Orthogonality of the coefficients of the Lorentz transformation

[Sigurdsson]

Homework Statement

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

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?

 

[sgd37]

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

[tex] \alpha^{\mu}_{\nu} (\alpha^T)^{\nu}_{\sigma} = \delta^{\mu}_{\sigma} [/tex]

 

[Sigurdsson]

... you are supposed to show that

[tex] \alpha^{\mu}_{\nu} (\alpha^T)^{\nu}_{\sigma} = \delta^{\mu}_{\sigma} [/tex]

This I have solved with your excellent advice my dear sir

[tex] \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} [/tex]


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.

 

[paranormal]

I would suggest approaching this problem by first understanding the concept of orthogonality in linear algebra. Orthogonality refers to the property of two vectors being perpendicular to each other, meaning their dot product is equal to zero. In this case, we are looking at the orthogonality of the coefficients of the Lorentz transformation, which can be thought of as vectors in a four-dimensional space.

To prove the orthogonality of these coefficients, we can use the definitions given in the problem and manipulate them using basic algebraic rules. For example, we can start by expanding the left side of the equation:

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

We can then use the chain rule to simplify this expression:

\frac{\partial x^{' \nu}}{\partial x^{\lambda}} \frac{\partial x^{' \lambda}}{\partial x^{\mu}} = \frac{\partial x^{' \nu}}{\partial x^{\mu}}

This is because the second derivative of a function with respect to a variable is equivalent to the first derivative of the function with respect to the same variable. Now, we can substitute this into our original equation:

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

We can then use the definition of the inverse Lorentz transformation (where the primed and unprimed coordinates are swapped) to write this as:

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

This shows that the coefficients of the Lorentz transformation are indeed orthogonal, as their product is equal to the Kronecker Delta symbol. This result is expected, as the Lorentz transformation is a rotation in four-dimensional space and rotations preserve orthogonality.

In conclusion, by using basic algebraic manipulations and the definitions given in the problem, we can prove the orthogonality of the coefficients of the Lorentz transformation. This result is important in understanding the properties of