- #1
Dmitry67
- 2,567
- 1
http://en.wikipedia.org/wiki/Scharnhorst_effect
I found one attempt to explain why it can't be used for FTL signalling:
http://arxiv.org/abs/gr-qc/0107091
but I don’t find that article convincing.
But I’ve got an alternative idea. Say, we make a very long tube. Because of the Casimir effect the smaller the tube is the stronger is the force vacuum applies to the tube and the stronger the Scharnhorst effect is at the same time. But in order to resist the Casimir effect the walls of the tube must resist pressure, hence, they create gravity (even if we forget about the atoms they made of). So light will propagate inside the tube FTL, but, it will be in the gravitational well. So, that effect (even it is tiny) can overweight the Scharnhorst effect which is even tinier.
In fact the gravity from the walls will be stronger for any realistic material, but I bet the pressure alone is enough and it is more fun to prove that it is independent of the material. Am I right?
I found one attempt to explain why it can't be used for FTL signalling:
http://arxiv.org/abs/gr-qc/0107091
but I don’t find that article convincing.
But I’ve got an alternative idea. Say, we make a very long tube. Because of the Casimir effect the smaller the tube is the stronger is the force vacuum applies to the tube and the stronger the Scharnhorst effect is at the same time. But in order to resist the Casimir effect the walls of the tube must resist pressure, hence, they create gravity (even if we forget about the atoms they made of). So light will propagate inside the tube FTL, but, it will be in the gravitational well. So, that effect (even it is tiny) can overweight the Scharnhorst effect which is even tinier.
In fact the gravity from the walls will be stronger for any realistic material, but I bet the pressure alone is enough and it is more fun to prove that it is independent of the material. Am I right?