Phys 7221 Homework #2 Gabriela Gonz´alez September 18, 2006

1. Derivation 1-9: Gauge transformations for electromagnetic potential If two Lagrangians differ by a total derivative of a function of coordinates and time, they lead to the same equation of motion: L0 (qi , q˙i , t) = L(qi , q˙i , t) +

∂F X ∂F dF (qi , t) = L(qi , q˙i , t) + + q˙j dt ∂t ∂qj j

as proven in the following lines: ∂L0 ∂F ∂L0 = + ∂ q˙i ∂ q˙i ∂qi ∂L0 ∂L ∂ dF = + ∂qi ∂qi ∂qi dt d ∂L0 ∂L0 − dt ∂ q˙i ∂qi

=

d ∂L ∂L d ∂F ∂ dF − + − dt ∂ q˙i ∂qi dt ∂qi ∂qi dt 

d ∂F dt ∂qi



∂ ∂  ∂F =  + q˙j ∂t ∂qj ∂qi j   ∂  ∂F X ∂F  = + q˙j ∂qi ∂t ∂qj X

j

=

∂ dF ∂qi dt

and thus the equations of motion are the same, derived from either Lagrangian: d ∂L0 ∂L0 d ∂L ∂L − = − dt ∂ q˙i ∂qi dt ∂ q˙i ∂qi 1

(What goes wrong if F = F (q, q, ˙ t)?) A particle moving in an electromagnetic field, with scalar potential φ and vector potential A, has a generalized potential function U = φ − A · v and a Lagrangian 1 L = T − U = mv 2 − q (φ − A · v) 2 Consider a gauge transformation for the scalar and vector potentials of the form φ0 = φ −

∂ψ(r, t) ∂t

A0 = A + ∇ψ(r, t) The change in potential energy under this transformation is U 0 = φ0 − A0 · v   ∂ψ = U −q + ∇ψ · v ∂t   ∂ψ dr = U −q + ∇ψ · ∂t dt   X ∂ψ ∂ψ dxi = U −q + ∂t ∂xi dt dψ = U −q dt and thus, the new Lagrangian differs from the original one by a total derivative: L0 = T − U 0 = T − U + q

dψ d(qψ) =L+ dt dt

and thus we know the equations of motion are the same. We can also see that the electromagnetic forces (and thus Newton’s equations) are invariant. The forces are derived from the potential, following (1.58): Qi = −

∂U d ∂U + ∂qi dt ∂ q˙i

If U = q(φ(r, t) − A(r, t) · v), then the force is ∂ d ∂ (φ − A · v) + q (φ − A · v) ∂xi dt ∂vi ∂ dAi ∂φ = −q +q (A · v) − q ∂xi ∂xi dt

Qi = −q

2





∂φ ∂ ∂ ∂  +q (A · v) − q  + x˙ j A ∂xi ∂xi ∂t ∂xj j   ∂A + q∇(A · v) − q(v · ∇)A F = −q ∇φ + ∂t   ∂A = −q ∇φ + + qv × (∇ × A) ∂t = q(E + v × B) X

= −q

and, as we know, the electric and magnetic fields are invariant under the gauge transformation: ∂A0 E0 = −∇φ0 − ∂t ∂ψ ∂ = −∇(φ − ) − (A + ∇ψ) ∂t ∂t ∂A ∂ψ ∂ = −∇φ − +∇ − ∇ψ ∂t ∂t ∂t ∂A = −∇φ − ∂t = E B0 = ∇ × A0 = ∇ × (A + ∇ψ) = ∇ × A + ∇ × ∇ψ = ∇ × A = B 2. Problem 1-15: A potential for spin interactions Consider a point particle moving in space under the influence of a force derivable from a generalized potential: U (r, v) = V (r) + σ · L We can derive the generalized forces using (1.58): Qi = −

d ∂U ∂U + ∂qi dt ∂ q˙i

using either Cartesian coordinates (x, y, z) or spherical polar coordinates (r, ψ, θ) for generalized coordinates qi . The first choice will produce the components of the force vector Fi ; the generalized forces for any other choice of generalized coordinates (such as spherical coordinates) will be related to the force vector as in (1.49): X ∂rj Qi = Fj · ∂qi j

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This problem is a marathon in vector calculus in different coordinate systems. We first calculate the force vector using Cartesian coordinates: U

= V (r) + σ · L = V (r) + σ · (r × mv) X = V (r) + m σl (r × mv)l l

= V (r) + m

X

σl

X

lmn xm x˙ n

m,n

l

= V (r) + m

X

lmn σl xm x˙ n

l,m,n

∂U ∂ x˙ i

= m

d ∂U dt ∂ x˙ i

= m

∂U ∂xi

X

lmi σl xm

l,m

X

lmi σl x˙ m = (σ × p)i

l,m

= =

X ∂V (r) +m lin σl x˙ n ∂xi ln X dV (r) ∂r −m iln σl x˙ n dr ∂xi ln

dV (r) xi − (σ × p)i dr r ∂U d ∂U Fi = − + ∂qi dt ∂ q˙i dV (r) F = − eˆr + 2(σ × p) dr =

