# Can Angles be Assigned a Dimension?

### 1. Some Background on Dimensional Analysis

… if you are not already familiar with it.

#### 1.1 Dimensions

Dimensional Analysis is a way of analysing physics equations that considers only the qualitative dimensions – mass, length, time, charge .. – of the quantities involved, not the values that they take in the given problem.  This is not to be confused with Euclidean dimensions.

The fundamental rule is that you can only add, subtract or equate terms that have exactly the same constituent dimensions, and each to the same degree.  An acceleration cannot be a force, or be added to a force, because the latter includes a mass dimension while the former does not.  An area cannot be compared to a distance because it is length squared.

For a detailed discussion see e.g.  http://web.mit.edu/2.25/www/pdf/DA_unified.pdf, or any text on the Buckingham Pi theorem.   (There are also Wikipedia and Khan academy links, but these unfortunately confuse DA with managing units as variables, which is a separate topic.)

#### 1.2 Notation

The standard notation in DA is that if x is a variable in an equation then [x] extracts its dimensionality.  The dimensions themselves are labelled M for mass, L for length, T for time, Q for charge, θ for temperature …

Thus, if F is a force then ##[F]=MLT^{-2}##.  The well-known equation F=ma would be analysed as ##MLT^{-2} = (M)(LT^{-2})##, which is clearly true.

#### 1.3 Uses

##### 1.3.1 Predicting the form of a relationship

DA can be a useful shortcut to establishing the general form of how one quantity depends on others.

Example: We presume that the pressure difference, ##\Delta P##, between the inside and outside of a bubble depends on the radius, ##r## and the surface tension ##S##.

## [ \Delta P]=ML^{-1}T^{-2}##

##[r]=L##

##[S]=MT^{-2}##

The only way to combine the pressure difference and surface tension to obtain a length is

## (MT^{-2})/(ML^{-1}T^{-2})=L##

Hence we can say

##r=kS/\Delta P##

for some constant k.

##### 1.3.2 Error checking

Many algebraic errors can be caught by checking dimensional consistency.

### 2. Angles

Angles have never been considered to have dimension.  Consider, for example, the formula for arc length ##s##:

##s = r\theta##

Since ##s## and ##r## each have dimension ##L##, the angle cannot have a dimension, or so it seems.

#### 2.1 Units Matter

Dimensionless combinations, like ##force/(mass \times acceleration)##, generally have the useful feature that they are invariant to the units used.  As long as the units used for the variables are consistent, the same number results whether you use SI or Imperial, or Babylonian.

This is not true of angles; many problems posted on the Homework forums are resolved when the student remembers to plug in the number of radians as argument to the sine function instead of degrees.

#### 2.2 Distinct Entities with Same Dimensionality

It is somewhat unsatisfactory that, when reduced to mere dimensions, some pairs of quite different entities appear to be the same.

Torque and energy are both force times distance.  In terms of vectors, the first is the vector product, the second the scalar.

Angular momentum and action are both ##ML^2T^{-1}##.  Again, a vector and a scalar.

### 3 Axioms for an “Imaginary” Dimension

Consider assigning angles the dimension ##\Theta##, with some unusual properties:

• ##\Theta^2=1##
• The cross product operator itself has dimension ##\Theta##
• ##i##, the square root of -1, has dimension ##\Theta##

#### 3.1 Vectors and Angles

The following table illustrates the use of the dimensionality of angles with cross and dot products.

