Orthogonality of the coefficients of the Lorentz transformation

[Sigurdsson]

Homework Statement

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

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?

 

[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

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

 

[Sigurdsson]

... you are supposed to show that

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

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?

 

[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.