What would the periodic table of a 4-Dimensional universe look like?

In four dimensions, you don't have a rotation axis (fixed straight line) but a rotation plane (fixed plane). However, not every 4D rotation has a rotation plane; there are rotations which have no fixed points (except for the origin). Indeed, rotations in 4 dimensions have six parameters instead of the three we know from 3D space. The corresponding rotation group is known as $SO(4)$ (as opposed to $SO(3)$ for 3D space). You can read everything about it on Wikipedia.

To obtain the corresponding quantum spin, we have to look for its universal covering group. The universal cover of $SO(4)$ is $S^3_L\times S^3_R$ (see the linked Wikipedia article) which is a double-cover of $SO(4)$ (just like in the $SO(3)$ case). Here $S^3$ is the group of unit quaternions. Since the group of unit quaternions is isomorphic to $SU(2)$, this means that the universal cover of $SO(4)$ is isomorphic to $SU(2)\times SU(2)$.

This gives a much richer structure to the spin of 4-dimensional particles. While in three dimensions, the representation is labelled by one number (the total spin), four-dimensional particles are classified by two numbers, which might be termed the left-spin and the right-spin, corresponding to the left and right Clifford rotations.

The simplest particle would still be the spinless particle, with spin $(0,0)$. However the lowest non-spinless particles would come in two sorts, with spin $(1/2,0)$ and $(0,1/2)$. Each of them would have only two levels. However it might be that there's an additional symmetry between left-spin and right-spin, in which case those two different particle types could indeed be seen as one type of particle with four different spin states. However that's not really necessary; it is just as well possible that particles with spin $(1/2,0)$ are distinguishable from particles with spin $(0,1/2)$.

However let's for simplicity assume that there is indeed such a symmetry. And let's assume that electrons are such $\{1/2,0\}$ particles (using curly braces to stress that the order no longer matters since all orders are included). Then you'd indeed get four electrons per level (ignoring fine structure effects).

However that would not be the only effect on the periodic table: Also the orbit angular momentum of the electrons would be guided by the two quantum numbers; however in analogy to in the 3D case you'd only get actual $SO(4)$ representations (that is, integer quantum numbers). Thus where you get one pair of quantum numbers $l$ and $m$ for 3D, you'd get two of them for 4D.

So assuming that the main quantum number is not affected, you'd get the following orbital quantum numbers:

(n; l1, m1; l2, m2; s1; s2)

Assuming that in leading order the energy is still dominated by $n$, and the restrictions on $l$ are individually as in 3D (one could explicitly check that, but that's more than I'm willing to do that late in the night), you'd therefore get the following lowest degeneracies for each $n$ (lifted by fine structure), given by (left angular momentum)×(right angular momentum)×(Spins):

• $n=1$: fourfold degeneracy ($1\times 1\times 4$)
• $n=2$: $64$-fold degeneracy ($4\times 4\times 4$)
• $n=3$: $324$-fold degeneracy ($9\times 9\times 4$)

Note that when the first three shells are filled, we are already at element number 392.

Unfortunately I don't know enough about nuclear physics to say what the magic numbers there would be, and up to which element number nuclei would be still stable.

Note also that even if you assume that $(1/2,0)$-spin particles and $(0,1/2)$-spin particles are inequivalent, and electrons are e.g. $(1/2,0)$ particles, this would only cut the numbers above in half.

Edit: I noticed that I overlooked the most crucial difference in four dimensions: Thanks to the Maxwell equation $\vec\nabla\cdot\vec E = \rho/\epsilon_0$ we get for a point charge in $d$ dimensions a field that falls off as $1/r^{d-1}$, and therefore a potential that falls off as $1/r^{d-2}$. For four dimensions, this has far-reaching consequences:

• In quantum mechanics, an attractive $1/r^2$ potential has no ground state. Now, the real potential will deviate from that in the nucleus, since the nucleus has finite size. However that means that the position of the ground state depends very much on the charge distribution of the nucleus; unlike in three dimensions, the approximation as point charge will not work well. This means that the periodic table might be quite messed up, since the charge distribution depends not only on the number of protons, but on the number of neutrons (because that number enters the size). This might cause noticeable differences between atoms which only differ in the number of neutrons (and therefore in our 3D world would have basically the same properties).
• Since the centrifugal potential also goes with $1/r^2$ (but is repulsive) irrespective of dimension, and thus in 4D would be of the same form as the attractive force of the nucleus, outside of the nucleus any angular momentum would simply act like a reduction of the nucleus charge. This especially means that there's a maximal angular momentum that can be achieved before electrons stop being bound.

posted about 9 years ago

celtschk‭

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