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My question has to do with the semantics of "proportionality". Often people say quantity y is proportional to quantity x if increasing x by a factor k increases y by the same factor. However, doesn't this simple "definition" lead to inconsistencies when it comes to differentiation/integration?

For example, the total charge Q in some volume is equal to the triple integral of the charge density over the volume. Doubling the charge density function doubles its integral over the volume and hence results in a total charge 2Q. Would it be appropriate to say Q is proportional to the charge density in that case? Other examples come to mind, such as the relationship between acceleration and velocity, and invoking the constant factor property of the derivative.

Thank you in advance.

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The following operators:

  • Proportionality: $g(x)=kf(x)$
  • Integration: $g(x)=\int^x f(x') dx'$
  • Differentiation: $g(x)=\frac{d}{dx}f(x)$

are all linear, which means aplying a proportionality in the first argument is the same than applying it in the second, which in words is the same that saying "amplifying by $k$ the $f(x)$ magnitude is the same than amplifying by $k$ the integral|derivative of $f(x)$"

In the case of notables physical identities, of course we must say if we double the charge density, the total charge gets doubled, or if we double the magnetic flux density, the magnetic flux gets doubled, and so on.

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  • $\begingroup$ But we can't say g(x) is proportional to f(x) in the case of differentiation can we? Because proportionality implies that we can write g/f = constant. $\endgroup$ – T.A. Sep 30 '17 at 11:48
  • $\begingroup$ If i understood proportionality as a k factor, yes we can. The magnetic flux density is proportional to the magnetic flux, or the velocity is proportional to the acceleration and to the displacement. We cannot say current is proportional to voltage.... $\endgroup$ – Brethlosze Sep 30 '17 at 12:29
  • $\begingroup$ If we understood proportionality as a ratio, we cannot, $a/v$ would be a different signal indeed. Under this sense, most of linear relations will fail. $\endgroup$ – Brethlosze Sep 30 '17 at 12:33

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