Trojan666ru said:
Will there be any deformations to a spinning disk?
There are arguably many more subtleties involved with a spinning disk than with a uniformly accelerating rod. See here for a more or less detailed analysis: http://www.projects.science.uu.nl/igg/dieks/rotation.pdf
Also, you have to be specific with the term "deformation". Take for example a disk spinning with constant angular velocity ##\omega##. Each point on the disk has its own worldline and the worldline is an integral curve of the vector field ##\xi^{\mu} = \gamma \partial_t^{\mu} + \gamma \omega \partial_{\phi}^{\mu}## where ##\gamma = (1 - \omega^2 r^2 )^{-1/2}## and we are working in cylindrical coordinates. So I can represent the worldtube of the entire disk by ##\xi^{\mu}##. If by "deformation" you mean shear and expansion stresses to the disk then in this case there will be none. The expansion tensor (which codifies both shear and expansion stresses) vanishes identically for this system: ##\theta_{\mu\nu} = h_{\mu}{}{}^{\alpha}h_{\nu}{}{}^{\beta}\nabla_{(\alpha}\xi_{\beta)} = 0## where ##h_{\mu\nu}## gives the
spatial distances between neighboring points on the disk relative to ##\xi^{\mu}##. In fact ##\theta_{\mu\nu} = 0## is equivalent to Born rigid motion so it is clear that there are neither shear nor expansion stresses in this system.
If, on the other hand, you are using "deformation" to mean a change in the "shape" of the disk then this is quite subtle. The "shape" of the disk is an aspect of its
spatial (i.e. not space-time) geometry and as such has no frame invariant meaning. In fact, going back to the case of the uniformly rotating disk, there is no natural way of defining the spatial geometry of the disk relative to a family of observers each comoving with a respective point on the disk because the vector field ##\xi^{\mu}## above which describes the worldlines of these obeservers is not irrotational: ##\omega^{\mu} = \epsilon^{\mu\nu\alpha\beta}\xi_{\nu}\nabla_{\alpha}\xi_{\beta} \neq 0## so what this means is that there is no
natural notion of simultaneity between these observers and hence no natural notion of spatial geometry of the disk relative to these observers. See the following link:
http://arxiv.org/pdf/gr-qc/0311058v4.pdf and take heed of the following quote from p.65 of the linked paper: "Every notion of simultaneity has associated a different notion of spatial length, and therefore a different radius and circumference length will a appear at each non-inertial observers, namely the disk 3-geometry will be simultaneity-dependent."
However there is a non-canonical notion of simultaneity that is well adapted to the above scenario that gets around the fact that the congruence of observers comoving with the disk is not twist-free. This is the simultaneity convention given by (local) radar distance. Under this the spatial geometry of the disk does become curved. See here:
http://en.wikipedia.org/wiki/Born_coordinates#Radar_distance_in_the_small
Have fun with the readings!
