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I recently explained multiplication of (non-zero) complex numbers to my Mathematics Fundamentals students, the usual bit about “multiply their lengths, and add their angles”. Of course, there is always at least one student who wants to know why the angles are added instead of multiplied – after all, this is multiplication, right? – and so I go through a non-trivial example to establish that it is at least highly plausible that addition, not multiplication, of angles is what occurs. This quiets their objection, but it always gets me to wondering: Indeed, are angles EVER multiplied anywhere in Mathematics? I don’t recall ever having seen such a thing. And since this wonderful wonderland of MSE now exists, I will pass this question on to the community.

Of course, there are exercises of the form “If cos(f(x)) = sin(g(x)), solve for x.”, where this could happen, but of course what I am asking is whether this happens as part of a significant theorem, or as part of solving a problem of real physical interest. The non-multiplication of angles seems all the more counter-intuitive because they are actually dimensionless, in spite of the use of dimension-sounding phraseology such as “degrees” and “radians”. So, angles are just numbers, and surely numbers can be multiplied together, right?

(Of course, not being dimensionless is no barrier to getting multiplied by objects of the same/different type. After all, we deal with the square of seconds in regard to acceleration, and with the square of grams in regard to the (statistical) variance of weights, and so on. I’m just saying that it would seem all the more plausible that they would occasionally get multiplied together if they are dimensionless.)

I’m going to go out on a limb and conjecture that there is no such case. The basis of my conjecture is simply that I have never seen it happen, and also the fact that angles do not exist for vectors of zero length. I know this is quite tenuous, but what I’m guessing is that only quantities that behave well for 0 are eligible, so to speak, to be multiplied together.

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The reason you add angles is because that's a group homomorphism from $\mathbb{R}$ to the rotation group; you don't have to discount the possibility of multiplying angles to say that. –  Qiaochu Yuan Jun 26 '11 at 23:22
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Multiplying angles would give some special meaning to an angle of 1 radian. I can't think of any sort of geometric construction in which an angle of 1 radian arises naturally. –  Corey Jun 26 '11 at 23:59
    
@QiaochuYuan, by the way, angles can also be subtracted: this is another way to form a field in $\mathbb{C}$, by complex conjugate multiplication. Is this operation also a group hom? –  alancalvitti Sep 19 '12 at 18:00
    
Solid angles, measured in steradians. –  dimensio1n0 Jun 8 '13 at 5:50

4 Answers 4

up vote 22 down vote accepted

Perhaps this is a bit too obvious to be a useful answer, but I will post it anyway:

Angles are only defined up to congruence $\!\!\mod2\pi$. Addition preserves this symmetry: $$ (\vartheta+2\pi k) + (\varphi+2\pi j) = \varphi + \vartheta + 2\pi(k+j) =: \xi + 2\pi l, \qquad k,j,l\in\mathbb{Z} $$ while multiplication doesn't. Therefore, the multiplication of two angles cannot be well-defined, unless you come up with some additional constraint on the range of the angles (in physics, you would call it a gauge). But this would not represent the mathematical/physical meaning of angles.

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Why would the product of two angles be measured in radians? See Gerry's answer; square-radians are not meaningless. –  user6701 Jun 27 '11 at 7:13
    
Why would the product of two angles be measured in radians? I did not say it would. — As for square radians, these are not really a unit of the product of two angles, but rather the unit of the surface spanned by the exterior product of two "infinitesimal angles" (differential forms). For these, the congruence problem does not arise. –  leftaroundabout Jun 27 '11 at 12:49
    
The question was whether angles are ever multiplied. If I understand your answer correctly, it argues that multiplication of angles wouldn't make sense if angles were closed under multiplication. Apparently angles are multiplied in some areas, and their product is not an angle. –  user6701 Jun 27 '11 at 12:53
    
Bingo. This is the I-should-have-thought-of-that answer that I was looking for. I have up-voted this answer, and accepted it. I love MSE! –  Mike Jones Jun 27 '11 at 21:44
    
@Tim: regardless of whether the product is again an angle or something else, there can be no equivalence relation on the space of results that takes into account any possible gauge of the factor angles. So as I said, you would need to choose a particular gauge, which would be something like $\vartheta\in[0,2\pi[$. But even then, the product of two angles would equal the solid angle of the area spanned by them only approximately, and only if they are small. –  leftaroundabout Jun 27 '11 at 22:26

For what it's worth, there is such a thing as a "square degree" and a "square radian", better known as a steradian. As GEdgar notes, these come up in spherical trig.

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I was wondering why no one mentioned Solid Angle and Steradians except for you. Is it not a topic in High School Physics everywhere? –  Galois Group May 9 '12 at 20:57
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Steradians do measure solid angles, but I have only ever seen trig functions of angles actually multiplied together, even in L'Huilier's formula it is tangents of angles that are multiplied together. This preserves the modular congruence that leftaroundabout mentions. –  robjohn Sep 20 '12 at 11:47

Maybe look in spherical trig for more. Area of a spherical triangle, and such things.

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One could define an alternate "multiplication" $*$ for complex numbers $u,v$, such that $uv$ corresponds to ordinary complex multiplication but $u*v = (|u|+|v|)e^{i \arg(u)\arg(v)}$, but I haven't seen anything like that in practice and am not sure what the utility of it would be.

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