In this question, I asked about a universe with (amongst other things) 4 large spatial dimensions.

In 3 dimensions, we have the familiar periodic table with its familiar arrangement of atoms in the S, P, D, F and the predicted G & H blocks.

However, in four dimensions (with the assumption that electrons, neutrons and protons exist), what would the periodic table look like, given that there is another dimension in which to put electrons? Also, since in four dimensions, objects can have two axes of spin, would electrons have four possible spin states rather than two as in 3-dimensions, resulting in yet more electrons per shell?

In addition, the atomic islands of stability are predicted in three dimensions by the nuclear shell model. Given an extra dimension, as well as the possible extra two spin-states, what would the magic numbers for protons and neutrons be in 4 dimensions?

  • 4
    $\begingroup$ Paging Greg Egan... here's what you can explore now that Orthogonal is finished. $\endgroup$
    – JDługosz
    Jun 12, 2015 at 6:24
  • 4
    $\begingroup$ Also related, from Wikipedia: Finally, Tangherlini showed in 1963 that when there are more than three spatial dimensions, electron orbitals around nuclei cannot be stable; electrons would either fall into the nucleus or disperse. I don't know what assumptions were made there, though. This paper suggests that in some cases, his analysis was incomplete or incorrect. However, it only considers dimensions $D \geq 5$. $\endgroup$
    – HDE 226868
    Sep 18, 2015 at 2:06
  • 2
    $\begingroup$ @HDE226868, I was aware of this. What I'm after is a 4D universe that is sufficiently similar to our 3D one while still capable of existing - if that takes a little modification (or dismissing) of the laws of physics as we know them, so be it, but I'd like to be able to point to atoms and say "The extra dimension does this...". $\endgroup$
    – Monty Wild
    Sep 18, 2015 at 2:12
  • 3
    $\begingroup$ @JDługosz I think you mean Diaspora. Although Diaspora (highly recommended) deals with 5D, some of the consequences are valid in 4D as well - in particular, there are no stable orbits, so there are no chemical elements, no solar systems, no galaxies. However, the uncertainty principle keeps electrons crashing into the nucleus, so you can have "atoms" of sorts - but the "chemical" energies comparable with nuclear binding energy, so the periodic table would be rather complicated (chemistry and nuclear physics blend together). $\endgroup$ Sep 20, 2015 at 7:25
  • 1
    $\begingroup$ @RadovanGarabík I don't think the uncertainty principal is "what keeps electrons from crashing". It's quantization of energy levels and exclusion. $\endgroup$
    – JDługosz
    Sep 20, 2015 at 7:45

1 Answer 1


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.
  • 22
    $\begingroup$ Wow. You could have just made all that up and I'd be none the wiser. $\endgroup$
    – Seth
    Mar 27, 2015 at 1:26
  • $\begingroup$ From my reading on this subject, I thought that the number of electrons in each shell was 2 x Floor([Dimensions]/2) x ([0-based shell number] x ([Dimensions] -1) +1), so in 3D, S=2, P=6, D=10 & F=14, while in 4D S=4, P=16, D=28 & F=40. $\endgroup$
    – Monty Wild
    Mar 27, 2015 at 2:23
  • 2
    $\begingroup$ wow... and... what does that mean? $\endgroup$
    – Burki
    Jun 12, 2015 at 8:12
  • 1
    $\begingroup$ @Seth & Burki: Do you guys really thought you could understand 4D particles while living in a 3D world? I don't think atoms (as we know them) would really exist as such in a 4D world. The primary structures of matter would be far far more complex. $\endgroup$ Sep 16, 2015 at 10:06
  • 5
    $\begingroup$ My understanding is that you can't have a stable orbit in a 4D world. In a classical world that would play havoc with the electron shells but in a QM world I'm utterly unqualified to address what might happen. $\endgroup$ Sep 21, 2015 at 23:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .