# Lorentz transformations

Givne the Lorentz transformations (LTs)}, $$x'^{\mu} = L_{\nu}^{\mu} x^{\nu}$$, between the coordinates, $$x^{\mu} = (ct , \vec{r})$$ of an event as seen by O, and coordinates, $$x'^{\mu} = (ct', \vec{r'})$$ of the same event as seen by an inertial observer O', show that if we write the inverse transformation as $$x^\alpha = \tilde{L}_{\beta}^{\alpha} x'^{\beta},\mbox{then} \ L_{\omega}^{\alpha} \tilde{L}_{\beta}^{\omega} = \delta_{\beta}^{\alpha}$$

WELL from the inverse transformation we ca figure out that
1... $$\frac{\partial x^{\alpha}}{\partial x'^{\omega}} = \tilde{L}_{\beta}^{\alpha} \frac{\partial x'^{\beta}}{\partial x'^{\omega}}$$

also
2... $$\frac{\partial x'^{\alpha}}{\partial x^{\omega}} =L_{\beta}^{\alpha} \frac{\partial x^{\beta}}{\partial x^{\omega}}$$

there is a notation problem here that i am trying to resolve as well...
do i simply rearrange for L and tilde L andmultiply out??
mroe to come as i type it out

Last edited:

Hurkyl
Staff Emeritus
Gold Member
You didn't finish the statement of the problem!

Anyways, why did you differentiate anything?

p.s. corrected now
there was wayyy too much typing involved with the statement of the problem! AHH

dextercioby
Homework Helper
Speaking of notation, since $\Lambda$ is a matrix, we use these conventions to denote its elements

$$\Lambda^{\mu}{}_{\nu} ^$$

for the direct transformation matrix.

and

$$\Lambda_{\nu}{}^{\mu}$$

for its transposed.

However, since both $\Lambda$ and its transposed belong to $\mbox{O(1,3)}$, then the inverse of $\Lambda$ is equal to its transposed.

Daniel.

dextercioby
Homework Helper
There's something spooky with the latex compiler today....

Daniel.

nrqed
Homework Helper
Gold Member
stunner5000pt said:
Givne the Lorentz transformations (LTs)}, $$x'^{\mu} = L_{\nu}^{\mu} x^{\nu}$$, between the coordinates, $$x^{\mu} = (ct , \vec{r})$$ of an event as seen by O, and coordinates, $$x'^{\mu} = (ct', \vec{r'})$$ of the same event as seen by an inertial observer O', show that if we write the inverse transformation as $$x^\alpha = \tilde{L}_{\beta}^{\alpha} x'^{\beta},\mbox{then} \ L_{\omega}^{\alpha} \tilde{L}_{\beta}^{\omega} = \delta_{\beta}^{\alpha}$$

WELL from the inverse transformation we ca figure out that
1... $$\frac{\partial x^{\alpha}}{\partial x'^{\omega}} = \tilde{L}_{\beta}^{\alpha} \frac{\partial x'^{\beta}}{\partial x'^{\omega}}$$

also
2... $$\frac{\partial x'^{\alpha}}{\partial x^{\omega}} =L_{\beta}^{\alpha} \frac{\partial x^{\beta}}{\partial x^{\omega}}$$

there is a notation problem here that i am trying to resolve as well...
do i simply rearrange for L and tilde L andmultiply out??
mroe to come as i type it out

there is no need to get into partial derivatives. You only need to use $$x'^{\mu} = L_{\nu}^{\mu} x^{\nu}$$ and $$x^\alpha = \tilde{L}_{\beta}^{\alpha} x'^{\beta}$$....Putting one into the other gives
$$x^\alpha = \tilde{L}_{\beta}^{\alpha} L_{\nu}^{\beta} x^{\nu}$$. But of course, when $$\alpha = \nu$$ the two sides are equal..Obviously, $$x^\alpha = \delta^\alpha_\nu x^\nu$$. Therefore....