Iyfjky

Question

Summary

The main idea behind the "cuboverse" is that spacetime distances are measured by (something close to) the sup norm or infinity norm. Under this norm, spheres (the set of points at a fixed distance from an origin) are the same as cubes, hence the name.

Other features that I have been able to derive from this are:

Geodesics are straight lines like in our world, but

Objects can basically only move at a certain constant speed, and only in one of eight special directions, $(pm 1, pm 1, pm 1)$ in Cartesian coordinates.

There is an attractive force of "gravity", and a second cohesive force that allows primordial material to form big planet-like bodies of liquid.

My question is:

Would planets be cubic in this universe? If not, what shape would they attain (octahedra, ordinary spheres, unstable, something else)?

I would like answers based on physical reasoning and supported by mathematical calculations if possible, taking into account the relevant changes to real-world physics (see the details section below).

Background

I recently discovered the science fiction writer Greg Egan. Many of his novels like Diaspora, the Orthogonal series and Dichronauts share the idea of changing one or more fundamental things about our world's physics (the number of dimensions, the metric signature of these dimensions, changes to particle physics etc.) and exploring the consequences of that change. The author keeps some science notes online relating to these works, and after reading them I got inspired to attempt to build one such world myself.

The cuboverse I imagined consists of big planets made of liquid (similar to water), one of them inhabited by a small intelligent species of eight-spiked "sea urchins", along with some other eel-like and carpet-like sentient creatures, all of them living near the surface. There are no stars in this world, so the necessary heat comes from the planet itself. I have already thought of a method of propulsion for the sea urchins and some rough details about their society. I still haven't developed the chemistry and particle physics, and I also have some questions about the biology, but first of all I would like to know whether the setting I imagined (specifically the shape of the planets and their stability) is realistic in the context of this modified physics.

Details

As a warning, I'm not at all experienced in exploring alternate world physics, it's my first time doing this, so some of the things I derived below could be wrong. Anyways, my basic idea is to change the Minkowski metric

$$ds = (-c^2 dt^2 + dx^2 + dy^2 + dz^2)^1/2$$

to a $lambda$-norm

$$ds = (-c^lambda |dt|^lambda + |dx|^lambda + |dy|^lambda + |dz|^lambda)^1/lambda,$$

where $lambda$ is a very big number (I decided not to choose the sup norm itself $lambda to infty$ because that would make geodesics non-unique). According to this, with this norm spacetime seemingly becomes a kind of Lorentzian analog of a Finsler geometry. We can calculate geodesics (which turn out to be straight lines) and define a four-momentum vector for point particles as usual, using the Lagrangian formalism.

After some calculations (I can provide details if needed), we arrive at equations for the momentum $p_i = mc: gamma^lambda-1 leftvertdfracv_icrightvert^lambda-1 operatornamesign(v_i)$ and energy $E = mc^2: gamma^lambda-1$, where $$gamma = dfrac1left(1-fracv_yc^lambdaright)^1/lambda.$$

Newton's second law $mathbfF = dfracdmathbfpdt$ still holds, so heuristically, for a generic set of particles at generic positions under the action of generic forces, the probability that the momenta have one or more components nearly equal to zero will be very small. Since $lambda$ is very big, the components of the velocity vector will generically be close to $$v_i = pm leftvertdfracp_i cErightvert^1/(lambda-1): c approx pm leftvertdfracp_i cErightvert^0: c = pm c,$$ as I claimed in the summary.¹ This implies, among other things, that it is virtually impossible for any object to stay still: its velocity will generically be one of the eight possible vectors pictured below (which one depends on which octant the momentum vector lies).

Eight possible velocity vectors in the "cuboverse"

For gravity, the most reasonable thing would be to work with a generalized² gravitational potential $V = dfracGm_1 m_2lVertmathbfrrVert_lambda$, where $lVertmathbfrrVert_lambda = (|r_x|^lambda + |r_y|^lambda + |r_z|^lambda)^1/lambda$ is the sup norm distance between two particles of masses $m_1$ and $m_2$. I managed to make some calculations but the orbits look too funky, so I decided to get rid of stars and planetary systems completely, having instead a single kind of astronomical body. If planets are cubic (that's my question!), I believe the gravity near the surface would be constant.

Since it is extremely easy to move a stationary object by applying a very small force, every big structure would become unstable under gravity alone, so I decided to have a secondary cohesive force that sticks particles of primordial matter together, while still allowing for some free fluid-like motion. I am not really sure how this force will look like since the chemistry isn't yet developed, so for the moment I'm forced to work with this rough description of how I want it to behave.

For the moment, as a starting point I'm assuming the primordial matter consists of small hard particles, let's say ordinary spheres, and studying the collapse of a cloud of this material under gravity and a perfectly inelastic contact force (these assumptions can be changed if needed for the answer). At this point the analysis becomes more difficult, and I've been unable to find whether the planets are cubes or not. Based on the form of the gravitational potential, I would expect a "yes" answer, but the weird restrictions on the velocity make me doubt. Also, I'm not completely sure that a cohesive force will fully solve the instability problem. On the other hand, things could get complicated by relativistic effects since the velocities are close to $c$.

Two final notes:

Just to prevent any possible confusion: my intention behind the question isn't to make a world with cubic planets, it's perfectly OK for me if the answer is "these planets can't ever exist in your world, even if you modify the gravitational/cohesive forces". The underlying intention is just to explore the consequences of the main premise; it's not necessary to keep the little sea urchins idea viable.

Although the context is worldbuilding, this is at its core a mathematical physics problem, and as such it most certainly has a unique right answer. I believe I have developed the basic physics enough for the question to be answerable with the information I provided. If that's not the case (if there's a free variable unaccounted for, or if the answer crucially needs concepts from the corresponding version of, for example, thermodynamics), please point out what else is needed and I'll try to rigorously work it out if it's feasible.

EDIT: Inspired by Aric's answer I tried to make a simulation myself. I'm not very good at coding so I haven't been able to make the inelastic collisions work yet. Applying the force of gravity only, it turns out that an initially static cloud of material does seem to collapse into an octahedral shape, as suggested by some of the answers. Here are the results for a spherical slice of primordial material (color represents height):

Collapse of matter under gravity only

However, since there are some points where particles tend to clump together heavily and the simulation doesn't take into account the cohesive force that would separate them, I think there's still a good chance that the real shape is more cube-like, or perhaps something in between similar to this, as suggested by JBH's answer. There's still the question of stability, that I don't know how to tackle.

Just for reference, I also found some slides online about a possible way to treat fluid dynamics in a Finsler spacetime. Most of it is over my head right now, but perhaps someone will find them useful.

