have a good understanding so far.

To address your questions about the operators in equations (4c) and (4d), the dot product (·) should be used for both cases. This is because the dot product is defined for two vectors and results in a scalar quantity. The cross product (×), on the other hand, is defined for two vectors and results in a vector quantity. Since the RHS of equations (4c) and (4d) should be a vector, the dot product is the appropriate operator to use.

Now, for the expressions in equations (6c) and (6d), you are correct that they need to be simplified. To do this, you can use the chain rule. For example, for the term ##\mathbf{\nabla_k} \mathbf{P} \star \mathbf{\nabla_P}##, you can write it as ##\mathbf{\nabla_k} ( \mathbf{P} \cdot \mathbf{\nabla_P})## and then use the chain rule to simplify it. Similar steps can be followed for the other terms in equations (6c) and (6d).

I hope this helps to nudge you in the right direction. Keep up the good work and keep practicing, and you will continue to improve your understanding of coordinate transformations.
 

Suggested for: Coordinate Transformation (multivariable calculus)

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