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

P: 46
 Quote by 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.
Could you provide details? I've been looking at Section IX of Siano's second paper, 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.
P: 3,014
 Quote by cortiver Could you provide details? I've been looking at Section IX of Siano's second paper, 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:

$$\hat{\sigma}_{i} \, \hat{\sigma}_{k} = \delta_{i k} \, \hat{1} + i \, \epsilon_{i k l} \, \hat{\sigma}_{l}$$

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 1z) and $\phi$ is a phase angle.
P: 789
 Quote by Dickfore 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 1z) 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;

$$\sin(\theta\,\,1x+\phi \,\,1x)=\sin(\theta \,\,1x)\cos(\phi\, \,1x)+\sin(\phi \,\,1x)\cos(\theta \,\, 1x) = 1x\, \sin(\theta)\cos(\phi)+1x \,\sin(\phi)\cos(\theta)$$

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

 Quote by 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.
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.
P: 3,014
 Quote by Rap I think this is the same as the $$\sin(\theta+\pi/2)=\cos(\theta)$$ example. With explicit orientational symbols; $$\sin(\theta\,\,1x+\phi \,\,1x)=\sin(\theta \,\,1x)\cos(\phi\, \,1x)+\sin(\phi \,\,1x)\cos(\theta \,\, 1x) = 1x\, \sin(\theta)\cos(\phi)+1x \,\sin(\phi)\cos(\theta)$$ 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.
 Quote by Rap 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.
P: 789
 Quote by Dickfore 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.

 Quote by Dickfore 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.

 Quote by Dickfore 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.

 Related Discussions Electrical Engineering 8 Atomic, Solid State, Comp. Physics 0 Introductory Physics Homework 3 General Physics 1