In Dimensional analysis why is Lenght/Lenght=1 (a dimensionless number)?

[ChARMELeOn]

In Dimensional analysis if L=Lenght why is L/L=1 (a dimensionless number), and not just nothing more than L/L ?

It started with that I was thinking of why degrees do not have any dimension, and then I thought of the definition of radians:
q = "angle"
s = "a length on the circles circumference"
r = "radius"
then:
s/r=q

But now we have the dimensions L/L in the equation s/r=q which then must be dimensionles if degrees is dimensionles. So to solve this problem I need to solve that.

It may have been a stupid question but I want to know, so I appreciate any help.

 

[sophiecentaur]

If you accept that dimensional analysis follows the rules of algebra then L/L is 1 - so no dimension.
If you can't accept that then how would you get on with resistivity, energy, stress, strain and all the others?

It's true that one person in one Lab would get the same value for a strain measurement on a given spring with an identical load, using mm, as another person, in another Lab, would get, using inches. That's what 'dimensionless' means.

 

[Leperous]

L is a number multiplied by a unit. The units cancel each other out when you divide because of simple algebra* and convenience - if you asked "how many metres in 100 metres", the answer "100 metres/metres" doesn't make any sense. It's even more insane to claim it is "100 degrees".

* You can and should treat units in an equation like algebraic terms - it avoids confusion. For example, speed = distance / time = x metres / y seconds = x/y metres/second.

 

[AlephZero]

Degrees and radians are not dimensionless. They are two different measures of something (angles), just like inches and meters are different measures of something (lengths).

The fact that you have chosen to define a radian by drawing a diagram that contains two equal lengths is irrelevant. You could just as well define a radian to be 1/(2 pi) of a complete revolution, just like you define a degree to be 1/360 of a revolution.

 

[Pythagorean]

Once you start using steradians, you recognize the need for dimensionality in radians/degrees.

Mass was originally a meaurement of distance ratios.

 

[sophiecentaur]

Degrees and radians are not dimensionless. They are two different measures of something (angles), just like inches and meters are different measures of something (lengths).

The fact that you have chosen to define a radian by drawing a diagram that contains two equal lengths is irrelevant. You could just as well define a radian to be 1/(2 pi) of a complete revolution, just like you define a degree to be 1/360 of a revolution.

So can you tell me how you can have a quantity with dimension if it's put into an exponential when defining the sine of that quantity?
You'll be telling me that you can have Exp(£35.50) next.

 

[cdotter]

Degrees and radians are not dimensionless. They are two different measures of something (angles), just like inches and meters are different measures of something (lengths).

How can that be if dθ=s/r. Arc lengths have units of meters, radii have units of meters.

If radians weren't dimensionless then how could you, for example, take the sine of an angle in radians?

 

[Pythagorean]

If radians were dimensionless, then it would be the same thing as a steradian... which it's not.

 

[nasu]

The fact that radian and steradian are both dimensionless does not necessarily mean they are "the same". There are many things that are dimensionless. It doe not follow that they all must be the same thing.
Work and torque have both units of N m and still are not the same.

 

[K^2]

It's the definition of division. L/L = L * (L^-1) = 1 by definition of inverse element or is undefined.

Keep in mind that this isn't a number 1. It's a unit. An element that you multiply by anything and get that same anything back. That's all it means.

 

[Dickfore]

Angles are dimensionless but not unitless. The radian is a unit of angle.

 

[Pythagorean]

Hrm... How's a dimension defined? If I have a model in which one of variables is an "angle" (not necissarily spatial) isn't it one of the dimensions of my model?

 

[Dickfore]

Hrm... How's a dimension defined? If I have a model in which one of variables is an "angle" (not necissarily spatial) isn't it one of the dimensions of my model?

From the Wikipedia article on http://en.wikipedia.org/wiki/Dimensional_analysis" :

The dimension of a physical quantity is the combination of the basic physical dimensions (usually mass, length, time, electric charge, and temperature) which describe it; for example, speed has the dimension length / time, and may be measured in meters per second, miles per hour, or other units. Dimensional analysis is based on the fact that a physical law must be independent of the units used to measure the physical variables.