EntitySample EquationDimension
Arc length element ##\vec{ds}=\vec r\times\vec{d\theta}#### L=(L)(\Theta)(\Theta)##
Torque ##\vec\tau=\vec r\times\vec F####ML^2T^{-2}\Theta=(L)(\Theta)(MLT^{-2})##
Work  ##E=\vec r.\vec F## ##ML^2T^{-2}=(L)(MLT^{-2})##
Angular momentum ##\vec L=\vec r\times\vec p## ##ML^2T^{-1}\Theta=(L)(\Theta)(MLT^{-1})##
Gyroscopic precession ##\vec \tau=\vec \Omega_p\times\vec L## ##ML^2T^{-2}\Theta=(\Theta T^{-1})(\Theta)(ML^2T^{-1}\Theta)##
Velocity ##\vec v=\vec r\times\vec{\omega}## ##LT^{-1}=(L)(\Theta)(\Theta T^{-1})##

E.g. to resolve the ##s=r\theta## case for arc length, we can argue that this should really be expressed as the integral of the magnitude of a vector:

##s=\int|\vec{dr}|=\int |\vec r\times\vec{d\theta}|##.  The ##\Theta## dimension of the angle is neutralised by that of the cross-product operator.

#### 3.2 Functions of Angles

Raising a dimensioned entity to a power is fine, because we can still express the dimensions of the result.  For other functions, such as exp, log and trig functions, it is more problematic.  If you ever find you have an equation of the form ##e^x##, where ##x## has dimension, you can be pretty sure you have erred.

For trig functions, it would be reasonable to require the argument to be an angle, but ##e^{i\theta}## would appear to create a difficulty for assigning angles a dimension.

This can be resolved by giving ##i## the ##\Theta## dimension also.  Based on the power series expansions, we see that the odd trig functions (those for which f(-x)=-f(x)) necessarily return the ##\Theta## dimension but the even functions, such as cosine, return a dimensionless value.

Thus, ##e^{i\theta}=\cos(\theta)+i\sin(\theta)## is entirely dimensionally consistent.

#### 3.3 Areas and Volumes

Areas are naturally generated as cross products of vectors, imbuing them with the dimension ##L^2\Theta##.  Since the vector is normal to the surface, this is analogous to rotations as vectors.

Since volumes arise from the triple scalar product, it would seem that these should also have the ##\Theta## dimension.  This is more surprising.

The solid angle element subtended at the origin by a surface element ##\vec {dS}## at position ##\vec r## is given by ##d\Omega=\frac{\vec r.\vec{dS}}{|r|^3}##.  Or, if we wish to make this a vector in the direction ##\vec r##, ##\vec{d\Omega}=\vec r\frac{\vec r.\vec{dS}}{|r|^4}##.

Alternatively, in polar coordinates, ##d\Omega=\sin(\theta).d\theta d\phi##

Whichever way, the dimension is again ##\Theta##, which feels consistent with the result for volumes.

#### 3.4 Complex Arithmetic

If ##i## is to be given dimension, what are we to make of ##1+i##?

Despite appearances, there is no difficulty.  ##1+i## is a convenient notation, but it is not addition in the same sense as in ##1+1##.  The 1 and the ##i## retain their separate existences.  We might just as easily, if less conveniently, have chosen to write complex numbers as an ordered pair, like <x, y>.  The real and imaginary parts never get crunched together in the same way as in normal addition, so they can have different dimensions without creating any inconsistencies.

#### 3.5 Frequency and Angular Frequency

In wave expressions, frequency, ##f##, is the number of cycles per unit time, while angular frequency, ##\omega##, is radians per unit time.  ##\omega=2\pi f##.

Clearly ##\omega## should have dimension ##\Theta T^{-1}##.  The dimension for ##f## depends on whether the factor ##\pi## is to be taken as an angle or as a dimensionless number performing a conversion of units.  Taking ##f## as having dimension ##T^{-1}## appears to be best.

#### 3.6 Planck’s Constants

Consider the equations

##E=hf##

##E=\hbar \omega##

Since ##h## has dimension of action, ##ML^2T^{-1}##, it has no angular dimension.  That is fine for the first equation since frequency is just ##T^{-1}##.

In the second equation, ##[\omega]=\Theta T^{-1}##.  ##\hbar## is defined as ##\frac h{2\pi}##.  Since ##\pi## is an angle here, that has dimension ##ML^2T^{-1}\Theta##, cancelling the ##\Theta## from ##\omega## and achieving dimensional consistency.

