# Can Angles be Assigned a Dimension?

Common Topics: dimension, vector, angular, dimensions, length

### 1. Some Background on Dimensional Analysis

… if you are not already familiar with it.

#### 1.1 Dimensions

Dimensional Analysis is a way of analyzing physics equations that consider 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.  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 labeled M for mass, L for length, T for time, Q for a charge, θ for temperature …

Thus, if F is a force then ##[F]=MLT^{-2}##.  The well-known equation F=ma would be analyzed 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 the 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 neutralized by that of the cross-product operator.

#### 3.2 Functions of Angles

Raising a dimensioned entity to 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 added 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.

Tags:
106 replies
1. Stephen Tashi says:

[QUOTE="robphy, post: 5624227, member: 9587"]In the series expansion for exp(x/a), all of those numbers are pure [dimensionless] numbers… they are part of the definition of exp(z), where z is dimensionless.[/QUOTE]The mathematical definition of the function ##f(x) = x^2## likewise assigns no dimension to ##x##.  So the lack of dimension in the mathematical definition of a function don't prevent us from giving the argument of the function a dimension when we employ it in physics.[QUOTE]Thus, the only thing that carries units is "x"[/QUOTE]I'm not making that assumption.[QUOTE]No, ## y = e^x## cannot be a position equation…You could have, say, ##y=Ae^{(-t/tau)}##, where ##A## has units of length, and ##t## and ##tau## have units of time.[/QUOTE]I disagree.  If an experimenter fits an equation of the form  ## y = e^t ## to his data where ##y## is in meters and ##t## is in seconds, he has described a physical relation unambiguously and a different experimenter who wishes to measure distance in centimeters and time in minutes can figure out how to create an equivalent equation using those units of measurement.It may be true that it would more convenient for the second experimenter if the first experimenter had written his results in a different form.

2. robphy says:

[QUOTE="Stephen Tashi, post: 5624223, member: 186655"]That depends on the units of the constants 1,2,6,24,120…For example,  the equation   ## y = 1 + 5x + 2x^2##  can describe a physical situation where ##y## is in units of newtons, x is in units of meters, 1 is in units of newtons, 5 is in units of newtons per meter and 2 is units of newtons per meter squared.Can an object have a position given by  ## y = e^x## ?[/QUOTE]In the series expansion for exp(x/a), all of those numbers are pure [dimensionless] numbers… they are part of the definition of exp(z), where z is dimensionless.Thus, the only thing that carries units is "x".So, what are the units of the right-hand side?No, ## y = e^x## cannot be a position equation…You could have, say, ##y=Ae^{(-t/tau)}##, where ##A## has units of length, and ##t## and ##tau## have units of time.edit:Your proposed equation: ## y = 1 + 5x + 2x^2##  with units as you specifiedpossibly should be written as## y = (1 rm{Newton})( 1 + 5 (frac{x}{m}) + 2(frac{x}{m})^2)##[trying to conform to the exponential series expansion].The point is… if there are units, they should be shown.

3. Stephen Tashi says:

[QUOTE="robphy, post: 5624218, member: 9587"]What are the units on the right-hand side?[/QUOTE]That depends on the units of the constants 1,2,6,24,120…   For example,  the equation   ## y = 1 + 5x + 2x^2##  can describe a physical situation where ##y## is in units of newtons, x is in units of meters, 1 is in units of newtons, 5 is in units of newtons per meter and 2 is units of newtons per meter squared.Can an object have a position given by  ## y = e^x## ?

4. Stephen Tashi says:

[QUOTE="robphy, post: 5624212, member: 9587"]Consider http://www.wolframalpha.com/input/?i=series(exp(x/a),x)##exp(frac{x}{a})=1+frac{x}{a}+(frac{x}{a})^2/2+(frac{x}{a})^3/6+(frac{x}{a})^4/24+(frac{x}{a})^5/120+…##with ##a## as a dimensionless quantity but ##x## with units of length.[/QUOTE]What do you want me to consider about it ?

5. Stephen Tashi says:

We haven't managed to state precise mathematical properties for a "dimension".   If we can't define what a "dimension" is, perhaps we can make definite statements about what it can't be.For example, traditional dimensional analysis insists that the arguments to transcendental functions must be dimensionless.   As a consequence, the transcendental functions  themselves are dimensionless.   Why is this assumed to be the case?    If we let an argument to a transcendental function have a dimension, what is supposed to go wrong ?