_{1. - If my calculations are correct, the true speed $lVertmathbfvrVert_lambda$ is actually not so close to $c$ but a bit lower for massive particles. For example, for $m = 1.8$ kg, $lambda = 100$ and a range of kinetic energies within $(1.5 - 180) mc^2$, the speed remains between $95-97%$ of $c$. I believe the most common relativistic phenomena like length contraction or time dilation aren't expected to play a big role though, because $gamma$ is practically $1$ for almost all speeds, but I may be wrong on this, I haven't given it much thought yet.}

_{2. - One can consider the analogue of a Klein-Gordon massless field and work out the corresponding "Coulomb" interaction potential as the Green function for a static field background. Dimensional analysis of the resulting integral suggests a law of the form $V propto lVertmathbfrrVert_lambda^lambda/(lambda - 1)-3$ (which is an inverse square of the distance for big $lambda$) rather than $V propto lVertmathbfrrVert_lambda^-1$ (someone asked for my reasoning and I put it here, in case anyone else's interested). For a true gravitational force I guess I would have to look into a suitable coupling between matter and a curved Finsler geometry, and it seems there is already some work being done on this. But I don't think the specifics matter much at this point, for the moment I just want a reasonable-looking attractive force.}

Comments are not for extended discussion; this conversation has been moved to chat. — Aug 23 at 18:26
The comments on the mathoverflow post you linked seem to suggest there might be problems defining the Finsler connection on such a manifold. That gives me some doubt as to whether your question is even well posed, but I must admit that I am slightly out of my element. — Aug 24 at 8:32
@TimSeguine Yeah, I've thought about this too. The "Hessian of the squared norm" referred to in the comments is really the metric HDE226868 defined in his answer. If I understood them correctly, in my Minkowskian context this means that the metric may not be always invertible, so one can't for example use $g^mu nu$ in any equation, or define a momentum with "up indices", without being extra careful. However, I believe the Lagrangian analysis I've been doing (including geodesics and canonical "down indices" energy-momentum) still works. — Aug 24 at 8:46

score 59 · Answer 1 · 2018-08-23 15:38:24Z

TMM;DR (Too Much Math, Didn't Read):

For anyone who doesn't want to go through the derivations and calculations below, here are the important points from my answer:

We're not working with the same space as normal, friendly, Euclidean space.

This means that while we can still integrate and differentiate scalar functions defined on this space, we need to make slight corrections.

These corrections can be calculated from something called the "metric tensor", which describes the curvature of space.

There are several methods that could be used to find an equipotential surface (and therefore the shape of the planet); they should require these corrections.

If you want more details, read on!

Calculating the metric tensor

To do certain calculations in this space, it is necessary to calculate the metric tensor $g$. This is commonly used in the case of (pseudo)-Riemannian manifolds, and plays a central role in general relativity, as I assume you're aware. To do computations involving the curvature of your space, we need to know $g_ij$, the components of $g$.

In the Riemannian case, we have the (smooth), non-negative norm $F$. From it we can derive the components of the metric tensor by
$$g_ij=frac12fracpartial^2F^2partial x^ipartial x^jtag1$$
In Euclidean three-space (and in Cartesian coordinates), $F$ is the standard Euclidean distance metric
$$F=sqrtx^2+y^2+z^2$$
which is your case where $lambda=2$, and you can probably convince yourself that $(1)$ yields the familiar
$$g=
beginbmatrix
1 & 0 & 0\
0 & 1 & 0\
0 & 0 & 1\
endbmatrix$$
Here, we can extent the Riemannian definition to Finsler spaces using your $p$-norm; in your case, we're dealing with the more general
$$F=sqrt[lambda]x^lambda+y^lambda+z^lambda,quad F^2=left(x^lambda+y^lambda+z^lambdaright)^2/lambda$$
Going back to $(1)$, we have
$$beginaligned
g_ij &=frac12fracpartialpartial x^jfracpartial F^2partial x^i\
&=frac12fracpartialpartial x^jleft(2(x^i)^lambda-1left(x^lambda+y^lambda+z^lambdaright)right)^-1+2/lambda
endaligned$$
And so
$$g_ij=begincases(2-lambda)(x^i)^lambda-1(x^j)^lambda-1left(x^lambda+y^lambda+z^lambdaright)^-2+2/lambda,quad ineq j\
beginaligned
&(2-lambda)(x^i)^2lambda-2left(x^lambda+y^lambda+z^lambdaright)^-2+2/lambda\
&+(lambda-1)(x^i)^lambda-2left(x^lambda+y^lambda+z^lambdaright)^-1+2/lambda,quad i=j
endaligned\
endcasestag2$$
You can again check that this works like normal in the case where $lambda=2$.

Volume elements

So, why do we really care about $g_ij$? Well, if we look at Danijel's answer, we can see that they define the potential by
$$V(mathbfx)=intfracGrho(mathbfx)d^3mathbfx'tag3$$
where $d^3mathbfx'$ is the volume element; notice that we've allowed $rho$ to vary, as it would for a real planet. For Cartesian coordinates in Euclidean space, $d^3mathbfx=dxdydz$. However, this is not always true. Indeed, in the general case,
$$d^3mathbfx=sqrtdxwedge dywedge dztag4$$
where $wedge$ denotes the wedge product, and we see that we need to calculate the determinant of $g$, $|g_ij|$. (Note again how $sqrt=1$ for Cartesian coordinates in Euclidean space.) The calculation of the determinant is ugly, and I won't do it here because there's little point to it. I assume symmetry allows it to be written somewhat neatly, though.

Gradient

As has already been noted, the surface of the planet should lie on an equipotential surface, that is, one where $|vecnablaV|=0$. The familiar Euclidean gradient is
$$vecnabla=left(fracpartialpartial x^1,fracpartialpartial x^2,fracpartialpartial x^3right)hatx_i$$
for generalized coordinates $x^1$, $x^2$, and $x^3$, which are are $x$, $y$ and $z$. It should come as no surprise, however, that this fails for non-Euclidean spaces. In the more general case,
$$vecnabla=fracpartialpartial x^icdot g^ijsqrtg_jjhatmathbfx^itag5$$
for unit vector $hatmathbfx^i$, and again $g_ij$ plays a role.

Poisson's equation and the shell theorem

Here, of course, is the problem - and it's been bothering me for about a day now. The core question you're asking here involves determining the potential corresponding to a given density distribution. We can do that easily enough in flat space by solving Poisson's equation; the solution can be found through $(3)$. Okay. But does Poisson's equation look the same (or even similar) in this Finsler space, and can we solve it the same way? The answer may be that it doesn't. In general relativity, on a Riemannian manifold, holds in the Newtonian limit of the Einstein field equations. However, it's not clear that we can make the same assumption in Finsler space.

Let's say Poisson's equation holds in some sort of form (here, I have set the appropriate constants to $1$):
$$nabla^2V=rhotag6$$
Then, indeed, $(3)$ holds in general, in some form, again with the appropriate volume element. Now, what I'm curious about - and I've talked about this in chat a bit - is the answer to a question posted by Dietrich Epp: Is the gravitational field outside a "sphere" identical to that outside a point particle with the same mass (the Shell Theorem)? If so, your given equation for the potential holds for a planet, and
$$V(mathbfx)=fracGm_1m_2_lambda$$
If not, we need to integrate $(3)$ properly before making any assumptions about the form of the potential. At any rate, I'd be interested in seeing a proof that a version of the shell theorem (or a version of Birkhoff's theorem) holds.

A path to a solution

I don't have an answer to your question yet. There are a couple of ways you could go find an equipotential surface:

Muck about with various surfaces and find one where $(text3)$ is constant for all $V(mathbfx)$ on the surface.

Set $|vecnablaV|=0$ and find the resulting surface.

Set $V(mathbfx)$ equal to some constant $C$ and again find the surface.

The third method seems like it should definitely work in the case of a Euclidean metric, for your chosen potential. I can't say for sure whether it works for all relevant Finsler metrics (and I suspect it doesn't), so I won't claim that the general case works just yet.

The naive third case

Okay, so let's set $V(mathbfx)=C$ for some constant $C$. Here, I assume that your potential is applicable for a planet. We then have
$$||mathbfr||_lambda=fracGMCimpliesleft(x^lambda+y^lambda+z^lambdaright)^1/lambda=fracGMC$$
Therefore, we're left with
$$z^lambda=left(fracGMCright)^lambda-x^lambda-y^lambdatag7$$
Interestingly enough, as $lambdatoinfty$, this becomes a cube!

