(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

For flat spacetime the coefficients of the Lorentz transformation are defined as

[itex] \alpha^{\nu}_{\mu} = \frac{\partial x^{' \nu}}{\partial x^{\mu}} [/itex]

Whereas the Lorentz transformation is

[itex] \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} [/itex]

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

[itex] \alpha^{\nu}_{\lambda} \alpha^{\lambda}_{\mu} = \delta^{\nu}_{\mu} [/itex]

The last term is the Kronicker-Delta symbol

3. The attempt at a solution

Let's define some arbitrary four vector [itex] \mathbf{A} [/itex]

[itex] \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} [/itex]

[itex] \delta^{\nu}_{\mu} A^{\mu} = \frac{\partial x^{' \lambda}}{\partial x^{\lambda}} A^{' \nu} [/itex]

Now according to Einsteins summation rule we should have

[itex] \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) [/itex]

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

Any ideas?

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# Homework Help: Orthogonality of the coefficients of the Lorentz transformation

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