Notes and examples:
Buckingham Π theorem
PX391- S C Chapman
Universality- 1 d.o.f.
Pendulum
d 2θ
F = mg , Ft = mg sin θ , at = l 2
dt
d 2θ
g
∂V
Ft = mat ; 2 = − sin θ = −ω 2
dt
l
∂θ
V (θ ) = 1 − cos(θ ) ~
θ2
2
+ ...
same behaviour at
any local minimum in V (θ )
(insensetive to details)
V (θ )
Buckingham π theorem
System described by F (Q1...Q p ) where Q1.. p are the relevant macroscopic variables
F must be a function of dimensionless groups π 1.. M (Q1.. p )
if there are R physical dimensions (mass, length, time etc.)
there are M = P − R distinct dimensionless groups.
Then F (π 1.. M ) = C is the general solution for this universality class.
To proceed further we need to make some intelligent guesses for F (π 1.. M )
See e.g. Barenblatt, Scaling, self - similarity and intermediate asymptotics, CUP, [1996]
also Longair, Theoretical concepts in physics, Chap 8, CUP [2003]
Example: simple (nonlinear) pendulum
System described by F (Q1...Q p ) where Qk is a macroscopic variable
F must be a function of dimensionless groups π 1.. M (Q1.. p )
if there are R physical dimensions (mass, length, time etc.) there are M = P − R dimensionless groups
Step 1: write down the relevant macroscopic variables:
variable dimension
description
angle of release
−
θ0
m
τ
g
l
[M ]
[T ]
[ L][T ]−2
[ L]
mass of bob
period of pendulum
gravitational acceleration
length of pendulum
Step 2: form dimensionless groups: P = 5, R = 3 so M = 2
π 1 = θ0 , π 2 =
τ 2l
g
and no dimensionless group can contain m
2
then solution is F (θ0 ,τ l ) = C
g
Step 3: make some simplifying assumption: f (π 1 ) = π 2 then the period:τ = f (θ 0 )
NB f (θ 0 ) is universal ie same for all pendulawe can find it knowing some other property eg conservation of energy..
l
g
Example: fluid turbulence, the Kolmogorov '5/3 power spectrum'
System described by F (Q1...Q p ) where Qk is a macroscopic variable
F must be a function of dimensionless groups π 1.. M (Q1.. p )
if there are R physical dimensions (mass, length, time etc.) there are M = P − R dimensionless groups
Step 1: write down the relevant variables (incompressible so energy/mass):
variable dimension
[ L]3 [T ]−2
[ L]2 [T ]−3
[ L ] −1
E (k )
ε0
k
description
energy/unit wave no.
rate of energy input
wavenumber
Step 2: form dimensionless groups: P = 3, R = 2, so M = 1
π1 =
E 3 ( k )k 5
ε 02
Step 3: make some simplifying assumption:
2
F (π 1 ) = π 1 = C where C is a non universal constant, then: E (k ) ~ ε 0 3 k
−5
3
Buchingham π theorem (similarity analysis)
universal scaling, anomalous scaling
System described by F (Q1...Q p ) where Qk is a relevant macroscopic variable
F must be a function of dimensionless groups π 1.. M (Q1.. p )
if there are R physical dimensions (mass, length, time etc.) there are M = P − R dimensionless groups
Turbulence:
variable dimension
E (k )
ε0
k
[ L]3 [T ]−2
[ L]2 [T ]−3
[ L]−1
description
energy/unit wave no.
rate of energy input
M = 1, π 1 =
E 3 ( k )k 5
M = 2, π 1 =
E 3 ( k )k 5
ε
2
0
2
, E (k ) ~ ε 0 3 k
−5
3
wavenumber
introduce another characteristic speed....
variable dimension
description
E (k )
ε0
k
v
[ L]3 [T ]-2
[ L]2 [T ]-3
[ L]-1
[ L][T ]−1
energy/unit wave no.
rate of energy input
wavenumber
characteristic speed
ε 02
−( 5+α )
v2
α
(3+α )
,π 2 =
let π 1 ~ π 2 , E ( k ) ~ k
Ek
Homogeneous Isotropic Turbulence and Reynolds Number
Step 1: write down the relevant variables:
variable dimension
description
driving scale
L0
[ L]
[ L]
−1
[ L ][T ]
−1
2
[ L ] [T ]
η
U
ν
dissipation scale
bulk (driving ) flow speed
viscosity
Step 2: form dimensionless groups: P = 4, R = 2, so M = 2
UL
L
L
π 1 = 0 = RE , π 2 = 0 and importantly 0 = f ( N ), where N is no. of d.o.f
ν
η
η
Step 3: d.o.f from scaling ie f ( N ) ~ N α here
L0
η
~ N 3 ,or N 3 β or
L0
η
~λ
N
3
or ...
Step 4: assume steady state and conservation of the dynamical quantity, here energy...
ur3
ν3
U3
transfer rate ε r ~ , injection rate ε inj ~
, dissipation rate ε diss ~ 4 - gives ε inj ~ ε r ~ ε diss
r
L0
η
⎛L ⎞
~⎜ 0⎟
this relates π 1 to π 2 giving: RE =
ν
⎝η ⎠
UL0
4
3
~ N α ,α > 0
thus N grows with RE