References:

Shen 2000

Lecture notes by Matias Dahl

Popvici 2017

Busemann 1950

This is an amazing answer, thanks for all the effort in writing it! It's a good point about the volume form, I have to think about whether it would invalidate Danijel's answer (though I think the heuristic observations in Logan's answer make it likely that it will still hold). — Aug 22 at 17:26
@pregunton Yes, I don't think it invalidates Danijel's answer at all; it just adds a slight layer of complexity. — Aug 22 at 17:35
Wow. What an incredible amount of effort. I wish I could upvote twice, but you'll have to take this verbose kudos instead! — Aug 25 at 20:11

score 27 · Answer 2 · 2018-08-22 18:06:10Z

up vote
27
down vote

No

The necessary condition for a liquid body to be in a gravitational equilibrium is that gravitational potential should be constant over its surface. For the cube with density $rho$, by integrating
$$
V(mathbf r) = int_cube fracG rholVert mathbf r - mathbf r' rVert_lambda d^3 mathbf r'
$$
for various locations $mathbf r$ on its surface, it is easy to see that e.g. the middle of a face of the cube is deeper in the gravitational well of than a point on its edge (or vertex).

The correct shape probably isn't a sphere, which can also be checked using the same method, but it is harder to integrate, so I didn't do it. It is still an interesting mathematical question what would be the correct shape.

edited Aug 22 at 18:06

answered Aug 22 at 14:21

Danijel

891718

Thanks for your answer! That makes sense. I tried with the alternate potential I mentioned in a footnote $V propto lVertmathbfrrVert_lambda^lambda/(lambda - 1)-3$ and it seems it's still not a cube. I guess that answers half my question, now I wonder what could be the right shape (if there's even a stable shape).
â€“Â pregunton
Aug 22 at 14:46

6

I disagree with Danijel on this point. The key is that because of the way that distance would need to be measured, the distance from the center of the cube to any of the surface points would actually be the same regardless of whether it is on the center of a face, on an edge, or on a vertex.
â€“Â Mathaddict
Aug 22 at 15:07

7

@Mathaddict The distance from a point on surface to the center of the cube is indeed the same, regardless of the choice of the point. But to integrate, you need all distances between a point on the surface and all points of the cube.
â€“Â Danijel
Aug 22 at 15:26

3

The metric still has radial symmetry and the potential will also have radial symmetry (at least for non-rotating bodies). Surely then the shape of a planet (if they can exist) should approach radial symmetry as far as possible ? I don't get your argument for "probably isn't a sphere".
â€“Â StephenG
Aug 22 at 15:57

8

@StephenG The problem does not exhibit spherical symmetry because $lVert mathbf r rVert_lambda$ isn't a spherically symmetric function of $mathbf r$, i.e. it is not a function of $r=sqrtx^2+y^2+z^2$, so a spherically symmetric solution would be quite unexpected.
â€“Â Danijel
Aug 22 at 18:11

Â |Â
show 2 more comments

Logan R. Kearsley 8,82012746 · Answer 3 · 2018-08-22 14:55:04Z

As Danijel has pointed out, the question can be reframed as "what is the shape of the equipotential surface under this force law?"

There are further two different ways to view the results: the external perspective, where we see what shapes these "planets" map onto if we overlay them on our normal Euclidean space, and the internal perspective--what would they "look like" to someone embedded in the same metric?

The external perspective is a lot easier to figure out. You just calculate the shape of the equipotential surface in Euclidean space using a different force law. Finding the exact shape is complicated, but the broad strokes are fairly clear: force reduces over Euclidean distance more slowly along the octants than it does normal to the faces, which means the potential at equal Euclidean distance is higher along the octants than the faces... and so the equipotential surface ends up dimpled inwards along the octants, rather than poking out to form cube vertices. If you actually wanted cubical worlds, you'd need a force law that decreased faster along the octants, not slower.

Exploring the internal perspective is considerably more complicated. Honestly, I'm not even certain what the best way to approach it is; trying to imagine living in a lambda-norm world breaks my brain! As a first step, I'd probably try to investigate the local curvature of the equipotential surface under that norm; is it roughly uniform, in which case the inhabitants of this universe would likely perceive their worlds to be approximately smooth, like we do ours, or does it have regular variations that would correspond to edges and vertices? With a little more formalizing (which it seems you are perfectly well qualified to do), that's a question you could probably take to the Mathematics StackExchange and get much more productive answers.

Thank you! Your observations make a lot of sense in hindsight; interestingly, this means that the planets would look more like octahedra than cubes. The question about the internal perspective sounds pretty interesting, I'll have to think about it. — Aug 22 at 15:09
I am not convinced that this is equivalent. With the simple inverse square law objects can be approximated by point masses, I am not sure this is true in this case. If it is not true, then planets would not be shaped like equipotential surfaces. They may in fact be shaped differently depending on their composition. — Aug 22 at 18:27
To clarify what I just said, the surface will still be equipotential but the equipotential surface could have a different shape for planets and point masses. — Aug 22 at 18:30

score 15 · Answer 4 · 2018-08-23 15:27:32Z

My answer isn't going to be anywhere near as in-depth as your question but I'm happy to supply a few inputs.

First, I'm going to take a look at your particles:

1. Particles can only move at one constant speed.

2. Particles can only move in six directions. (Or four in my 2D model.)

3. Particles have "gravity".

I'm going to imagine two particles in a two-dimensional plane for now:

enter image description here

We could represent gravity as a vector from one particle to another, so the top left particle would have (3,-2). To make this pull weaker with distance, we can multiply the vector by the reciprocal of its magnitude:

enter image description here

If there were other particles, we'd have to calculate the force vector for each. The resultant gravity would be the sum of the vectors, which in this example is just (3/Ã¢ÂˆÂš13, -2/Ã¢ÂˆÂš13)

Now, the particle can only move in one of the four directions in this 2D example, so let's take the vector component with the largest magnitude: x = 3/Ã¢ÂˆÂš13 and move one square in that direction. The result would be that both particles zigzag towards each other until they meet. At this point, two things can happen:

The particles stop moving.

This violates your rule that the particles "basically only move at a certain constant speed". You could change this rule to The particles are either stationary, or moving with a set speed, having a binary value for their velocity, OR:

The particles oscillate between their two positions.

This means that the particles will be fixed into an "orbit" around each other that lasts 2 time ticks, however it lets them pass through each other. This can be explained by having the particles move from one point in space to the adjacent point without interacting with, or passing through, the space between (BECAUSE QUANTUM!).

This is all fun and dandy until you hit a situation like this:

enter image description here

Provided there are no other particles in the entire universe, these two particles will have a force vector where two or more components are equal. In this case, how does the particle choose which way to go?

Solution 1: Randomly decide

Let's forget all of those God Does Not Play Dice with the Universe shenanigans, because we can't simply "move diagonally" in this system. If there's two possible directions to move in, each with an equal likelihood of occurring, one of them must happen. Just randomly pick one of the tied directions and go that way.

Solution 2: QUANTUM STUFF???

Just make the particles go into superposition. Ya know, go both ways at once. Why not? After one time tick it'll look like this (particles now coloured red and blue)

enter image description here

Now there's three possible positions for each particle, with probabilities of 25%, 50%, 25% of existing in each position.

enter image description here

Oh darn those sundries! Now we have to perform a measurement, since each superposition may be colliding with something. And the probabilities of both red and blue existing in a specific position is a binomial distribution! That's fine, just randomly decide which of the paths the two particles took... But wait! What happens if they both end up on the same square? Well, SOMETHING would have to happen. I can't for the life of me figure out a way to work around this. Maybe you can, I don't know.

The advantage of Solution 2: QUANTUM STUFF??? is that it allows the superimposed particles to interact with other, third party particles that may cause a measurement to happen early. This would, again, cause two particles to exist in the same place. The advantage of method 1: No nasty probabilities.

