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Langrangian Density, Wave Equation and Separation of variables, Fluid Mechanics, Dynamics, Perturbation Theory
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Recall the definition of the Lagrangian density as the Lagrangian per unit length.
0
dxL
y, ∂y ∂x
∂y ∂t , x, t
Definition (Canonical momentum density). The canonical momentum den- sity is defined as
P =
(∂y ∂t
Definition (Hamiltonian density). The Hamiltonian density is defined as
H = P ∂y ∂t
1.3.1 Time derivative of H
We seek a conservation equation, so take the time derivative of H,
∂H ∂t
∂t
∂y ∂t
P ∂^2 y ∂t^2
∂y
∂y ∂t
∂ ∂y∂t
∂^2 y ∂t^2
( (^) ∂y ∂x
∂t
∂y ∂x
∂t
∂y ∂t
∂y
∂y ∂t
( (^) ∂y ∂x
∂t
∂y ∂x
Computing the time derivative of P explicitly, ∂ ∂t
∂t
(∂y ∂t
∂y
∂y
( (^) ∂y ∂x
1.3.2 Euler-Lagrange equations
We also have the Euler-Lagrange equations,
d dt
( (^) ∂y ∂x
∂x
( (^) ∂y ∂x
∂y
In higher dimensions,
d dt
∂ ∇~y
∂ ∇~y
∂y
1.3.3 Substituting into ∂ ∂tH
∂t
∂x
( (^) ∂y ∂x
∂y ∂t
where ρ is the mass density and τ is the tension per unit length in the string. The momentum density is
P =
(∂y ∂t
) = ρ ∂y ∂t
The Hamiltonian density is then,
H = P ∂y ∂t
= ρ
∂y ∂t
ρ
∂y ∂t
τ
∂y ∂x
ρ
∂y ∂t
τ
∂y ∂x
S^ ~ is equal to
S = −
∂ ∂y∂x
∂y ∂t = −τ ∂y ∂x
∂y ∂t
To find an explicit form of ~S, we need an expression for y. We expect wave solutions which satisfy ( −ω^2 ←→ t + ←→v
The solutions are of the form
y(x, t) = A cos (kx − ωt)
where the velocity of the waves is given by
v = ω k Computing derivatives, ∂y ∂t
= ωA sin (kx − ωt) ∂y ∂x = −kA sin (kx − ωt)
We have then,
S = A^2 τ ωk sin^2 (kx − ωt) = τ vk^2 A^2 sin^2 (kx − ωt)
Taking the average value,
〈S〉 =
τ A^2 vk^2
Note. It is often the average energy flow that is of interest, and so 〈S〉 is the more useful quantity.
2 Wave equation and separation of variables
The discussion of waves on a massive string leads naturally to the discussion of waves. Systems involving wave modes are almost always solved using separation of variables. The technique is powerful and general.
The general form of the wave equation is
v^2
∂t^2
where v is the phase velocity of the wave. Assume the solution is separable into two independent functions, one only of position and the other only of time.
Ψ(x, t) = ψ(t)χ(t)
With this assumption, take derivatives to insert into the wave equation, ∂Ψ ∂t
= ψ(x) ∂χ ∂t ∂^2 Ψ ∂t^2 = ψ(x) ∂^2 χ ∂t^2
∂x^2
ψ =
∂x^2
∂y^2
ψ = X
∂y^2 The Helmholtz equation becomes ∂^2 X ∂x^2
∂y^2
As usual, divide by ψ = XY ,
1 X
∂x^2
∂y^2
The second and third terms in the above equation are independent of x. Thus we can define an effective wave number which is constant in the x coordinate.