 

[Pythagorean]

I was hoping for a definition that excluded angle on principle. I don't see it if it's there.

 

[Pythagorean]

The only significant difference I can think of is the bounds (2pi rad = 0 rad)

 

[Dickfore]

That is irrelevant and not always true.

 

[sophiecentaur]

Angles are dimensionless but not unitless. The radian is a unit of angle.

A radian is just a ratio. Would you also say that a Sine (which is still a ratio of two lengths (cm / inches / furlongs / chains) is a unit?

Is there any difference between the two 'units'?

 

[Pythagorean]

That is irrelevant and not always true.

Well yes, that's my point! What makes an angle not a dimension?

Just like x,y are an eigenbasis of the Cartesian coordinate system, r,theta represent the eigenbasis of the polar coordinate system.

That's what I thought dimensionality was all about.

 

[Rap]

Keep reading the Wikipedia article on dimensional analysis, especially the Siano extensions. In this formulation, lengths in different directions are incompatible, and have their own algebra. A radian is the ratio of two differently directed lengths and is therefore not dimensionless, but a radian squared IS dimensionless. Thus we can say, for example, that sin(θ)=θ + θ³/6 + ... in which each term is compatible, and cos(θ)=1+θ²/2+... in which each term is compatible. The extension is self-consistent and may be used to derive more information from the dimensional analysis of a problem than simply assuming the radian to be dimensionless. For example, you will never find a physically meaningful equation which contains cos(θ)+sin(θ) because the two are dimensionally incompatible.

 

[Pythagorean]

Keep reading the Wikipedia article on dimensional analysis, especially the Siano extensions. In this formulation, lengths in different directions are incompatible, and have their own algebra. A radian is the ratio of two differently directed lengths and is therefore not dimensionless, but a radian squared IS dimensionless. Thus we can say, for example, that sin(θ)=θ + θ³/6 + ... in which each term is compatible, and cos(θ)=1+θ²/2+... in which each term is compatible. The extension is self-consistent and may be used to derive more information from the dimensional analysis of a problem than simply assuming the radian to be dimensionless. For example, you will never find a physically meaningful equation which contains cos(θ)+sin(θ) because the two are dimensionally incompatible.

Actually, I think this is what might make it dimensionless. That the length's are independent.

In my argument for dimensionality above, I neglected that r and theta are coupled (you can't integrate over theta without defining some r, for instance), whereas in the cartesian coordinate system, x and y are completely independent.

and the infinite sum argument of course: sin(x) = x -x^3/6 + ...

x^m + x^n with m != n, x must be dimensionless.

I don't know, I'm beginning to regret not double-majoring with the maths.

 

[Rap]

Actually, in the infinite sum argument for the sine, the x need not be dimensionless, all you need is that x^(2n+1) have the same dimension for any n. By the Siano extension, they do, yet x is not dimensionless.

 

[Dickfore]

Actually, in the infinite sum argument for the sine, the x need not be dimensionless, all you need is that x^(2n+1) have the same dimension for any n. By the Siano extension, they do, yet x is not dimensionless.

This can only be true if the dimension of the angle is zero, -1 or 1.

 

[Rap]

That's assuming that dimensions multiply according to a group algebra, similar to integers. If Lx is a length in the x direction, then successive powers are (L_x, L_x², L_x³ L_x⁴ ...) According to the Siano extension, lengths have a "directed dimension" and combine according to a different group (Klein group) i.e. sucessive powers are (Lx, 1, Lx, 1,... ). Its sort of like the cross product of vectors: Lx Ly -> Lz, but Lx Lx -> 1 and Lx Lx Lx -> (1) Lx -> Lx.

 

[gmax137]

... you will never find a physically meaningful equation which contains cos(θ)+sin(θ) because the two are dimensionally incompatible.