Okay, so you talked a lot about gravity. Stuff can clump together. But nothing can orbit without momentum!

Great point, Jenkins! Momentum can be represented as another vector. Each time tick, The resultant force of the gravitational attraction to every other particle in the entire universe is added to the momentum of the particle, forming a new momentum vector. Now, instead of moving in the direction of the strongest pull of gravity, the particles move in the direction of the strongest component of their momentum vector.

Now, all of this is great, but does it work? I will attempt to simulate this in Python and then update my post in due course.

My predictions:

Unless I find a way to prevent multiple particles moving into the same space, the whole system may collapse into a singularity.

If momentum works correctly, there MAY be some form of orbital mechanic possible.

Clumps of particles may form. If they do, they will either have a roughly circular shape, or may create "crystal" shapes such as this:

enter image description here

My theory is that a particle may be more likely to move towards another body of particles if it is vertically or horizontally in line with it, resulting in protrusions from the clump along the axes and possible diagonal edges.

The simulation

Here's what I've got so far:

import pygame
import random

(width, height) = (1000, 1000)
screen = pygame.display.set_mode((width, height))


class Particle:
 def __init__(self):
 self.position = (random.randint(-50, 50), random.randint(-50,50))
 self.momentum = (random.randint(-10, 10)/10, random.randint(-10,10)/10)

 def move(self, positions):
 resultant = (0, 0)
 for position in positions:
 vector = (position[0] - self.position[0], position[1] - self.position[1])
 magnitude = (vector[0] ** 2) + (vector[1] ** 2)
 if magnitude == 0:
 return
 resultant += (vector[0]/magnitude, vector[1]/magnitude)
 self.momentum += resultant

 if abs(self.momentum[0]) > abs(self.momentum[1]):
 self.position[0] += self.momentum[0]/abs(self.momentum[0])
 elif abs(self.momentum[0]) < abs(self.momentum[1]):
 self.position[1] += self.momentum[1]/abs(self.momentum[1])
 else:
 if random.randint(0, 1) == 1:
 self.position[0] += self.momentum[0]/abs(self.momentum[0])
 else:
 self.position[1] += self.momentum[1] / abs(self.momentum[1])

 def render(self):
 pygame.draw.rect(screen, (255, 255, 255),
 (int((self.position[0] * 10) + 495), int((self.position[1] * 10) + 495), 10, 10),
 0)


def run():
 screen.fill((0, 0, 0))

 particles = 
 for x in range(10):
 particles.append(Particle())

 running = True
 while running:
 for event in pygame.event.get():
 if event.type == pygame.QUIT:
 running = False

 positions = 
 for particle in particles:
 positions.append(particle.position)

 for particle in particles:
 particle.move(positions)

 for particle in particles:
 particle.render()

 pygame.display.flip()


run()

It either takes incredibly long to calculate the next frame, or it doesn't work. I haven't worked that out yet...

enter image description here

Nice answer. Your setup looks a lot like how I first approached the problem. Looking forward to your simulation! — Aug 22 at 18:47
According to the question, if you move in one dimension, you must move the same distance in the other two, but in either direction (positive or negative), giving 8 directions, not 6. This means a 2D analogy would only have diagonal movement. — Aug 24 at 23:19
@CJDennis this means that a 2D analogy would only have four diagonal directions to move in, so I am still correct despite my inaccuracies. — Aug 25 at 19:14

JBH 32.4k578155 · Answer 5 · 2018-08-23 19:43:05Z

I wish I had more time to play with the math, but let's use some logic.

Velocity vectors must work in both directions. If your particle can only move in an octant, then gravity can only draw in an octant. This suggests...
1. That gravity is functionally discrete. Unlike in our universe, where we could simplify and simulate this issue with two ball bearings that roll around each other's surface continuously with equal forces of gravity at any point, yours looks like the diagrams below.
2. The force of gravity over distance is wild wonky. Rather than simulating it as a growing sphere of diminishing influence, it's more like a simple star you'd plop on top of a Christmas tree. The "arms" of the star would diminish with the square of distance, but between the arms it would diminish with something more like the bazillionth exponent of distance. Random attraction in your universe would be a much, much slower process.

Let's look at the pretty pictures, which I ruthlessless appropriated from your post. I hope you don't mind.

The weakest gravitational force, the single-vertex force

enter image description here

Double the gravitational force, the dual-vertex force

enter image description here

The strongest gravitational force, the quad-vertex force

enter image description here

Unless you change the various lowest-energy-state/conservation-of-X rules in your universe, your planets will be formed predominantly by particles snapping into the quad-vertex force position. They can form in the single-vertex condition, but any little "bump" will slap them into one of the two higher gravity conditions, eventually into the quad.

At first blush, this would suggest your planets are shaped like that Christmass star or an 8-pointed pinata.

enter image description here

Except there's all those spaces in between with 6-vertex and 8-vertex connections... In the end, I believe you'd get planets that are roundy-with-corners. Think cube balloon, where the sides bulge from the internal pressure:

enter image description here

Finally, a word about orbits

I predict that all your orbits will be eclipical due to the descrete nature of the gravity vectors. You may have small/light objects that orbit on an elliptic plane tilted 45Ã‚Â° to the ecliptic, but that's the weakest gravitational path. I also suspect you won't have much in the way of planetary axial tilts (0Ã‚Â°, 45Ã‚Â°, 90Ã‚Â°).

If you'll fogive the pun, those gravity vectors will have a tendency to keep things "boxed."

Jay 1513 · Answer 6 · 2018-08-22 23:24:56Z

Edit: I think I misread the question, but the gist is: If your cube shape is equidistant from the center-of-mass, and gravity depends on distance, then the shape of the planet will be all "same-distance" from the center.

A good first approximation is an octohedron.

Why is it an octohedron and not a cube? It's because your universe has 6 special directions and they correspond to the 6 points of the polyhedron.

In this co-ordinate system, all points on the surface of an octohedron are equidistant from the center of mass.

enter image description here

It's worth noting that cubes and octohedra are "dual polygons" and the faces of one correspond to the points of the other.

Welcome to Worldbuilding, Jay! If you have a moment, please take the tour and visit the help center to learn more about the site. You'll also find Worldbuilding Meta and The Sandbox useful. Here is a meta post on the culture and style of Worldbuilding.SE, just to help you understand our scope and methods. Have fun! — Aug 23 at 0:19

Ash 20.7k252127 · Answer 7 · 2018-08-22 15:24:03Z

Material science answer; it'll always be a "gravitational sphere" (a smooth round object with a gravitationally flat surface) because that's how stable mass assemblages always come to rest, lowest potential energy state. For the sake of clarity that's "smooth", "round" and "flat" on a planetary scale.

I'm not at all sure what a gravitational sphere actually looks like to a human equivalent observer in the proposed universe but looking at it through the euclidean filter of our physical senses I'm pretty sure it'll still be a sphere.

Ã¢Â€ÂœIt will alwaysÃ¢Â€Â is not necessarily true given that OP is proposing a universe with DIFFERENT physical laws than our own. — Aug 22 at 19:36
@WGroleau Not given what he isn't changing; matter exists and is subject to something akin to gravity, ergo gravitationally spherical. — Aug 22 at 19:40
..isn't the question asking "what shape is gravitationally spherical in this universe" though? — Aug 26 at 18:37

David Thornley 4314 · Answer 8 · 2018-08-23 18:15:10Z

I'm going to try a short and simple answer.

I assume gravity works similarly to our universe, except for the calculation of distance. A gravitationally stable form is where the surface points are equidistant from the center. If some points were farther from the center than others, smoothing them out would reduce the gravitational potential energy.

