lorentz transformations

[stunner5000pt]

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

[Hurkyl]

You didn't finish the statement of the problem!

Anyways, why did you differentiate anything?

[stunner5000pt]

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

[dextercioby]

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]

There's something spooky with the latex compiler today....


Daniel.

[nrqed]

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....