3 ms·
I think violation of conservation of momentum is more likely than new physics. Basically, if this thing produces a general relativity effect, then it is possibl
by yk 11y ago
I think violation of conservation of momentum is more likely than new physics. Basically, if this thing produces a general relativity effect, then it is possible that momentum is not conserved because the underlying metric does not have translation symmetry. (GR Energy Momentum conservation would still be in effect in that case, but classical momentum conservation would not.)
Apart from that, I am unhappy about your percentages, I think the non experimental error explanations are an order of magnitude less likely.
- Symmetry 11y agoMy physics education topped out at special relativity and statistical mechanics so I'll take your word for it.
- AnimalMuppet 11y ago> Basically, if this thing produces a general relativity effect, then it is possible that momentum is not conserved because the underlying metric does not have translation symmetry. Given the setup, it seems highly unlikely that this thing produces something that requires a GR explanation. The energies aren't high enough; there's no high gravitational field or high acceleration. > I think violation of conservation of momentum is more likely than new physics. But if GR is out, then violation of conservation of momentum is new physics.
- yk 11y agoI agree, or actually I assume that the disagreement is only about words. I would not call it new physics, if, for example, we have to couple quantum fields to GR in a not straightforward way, so in the case were we already know all the parts but not the specific combination of the parts. I would only call it new physics if we have to rewrite basic textbooks.
- AnimalMuppet 11y agoIf this showed us how to couple quantum fields to GR, that would be pretty good. Personally, I would consider that "new physics", given how unclear it is how to do so at the moment. If doing so doesn't significantly change either GR or quantum, however (and so the basic texts are still valid), I could go with your view.
- widdma 11y agoDoesn't GR conserve local momentum by construction? Since Spacetime is a Lorentzian manifold, the metric is always locally translation invariant.
- yk 11y agoYes, up to first order. ( So it seems implausible that a rather small EM thruster would provide trust, but the thing is already implausible in the first place.)