In the metric you're using, the distance is the largest of the x, y, and z differences. If we take a radius of 3 for an example, the equidistant surface is that which is 3 away in one dimension, and no more than 3 in any other dimension. That gives us a cube.

If the distance was x + y + z, we'd get an octahedron, which I'll leave as an exercise for the reader.

Someone Else 37 3,4841416 · Answer 9 · 2018-09-01 06:29:59Z

Your movement rules... are very confusing, what with the eight special directions and the tendency of matter to constantly travel at the speed of light along one of them. If I'm even understanding that correctly. So, for the remainder of this answer, I will assume that the "cohesive force" that you mention causes primordial particles to group up into clusters ("atoms") that can move more freely. The primordial particles within these atoms bounce around in the eight special directions at the speed of light, but the whole atom can move in any direction at any speed, depending on what fraction of the primordial particles therein are moving in each of the special directions.

As a side note, this actually reminds me of the strong nuclear force. Protons and neutrons are chaotic messes of quarks and gluons bound together by the incredibly strong forces of quantum chromodynamics. Smaller hadrons occasionally leak out of them and can interact with any other nucleons that happen to be nearby, resulting in the forces that bind atomic nuclei together. The atoms that I describe above are chaotic messes of primordial particles held together by an incredibly strong force. Individual primordial particles (or, if you prefer, smaller "atomic" clusters) could occasionally leak out of them and interact with other nearby atoms, resulting in something analogous to either nuclear physics or chemistry.

Anyway. With that out of the way, let's look at gravity.

You give a definition right in the question:

$$ V = G m_1 m_2 over Ã¢Â€Â–mathbfrÃ¢Â€Â–_ÃŽÂ» $$

which decreases with the inverse of the sup norm of the position, presumably in analogy with gravitational potential in our universe, which decreases with the inverse of the Euclidean distance. Thus, escape velocity will work the same way, and orbits should at least exist, although their shapes will be pretty funky and may or may not be stable.

The answer to the question of "what shape will a liquid planet have" is always "whatever shape has constant gravitational potential across its whole surface." For a planet consisting of a volume of massless liquid, this is quite simple: It's a cube. In the sup norm, the entire surface of a cube is the same distance from its center, and the sup norm distance is exactly what matters here.

However, when dealing with planets, point masses covered in massless liquids make for a bit of an oversimplification. It works in our universe, thanks to symmetry and Gauss's Law, but the physics here are very different. To figure out what planets in this universe actually look like, we need to dig a little deeper.

Specifically, we need to figure out how gravity actually works. The force, I mean; not just the potential. For point masses, this is fairly simple: the direction is straight down the potential gradient, and the magnitude is equal to the slope of that gradient. Thus, the gravitational force between two point masses will always be in one of the six Cartesian cardinal directions. That is, perpendicular to the faces of a cubic planet. And the magnitude will, as in our universe, drop off with the inverse square of the (sup norm) distance.

With that, we can start to explore the gravitational fields of things that aren't point masses. Note that for computing the gravitational force along each axis, we only care about matter in a 45-degree pyramid centered on that axis. If you imagine a cube broken up into six identical square-based pyramids, such that the apex of each is at the center of the cube, each pyramid gives the volume in which matter can pull the center of the cube in one of the six axial directions.

... I'm not sure if that description made any sense at all, but no matter. Onward!

Case study 1: An infinite, flat, zero-thickness plane of uniform density, perpendicular to one of the axes. An object on either side of the plane will feel a gravitational attraction toward it. Specifically, it'll feel the gravity of a square region of the plane, with side length equal to twice the distance to the plane. The gravitational force will be proportional to the area of the square (which goes up with the square of the distance) divided by the square of the distance... so the plane creates a uniform gravitational field on either side of it. Just like in real life, with out Euclidean norms. This also means that infinite planes with non-zero thickness will also behave the same way.

Case study 2: An infinitely long, infinitely thin rod, aligned with one of the axes. An object will feel the gravity of a section of this rod with length proportional to the sup norm distance to it- and, given the inverse square law, this results in a force of gravity proportional to the reciprocal of the sup norm distance to the rod. Again, much like real life.

Case study 3: A solid cube. At extreme distances, this will look much like a point mass, but this question is asking about the shape of a liquid planet, so what really matters is the surface. At the center of each face, gravity will point directly toward the center, perpendicular to the face. At the center of each edge, half of the mass of the cube will be pull along one axis, while the other half will pull along another axis, resulting in a net force at a 45-degree angle. Points near the edge will experience only marginally different gravity, so liquids will flow down from the vertices and edges and pool up in the centers of the faces. Thus, uniform-density liquid planets won't be cubes.

Case study 4: Maybe an octahedron will do it, then? At each vertex, the entire mass of the octahedron will pull matter in the direction of one of the axes. At points near each vertex, almost all of the mass of the octahedron will exert a force in that same direction... once again, causing liquids to flow toward the center of the nearest face.

So, clearly, liquid planets in this universe will be neither cubes nor octahedra. They will be a smooth surface of some kind. The gravitational field in the vicinity of an object with finite density (i.e. if we exclude things like point masses and the zero-thickness planes and rods discussed above) will always be continuous and differentiable; therefore, for its surface to be perpendicular to its own gravitational field, its surface must also be differentiable. So what shape will the planets be? Spheres? Maybe. I don't know. They will have at least as much symmetry as cubes and octahedra, I can tell you that much.

In any case, whatever shape uniform-density planets turn out to have, the denser the planet's core, the more cube-like it will be. In the limiting case where all of the planet's mass is concentrated in a point at its center, the planet will be a perfect cube.

score -1 · Answer 10 · 2018-08-22 16:54:20Z

Yes

This is, in fact, only a matter of perspective (as well as altering the way gravity works). If distance is defined as you suggest you could even argue that our earth is also a cube if that is the way that you actually perceive distance. What we're really talking about is the definition of a metric, and what you're describing is called a Taxicab metric as opposed to a euclidean metric. And by stating that gravity uses the Taxicab metric instead of a euclidean one (gravity would operate by measuring the distance the Taxicab way), that would make our earth into a cube (by euclidean standards), but they might still call it a sphere because by the taxicab metric those things are the same. This also contradicts your other premise that this space follows Minkowski space principles, as this space is defined from a euclidean basis. So maybe I'm misunderstanding your premise.

One problem you will face is that of orientation (what's the cause of a certain direction being a cardinal one). Then vertices of each of your cubic planets will actually be on the 8 cardinal directions. So, if an object is limited to movement only in those directions, they couldn't move along the surface of the planet, they would need to go in the direction of one of the vertices of the planet they are on away from the surface, and then "fall" back down along another vertex's direction.

Hmm, I'm not sure it's only a matter of perspective. After all, the choice of norm has physical consequences (the eight velocities), and it is plausible that these consequences also physically affect the shape of planets. About the problem you mention, yeah, that's basically why I wanted the planets to be "liquid". Things also get complicated by the fact that the entire planet would also be moving at constant speed in one of the directions. — Aug 22 at 15:30
I've edited the post to be more clear about the context of my answer. Also, it is worth considering what you mean by liquid. If everything can only move in the directions that you mentioned, does that apply to individual atoms as well? If so then the properties of liquids would be fundamentally changed, as would gasses and temperature and a host of other things which are predicated on the degrees of freedom given by being able to move in euclidean geometries. — Aug 22 at 16:49
That's a good question! I'm actually trying to develop thermodynamics in parallel to this. My results suggest that there is indeed something that could be called liquid, although it acts more like a gas and has some weird properties w.r.t. pressure; I want to make a separate question about it, probably in Math Stack Exchange. But for this question I didn't consider the material to be important to determine the final shape. — Aug 22 at 17:14
A minor point about your edit: the Taxicab norm would correspond to $lambda=1$, and is different from the one I'm using, which corresponds to taking a big $lambda$ and is more similar to the sup norm (although they are related because they are duals of each other). The norm I'm using is a generalization of the Minkowski norm in the sense that it keeps the minus sign next to the $|dt|$. Hope I explain myself well. — Aug 22 at 17:14

score 59 · Answer 11 · 2018-08-23 15:38:24Z

TMM;DR (Too Much Math, Didn't Read):

For anyone who doesn't want to go through the derivations and calculations below, here are the important points from my answer:

We're not working with the same space as normal, friendly, Euclidean space.