6. haruspex says:

[QUOTE="robphy, post: 5623336, member: 9587"]this abstract discussion might be useful about what may be going on with regard to units (and dimensional analysis) in general:https://golem.ph.utexas.edu/category/2006/09/dimensional_analysis_and_coord.html[/QUOTE]That's a fascinating thread.  The comments most relevant to my article concern cycles.I feel those parts get confused because we use the term both in a generic sense of repeating events and in the more physical sense of rotation. This is similar to the way distance was originally used in a Euclidean sense, but now is generalised to such as graphical distance, emotional distance, …  We are comfortable using the dimension L in the former but not the latter, so there is precedent for saying cycles as rotation can have dimension but not in the other uses.  Admittedly, this could lead to some tangled terminology.  That could be avoided by agreeing that "cycle" always has the generic sense, and if we want to refer to a cycle in the rotational sense we should write "revolution".  Thus, a rotating body rotates at one revolution per cycle, or 2π radians per cycle.  Each of those would have dimension ϑ.  This angular sense would also apply to phase angles in trig functions.

7. haruspex says:

[QUOTE="robphy, post: 5623336, member: 9587"]There may be other not-so-slight modifications.[/QUOTE]I'm not suggesting any modification to the way we represent or perform complex addition.  The consideration of alternative representations was to illustrate that, unlike regular addition, adding a real to an imaginary can cope with their having different dimensions.

8. robphy says:

[QUOTE="haruspex, post: 5623208, member: 334404"]The ability to represent angles as a dimension slightly increases the power of DA.[/QUOTE]So, it may be worthwhile exploring this, as you have done in your Insight.However, it seems that to do so following your definitionsleads to not-so-slight modifications of how to do addition (in response to my question about 1+i in relation to your definitions).[quote]This kind of addition can cope with adding items of different dimension. That is, to fit with the ϑ Dimension concept, I could define a complex number as an ordered pair, one of 0 dimension and one of dimension ϑ.[/quote]There may be other not-so-slight modifications.So, maybe this isn't the way to do it [if it is at all possible to do it "slightly"].As a possible guide to a better approach,this abstract discussion might be useful about what may be going on with regard to units (and dimensional analysis) in general:https://golem.ph.utexas.edu/category/2006/09/dimensional_analysis_and_coord.htmlIn an abstract sense, it seems that our physically-dimensionful formulasare mapping values from different spaces (somehow each associated with a "unit")into another space of values (with a unit consistent with the algebraic operations).

9. haruspex says:

[QUOTE="anorlunda, post: 5622971, member: 455902"]I normally associate dimensions with degrees of freedom, as in 3D space or 4D spacetime.A point-like particle can be described in 3D space with three coordinates.   An asymmetric object needs 3 coordinates, plus 3 angular rotations to describe it's position-orientation.  Aren't those rotations on an equal footing with translations as being dimensions?p.s. I normally eschew semantic discussions, but this one caught my fancy.  Nice thought provoking Insights article [USER=334404]@haruspex[/USER][/QUOTE]Thanks for the appreciation.That's really a different usage of the term dimension.  Dimensional analysis concerns what might be termed qualitative dimensions.  All lengths are qualitatively the same, so just L.  Area is different from length, but in a quantifiable way, as L[SUP]2[/SUP], etc.It is not just a semantic issue.  The ability to represent angles as a dimension slightly increases the power of DA.

10. anorlunda says:

I normally associate dimensions with degrees of freedom, as in 3D space or 4D spacetime.A point-like particle can be described in 3D space with three coordinates.   An asymmetric object needs 3 coordinates, plus 3 angular rotations to describe it's position-orientation.  Aren't those rotations on an equal footing with translations as being dimensions?p.s. I normally eschew semantic discussions, but this one caught my fancy.  Nice thought provoking Insights article [USER=334404]@haruspex[/USER]

11. haruspex says:

[QUOTE="Stephen Tashi, post: 5622078, member: 186655"]I don't see arguments of the form "You can rewrite …" as having any bearing on question.[/QUOTE]Then let me put it a different way.  In the post in which you brought up this issue, λ was the average number of events in a specific, fixed time interval, and the algebraic expression featured e[SUP]λ[/SUP].  It seems to me that this way of defining λ makes it a pure number, so dimensionless, so no problem.  It only becomes a problem if you then say, oh, but clearly it is really a rate, i.e. λ per that interval.  But if it is to be thought of as a rate then that is how it should appear in the equation, e[SUP]λt[/SUP].Otherwise, you could apply the same thinking to e.g. KE: 1/2 ms[SUP]2[/SUP], where s is the distance travelled per second.  Dimension=ML[SUP]2[/SUP].

