# Metric orthonormal basis

1. Dec 19, 2013

### PhysicsDude1

1. The problem statement, all variables and given/known data

For the orthonormal coordinate system (X,Y) the metric is

\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

Calculate G' in 2 ways.

1) G'= M$^{T}$*G*M
2) g$\acute{}$$_{ij}$ = $\overline{a}\acute{}_{i}$ . $\overline{a}\acute{}_{j}$

2. Relevant equations

\begin{pmatrix} \overline{a}\acute{}_{1} \\ \overline{a}\acute{}_{2} \end{pmatrix}
= \begin{pmatrix} -cos(\phi).\overline{a}\acute{}_{1} -\overline{a}\acute{}_{2} \\
cos(\phi).\overline{a}\acute{}_{2}\end{pmatrix}

M= \begin{pmatrix} -cos(\phi) & 0 \\ -1 & cos(\phi) \end{pmatrix}
M$^{T}$ = \begin{pmatrix} -cos(\phi) & -1 \\ 0 & cos(\phi) \end{pmatrix}
G=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
=> G' = \begin{pmatrix} cos²(\phi) +1 & -cos(\phi) \\ -cos(\phi) & cos²(\phi) \end{pmatrix}

3. The attempt at a solution

So I'm having problems with the 2nd method i.e. g$\acute{}$$_{ij}$ = $\overline{a}\acute{}_{i}$ . $\overline{a}\acute{}_{j}$

g$\acute{}$$_{11}$ = (-cos$(\phi)$ . $\overline{a}_{1}$ -$\overline{a}_{2}$) . (-cos$(\phi)$ . $\overline{a}_{1}$ -$\overline{a}_{2}$) = ?

What are the values for $\overline{a}_{1}$ and $\overline{a}_{2}$ ?
I think they're both 1 because they're both unit vectors of length 1 but I'm not sure.
Also, this is the first time ever I have used LaTeX so sorry if it's a bit sloppy.

2. Dec 19, 2013

### PhysicsDude1

Also, I can't fall asleep because of this. I REALLY want to know the answer :p

3. Dec 19, 2013

### PhysicsDude1

If there's something confusing, just ask :)

4. Dec 19, 2013

### Dick

I don't know what problem you are trying solve. I know what "orthogonal coordinates" means and I know what "orthonormal vector fields" are. I not sure I know what "orthonormal coordinates" are. I can guess but could you explain that? Might be best to start with the definitions before you plow into trying to prove things. That might make it a lot easier than you think. And what IS this G' you are trying to calculate anyway?

Last edited: Dec 19, 2013
5. Dec 20, 2013

### PhysicsDude1

I'm sorry, I meant orthogonal coordinate system :)
Basically I'm trying to find G' which is the metric of the coordinate system (X', Y').
There are 2 ways to find this :
1) G' = M^T * G * M (M=inverse transformationformula // M^T = the transposed matrix of M // G = metric of the coordinate system (X,Y) )

2) Via the multiplication of the primed basic vectors a1 and a2 (see first matrix) and that's where I'm stuck. I don't know if I'm supposed to fill in those basis vectors with the value 1 or just leave them like that BUT that would mean that the 2nd method has a different result from the 1st method which can't be possible. I think I have to fill it in.

Sorry for all the confusion!