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

[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.

 

[Rap]

What about:

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

Just as in the Newtons Law example, if we had F^{\sqrt{1/2}}=m^{\sqrt{1/2}}a^{\sqrt{1/2}}, it would be something that could be "fixed". If not, then its not a good physics equation.

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 />

True, but if do find an example, let me know. About the B(p,q)=B(q,p), the p and q are exponents of the trig functions in the integral, so its very similar to the \sin^x(\theta) problem. Just because the exponents are "bad" doesn't mean they cannot be "fixed". Take a definition of the Beta function:

B(p,q)=\int_0^{\pi/2} \sin^{2p+1}(\theta)\cos^{2q+1}(\theta)\,d\theta

If we treat theta as a directed quantity, we write it as \theta 1_x and the sine is directed the same way, the cosine is dimensionless, the d\theta is directed the same way and the direction of the Beta function is 1_x^{2p+2}=(1_x^2)^{p+1} which is dimensionless. Defining \theta =\phi +\pi/2 we get \sin(\theta 1_x)=1_x \cos(\phi 1_x) and \cos(\theta 1_x)=-1_x \sin(\phi 1_x) (i.e. still dimensionless) so that B(p,q) is equal in value to B(q,p) and is again directed as 1_x^{2p+2} (i.e. dimensionless). But if we take your original definition (replacing p with (p+1)/2 etc.)

B\left(\frac{p+1}{2},\frac{q+1}{2}\right)=2\int_0^{\pi/2} \sin^p(\theta)\cos^q(\theta)\,d\theta

And now its directed as 1_x^{p+1} which is not well defined unless p+1 takes on certain values (like 1/3, but not 1/2, as in the sin(alpha) example). I have not checked, but I bet the directions are not consistent for the second case either. But the point is, a "bad" statement like the second case can be "fixed" to be good, like the first case. If Siano's extension is valid, then the Beta function will not occur in any physically meaningful equation in such a way that it cannot be "fixed" somehow, just like the Newtons law case.

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

A cycle is 2 pi radians, so if radians are not dimensionless, then neither are cycles. Not an invention, just a consequence of saying radians are not dimensionless.

 

[cortiver]

Could I repeat my question from before: why are you so sure that Siano's extension is valid? Does it follow from rotational invariance somehow? I would certainly not be comfortable making use of some ad-hoc set of rules unless I understand how they come about.

 

[Rap]

Could I repeat my question from before: why are you so sure that Siano's extension is valid? Does it follow from rotational invariance somehow? I would certainly not be comfortable making use of some ad-hoc set of rules unless I understand how they come about.

I am not sure that Siano's extension is valid, but try as I might, I cannot break it. Every time I think I have, further study convinced me that I have not, and that further study gave me insights into dimensional analysis in general. I have not investigated it on any deeper level than that. Trying to show that all the challenges to it given, esp. by Dickfore, can be dealt with is very interesting to me, but what would be excellent is to find one that really did not work. That would give some real insight. Thats why I am happy to try to answer all challenges to it, but I will not defend it to the death. Thats why I challenge anyone to break it. If it cannot be broken, then maybe its time to go deeper into the theory and ask why does it work? That means going deeper into dimensional analysis in general. The Buckingham Pi theorem is about as far as I have gone in this direction. Its evidently not enough. Any guidance would be welcome.

 

[Dickfore]

I think I understand why Siano's extension to dimensional analysis works! In fact, if my logic is correct, I would propose an extension to Siano's approach to space-time that might be useful in relativistic physics.

 

[cortiver]

I think I understand why Siano's extension to dimensional analysis works! In fact, if my logic is correct, I would propose an extension to Siano's approach to space-time that might be useful in relativistic physics.