12. haruspex says:

[QUOTE="Stephen Tashi, post: 5622078, member: 186655"]From my point of view the probability density function f(x) is not dimensionless. L[/QUOTE]You are right.

13. Stephen Tashi says:

[QUOTE="haruspex, post: 5621999, member: 334404"]You don't need to assign units to either side of the f(x)= equation, they're dimensionless.  But you can rewrite the ##e^x## as ##e^{lambda x}## where ##lambda=1m^{-1}##.[/QUOTE]In relation to issue of whether ##e^x## is an "error" in an equation describing a physical process when ##x## has a dimension:In the first place,   I don't see arguments of the form "You can rewrite …"  as having any bearing on question.   Yes, an equation representing a physical process can be transformed to an equation in dimensionless form, but that doesn't show the original form of the equation is invalid.Perhaps your complete thought is "Your original equation is wrong or meaningless and you should rewrite it as ….##. ##  In the example, I don't see that the original equation is wrong or meaningless in the sense of being uninterpretable or so ambiguous that a person doing in measurements in cm instead of meters couldn't figure out how to rewrite it as different equation ##g(y)## where ##y## has units of cm.    My equation may be wrong in the sense that the task of defining ##g(y)## can't be accomplished by the straightforward use of conversion factors.  That's a topic we should investigate!  Let's pursue your suggestion of stating the equation as  ##p = f(x) = C( 2 – e^{lambda x} ) ## where ##lambda## has units of ##m^{-1}##   and ##x## has units of ##m##.    Can we convert that equation to a formula ##p = g(y) ## where ##y## has units of cm by using conversion factors?To convert to cm, we must convert both ##lambda## and ##x## using the conversion factor (m/100 cm).  We have ##100 y   (cm) = x  (m) ## and ##lambda  (m^{-1}) = lambda  (100  cm)^{-1})##   So the equation converts to  ##p = g(y) = C(2 – e^{ frac{lambda}{100}100 y}) = C(2 – e^y)##   But the correct equation (for ##y## in cm) should be something like ##p = g(y) = C(2 – e^{frac{y}{100}}) ##.  I said "something like" that because we must change the value of ##C## from ##int_{0}^{ln(2)} {(2 – e^x)} dx ## to ##C_2 = int_0^{ln(200)} { ( 2 – e^{frac{y}{100}} ) } dy ## in order to normalize the probability distribution.   We also must convert the interval on which the equation applies from ##[0, ln(2)]## to ##[0, ln(200)]##.   Are we opposed to letting the function ##ln(.)## have an argument with a dimension?  If so, how can we justify converting ##ln(2)## to ##ln(200)## ?   A dimensionless constant like "C" or "2" can be converted to a different numerical value if it depends on several different dimensions.  For example  the "1" in F = (1)MA   can convert to a different constant if we don't use MKS units.  However, the only dimension that has been mentioned in this problem is length [L].   I don't see any way that a dimensionless constant that is define only in terms of lengths can be converted to a different numerical value by changing the unit of measure for length.In contrast to the above difficulties if we take the viewpoint that the ##x## in ##e^x## and the "2" in ##ln(2)## have dimension [L] length given in meters then the conversion from  meters to cm gives results we need, namely ##e^{frac{y}{100}}## and ##ln(200)##.From my point of view the probability density  function ##f(x)## is not dimensionless.   Like a linear density function for the density of physical mass, it represents "per unit length", so in my equation ##f(x)## has units of  (1/meter).    However, that consideration still leaves length as the only dimension represented in the equation.

14. Demystifier says:

[QUOTE="Krylov, post: 5621293, member: 571630"]Otherwise, I am afraid that your next question will be: "To which category of mathematics does it belong?" :wink:[/QUOTE]It would be definitely algebra. :smile:

15. Krylov says:

[QUOTE="Demystifier, post: 5621249, member: 61953"]I have a related question for everybody. Does the dimensional analysis belongs to mathematics? Or should it be considered as a part of physics? Can the notion of dimension (like meter or second) make sense without referring to a physical measurement?[/QUOTE]Preferably it belongs to physics. Otherwise, I am afraid that your next question will be: "To which category of mathematics does it belong?" :wink:

16. Demystifier says:

I have a related question for everybody. Does the dimensional analysis belongs to mathematics? Or should it be considered as a part of physics? Can the notion of dimension (like meter or second) make sense without referring to a physical measurement?

