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We started the first day of our semester today by having a review of dimensional analysis. Viewing it afresh, I began wondering how it all “works”, i.e. what is the physics behind it all?

Nature sure doesn’t give a damn about how (or in which units) we choose to measure its properties. But anyway, we, according to our whatsoever knowledge, decided to “label” the observables in the universe with a finite set of “fundamental” units including meters, seconds, grams, etc.

And then the dimensional analysis applied to get time period ($T$) of a pendulum, assuming (reasonably) that it might only depend on gravity ($g$), length ($l$), mass ($m$) and amplitude ($\theta$), we neatly (and correctly) get that $$T=f(\theta)\sqrt{{l\over g}},$$

which I find humongously nontrivial.

How on earth can a “labelling system” devised by us put such severe limits on how a pendulum can oscillate and thus tell Nature how to behave? Just by noticing that adding grams to seconds is meaningless as adding apples to oranges?

(Note that the example of pendulum was just representative.)

The question which has been claimed mine to be the duplicate of is entirely different indeed, except the similar title. Let me show how.

That question asks why it is not justified to add quantities of differing dimensions. While I ask why dimensional analysis gives the correct physical form of the answer in terms of the observables.

Atom
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    Minor nuance: the labeling system doesn't "tell Nature how to behave." It instead describes how Nature behaves. – Kyle Kanos Jan 01 '20 at 16:44
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    The question seems vague and tendentious. It's not clear what set of premises you're willing to admit, so it's not clear what would satisfy you as an answer. One thing I'd suggest is that you distinguish between abstract dimensions like length, time, and mass, and concrete ones like meters, seconds, and kilograms. The concrete ones are clearly arbitrary human inventions. Not so for the abstract ones. –  Jan 01 '20 at 16:45
  • @KyleKanos Yep. But I wanted to sound dramatic ;) – Atom Jan 01 '20 at 16:45
  • @BenCrowell Yes, I meant what you mean. I jaunt prefer to call the abstract dimensions as units instead. – Atom Jan 01 '20 at 16:48
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    Does this answer your question? What justifies dimensional analysis? – Kyle Kanos Jan 01 '20 at 17:01
  • @KyleKanos No, it doesn’t. Already read it before posting mine. – Atom Jan 01 '20 at 17:18
  • define 'fruit' as apples plus oranges – user45664 Jan 01 '20 at 17:29
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    Okay, then you need to explain (via editing the post, not a comment) about how this question differs. Because to me, and apparently two other people, this is answered in the proposed duplicate. – Kyle Kanos Jan 01 '20 at 17:35
  • @KyleKanos Now? – Atom Jan 01 '20 at 17:41
  • @KyleKanos: make that three. – Gert Jan 01 '20 at 18:42
  • I find it a little funny that despite all the discussion, the formula in the question is actually dimensionally wrong :P since $T \propto \sqrt{l/g}$. – Philip Jan 01 '20 at 18:44
  • @Philip No, it is absorbed in $f(\theta)$. – Atom Jan 01 '20 at 18:52
  • @Atom Not sure what you mean, I wasn't talking about the numerical constants, I was talking about $\sqrt{g/l}$ (as mentioned in the question) not having dimensions of time. – Philip Jan 01 '20 at 18:56
  • @Philip Oh my! I made a blunder! Thanks for pointing! Fixed now. – Atom Jan 01 '20 at 19:10
  • Otherwise a possible duplicate of Why does dimensional analysis to find a characteristic length, time, etc work? which contains further suggestions. See also the 'linked questions' column in each of these. – sammy gerbil Jan 01 '20 at 20:06
  • If you remain dissatisfied with all of the answers given in the suggested questions, you could start a bounty on one of the questions which remains open. The procedure allows you to indicate in what way the answers are unsatisfactory, and what kind of an answer you are looking for. – sammy gerbil Jan 01 '20 at 20:10
  • @Atom: even though you feel that the question is different, go check out the answers and the links there nevertheless (some of them in the comments), as they may answer your particular question as well. BTW, as far as I'm concerned, dimensional analysis is really a bookkeeping device. We tend to get philosophical about the meaning of dimensions and units, but mathematically, a unit is just a placeholder for an unspecified constant; in that view, dim. analysis comes down to following equation manipulation rules. – Filip Milovanović Jan 01 '20 at 20:23

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I think you believe the "labeling system" is arbitrary, frivolous, and of no consequence, so how can dimensional analysis predict anything about the period of a pedulum? Using Ben Crowell's language of "abstract dimension", the discovery of dimension is a fundamental empirical measurement made on the real world. Someone first observed that when you wait the world changes and quantified it's passage with the unit time. Someone first noticed an object can be moved over in 3 different directions and quantified the movement with the unit distance. No one has observed any movement that could be called a fifth dimension so we don't need it's unit when we do dimensional analysis. Identifying these transformations we can do to real world objects is a fundamental observation and provides the nontrivial units for dimensional analysis.

Further experimentation has led to the discovery that waiting, translations, rotations, and boosts form a particular mathematical group. This has allowed even more predictions about nature (eg: spin) than dimensional analysis.

Gary Godfrey
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  • This is the kind of answer I'm looking for! Thanks. But I'm still not completely satisfied. Can you add on? – Atom Jan 02 '20 at 04:13