# Why 3+1 dimensions?

1. May 8, 2006

### hossi

Hi there,

last week I got stuck on a paper by Randall and Karch, addressing the question why we live in 3 dimensions

"[URL [Broken] to Three Dimensions
http://xxx.lanl.gov/abs/hep-th/0506053

We propose a new selection principle for distinguishing among possible vacua that we call the "relaxation principle". The idea is that the universe will naturally select among possible vacua through its cosmological evolution, and the configuration with the biggest filling fraction is the likeliest. We apply this idea to the question of the number of dimensions of space. We show that under conventional (but higher-dimensional) FRW evolution, a universe filled with equal numbers of branes and antibranes will naturally come to be dominated by 3-branes and 7-branes. We show why this might help explain the number of dimensions that are experienced in our visible universe.

Though I wanted to put the question aside, it kept running through my head the whole weekend (okay, among one billion other things). Now I wonder whether anyone has some further references or remarks on the question. (The question of why 3 as well as why Lorentzian.)

http://backreaction.blogspot.com/2006/05/why-do-we-live-in-31-dimensions.html" [Broken]

If possible, don't go anthropic

Best,

B.

Last edited by a moderator: May 2, 2017
2. May 9, 2006

### arivero

When younger, I expected the answer to be some Galois-based trick.

3. May 9, 2006

### arivero

Galois is about the small permutation groups and their subgroups, and how for low dimensions they are more peculiar than usual, and at high dimensions all the hisotry is trivial. Another thing I do not know if it is related (as subgroups of rotation groups, perhaps) is the regular polyhedra, whose number is trivial for dim>3 and infinite for dim=2, while finite -exactly, five- for dim=3. (hmm I see from MathWorld that the result depends a bit of how do you define platonic properties, but still the low dimensions are seen to be peculiar)

For arguments closer to the ones in the articles you mention, I think I had some in my bag. The point is to notice that space time dimensionality D implies long range forces going as 1/r^(D-2) and then instead asking why the space time dimension is D we can try to ask why detected forces are inverse of the square of distance instead of other power. And for this question we are more into physics:

-We can notice that the centrifugal force for fixed (preserved) angular momentum goes always as inverse of the cube of the distance, and then we could ask physical forces to decrease slower than this.

-We can inquire about stability of relativistic orbits (I saw a work of Martin from the Basque University about this) and then notice that the inverse square force is singled out.

-Probably related to the stability issue, we can check for the third kepler law and see how the periods relate to the radiuses. We have that $$T^2 \propto R^{D-1} \propto A^{(D-1)/2}$$ so that for space time D=5 there is a sort of criticality. This is seen in the second law, which for $$F=K r^{-n}$$ gives $$A(t) = \sqrt{\frac Km} R^{({(-n-1)\over 2}+2)} \ t = \sqrt{\frac Km} R^{(2-{(D-1)\over 2})} \ t$$

Last edited: May 9, 2006
4. May 9, 2006

### marcus

In the past when we have talked about "Why is the world 3D or 3+1D?" here at PF this Hendryk Pfeiffer paper has come up.

http://arxiv.org/abs/gr-qc/0404088
Quantum general relativity and the classification of smooth manifolds
Hendryk Pfeiffer
41 pages
DAMTP 2004-32 (Cambridge)
"The gauge symmetry of classical general relativity under space-time diffeomorphisms implies that any path integral quantization which can be interpreted as a sum over space-time geometries, gives rise to a formal invariant of smooth manifolds. This is an opportunity to review results on the classification of smooth, piecewise-linear and topological manifolds. It turns out that differential topology distinguishes the space-time dimension d=3+1 from any other lower or higher dimension and relates the sought-after path integral quantization of general relativity in d=3+1 with an open problem in topology, namely to construct non-trivial invariants of smooth manifolds using their piecewise-linear structure. In any dimension d<=5+1, the classification results provide us with triangulations of space-time which are not merely approximations nor introduce any physical cut-off, but which rather capture the full information about smooth manifolds up to diffeomorphism. Conditions on refinements of these triangulations reveal what replaces block-spin renormalization group transformations in theories with dynamical geometry. The classification results finally suggest that it is space-time dimension rather than absence of gravitons that renders pure gravity in d=2+1 a `topological' theory."

good slogan to say no to the anthropic principle (it hardly seems necessary to accept it at this stage of the game!)