17. Stephen Tashi says:

[QUOTE="haruspex, post: 5620803, member: 334404"]I'm sorry, I am not grasping your point.[/QUOTE]Lets try this:  Suppose there is a random variable X , measured in meters, that has its density defined on interval ## [0,  ln (2) ] ## by  ##f(x) = C( 2 – e^{x}) ## where ##C## is the normalizing constant ##int_{ 0}^{ln (2)} {(2 – e^ {x})} dx##.   A experimenter who measures ##X## in centimeters can convert the above density function to the appropriate density for##X## when ##X## is measured in centimeters.     I agree that assigning units to the left and right hand sides of ##f(x) = C( 2 – e^{x})## is a confusing or impossible task.   But I don't agree that the ##e^x## in the equation implies that the equation describes a physically impossible situation or that it makes it impossible for a experimenter measuring X in different units to convert the above density to his system of measurement.

• Derek Bolton says:

You don't need to assign units to either side of the f(x)= equation, they're dimensionless.  But you can rewrite the ##e^x## as ##e^{\lambda x}## where ##\lambda=1m^{-1}##.

18. haruspex says:

[QUOTE="Stephen Tashi, post: 5620643, member: 186655"]But if I am stating an equation that describes a physical situation, I can't get away with giving an equation that applies to an unstated interval. Suppose the equation that fits my experimental data is  ##f(k) = frac{ (2.3)^k e^{-2.3}}{k!} ## and an experimenter attempts to duplicate my results.  He uses an interval of 10 seconds to define ##lambda##.  In order to compare his results to mine, he needs to know what interval I used.    He asks me and I tell him "My interval was 5 seconds long".   The version of my equation that he can check against his data is ##f(k) = frac{(4.6)^k e^{-4.6}}{k!} ##.  Are we to say that this conversion of equations takes place by some method other than by converting units using conversion factors ?One may object: "You should have reported your equation in dimensionless form".   That would side-step the need to convert units.  However,  reporting results in dimensionless form isn't a requirement in science.[/QUOTE]I'm sorry, I am not grasping your point.If the two experiments concern the same underlying process, presumably the rates should be the same.  Therefore the "correct" version of the equation would make λ that rate and have λt everywhere that your equation has just λ.  λt is dimensionless, as required.The version of the equation in your post #25 can be likened to rating the top speed of a car as the number of kilometres it can go in a standard interval of one hour.  That does not mean its speed has only a length dimension.

19. Stephen Tashi says:

[QUOTE="haruspex, post: 5620464, member: 334404"]In your Poisson example, yes.  λ was specified as the average number of events in some unstated but fixed interval. [/QUOTE]But if I am stating an equation that describes a physical situation, I can't get away with giving an equation that applies to an unstated interval.  Suppose the equation that fits my experimental data is  ##f(k) = frac{ (2.3)^k e^{-2.3}}{k!} ## and an experimenter attempts to duplicate my results.  He uses an interval of 10 seconds to define ##lambda##.  In order to compare his results to mine, he needs to know what interval I used.    He asks me and I tell him "My interval was 5 seconds long".   The version of my equation that he can check against his data is ##f(k) = frac{(4.6)^k e^{-4.6}}{k!} ##.   Are we to say that this conversion of equations takes place by some method other than by converting units using conversion factors ? One may object: "You should have reported your equation in dimensionless form".   That would side-step the need to convert units.  However,  reporting results in dimensionless form isn't a requirement in science.

20. haruspex says:

[QUOTE="Stephen Tashi, post: 5620434, member: 186655"]does "fixing upon" an interval of 1 meter give different numerical results than fixing upon an interval of 1 kilometer ?[/QUOTE]In your Poisson example, yes.  λ was specified as the average number of events in some unstated but fixed interval.  It was not the rate of events, so was indeed dimensionless.  If you change the interval (but keep the same process) then λ will change.[QUOTE="Stephen Tashi, post: 5620434, member: 186655"]Doesn't "varying" the interval require having some reference length stated in particular units to vary it from? If λ is dimensionless, then λt presumably has a dimension of time [T][/QUOTE]If you allow for different intervals then, as I posted, you must change the definition of λ to be a rate.  So λt is dimensionless.

21. Stephen Tashi says:

[QUOTE="atyy, post: 5620414, member: 123698"]I agree with Baluncore – angles are ratios, so they do not have a dimension.[/QUOTE] If we consider an angle as some sort of physical object, it is more than a ratio.  For example, it has a vertex and sides.   According the book by A. Sonin,  a dimension is a property of an object.  There can be properties of an object that have "dimension 1", meaning that in manipulations with the dimension of that property we use "1" rather than [M],[L],[T] etc.   (So far, although thread participants are willing to take definite sides on the question of whether the particular property of angles that we measure in degrees has (or does not have) a dimension of 1 ,  nobody has ventured to state what the criteria are for something to be "a dimension" or what criteria determine whether a dimension must be "1".  So I'm still going by Sonin's book even though I find it unclear on the mathematical axioms.)