Now we want to calculate the generalized forces in spherical polar coordinates r, ψ, θ. We first need to express the potential in terms of the coordinates and their derivatives. ˆr , e ˆψ , e ˆθ , which are themselves functions of the coordinates We will use unit vectors e and time: ˆr = sin ψ cos θˆi + sin ψ sin θˆj + cos ψ kˆ e ˆψ = cos ψ cos θˆi + cos ψ sin θˆj − sin ψ kˆ e ˆθ = − sin θˆi + cos θˆj e ˆi × e ˆj = ijk e ˆk but they are not constant: in They are orthogonal and cyclical: e general ∂ˆ ei /∂qj 6= 0, and also dˆ ei /dt 6= 0. 4

We first calculate the angular momentum in spherical coordinates: r = rˆ er d ˙ eψ + r sin ψ θˆ ˙ eθ v = (rˆ er ) = rˆ ˙ er + rψˆ dt L = r × p = r × mv ˙ eψ + r sin ψ θˆ ˙ eθ ) = mrˆ er × (rˆ ˙ er + rψˆ ˙ eθ − sin ψ θˆ ˙ eψ ) = mr2 (ψˆ so the potential, in terms of generalized coordinates r, ψ, θ is ˙ eθ − sin ψ θˆ ˙ eψ ) U = V (r) + σ · L = V (r) + mr2 σ · (ψˆ We can now calculate the generalized forces: d ∂U ∂U ∂U + =− ∂r dt ∂ r˙ ∂r dV ˙ eθ − sin ψ θˆ ˙ eψ ) = − − 2mrσ · (ψˆ dr dV 2 dV = − − σ·L=− + 2ˆ er · (σ × p) dr r dr

Qr = −

∂U d ∂U + ∂ψ dt ∂ ψ˙ ∂ ˙ ˙eψ ) + d (mr2 σ · e ˆθ ) = −mr2 σ · (ψˆ eθ − sin ψ θˆ ∂ψ dt ˙eψ + sin ψ θˆ ˙er ) + 2mrrσ ˙ ˆθ − mr2 σ · θ(sin = −mr2 σ · (− cos ψ θˆ ˙ ·e ψˆ er + cos ψˆ eψ ) 2 ˙ ˆr + 2mrrσ ˆθ = −2mr sin ψ θσ · e ˙ ·e ˙ = 2mrσ · (rˆ ˙ eθ − r sin ψ θˆ er )

Qψ = −

= −2mrσ · (ˆ eψ × v) = 2rˆ eψ · (σ × p)

d ∂U ∂U + ∂θ dt ∂ θ˙ ∂ ˙ d ˆψ ) = −mr2 σ · (ψˆ eθ − θ˙ sin ψˆ eψ ) + (−mr2 sin ψσ · e ∂θ dt ˙ sin ψˆ = −mr2 σ · (ψ(− er − cos ψˆ eψ ) − θ˙ sin ψ(cos ψˆ eθ ))

Qθ = −

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˙ er + θ˙ cos ψˆ ˆψ + r2 ψ˙ cos ψσ · e ˆψ + r2 sin ψσ · (−ψˆ −m(2rr˙ sin ψσ · e eθ )) 2 ˙ ˆr − 2mrr˙ sin ψσ · e ˆψ = 2mr ψ sin ψσ · e ˙ er − rˆ = 2mr sin ψσ · (rψˆ ˙ eψ ) = −2mr sin ψσ · (ˆ eθ × v) = 2r sin ψˆ eθ · (σ × p) ˆ and the generalized Now, we will prove that the force vector F = Fxˆi + Fyˆj + Fz k forces Qr , Qψ , Qθ as obtained above, are related by the expression (1.49), using the expression we derived earlier for the force vector: ∂er Qi = F · ∂qi   dV (r) ∂ ˆr + 2(σ × p) · = − (rˆ er ) dr ∂qi 

Qr

 dV (r) ∂ = − eˆr + 2(σ × p) · (rˆ er ) dr ∂r   dV (r) ˆr ˆr + 2(σ × p) · e = − dr dV (r) = − + 2ˆ er · (σ × p) dr 

Qψ

 dV (r) ∂ = − eˆr + 2(σ × p) · (rˆ er ) dr ∂ψ   dV (r) ˆr + 2(σ × p) · (rˆ = − eψ ) dr = 2rˆ eψ · (σ × p)



