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Since the fine structure constant (denoted alpha) is a pure real number, it just occured to me to ask if it is a rational number or not.

Qmechanic
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user107952
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    Pretty sure there's a pi in there somewhere, making it irrational. Also I think it's defined in terms of experimentally measured values, so what you're asking is like asking if we have measured the charge of an electron to a high enough accuracy to say that it's this number of Coulombs (which of course isn't possible). – Physics Llama Sep 10 '14 at 01:00
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    There is no difference in physics between rational and irrational numbers. That's a purely mathematical concept that is completely irrelevant "in reality". – CuriousOne Sep 10 '14 at 01:28
  • @PhysicsLlama Indeed it's defined in terms of experimentally measured values. So your first comment that it's irrational since it contains a factor of $\pi$ is not correct. – gj255 Oct 21 '16 at 19:00
  • Related: http://physics.stackexchange.com/q/2010/2451 , http://physics.stackexchange.com/q/52273/2451, http://physics.stackexchange.com/q/251347/2451 and links therein. – Qmechanic Oct 21 '16 at 19:07
  • Why is there no difference in physics between rational and irrational numbers? –  Nov 01 '18 at 22:18
  • It is close to the rational number $137.036$ – Jaume Oliver Lafont Jan 14 '21 at 11:56

2 Answers2

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The fine structure constant is given as: $$\alpha = \frac {k_{e} e^2} {\hbar c^2}$$ Immediately we have a problem in determining the rationality or otherwise of $\alpha$. The Coloumb constant, Planck constant (maybe not?) and speed of light are all either exact numbers or pre-defined. Since the elementary charge $e$ is an empirically derived constant we can only measure a finite number of digits of this number. This is true for two reasons: First any measurement device will have $\it{some}$ degree of imprecision. Second, the uncertainty principle tells us we cannot measure electron motion+position simultaneously with perfect precision anyway.

Taking this into account we can only know a finite number of significant figures of $\alpha$ and so it must be rational empirically.

As @CuriousOne points out though, this is a rather meaningless concept.

Arthur Suvorov
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  • Nice answer, but the part about the uncertainty principle leaves me a little puzzled. Why would you need to know position and momentum with perfect precision to measure $e$ with perfect precision? – Lorenzo Pompili Nov 01 '23 at 21:17
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As other people have stated, we are currently getting this quantity experimentally, so thus far this is not an appropriate question.

The origin of the value ~1/137 remains unexplained; There is currently no theoretical explanation for why is has the value that is has. In the future, physicists might be able to explain its origin, perhaps by deriving it from other more fundamental constants. Then we would be able to answer your question.

Contrast this with, say, the gyroscopic factor, which characterizes the magnetic moment of quantum particles. There is a theoretical derivation for this dimensionless quantity which can be derived from the theory of quantum electrodynamics.

Someday, we hope, the same will be true of $\alpha$.

psitae
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