Can someone help me understand this? seems like cos(θ)+sin(θ) is just another sine function:

\cos{x}+\sin{x}=\ \sqrt{2} \ \sin(x+\frac{\pi}{4})

I must really be missing the boat here...

Edit - fix LaTex

 

[nasu]

It's even more generally valid.
Any combination of sin and cos of an angle can be written in terms of a phase shift of a single function.

a Sin(x) + b Cos(x) = A Sin(x+f)

where f is a phase shift.
You can do it with Cos as well.
In order to find the relationship between (a,b) and (A,f) you can use the formula for the Sin of the sum.

 

[Rap]

Not missing the boat as much as you think. This is a good challenge to the Siano extension. But the sum-of-angle formula is

<br /> \sin(a+b) = \sin(a) \cos(b) + \sin(b) \cos(a)<br />

which IS dimensionally consistent (sin(x) has dimension radians, cos(x) is dimensionless). So if b=π/4 then

<br /> \sin(a+\pi/4) = \sin(a) \cos(\pi/4) + \sin(\pi/4) \cos(a)<br />

which means \cos(\pi/4)=\sqrt{1/2} is dimensionless, while \sin(\pi/4)= \sqrt{1/2} radians. The dimensional analysis of a physically meaningful equation may give the result A*sin(a)+B*cos(a) where B/A has the dimensions of radians, and the numerical values of A and B may both be unity.

 

[timthereaper]

Measurements are nothing more than ratios of objects to other objects. They are there to facilitate comparisons between two different types of things. We can't compare a unit of length to a unit of momentum or a unit of magnetic field to a unit of time, so we use dimensions such as meters, teslas, coulombs, seconds, etc. to give us a common base for comparison.

Our base measurement systems are defined by certain objects or physical phenomena anyway. It only provides a reference point so we can compare a set of objects. Really, we say that a 2 kg billet of steel weighs 2 times as much as a weight locked up in a vault in France. If we compare that to a 4 kg rock, we are saying that the rock weighs twice as much as the steel. That's all we're really doing when we measure things.

We can cancel out units because it's basically stating a ratio anyway. Strain is a great way to see that. We measure elongation of a bar with a nominal length and we are dividing the former by the latter. The lengths cancel and we get a dimensionless unit. Angles do this too. We define an angle (in radians) as the ratio of a portion of a circle's arc length to its radius (s=r*theta). Since length divided by length cancels, the angle is dimensionless.

As far as the steradian and the radian, they are not the same thing because they are different ratios. The radian is used to compare arc length of a circle to its radius whereas the steradian is used to compare areas of a sphere to the square of the radius.

 

[gmax137]

...sin(x) has dimension radians, cos(x) is dimensionless...

This is something strange to me, I can't remember if I ever heard, or thought about this before. Does this follow from the series:

\sin{x}=x-...

whereas,

\cos{x}=1-...

That's kind of weird, let me think about that for awhile...

 

[Dickfore]

Question:

What is the dimensionality of:

<br /> \exp{(x)}<br />

if x is in radians?

 

[Rap]

Question: What is the dimensionality of: \exp{(x)} if x is in radians?

Using the Siano extension, expanding \exp{(x)} you get 1+x+x^2/2 + ... and these terms are incompatible (1 is dimensionless, x is radians, x^2 is dimensionless ...)

That means that no physically meaningful equation will yield this expression. On the other hand, \exp{(ix)} IS meaningful, if x is in radians. That means that "ix" is dimensionless, as is exp(ix). According to the Wikipedia article "Dimensional analysis / Siano's extension", that means that "the complex quantity i has an orientation equal to that of the angle it is associated with". I think maybe a better way to say it is that expressions of the form exp(Ax), where x is in radians, are physically meaningful as long as A has an imaginary numerical value and has units of radians, oriented in the same direction as x. Note that exp(ix)=cos(x)+i sin(x) is dimensionally consistent, yielding a dimensionless value.

