Lorentz transformations

stunner5000pt · Mar 29, 2006

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

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

also
2... [tex] \frac{\partial x'^{\alpha}}{\partial x^{\omega}} =L_{\beta}^{\alpha} \frac{\partial x^{\beta}}{\partial x^{\omega}} [/tex]

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 · Mar 29, 2006

You didn't finish the statement of the problem!

Anyways, why did you differentiate anything?

stunner5000pt · Mar 29, 2006

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

dextercioby · Mar 30, 2006

Speaking of notation, since [itex] \Lambda [/itex] is a matrix, we use these conventions to denote its elements

[tex]\Lambda^{\mu}{}_{\nu} ^[/tex]

for the direct transformation matrix.

and

[tex] \Lambda_{\nu}{}^{\mu} [/tex]

for its transposed.

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

Daniel.

dextercioby · Mar 30, 2006

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

Daniel.

nrqed · Mar 30, 2006

stunner5000pt said:

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

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

also
2... [tex] \frac{\partial x'^{\alpha}}{\partial x^{\omega}} =L_{\beta}^{\alpha} \frac{\partial x^{\beta}}{\partial x^{\omega}} [/tex]

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 [tex] x'^{\mu} = L_{\nu}^{\mu} x^{\nu} [/tex] and [tex] x^\alpha = \tilde{L}_{\beta}^{\alpha} x'^{\beta} [/tex]....Putting one into the other gives
[tex] x^\alpha = \tilde{L}_{\beta}^{\alpha} L_{\nu}^{\beta} x^{\nu} [/tex]. But of course, when [tex] \alpha = \nu [/tex] the two sides are equal..Obviously, [tex] x^\alpha = \delta^\alpha_\nu x^\nu [/tex]. Therefore....

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Lorentz transformations

stunner5000pt

Hurkyl

stunner5000pt

dextercioby

dextercioby

nrqed

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