22. Stephen Tashi says:

[QUOTE="haruspex, post: 5620421, member: 334404"]That is because some interval has been fixed upon[/QUOTE]But what would we mean by "fixed upon"?   If we are doing a physics problem, does "fixing upon" an interval of 1 meter give different numerical results than fixing upon an interval of 1 kilometer ?   If someone determines an equation with a given ##lambda## applies when the units of length are meters, shouldn't we be able to to deduce what equation applies when the units of length are kilometers by the usual conversion of units ?   [QUOTE]If you want to vary the interval you can make λ a rate:  ##f(k, t) = frac{ (lambda t)^k e^{-lambda t}}{k!}##[/QUOTE]  Doesn't "varying" the interval require having some reference length stated in particular units to vary it from?   If ##lambda## is dimensionless, then ##lambda t## presumably has a dimension of time [T].

23. haruspex says:

[QUOTE="robphy, post: 5618407, member: 9587"]If i carries units, is there any meaning to (say) 1+i?[/QUOTE]The + in 1+i is a different beast from that in 1+1.  The 1 and the i retain their separate identities.  That we write the sum of a real and an imaginary that way is mere convenience.  We could instead have the notation <x,y> to represent complex numbers.  Addition would be just like vectors, but a unique rule for multiplication.So the fact that we write 1+i creates no diffiiculty.  This kind of addition can cope with adding items of different dimension.  That is, to fit with the ϑ Dimension concept, I could define a complex number as an ordered pair, one of 0 dimension and one of dimension ϑ.

24. haruspex says:

[QUOTE="atyy, post: 5620414, member: 123698"]angles are ratios[/QUOTE]As I thought I showed, you can think of them as fractions of a standard angle, but that does not make them ratios.Baluncore's argument could equally well be applied to masses: All masses can be thought of as a fraction of a standard kilogram mass.  If that makes it a ratio then masses are dimensionless.

25. haruspex says:

[QUOTE="Stephen Tashi, post: 5620356, member: 186655"]Is it also an error to have a term ##e^c## where c is a constant with dimensions?The Poisson distribution has density  ##f(k) = frac{ lambda^k e^{-lambda}}{k!}## where ##lambda## is "The average number of events in the interval".   So I assume ##lambda## has a dimension since "the interval" might mean 1 second or 1 hour etc..   How are the dimensions going to work out in that formula? .[/QUOTE]That is because some interval has been fixed upon, making λ purely a number.  If you want to vary the interval you can make λ a rate:  ##f(k, t) = frac{ (lambda t)^k e^{-lambda t}}{k!}##

26. atyy says:

I agree with Baluncore – angles are ratios, so they do not have a dimension.

27. Stephen Tashi says:

[QUOTE]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.[QUOTE="Drakkith, post: 5620278, member: 272035"]I was not aware of this fact. Very interesting.[/QUOTE][/QUOTE]Is it also an error to have a term ##e^c## where c is a constant with dimensions?The Poisson distribution has density  ##f(k) = frac{ lambda^k e^{-lambda}}{k!}## where ##lambda## is "The average number of events in the interval".   So I assume ##lambda## has a dimension since "the interval" might mean 1 second or 1 hour etc..   How are the dimensions going to work out in that formula?  Oh well, maybe the whole idea of probability is an error –  God doesn't play dice etc.

28. Drakkith says:

[quote] 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.[/quote]I was not aware of this fact. Very interesting.

29. haruspex says:

[QUOTE="Stephen Tashi, post: 5620006, member: 186655"]It's interesting to consider the distinction between a mathematical definition of a function and a physical definition of a function.   To define ##sin(theta)## mathematically (i.e. a mapping from real numbers to real numbers)  one would have to unambiguously answer questions like "What is ##sin(0.35)##?" without any discussion of "units of measure" – e.g. 0.35 deg vs 0.35 radians.  From a mathematical point of view,  ##sin(theta  deg)## and ##sin(theta  radians)## are different functions, even though we use the ambiguous notation ##sin(theta)## to denote both of them.  Only the family of trig functions where ##theta## is measured in radians satisfy mathematical laws like ##D sin(theta) = cos(theta)##.To give a physical law in the form of a function we may do it by assuming certain units of measure. Then it is assumed that changing the units of measure appropriately produces a new mathematical function which states the same physical law.  So a physical definition of a function defines a set of different mathematical functions that are regarded as physically equivalent.The physical definition of ##sin(theta)## defines  a set of different, but physically equivalent mathematical functions.[/QUOTE]Yes, I think that is why I have never been satisfied with the view that angles are utterly dimensionless.[QUOTE="robphy, post: 5619841, member: 9587"]Which circle are you referring to?[/QUOTE]Whatever circle Baluncore had in mind.[QUOTE="robphy, post: 5619841, member: 9587"]At this stage, my question of the consistency of "1+i" in post 2 stands out as still unresolved, despite your reply in post 5.[/QUOTE]I haven't forgotten this.  I want to take a look at the Brownstein article first.

