Lecture 19 - Buckingham's $$\Pi$$ theorem for dimensionless symmetry groups

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Dimensional analysis summary --

• Fundamental units are choosen
• Derived units are expressed in terms of fundamental units
• Hypothesis: If the world is a continuum, its laws should not depend on the units with which we describe it.
• By picking convenient units, we can arrive at general symmetry forms for these laws.

Dimensionless groups so far

• Pressure in a soap bubble: $$P \propto \dfrac{s}{R}$$
• Pendulum group: $$\dfrac{T^2 g}{L} = f(\theta_0)$$

Buckingham $$\Pi$$ theorem

If there are M dimensional parameters involving N unit dimensions, then system has M-N dimensionless groups.

If we have a dimensioned equation

$f(q_1,q_2, ... ,q_n) = 0$

where the $$q_i$$ are the $$n$$ physical variables, with $$k$$ independent physical units, then the equation can be stated as

$F(r_1,r_2, ... ,r_p)=0$

where the $$r$$ are dimensionless values constructed from the $$q$$ by $$p = n - k$$ dimensionless equations --the so-called $$\Pi$$ groups-- of the form

$\Pi_i = q_1^{a_1} q_2^{a_2} \ldots$

where the exponents $$a_i$$ are rational numbers (they can always be taken to be integers: just raise it to a power to clear denominators).

Linear algebra approach to construction of dimensionless groups

Ship-speed example

Parameters:

• $$g$$ = gravity has dimensions of length per time per time
• $$r$$ = size can be represented in dimensions of length
• $$v$$ = velocity has dimensions of length per time
• $$\rho$$ = density of water has units of mass per length cubed
• $$\mu$$ =viscosity has dimensions of mass per length per time.

If we conveniently arrange our table of variables and units,

 r g v $$\mu$$ $$\rho$$ distance 1 1 1 -1 -3 time 0 -2 -1 -1 0 mass 0 0 0 1 1

Applying the $$\Pi$$ theorem,

• The dimensions of the row space = 3
• The dimensions of the row nullspace = 2
• The dimensions of the col space = 3
• The dimensions of the col nullspace = 0

So, from the $$\Pi$$ theorem, we deduce we must have 2 linearly independent dimensionless parameters. In RREF, the dimension matrix becomes

$\left[\begin{matrix}1 & 0 & \frac{1}{2} & 0 & - \frac{3}{2}\\0 & 1 & \frac{1}{2} & 0 & - \frac{1}{2}\\0 & 0 & 0 & 1 & 1\end{matrix}\right]$

from sympy import *
from sympy.abc import *
vs=(r,g,v,mu,rho)
A = Matrix([[1,1,1,-1,-3],[0,-2,-1,-1,0],[0,0,0,1,1]])
ns = A.nullspace()
F = lambda k : Mul( * [i**j for i,j in zip(vs, ns[k])])**2
print latex((F(0),F(1)))

This leads to an equation $F\left( \frac{v^{2}}{g r}, \frac{g r^{3}\rho^{2}}{\mu^{2}} \right)=0$ If we multiply the second by the first and take a square root, we get the more standard (and slightly simpler) version $F\left ( \frac{v^{2}}{g r}, \quad \frac{r \rho v}{\mu} \right )=0$

• The Froude number $$\frac{v^{2}}{g r}$$.
• The Reynolds number $$\frac{r \rho v}{\mu}$$.

For a ship in water, $$\rho \approx 10^3$$ kg/$$m^3$$, $$\mu \approx 10^{-3}$$ kg/s/m, $$r \approx 10$$ m, and $$v \approx 1$$ m/s, so the reynolds number is about $$10^7$$, very large. If the Reynolds number is large, we can asymptotically expand around it's reciprical being small, and hope to find

$v \propto \sqrt{gr}$

Thus, the speed is proportional to the square root of water-line length, and longer boats will be faster that shorter boats, all else equal. Also, ships will be faster on planets will lower gravity, and slower on planets with higher gravity.

Frequency of oscillations in a water drop

A drop of water falls from a facet. It oscillates from ovoid to sphere to ovoid as it falls, as the surface tension holding it together fights with the momentum of the water inside it. (The oscilations of big water balloons are similar, though less regular) Can we find scaling-law equations for this frequency?

• $$f$$ is frequency of oscillations
• $$\rho$$ is the mass density
• $$V$$ is volume of the drop
• $$s$$ is the surface tension
 f V s $$\rho$$ distance 0 3 0 -3 time -1 0 -2 0 mass 0 0 1 1

1 dimensionless group, by the $$\Pi$$ theorem.

from sympy import *
from sympy.abc import *
vs=(f,V,s,rho)
A = Matrix([[0,3,0,-3],[-1,0,-2,0],[0,0,1,1]])
ns = A.nullspace()
F = lambda k : Mul( * [i ** j for i,j in zip(vs, k)])
print latex([ F(i) for i in ns])

This leads to the single group symmetry equation $\frac{V \rho f^{2}}{s} = C$ Thus, as drops get bigger in volume, their frequency of oscillation decreases.