k x^2 =
∂y^2
With this substitution, ∂^2 X ∂x^2
This is the familiar differential equation of simple harmonic oscillation. Now apply the boundary conditions to quantize the X wave number. ψ(x, y, t) = X(x)Y (y)eıωt X(x) = A cos kxx + B sin kxx X(0) = 0 ⇒ A = 0 X(L) = 0 ⇒ kxL = nπ X(x) = B sin
(nπ L
x
In the same way, kyL = mπ
And by the way in which k is defined,
k =
n^2 π^2 L^2
m^2 π^2 L^2
ω v The general solution is of the form ψ(x, y, t) = A sin
(nπ L x
sin
(mπ L y
eıωt
The same separation of variables procedure works in other coordinate sys- tems. The method requires the Laplacian operator in the other coordinate system. For example, in spherical coordinates,
∇^2 ψ =
r
∂r
r ∂ψ ∂r
r^2
∂^2 ψ ∂θ^2
∂^2 ψ ∂z^2
Assume a solution of the form
ψ(r, θ, z) = R(r)Θ(θ)Z(z) The Helmholtz equation becomes ∂^2 R ∂r^2
r
∂r
r^2
∂θ^2
∂z^2 RΘ + k^2 RΘZ = 0
Dividing by r 2 RΘZ , ( r^2 R
∂r^2
r R
∂r
∂θ^2
r^2 Z^2
∂z^2
The second term is only a function of θ, so write the rest of the equation is m^2 , which is a constant with respect to θ. The modes are those of a simple harmonic oscillator,
Θ(θ) = C cos mθ + D sin mθ Divide the remaining equation by r^2 , 1 R^2
∂r^2
rR
∂r
m^2 r^2
∂z^2
As in the case of the θ coordinate, define the rest of the equation which is independent of z to be a constant, −n^2. Because of the sign of n, the solutions for z are exponential
Z(z) = Ee−nz^ + F enz The solution to the remaining R component is of the form of a Bessel function. ∂^2 R ∂r^2
r
∂r
n^2 + k^2 − m^2 r^2
Note. Bessel functions are like surface waves on a cup of coffee that are peaked in the center and confined to be 0 at the boundary.
3.2.2 External forces
There could also be external forces acting on the differential cube of fluid. These forces can be written in terms of the density,
fextρdV
where fext is the force per unit density per unit volume.
3.2.3 Mechanical equilibrium
If the fluid is in equilibrium, Newton’s laws ensure that there is no net force on the cube. ( ρ f~ext − ∇~p
dV = 0
The condition for hydrostatic equilibrium is then
f^ ~ext = − 1 ρ ∇~p
3.3.1 Convective derivative
To obtain dynamics, we need to insert an inertial term into the equation for the balance of forces above. The difficult part is to find d~dtv , since the volume of any given mass element is not constant as the fluid changes shape. We seek d~ dtvi , the acceleration of the ith^ mass element.
d~vi dt
∂vi ∂x
dx dt
∂vi ∂y
dy dt
∂v ∂z
dz dt
∂vi ∂t
This is called the convective derivative,
d~v dt
∂~v ∂t
~v · ∇~
~v
Newton’s second law for fluids then becomes
∂~v ∂t
~v · ∇~
~v = f~ext −
ρ
∇~p
3.3.2 Continuity equation
We assume that the amount of fluid in the system is constant.
∂ρ ∂t
From the continuity equation along with the expression of Newton’s laws above, it is possible to derive conservation of energy and momentum.
3.3.3 Kinetic description
The continuous description of the fluid employed above is called the fluid description of a fluid. It is also possible to obtain a kinetic description of the fluid by treating it as a number of point particles.
f (x, v, t) =
unit space unit time Integrating,
f (v) = f 0 e−^ vv^2 th
where vth is like a thermal characteristic speed,
vth =
2 kT m We then have (^) ∫
v^2 f (v)dvxdvydvz = ρ
3.4.1 Linearize the equations
The usual procedure for fluid equations is to linearize the equations by as- suming the presence of constant parameters perturbed by a small amount. The equations are linearized if we keep only first order correction terms. Assume the parameters of the system are given in terms of constants ( terms) plus perturbations (1 terms).
ρ = ρ 0 + ρ 1 v = v 0 + v 1 p = p 0 + p 1
The derivative of all of the constants is 0, so the derivative of any of the above quantities is simply the derivative of the perturbation.