Qθ

 dV (r) ∂ ˆ = − er + 2(σ × p) · (rˆ er ) dr ∂θ   dV (r) ˆr + 2(σ × p) · (r sin ψˆ = − eθ ) dr = 2r sin ψˆ eθ · (σ × p)

Part(c) of this problem was not included in the homework set, but it maybe the most interesting one: the equations of motion of the system are, using spherical polar coordinates: 6

1 ˙ eθ − sin ψ θˆ ˙ eψ ) L = T − U = m(r˙ 2 + r2 ψ˙ 2 + r2 sin2 ψ θ˙2 ) − V (r) − mr2 σ · (ψˆ 2

0 = = Qj

=

∂L d ∂L − dt ∂ q˙j ∂qj d ∂T ∂T d ∂U ∂U − − + dt ∂ q˙j ∂qj dt ∂ q˙j ∂qj d ∂T ∂T − dt ∂ q˙j ∂qj

We can use the expressions we obtained earlier for the generalized forces. We need to express the equations in terms of the cartesian components of the vector σ, since those are constants (the spherical components are not constants, since the spherical unit vectors themselves are not constant). Since the vector σ is constant, we can choose the z-axis along the σ vector without any loss of generality. Then ˆj · (σ × p) = mσ · (v × e ˆj ) e ˙ eψ + r sin ψ θˆ ˙ eθ ) × e ˆj = mσ · (rˆ ˙ er + rψˆ ˙ eθ + r sin ψ θˆ ˙ eψ ) ˆr · (σ × p) = mσ · (−rψˆ e 2 = −mσrθ˙ sin ψ ˙ er ) ˆψ · (σ × p) = mσ · (rˆ e ˙ eθ − r sin ψ θˆ = −mσrθ˙ sin ψ cos ψ ˙ er ) ˆθ · (σ × p) = mσ · (−rˆ e ˙ eψ + rψˆ = mσ(−r˙ sin ψ + rψ˙ cos ψ)

Qr =

d ∂T ∂T − dt ∂ r˙ ∂r

dV (r) + 2ˆ er · (σ × p) = m¨ r − mr(ψ˙ 2 + sin2 ψ θ˙2 ) dr dV (r) − − 2mσrθ˙ sin2 ψ = m¨ r − mr(ψ˙ 2 + sin2 ψ θ˙2 ) dr dV (r) ˙ θ˙ − 2σ) = m¨ mrψ˙ 2 + mr sin2 ψ θ( r+ dr −

Qψ =

d ∂T ∂T − dt ∂ ψ˙ ∂ψ 7

¨ − mr2 sin ψ cos ψ θ˙2 2rˆ eψ · (σ × p) = (2mrr˙ ψ˙ + mr2 ψ) −2mσr2 θ˙ sin ψ cos ψ = 2mrr˙ ψ˙ + mr2 ψ¨ − mr2 sin ψ cos ψ θ˙2 rθ˙ sin ψ cos ψ(θ˙ − 2σ) = 2rψ ˙ + rψ¨

d ∂T ∂T − dt ∂ θ˙ ∂θ 2 ˙ 2rˆ eθ · (σ × p) = 2mrr˙ sin ψ θ + 2mr2 sin ψ cos ψ θ˙ + mr2 sin2 ψ θ¨ mσ(−r˙ sin ψ + rψ˙ cos ψ) = mr sin ψ(2r˙ θ˙ sin ψ + 2rθ˙ cos ψ + rθ¨ sin ψ) Qθ =

3. Problem 1-19: Spherical pendulum A spherical pendulum is a mass point m suspended by a rigid weightless rod of length l. We set up a coordinate system with the origin at the attachment point, and the z-axis pointing down, along the gravitational force. In spherical polar coordinates, the position vector of the mass is r = lˆ er , its velocity is ˙ eψ + l sin ψ θˆ ˙ eθ , v = lψˆ its kinetic energy is 1 1 T = mv 2 = ml2 (ψ˙ 2 + sin2 ψ θ˙2 ), 2 2 and the potential energy is V = −mg · r = −mgl cos ψ. The Lagrangian is 1 L = T − V = ml2 (ψ˙ 2 + sin2 ψ θ˙2 ) + mgl cos ψ 2 There are two generalized coordinates, ψ and θ. Lagrange’s equations of motion are d ∂L ∂L − dt ∂ θ˙ ∂θ d ∂L ∂L − dt ∂ ψ˙ ∂ψ

= ml2

d ˙ 2 (θ sin ψ) = 0 dt

= ml2 ψ¨ − ml2 θ˙ sin ψ + mgl sin ψ = 0

The first equation is another way of saying the z-component of the angular momentum is constant. 8