=====================

there must be a bunch of different things that are special about 3D and it might be worth reviewing them because one of them might be the reason the world is that.

Pfeiffer has a line on at least one way that 3D or 4D is special and B and arivero suggested other ways that dimensionality might be special. I'd like to know still others

Last edited: May 9, 2006
5. May 9, 2006

### arivero

Another guide is renormalizability of the forces, or perhaps more classically -and more general- just to look to the physical units (er, dimensions) of the coupling constants. Again, from $$F= k q^2 r^{-n}$$ we have that
$$[k q^2]= [L]^{n+1} [M] [T]^{-2} = [h][c] [L]^{n-2} = [h][c][L]^{D-4}$$

EDITED: Here I will confess a thing that puzzles me: The gravity coupling constant as coming from geometry has always dimension mass^2 (coming eg from the Riemann Tensor or even from the string tension) independently of the dimension of space time, but the dimension of the coupling for a force proportional to the product of the masses (ie q=m above) depends of the dimension of the space time.

Last edited: May 9, 2006
6. May 9, 2006

### hossi

The coupling constant of gravity has dimenson mass^(2+d). You see that as follows: couple the curvature scalar R to a QFT - Lagrangian, integrate over the d+4 dimensional spacetime volume. The result is the action and has to be dimensionless. The dimension of R is independent of d and 1/length^2 = mass^2. The volume of spacetime makes a factor length^d+4, or mass^-d-4. Togehter you have mass^(-d-2), so you need a coupling constant of dimension mass^(d+2). Best,

B.

7. May 9, 2006

### arivero

Yep, that is the amusing thing. One could expect that as we have this dimension fixed, some argument for the dimension of space time could be derived from here.

8. May 9, 2006

### Tyris

My best answer runs as follows: dimensions have to be at 90º to each other, otherwise they intrude onto each other.
There's only room for two 90º rotations, making three dimensions.
So three is the maximum you can have.

As long as I'm here, this seems the perfect place to slip in a quotation from Sid Mier's Alpha Centauri: "A brave little theory, and actually quite coherent for a system of five or seven dimensions - if only we lived in one." - Academician Prokhor Zakarhov on Superstring Theory

9. May 9, 2006

### arivero

It could be of some value to keep different "lenght constants" for R and spacetime, so that R goes as $$[l_R]^{-2}$$, ST volume goes as $$[l_{ST}]^{d+4}$$, and Newton constant is made of
$$G_d= {l_{ST}^{d+4} \over l_R^{2}}= {l_{ST}^{4} \over l_R^2} {l_{ST}^{d}}= ({l_{ST}^{2} \over l_R})^2 {l_{ST}^{d}}= l_P^2 l_{ST}^d$$

Last edited: May 9, 2006
10. May 9, 2006

### garrett

Here's a related question: do the manifold coordinates carry units, or are they in the metric?

I think it makes more sense, mathematically, to have them in the metric.

Of course, you can associate units with the manifold coordinates for convenience, but I don't think they can mean anything physically.

11. May 9, 2006

### hossi

Hi arivero,

I am afraid that doesn't make any sense. I was talking about the dimension of the coupling, not its actual value. You have mixed up things. However, one gets something simlar to what you find when the d extra dimensions are compactified. In that case you can integrate them out, lets say the Volume is V_d, and get an effective 4-dimensional coupling G_4 V_d = G_(d+4). Best,

B.

12. May 10, 2006

### arivero

Another "classical" quantity suggested by Kepler laws is the areal speed, and it becomes interesting when we try to put it in terms of natural units. Consider again a generic-dimension gravity force between two particles M m, as usual M>>m,
$$F=G {m M \over r^{2+d}}$$
(... and we are starting to need an standard to label G according dimensionality).
The areal speed is from the angular momentum
$$\dot A(t) = \frac 12 \frac Lm$$
but also from the force equation
$$\dot A(t) = \frac 12 \sqrt{G M r^{1-d}}$$

The natural unit of lenght we can extract from a coupling G having d extra dimensions is
$$l_d= ( {G \hbar \over c^3} ) ^{1\over 2+d}$$

And the "adimensional" (i.e. natural units) areal speed is
$$\hat \dot A(t)= c^{-1} l_d^{-1} \dot A= G^{-1\over 2+d} \hbar ^{-1\over 2+d} c^{{3\over 2+d}-1} \ G^{1\over 2} M^{1\over 2} r^{1-d\over 2}$$

