- #1
Karol
- 1,380
- 22
Homework Statement
[itex]f[/itex] is a function of two variables: y, z. I want to show that the gradient:
[tex]\nabla f=\frac{\partial f}{\partial y}\hat y + \frac{\partial f}{\partial z}\hat z[/tex]
Transforms as a vector under rotation of axes.
Homework Equations
The rotation of axes:
[tex]A \left\{\begin{array}{l}\bar {y} = y \cdot \cos \phi + z \cdot \sin \phi \\ \bar {z} = -y \cdot \sin + z \cdot \cos \end{array}\right.[/tex]
The Attempt at a Solution
I use the chain rule:
[tex]\frac{\partial f}{\partial \bar {y}}=\frac{\partial f}{\partial y}\frac{\partial y}{\partial \bar {y}}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial \bar {y}}[/tex]
And the similar for [itex]\frac{\partial f}{\partial \bar {z}}[/itex]. In order to find [itex]\frac{\partial y}{\partial \bar {y}}[/itex] and [itex]\frac{\partial z}{\partial \bar {z}}[/itex] i solve A for y and z and get:
[tex]\left\{\begin{matrix} y= \bar {y}\cdot \cos \phi - \bar {z}\cdot \sin \phi \\ z= \bar {y}\cdot \sin \phi +\bar {z}\cdot \cos \phi\end{matrix}\right.[/tex]
Now i get:
[tex]\left\{\begin{array}{l}\frac{\partial f}{\partial \bar {y}}=\frac{\partial f}{\partial y}(-\sin \phi)+\frac{\partial f}{\partial z}\cos \phi \\ \frac{\partial f}{\partial \bar {z}}=\frac{\partial f}{\partial y}(-\cos \phi)+\frac{\partial f}{\partial z}(-\sin \phi)\end{array}\right.[/tex]
Not as i expected, in the form of A.
It is closer to switching between the axes, but that's all i understand from this result.