Could you provide details? I've been looking at Section IX of Siano's http://dx.doi.org/10.1016/0016-0032(85)90032-8" , and it turns out he does give a partial justification of why his method works, by assuming that physical laws are tensor equations (as they must be from rotational invariance). This helps me understand what was going on in the example given in #31 (namely, if you're going to treat the forces as vectors, then there will be a rotation matrix involved to to relate them, whose entries have the right dimensions to ensure everything works out right), but I still can't see how assigning dimensions to angles can be justified.

EDIT: He also admits that his method doesn't work for equations with fractional exponents.

 

[Dickfore]

Could you provide details? I've been looking at Section IX of Siano's http://dx.doi.org/10.1016/0016-0032(85)90032-8" , and it turns out he does give a partial justification of why his method works, by assuming that physical laws are tensor equations (as they must be from rotational invariance). This helps me understand what was going on in the example given in #31 (namely, if you're going to treat the forces as vectors, then there will be a rotation matrix involved to to relate them, whose entries have the right dimensions to ensure everything works out right), but I still can't see how assigning dimensions to angles can be justified.

EDIT: He also admits that his method doesn't work for equations with fractional exponents.

Yes, my ideas also started similarly, although I was looking at the rotation matrices in Cartesian coordinates (no curvilinear coordinates for now). I think his method works well because of the "well known" fact that in 3d the dual of an antisymmetric tensor of rank 2 is a vector (albeit an axial one). I don't want to say too much. Let me just say that Pauli matrices and the unit matrix:

<br /> \hat{\sigma}_{i} \, \hat{\sigma}_{k} = \delta_{i k} \, \hat{1} + i \, \epsilon_{i k l} \, \hat{\sigma}_{l}<br />

obey a similar algebra as V except that it is anti-Abellian for the Pauli matrices that are not equal. Perhaps I need more group theoretical knowledge to refine this point. One might ask, what do Pauli matrices have to do with rotations. Again, there is a very "convenient" coincidence in 3d.

BTW, at one point in his first paper (third paragraph on the sixth page), he says that \sin{(\theta)} is orientationally quite different from \sin{(\phi)}, where \theta is an angle (with orientational symbol 1_z) and \phi is a phase angle.

 

[Rap]

BTW, at one point in his first paper (third paragraph on the sixth page), he says that \sin{(\theta)} is orientationally quite different from \sin{(\phi)}, where \theta is an angle (with orientational symbol 1_z) and \phi is a phase angle.

I think this is the same as the \sin(\theta+\pi/2)=\cos(\theta) example. With explicit orientational symbols;

\sin(\theta\,\,1x+\phi \,\,1x)=\sin(\theta \,\,1x)\cos(\phi\, \,1x)+\sin(\phi \,\,1x)\cos(\theta \,\, 1x) =<br /> 1x\, \sin(\theta)\cos(\phi)+1x \,\sin(\phi)\cos(\theta)

so that \sin(\theta\,\,1x+\pi/2\,\,1x)=1x \cos(\theta) and the discrepancy disappears.

I think I understand why Siano's extension to dimensional analysis works! In fact, if my logic is correct, I would propose an extension to Siano's approach to space-time that might be useful in relativistic physics.

Yes! You can derive the algebra of the directional symbols by the dot product, which is dimensionless (e_i are unit vectors):

(A_j\mathbf{e}_j 1_j)\cdot(B_k \mathbf{e}_k 1_k)=A_j B_j 1_j^2 = A_j B_j 1_0

which proves 1_j^2=1_0 (dimensionless). Then do the cross product:

(A_j\mathbf{e}_j 1_j) \mathrm{X} (B_k \mathbf{e}_k 1_k)=\varepsilon_{ijk}A_j B_k \mathbf{e}_i (1_j 1_k) = \varepsilon_{ijk}A_j B_k \mathbf{e}_i 1_i

which proves that 1_j 1_k= 1_i where i,j, and k are all different. The same procedure could be carried out for the invariant analogs in relativity for the direction symbols 1x, 1y, 1z, 1t, 10.

