I've been reading about dimensions analysis and at one point it mentions that there could be constants that dimensional analysis fails to define and dimensional analysis is never off by more that $(2\pi)^{\pm1}$? why the factor $(2\pi)^{\pm1}$? how does this come about?

because people are saying this is untrue I thought I should link the resource: what I'm reading is here: http://www.physics.drexel.edu/~bob/Chapters/dimensional3.pdf

1 line above example 1b

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    $\begingroup$ This is not true. Your source is probably writing about a specific problem/topic. $\endgroup$
    – valerio
    Jan 20, 2018 at 14:54
  • $\begingroup$ That's totally wrong! You should throw that book in the trash. $\endgroup$
    – knzhou
    Jan 20, 2018 at 14:57
  • $\begingroup$ Factor $2\pi$ is a difference between linear and angular frequency. Or between wavelength and inverse of wavenumber. Maybe thats where this came from. But yes, the statement is too bold. $\endgroup$
    – A.V.S.
    Jan 20, 2018 at 15:02
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    $\begingroup$ The quote is specific to that example, where it's using a pendulum - which relates to the linear vs. angular frequency thing. But note that dimensional analysis normally just tells you dimensions and you're guessing the form based on that. The errors can be arbitrarily large in doing that if you guess the form completely wrong, even if your guess has the right dimensions. $\endgroup$ Jan 20, 2018 at 15:16

1 Answer 1


Counterexample: what is the surface area of a sphere? Dimensional analysis says $r^2$, which is off by $4\pi$.

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    $\begingroup$ I thought dimensional analysis says it goes like $d^2$, which is only off by $\pi$. $\endgroup$
    – JEB
    Jan 20, 2018 at 18:35
  • $\begingroup$ @JEB there you go letting the facts get in the way of a good story $\endgroup$
    – Ben51
    Jan 20, 2018 at 19:12

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