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- From: jbaez@riesz.mit.edu (John C. Baez)
- Subject: Physics puzzle - antigravity device!
- Message-ID: <1992Nov22.234125.13222@galois.mit.edu>
- Sender: news@galois.mit.edu
- Nntp-Posting-Host: riesz
- Organization: MIT Department of Mathematics, Cambridge, MA
- References: <1992Nov20.075008.9708@Princeton.EDU>
- Date: Sun, 22 Nov 92 23:41:25 GMT
- Lines: 35
-
- I just got back from the University of Utah, where the physics
- department provided excellent hospitality. In particular, Richard Price
- told me the following puzzle (our of a book of his, apparently).
-
- Part 1 - Place a conducting hollow (and evacuated) spherical
- shell on the table. Since (at least some) of the electrons are free to
- move in the conductive shell, gravity will pull them downwards a little
- bit. Of course, when there are more electrons near the bottom of the
- shell, they produce a repulsive electric field. Thus the electrons will
- come into an equilibrium where there are slightly more electrons near
- the bottom of the shell - distributed in such a way that the electric
- field they produce is just enough to cancel the gravitational pull on
- the electrons. (This effect will be very small.) Show that the
- resulting configuration has the following nice property: for any
- electron placed *inside* the evacuated shell, the gravitation downwards
- force will be equal and opposite to the electrostatic force created by the
- electrons on the shell.
-
- This part is for real, and according to Price this effect has actually
- been observed. Note that protons or neutrons would fall in this shell
- since they have a different charge/mass ratio.
-
- Part 2 - Now *imagine* that there are 2 kinds of mobile charge-carrying
- particles on the surface of the shell, the electron and the "selectron"
- say, with different charge-mass ratios. One can use the same argument
- as in the answer to part 1 to show that a particle with *any*
- charge-mass will levitate in the shell. Thus one has an anti-gravity
- chamber!
-
- This part is bogus. But when someone gives the answer to number 1 I
- will show how to use that answer to "prove" part 2, and then you can
- have fun trying to spot the error in my argument.
-
- So try part 1 first.
-
-