Infinite Grid of Resistors
niklasbuschmann | 215 points | 35day ago | www.mathpages.com
neepi|35day ago
I'm a bit mathematician and a bit electrical engineer.
The electrical engineer suggests it's not measurable unless you apply current and also asks "when" after the current is applied referring to the distributed inductive and capacitive element and the speed of field propagation. The mathematician goes to a bar and has a stiff drink after hearing that.
Taniwha|35day ago
Eventually you need to pullin a physicist too who will point out that at an appropriate distance quantum effects will dominate - because eventually at a far enough distance the number of electrons moving per second (ie current flow) will be either 0 or 1 at some nodes
__MatrixMan__|35day ago
I've not studied QED directly, so by all means correct me if I'm wrong, but it seems to me that we'd get a double-slit like scenario where it's as if a partial electron went through either path. We might want to say that surely a whole electron took one path and not the other but we couldn't say which and if we tried to instrument to and find out we'd affect the resistance.
But that's fine because knowing which path the electron took is not part of the problem. Both paths contributed to the resistance even if one was not taken.
We only have to worry about quantum effects if the probabilities are not a decent proxy for the partial-particles that we suspect don't exist. In this case, the Physicist can probably proceed directly to the bar and have a drink with the Mathematician.
dgfl|34day ago
Resistance is inherently dissipative, so there is no coherent path the electron can take. No quantum effects here, the electron is always interacting with the resistor lattice.
viraptor|34day ago
I don't think you'll ever get 0/1. You get a difference in voltage that influences all electrons to move slightly more in one direction than another in electric current. They'll just drift very very slightly as a group, not measurable when you get far enough. But they're always all affected, rather than individually.
mjevans|35day ago
Intuitively I knew this class of problem was theoretical only BS when it came up in college...
I hadn't considered that sort of strange effect though! Makes me feel not so bad for 'never really getting it' because I just couldn't wrap my mind around the problem description's obvious inanity and the infinite edges.
aydyn|34day ago
The question is not pretending to be realistic. No one is thinking it is possible to build an infinite grid of resistors.
It's simply an evaluation of your mathematical ability to manipulate the equations and overall understanding of them, wrapped up in a cute little thought experiment. This evaluation IS relevant to more realistic scenarios and therefore your grade and engineering ability.
sandworm101|35day ago
And the electrician knows he can get a 99% answer out of a 10x10 grid on a workbench. The engineer is free to then add more resisters to the periphery until either the grant money runs out or the physicist's publishing deadline approaches.
A really difficult question: At each distance, what percentage of soldering errors in the grid can be tolerated before the fluke meter across the center square detects the fault? (That might actually be a thing as I've heard people talk about using changes of local resistance to detect remote cracks in conductive structures ... like maybe in a carbon fiber submarine hull.)
nerdsniper|35day ago
For measuring corrosion in conductive surfaces, “eddy current” testing is often used. It uses AC current of some frequency, so it’s technically measuring inductance rather than resistance.
repiret|35day ago
I think there are two interpretations of schematics.
One is where the components on the schematic represent physical things, where the resistors have some inductance and some non-linearity, and some capacitance to the ground plane and so on. This is what we mean by schematics when we’re using OrCad or whatever.
There is another interpretation where resistors are ideal ohms law devices, the traces have no inductance or propagation delay or resistance. Where connecting a trace between both ends of a voltage source is akin to division by zero.
Sometimes you translate from the first interpretation to the second, adding explicit resistors and inductors and so on to model the real world behavior of traces etc. if you don’t, then maybe SPICE does for you.
Infinite resistor lattices exist only in the second interpretation.
red75prime|34day ago
> The electrical engineer suggests it's not measurable unless you apply current and also asks "when"
Just wait infinite time for all the transient responses to die down. The grid to enter steady state and to became true to the schematic.
eternauta3k|34day ago
An infinitely large grid never reaches equilibrium, like it says in the article.
red75prime|34day ago
That's why you need to wait longer than ever.
