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Suppose we were able to build up a diamond (it could also be another material, but the structure of a diamond is very solid, literally) with the form of a cube in outer space. Will the diamond remain cubical if we make it bigger and bigger? I suppose the diamond’s mass will become big enough at a certain point to transform its cubical shape into a spherical shape, and if so how can one calculate the length of the cube's edges when this happens?

EDIT
Maybe it's easier to ask, as Anders Sandberg commented, to ask if, and when, if so, a spherical diamant will start to contract if we increase the mass of the diamant symmetrically and (almost) continuously. Will the increase in mass of the diamant (or some other solid) cause a gravitational force at the surface which will overcome the force of the solid pushing outward, and, if so, at which radius will this occur? Or will the spherical diamant grow without limit?

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    $\begingroup$ Looking at the expressions in arxiv.org/abs/1206.3857 suggests that one could perhaps find a closed form formula of when the central pressure becomes larger than the compressive strength of diamond, but it is going to be a messy formula. One can also get an upper bound by considering an already spherical body and consider when it would start compressing. Some tricky issues in the cube case with crystal anisotropy. $\endgroup$ – Anders Sandberg Mar 20 at 14:50
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    $\begingroup$ Another thing: the title of the question made me think it related to the prediction that over very long timespans solid matter behaves like a quantum liquid due to tunneling. However, a smart friend pointed out that while tunneling to lower total energy states is allowed it is not going to happen to higher states, so if there are surface forces like in crystals the equilibrium shape may well be non-spherical. $\endgroup$ – Anders Sandberg Mar 20 at 14:53
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    $\begingroup$ There probably isn't a well defined point where the diamond cube suddenly converts to a sphere. $\endgroup$ – David White Mar 20 at 15:50
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    $\begingroup$ @descheleschilder It will collapse to a white dwarf, and if you keep adding mass it will collapse to a neutron star before it's massive enough to become a black hole. Another possible option is carbon fusion, but that requires a temperature around 500 million kelvin, and very high pressure. But I don't think you can get that temperature & pressure before creating degenerate matter if you start with pure carbon. $\endgroup$ – PM 2Ring Mar 20 at 21:56
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    $\begingroup$ @descheleschilder - Yes, 2Ring is right, it will undergo several transformations before eventually becoming a black hole. And yes, carbon fusion will almost certainly set in as parts compress to neutronium (after all, that is a kind of weird fusion too). The black hole triggering issue is not the gravitational acceleration at the centre (it is always zero) but having too much mass inside a certain radius. $\endgroup$ – Anders Sandberg Mar 20 at 22:32
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If we look at the spherical case, the pressure at the core of a sphere of constant density is $$P=\frac{2\pi}{3}G\rho^2 r^2.$$ At some point this becomes higher than what the diamond lattice will support, and adding more radius will cause the core to contract. This happens at $$r_{max}=\sqrt{\frac{3P_{max}}{2\pi G\rho^2}} .$$ For diamond the compressive strength is somewhere north of 110 GPa (with simulations suggesting 223.1, 469 and 470 Gpa along different directions), so if we plug in $\rho=3.5\cdot 10^3$ kg/m$^3$ we get a max size of 8015 to 16567 km depending on whether we go with 110 or 470 GPa.

Note that this does not mean nothing happens before. Diamond has a bulk modulus of 443 GPa, which means that at 100 GPa core pressure you would get a volume contraction of about 22%. So the crystal lattice would have to accommodate some strain, and I have no doubt there would be some risk of of cracking.

After reaching $P_{max}$ other forces will come into play, in particular electron degeneracy pressure. Basically one can model the further evolution of the diamond ball as a carbon white dwarf star at zero temperature. One of the interesting things with such degenerate objects is that they become smaller with increasing mass, so it is a reasonable bet that $r_{max}$ represents the peak size of the diamond object.

What about the cubical case? It is possible to get closed-form expressions of gravity in and around a cube. They are not particularly simple, and integrating them again to get the central pressure looks like it will give very messy algebra (or maybe I am just too lazy). Doing a numerical integration in Matlab along a line towards a face, towards a corner and towards an edge midpoint (plus comparing with the spherical case of the same volume) gives the following forces and pressures for a $2\times2\times 2$ meter cube: enter image description here

The sphere case gives a slightly smaller force but it extends further than the center-side distance, while giving more force than the corner directions that extend even further. Approximating the cube with a sphere does not give a too off answer.

We can see that the cube corners experience less force inwards than the faces, but they are significantly taller. The height wins: the pressure will be about twice bigger in their direction, and the alignment with the crystal lattice will really matter.

(EDIT: I redid my calculations, this time hopefully getting all coordinate lengths right.)

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  • $\begingroup$ Great answer! One question though for the spherical diamond. Can the surface gravity (if the mass of the diamond is big enough) cause the diamond to contract on the surface after which the process proceeds inward? $\endgroup$ – descheleschilder Mar 22 at 23:40
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    $\begingroup$ The gravity can become big enough on a white dwarf-density carbon ball to turn diamond into a so-called Coulomb crystal. But at this point the interior will be much more compressed. Generally the compression is largest at the core, which causes contraction to spread from the interior outward. In fact, there are models suggesting there might even be "cracking" at certain radial distances. $\endgroup$ – Anders Sandberg Mar 23 at 22:23

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