30. robphy says:

[QUOTE="Stephen Tashi, post: 5619988, member: 186655"]Let me see if I understand you viewpoint.In the PDF linked in the Insight and post #10, the author, A. Sonin,  makes a distinction among:1) A physical object or phenomena (e.g. a stick)2) A "dimension", which is a property of a physical object or phenomena (e.g. length)3) A "unit of measure", which is a way to quantify a dimension (e.g. meters)The author is careful to point out that a "dimension" is not a physical phenomena.  It is a property of a physical phenomena.You describe "an angle" in mathematical terms, but since you say an "angle" can have various properties, I think you mean an "angle" to denote a physical phenomena, which is alternative 1)…[snip]…[/QUOTE]I didn't read the PDF. So, I can't answer your questions using that author's distinctions.I think the bottom line here is: clearly define terms, especially when one is trying to change definitions.

31. Stephen Tashi says:

[QUOTE="haruspex, post: 5618575, member: 334404"]Not if you redefine trig functions as taking arguments of dimension Θ, as I did.[/QUOTE]It's interesting to consider the distinction between a mathematical definition of a function and a physical definition of a function.   To define ##sin(theta)## mathematically (i.e. a mapping from real numbers to real numbers)  one would have to unambiguously answer questions like "What is ##sin(0.35)##?" without any discussion of "units of measure" – e.g. 0.35 deg vs 0.35 radians.  From a mathematical point of view,  ##sin(theta  deg)## and ##sin(theta  radians)## are different functions, even though we use the ambiguous notation ##sin(theta)## to denote both of them.  Only the family of trig functions where ##theta## is measured in radians satisfy mathematical laws like ##D sin(theta) = cos(theta)##.To give a physical law in the form of a function we may do it by assuming certain units of measure. Then it is assumed that changing the units of measure appropriately produces a new mathematical function which states the same physical law.  So a physical definition of a function defines a set of different mathematical functions that are regarded as physically equivalent.The physical definition of ##sin(theta)## defines  a set of different, but physically equivalent mathematical functions.

32. Stephen Tashi says:

[QUOTE="robphy, post: 5619952, member: 9587"]  Before somehow specifying an angle-measure, one could talk about all sorts of properties of angles at this stage. Then, when introducing an angle-measure, it probably should be explicitly defined—maybe operationally.[/QUOTE][QUOTE]But all of this "angle-measure" discussion is distinct from the "angle" discussion in the previous paragraph.[/QUOTE]Let me see if I understand you viewpoint. In the PDF linked in the Insight and post #10, the author, A. Sonin,  makes a distinction among:1) A physical object or phenomena (e.g. a stick) 2) A "dimension", which is a property of a physical object or phenomena (e.g. length)3) A "unit of measure", which is a way to quantify a dimension (e.g. meters)The author is careful to point out that a "dimension" is not a physical phenomena.  It is a property of a physical phenomena.You describe "an angle" in mathematical terms, but since you say an "angle" can have various properties, I think you mean an "angle" to denote a physical phenomena, which is alternative 1)When you say "angle measure", I'm not sure whether you mean alternative 3) or alternative 2).  But does alternative 3) (units of measure)  make any sense without the existence of alternative 2) (dimension) ?As I mentioned previously,  I haven't yet seen a precise statement of what mathematical or physical properties a "dimension" must have.   I don't know whether other thread participants agree with those listed by A. Sonin.  In regards to "dimensionless ratios",  a dimensionless ratio can associated with a property of a physical object.  Different dimensionless ratios can be associated with different properties (e.g. height of a person/ length of that persons right leg,  weight of a person now / weight of that person at birth).     "Dimensionless ratios" can obviously be quantified.    So it is rather confusing to consider the question of whether a "dimensionless ratio" is (or isn't) a associated with a "dimension".