We can reorder thing a bit
$$\hat \dot A= G^{d\over 2 (2+d) } \hbar ^{-1\over 2+d} c^{{3\over 2+d}-1} \ M^{1\over 2} r^{1-d\over 2} = G^{d\over 2 (2+d) } \hbar ^{d \over 2 (2+d)} c^{-3d \over 2 (2+d)} ({M c \over \hbar })^{1\over 2} r^{1-d \over 2}$$

and then we notice the peculiar thing: that for the usual space-time D=4 (d=0) this quantity does not show explicitly the gravitational constant, while for any other dimension it appears. Also we see that the case D=5 (number of extra dimensions equal to one, d=1) the areal speed (and then the angular momentum) does not depend of the orbit radious, as we had already noticed some posts above. Both observations are purely classical, but I keep wondering if it could be related to the mechanism that singles out 3branes in Randall's work. After all, the classical limit always live buried depth inside any relativistic quantum theory.

(Please forgive the heavy edition of this post, it took a bit to put all the signs right)

Last edited: May 10, 2006
13. May 10, 2006

### arivero

I'd not say I "mixed up", I'd prefer to think I "added", or even "build upon". The dimension of the coupling is the number of powers of a lenght unit, which you need to use to sit in right in the action. I just investigated (a short, one line investigation ) if such unit of length needs to be the same for the volume piece and for the riemann tensor piece.

I must confess that my subsconscient mind was probably aware of it, as I have perused a lot of stringy literature at fastforward reading speed (sort of clockwork orange, but no music)

What boils down, in all this series of arguments, is that Newton constant in usual spacetime can be visualized (when h=c=1) either as an area, or as a inverse of mass square, or as a inverse of force. Of all these, only the fact that G_N has dimensions of inverse of Force does not generalise to extra dimensions (we can easily think of n-volumes as generalising areas, but a force is a force here and at D=26). And if we want to single out a 3 brane (or usual ST) a good approach is to try to exploit the relationships that do not generalise to arbitrary D.

Last edited: May 10, 2006
14. May 10, 2006

### hossi

did your post just get longer while I wrote?

Anyway, I just wanted to point out that your argument is of course right, but it's the same argument as with renormalizability of QFTs. As it happens, dimensional regularization works only in d=4. The reason is that in higher dimensions the coupling constants (in the SM) acquire a dimension. I.e. whatever result you get through regularization (which you can still apply) it will depend on the cutoff.

I admit that I find that very intriguing, just that it's not useful unless on knows what to do with a non-renormalizable theory. If one knew, maybe one could figure out a dynamical mechanism to drive the system to d=4 or so. Well. That was rather vague, but I am a dreamer

Best,

B.

15. May 10, 2006

### arivero

Indeed it did. Here is where the quote utility becomes handy

.

Hmm I think you are right here. I had neglected to compare with the usual renormalizability argument because gravity is not renormalisable and I was relying heavily in G for my examples, but at the end both of them work because of some shared reason.

At the end we must blame Newton who defined $$F= m {d^2x\over dt^2}$$ instead of $$m {d^qx\over dt^q}$$. We live in 3+1 dimensions because Newton did it, look upon it, and saw it was good. Blessed Newton will be

16. May 10, 2006

### Hans de Vries

The number 3 comes out as the only dimension of space which allows
Yukawa potentials as solutions of the Klein Gordon equation (The latter
being the mother of all wave-equations). As a bonus all solutions are

If you want to find a localized solution of Q in:

$$\Psi \ \ \ \equiv \ \ \ Q \ e^{-iE\mathbf{t}/\hbar + ip\mathbf{x}/\hbar}$$

Where Q is a wavepacket of finite size and moving in the direction
of $x_1$ in a d-dimensional space then we find that Q is a solution of:

$$\left(1-\frac{v^2}{c^2}\right) \frac{\partial^2 \Psi}{\partial x_1^2}\ +\ \frac{\partial^2 \Psi}{\partial x_2^2}\ + \....\ + \ \frac{\partial^2 \Psi}{\partial x_d^2}\ =\ 0 \ \mbox{or}\ C\Psi$$

