I Michelson-Morley Experiment: Objection Explained

  • I
  • Thread starter Thread starter Sonuz
  • Start date Start date
  • Tags Tags
    Experiment
Sonuz
Messages
2
Reaction score
1
Could you please elucidate the below statement which is given as an objection to one of the possible explanation(earth drags the ether surrounding to it) for the negative result of Michelson-morely inferometer experiment?

A second objection arises from the fact that a transparent object of laboratory size does not drag the light waves with the full velocity of the moving matter, as it necessarily would do if it completely dragged the ether along with it; and the observed partial drag is fully accounted for by current electromagnetic theory

Thank you.
 
Physics news on Phys.org
You will get more helpful answers if you provide the source of that quotation and tell us what you’re finding unclear
 
The usual objection to ether dragging is that if the air can drag ether 100% (necessary for a null Michelson-Morley) then glass or water ought to, but Fizeau's experiments showed that it doesn't.

As Nugatory says, more detail on what you are reading and what you don't understand would help.
 
Last edited:
  • Informative
  • Like
Likes vanhees71, FactChecker and Dale
I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté ! If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees? How about if I'm looking at this paper outside of the (reasonable)...
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...
Back
Top