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## Homework Statement

A scalar function can be represented as a position on the x-y plane, or on the u-v plane, where u and v are axes rotated by θ from the x and y axes.

Prove that the 2-dimensional [tex]\nabla^2[/tex] operator is invariant under a rotation of axes.

ie,

[tex]\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = \frac{\partial^2 f}{\partial u^2} + \frac{\partial^2 f}{\partial v^2}[/tex]

## Homework Equations

u = x cos θ + y sin θ

v = - x sin θ + y cos θ

I think I got the answer, *but* I wonder if there is an easier and more elegant way to do it. My method was to write out (this is just for [tex]\partial^2 f / \partial x[/tex]

[tex]\frac{\partial^2 f}{\partial x^2} = \frac{\partial}{\partial x}(\frac{\partial f}{\partial x}) = \frac{\partial}{\partial u}(\frac{\partial f}{\partial x})\frac{\partial u}{\partial x} + \frac{\partial}{\partial v}(\frac{\partial f}{\partial x})\frac{\partial v}{\partial x} [/tex]

And then substitute in

[tex]\frac{\partial f}{\partial x} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial x } + \frac{\partial f}{\partial v}\frac{\partial v}{\partial x} = \frac{\partial f}{\partial u} cos \theta } + \frac{\partial f}{\partial v}(-sin \theta)[/tex]

Eventually I get

[tex]\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} = \frac{\partial^2 f}{\partial u^2}({sin}^2 \theta + {cos}^2 \theta) + \frac{\partial^2 f}{\partial v^2}({sin}^2 \theta + {cos}^2 \theta)[/tex]

to get the result needed.

But this becomes incredibly long and tedious, like one to two pages of working and it's pretty easy to make a careless mistake somewhere. Is there an easier/shorter way to do this? For instance, is there a way that I could do it without the rotation of axes formula?