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....