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PHYS 223 Notes
Max Christie
Newtonian Mechanics
Newton’s First Law - The Law of Inertia If a particle does not interact with other objects then it is possible to identify inertial reference frames where it moves with constant velocity. Newton’s Second Law - Force is proportional to the change in momentum: F = dpdt Newton’s Third Law - Action/Reaction If an object 1 exerts a force F 21 on an object 2, then object 2 exerts a force F 12 on 1 which fulfils F 12 = −F 21 Conservative Forces and Potential Energy If the force on a particle depends only on the particle position and the amount of work done to move the particle between any two points is the same, then that force is conservative. A conservative force can be expressed as:
F = −
∑^ n
i=
δV δxi
The work done by a conservative force is:
W =
∫ (^) r 2
r 1
F dr = −
∫ (^) r 2
r 1
∇V dr = V (r 1 ) − V (r 2 )
The work done by a conservative force can also be expressed by:
W =
∫ (^) r 2
r 1
F dr =
∫ (^) t 2
t 1
F v(t)dt = m
∫ (^) t 2
t 1
vvdt ˙ =
∫ (^) t 2
t 1
d dt
[ 1
2 mv
(^2) (t)
]
dt = T (t 2 ) − T (t 1 )
Which means that T 1 + V 1 = T 2 + V 2
Calculus of Variations
A functional is defined by
I[y] :=
∫ (^) x 2
x 1
F [y(x), y′(x), x]dx
If y(x) is an extremal function then curves in the vicinity of y(x) can be written as
yˆ(x) = y(x) + η(x)
with η(x 1 ) = η(x 2 ) = 0 Inserting this into the functional gives
Iη() =
∫ (^) x 2
x 1
F [y + η, y′^ + η′, x]dx
To have an extremum = 0, so dIη d
So dIη d
∫ (^) x 2
x 1
[
η δF δy
]
dx =
∫ (^) x 2
x 1
[ (^) δF δy
− d dx
δF δy′
]
ηdx = 0
Since this must hold for all η(x), this leads to the Euler-Lagrange Equation:
d dx
δF δy′^
− δF δy
Lagrangian Mechanics
Advantages of Lagrangian Mechanics
- Lagrange equations take the same form in any coordinate system.
- Knowledge of the forces due to constraints is not required.
- Setting up the equations of motion is often easier than with Newton’s second law.
For a system of particles subject only to holonomic constraints and conservative forces, we define the Lagrangian as L(q, q, t˙ ) = T − V
The motion of this system from time t 1 to t 2 is such that the action integral below is stationary
S =
∫ (^) t 2
t 1
L(q, q, t˙ )dt = 0
Which gives the Lagrange equations d dt
δL δ q˙i^ −^
δL δqi^ = 0 Cartesian Coordinates L =
2 m( ˙x
(^2) + ˙y (^2) + ˙z (^2) ) − V (x, y, z)
Cylindrical Coordinates
~r =
ρ cos ϕ ρ sin ϕ z
(^) = ~r(ρ, ϕ, z)
L =^1
m( ˙ρ^2 + ρ^2 ϕ˙^2 + ˙z^2 ) − V (ρ, ϕ, z)
Spherical Coordinates
~r =
x y z
r sin θcosϕ r sin θ sin ϕ r cos θ
(^) = ~r(r, θ, ϕ)
L =^1
m( ˙r^2 + r^2 θ˙^2 + r^2 ϕ˙^2 sin^2 θ) − V (r, θ, ϕ)
The real part of this solution - what we care about - is
xn = ~an cos (ωnt − δn)
The general real solution takes the form
x(t) =
n
cnxn
We can also use the matrix notation to represent kinetic and potential energies before the equations of motion are found. These take the general forms
T =^1 2
jk
Mjk q˙j q˙k
with Mjk := Ajk(0)
and V =^1 2
jk
Kjkqj qk
with
Kjk := δ
2 V
δqkδqj
The Lagrangian takes the form L = T − V
Which gives equations of motion of the form similar to above
M q¨ = −Kq
Hamiltonian Mechanics
The Legendre Transform is defined as
g(μ, y) := f (x(μ, y), y) − μx(μ, y)
And the differential of this is dg = df − μdx − xdμ
But μ = δf δx
and df = δf δx
dx + δf δy
dy
So dg = df − δf δx
dx − xdμ
dg =
δf δy dy^ −^ xdμ
Because g is a function of μ and y, we also have
dg =
δg δy dy^ +^
δg δμ dμ
And by comparison, we get
x = − δg δμ
and δf δy
= δg δy
The Hamiltonian is defined as
H :=
∑^ n
i=
q ˙iPi − L(q, q, t˙ )
Which can be thought of as the negative of the Legendre Transform with H ∼ g and L ∼ f. dH can then be found by the same process
dH =
∑^ n
i=
q ˙idPi + Pid q˙i −
δL δqi^ dqi^ −^
δL δ q˙i^ d^ q˙i
δL δt dt
But δL δ q˙i
= Pi and δL δqi
= P˙i
So
dH =
∑^ n
i=
q ˙idPi − P˙idqi
− δL δt
dt
Because H is a function of qi, Pi, and t, we also have
dH =
∑^ n
i=
( (^) δH δqi^ dqi^ +^
δH δPi^ dPi
δH δt dt
And by comparison, we get Hamilton’s equations
q˙i = δH δPi
and P˙i = − δH δqi
and − δL δt
= δH δt
Poisson Brackets A differentiable function f (q, P, t) is an observable
df dt =^
i
( (^) δf δqiq^ ˙i^ +^
δf δPi P^ ˙i
δf δt dt
Inserting Hamilton’s equations yields
df dt
i
( (^) δf δqi
δH δPi
− δf δPi
δH δqi
dt
The Poisson Bracket is defined as
[f, g]q,P =