Mostly joking, but an idealized schematic requires idealized conditions to observe.
bravesoul2|35day ago
Given an infinite grid of resistors... would you expect planets to form?
corysama|35day ago
They say hydrogen is an odorless colorless gas which, in sufficient quantities, given enough time, turns into people. I’m sure the same could be true of resistors.
bravesoul2|35day ago
Resistors are made of heavier elements though. And I remember something like everything wants to become iron (fuse if lighter, decay if heavier)
That said there might be enough energy (infinity!) for anything to be possible.
jfengel|35day ago
You get stable things heavier than iron, and they're more common than you'd expect. It's possible that they form in neutron star collisions, which are complete anarchy in atomic terms.
inopinatus|35day ago
People are resistors too.
neepi|34day ago
Just checked. Around 1.5Mohms today.
temp0826|35day ago
I resist that statement
taneq|35day ago
Gee, I hope nobody ever puts you in charge of a train!
farhaven|34day ago
Well they can still be a conductor, even if they're not a resistor. Actually they'd be a pretty good conductor. A super-conductor, if you will.
taneq|34day ago
So wait, if they resist the statement that they're a resistor, are they a resistor or not? I think they're a semiconductor. Maybe they work at a railway junction. :D
QuadmasterXLII|35day ago
I think it collapses into a black hole. black hole mass scales with radius, grid mass scales with radius squared
jfengel|35day ago
It becomes a black hole, but it doesn't necessarily collapse, at least not at first. A supermassive black hole has very low density and a very gentle gravitational gradient.
All of the mass does end up in the singularity, in finite time (at least for any finite subset of the black hole), but it doesn't automatically become super dense just because it's a black hole. It can remain quite ordinary for a very long time.
kqr|34day ago
Wait, can you take this a little slower? I was not aware black holes could have sensible density.
jfengel|34day ago
In the formula for a Schwarzchild black hole, the mass is proportional to the radius. Since volume goes up with the cube of the radius, the density drops quickly.
The kind of black hole that forms from a collapsing star is super dense. But a supermassive black hole forms differently, as a denser region of gas and stars during galaxy formation. The density can be lower than that of water. You could be inside it without even realizing anything is odd.
There is no known way to form black holes bigger than that. We had been discussing a pure thought exercise. Though it is possible that the universe as a whole has enough density to be a black hole. (Signs point to no, but it's an open question.)
pixl97|35day ago
Thats what I'm not sure about here...
If we assume this is an infinite grid in its own universe then nothing can actually move. The gravitational pull should be the same from every direction. If we assume the grid is perfect then there is no nucleation sites to start a collapse. The grid would be in perfect balance.
The same is thought about our universe. If there hadn't been small quantum fluctuations during the inflationary stage it would have taken much longer for what we see in the modern universe to form.
staplung|35day ago
In fact, we might have a different problem: dark energy should tear the grid up into (very large) bits. I guess the question is then would the bits then collapse into black holes or not. I assume so since the mass would not longer be perfectly balanced.
pixl97|35day ago
Guess we'll have to wait for an actual answer on dark energy and universal expansion.
What it collapses into depends on the time for the bits to coalesce into. If it's a slow collapse you could get a mass large enough to form a neutron star (like thing) instead of black holes.
If those neutron stars crash into each other they can release a large amount of 'recycled' matter from all over the atomic spectrum back into this universe.
raattgift|35day ago
In 1+1 dimensions one can analyse the gravitational behaviour of an infinite line of ...-wire-resistor-wire-resistor-... with an adaptation of Bell's spaceship. Throwing away two dimensions eliminates shear and rotation (and all sorts of interesting matter-matter interactions) so we can take a Raychaudhuri approach.
We impose initial conditions so that there is a congruence of motion of the connected resistors, so that we have a flavour of Born rigidity. Unlike in the special-relativistic Bell's spaceship model (in which the inertial motion of each spaceship identical save for a spatial translation), in our general-relativistic approach none of the line-of-connected-reistors elements' worldlines is inertial, and each worldline's proper acceleration points in a different direction but with the same magnitude. This gives us enough symmetry to grind out an expansion scalar similar to Raychaudhuri's, Θ = ∂_a v^a (<https://en.wikipedia.org/wiki/Raychaudhuri_equation#Mathemat...>). As an aid to understanding, we can rewrite this as 1/v \frac{d v}{d \tau}, and again in terms of a Hubble-like constant, 3H_0.
We can then understand Θ as a dark energy, and with Θ > 0 the infinitely connected line of ...-wire-resistor-wire-resistor-... is forced to expand and will eventually fragment. If Θ < 0, the line will collapse gravitationally.
