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I was recently told that in physics, if your equation contains an infinity, it usually suggests that something is wrong. As an example, the infinity around black holes (also) tells us that general relativity isn't the full picture of gravity. How true is this? And what logic underlies this methodology of verifying a theory?

Edit to suffice this post The answers don't seem to explain why; there is no analytical or empirical grounding as to why 'limits are(n't) okay'.

yolo
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7 Answers7

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There's no rule against infinity, only a rule against being empirically wrong. Historically, the infinite has hinted we're missing something, but so have plenty of other things too. Let's discuss some historical examples.

The ultraviolet catastrophe and its resolution respectively come down to physical reasons to care about $\int_0^\infty x^2dx=\infty$ and $\int_0^\infty\frac{x^3dx}{e^x-1}=\frac{\pi^4}{15}$. To which is the power output of a glowing object (be it black or grey) proportional? If it were the former, our power sources couldn't make anything hot enough to glow in the first place. The catastrophe was so-called not because mathematicians, philosophers or physicists are prejudiced against infinite quantities, but because it's an empirical fact that finite amounts of heat are convertible to light.

Galilean relativity was at odds with Maxwell's equations until Einstein argued it was the former, not the latter, that had to change. The latter's Lorentz-invariance is the only alternative this group theory argument allows for. Long story short, a special speed appears, which is infinite in Galilean physics. No prejudice ruled against this, only the empirical success of a theory instead claiming that speed would be $c:=(\mu_0\epsilon_0)^{-1/2}$. We could get around this by setting $\epsilon_0=0$ so $\rho=\epsilon_0\nabla\cdot E=0$, i.e. electric charges cannot exist, or $\mu_0=0$ so currents can't generate $\nabla\times B$, but those are also empirically untrue.

Newton had been dissatisfied with the infinite-speed action-at-a-distance nature of his own theory of gravity, but he and many successors accepted it as a brute fact about the universe, and it seemed like they always would. Einstein spent a few years working out how a speed-$c$ gravity would work, not because nothing can be infinite, but because special relativity's implications seemed so far-reaching. Eventually, general relativity was not only theoretically decent, but empirically vindicated. But GR gave rise to a different source of infinite quantities: singularities. To this day, we don't know exactly how if at all post-GR gravitational physics will disprove that (but see also here).

I've discussed three problems so far. Quantum mechanics solved the first, but its unification with special relativity, quantum field theory, introduced infinities of its own, for each fundamental interaction in nature. The resolution to these refines an objection you might have had to my first example: what if $\int_0^\Lambda x^2dx$ is the integral that matters, i.e. a glowing object doesn't release arbitrarily high frequencies? The value of $\Lambda$ you'd get from that idea contradicts observed spectra, i.e. we really do need the integrand to taper off in a way $x^2$ doesn't. But it's not a bad idea in theory, which is why analogous ideas can be more successful. See here for a discussion of such techniques.

For a reason the above link discusses (although it can also be explained in other ways), gravity's version of these problems has proven tougher. Quantum gravity isn't renormalizable (T & Cs apply, hence the multiple approaches or overlaps thereof to massaging the infinities we get this time). But issues of renormalization, regardless of the decade they're solved in, lead to a different, subtler "is it infinite & does it matter?" question: how many degrees of freedom does it take to nail down your theory? While infinitely many do present a problem (see e.g. Sec. 7.1 here), not all useful theories are renormalizable, because the first few parameters can do most of the work in explaining or predicting empirical reality.

As for examples where finite quantities are too large or too small in theory to match the data, we could be here with those all day.

J.G.
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Infinity is a shorthand for unbounded. When we say that $\frac 1 x$ goes to infinity as $x$ approaches $0$, what we really mean is that we can make $\frac 1 x$ as large as we like by using a value of $x$ that is sufficiently close to $0$. The actual value of $\frac 1 x$ when $x=0$ is undefined. If we restrict ourselves to real numbers then there is no such value as infinity.

But real numbers (and, sometimes, complex numbers too, but a complex number is a pair of real numbers) are what we use in physics to model all aspects of reality. Everything we measure in physics, whether it is a mass or a length or a temperature or a density etc. etc. is a real number.

