Central Forces

Question 1

For planar motion in a central field, what is the angular momentum L equal to?

Answer 1

L = r x p, where r & p define a plane

Question 2

What is the Lagrangian equal to for planar motion?

Answer 2

L = 1/2 m(r’^2 + r^2 *θ’^2) - V(r)

Question 3

What does the canonical momenta p(θ) equal for the planar motion? What is the case for circular and elliptical orbit?

Answer 3

p(θ) = dL/dθ’ = mr^2 *θ’

Circular orbit: r fixed, θ’ const, elliptical orbit: r varies, θ’ varies to make r^2 *θ’ const

Question 4

How do we find equations for the central and centrifugal forces for the planar motion?

Answer 4

• Use E-L equation in r, and rearrange to make m*r’’ the subject
• dV/dr term if the central force and mr*θ’^2 term is the centrifugal force (not real force, consequence of the rotational motion)

Question 5

How can we write the centrifugal force in terms of the angular momentum?

Answer 5

Fc = |L|^2/(m*r^3) = -dUc/dr, where Uc(r) is a potential energy

Question 6

How can we write the equation of motion for planar motion in terms of Uc(r)?

Answer 6

m*r’’ = -d/dr(Uc + V) = -d/dr Ueff, where Ueff is the effective potential energy.= Uc + V

Question 7

What is the equation for the Hamiltonian in terms of Ueff for planar motion?

Answer 7

H = p(r)^2/2m + Uc(r) + V(r) = p(r)^2/2m + Ueff(r) = E

Question 8

What is the equation for p(r)?

Answer 8

p(r) = sqrt(2m*(E-Ueff(r)))

Question 9

What is the equation for E if we look at p(r) = 0 and what can we learn from this?

Answer 9

E-Ueff(r) = 0. From this we learn that there is no radial motion when p(r) = 0 - closest or furthest approach in an orbit

Question 10

What does a sketch of Ueff look like?

Answer 10

Like the strong nuclear force graph - high at first then comes down past x axis then back up and tends to -0.

Question 11

On the Ueff graph, where are E1, E2 and E3 and their corresponding radii?

Answer 11

y-axis is E, x-axis is r, E1 is at start, E2 is halfway between crossing x-axis and tunring point, and E3 is just after turning point. r values can be read off from that.

Question 12

What happens at E = 0, E=E2 and E = E3?

Answer 12

E=0 is parabola, E=E2 has 2 rotts, r2 and r2’ where p(r) = 0, E=E3 is a circular orbit, here 2 repeated roots at r = r3

Question 13

For 2 masses, what do we make the Lagrangian a function of?

Answer 13

The centre of mass coordinates: L(x0, y0, r, θ, x0’, y0’, r’, θ’), where θ is the angle from horizontal the line joining the masses is and (x0, y0) is the centre of mass

Question 14

What is the equation for the kinetic energy of the centre of mass for the 2mass system?

Answer 14

T(CoM) = 1/2 (2m)(x0’^2 + y0’^2)

Question 15

What is the equation for the potential energy of the 2 mass system?

Answer 15

V = 1/2 k(r-r0)^2, where r0 is the equilibrium value of r

Question 16

What is the kinetic energy o the 2 mass system relative to the CoM?

Answer 16

T(rel) = 1/4mr^2 θ’^2 + 1/4m*r’^2

Question 17

What is the equation for the Lagrangian of the 2 mass system?

Answer 17

L = m(x0’^2 + y0’^2) + 1/4m(r’^2 + r^2 θ’^2) - 1/2k(r-r0)^2

Question 18

What do we find if we look at the E-L equation in x0 for the 2 masses connected by a spring system?

Answer 18

No x0 dependence in L, so total momentum along x-direction is conserved.

Question 19

What do we find if we look at the E-L equation in y0 for the 2 masses connected by a spring system?

Answer 19

Momentum along y-direction is conserved.

Question 20

What do we find if we look at the E-L equation in r for the 2 masses connected by a spring system?

Answer 20

Find equation of motion r’’ = -2k/m (r-r0) +rθ’^2

Question 21

What do we find if we look at the E-L equation in θ for the 2 masses connected by a spring system?

Answer 21

Find that angular momentum is conserved

Question 22

If we set the angular momentum θ’ to zero, what do we find for the 2 mass system?

Answer 22

Motion split into several ‘modes’: 2 translational modes up and across, 1 vibrational mode (side to side) and 1 rotational mode

Question 23

What are normal modes?

Answer 23

A pattern of motion where all particles are moving with same frequency.

Question 24

How can we build up the overall motion of the system?

Answer 24

By using a superposition of normal modes.

Question 25

How do we determine how many normal modes we have?

Answer 25

If we have n generalised coordinates, we have n normal modes with n frequencies (in total).

Question 26

What is a good example of normal modes?

Answer 26

2 coupled oscillators (2 masses, 3 springs, connect to walls either side.

Question 27

What is the Lagrangian for the 2 coupled oscillator problem?

Answer 27

L = 1/2mx1’^2 + 1/2mx2’^2 - (1/2kx1^2 + 1/2kx2^2+1/2*k12(x1-x2)^2)

Question 28

Which 2 variables for we apply the E-L equation to for the 2 coupled oscillator problem?

Answer 28

x1 and x2.

Question 29

What solutions do we try for x1 and x2 for the 2 coupled oscillator problem?

Answer 29

x1(t) = A1exp(iωt), x2(t) = A2exp(iωt)

Question 30

What do we do with the trial solutions for x1 and x2 for the 2 coupled oscillator problem?

Answer 30

Sub them into E-L equations and rearrange to make A1 the subject in each. Then set these equal to eachother and rearrange to fine ω1 and ω2.

Question 31

How do we compare the motion of ω1 and ω2?

Answer 31

Use equation found for A1 (relative displacement of particle 1 & 2), and find that ω1 is symmetric mode (moves in same direction) whereas ω2 is anti-symmetric mode (moves towards eachother)

Question 32

What is another expression for T to find the inertia matrix?

Answer 32

T = 1/2 sum over i,j of qi’mij*qj’, where mij = m = (m11, m12, m21, m22) = 4x4 matrix

Question 33

What can we write mij as? What does this mean?

Answer 33

mij = d^2T/dqi’dqj’ = d^2T/dqj’dqi’ = mji -> matrix m is symmetric about diagonal: is the inertia matrix

Question 34

What is a second example we can use for the inertia matrix?

Answer 34

Mass m2 moving at x2’ to a mass m1 (moving at x1’), so T = 1/2m1x1’^2 + 1/2m2(x2’-x1’)^2

Question 35

How do we find the inertia matrix M from T?

Answer 35

Using d^2T/dqi’dqj’, so m11 = d^2T/dx1’^2, m12 = d^2T/dxq’dx2’, m22 = d^2T/dx2’^2

Question 36

What does M equal for the two masses moving problem?

Answer 36

M = (m1+m2, -m2, -m2, m2) = matrix