i
( (^) δf δqi
δH δPi^ −^
δf δPi
δH δqi
And Hamilton’s Equations can be written in this form
q˙i = [qi, H] and P˙i = [Pi, H]
The isotherms that come about from this equation look like:
The First Law of Thermodynamics The change in internal energy of a system from one state to another state will be the same regardless of the process undergone to get between the states. In equation form, this is dU = dQ + dW
Also, dU = CV ∆T
W = −
P dV
Enthalpy is defined as H = U + P V
An isothermal change is one in which the internal energy U of the system remains unchanged. An adiabatic change is one in which no heat enters or leaves the system. This gives a relationship derived below: dU = dQ − dW ⇒dQ=0 CV dT = −P dV
CV dT = −nRT dV V ∫ (^) Tf
Ti
dT T
= − nR CV
∫ (^) Vf
Vi
dV V
ln
Tf Ti
= − nR CV
ln
Vf Vi
Tf V (^) fnR/C V = TiV (^) inR/CV nR CV^ =^
CP
CV^ −^ 1 =^ γ^ −^1 T V γ−^1 = constant
With the ideal gas law this can be manipulated to
P V γ^ = constant
The Second Law of Thermodynamics There are two equivalent definitions for this law. The first is that a cycle that produces work using only one heat reservoir is impossible. The second is that a cycle that transfers heat from a cold reservoir to a hot reservoir with no work being done is impossible. The efficiency of a cycle is
η = (^) QW 1 Because in a cycle the initial and final states are the same, ∆U = 0, which means that
|Q 1 | = |Q 2 | + W
So the efficiency can also be written as
η =
|Q 1 | − |Q 2 |
|Q 1 | = 1^ −
|Q 2 |
|Q 1 |
The Carnot cycle involves two isothermal changes and two adiabatic changes. The efficiency of this cycle can be shown to be
η = 1 − T^2 T 1 Carnot’s Theorem states that no engine operating between two temperatures can be more efficient than a Carnot engine operating between the same temperatures. He reaches the conclusion by considering a scenario with two heat engines operating between the same two temperatures and showing that if one has a greater efficiency than the other then no work is done and heat is transferred from the cold reservoir to the hot, which is forbidden by the second law. Entropy The change in entropy of a system is defined as
dS = dQ T
For an adiabatic or cyclic process, ∆S = 0
For a isothermal change,
∆S = nR ln
Vf Vi
For an isobaric change,
∆S = CP ln Vf Vi
Kinetic Theory of Gases
P V =^13 N mv^2 =^23 N 12 mv^2
nRT =
3 N^
2 mv
2
1 2
mv^2 =^3 2
kB T
The flux of molecules on a wall is φ =
4 nv The mean free path length is
λ =
√^1
2 πσ^2 n
RT
P NA
√^1
2 πσ^2 Thermal conductivity is
k =
3 nCV^ λv Dynamic viscosity is
η =^13 ρλv
Diffussion coefficient is η =^1 3
λv
The Equipartition of Energy Any degree of freedom has an energy of 12 kB T associated with it. The Boltzmann Distribution Consider a container with a movable lid. The lid increases in height by dh, causing the pressure to decrease by dh. The pressure upwards on the gas layer between the new lid height and the old is balanced by the gravitational force downwards on the layer
dP = −nAdh
mg A =^ −nmgdh
Kinetic theory tells us that
P =^13 nmv^2
So deriving this gives
dP =^1 3
dnmv^2
And equating the two gives
−nmgdh =
3 dnmv
2
Then rearranging dn n =^
−mg 1 3 mv^2
dh
We have also seen that RT = NA
2 mv
2
And know that NAm = Mr
So substituting these in gives dn n
= −Mrg RT
dh
Integrating results in
n = n 0 exp(^
−Mr ghRT ) = n 0 exp(^
−kmgh BT^ )
This can be generalised to show the relationship for any distribution of molecules between two energy states n 1 n 2
= exp(^
−k∆ε BT^ )
To make this a probability distribution, we think of the fraction of energies between ε and ε + dε, which is given by dn n =^ A^ exp
( (^) k−BT ) (^) dε
The fraction of molecules with speeds between v and v + dv is given by
dn n
= B exp(^
− 2 kmv^2 BT^ )^4 πv^2 dv
Because dn/n is the fraction of molecules with speeds between v and v + dv, integrating over all speeds must result in 1
1 =
0
B exp(^
− 2 kmv^2 BT^ )^4 πv^2 dv
To solve this, let
x = v
m 2 kBT
We then have 1 = 4πB
( (^2) kB T m
) 3 / 2 ∫^ ∞
0
x^2 exp−x^2 dx
Consider I =
0
exp−x^2 dx
Then I^2 =
0
0
exp−(x^2 +y^2 )^ dxdy
Let x^2 + y^2 = r^2
Then express the equation in polar coordinates
I^2 =
∫ π 2
0
0
exp−r^2 rdrdθ
Which integrates to
I =^12
π
With I known, the original integral can be integrated by parts. B can then be solved and is found to be B =
( (^) m 2 πkB T