So if a model of reality gives a result that is infinite (or, more formally, unbounded) in a particular scenario then we know that the model must break down in that scenario. So the appearance of "infinities" in a model tell us the model is incomplete.

gandalf61
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  • Comments are not for extended discussion; this conversation has been moved to chat. – tpg2114 Mar 19 '21 at 16:46
  • It is the same in cosmology. When people say that the universe is infinite, they mean that there is no maximum distance. But all distances in such a universe are finite. – Arman Armenpress Jul 27 '21 at 11:57
  • But, can't "infinity" also mean, "no number is big enough to answer the question." And, when you get that as an answer, does it not sometimes mean that you've asked the wrong question? – Solomon Slow Apr 06 '22 at 20:05
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For any quantity $x$ there are an unlimited number of related quantities like $y=1/x$ and $z=1/(x-3)^2$ and so on that have infinities in different places. When $x=\infty$, $y=0$. When $x=3$, $z=\infty$, and so on. So finding a particular quantity $x$ in an equation being infinite may just mean that it is more appropriate to talk about $1/x$.

There is a particularly famous example in black hole physics: the Schwarzschild singularity at the event horizon. When Schwarzschild first solved Einstein's equation to find the gravitational field of a point mass in a vacuum, there was a radius at which the expression went to infinity.

Schwarzschild's solution looked like this:

$$ds^2=\left(1-{2Gm\over c^2 r}\right)c^2 dt^2-{1\over 1-{2Gm\over c^2 r}}dr^2-r^2 d\theta^2-r^2\sin^2\theta \ d\phi^2$$

The part in the middle ${1\over 1-{2Gm\over c^2 r}}$ goes to infinity when $r={2GM\over c^2}$.

For a long time, people thought this meant the physics became meaningless and invalid at the event horizon. However, it later transpired that this was just an artifact of the coordinate system used. It's like the way the latitude and longitude coordinates go crazy at the north and south poles. Nothing strange or unphysical actually happens there - the sphere looks the same at the north pole as everywhere else. It's just the coordinate system that doesn't work there. Just as there are other coordinate systems where the north pole on Earth looks normal, so there are coordinate systems where the event horizon of a black hole looks normal and unremarkable.

The singularity at the centre of a black hole, however, cannot be removed by any coordinate change. Spacetime 'forms a sharp point' there, like pressing into a rubber sheet with a sharp pencil. If you look at any other portion of spacetime and magnify it sufficiently, it always starts to look like a tiny microscopic patch of flat space, in which we can use flat-space physics to predict what will happen. Surfaces with this property everywhere are called "manifolds". However, if you magnify the space around the pointy singularity, it never flattens out - it looks pointy at every scale. Spacetime is no longer a manifold, as general relativity assumes. This means we can never apply our well-understood flat-space physics.

In a manifold, there are $360^\circ$ around any point (thinking two-dimensionally). But at the point of a cone there are less, and as the point gets sharper, the range of angles around it approaches zero. In addition, the time axis points inwards towards the singularity, so it is the future for every possible path. Nobody has any idea what to do with that, or what physics would look like in such strange circumstances. What effect does having less than $360^\circ$ around a point have on sub-atomic physics? If a particle is spinning, does it go round faster, or does it try to overlap itself, or what? Since we don't know how elementary particles work at that scale, anyway, we can't even guess.

(I'm simplifying hugely, of course. I'm just trying to give a vague intuitive impression of the sort of problem it poses.)

Saying that the curvature goes to infinity and thus becomes nonsensical is a simplification, to avoid having to talk about the real issues. Curvature going to infinity just means 1/curvature goes to zero, and if you can explain your physics using 1/curvature instead, maybe nothing will go wrong. In the case of the Schwarzschild black hole's central singularity, the problem is that all the matter gets packed into an ever smaller space, the energy density increases without limit, and we just don't know what happens at energies far higher than those we have explored experimentally. (Like, $10^{1,000,000}$ times higher is still tiny compared to infinity!) Do the known rules of physics still apply? Almost certainly not, but nobody knows what to replace them with.

