Reversing transformations of displacements

[beowulf.geata]

Homework Statement

I've just started (self-studying) Neuenschwander's Tensor Calculus for Physics and I got stuck at page 23, where he deals with transformations of displacements. I've made a summary of page 23 in the first part of the attached file.

Homework Equations

I want to use the coordinates x(x', y', z'), y(x', y', z'), z(x', y', z') and x'(x,y,z), y'(x,y,z), z'(x,y,z) and try to reverse the original transformation from unprimed to primed, see attached file.

The Attempt at a Solution

[/B]
I get a result which seems to say that dx'= 3dx', see attached file. So I don't understand where I'm going wrong. I'd be very grateful if anyone could help.

 

[Samy_A]

Homework Statement

I've just started (self-studying) Neuenschwander's Tensor Calculus for Physics and I got stuck at page 23, where he deals with transformations of displacements. I've made a summary of page 23 in the first part of the attached file.

Homework Equations

I want to use the coordinates x(x', y', z'), y(x', y', z'), z(x', y', z') and x'(x,y,z), y'(x,y,z), z'(x,y,z) and try to reverse the original transformation from unprimed to primed, see attached file.

The Attempt at a Solution

[/B]
I get a result which seems to say that dx'= 3dx', see attached file. So I don't understand where I'm going wrong. I'd be very grateful if anyone could help.

Look carefully at the expression multiplying dx', and compare it with the fourth equation in your text, the one with the Kronecker delta.
You seem to assume that $\frac{\partial x'}{\partial x}\frac{\partial x}{\partial x'}=1$ and similarly for the other two terms. That is not necessarily the case.

You can test it with a simple transformation (for two dimensions):
$x'=x+y$
$y'=x-y$

 

[beowulf.geata]

Thanks very much! I now see the error of my ways :-)