3 ms·

I can't read this in any way that doesn't just mean, "Destructive interference". Given that... NO EXHAUST, no reaction mass, no momentum exchange.

• T-A 10y ago
  I've only scanned the paper quickly (and it does look a bit unusual for a physics paper) but I think there is an obvious answer to that: it's not perfect destructive interference. You can imagine two photons with opposite polarization starting out together and propagating in almost exactly the same direction. As they go along their merry way, they diverge. Eventually, far away, you have two well-separated, ordinary photons. But close to the source, you have near-perfect cancellation, and so negligible interaction with the cavity walls, as the authors would have it.
  • amluto 10y ago
    Sorry for the accidental down vote. Button too small for finger :(
  • Aelinsaar 10y ago
    ...Then they should be detectable, and this thing should be producing light as exhaust, which it isn't. The only way this works is if the cancellation is total, which again...
  • gus_massa 10y ago
    If this were the explanation, they should get a thrust to powered ratio that is lower than the theoretical maximum with the currently accepted physics laws. The theoretical maximum is 1/c = 0.0033 mN/KW, but all the "interesting" experiments get more, for example 0.4 N/KW = 400 mN/KW.