This shows that it is Lorentz contracted in the direction of the speed.
For v is 0 we get the d-dimensional Laplacian which must either be 0
or constant (The constant will modify the effective mass). For sperical
symmetric functions we take the Radial Laplacian in d dimensions:

$$\nabla_d^2\ =\ \frac{\partial^2}{\partial r^2}\ + \frac{(d-1)}{r}\frac{\partial}{\partial r}$$

For the solutions where the Laplacian is 0 we get:

$$\nabla_d^2\ \left( \frac{1}{r^{d-2}}\right)\ =\ 0$$

With the ordinary 1/r potential as the solution in 3d.
For the Generalized Yukawa Potential we can write:

$$\nabla_d^2\ \left( r^n e^{-r/r_o}\right)\ =\ \left( \frac{1}{r_o^2}\ -\ \frac{2n+d-1}{r_o r}\ +\ \frac{n(n+d-2)}{r^2}\ \right) \left( r^n e^{-r/r_o}\right)$$

This must be a constant so the 2nd and 3rd term must be zero, thus:

$$2n+d-1 = 0, \qquad \mbox{and} \qquad n(n+d-2)=0$$

There is only one solution: n=-1 and d=3, giving the well known
Yukawa Potential in 3 dimensions.

Regards, Hans

Last edited: May 11, 2006
17. May 10, 2006

### robphy

Last edited: May 10, 2006
18. May 10, 2006

### hossi

Hmm, that looks nice, but how do we know the potential should be Yukawa? I.e. why not $$\sim \exp (- (r/r_o)^k)$$. I suspect that would spoil the whole argument. Best.

B.

19. May 11, 2006

### arivero

Hmm for contact interactions (dirac delta potentials) there are similar arguments, it is easy to work out the solutions in one dimenstion, touchy in two, and you definitively need to do something about the coupling when in three dimensions; I guess that beyond three every hope is to be abandoned. And if you consider that Yukawa shares properties with the delta potentials (goes infinity at origin, and decays very very fast), not surprising to have related results.

Work on delta potencials was done time ago by the team of Rolf Tarrach in Barcelona, as well as some other groups, of course.

Last edited: May 11, 2006
20. May 11, 2006

### arivero

Is this a guess, or is it the real shape of the potential associated (say via Born approximation) to a masive propagator in k-1 extra dimensions?

21. May 11, 2006

### Hans de Vries

Well, It so happens that I have the particular case of

$$r^n \exp (- (r/r_o)^k)$$

with my notes:

$$\nabla_d^2\ \left\{ r^n \exp{(-(r/r_o)^k)}\right\}\ =\$$

$$\left( \frac{k^2}{r_o^{2k} r^{2-2k}}\ -\ \frac{k(2n+k+d-2)}{r_o^k r^{2-k}}\ +\ \frac{n(n+d-2)}{r^2}\ \right)\ r^n \exp{(-(r/r_o)^k)}$$

and in this case there still is only the n=-1, d=3 (and k=1) solution.
It would be interesting though to have a look at the solutions where
the Laplacian of the potential is proportional to the potential itself.

Regards, Hans

22. May 11, 2006

### hossi

Indeed. I see, it's the first term that kills the other k's. That is interesting, I will keep it in mind, thanks,

B.

23. May 12, 2006

A 4D spacetime is the natural background for interesting representations of the algebras of diffeomorphisms and gauge transformations, see http://www.arxiv.org/abs/math-ph/0210023 . Despite the obvious flaws in this paper, the preference for a 4D spacetime seems rather robust.

24. May 12, 2006

### Careful

Nobody seems to have mentioned Weyl gravitation yet where D = 4 rolls out very nicely.

Careful

25. May 12, 2006

### arivero

Yes, this can be done from a handbook on differential equations, reshaping the equation into a modified Bessel's differential equation, and then getting the answer in terms of modified Bessel's functions. The Yukawa potential appears for the 1/2 functions, which carry some sinh and cosh around.

I suppose that one could also try an ansatz $$\Phi= \sum_i {c_i \over r^i} e^{-iR}$$, because basically higher Bessel functions provide additional inverse power terms. If we are string-minded, we could claim that the radial Laplacian with extra dimensions ask for force fields expanding simultaneusly on different branes (different $$1/r^i$$ terms). One interesting point is if there is a Yukawa 1/r term dominating at large r, because in such case we could claim that the 3brane is the one that dominates for large distances.

Last edited: May 12, 2006