Legendre Transformations - Essay - Physics
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Legendre Transformations - Essay - Physics

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The time average of P i piri is 0. The quantity could be periodic, as in the case of central potentials. Otherwise, assume it is bounded, so its time average is 0. It is not meaningful to talk about the time average of a...
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Legendre Transformations

Adrian Down

October 20, 2005

1 Virial theorem

1.1 Motivation

The Virial theorem is a statistical relationship that relates long-term averages of T and V .

1.2 Setup

Consider

d

dt

(∑ i

piri

)

Assume a conservative force, so that ~p can be written,

~̇p = −~∇V

From the chain rule,

d

dt

(∑ i

piri

) = ∑

piṙi + ∑ i

ṗiri

Make the substitutions,

ṙ = v p = mv

1

The equation becomes,∑ piṙi = mv

2 = 2T

⇒ 2T = d dt

(∑ i

piri

) − ∑ i

ṗiri

1.3 Long-term averages

Long term averages are defined as

f = lim τ→∞

1

τ

∫ τ 0

f(t)dt

Using the fundamental theorem of calculus, if f(t) = dF dt , then

f = lim τ→∞

1

τ

∫ τ 0

dF

dt dt = lim

τ→∞

F (τ)− F (0) τ

Note. • If f(t) is bounded, then f → 0.

• If the system is periodic with period τ , then F (τ) = F (0).

1.4 Result

The time average of ∑

i piri is 0. The quantity could be periodic, as in the case of central potentials. Otherwise, assume it is bounded, so its time average is 0. It is not meaningful to talk about the time average of an unbounded system, so the Virial theorem would not be applicable.

2T = − ∑ i

ṗiri = ∑ i

ri

( ~∇V ) i

2T = ∑ i

ri

( ~∇V ) i

Note. • The terms on the right are the components of the force.

• The force must be conservative for a potential function to exist.

2

1.5 Central forces

Applying the virial theorem to the case of a central force,

V = αrn

~∇V = ∂V ∂r

r̂ = αnrn−1r̂

~r · ~∇V = αnrn = nV 2T = nV

In the case of gravity, n = −1.

2T + V = 0

2 Legendre Transformations

2.1 Motivation

We have been considering the Lagrangian,

L(q, q̇, t) = T − V

In Hamiltonian mechanics, we consider

H = ∑ i

piq̇i − L

The general method discussed before is to compute the canonical momenta,

pi = ∂L

∂q̇i

Invert these equations to get

pi = ∂L

∂q̇i

The Hamiltonian can then be written in terms of q, p, and t by substitution.

H = ∑ i

piq̇i(q, p, t)− L(q, q̇(q, p, t), p, t)

3

q and p are then assumed to be independent variables. However, it is not clear that they are really independent.

The Legendre transformation is a mapping from the graph of a function to a family of tangents. The parameters that describe the tangent curve are the slope and the y-intercept. If the slop is written as y = mx + b, the the Legendre transformation is a mapping,

y(x) 7→ b(m)

Note. • The new expression for y(x) is completely equivalent to the orig- inal expressed in different variables.

• The function being transformed must be differentiable.

• The Legendre transformation is used in thermodynamics to convert between variables such as entropy and enthalpy.

Example.

y(x) = ex

m = dy

dx = ex

⇒ x = ln x b = y −mx = y −m lnm

b(m) = m−m lnm

2.2 General Case

Consider a differentiable function f(x, y). The derivative is

df = ∂f

∂x dx+

∂f

∂y dy

= udx+ vdy

where

u = ∂f

∂x v =

∂f

∂y

Our goal so to change variables to get an expression of the form,

f(x, y) 7→ f(u, y)

4

Consider the function,

g = f − ux

The derivative is,

dg = df − udx− xdu =udx+ vdy −udx− xdy = vdy − xdu

If g is a function of the form g(y, u), the derivative would be,

dg = ∂g

∂y dy +

∂g

∂u du

These two expressions for the derivative will be equal if we make the following identifications,

