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While deriving new equations , how do theoretical physicists know whether the proportionality constant in their equation will be dimensional or dimensionless?

I mean, say for example, we consider Stokes' Force in hydro-statics. There the experimenter saw that $F$ varies as $\eta$,$r$ and $v$ where the symbols have the usual meanings. Hence he got the equation $F = k\eta^{x} r^{y} v^{z}$ where $k$ is the proportionality constant and it is strictly dimensionless. So he found $F = 6\pi r\eta v$ where $k=6\pi$ by statistical analysis. (He could have gone for other powers of the variables so that $k$ has a unit.) Same is for the expression for Reynold's Number.

Again if we consider Newton's Law of Gravitation, there the experimenter saw that $F$ varies as $m_1$, $m_2$ and $\frac{1}{r^2}$ and concluded the relation $F = \frac{Gm_1m_2}{r^2}$ where $G$ is the proportionality constant and it is strictly dimensional with proper units. (He could have gone for other powers of the variables so that $G$ has no unit.) Same is for the expression for Resistivity.

I don't know if I have been able to express my question through the examples. But if I have , please help.

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  • $\begingroup$ Stoke's force is found by considering a special approximate solution of the Navier-Stokes equation, not by experiment, the Reynolds number is found by a scaling argument of the Navier-Stokes equaiton! Usually you do not set out in theoretical physics to find an equation of a certain form, but you set out to calculate something from a certain set of assumptions (e.g. the validity of some set of equations). The law of gravitation was found by Newton as a hypothesis, because is could consistently explain orbital motion. The exact proportionality constant was only measured later. $\endgroup$
    – Sebastian Riese
    Commented Sep 16, 2015 at 15:03
  • $\begingroup$ I think it depends on whether or not the equation is valid experimentally. In case of Stokes' Law, he assumed it was dimensionless, derived the powers and found that the real world did follow the equation. If it hadn't, he would have gone back and changed the equation. $\endgroup$
    – Goobs
    Commented Sep 16, 2015 at 15:08
  • $\begingroup$ @Goobs Stoke's law was found analytically, see Wikipedia, although I guess some of the scaling laws where known experimentally before. (The factor of $6\pi$ is strong indicator, btw). But then you usually check the scaling in one of the variables, leaving the others constant (and if you have checked all relevant scales, you can hypothesize that there is no complex interplay of the degrees of freedom and write down a scaling equation, where the units of the factor are obvious as the powers are fixed – this of course you have to verify again). $\endgroup$
    – Sebastian Riese
    Commented Sep 16, 2015 at 15:15
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    $\begingroup$ I just don't understand the downvotes. This is a beginner's question but it's a perfectly good one. I would urge downvoters, and especially anonymous downvoters, to examine their conscience and ask themselves if they're really making this site a better place. $\endgroup$
    – John Rennie
    Commented Sep 16, 2015 at 15:48
  • $\begingroup$ When developing a new equation, you have a bit of influence over the form of that equation. For the fluid mechanics example, if you deliberately formulate your equation as the ratio of viscous forces vs. inertial forces, you know that the ratio will be dimensionless, so any constants should be dimensionless. Note that there is no guarantee that you will get a dimensionless relationship, but doing so guarantees a dimensionless constant. $\endgroup$
    – David White
    Commented Jan 28, 2020 at 15:52

2 Answers 2

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There is no answer to this. When you are taught to use dimensional analysis at school the teacher invariably selects an easy example (it's almost always the pendulum) to keep things simple. In the real world there is no guarantee that you have a dimensionless constant.

It's actually quite rare to use dimensional analysis to derive equations in the real world. The sorts of simple systems that are amenable to dimensional analysis are usually already well known. However it's very, very useful to use dimensional analysis to check that an equation you derive is dimensionally consistent.

For example suppose you're working through a differential equation for some quantity, and after covering many sheets of paper with scribbles you end up with a final equation. It's very easy to make a minor mistake along the way, so the first thing you check is that your final equation is dimensionally consistent, i.e. the dimensions of the left and right sides are the same. If they aren't that means you've made a mistake somewhere. I routinely do this in my answers to questions on this site!

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    $\begingroup$ Thanks for your answer. But is there really no way how one can decide if one's equation will have a proportionality constant with or without dimension while deriving equations? I mean , my intuition says there must be a way! However, correct me if I am wrong. $\endgroup$
    – SchrodingersCat
    Commented Sep 16, 2015 at 18:27
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He could have gone for other powers of the variables so that $G$ has no unit.

No he couldn't have! Then the equation wouldn't have accurately described the force of gravitational attraction between two objects. Scientists can't just "choose" the equation to look like whatever they want; it has to accurately describe experiment.

Dimensionless constants don't need units in order to get units to play nice; constants with dimensions do.

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