Nihar Karve
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  • "The singularity at the centre of a black hole, however, cannot be removed by any coordinate change." - Yes it can. See this paper at CERN: https://indico.cern.ch/event/382362/contributions/904877/attachments/1171153/1690816/Stoica.pdf – safesphere Mar 19 '21 at 05:25
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    @safesphere That paper is invoking a nonstandard use of the term "coordinates" in this context. When extended to included $r=0$, the given transformation is not a coordinate system as defined in differential geometry, i.e. a homeomorphism onto an open subset of Euclidean space. The extension there doesn't have the property that SR applies locally, for example. This is not to say that the paper's results are irrelevant, but your claim is misleading. Schwarzschild cannot be extended as a $C^2$ semi-Riemannian manifold, which is the standard context of GR. – jawheele Mar 19 '21 at 09:43
  • @safesphere Again, I wasn't saying that the paper's results aren't of interest, only that your terse contradiction to this answer was misleading without a clarification that you're using different definitions of the critically significant term. I expect that, as it stood, your comment was more confusing than helpful to the average reader trying to parse what is true. – jawheele Mar 20 '21 at 17:32
  • @safesphere The point of the "standard" isn't to impose arbitrary restrictions, but to aid in communication so that each in a pair of people can know what the other is trying to say when using these terms. – jawheele Mar 20 '21 at 17:37
  • @jawheele Sorry if my statement did not match the definitions that you had in mind. I am not endorsing this paper. It was just an example that preconceptions are not always right. Perhaps as a better example, the $r=0$ singularity can be removed by a simple coordinate transformation $t^\prime\equiv 4M^2/r$ inside the horizon. This way you get a spherinder universe expanding forever with acceleration. I guess, this transformation is also "non-standard" by your definitions, but it removes the singularity as a geodesic incompleteness at $r=0$. – safesphere Mar 20 '21 at 20:07
  • @jawheele To be clear, I am playing devil's advocate here. My actual point is that singular transformations are invalid. If I decide to use them, I could remove - or create - any singularity anywhere at will, at $r=2M$, at $r=0$, or in my kitchen by a singular coordinate transformation. – safesphere Mar 20 '21 at 20:25
  • @safesphere Of course you can come up with any number of perfectly acceptable coordinate transformations that move the $r=0$ point away to "infinity" in some sense, but none of these make the spacetime geodesically complete. The singular nature of the $r = 0$ boundary has nothing to do with it occurring at finite values of the Schwarzschild $t$ coordinate. The proper time associated to a timelike curve ending at $r = 0$ is a quantity invariant under coordinate transformations, so it is always finite, even if it extends infinitely in your new coordinate "time". – jawheele Mar 20 '21 at 22:20
  • @jawheele Ping… – safesphere May 08 '21 at 21:17
  • @jawheele Got it (you can remove it now.) Thanks and sorry for the confusion :) – safesphere May 26 '21 at 03:05
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How true it is that core of a black hole has an infinite curvature, is impossible to say without any experimental data. I think that what you called a methadology is nothing more than an expectation people have that infinities are unphysical, because we have not encountered them in reality. So since we haven't measured any infinites (however you would measure them) the expectation is, that the infinity should just arise because our theory does not capture the whole picture. Which is also a kind of magical thing that the theory "knows" where other things come in. You already get the issue in classical electrodynamics when you compute the energy of the electromagnetic field of an electron outside a ball of radius $r_0$ $E = \int d^3x \frac{\epsilon}{2} \vec{E}^2 $ $ = \frac{e^2}{8 \pi \epsilon r_0}$ With $\vec{E}(r) = \frac{e}{4\pi\epsilon r^2}\vec{e}_r$ the coloumb force,$\epsilon$ the vacuum permitivity (https://en.wikipedia.org/wiki/Vacuum_permittivity) and $e$ the elementary charge of the electron. You see that for zero radius, the energy diverges. But this assumes that all of the electron can be found at a single point and the electric field of the electron therefore has an infinite energy density at the position of the electron. I can not do the calculation but I would expect, as the people above do for quantum gravity, that the singularity in the energy should disappear in quantum mechanics, because there the electron would not be in a single place but "spread out" and what we compute classically is not the whole picture, because we do not incorporate the quantum mechanics of the electron.

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Infinite values are fine in contexts where they make sense (say, an infinite universe), but problematic where they do not. So, if a nonsensical infinity appears in a theory, it must logically be wrong.

Space can be infinite, as there is no known reason the universe must be bounded. Similarly, the total energy or number of particles in the universe may also be infinite. But in any finite volume, there should be a finite number of particles, or we would have an infinite mass with an infinite density, which would produce a black hole with an infinite radius. Note here that we don't get this problem with an infinite number of particles in an infinite volume, as the density can still be finite and below the threshold for black hole formation.

A superconductor can have an infinite conductance, but all this means is that there is no voltage drop across it (recall Ohm's Law: V = IR, where R is the inverse of conductance). You cannot get an infinite current passing through a superconductor, because that conductor only has a finite number of electrons, which can only move at a finite speed.

Strictly speaking, no one can really rule out singularities in black holes. However, we know from quantum mechanics that bound particles have higher zero-point energies. So a singularity with a zero radius would have an infinite zero-point energy, which isn't a sensible solution. Without a theory of quantum gravity it's hard to really say what all that means, but given how quantum mechanics tends to spread out mundane particles like electrons it seems reasonable to assume singularities may be spread out as well, avoiding the problem entirely.