 

[Dickfore]

That means that no physically meaningful equation will yield this expression.

Watch this:

http://www.youtube.com/watch?v=lJfuU7D1DOA#t=21m08s"

 

[Pythagorean]

i is an interesting number. Particularly useful to this discussion is it's geometric definition: as an orthogonal transform.

Still not sure what it means for i to have a spatial orientation. Does that mean the imag axis ceases to be orthogonal to the real axis?

 

[Rap]

Watch this:

http://www.youtube.com/watch?v=lJfuU7D1DOA#t=21m08s"

The coefficient of static friction is \mu=F_r/F_t where F_r is the force directed normal to the surface (i.e radially) and F_t is the frictional force, directed parallel to the surface (i.e. tangentially). Since these forces act in different directions, they have incompatible dimensions, and so the coefficient of friction is not dimensionless. Let's call the direction units 1_r, 1_t, and 1_n for radial, tangential, and normal (normal to the plane of 1_r and 1_t), and let's call 1₀ dimensionless. Using Siano's extension, the angle direction is perpendicular to the plane of the angle, that is 1_n. By Siano, the coefficient of static friction is directed as 1_r/1_t = 1_n and so the product of the coefficient of friction times the angle is directed as 1_n 1_n = 1₀ (i.e. dimensionless), and so the term in the exponential is dimensionless and everything is ok.

i is an interesting number. Particularly useful to this discussion is it's geometric definition: as an orthogonal transform.

Still not sure what it means for i to have a spatial orientation. Does that mean the imag axis ceases to be orthogonal to the real axis?

No, I think it means that it is oriented in the same direction as the angle it is multiplying, which is perpendicular to the plane of the angle. I don't have a clear mental picture of how it works yet. In the case of directed quantities, I think I understand the Siano extension, but in this case, I am just using the rules and noting that I cannot find an inconsistency.

 

[Dickfore]

The coefficient of static friction is \mu=F_r/F_t where F_r is the force directed normal to the surface (i.e radially) and F_t is the frictional force, directed parallel to the surface (i.e. tangentially). Since these forces act in different directions, they have incompatible dimensions, and so the coefficient of friction is not dimensionless. Let's call the direction units 1_r, 1_t, and 1_n for radial, tangential, and normal (normal to the plane of 1_r and 1_t), and let's call 1₀ dimensionless. Using Siano's extension, the angle direction is perpendicular to the plane of the angle, that is 1_n. By Siano, the coefficient of static friction is directed as 1_r/1_t = 1_n and so the product of the coefficient of friction times the angle is directed as 1_n 1_n = 1₀ (i.e. dimensionless), and so the term in the exponential is dimensionless and everything is ok.

How about the result of the exponential? The left hand side is a ratio of the magnitudes of two forces directed at an angle \theta.

 

[Rap]

Sorry, I don't understand "result of the exponential". exp(\mu \theta) is dimensionless because \mu \theta is dimensionless. Also, I don't understand, the left side of what equation? \mu=F_r/F_t? The left hand side of that is the directed radial force divided by the directed tangential force, which are at right angles.

 

[Dickfore]

The equation given in the lecture is:

<br /> \frac{T_{2}}{T_{1}} = \exp{(\mu \, \theta)}<br />

 

[Rap]

Aha! Good question. Let me think about it.

 

[Rap]

I think the answer is that T1 and T2 are tensions, not forces, and are not directed. If they were directed, they would be vectors, and I am not sure T1/T2 would then have any meaning. Tension is a force along the axis of the string, whatever that axis is. If you take a directed plane perpendicular to the axis of the string, you will define a directed area equal to the area of the cross sectional area of the string. The force on that directed area will be equal to the tension times a unit vector in the direction of that area, which will be along the axis of the string. If you reverse the direction of the area, you get the same force in the opposite direction. In other words (directed force) = (tension) x (direction), so the tension itself has no direction. You can see this by cutting the rope and bringing the two ends together. The two cross sectional areas resulting from the cut are in opposite directions, and so you will have to pull to the right with the left hand and pull to the left with the right hand with equal forces in order to bring the two ends together. This is the special case of the more general case of (directed force)=(stress tensor)(directed area) where the directed area can be in any direction, not just along the axis of the string. If a coordinate system is used in which one of the coordinate directions is along the axis of the string, then the diagonal element of the stress tensor in that direction will be the tension divided by the area, and it will be undirected.