But note that this gives ##\hbar## the units of angular momentum, not action. An implication is that ##\hbar## should perhaps be considered a vector, and we should write the photon energy as

##E=\vec{\hbar}.\vec{ \omega}##

though how one is to justify that ##\vec {\hbar}## is necessarily in the direction of the velocity is unclear.

Likewise for momentum

##\vec p = k\vec{\hbar}##

where k is the wavenumber.  Note that this provides momentum as the vector it should be, rather than just defining its magnitude.

The Heisenberg Uncertainty relations, such as

##\frac 12\hbar\leqslant \Delta \vec p.\Delta \vec x##

would appear to violate both the dimensionality and the notion of making ##\hbar## a vector.  But if we follow the steps in the proof of the uncertainty relation we come to this penultimate statement:

##\frac 14\hbar^2\leqslant (\Delta \vec p.\Delta \vec x)^2##

At this point, giving ##\hbar## an angular component of dimensionality creates no problem.   Neither is there an issue with thinking of it as a vector.  These problems only appear when we overlook the ambiguities that so commonly arise when taking square roots.

### 4. Postscript

Subsequent to penning the original article, I have become aware of numerous prior attempts, dating back as far as 1936.  An excellent summary is by Quincey and Brown at https://arxiv.org/ftp/arxiv/papers/1604/1604.02373.pdf.

But their list misses a key one:

C. H . Page, J. Research National Bureau of Standards 65 B (Math. and Math. Phys.) No. 4, 227-235; (1961). http://nvlpubs.nist.gov/nistpubs/jres/65B/jresv65Bn4p227_A1b.pdf

It appears that most of my work above is a rediscovery of Page’s:

ResultPage in “Page”
Cross product has angular dimension, dot product does not 231
##\Theta^2=1##Appendix 2
sine has angular dimension, cosine does notAppendix 2
solid angles have angular dimension (i.e. not squared)Appendix 2
whole cycles, as a unit, are dimensionlessAppendix 3

Curiously, Page drew attention to the difficulty posed by ##e^{i\theta}=\cos(\theta)+i\sin(\theta)##, but overlooked the remedy of assigning angular dimension to ##i##.

He did not consider Planck’s constants.

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106 replies
1. David Reeves says:

I think this is an interesting topic. What is more basic to physics than how we measure things?In SI units the angle is not a base quantity. We measure the angle in radians, which is a ratio of two lengths, namely the length of the subtended arc to the length of the radius, giving a dimensional ratio of L/L = 1. So we say the angle is dimensionless.Some people have argued for making the angle a base quantity. See for example https://arxiv.org/ftp/arxiv/papers/1604/1604.02373.pdf. This may be an intriguing proposal but I don't see much interest at this time. It seems we are happy with the SI system.It may just be more convenient to keep angles dimensionless. Consider two similar triangles. Perhaps they are both 30-60-90 triangles but the hypotenuse of triangle #1 is twice the length of that of triangle #2. The corresponding angles are equal, but the corresponding sides are not. I suppose it's fair to ask why this is more convenient? It certainly seems convenient to me.I think this question is related, at least subjectively, to time. The ancients came up with 360 degrees, I believe, because it corresponds to a 360-day year in some ancient calendar. You can associate a point moving around on the circumference of a circle with the passage of time. We don't care how long that circumference is. We just want to know how many units of time have passed.For example, consider our standard mechanical 12-hour clock with a round face. It may be a wristwatch or Big Ben. In either case, we know that when the little hand is at a certain angle from straight up, it means 20 minutes past the hour.In addition, in physics, the study of waves is an enormously important topic. So we want our system to be convenient for handling the mathematics of periodic motion.

2. haruspex says:

[QUOTE="Stephen Tashi, post: 5653780, member: 186655"]What would the topic of the new thread be?[/QUOTE]It's your topic.  Maybe "why does dimensional analysis work?"

