This is a purely theoretical question about "perfect" solids under "perfect" conditions.

Assume you have a crystalline solid with a perfect crystal lattice (i.e. no defects). Let's imagine a cube of pure diamond, perhaps 2 cm^3 in volume.

Now imagine that this solid was perfectly cleaved along a perfect single plane, with no damage to the crystal lattice on the exposed surfaces. (i.e. The cube is split into two halves, exposing two perfectly planar surfaces with no damage or loss of material.)

Furthermore, for the sake of discussion, assume that the plane of cleaving is also perfectly aligned with one of the "natural" axes of the crystal lattice.

Given these two halves of the crystalline solid, would it be possible to "reassemble" them into the original solid by perfectly aligning the two halves and pressing them together?

Given that the "cleaving" of the original solid involves breaking the inter-atomic or inter-molecular bonds holding the solid together, it seems reasonable to suggest that the solid can be reassembled by placing the two halves in their original positions, and, by exerting sufficient pressure (or energy?), cause these bonds to re-form.

In essence, we are talking about re-assembling a solid crystal without any "glue", and ending with with a single solid indistinguishable, and just as "strong" as the original.

What physical properties would prevent this from working (under ideal conditions), or is it theoretically possible? If possible, how would the required force compare to the force of cleaving? Where does the energy expended during cleaving go, and how is energy re-absorbed during re-assembly?

  • 1
    $\begingroup$ They would have to be perfectly clean - look up "vacuum welding". $\endgroup$ Oct 7, 2016 at 23:53
  • $\begingroup$ This happens in one of Asimov's Foundation stories. Not that that makes it true, just feel the need to chime in. $\endgroup$
    – zeldredge
    Oct 7, 2016 at 23:54
  • $\begingroup$ Vacuum welding is not what I would think of first, but rather cold welding where very clean metals in a vacuum can end up welded together through mere contact. (Perhaps vacuum welding is just another name for this, but searching for vacuum welding turned up vacuum cementing, an entirely different thing) $\endgroup$
    – tpg2114
    Oct 7, 2016 at 23:58
  • $\begingroup$ Cold welding does seem to be very similar to what I am describing, thanks! However, it seems to be a phenomenon observed in metals particularly, and I am also curious about non-metallic solids like the crystal mentioned. $\endgroup$ Oct 8, 2016 at 0:01
  • $\begingroup$ Optical contact bonding may be the phenomenon I am describing, but I'm afraid to "answer" the question because I don't know enough about the phenomenon to respond in detail. @tpg2114 $\endgroup$ Oct 8, 2016 at 0:09

3 Answers 3


For example, see cold or contact welding of ultraclean, similar metallic surfaces under ultrahigh vacuum conditions.

After a few such experiences with what I thought were cleverly designed friction fittings for some electron beam optics, I soon learned to either use different metals, or sprinkle a bit of dry molybdenum disulfide on the joints to dirty them up!

Of course, as the article makes clear, these are nano or micro crystals. You could probably get an article in Science if you could do it with silicon. Or maybe a patent. Entropy is against you, and gets worse as size goes up. You can actually calculate this. Expect to test for all of the many dislocation types known to crystallographers.


One word answer: Yes! To add a little to Peter Diehr's reference to cold welding, here is a physically insightful argument that the answer to your question is yes due to Richard Feynman. He asks the rhetorical question: suppose we re-align the two halves so that all the atoms on either side of the cleave are in the same positions as they would be in the periodic lattice of a single connected piece of the same metal. Where then is the information that tells which former half of the rejoined piece each atom formerly belonged to? Or, more "teleologically": how does each atom know which half it formerly belonged to? Feynman deduces that there can be no difference between the re-aligned two halves and the single connected metal block.

Of course, this argument tacitly assumes that the states of all the delocalized electrons have reached the steady state that they would have in the single, rejoined piece of metal - witness that the steady states for the rejoined and re-aligned systems are indeed identical - and that there is no energy or other barrier to the electrons' reaching this steady state. If there were, then there would be an encoding of where the barrier lay.

But, assuming there is nothing stopping progress to the steady state for the electrons, the argument holds good. So it's not a perfect argument, but it is a simple, compelling one.

  • 2
    $\begingroup$ one maybe should add that the joined energy level is lower than the split, there is an attraction; also perfect alignment would only need to bring the geometry together for a single unit cell in space on the two surfaces, i.e. a rotation, and a tiny slide, not the total crystal face. $\endgroup$
    – anna v
    Oct 8, 2016 at 3:23

What physical properties would prevent this from working (under ideal conditions), or is it theoretically possible?

Yes it is theoretically possible. The things that prevent it from working are (1) Dirt / oxidation / etc. obviously; (2) Depending on the cleave plane and material, there may be surface reconstruction, which would most likely mean that the reassembled block will have a bunch of defects at the interface.

If possible, how would the required force compare to the force of cleaving? Where does the energy expended during cleaving go, and how is energy re-absorbed during re-assembly?

There should be no required force; they will rejoin spontaneously and release heat when doing so. The amount of heat released is (ideally) equal to the work expended when cleaving it earlier.

  • $\begingroup$ Actually this seemed like the most direct and complete answer, other than lack of references for some of your assertions. :) Richard Feynman (see above) qualifies in this case. $\endgroup$ Dec 4, 2017 at 0:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.