 

[Dickfore]

I think one should not pay too much attention to Siano's extension of dimension.

 

[Rap]

I think one should not pay too much attention to Siano's extension of dimension.

I think paying attention to dimensional analysis without Siano's extension is very productive. I think Siano's extension is a consistent and natural extension of dimensional analysis, it can provide deeper insight into a problem. It does require some deeper thinking, and if you get satisfactory results without it, then fine, don't use it. But every time I do dimensional analysis without it, I can't help but be curious to know if using Siano's extension will give me even more information. Sometimes it does, sometimes it does not. But to not pay too much attention to it because its too hard to understand, or not mainstream... well I find that a boring reason. Why do you think one should not pay too much attention to it?

 

[Dickfore]

<br /> \sin{(\alpha)} = \cos{(\frac{\pi}{2} - \alpha)}<br />

 

[Rap]

That was covered in #26

 

[Dickfore]

How is the following identiy:

<br /> \int_{0}^{\frac{\pi}{2}}{\sin^{p}{x} \, \cos^{q}{x} \, dx} = \frac{1}{2} \, B\left(\frac{p + 1}{2},\frac{q + 1}{2}\right), \; p, q &gt; -\frac{1}{2}<br />

dimensionally consistent according to Siano's extension?

 

[Dickfore]

The period of oscillation T of a mathematical pendulum with length L in a uniform gravitational field with acceleration due to gravity g and an amplitude angle \alpha with the vertical is given by:

<br /> T = 4 \sqrt{\frac{L}{g}} \, \int_{0}^{\frac{\pi}{2}}{\frac{dt}{\sqrt{1 - \sin^{2}{(\frac{\alpha}{2})} \sin^{2}{(t)}}}}<br />

How is this dimensionally consistent according to Siano's extension of dimensions?

 

[cortiver]

The way I like to think about the dimensionality of a quantity is that it is characterizes how the numerical value of the quantity changes when we make use of our freedom to arbitrarily scale units.

Since all the SI units like metre, second, and so forth are defined completely arbitrarily [e.g. 1 metre is the distance traveled by light in 1/(299792458) of a second -- there's clearly nothing special about this number], we don't expect the form of any of our equations to change if we instead measure length in half-metres, time in minutes, and so forth. So suppose we change our units so that the numerical value of lengths are multiplied by some factor L, whereas numerical value of times are scaled by another factor T. Then since velocity = (distance)/(time), the numerical value of velocity is scaled by a factor LT^-1, so this is the dimensionality of velocity. The requirement for "dimensional correctness" (i.e. the two sides of a physically meaningful equation must have the same dimension) then follows from the fact that, for the equation to still be valid when we rescale our units, the two sides of the equation have to transform in the same way.

When you think about it this way, a dimensionless quantity is one whose numerical value doesn't change when you scale your arbitrary units. The reason we think of an angle measured in radians as being dimensionless, is that, unlike metres and seconds, radians are not an arbitrarily defined unit. They are a very special unit of angle which are chosen to make our equations as simple as possible (for example, (d/dx)sin(x) = cos(x) if we measure angles in radians, but not otherwise). We can't change units of angle and expect our equations to retain the same form, so an angle measured in radians is dimensionless (of course, we can say the same kind of thing about a solid angle measured in steradians). On the other hand, degrees are a completely arbitrary unit of angle, so if we measure angles in degrees then we should no longer think of them as dimensionless.