So, yes, physicists tend not to like infinities appearing where they don't make sense, but no one has an issue with their existence in contexts where they do.

Quick addition: To address your point about general relativity being incomplete, we know this because it does not incorporate quantum mechanics, and there have been significant theoretical difficulties uniting the two. So, we know it is incomplete, because there are many things in the quantum world it does not explain. Our understanding of black hole physics is unnecessary here.

  • "Note here that we don't get this problem with an infinite number of particles in an infinite volume, as the density can still be finite and below the threshold for black hole formation." For any density above zero, no matter how low, isn't there some finite volume of that density that would collapse into a black hole? In other words, if the universe is infinitely big, wouldn't its density have to be zero? The Schwarzschild radius increases linearly with mass, but the radius of a uniform-density sphere only increases with the cube root of mass. – Joseph Sible-Reinstate Monica Mar 19 '21 at 04:59
  • The existence of finite black holes is fine, and we observe these in nature so we know they must exist. We don't observe infinitely large black holes anywhere, so these are a problem. Now, I guess you could argue they might lie outside the observable universe, but the concept is pretty ugly and there are statistical reasons as to why we can be reasonably confident they don't. – Matt Thompson Mar 19 '21 at 05:05
  • Never mind; I was making the same mistake as this question. – Joseph Sible-Reinstate Monica Mar 19 '21 at 05:30
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    "Infinite values are fine in contexts where they make sense (say, an infinite universe)" - An "infinite universe" makes no sense whatsoever. – safesphere Mar 19 '21 at 05:35
  • A universe that never ends is infinite in time, is it not? – Matt Thompson Mar 19 '21 at 05:45
  • To add to that, is there any logical contradiction in a universe that is infinite? I tend to prefer finite things to infinite things myself, but I'm not aware of any reason the universe must be bounded. – Matt Thompson Mar 19 '21 at 05:50
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    A universe that never ends is infinite in time, is it not? - No, because the future does not exist yet. It is the same as with the number line. It has no bound, but there is no point of infinity on the number line. Reality or existence is described by observation, which is another word for measurement. Infinity is not a number, cannot be measured or observed, and so is not a part of reality. So nothing that exists can be infinite by the very definition and meaning of existence. – safesphere Mar 19 '21 at 14:54
  • @safesphere I love my girl infinitely much! A real observation. How real does that feel! No number involved. There is no bound on my love. – Deschele Schilder Mar 19 '21 at 15:29
  • @safespace - Ah, I think I understand now. If you're defining the size of the universe in terms of your past light cone (the "observable universe") then yes, it will be finite. So everything that is measurable is finite, and always will be. I was viewing it at a more conceptual level, as we often do want to make predictions about the future in our models, and we can choose any arbitrary point in time or space for these models. If your objection is purely semantic, it may be best to end the discussion here. – Matt Thompson Mar 22 '21 at 04:32
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Superconductivity and insulation are examples of an infinite, measurable quantity in physics. For a superconductor the conductivity is infinite. For an insulator the resistivity is infinite.

my2cts
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Why not say that equations which don't contain (or that result in) infinities should be abandoned? Why shouldn't Nature like infinities, to put anthropomorphically? Who knows?
It depends on the theory and what we like. For example, in quantum field theory, renormalization is introduced to avoid an infinite mass. We don't like infinite masses (though the difference between two infinite masses can be finite), so the theory is renormalized with the aim of obtaining finite masses.
It's totally normal though to introduce an infinite potential well. Does this correspond to a real infinity? Of course. Infinite potential walls do exist (quark confinement).
Going near an electrical point charge, the force you feel will grow to an infinite value. Will it reach infinity? Yes, why not? Is this a sign that classical electrodynamics is wrong? No. You can consider it even a sign for CED to be right on track.
According to quantum field theory though, it doesn't even make sense to speak of forces and particles. But in QFT infinite masses and coupling constants occur, which are doctored up by some infinite procedure to fit our expectations of masses and coupling constants to be finite.
What about the infinite curvature of spacetime. Is it really infinite? I think you'll be not surprised to see me answer yes. If not, then general relativity would be in trouble. You can use this infinity to herald the event of some theory that evades this infinity, like quantum loop gravity. And as such the avoidance of infinities is a good thing! It can even inspire new models of particles (beside strings).
To say that Nature doesn't like infinities intrinsically is exactly the same as saying that it does like infinity. Maybe Nature conforms to an infinity of laws. Wouldn't that be nice? Every physicist to come can have a law named after her!
The existence of zero values automatically implies that infinite "values" exist too. The two most distant (wrt each other) "values" imply the existence of each other.