This means that while we can still integrate and differentiate scalar functions defined on this space, we need to make slight corrections.

These corrections can be calculated from something called the "metric tensor", which describes the curvature of space.

There are several methods that could be used to find an equipotential surface (and therefore the shape of the planet); they should require these corrections.

If you want more details, read on!

Calculating the metric tensor

To do certain calculations in this space, it is necessary to calculate the metric tensor $g$. This is commonly used in the case of (pseudo)-Riemannian manifolds, and plays a central role in general relativity, as I assume you're aware. To do computations involving the curvature of your space, we need to know $g_ij$, the components of $g$.

In the Riemannian case, we have the (smooth), non-negative norm $F$. From it we can derive the components of the metric tensor by
$$g_ij=frac12fracpartial^2F^2partial x^ipartial x^jtag1$$
In Euclidean three-space (and in Cartesian coordinates), $F$ is the standard Euclidean distance metric
$$F=sqrtx^2+y^2+z^2$$
which is your case where $lambda=2$, and you can probably convince yourself that $(1)$ yields the familiar
$$g=
beginbmatrix
1 & 0 & 0\
0 & 1 & 0\
0 & 0 & 1\
endbmatrix$$
Here, we can extent the Riemannian definition to Finsler spaces using your $p$-norm; in your case, we're dealing with the more general
$$F=sqrt[lambda]x^lambda+y^lambda+z^lambda,quad F^2=left(x^lambda+y^lambda+z^lambdaright)^2/lambda$$
Going back to $(1)$, we have
$$beginaligned
g_ij &=frac12fracpartialpartial x^jfracpartial F^2partial x^i\
&=frac12fracpartialpartial x^jleft(2(x^i)^lambda-1left(x^lambda+y^lambda+z^lambdaright)right)^-1+2/lambda
endaligned$$
And so
$$g_ij=begincases(2-lambda)(x^i)^lambda-1(x^j)^lambda-1left(x^lambda+y^lambda+z^lambdaright)^-2+2/lambda,quad ineq j\
beginaligned
&(2-lambda)(x^i)^2lambda-2left(x^lambda+y^lambda+z^lambdaright)^-2+2/lambda\
&+(lambda-1)(x^i)^lambda-2left(x^lambda+y^lambda+z^lambdaright)^-1+2/lambda,quad i=j
endaligned\
endcasestag2$$
You can again check that this works like normal in the case where $lambda=2$.

Volume elements

So, why do we really care about $g_ij$? Well, if we look at Danijel's answer, we can see that they define the potential by
$$V(mathbfx)=intfracGrho(mathbfx)d^3mathbfx'tag3$$
where $d^3mathbfx'$ is the volume element; notice that we've allowed $rho$ to vary, as it would for a real planet. For Cartesian coordinates in Euclidean space, $d^3mathbfx=dxdydz$. However, this is not always true. Indeed, in the general case,
$$d^3mathbfx=sqrtdxwedge dywedge dztag4$$
where $wedge$ denotes the wedge product, and we see that we need to calculate the determinant of $g$, $|g_ij|$. (Note again how $sqrt=1$ for Cartesian coordinates in Euclidean space.) The calculation of the determinant is ugly, and I won't do it here because there's little point to it. I assume symmetry allows it to be written somewhat neatly, though.

Gradient

As has already been noted, the surface of the planet should lie on an equipotential surface, that is, one where $|vecnablaV|=0$. The familiar Euclidean gradient is
$$vecnabla=left(fracpartialpartial x^1,fracpartialpartial x^2,fracpartialpartial x^3right)hatx_i$$
for generalized coordinates $x^1$, $x^2$, and $x^3$, which are are $x$, $y$ and $z$. It should come as no surprise, however, that this fails for non-Euclidean spaces. In the more general case,
$$vecnabla=fracpartialpartial x^icdot g^ijsqrtg_jjhatmathbfx^itag5$$
for unit vector $hatmathbfx^i$, and again $g_ij$ plays a role.

Poisson's equation and the shell theorem

Here, of course, is the problem - and it's been bothering me for about a day now. The core question you're asking here involves determining the potential corresponding to a given density distribution. We can do that easily enough in flat space by solving Poisson's equation; the solution can be found through $(3)$. Okay. But does Poisson's equation look the same (or even similar) in this Finsler space, and can we solve it the same way? The answer may be that it doesn't. In general relativity, on a Riemannian manifold, holds in the Newtonian limit of the Einstein field equations. However, it's not clear that we can make the same assumption in Finsler space.

Let's say Poisson's equation holds in some sort of form (here, I have set the appropriate constants to $1$):
$$nabla^2V=rhotag6$$
Then, indeed, $(3)$ holds in general, in some form, again with the appropriate volume element. Now, what I'm curious about - and I've talked about this in chat a bit - is the answer to a question posted by Dietrich Epp: Is the gravitational field outside a "sphere" identical to that outside a point particle with the same mass (the Shell Theorem)? If so, your given equation for the potential holds for a planet, and
$$V(mathbfx)=fracGm_1m_2_lambda$$
If not, we need to integrate $(3)$ properly before making any assumptions about the form of the potential. At any rate, I'd be interested in seeing a proof that a version of the shell theorem (or a version of Birkhoff's theorem) holds.

A path to a solution

I don't have an answer to your question yet. There are a couple of ways you could go find an equipotential surface:

Muck about with various surfaces and find one where $(text3)$ is constant for all $V(mathbfx)$ on the surface.

Set $|vecnablaV|=0$ and find the resulting surface.

Set $V(mathbfx)$ equal to some constant $C$ and again find the surface.

The third method seems like it should definitely work in the case of a Euclidean metric, for your chosen potential. I can't say for sure whether it works for all relevant Finsler metrics (and I suspect it doesn't), so I won't claim that the general case works just yet.

The naive third case

Okay, so let's set $V(mathbfx)=C$ for some constant $C$. Here, I assume that your potential is applicable for a planet. We then have
$$||mathbfr||_lambda=fracGMCimpliesleft(x^lambda+y^lambda+z^lambdaright)^1/lambda=fracGMC$$
Therefore, we're left with
$$z^lambda=left(fracGMCright)^lambda-x^lambda-y^lambdatag7$$
Interestingly enough, as $lambdatoinfty$, this becomes a cube!

References:

Shen 2000

Lecture notes by Matias Dahl

Popvici 2017

Busemann 1950

This is an amazing answer, thanks for all the effort in writing it! It's a good point about the volume form, I have to think about whether it would invalidate Danijel's answer (though I think the heuristic observations in Logan's answer make it likely that it will still hold). — Aug 22 at 17:26
@pregunton Yes, I don't think it invalidates Danijel's answer at all; it just adds a slight layer of complexity. — Aug 22 at 17:35
Wow. What an incredible amount of effort. I wish I could upvote twice, but you'll have to take this verbose kudos instead! — Aug 25 at 20:11

score 27 · Answer 12 · 2018-08-22 18:06:10Z

up vote
27
down vote

No

The necessary condition for a liquid body to be in a gravitational equilibrium is that gravitational potential should be constant over its surface. For the cube with density $rho$, by integrating
$$
V(mathbf r) = int_cube fracG rholVert mathbf r - mathbf r' rVert_lambda d^3 mathbf r'
$$
for various locations $mathbf r$ on its surface, it is easy to see that e.g. the middle of a face of the cube is deeper in the gravitational well of than a point on its edge (or vertex).