If we wanted to, we could also decide to measure quantities such as time and distance in http://en.wikipedia.org/wiki/Planck_units" , so that c = hbar = G = 1. As these are "natural", not arbitrary units (as reflected in the fact that formulas take simpler forms in these units, since lots of physical constants are just equal to 1), distances and times would then become dimensionless as well.

---

A question for those advocating Siano's "orientational analysis": is it possible to derive the rules of orientational analysis from some symmetry of physical laws in the same as way as I did above for ordinary dimensional analysis? Presumably from rotational symmetry? Siano seems to suggest this is possible in his papers, but I can't see where he actually does it. Also, that would imply that orientational analysis is no longer valid in problems with a preferred direction, due to, e.g., the Earth's gravitational field.

 

[ChARMELeOn]

Thanks for all your answers it's my first time posting here so I didn't expect this much ^^
But I believe I have already solved it on my own.

If you for example have a graph where both x and y have the same dimension (let's say its lenghth) and you find in the graph that y/x=a, which is in dimensions L/L=[a] then you will also find that y=a*x, which is in dimensions L=[a]*L and that must mean that a is dimensionless or else there won't be the same dimension on both sides in the equation that is shown to be correct through the graph.

So L/L=1 and through the definition of radians (s/r=q where s and r have the dimension length) we find that radians also must be dimensionless.You are free to correct me if I'm wrong.
Thanks.

 

[Rap]

How is the following identiy:

<br /> \int_{0}^{\frac{\pi}{2}}{\sin^{p}{x} \, \cos^{q}{x} \, dx} = \frac{1}{2} \, B\left(\frac{p + 1}{2},\frac{q + 1}{2}\right), \; p, q &gt; -\frac{1}{2}<br />

dimensionally consistent according to Siano's extension?

Its a definition of the Beta function, so it is consistent by definition. For example, if p is odd, the Beta function will be dimensionless. If not, it won't be. This is not strange, the sine function is a dimensioned function.

The period of oscillation T of a mathematical pendulum with length L in a uniform gravitational field with acceleration due to gravity g and an amplitude angle \alpha with the vertical is given by:

<br /> T = 4 \sqrt{\frac{L}{g}} \, \int_{0}^{\frac{\pi}{2}}{\frac{dt}{\sqrt{1 - \sin^{2}{(\frac{\alpha}{2})} \sin^{2}{(t)}}}}<br />

How is this dimensionally consistent according to Siano's extension of dimensions?

The dimensions of T are seconds per cycle or seconds per 2 pi radians. Radians are pure direction (e.g. 1x), unlike directed lengths which are directed and have units of, say, meters. (meters * 1x). That means T has units of (sec/1x = sec*1x). dt in the integral is radians (1x), the term under the square root in the integral is dimensionless, the term in the square root outside the integral has units of seconds, so the equation is consistent.

Thats a bit glib, I think, to say T is seconds per 2 pi radians, so I will think about that. For sure T is not seconds, but in seconds per cycle, and that's where the consistency occurs.

 

[Dickfore]

Its a definition of the Beta function, so it is consistent by definition. For example, if p is odd, the Beta function will be dimensionless. If not, it won't be. This is not strange, the sine function is a dimensioned function.

The problem is that both p and q are any real numbers. What is the dimensions of, let's say:

<br /> \sin^{\frac{1}{3}}{(\alpha)}<br />

Also, don't you remember that:

<br /> B(a, b) = B(b, a)<br />

How is this consistent with what you said?

The dimensions of T are seconds per cycle or seconds per 2 pi radians. Radians are pure direction (e.g. 1x), unlike directed lengths which are directed and have units of, say, meters. (meters * 1x). That means T has units of (sec/1x = sec*1x). dt in the integral is radians (1x), the term under the square root in the integral is dimensionless, the term in the square root outside the integral has units of seconds, so the equation is consistent.

Thats a bit glib, I think, to say T is seconds per 2 pi radians, so I will think about that. For sure T is not seconds, but in seconds per cycle, and that's where the consistency occurs.

But, the integral on the rhs has dimensions:

<br /> \mathrm{T} \cdot [\mathrm{angle}]<br />

so everything you said actually contradicts that equation.