 

[Dickfore]

I think this is the same as the \sin(\theta+\pi/2)=\cos(\theta) example. With explicit orientational symbols;

\sin(\theta\,\,1x+\phi \,\,1x)=\sin(\theta \,\,1x)\cos(\phi\, \,1x)+\sin(\phi \,\,1x)\cos(\theta \,\, 1x) =<br /> 1x\, \sin(\theta)\cos(\phi)+1x \,\sin(\phi)\cos(\theta)

so that \sin(\theta\,\,1x+\pi/2\,\,1x)=1x \cos(\theta) and the discrepancy disappears.

The sine function, being a Taylor series of only odd powers of its argument, has the same orientational symbol as its argument. If the argument had orientational symbol 1_{0}, then the value of the sine has the dimension 1_{0} as well. If the argument has orientational symbol 1_{z}, then, so does the value of the sine function.

Siano actually talks about this. Due to orientational analysis, we can distinguish between angular velocity (which has orientation) and circular frequency (which does not); torque (which is oriented) and work (which is not) and so on. Perhaps a very drastic example is the case of coefficient of surface tension. It is defined as the energy per unit area and, thus it has the same orientation as the area. It's dimension is, however \mathrm{M} \mathrm{T}^{-2}, just as the rate of change of growth of, e.g. an animal, which, of course, is orientationless.

Yes! You can derive the algebra of the directional symbols by the dot product, which is dimensionless (e_i are unit vectors):

(A_j\mathbf{e}_j 1_j)\cdot(B_k \mathbf{e}_k 1_k)=A_j B_j 1_j^2 = A_j B_j 1_0

which proves 1_j^2=1_0 (dimensionless). Then do the cross product:

(A_j\mathbf{e}_j 1_j) \mathrm{X} (B_k \mathbf{e}_k 1_k)=\varepsilon_{ijk}A_j B_k \mathbf{e}_i (1_j 1_k) = \varepsilon_{ijk}A_j B_k \mathbf{e}_i 1_i

which proves that 1_j 1_k= 1_i where i,j, and k are all different. The same procedure could be carried out for the invariant analogs in relativity for the direction symbols 1x, 1y, 1z, 1t, 10.

Cross product is defined only in 3d.

 

[Rap]

The sine function, being a Taylor series of only odd powers of its argument, has the same orientational symbol as its argument. If the argument had orientational symbol 1_{0}, then the value of the sine has the dimension 1_{0} as well. If the argument has orientational symbol 1_{z}, then, so does the value of the sine function.

Agreed. Do you have in mind a physical situation in which the argument is dimensionless? I will try to think of one.

Siano actually talks about this. Due to orientational analysis, we can distinguish between angular velocity (which has orientation) and circular frequency (which does not); torque (which is oriented) and work (which is not) and so on. Perhaps a very drastic example is the case of coefficient of surface tension. It is defined as the energy per unit area and, thus it has the same orientation as the area. It's dimension is, however \mathrm{M} \mathrm{T}^{-2}, just as the rate of change of growth of, e.g. an animal, which, of course, is orientationless.

I would say that the dimension of the energy per unit area is m\,\,1x/t^2 (or 1y or whatever) rather than say its dimension is m/t^2 and is oriented. This is a semantic disagreement, so its not super critical.

Cross product is defined only in 3d.

I didn't say "cross product", I said "invariant analogs". The cross product in 3d is

(\mathbf{A} \mathrm{x} \mathbf{B})_i = \varepsilon_{ijk}A_jB_k

where \varepsilon_{ijk} is the permutation symbol (=1 for even permutations of 123, -1 for odd, and zero otherwise). For a 4-d Euclidean space the analog is

\varepsilon_{ijkl}A_kB_l

where \varepsilon_{ijkl} is the permutation symbol (=1 for even permutations of 1234, -1 for odd, and zero otherwise). For Minkowski space, we need to make the covariant/contravariant distinction and the analog is

\varepsilon_{ijkl}A^kB^l

Note that the analog is a 2nd rank tensor rather than a vector.