The correct shape probably isn't a sphere, which can also be checked using the same method, but it is harder to integrate, so I didn't do it. It is still an interesting mathematical question what would be the correct shape.

edited Aug 22 at 18:06

answered Aug 22 at 14:21

Danijel

891718

Thanks for your answer! That makes sense. I tried with the alternate potential I mentioned in a footnote $V propto lVertmathbfrrVert_lambda^lambda/(lambda - 1)-3$ and it seems it's still not a cube. I guess that answers half my question, now I wonder what could be the right shape (if there's even a stable shape).
â€“Â pregunton
Aug 22 at 14:46

6

I disagree with Danijel on this point. The key is that because of the way that distance would need to be measured, the distance from the center of the cube to any of the surface points would actually be the same regardless of whether it is on the center of a face, on an edge, or on a vertex.
â€“Â Mathaddict
Aug 22 at 15:07

7

@Mathaddict The distance from a point on surface to the center of the cube is indeed the same, regardless of the choice of the point. But to integrate, you need all distances between a point on the surface and all points of the cube.
â€“Â Danijel
Aug 22 at 15:26

3

The metric still has radial symmetry and the potential will also have radial symmetry (at least for non-rotating bodies). Surely then the shape of a planet (if they can exist) should approach radial symmetry as far as possible ? I don't get your argument for "probably isn't a sphere".
â€“Â StephenG
Aug 22 at 15:57

8

@StephenG The problem does not exhibit spherical symmetry because $lVert mathbf r rVert_lambda$ isn't a spherically symmetric function of $mathbf r$, i.e. it is not a function of $r=sqrtx^2+y^2+z^2$, so a spherically symmetric solution would be quite unexpected.
â€“Â Danijel
Aug 22 at 18:11

Â |Â
show 2 more comments

Logan R. Kearsley 8,82012746 · Answer 13 · 2018-08-22 14:55:04Z

As Danijel has pointed out, the question can be reframed as "what is the shape of the equipotential surface under this force law?"

There are further two different ways to view the results: the external perspective, where we see what shapes these "planets" map onto if we overlay them on our normal Euclidean space, and the internal perspective--what would they "look like" to someone embedded in the same metric?

The external perspective is a lot easier to figure out. You just calculate the shape of the equipotential surface in Euclidean space using a different force law. Finding the exact shape is complicated, but the broad strokes are fairly clear: force reduces over Euclidean distance more slowly along the octants than it does normal to the faces, which means the potential at equal Euclidean distance is higher along the octants than the faces... and so the equipotential surface ends up dimpled inwards along the octants, rather than poking out to form cube vertices. If you actually wanted cubical worlds, you'd need a force law that decreased faster along the octants, not slower.

Exploring the internal perspective is considerably more complicated. Honestly, I'm not even certain what the best way to approach it is; trying to imagine living in a lambda-norm world breaks my brain! As a first step, I'd probably try to investigate the local curvature of the equipotential surface under that norm; is it roughly uniform, in which case the inhabitants of this universe would likely perceive their worlds to be approximately smooth, like we do ours, or does it have regular variations that would correspond to edges and vertices? With a little more formalizing (which it seems you are perfectly well qualified to do), that's a question you could probably take to the Mathematics StackExchange and get much more productive answers.

Thank you! Your observations make a lot of sense in hindsight; interestingly, this means that the planets would look more like octahedra than cubes. The question about the internal perspective sounds pretty interesting, I'll have to think about it. — Aug 22 at 15:09
I am not convinced that this is equivalent. With the simple inverse square law objects can be approximated by point masses, I am not sure this is true in this case. If it is not true, then planets would not be shaped like equipotential surfaces. They may in fact be shaped differently depending on their composition. — Aug 22 at 18:27
To clarify what I just said, the surface will still be equipotential but the equipotential surface could have a different shape for planets and point masses. — Aug 22 at 18:30

score 15 · Answer 14 · 2018-08-23 15:27:32Z

My answer isn't going to be anywhere near as in-depth as your question but I'm happy to supply a few inputs.

First, I'm going to take a look at your particles:

1. Particles can only move at one constant speed.

2. Particles can only move in six directions. (Or four in my 2D model.)

3. Particles have "gravity".

I'm going to imagine two particles in a two-dimensional plane for now:

enter image description here

We could represent gravity as a vector from one particle to another, so the top left particle would have (3,-2). To make this pull weaker with distance, we can multiply the vector by the reciprocal of its magnitude:

enter image description here

If there were other particles, we'd have to calculate the force vector for each. The resultant gravity would be the sum of the vectors, which in this example is just (3/Ã¢ÂˆÂš13, -2/Ã¢ÂˆÂš13)

Now, the particle can only move in one of the four directions in this 2D example, so let's take the vector component with the largest magnitude: x = 3/Ã¢ÂˆÂš13 and move one square in that direction. The result would be that both particles zigzag towards each other until they meet. At this point, two things can happen:

The particles stop moving.

This violates your rule that the particles "basically only move at a certain constant speed". You could change this rule to The particles are either stationary, or moving with a set speed, having a binary value for their velocity, OR:

The particles oscillate between their two positions.

This means that the particles will be fixed into an "orbit" around each other that lasts 2 time ticks, however it lets them pass through each other. This can be explained by having the particles move from one point in space to the adjacent point without interacting with, or passing through, the space between (BECAUSE QUANTUM!).

This is all fun and dandy until you hit a situation like this:

enter image description here

Provided there are no other particles in the entire universe, these two particles will have a force vector where two or more components are equal. In this case, how does the particle choose which way to go?

Solution 1: Randomly decide

Let's forget all of those God Does Not Play Dice with the Universe shenanigans, because we can't simply "move diagonally" in this system. If there's two possible directions to move in, each with an equal likelihood of occurring, one of them must happen. Just randomly pick one of the tied directions and go that way.

Solution 2: QUANTUM STUFF???

Just make the particles go into superposition. Ya know, go both ways at once. Why not? After one time tick it'll look like this (particles now coloured red and blue)

enter image description here

Now there's three possible positions for each particle, with probabilities of 25%, 50%, 25% of existing in each position.

enter image description here

Oh darn those sundries! Now we have to perform a measurement, since each superposition may be colliding with something. And the probabilities of both red and blue existing in a specific position is a binomial distribution! That's fine, just randomly decide which of the paths the two particles took... But wait! What happens if they both end up on the same square? Well, SOMETHING would have to happen. I can't for the life of me figure out a way to work around this. Maybe you can, I don't know.

The advantage of Solution 2: QUANTUM STUFF??? is that it allows the superimposed particles to interact with other, third party particles that may cause a measurement to happen early. This would, again, cause two particles to exist in the same place. The advantage of method 1: No nasty probabilities.

Okay, so you talked a lot about gravity. Stuff can clump together. But nothing can orbit without momentum!

Great point, Jenkins! Momentum can be represented as another vector. Each time tick, The resultant force of the gravitational attraction to every other particle in the entire universe is added to the momentum of the particle, forming a new momentum vector. Now, instead of moving in the direction of the strongest pull of gravity, the particles move in the direction of the strongest component of their momentum vector.

