Is the gravitational constant $G$ a rational number or an irrational number?
Not this only, what about such similar questions about all the universal constants?
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6 Answers
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Fundamental constants have arbitrary values since they depend on what measure you use to measure them. So we could make $G$ rational by using units where $G=1$. So the question does not have meaning there.
However, in physics people usually prefer dimensionless numbers to denote the fundamental parameters of the universe for this reason. Perhaps the most well known is the fine structure constant $$\alpha = \frac{e^2}{4\pi\epsilon_0 \hbar c}\approx 0.0072973525693(11) \approx \frac{1}{137}.$$ For a while it was believed that it might actually be exactly 1/137, but measurements have disproven this. At present there is no theory saying it has to be rational or not.
answered Jul 17, 2021 at 17:33
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This is unknowable, because all physical constants are measured quantities, and all measured quantities come with some uncertainty, but you need to know the complete decimal expansion for a quantity to know whether it is rational or irrational.
The other question, of course, is "in what units?". But the first thing is true no matter what units you use, so it's moot.
The exception to this is artificats like the speed of light in SI units, which is fixed to be exactly 299792458 m/s by the definitions of the meter and second, and thus, a "rational number", in a trivial sort of way.
answered Jul 17, 2021 at 17:26
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$\begingroup$ Relevant is that it is a measured quantity, not a calculated one, like for example π (pi, which starts as 3.14159265358979323846264338327950.... etc. $\endgroup$– PcManJul 17, 2021 at 17:33
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$\begingroup$ @HTNW: we can never compute ALL of the decimals of any irrational number, though. We can only come up with some principle, usually an iterative one, that lets us comute the 'n'th element. $\endgroup$ Jul 17, 2021 at 18:20
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$\begingroup$ Knowing a decimal expansion doesn't tell you whether a number is rational or not; e.g. we know the expansion of $e+\pi$ but we don't know if it's rational. Conversely, e.g. the standard proof for the irrationality of $\sqrt{2}$ does so without a decimal expansion. And finally, I think you are wrong to gloss over the "units" thing. The constant $G$ (and many others) isn't a number! It is a dimensionful quantity. Rationality is meaningless for it. In particular, you can choose units to make it look rational or irrational. This answer fails to discuss that important point. $\endgroup$– HTNWJul 17, 2021 at 18:20
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$\begingroup$ @JerrySchirmer And the same is true for most of the irrational numbers we come across, like $e$, $\pi$, and $\sqrt{2}$, and yet we know all of them to be irrational without inspecting their entire, infinite decimal expansions. What I'm trying to say is that bringing up decimal expansions at all is simply the wrong way to discuss this topic. $\endgroup$– HTNWJul 17, 2021 at 18:22
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$\begingroup$ @HTNW most irrational numbers are not special numbers with names like that though. The disytinction between "rational" and "irrational" is "expressible as a fraction", and there is a one to one correspondence between "expressible as a fraction" and "the sequence of decimals is nonterminating and nonrepeating". And a measured quantity in the world must be known with zero uncertainty to make that distinction, and generally, we can think of the precision of measurment of a measured quantity as "the number of decimal places we know" $\endgroup$ Jul 17, 2021 at 20:14
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I would like to add a few comments to this discussion.
Newton's gravitational constant has dimensions. This is the point of Gandalf61's answer: We can either define the dimension of $G_N$ by setting $G_N = 1$ or making some other convention, or we can try to compute $G_N$ in terms of another dimensionful parameter. In the latter case, Firdous' original question makes sense. If we have a dimensionful parameter $A$, we can try to prove an equation $G_N = C \cdot A$ and ask whether $C$ is an integer.
In quantum field theory, we usually work in "natural units" in which $c = 1$ and $\hbar = 1$. Then $G_N$ has the dimensions of (mass)$^{-2}$. If $M$ is some mass in the theory, we can ask if the theory can predict $G_N = C \cdot M^{-2}$, where $C$ is an integer.
There is actually a theory in which this is possible. There is a quantum field theory model called the "nonlinear sigma model" in which the fields take values a sphere instead of, as usual, a flat Euclidean space. The radius of the sphere is a quantity called $F_\pi$, which has the dimensions of mass. Next, consider the supersymmetric generalization of this model. In 1982, Edward Witten and Jonathan Bagger studied the coupling of this model to supergravity. They found that there are topological constraints that can only be satisfied if $G_N = m/F_{\pi}^2$, where $m$ is an integer! See Witten and Bagger, Physics Letters B115, 202 (1982).
