I'm sticking my neck out because I just worked this out and may regret this post later.
The spaceship paradox arises because the expansion scalar of the 'Bell' congruence is always positive, implying that the thread will break, even when the ships are not separating.
For this vector field ##V=\gamma \partial_t + \gamma\beta \partial_x## where ##\beta## is a function of ##t## I find
##\Theta=\frac{d\gamma}{dt}=\gamma^3\ B\,\left( \frac{dB}{d\,t}\right) ##.
I suggest that there are three congruences here, corresponding to
##\Theta<0,\ \Theta=0,\ \Theta>0##.
The 'Bell' congruence is ##V## with ##\Theta > 0##, and the case where there is no separation is obviously ##\Theta=0##. This ties in with a number of other calculations.
If a congruence forms a group closed under Lorentz transformation then the above division is necessary, because there is no LT which takes a member of the ##\Theta>0## subset of ##V## to the ##\Theta =0## subset.
I rest my case.
The spaceship paradox arises because the expansion scalar of the 'Bell' congruence is always positive, implying that the thread will break, even when the ships are not separating.
For this vector field ##V=\gamma \partial_t + \gamma\beta \partial_x## where ##\beta## is a function of ##t## I find
##\Theta=\frac{d\gamma}{dt}=\gamma^3\ B\,\left( \frac{dB}{d\,t}\right) ##.
I suggest that there are three congruences here, corresponding to
##\Theta<0,\ \Theta=0,\ \Theta>0##.
The 'Bell' congruence is ##V## with ##\Theta > 0##, and the case where there is no separation is obviously ##\Theta=0##. This ties in with a number of other calculations.
If a congruence forms a group closed under Lorentz transformation then the above division is necessary, because there is no LT which takes a member of the ##\Theta>0## subset of ##V## to the ##\Theta =0## subset.
I rest my case.
0 commentaires:
Enregistrer un commentaire