Wireless power with non-radiative magnetic resonances !

[Super God]

Right on.  Tesla did nearly everything we are 'discovering' about a hundred years ago.  He was a true genius!

[Paul-R]

Soljacic seems to be saying that the radiator radiates power, and the
receiver receives power by virtue of being tuned. But the power which
radiates out and does not hit the receiver is NOT wasted. How can
this be?
Paul.

[omnispace]

Soljacic seems to be saying that the radiator radiates power, and the
receiver receives power by virtue of being tuned. But the power which
radiates out and does not hit the receiver is NOT wasted. How can
this be?
Paul.

I believe it is re-absorbed by the radiator.  The physics might be a little beyond my capabilities but I will try to explain it anyway:

If there is no receiver present, the radiator acts like a normal inductor.  It doesn't have any significant resistance; it just stores energy and returns it later.

If a receiving coil (with the same resonant frequency) is within a few meters, the magnetic field energy isn't just stored, it is transferred to the receiver.  The receiver isn't close enough for the coils to act like a transformer, though.  Instead the waves (photons?) are able to "jump" across a few meters of air, similar to quantum tunneling.  Here's an informative link:

http://en.wikipedia.org/wiki/Evanescent_wave_coupling

Evanescent waves are also mentioned in my undergrad-level electromagnetics book, so I don't think this is fringe science, just new technology.

[hartiberlin]

This sounds right, as a parallel L C circuit has a very high ohmical resistance, so at the resonance frequency the transmitter just sees a very high output resistance only and thus the transmitter does not draw much power...
I wonder, how this changes, if a receiver L C circuit with the same resonance frequency is near by. Then probably the Q of the transmitter LC tank will change and thus it will draw more power from the power supply.