Orthogonality of the coefficients of the Lorentz transformation

1. Oct 20, 2011

Sigurdsson

1. The problem statement, all variables and given/known data
For flat spacetime the coefficients of the Lorentz transformation are defined as

$\alpha^{\nu}_{\mu} = \frac{\partial x^{' \nu}}{\partial x^{\mu}}$

Whereas the Lorentz transformation is

$\begin{pmatrix} x_1' \\ x_2' \\ x_3' \\ x_4' \end{pmatrix} = \begin{pmatrix} \alpha_0^0 & \alpha_1^0 & \alpha_2^0 & \alpha_3^0 \\ \alpha_0^1 & \alpha_1^1 & \alpha_2^1 & \alpha_3^1 \\ \alpha_0^2 & \alpha_1^2 & \alpha_2^2 & \alpha_2^3 \\ \alpha_0^3 & \alpha_1^3 & \alpha_2^3 & \alpha_3^3 \end{pmatrix} \begin{pmatrix}x_1 \\ x_2 \\ x_3 \\ x_4 \end{pmatrix}$

Here's the problem, show that the coefficients are orthogonal. That is

$\alpha^{\nu}_{\lambda} \alpha^{\lambda}_{\mu} = \delta^{\nu}_{\mu}$

The last term is the Kronicker-Delta symbol

3. The attempt at a solution

Let's define some arbitrary four vector $\mathbf{A}$

$\delta^{\nu}_{\mu} A^{\mu} = \alpha^{\nu}_{\lambda} \alpha^{\lambda}_{\mu} A^{\mu} = \frac{\partial x^{' \nu}}{\partial x^{\lambda}} \frac{\partial x^{' \lambda}}{\partial x^{\mu}} A^{\mu}$

$\delta^{\nu}_{\mu} A^{\mu} = \frac{\partial x^{' \lambda}}{\partial x^{\lambda}} A^{' \nu}$

Now according to Einsteins summation rule we should have

$\mathbf{A} = \sum_{ \nu = 0}^{3} \left( \frac{\partial x^{' 0}}{\partial x^{0}} A^{' \nu} + \frac{\partial x^{' 1}}{\partial x^{1}} A^{' \nu} + \frac{\partial x^{' 2}}{\partial x^{2}} A^{' \nu} + \frac{\partial x^{' 3}}{\partial x^{3}} A^{' \nu} \right)$

This is where I get stuck. I have no idea how the sum on the right side should be equal to our 4-vector $\mathbf{A}$.

Any ideas?

2. Oct 21, 2011

sgd37

please note that the transpose of $$\alpha^{\mu}_{\nu} = \frac{\partial x'^{\mu}}{\partial x^{\nu}}$$ is $$(\alpha^T)^{\mu}_{\nu} = \frac{\partial x^{\mu}}{\partial x'^{\nu}}$$ and you are supposed to show that

$$\alpha^{\mu}_{\nu} (\alpha^T)^{\nu}_{\sigma} = \delta^{\mu}_{\sigma}$$

3. Oct 21, 2011

Sigurdsson

This I have solved with your excellent advice my dear sir

$$\mathbf{A} = \delta^{\nu}_{\mu} A^{\mu} = \alpha^{\nu}_{\lambda} (\alpha^T)^{\lambda}_{\mu} A^{\mu} = \frac{\partial x^{\nu}}{\partial x^{' \lambda} } \frac{\partial x^{' \lambda} }{\partial x^{\mu}}A^{\mu} = \frac{\partial x^{\nu}}{\partial x^{\mu}} A^{\mu} = \mathbf{A}$$

But this is of course different from my original problem where there was no transpose. Could it be that my textbook is wrong?

4. Oct 21, 2011

dextercioby

The standard proof is to start off with the requirement that the norm of a 4-vector be a scalar/invariant wrt a Lorentz transformation. So x'2=x2.

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