3. Baluncore says:

[QUOTE="haruspex, post: 5653671, member: 334404"]How would you feel about having it moved to a different thread?[/QUOTE]Sorry about being off topic, I plead a little bit guilty. Go ahead and split or fork the thread.We need more discussion about those interesting bundles of {numbers, units and dimensions}.

4. Stephen Tashi says:

5. haruspex says:

[USER=186655]@Stephen Tashi[/USER] , [USER=447632]@Baluncore[/USER] :I am pleased to have engendered such an interesting debate, but it does seem to have wandered a long way from my original article.  How would you feel about having it moved to a different thread?Meanwhile, you might be interested in http://nvlpubs.nist.gov/nistpubs/jres/65B/jresv65Bn4p227_A1b.pdf.  Seems like much of my endeavour is a rediscovery of Page's work.  See p 231 and Appendices 2 and 3.  He did miss the trick of giving i angular dimension, and did not discuss Planck's constants.  The rest of his article might be relevant to your own discussion.

6. Baluncore says:

[QUOTE="Stephen Tashi, post: 5651861, member: 186655"]By contrast, empirically, there are few situations in physics where knowing the sum x + y of units of different dimensions is useful.[/QUOTE]Can you please give examples of those few situations in physics.

7. DaTario says:

[QUOTE="Stephen Tashi, post: 5649802, member: 186655"]In general, changing units changes equations by multiplying one or both sides of the equation by constant factors.So the prohibition against adding unlike units can't be explained by the invariance of equations.  We have to explain why a particular type of variation is the only permissible kind.[/QUOTE] Yes, I agree. Admissible invariances seems to be those which involves solely a scale factor.

8. Stephen Tashi says:

9. haruspex says:

[QUOTE="Stephen Tashi, post: 5650730, member: 186655"]Dimensional analysis does not say that a physical unit can take a values corresponding to any given real number.    So I see nothing wrong with your analogy.I'd put it this way:  Changing the units in a product and "knowing how to convert" the result (using conversion factors in the standard manner" introduces no new ambiguity in the description of the physical situation. We "know how to convert" because we know to use conversion factors.  But the conventions of using conversion factors doesn't explain why those conventions are physically useful.  In my opinion, it is just an empirical fact that the ambiguity in knowing the product of dimensions, but not knowing the individual factors often doesn't detract from the usefulness our knowledge in making physical predictions.That depends on what you mean by "equivalent".   Perhaps you mean we have no obvious rules to convert it to a unique number of different units.  For  example,  the conversion  4 apples+oranges to units of half_apples+oranges in a naive fashion is ambiguous because it might  be done by converting 4 apples+oranges as (3 apples + 1 oranges) which is converted to (6 half_apples + 1 oranges) = 7 half_apples+oranges.  Or it might be converted from (1 apples + 3 oranges) as ( 2 half_apples + 3 oranges) = 5 half_apples+oranges.    So , if there is an empirical difference between the physical situation producing the measurement 7 half_apples+oranges and the situation producing the measurement 5 half_apples+oranges, this is a argument that converting units in such a manner introduces a harmful ambiguity.[/QUOTE]Maybe something along these lines…We have two systems characterised by vectors of attributes, S=(x[SUB]1[/SUB], x[SUB]2[/SUB], ..) , S'=(x[SUB]1[/SUB]', x[SUB]2[/SUB]', …).  We can invent measures of attributes, i.e. maps to reals, m[SUB]1[/SUB], m[SUB]2[/SUB],…  We believe (and this is the physics) that the attributes have an additive property such that in some sense we can partition an attribute and recover the whole by summing its parts.  This being so, we require our measures to be linear in this sense.Given two measures, m[SUB]1[/SUB] and n[SUB]1[/SUB] for X[SUB]1[/SUB], if one is linear and there is a nonzero constant μ such that n[SUB]1[/SUB](x)=μm[SUB]1[/SUB](x) for all x, then clearly theother is linear.  [Can we show that for two linear measures such a ratio exists?]Anyway, if we want the product of measures of two attributes, we may choose m[SUB]1[/SUB](x[SUB]1[/SUB])m[SUB]2[/SUB](x[SUB]2[/SUB]) or n[SUB]1[/SUB](x[SUB]1[/SUB])n[SUB]2[/SUB](x[SUB]2[/SUB]).  Armed with the knowledge of the measure ratios, μ, ν, we can compute the measure ratio μν for the measure products.If we try to do the same summing two measures, there is no constant ratio that can convert one sum of linear measures to another.This is much the same as I wrote before, but highlighting the dependence on an essential additivity in the underlying attributes, and the consequent requirement of linearity in the measures.

