Path: utzoo!utgpu!news-server.csri.toronto.edu!rpi!usc!cs.utexas.edu!bcm!dimacs.rutgers.edu!aramis.rutgers.edu!athos.rutgers.edu!nanotech From: Howard.Landman@eng.sun.com (Howard A. Landman) Newsgroups: sci.nanotech Subject: Re: Diamonds? Keywords: diamonds, bones Message-ID: Date: 19 Apr 91 04:30:49 GMT Sender: nanotech@athos.rutgers.edu Organization: Sun Microsystems, Mt. View, Ca. Lines: 48 Approved: nanotech@aramis.rutgers.edu In article neufeld@aurora.physics.utoronto.ca (Christopher Neufeld) writes: >I'd design my diamond-building machines to be dominantly atomic step >extenders. They would come in with a methyl group on a long stick, and >would insert it at the edge of a step from one crystal plane to the >other. In doing so they would trade the methyl for the hydrogen on the >surface of the crystal, ie. the surface of the crystal would always be >fully hydrogenated, with the crystal being built by substituting a >methyl group for one or more hydrogens on the surface. ^^ ^^^^ The two words "or more" are critical here. Without their presence, I can prove mathematically that the above process doesn't work, i.e. is incapable of producing a diamond lattice. There are at least two approaches, one graph-theoretical, one combinatorial. Before you look at my proofs, you should pause to try to think this through. It's actually pretty obvious. THE GRAPH-THEORETICAL PROOF A diamond lattice, considered as a graph, contains many cycles. Substituting a methyl radical for a hydrogen does not create a cycle. Therefore, if we define a "diamond" carbon atom to be one that is contained in at least one cycle, such a step is incapable of creating a new "diamond" atom. In fact, if we start with a tree (connected acyclic graph), each step leaves us with a tree. Starting from methane, all you can build are saturated hydrocarbons. THE COMBINATORIAL PROOF Diamond is pure carbon. Therefore, any process to produce diamond must end up with mostly pure carbon. When you replace a hydrogen with a methyl radical, you add one carbon atom but also add (net) two hydrogen atoms. Thus the asymptotic composition of the product must be two hydrogens for every carbon, which is not diamond, but something more like polyethylene. The first proof implies that we need to have at least one step in any diamond building procedure which closes loops. The second proof implies that we need to have at least one step in any such procedure which removes (net) hydrogen without removing carbon (this could of course be the same step that closes loops), OR that the step which adds a methyl carbon removes ON AVERAGE about three hydrogen atoms. Not so simple sounding anymore, is it? -- Howard A. Landman landman@eng.sun.com -or- sun!landman