Now, all of this is great, but does it work? I will attempt to simulate this in Python and then update my post in due course.

My predictions:

Unless I find a way to prevent multiple particles moving into the same space, the whole system may collapse into a singularity.

If momentum works correctly, there MAY be some form of orbital mechanic possible.

Clumps of particles may form. If they do, they will either have a roughly circular shape, or may create "crystal" shapes such as this:

enter image description here

My theory is that a particle may be more likely to move towards another body of particles if it is vertically or horizontally in line with it, resulting in protrusions from the clump along the axes and possible diagonal edges.

The simulation

Here's what I've got so far:

import pygame
import random

(width, height) = (1000, 1000)
screen = pygame.display.set_mode((width, height))


class Particle:
 def __init__(self):
 self.position = (random.randint(-50, 50), random.randint(-50,50))
 self.momentum = (random.randint(-10, 10)/10, random.randint(-10,10)/10)

 def move(self, positions):
 resultant = (0, 0)
 for position in positions:
 vector = (position[0] - self.position[0], position[1] - self.position[1])
 magnitude = (vector[0] ** 2) + (vector[1] ** 2)
 if magnitude == 0:
 return
 resultant += (vector[0]/magnitude, vector[1]/magnitude)
 self.momentum += resultant

 if abs(self.momentum[0]) > abs(self.momentum[1]):
 self.position[0] += self.momentum[0]/abs(self.momentum[0])
 elif abs(self.momentum[0]) < abs(self.momentum[1]):
 self.position[1] += self.momentum[1]/abs(self.momentum[1])
 else:
 if random.randint(0, 1) == 1:
 self.position[0] += self.momentum[0]/abs(self.momentum[0])
 else:
 self.position[1] += self.momentum[1] / abs(self.momentum[1])

 def render(self):
 pygame.draw.rect(screen, (255, 255, 255),
 (int((self.position[0] * 10) + 495), int((self.position[1] * 10) + 495), 10, 10),
 0)


def run():
 screen.fill((0, 0, 0))

 particles = 
 for x in range(10):
 particles.append(Particle())

 running = True
 while running:
 for event in pygame.event.get():
 if event.type == pygame.QUIT:
 running = False

 positions = 
 for particle in particles:
 positions.append(particle.position)

 for particle in particles:
 particle.move(positions)

 for particle in particles:
 particle.render()

 pygame.display.flip()


run()

It either takes incredibly long to calculate the next frame, or it doesn't work. I haven't worked that out yet...

enter image description here

Nice answer. Your setup looks a lot like how I first approached the problem. Looking forward to your simulation! — Aug 22 at 18:47
According to the question, if you move in one dimension, you must move the same distance in the other two, but in either direction (positive or negative), giving 8 directions, not 6. This means a 2D analogy would only have diagonal movement. — Aug 24 at 23:19
@CJDennis this means that a 2D analogy would only have four diagonal directions to move in, so I am still correct despite my inaccuracies. — Aug 25 at 19:14

JBH 32.4k578155 · Answer 15 · 2018-08-23 19:43:05Z

I wish I had more time to play with the math, but let's use some logic.

Velocity vectors must work in both directions. If your particle can only move in an octant, then gravity can only draw in an octant. This suggests...
1. That gravity is functionally discrete. Unlike in our universe, where we could simplify and simulate this issue with two ball bearings that roll around each other's surface continuously with equal forces of gravity at any point, yours looks like the diagrams below.
2. The force of gravity over distance is wild wonky. Rather than simulating it as a growing sphere of diminishing influence, it's more like a simple star you'd plop on top of a Christmas tree. The "arms" of the star would diminish with the square of distance, but between the arms it would diminish with something more like the bazillionth exponent of distance. Random attraction in your universe would be a much, much slower process.

Let's look at the pretty pictures, which I ruthlessless appropriated from your post. I hope you don't mind.

The weakest gravitational force, the single-vertex force

enter image description here

Double the gravitational force, the dual-vertex force

enter image description here

The strongest gravitational force, the quad-vertex force

enter image description here

Unless you change the various lowest-energy-state/conservation-of-X rules in your universe, your planets will be formed predominantly by particles snapping into the quad-vertex force position. They can form in the single-vertex condition, but any little "bump" will slap them into one of the two higher gravity conditions, eventually into the quad.

At first blush, this would suggest your planets are shaped like that Christmass star or an 8-pointed pinata.

enter image description here

Except there's all those spaces in between with 6-vertex and 8-vertex connections... In the end, I believe you'd get planets that are roundy-with-corners. Think cube balloon, where the sides bulge from the internal pressure:

enter image description here

Finally, a word about orbits

I predict that all your orbits will be eclipical due to the descrete nature of the gravity vectors. You may have small/light objects that orbit on an elliptic plane tilted 45Ã‚Â° to the ecliptic, but that's the weakest gravitational path. I also suspect you won't have much in the way of planetary axial tilts (0Ã‚Â°, 45Ã‚Â°, 90Ã‚Â°).

If you'll fogive the pun, those gravity vectors will have a tendency to keep things "boxed."

Jay 1513 · Answer 16 · 2018-08-22 23:24:56Z

Edit: I think I misread the question, but the gist is: If your cube shape is equidistant from the center-of-mass, and gravity depends on distance, then the shape of the planet will be all "same-distance" from the center.

A good first approximation is an octohedron.

Why is it an octohedron and not a cube? It's because your universe has 6 special directions and they correspond to the 6 points of the polyhedron.

In this co-ordinate system, all points on the surface of an octohedron are equidistant from the center of mass.

enter image description here

It's worth noting that cubes and octohedra are "dual polygons" and the faces of one correspond to the points of the other.

Welcome to Worldbuilding, Jay! If you have a moment, please take the tour and visit the help center to learn more about the site. You'll also find Worldbuilding Meta and The Sandbox useful. Here is a meta post on the culture and style of Worldbuilding.SE, just to help you understand our scope and methods. Have fun! — Aug 23 at 0:19

Ash 20.7k252127 · Answer 17 · 2018-08-22 15:24:03Z

Material science answer; it'll always be a "gravitational sphere" (a smooth round object with a gravitationally flat surface) because that's how stable mass assemblages always come to rest, lowest potential energy state. For the sake of clarity that's "smooth", "round" and "flat" on a planetary scale.

I'm not at all sure what a gravitational sphere actually looks like to a human equivalent observer in the proposed universe but looking at it through the euclidean filter of our physical senses I'm pretty sure it'll still be a sphere.

Ã¢Â€ÂœIt will alwaysÃ¢Â€Â is not necessarily true given that OP is proposing a universe with DIFFERENT physical laws than our own. — Aug 22 at 19:36
@WGroleau Not given what he isn't changing; matter exists and is subject to something akin to gravity, ergo gravitationally spherical. — Aug 22 at 19:40
..isn't the question asking "what shape is gravitationally spherical in this universe" though? — Aug 26 at 18:37

David Thornley 4314 · Answer 18 · 2018-08-23 18:15:10Z

I'm going to try a short and simple answer.

I assume gravity works similarly to our universe, except for the calculation of distance. A gravitationally stable form is where the surface points are equidistant from the center. If some points were farther from the center than others, smoothing them out would reduce the gravitational potential energy.

In the metric you're using, the distance is the largest of the x, y, and z differences. If we take a radius of 3 for an example, the equidistant surface is that which is 3 away in one dimension, and no more than 3 in any other dimension. That gives us a cube.

If the distance was x + y + z, we'd get an octahedron, which I'll leave as an exercise for the reader.

Someone Else 37 3,4841416 · Answer 19 · 2018-09-01 06:29:59Z