> no nucleation sites
If Θ = 0 initially, we have a Jeans instability problem to solve. Any small perturbation will either break the infinite ...wire-resistor-wire-..., leading to an evolution comparable to Bell's spaceship: the fragments will grow more and more separated; or it will drive the gravitational collapse of the line. The only way around this is through excruciatingly finely balanced initial conditions that capture all the matter-matter interactions that give rise to fluctuations in density or internal pressure. It is those fluctuations which break the initial worldline congruence.
This is essentially the part of cosmology Einstein struggled with when trying to preserve a static universe.
In higher dimensions (2+1d, 3+1d) the evolution of rotation and shear (instead of just pressure and density) becomes important (indeed, we need an expansion tensor and take its trace, rather than use the expansion scalar above). A different sort of fragmentation becomes available, where some parts of an infinite plane or infinite volume of connected resistors can undergo an Oppenheimer-Snyder type of collapse (probably igniting nuclear fusion, so getting metal-rich stars in the process) and other parts separate; the Lemaître-Tolman-Bondi metric becomes interesting, although the formation of very heavy binaries early on probably mitigates against a Swiss-cheese cosmological model: too much gravitational radiation. The issue is that the chemistry is very different from the neutral-hydrogen domination at recombination during the formation of our own cosmic microwave background, but grossly a cosmos full of luminous filaments of quasi-galaxies and dim voids is a plausible outcome. (It'd be a fun cosmology to try to simulate numerically -- I guess it'd be bound to end up being highly multidisciplinary).
rzzzt|35day ago
Assume perfectly spherical through-hole resistors soldered on an infinite PCB.
divbzero|34day ago
An infinite grid of resistors is clearly a toy scenario, but the infinite universe is a reality that astrophysicists try to reason about. I wonder if there are blindspots in astrophysics because we lack intuition about the universe at that scale and are forced to approach it from theory.
nofunsir|34day ago
Does Schrodinger's cat study Fourier transforms?
voxelghost|34day ago
I think they are called furrier transforms, you lick something that has a certain shaped that can be described frequency curves, then you throw up a furball that to any cat accurately describes the licked object.
Koshkin|34day ago
But you already know that it does and it doesn’t at the same time.
Koshkin|34day ago
> apply current
Going on something of a tangent: in engineering, it seems unusual to talk about “applying current,” it’s usually voltage (say, across a resistor) or some sort of an “electromotive force.”
gugagore|34day ago
In idealizations, there are both voltage sources and current do m sources.
bgnn|34day ago
People think this is not relevant to real world problems but it actually is, albeit all the calculations aren't that relevant. Silicon substrate's resistance is basically an infinitely large grid of unut resistances at the distances relevant for a local point of an IC. Note that silicon substrate is often heavily doped (p-type) and all info you get from the fab is it's resistivity (often somewhere between 1 to 100 ohm per cm). For the most advanced tech nodes its often 10 ohm/cm. If you need to develop some intuition about noise coupling via the substrate you have to think that it's a grid instead of just calculating the resustance between point A and B. We need to distribute a grid of substrate contacts to collect the noisy currents too. So the grid shows up again!
ChoGGi|34day ago
My vague understanding of photolithography is that it's hard, though I didn't realise it's bad enough to evoke an egyptian goddess.
I'll see myself out.
em3rgent0rdr|34day ago
The word used to pray to Unut is 'ohm', I believe...
bgnn|33day ago
Sorry for the uncut unut units.
eternauta3k|34day ago
I'd argue the case you're describing is mathematically simpler precisely because it is continuous.
bgnn|34day ago
True, but the continuous solution is just a limit condition of tge discrete one. It doesn't make it any harder or easier, at least from what I know fron calculus. The software tools use numerical methods to solve this type of problems and they tend to divide the continuous substrate into a mesh of discrete elements to model them as lumped circuit elements so that we can represent them in a matrix and simulate the circuit using linear algebra. They often use random walk in their algorithm to find a mesh which introduces a minimum error.
gugagore|34day ago
Right, why is it a 4-connected grid instead of 8-connected, or any other topology, like a hex grid.
Den_VR|34day ago
You’re practically describing the invention of Calculus.
fraserphysics|34day ago
The units of resistivity are ohm * cm not ohm/cm. (I worked at Fairchild a long time ago.)
bgnn|33day ago
Correct. Sorry for the typo. I was very sleepy.