- #36
Dickfore
- 2,987
- 5
The equation given in the lecture is:
[tex]
\frac{T_{2}}{T_{1}} = \exp{(\mu \, \theta)}
[/tex]
[tex]
\frac{T_{2}}{T_{1}} = \exp{(\mu \, \theta)}
[/tex]
Dickfore said:I think one should not pay too much attention to Siano's extension of dimension.
Dickfore said:How is the following identiy:
[tex]
\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 > -\frac{1}{2}
[/tex]
dimensionally consistent according to Siano's extension?
Dickfore said:The period of oscillation [itex]T[/itex] of a mathematical pendulum with length [itex]L[/itex] in a uniform gravitational field with acceleration due to gravity [itex]g[/itex] and an amplitude angle [itex]\alpha[/itex] with the vertical is given by:
[tex]
T = 4 \sqrt{\frac{L}{g}} \, \int_{0}^{\frac{\pi}{2}}{\frac{dt}{\sqrt{1 - \sin^{2}{(\frac{\alpha}{2})} \sin^{2}{(t)}}}}
[/tex]
How is this dimensionally consistent according to Siano's extension of dimensions?
The problem is that both p and q are any real numbers. What is the dimensions of, let's say:Rap said: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.
Rap 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.
Dickfore said:The problem is that both p and q are any real numbers. What is the dimensions of, let's say:
[tex]
\sin^{\frac{1}{3}}{(\alpha)}
[/tex]
Dickfore said:Also, don't you remember that:
[tex]
B(a, b) = B(b, a)
[/tex]
How is this consistent with what you said?
Dickfore said:But, then, how is this consistent with:
[tex]
\omega = \frac{2 \pi}{T}
[/tex]
If anything, I would expect that [itex][\omega] = [\mathrm{angle}] \cdot \mathrm{T}^{-1}[/itex].
Rap said:Since alpha is dimensioned 1x (i.e. radians), sin(alpha) is dimensioned 1x, so we are looking for some direction when cubed gives you 1x, and that's 1x since 1x^2=10 (dimensionless). But what about the square root of [tex]\sin(\alpha)[/tex]? Thats harder, because there is no direction when squared that gives you 1x. But this problem might be encountered in regular dimensional analysis. Newtons law is F=ma but it could be written F1/2=m1/2 a1/2, 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 [tex]\sqrt{\sin(\alpha)}[/tex] in a physical equation in such a way that things cannot be "fixed" as in the Newton law example.
Rap 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"?
Rap said:[tex]2\pi[/tex] has units of radians per cycle, T has units of seconds per cycle, and [tex]\omega[/tex] has units of radians per second.
Dickfore said:What about:
[tex]
\sin^{\frac{1}{\sqrt{2}}}{(\alpha)}
[/tex]
Dickfore said: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:
[tex]
B(p, q) = B(q, p)
[/tex]
Dickfore said:I just noticed you `invented' a new dimension - cycle.
cortiver said: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.
Dickfore said: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 said: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" [Broken], 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 said:BTW, at one point in his first paper (third paragraph on the sixth page), he says that [itex]\sin{(\theta)}[/itex] is orientationally quite different from [itex]\sin{(\phi)}[/itex], where [itex]\theta[/itex] is an angle (with orientational symbol 1z) and [itex]\phi[/itex] is a phase angle.
Dickfore said: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.
Rap said:I think this is the same as the [tex]\sin(\theta+\pi/2)=\cos(\theta)[/tex] example. With explicit orientational symbols;
[tex]\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) [/tex]
so that [tex]\sin(\theta\,\,1x+\pi/2\,\,1x)=1x \cos(\theta)[/tex] and the discrepancy disappears.
Rap said:Yes! You can derive the algebra of the directional symbols by the dot product, which is dimensionless ([tex]e_i[/tex] are unit vectors):
[tex](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[/tex]
which proves [tex]1_j^2=1_0[/tex] (dimensionless). Then do the cross product:
[tex](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[/tex]
which proves that [tex]1_j 1_k= 1_i[/tex] 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.
Agreed. Do you have in mind a physical situation in which the argument is dimensionless? I will try to think of one.Dickfore said: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 [itex]1_{0}[/itex], then the value of the sine has the dimension [itex]1_{0}[/itex] as well. If the argument has orientational symbol [itex]1_{z}[/itex], then, so does the value of the sine function.
Dickfore said: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 [itex]\mathrm{M} \mathrm{T}^{-2}[/itex], just as the rate of change of growth of, e.g. an animal, which, of course, is orientationless.
Dickfore said:Cross product is defined only in 3d.
Dimensional analysis is a mathematical method used to convert between different units of measurement. In this method, we use the fundamental dimensions of a physical quantity (such as length, time, mass, etc.) to create a dimensional equation. When we divide a quantity by itself, the resulting value is always 1. Therefore, length/length is equal to 1 in dimensional analysis.
A dimensionless number is a numerical value that does not have any units associated with it. In dimensional analysis, we use dimensionless numbers to compare and analyze physical quantities without being affected by their units. This allows us to focus on the underlying relationships between different quantities and make accurate predictions and calculations.
No, we cannot use any units for length in dimensional analysis. The units used for length must be consistent throughout the entire equation. For example, if we are using meters for length, then all other quantities in the equation must also be expressed in meters. This ensures that the final result is in the correct units and has a dimensionless value.
Dimensional analysis is used in scientific research to simplify complex equations and make predictions about physical phenomena. It is particularly useful in fields such as physics, chemistry, and engineering, where precise measurements and unit conversions are crucial. By using dimensional analysis, scientists can identify relationships between different quantities and make accurate predictions without the need for extensive experimentation.
While dimensional analysis is a powerful tool, it does have some limitations. It can only be used for physical quantities that have a measurable dimension, such as length, time, mass, etc. It also cannot account for any non-dimensional factors that may affect the outcome of an experiment. Additionally, dimensional analysis is only applicable to linear relationships between quantities and may not be suitable for more complex systems.