# Metric and Theorema Egregium

1. Jun 29, 2015

### Simone Furcas

I have $ds^2=\cos^2(v)du^2 + dv^2$ , i take a coordinate transformation x=u and cos(v)=$\frac{1}{(cosh(y))}$, I have to find a new metric with this coordinate transformation and proof it is in agreement with Theorema Egregium. $ds^2=\frac{dx^2}{(cosh^2(y)}) +\frac{dy^2 }{(y^2(1-y^2))}$ is my new metric . I used the jacobian to proof the second point, $J=\begin{pmatrix} 1 & 0 \\ 0 & \frac{-1}{y(1-y^2)^(1/2)} \end{pmatrix}$, so $j^-1 * \begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 & \frac{1}{y^2(1-y^2)} \end{pmatrix} *j$ = $\begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 & (\frac{1}{y^2(1-y^2)})^2 \end{pmatrix}$
I know cos(v)=$\frac{1}{(cosh(y))}$ so the first part is ok, the second $(\frac{1}{(y^2(1-y^2)})^2$=1 i think is not true...
What is my mistake? Does someone could help me?

Last edited: Jun 29, 2015
2. Jun 30, 2015

### Simone Furcas

I did a mistake, $ds^2=\frac{dx^2}{(cosh^2(y)}) +\frac{dy^2 }{cosh{y}}$, $J=\begin{pmatrix} 1 & 0 \\ 0 & \frac{1}{cosh{y}} \end{pmatrix}$, so $j^-1 * \begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 & \frac{1}{cosh{y}}\end{pmatrix} *j$ = $\begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 & (\frac{1}{cosh{y}})^4 \end{pmatrix}$
Does someone have any ideas?

3. Jun 30, 2015

### HallsofIvy

Staff Emeritus
That matrix multiplication is wrong. Since these are diagonal matrices multiplication is commutative so $J^{-1}MJ= (J^{-1}J)M= M$ for any matrix, M.

4. Jun 30, 2015

### Simone Furcas

$j^T * \begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 & \frac{1}{cosh{y}}\end{pmatrix} *j$ = $\begin{pmatrix} \frac{1}{cosh^2{y}} & 0 \\ 0 & (\frac{1}{cosh{y}})^4 \end{pmatrix}$ there was a mistake, it was the transpose not the inverse.

5. Jun 30, 2015

### Simone Furcas

I solved it.
I solved the exercise.