∂g

∂y = v

∂g

∂u = −x

2.3 Application to the Hamiltonian

We want to use the Legendre transformation to get a a transformation,

L(q, q̇) 7→ H(q, p)

As above, take the derivative of H,

dL = ∂L

∂qi dqi +

∂L

∂q̇i dqi +

∂L

∂t dt

Define the quantity,

pi = ∂L

∂qi

The physics comes from the Euler-Lagrange equations, which give

d

dt

( ∂L

∂q̇

) =

∂L

∂q

5

Substitute for the momentum,

dL = ṗidqi + pidq̇i + ∂L

∂t td

Now introduce the function,

H(q, p, t) = q̇ipi − L(q, q̇, t) The derivative of H is

dH = q̇idpi + pidqi − dL

= q̇idpi + pidq̇i −

( ṗidqi +

pidq̇i + ∂L

∂t td

) = q̇idpi − ṗidqi −

∂L

∂t dt

From the definition of H, the derivative is

dH = ∂H

∂qi dqi +

∂H

∂pi dpi +

∂H

∂t dt

The Legendre transformation holds if we can make the assignments,

ṗi = − ∂H

∂dqi q̇i =

∂H

∂pi

∂H

∂t = −∂L

∂t

2.4 General Lagrangian method

1. Choose coordinates and construct the Lagrangian.

L(q, q̇, t) = T − V

2. Compute the generalized momenta.

pi = ∂L

∂q̇i

3. Construct the Hamiltonian.

H(q, q̇, p, t) = q̇ipi − L(q, q̇, t)

4. Invert the expression for pi to get q̇(q, p, t).

5. Eliminate q̇ from H to get H(q, p, t).

6. Apply Hamilton’s equations of motion to get the dynamics of the sys- tem in phase space.

6

3 Phase space

3.1 Motivation

Phase space has some properties of fluid dynamics which give useful conser- vation properties. We investigate how phase space behaves like an incom- pressible fluid.

3.2 Flux of states

Consider a box in phase space with area A, width ∆q and height ∆p. Assume the density of points in phase space is ρ, which could be a function of q, p, and t. We want to know the change in the number of phase space points in the this box in a time dt.

Consider a vertical strip of width δq at the left vertical side. The number of points entering the left vertical side of the box is given by the number of points in this area.

δq = q̇dt

⇒ dN1 = ρδq∆p = ρq̇|q∆pdt

The number of points entering from the right vertical face is found in the same way, only with the sign reversed,

dN2 = −ρq̇|q+∆q∆pdt

The total number of points entering the area on the vertical sides is given by

dN12 = dN1 − dN2 = dt∆p (ρq̇|q − ρq̇|q+∆q)

The quantity in parentheses can be written using the partial derivative,

∂q (ρq̇) =

ρṗ|q − ρṗ|q+∆q ∆q

⇒ dN12 = ∂

∂q (ρq̇)∆p∆qdt

The same arguments can be used on the horizontal sides,

dN34 = − ∂

∂p (ρṗ)∆p∆qdt

7

The change in the number of points resulting from the change in time of the density is,

dNt = ∂ρ

∂t dt∆q∆p

Thus the number of points lost or gained during flow through the box is equal to the change in density inside the box. This is a statement of continuity. Mathematically, it is equivalent to,(

− ∂ ∂q

(ρq̇)− ∂ ∂p

(ρṗ)

) ∆p∆qdt =

∂ρ

∂t ∆p∆qdt

⇒ ∂ρ ∂t

+ ∂

∂q (ρq̇) +

∂p (ρṗ) = 0

This is a continuity equation. Defining the quantity

~J = ρ~v

the continuity equation can be written

∂ρ

∂t + ~∇ · ~J = 0

The continuity equation implies that area in phase space is conserved.

8

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