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Dimensional and similarity analysis
Strong explosion (atomic bomb)
Most of the energy input is at the initial instant t = 0.
Assume that the fireball radius R depends only on:
input energy E0 ,
time t,
background air density  0 .
R has dimension of length
t
L
has dimension of time T
E0 has dimension of
1 2
mv ,
2
 0 has dimension of
m
L3
or
m
L2
T2
We can then write [R] = L1 = [E0 ]a [0 ]b [t ]c ,
where […] means “the dimension of”.
Substituting the dimensions of energy, time and density, one gets
[R] = L1 = ma b L2 a 3bT c2 a
Therefore
a
+b
=0
2a
-3b
=1
-2a
+c = 0
Solution is a = 1/5, b = -1/5, c = 2/5 .
R =  E0 / 0  t 2/5
1/5
We can then obtain
or
E0 = R5 0 /t 2
This is the well-known Sedov-Taylor similarity solution for strong spherical explosion.
Note that the crucial point is to identify what variables are relevant!
Example: Atomic bomb fireball
TRINITY
radius of the blast sphere
First atom bomb test, 1945, New Mexico, USA
~160/2=80 m
density of air ~1.2 kg/m3
E = 85 0
get
~
1 4
10
1 . 2 / 0 . 0 0 26
2
kg m /s
k2 g 2 m / s2 1 ~ 1 0
NRL
erg
1 kiloton TNT (Tri-nitro-toluene) ~ 109 Cal ~ 4.2  1019 erg
NRL
Therefore the Trinity atom bomb has an equivalent energy of E = 24 kiloton TNT
Compare: your daily intake of energy is about 2000 Cal = 2 103 ton = 2 kg TNT . (Please check
this!) But only a part of this energy is used (and over a 24 hour period instead of a fraction of a
second.)
We can further extend the similarity solutions.
[ P]  0V 2
Pressure P
 0  L / T 
2
 0  E0 / 0  t 2/5 / t 


3/5
2/5 -6/5
P  0 E0 t .
1/5
2
In the fireball, we can define the dimensionless radius   r / R = r
 E0 / 0 
1/5 2/5
t
, so that the
pressure inside the fireball can be expressed as P  03/5 E0 2/5t -6/5 f ( ) , where f ( ) is any
function of  .
Similarly, the radial velocity of the fluid inside the fireball can be written as
[u ]  L / T   E0 /  0  t 2/5 / t   E0 /  0  t -3/5 ,
1/5
1/5
so that we can write u (r , t )   E0 / 0  t -3/5 g ( ) , where g is any function of  .
1/5
The density is given by  (r , t )  0 h( ) . Here h( ) is any function of  .
f ( ), g ( ) and h( ) must be obtained by substituting the above dimensionless variables into the
governing equations and solving them.
Normalization (non-dimensionalization) of physical equations
Assume that a physical phenomenon can be described by
where qi (i = 1, 2, …, n) are the physical variables in your governing equations, such as density,
velocity, temperature, etc.
If there are k independent basic physical units such as x, t, m, e, etc., (where m and e are mass and
charge,) then the equation can be written as
where πi (i = 1, 2, …, p) are the p = n − k dimensionless variables consisting of combinations of
qi , e.g.,
where mi are determined by solving the above p equations.
In principle, mi can be any rational number and is not unique.
This is the -theorem (Buckingham, 1915).
But normally, one can use trial and error to get the self-similar variables.
You can try to obtain the R in the Sedov-Taylor problem for the case when the energy input is over
the whole sphere (say, due to burning of the air as the fireball expands) instead of only at the
beginning. Hint: now the relevant variable is the energy input rate dE/dt instead of E.
A trivial example: the simple pendulum
The variables involved are the length l, the mass m, and the gravity g. We are interested in the
period t0 of the pendulum
The basic quantities involved in the motion of the pendulum are the time T, the length L, and the
mass M.
Thus we should have only one (4−3=1) dimensionless parameter.
The units of the dimensional quantities are:
So that [t0 ] = l1/2m0g1/2, or
[t0] = T, [m] = M, [l] = L, [g] = L/T2
t0 = ( l/g)1/2. Note that m does not enter since nothing can match it!
This turns out to be the linear solution for the period of the pendulum.
Nonlinear solutions of the full pendulum problem are obtained by normalizing the time with t0, and
solving the equation of motion.