EDIT:

Ok, I see you used the rule [\mathrm{angle}]^{2} = 1 to conclude that [\mathrm{angle}]^{-1} = [\mathrm{angle}]. So, now you are saying that period of oscillation does not have pure units of time?! But, then, how is this consistent with:

<br /> \omega = \frac{2 \pi}{T}<br />

If anything, I would expect that [\omega] = [\mathrm{angle}] \cdot \mathrm{T}^{-1}.

 

[Rap]

The problem is that both p and q are any real numbers. What is the dimensions of, let's say:

<br /> \sin^{\frac{1}{3}}{(\alpha)}<br />

Since alpha is dimensioned 1_x (i.e. radians), sin(alpha) is dimensioned 1_x, so we are looking for some direction when cubed gives you 1_x, and that's 1_x since 1_x^2=1₀ (dimensionless). But what about the square root of \sin(\alpha)? Thats harder, because there is no direction when squared that gives you 1_x. But this problem might be encountered in regular dimensional analysis. Newtons law is F=ma but it could be written F^1/2=m^1/2 a^1/2, and we would be taking the square root of a direction here too. But it can be "fixed" by squaring both sides. I would say that you will never encounter \sqrt{\sin(\alpha)} in a physical equation in such a way that things cannot be "fixed" as in the Newton law example.

Also, don't you remember that:

<br /> B(a, b) = B(b, a)<br />

How is this consistent with what you said?

Well, I guess its not. But the above example shows that some seemingly "bad" equations are "fixable". I think that any physical equation which involves the Beta function would have to be "fixable" if they were used in the analysis of any physical problem. I would be very interested to know if there is any physical equation involving the Beta function that was neither dimensionally consistent or "fixable". I don't know of any, but I believe that probability problems must also be dimensionally consistent, and the Beta function is often used in probability problems. Can you think of a physical or probability problem where the Beta function is used, or even the square root of a sine function, that is not "fixable"?

But, then, how is this consistent with:

<br /> \omega = \frac{2 \pi}{T}<br />

If anything, I would expect that [\omega] = [\mathrm{angle}] \cdot \mathrm{T}^{-1}.

2\pi has units of radians per cycle, T has units of seconds per cycle, and \omega has units of radians per second.

 

[Dickfore]

Since alpha is dimensioned 1_x (i.e. radians), sin(alpha) is dimensioned 1_x, so we are looking for some direction when cubed gives you 1_x, and that's 1_x since 1_x^2=1₀ (dimensionless). But what about the square root of \sin(\alpha)? Thats harder, because there is no direction when squared that gives you 1_x. But this problem might be encountered in regular dimensional analysis. Newtons law is F=ma but it could be written F^1/2=m^1/2 a^1/2, and we would be taking the square root of a direction here too. But it can be "fixed" by squaring both sides. I would say that you will never encounter \sqrt{\sin(\alpha)} in a physical equation in such a way that things cannot be "fixed" as in the Newton law example.

What about:

<br /> \sin^{\frac{1}{\sqrt{2}}}{(\alpha)}<br />

Well, I guess its not. But the above example shows that some seemingly "bad" equations are "fixable". I think that any physical equation which involves the Beta function would have to be "fixable" if they were used in the analysis of any physical problem. I would be very interested to know if there is any physical equation involving the Beta function that was neither dimensionally consistent or "fixable". I don't know of any, but I believe that probability problems must also be dimensionally consistent, and the Beta function is often used in probability problems. Can you think of a physical or probability problem where the Beta function is used, or even the square root of a sine function, that is not "fixable"?

Me not finding an example does not constitute a proof of your claims. It is up to you to show that there is no such law. Also, you still had not addressed the question of:

<br /> B(p, q) = B(q, p)<br />

2\pi has units of radians per cycle, T has units of seconds per cycle, and \omega has units of radians per second.

I just noticed you `invented' a new dimension - cycle.