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Science: Can diamond now be a superconductor?
Physicsweb ^ | 4 April 2003 | Peter Rodgers is Editor of Physics World

Posted on 04/16/2003 5:00:15 PM PDT by Ernest_at_the_Beach

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To: blam
"It would be a very big deal.
My son did some post-grad work in this area at the University of Houston with Paul Chu. "

Paul Chu's (team's) discovery of the high temperature superconductor was a great triumph - for the process as well as the substance of the discovery.

The room-temperature superconductor has been the holy grail since '87. Even if it is as expensive as a diamond, it will be put to use in a million ways.

PS.
The Yttrium-barium-copper superconductors have replaced tons of copper cable in a test application for power delivery, for 400 feet in length under the streets of Detroit. Superconducting cables have five times the power density of copper.


21 posted on 04/16/2003 7:00:06 PM PDT by edwin hubble
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To: Ernest_at_the_Beach
"Why, it is just a rock! "

Yes, and one that burns. A neat and expensive experiment.
(It is carbon, after all).
22 posted on 04/16/2003 7:07:58 PM PDT by edwin hubble
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To: Ernest_at_the_Beach; Jack of all Trades
" the density of electrons in this layer reaches a critical value at which a Bose-Einstein-type condensate of electron pairs forms.

Why should electrons with spin of +-1/2 pair up to essentially form a particle with spin zero and a charge of -2? Where's the driving force? They're both negatively charged, the repulsive force is significant and pairing these fermi particles in this gas involves a higher energy state.

Current continues to flow from the diamond cathode through this layer to the anode, even though there is no voltage across the layer - a sign of superconductivity.

The diamond is acting like a leaky dielectric here. The only reason there is emission from the surface of the diamond is, because the potential of the gold electrode, the anode, is different from that at the surface of the diamond. The field between that surface will be low, because most of it is across the diamond, but it's still there. All that's being observed here is that the mean free path of the electrons in any other direction than the field is much larger than the gap. ie. The electrons don't bump into each other, or into anything else, on the way to the gold plate. Low density electrons, flying though a vacuum, is not a superconducting system.

Also the gap can't be at the same potential as the gold. The gold is positive and the gap is less positive, by there respective makeups and geometry.

It's an expensive 2Mohm resistor.

23 posted on 04/16/2003 7:53:35 PM PDT by spunkets
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To: BraveMan
FYI
24 posted on 04/16/2003 8:35:37 PM PDT by LostThread (no way am I givin' up my rocks for science...lol)
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To: HuntsvilleTxVeteran
They are building on the work of Dr. Emmett Brown, inventor of the Flux Capacitor rated at 1.21Gigawatts.
25 posted on 04/16/2003 8:46:05 PM PDT by ffusco ("Essiri sempri la santu fora la chiesa.")
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To: edwin hubble
Di-Lithium Crystal?
26 posted on 04/16/2003 8:47:13 PM PDT by ffusco ("Essiri sempri la santu fora la chiesa.")
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To: spunkets
Why should electrons with spin of +-1/2 pair up to essentially form a particle with spin zero and a charge of -2? Where's the driving force? They're both negatively charged, the repulsive force is significant and pairing these fermi particles in this gas involves a higher energy state.

I think Bardeen, Cooper and Schrieffer answered this question back in, oh, 1957. Long range interactions, y'know...not only that, once you're dealing with bosons, not fermions, you don't have to follow the same statistics for the distribution in energy states--that is, with fermions, you're limited to 2 per state, but you can have as many bosons as you want in a given state.

27 posted on 04/16/2003 9:58:22 PM PDT by HassanBenSobar (I now inform you that you are too far from reality!)
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To: HassanBenSobar
Yes, Cooper pairs. Those rely on the presence of lattice nuclei though. It involves phonon exchanges that are only possible, because of electrostatic effects of those nuclei and the temperature related regularity and preponderance of lattice's characteristic vibrational modes. In this case there's just a field over a range long enough to always have a measurable potential difference, unless the field between the diamond and the gold has a standing wave pattern from some regular vibrations, or movements of electrodes, or charges.
28 posted on 04/16/2003 11:16:30 PM PDT by spunkets
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To: Ernest_at_the_Beach
Gold is just a weak metal. But it sure is purtee.
29 posted on 04/17/2003 1:04:36 AM PDT by rmlew ("Millions for defense, but not one cent for tribute.")
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To: Ernest_at_the_Beach; Psycho_Bunny; big bad easter bunny
psycho bunny knows the real deal about diamonds and supeconducting!!! ; )

now bunny, i will ask you the same thing that i asked your relative, big bad easter bunny, shouldn't you be coloring eggs somewhere instead of Freeping? ; )

30 posted on 04/17/2003 3:06:18 AM PDT by xsmommy
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To: LostThread
hey Lost Thread, knew i would find you here! we are big fans of diamonds, huh??
31 posted on 04/17/2003 3:07:42 AM PDT by xsmommy
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To: xsmommy
I am the top bunny, I have employees!
32 posted on 04/17/2003 9:22:11 AM PDT by big bad easter bunny
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To: big bad easter bunny
can you at least tell me what i am getting in my easter basket this year? a nice shiny semi-conducting diamond would be nice : )
33 posted on 04/17/2003 9:23:27 AM PDT by xsmommy
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To: Ernest_at_the_Beach
BTFL
34 posted on 04/17/2003 9:54:00 AM PDT by techcor
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To: xsmommy
the bigger the semi-condutor the better, eh?....lol

(I figured, you'd be on this thread too!!)
35 posted on 04/17/2003 1:27:03 PM PDT by LostThread (no way am I givin' up my rocks for science...lol)
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To: Ernest_at_the_Beach
Ah yes...

The advent of the Dilithium crystal.

Damn it Jim
36 posted on 04/17/2003 1:31:22 PM PDT by pray boy
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