33. robphy says:

[QUOTE="Stephen Tashi, post: 5619931, member: 186655"]You could also ask "The unit circle with center (0,0)?  The unit circle with center (15,12)?"You have to use a circle with its center at the vertex of the angle, so the measurement process isn't really independent of which circle is used unless we think of  "a circle" as a portable measuring instrument, just as we think of a meter stick as portable measuring instrument.[snip][/quote]Yes, but I didn't think I had to make further clarification on this. Shall we bring up issues of parallel transport on a non-Euclidean space as well?I would hope that when one says "arc-length divided by radius" that the rest of this is assumed.[quote][snip]I agree.  It's the distinction between "a dimension" (e.g. length) and "a unit of measure" (e.g. meters).[/QUOTE]My distinction is this… If two lines (or two segments) meet at a point, then one can talk about the angle [or an angle] at the location where the two lines meet, labeled by the vertex (call it) C or that vertex with a two points, one on each segment–like ACB.  Before somehow specifying an angle-measure, one could talk about all sorts of properties of angles at this stage. Then, when introducing an angle-measure, it probably should be explicitly defined—maybe operationally.Given two lines (or line segments) meeting at a point, one could define an angle-measure the usual way (essentially with a circular protractor, appropriately calibrated in the likely possibility that protractors have different radii), or maybe in a different way (e.g. https://mathnow.wordpress.com/2009/11/06/a-rational-parameterization-of-the-unit-circle/ ) although it might not give your angle-measure the desired properties of additivity, or maybe using a hyperbolic-protractor (as one might use in special relativity). Issues of "units of angle-measure" come into play here. But all of this "angle-measure" discussion is distinct from the "angle" discussion in the previous paragraph.

34. Stephen Tashi says:

[QUOTE="robphy, post: 5619841, member: 9587"]Which circle are you referring to?The unit circle? Or maybe the circle of radius 7?[/QUOTE]You could also ask "The unit circle with center (0,0)?  The unit circle with center (15,12)?"[QUOTE]One feature of the angle measure (defined as the ratio of circular-arc-length to radius) is that it is independent of the circle used to make that measurement.  [/QUOTE]You have to use a circle with its center at the vertex of the angle, so the measurement process isn't really independent of which circle is used unless we think of  "a circle" as a portable measuring instrument, just as we think of a meter stick as portable measuring instrument.If we have an object that moves along a path, to measure the property of the path called its "total length" with a meter stick, we have to move the meter stick to various locations on the path.  If we are dealing with an object moving in a circular path and want to measure a property of the path called the "total angle swept out", we may need a measuring instrument that can produce results greater than 360 deg.  Such a measuring instrument could involve a circle, but it would have to have the added feature of keeping track of arc lengths greater than ##2pi##.[QUOTE]In this general discussion, one needs to distinguish an "angle" from an "angle measure".[/QUOTE]I agree.  It's the distinction between "a dimension" (e.g. length) and "a unit of measure" (e.g. meters).

35. Stephen Tashi says:

[QUOTE="Baluncore, post: 5619648, member: 447632"]Polynomials and dimensions are incompatible. Transcendental functions that are approximated by polynomials must have dimensionless inputs and outputs.[/QUOTE]I'm curious why you say that polynomials are incompatible with dimensions.  Coefficients of different powers of x in a polynomial can be assigned different dimensions, so that each power of x is converted to the same dimension.If we have an equation that describes a dimensioned physical quantity as a power series, aren't we assigning different dimensions to each coefficient in the power series ?

36. robphy says:

[QUOTE="haruspex, post: 5619650, member: 334404"]A ratio can have no dimension since it must be a ratio of two things of the same dimension.  But at an angle is not a ratio.  You can say it is a certain fraction of a complete circle, but whether that has dimension depends on whether you consider the complete circle as having a dimension.  You are not used to thinking of it that way, but that does not mean it cannot be done.[/QUOTE]Which circle are you referring to?The unit circle? Or maybe the circle of radius 7?One feature of the angle measure (defined as the ratio of circular-arc-length to radius) is that it is independent of the circle used to make that measurement.In this general discussion, one needs to distinguish an "angle" from an "angle measure".Certainly, you can try to make definitions… but they have to lead to a consistent system.At this stage, my question of the consistency of "1+i" in post 2 stands out as still unresolved, despite your reply in post 5.