This model is very artificial. It also requires a lot of sophisticated theoretical baggage to understand. That is probably to be expected; finding an equation for Newton's constant is not an easy problem to solve. Nevertheless, it provides a concrete existence proof that a solution to Firdous' problem may be possible.
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First, the numerical value of a dimensionful quantity in physics is not particularly meaningful, because I can always change to a different system of units where it has any numerical value I want. For example, it is very common in gravitational physics to work in units where the numerical value of $G$ is exactly equal to 1.
Still, you can ask your question about dimensionless quantities. An example where $G$ makes an appearance would be the ratio of half the Schwarzschild radius of the Earth, $2 GM_\oplus/c^2 \approx 1\ {\rm inch}$ (where $M_\oplus$ is the mass of the Earth), to the actual radius of the Earth, $R_\oplus=6400\ {\rm km}$. The ratio (call it $C_\oplus$) is approximately $\alpha=1.4 \times 10^{-9}$, and tells us we would need to compress the Earth by about a billion times its current (linear) size to form a black hole. Is $C_\oplus$ rational or irrational?
In order for $C_\oplus$ to be irrational, we would need to know its value to infinitely many decimal places, and show they don't repeat. There are a few problems here.
- Eventually when we start computing $C_\oplus$ to many decimal places, we will find that assumptions in the simple model we started with (such as that the Earth is a sphere) break down. At a microscopic level, it's not even really possible to define the size of the Earth along one axis, since the molecules in the Earth's crust are moving randomly, and there are quantum effects that mean the positions of these molecules spread out. There are many physical sources of uncertainty that mean this number is not even really well-defined beyond a certain number of decimal places! Still, there dimensionless quantities that (as far as we can tell with our current state of knowledge) are more fundamental, such as the fine structure constant in electromagnetism. We could ask if a fundamental constant is rational or irrational.
- Even for fundamental constants like the fine structure constant, we only know these constants experimentally. Therefore we only ever know a finite number of digits of the constant. So we can never really know the answer to whether the constant is irrational or rational, but in practical terms any number in physics is always rational (or more precisely, its "best guess" value is rational, and also comes with a rational-valued estimate of our uncertainty on the best guess value).
- Finally, most abstractly and weirdly, fundamental constants are not really constants but change with energy due to the renormalization group, so even if we could answer this question for one constant and one value of energy, the answer itself may depend on the energy scale at which you ask the question.
I think a takeaway is that irrational numbers are actually extremely strange! Any number we can construct by typing it into a calculator or applying a series of arithmetic operations, is by definition a rational number. Having said that, they are an extremely useful abstraction for describing continuous functions, even if in practice due to our finite precision, in physics we never really deal with continuous functions in a mathematical sense [let's see the comments I get on that one...]
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$\begingroup$ I think taking "what a calculator computes" as a definition for what a "realistic number" is is deeply limiting, because "what you can do on a calculator" is inherently limited by the representation of the number on a finite set of bytes, and you already know it's all approximate, because it carries floating point errors along with it for the ride. $\endgroup$ Jul 17, 2021 at 17:47
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$\begingroup$ @JerrySchirmer By "number you can type into a calculator" I really mean "any number you can construct with integers and a finite number of algebraic operations." I agree the notion of a "realistic" number is subjective and would not want to even attempt to actually do any physics or math this this restriction. But I think often people have conceptual issues with complex numbers while (in my very subjective opinion) irrational numbers are less intuitive and have more paradoxes. Anyway I mainly meant this as a throwaway comment :) Definitely not suggesting to do math without irrational numbers. $\endgroup$– AndrewJul 17, 2021 at 17:54
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The value of $G$ depends on the units in which you measure it, so this question is not well defined. In SI units $G$ is almost certainly irrational since almost all real numbers are irrational. But you could choose to use a system of units in which $G=1$, in which it would of course be rational.
It makes slightly more sense to ask this question about physical constants which are dimensionless. As far as we know, none of the naturally occurring dimensionless physical constants are rational (although you could no doubt construct some artificial examples). Physicist Arthur Eddington had a theory that the reciprocal of the fine structure constant was a whole number, but more precise measurements of its value showed that to be incorrect.
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what about such similar questions about all the universal constants?
All universal constants are merely numbers, and as such are very likely to be irrational, but this would be almost impossible to prove.
Note that many things that are perceived as constants are nothing of the sort.
For example π and e are not "constants", but merely fixed mathematical ratios, and subject to calculation to any arbitrary accuracy you wish.