10. Stephen Tashi says:

[QUOTE="Baluncore, post: 5650815, member: 447632"]If x was the unit metres and y was the unit seconds, what would you do then?[/QUOTE]Are you trying to formulate an argument that says "We can't add unlike dimensions because we can't add unlike dimensions"?  As I said, I'm not advocating adding unlike dimensions.  I'm investigating what justification we can give for declaring that adding  3 meters and 2 seconds to obtain a composite unit of  5 meters+seconds can never be done  ( or should never be done) – no matter what interpretation we give to "meters+seconds". Declaring  "we can't add 3 meters to 2 seconds" or saying "adding 3 meters to 2 seconds doesn't make sense" isn't an explanation.   If we want to demonstrate that there is no way to do something, then issuing the challenge "Somebody show me how to do it" isn't a demonstration.   For example,  "Show me some positive integers ##a,b,c## such such that ##a^3 + b^3 = c^3##" isn't a proof of Fermat's theorem.

11. Baluncore says:

[QUOTE="Stephen Tashi, post: 5650812, member: 186655"]We don't "evaluate" 3x + 4y unless we are give numerical values for x and y. Likewise we don't evaluate 6xy unless we are given numerical values for x and y.[/QUOTE]If x was the unit metres and y was the unit seconds, what would you do then?

12. Stephen Tashi says:

[QUOTE="Baluncore, post: 5650787, member: 447632"]So how do you evaluate 3x + 4y = ?[/QUOTE]We don't "evaluate" 3x + 4y unless we are give numerical values for x and y.  Likewise we don't evaluate 6xy unless we are given numerical values for x and y.

13. Baluncore says:

[QUOTE="Stephen Tashi, post: 5650782, member: 186655"]Functions that can't be "simplified" can still be "evaluated".[/QUOTE]So how do you evaluate 3x + 4y = ?

14. Stephen Tashi says:

[QUOTE="Baluncore, post: 5650756, member: 447632"]Yes it does. Physics requires that a mathematical equation be evaluated. If you cannot evaluate an equation to a single numerical value then you do not need to add the dimensions. The mathematical reason why differing dimensions are not added in physics is the same reason that 3x + 4y cannot be simplified. 3x + 4y = 3x + 4y.[/QUOTE]"Equations" are mathematical statements that two functions are equal.  Functions that can't  be "simplified" can still be "evaluated".

15. Baluncore says:

[QUOTE="Stephen Tashi, post: 5650713, member: 186655"]That makes no connection to physics.[/QUOTE]Yes it does. Physics requires that a mathematical equation be evaluated. If you cannot evaluate an equation to a single numerical value then you do not need to add the dimensions. The mathematical reason why differing dimensions are not added in physics is the same reason that 3x + 4y cannot be simplified. 3x + 4y = 3x + 4y.