A not-so-trivial example: energy transfer in gas or liquid
When you shake a tea bottle, the fluid acquires a complex vortex-like rotational structure. How
does this structure evolve when you stop the shaking?
Assume that the fluid is represented by the different-sized vortex-like sub-structures. Shortly after
you stop the shaking there is a period of time in which the sub-structures go from big to small, and
are homogeneously distributed. This roughly means that any part of the fluid is physically similar
to any other part.
Assume that the energy transfer among the different-sized vortices is by a rate of transfer of the
specific (= per unit mass) energy , which has the dimension V2/T or L2/T3.
Let the size r of the vortices be denoted by k=1/r, which has the dimension 1/L.
Let the distribution of the specific energy in the vortices of size k be given by Ek. It has the
dimension [specific energy/k], or V2L or L3/T2.
Then the relevant variables are , k, Ek. We want to express Ek in terms of k and . The basic
parameters are just T and L. Thus, one has only one (3-2=1) dimensionless variable.
Noting that (L2/ has the dimension of T, we can write
[Ek] = L3/T2 = L3/(L2/)2/3=L5/32/3
Therefore the distribution Ek of the specific energy of vortices of different sizes k is given by
Ek = k -5/32/3.
Note that this applies to homogeneous fluids in any state, including turbulent states.
This is the famous Kolmogorov spectrum of energy cascade (1941) in homogeneous turbulence:
the energy flows from large vortices (created by the shaking) to smaller and smaller vortices. The
energy in the very small vortices is dissipated and transformed into heat. That is, the energy is
transferred into even smaller scales: molecular motion.
The scale (k) at which dissipation sets in can also be estimate by dimensional analysis:
This is when the viscosity becomes important. The relevant parameters are now k, and . The
dimension of  can be obtain from the Navier-Stokes equation which is of the form
 t v + ...  ... + 2 v ,
so that
[] = L2/T.
We can then write [] = L2/T3 = L2/(L2/)3 = L- 4/or  k4 Therefore dissipation becomes
important when the vortex size is so small such that
k > kD = (1/43/4.
Note that we cannot include the energy distribution Ek since otherwise there are too many variables
(but there are still only two basic parameters: L and T.)
Guessing the transport coefficients
Here we consider as example the diffusion coefficient appearing in the diffusion equation
 t n  D 2 n
This linear equation has the solution n  n0 exp(k 2 Dt  ik  x ) , which shows that small structures
(large k) diffuse faster. Diffusion means the plasma or gas expands outward and gets lost!
How to guess the diffusion coefficient D by dimensional analysis?
From the diffusion equation, we can see that [D] = L2/T.
We can estimate D by guessing appropriate values for L and T.
1. In neutral gas when collisions are important, we can put L = lmfp and T = 1/, where lmfp= Vth/ is
the mean free path and  is the collision frequencyWe then have
D  L2 / T  lm2 fp  Vth 2/ .
2a. If there is external magnetic field B as well as collisions, we can put L = s and T = 1/ Then
we have
D  D  2L/ T s 2 ( t hV2 / i 2 )
 T/ B2
where T in the last part is the temperature. Earlier magnetic confinement people used this relation
to argue that if the magnetic field is strong enough, diffusion can be made very small.
2b. But if you let L = lmfp and T = 1/i Then you get D  BT / 2 ???
3. If there is strong external magnetic field and weak collisions, like in hot confined plasmas, then
we should put L = s and T = 1/i, and get
T
D  D  L / T  V / i 
M
2
2
th
1
cT
 eB 
,

 
eB
 Mc 
This is the famous Bohm diffusion coefficient. It means that you need very strong magnetic field
in order to make diffusion (or plasma loss) small.
4. If Alfven wave turbulence is dominant, we can use VA as the typical velocity. Then you get
D  B 2
for collisional plasma ???
D  B
for collisionless plasma ???
The diffusion then increases with B, like the case 2b! (Need more detailed analysis.)
Self-similar solutions of the diffusion equation
From the form of the diffusion equation, we can guess that   x 2 / t may be a self-similar
variable. To check, we find
2
x
 t   dt    -   
t
2
2x
2
 2x 
  and  2x        2
t
t
 t 
So that the diffusion equation can be expressed as
 x   d x    
2 n 
  2D
 n  0
4 D
which is an ordinary differential equation (ODE) in the self-similar variable . That is, x and t can
be completely eliminated. This equation can be readily integrated.
Note that in the self-similar space the boundary condition is not arbitrary. It must fit the
self-similar variable.