37. haruspex says:

[QUOTE="Baluncore, post: 5619648, member: 447632"]Angles are ratios, parts of a circle.[/QUOTE]A ratio can have no dimension since it must be a ratio of two things of the same dimension.  But at an angle is not a ratio.  You can say it is a certain fraction of a complete circle, but whether that has dimension depends on whether you consider the complete circle as having a dimension.  You are not used to thinking of it that way, but that does not mean it cannot be done.[QUOTE="Baluncore, post: 5619648, member: 447632"]Polynomials and dimensions are incompatible.[/QUOTE]Not if the dimension has the unusual property that it becomes dimensionless when raised to some finite power.  The ϑ[SUP]2[/SUP]=1 axiom means that a polynomial function of an angle is fine if all the terms are even powers (dimensionless result) or all odd powers (result of dimension ϑ).

38. Baluncore says:

Angles are ratios, parts of a circle. Neither angles nor ratios have a dimension.Polynomials and dimensions are incompatible. Transcendental functions that are approximated by polynomials must have dimensionless inputs and outputs.

39. Stephen Tashi says:

40. haruspex says:

[QUOTE="Ygggdrasil, post: 5618633, member: 124113"]I guess I don't get the point of trying to give angles a dimension.  Angles are defined as a ratio (arc length : circumference) which is a dimensionless quantity and seems fundamentally different than something like mass.[/QUOTE]Fair question.I have always found it a bit unsatisfactory that some quite different pairs of physical entity are dimensionally indistinguishable.  I mentioned some in the post.  There seemed to be something distinct about rotational entities, such as angular momentum, that was not captured by DA.As regards utility, as I showed in the table, it can be added to normal DA for an extra bit (literally) of information.

41. Ygggdrasil says:

[QUOTE="haruspex, post: 5618594, member: 334404"]I don't understand your point.  They are normally considered dimensionless anyway; I'm looking for a way to give them dimension. Would thinking of all masses as fractions of 1kg make mass dimensionless?[/QUOTE]I guess I don't get the point of trying to give angles a dimension.  Angles are defined as a ratio (arc length : circumference) which is a dimensionless quantity and seems fundamentally different than something like mass.

42. haruspex says:

[QUOTE="Ygggdrasil, post: 5618591, member: 124113"]Angles can be defined as dimensionless quantities if one thinks of them as fractions of a circle (multiplied by the constant 2π).[/QUOTE]I don't understand your point.  They are normally considered dimensionless anyway; I'm looking for a way to give them dimension.  Would thinking of all masses as fractions of 1kg make mass dimensionless?

43. Ygggdrasil says:

Angles can be defined as dimensionless quantities if one thinks of them as fractions of a circle (multiplied by the constant 2π).

44. haruspex says:

[QUOTE="robphy, post: 5618407, member: 9587"]What is the justification of the claim: "The cross product operator also has dimension Θ"?[/QUOTE]That is not a claim, it is part of the definition of Θ.[QUOTE="robphy, post: 5618407, member: 9587"]If i carries units, is there any meaning to (say) 1+i?[/QUOTE]I'm not especially attached to the part relating to i.  It is independent of the rest and probably needs more thought.  There might be a way around the 1+i problem similar to how I resolved s = rθ, i.e. one would have to agree to treating complex algebra in a slightly different way.[QUOTE="robphy, post: 5618407, member: 9587"]Note: in exp (x), the x must be dimensionless[/QUOTE]That's why I assigned i the dimension Θ, to make iθ dimensionless.[QUOTE="robphy, post: 5618409, member: 9587"]You might be interested in this old article from the American Journal of Physics.http://scitation.aip.org/content/aapt/journal/ajp/65/7/10.1119/1.18616"Angles—Let’s treat them squarely" by K. R. Brownstein[/QUOTE]That sounds very much as though it is not a new idea, which is at once heartening and disappointing.  Thanks for the reference.[QUOTE="A. Neumaier, post: 5618432, member: 293806"]Angles have the dimension of 1. That this is a true dimension[/QUOTE]By definition, real numbers are dimensionless, so I do not understand what you mean by saying it is a true dimension.  Indeed, the fact that angles have units but not dimension is somewhat awkward, as I mentioned in the article.[QUOTE="A. Neumaier, post: 5618432, member: 293806"]for s=sinθ to make sense, the dimension of θ must be 1.[/QUOTE]Not if you redefine trig functions as taking arguments of dimension Θ, as I did.

45. Arnold Neumaier says:

Angles have the dimension of 1. That this is a true dimension can be seen from the fact that one measures angles in different units , namely either degrees or radians, and they convert into each other just like units for other dimensional quantities. Your ##\Theta## doesn't make sense unless it equals ##1## since for ##s=\sin\theta## to make sense, the dimension of ##\theta## must be ##1##.

46. robphy says:

What is the justification of the claim: "The cross product operator also has dimension Θ"?