16. Stephen Tashi says:

[QUOTE="haruspex, post: 5650678, member: 334404"]It can.  Tommy is allowed to take one apple and one orange from 3 apples and 4 oranges.  What is the set of possible choices?  12 apple-orange pairs.But fruit does not make a good analogy because you can also argue for adding apples and oranges.  The discreteness creates a natural unit of measure.[/QUOTE]Dimensional analysis does not say that a physical unit can take a values corresponding to any given real number.    So I see nothing wrong with your analogy.[QUOTE]Here is a possibility… Mutiplying makes sense because you can create new units to match.  In some scenario, I take K kg and M metres to compute the product KM kgm.  If you prefer to work in pound-inches, you know how to convert that without having to know K and M separately.  [/QUOTE]I'd put it this way:  Changing the units in a product and "knowing how to convert" the result (using conversion factors in the standard manner" introduces no new ambiguity in the description of the physical situation.   We "know how to convert" because we know to use conversion factors.  But the conventions of using conversion factors doesn't explain why those conventions are physically useful.  In my opinion, it is just an empirical fact that the ambiguity in knowing the product of dimensions, but not knowing the individual factors often doesn't detract from the usefulness our knowledge in making physical predictions.  [QUOTE]  But if we try to invent the concept of mass plus distance, and I tell you (K+M) "kg+m", you cannot convert the single number K+M to an equivalent number of pound+inches.[/QUOTE] That depends on what you mean by "equivalent".   Perhaps you mean we have no obvious rules to convert it to a unique number of different units.  For  example,  the conversion  4 apples+oranges to units of half_apples+oranges in a naive fashion is ambiguous because it might  be done by converting 4 apples+oranges as (3 apples + 1 oranges) which is converted to (6 half_apples + 1 oranges) = 7 half_apples+oranges.  Or it might be converted from (1 apples + 3 oranges) as ( 2 half_apples + 3 oranges) = 5 half_apples+oranges.    So , if there is an empirical difference between the physical situation producing the measurement 7 half_apples+oranges and the situation producing the measurement 5 half_apples+oranges, this is a argument that converting units in such a manner introduces a harmful ambiguity.

17. Stephen Tashi says:

[QUOTE="haruspex, post: 5650678, member: 334404"]It can.  Tommy is allowed to take one apple and one orange from 3 apples and 4 oranges.  What is the set of possible choices?  12 apple-orange pairs.But fruit does not make a good analogy because you can also argue for adding apples and oranges.  The discreteness creates a natural unit of measure.Here is a possibility… Mutiplying makes sense because you can create new units to match.  In some scenario, I take K kg and M metres to compute the product KM kgm.  If you prefer to work in pound-inches, you know how to convert that without having to know K and M separately. But if we try to invent the concept of mass plus distance, and I tell you (K+M) "kg+m", you cannot convert the single number K+M to an equivalent number of pound+inches.[/QUOTE]

18. Stephen Tashi says:

[QUOTE="Baluncore, post: 5650556, member: 447632"]We can simplify 3x*4y to 12xy.Now explain why 3x+4y cannot be simplified.That should satisfy your interest.[/QUOTE]That makes no connection to physics.

19. haruspex says:

[QUOTE="Stephen Tashi, post: 5650481, member: 186655"]Yet it does make sense to multiply apples by oranges ?)[/QUOTE]It can.  Tommy is allowed to take one apple and one orange from 3 apples and 4 oranges.  What is the set of possible choices?  12 apple-orange pairs.But fruit does not make a good analogy because you can also argue for adding apples and oranges.  The discreteness creates a natural unit of measure.Here is a possibility… Mutiplying makes sense because you can create new units to match.  In some scenario, I take K kg and M metres to compute the product KM kgm.  If you prefer to work in pound-inches, you know how to convert that without having to know K and M separately. But if we try to invent the concept of mass plus distance, and I tell you (K+M) "kg+m", you cannot convert the single number K+M to an equivalent number of pound+inches.

20. Baluncore says:

[QUOTE="Stephen Tashi, post: 5650481, member: 186655"]My interest is in finding a correct justification for why conventional dimensional analysis allows multiplication of different dimensions, but not addition of different dimensions.[/QUOTE]We can simplify 3x*4y to 12xy. Now explain why 3x+4y cannot be simplified. That should satisfy your interest.