4. Problem 21: A table with a hole A mass m1 on a table is connected to another mass m2 by a string of length l passing through a hole, with m2 moving a vertical direction. We set up a coordinate system with the origin at the center of the hole, the x-y plane on the table and the z-axis pointing down. We use cylindrical coordinates r, θ, z. The constraints are easily solved, with mass m1 moving in the plane z = 0 and mass m2 moving in the z-axis: r1 = r cos θˆi + r cos θˆj ˆ r2 = z k The length of the string is L = r + z, so the constraint is L = r + z, or r˙ = −z. ˙ We can choose r, θ as generalized coordinates for the system. The kinetic energy is 1 1 1 1 1 1 T = m1 v12 + m2 v22 = m1 (r˙ 2 + r2 θ˙2 ) + m2 z˙ 2 = (m1 + m2 )r˙ 2 + m1 r2 θ˙2 2 2 2 2 2 2 ˆ is The gravitational potential energy, using g = g k, V = −m1 g · r1 − m2 g · r2 = −m2 gz = −m2 g(L − r) The Lagrangian is 1 1 L = T − V = (m1 + m2 )r˙ 2 + m1 r2 θ˙2 − m2 gr 2 2 where we have omitted the constant term m2 gL, since we know it doesn’t change the equations of motion. Lagrange’s equation for θ is: d ∂L ∂L − dt ∂ θ˙ ∂θ d ˙ (m1 r2 θ) dt

0 = 0 =

˙ Since θ is a cyclical coordinate, we see that there is a conserved quantity, l = m1 r2 θ. This is the angular momentum of the mass m1 on the table, which only has a vertical component: r1 = r(cos θˆi + cos θˆj) 9

˙ sin θˆi + cos θˆj) v1 = r(cos ˙ θˆi + sin θˆj) + rθ(− ˆ = lk ˆ L1 = r1 × m1 v1 = m1 r2 θ˙k Lagrange’s equation for the coordinate r is: 0 = 0 = 0 = 0 − 0 = 0 = 0 = 0 = 0 =

d ∂L ∂L − dt ∂ r˙ ∂r (m1 + m2 )¨ r − m1 rθ˙2 + m2 g 2  l + m2 g (m1 + m2 )¨ r − m1 r m1 r2 l2 = (m1 + m2 )¨ r− + m2 g m1 r3 l2 = (m1 + m2 )¨ rr˙ − r˙ + m2 g r˙ m1 r3   d 1 l2 2 (m1 + m2 )r˙ + + m2 gr dt 2 2m1 r2   d 1 1 2 2 ˙2 (m1 + m2 )r˙ + m1 r θ + m2 gr dt 2 2 d (T + V ) dt dE dt

(1)

(2)

By transforming the equation into a total derivative in numbered line (2), we recognize in this equation the conservation of mechanical energy. In numbered line (1), we also recognize the term m2 g representing the magnitude of the total external force in the system, equal to the rate of change in kinetic energy, dT /dt = (m1 +m2 )¨ r −m1 rθ˙2 . Notice that in polar coordinates, the rate of change in kinetic energy does not just have second time derivatives (m1 + m2 )¨ r, but also an “angular momentum” term −m1 rθ˙2 . If we wanted to use Newton’s laws F = ma, we need to introduce the tension force, and then combine equations to eliminate it:

ˆ = m2 ¨r2 m2 g − T k ˆr = m1 ¨r1 −T e We need the components of m1 ’s acceleration in polar coordinates: r1 = = r cos θ i + r cos θ j = rˆ er 10

˙ θ r˙ 1 = re ˙ r + rθe ˙ θ + rθe ¨ θ − rθ˙2 er ¨r1 = r¨er + 2r˙ θe Newton’s equations are then: m2 g − T −T

= m2 z¨ = −m2 r¨ = m1 (¨ r − rθ˙2 )

1 d (mr2 θ˙2 ) 0 = 2r˙ θ˙ + rθ¨ = rθ˙ dt We can combine the first two equations and obtain Lagrange’ equation for r; the third equation is Lagrange’s equation for θ. Newton’s laws give us a way to calculate the tension if we know r(t): T = m2 (g + r¨); we don’t get the tension from Lagrange’s equations unless we use Lagrange’s multipliers. 5. Free fall with frictional forces A particle falls vertically under the influence of gravity, with fricitonal forces derivable from a dissipation function F = 12 kv 2 . The Lagrangian is (choosing the z-axis pointing down) 1 L = T − V = mz˙ 2 + mgz 2 Lagrange’s equation is 0=

d ∂L ∂L ∂F − + = m¨ z − mg + k z˙ dt ∂ z˙ ∂z ∂ z˙

We can write this as a differential equation for v = z: ˙ mv˙ + kv − mg = 0 We take another derivative and get an equation for a = v: ˙ v¨ + k v˙ = ma˙ + ka = 0 which has an exponential solution a(t) = a0 e−λt , with λ = k/m. We integrate once to get the velocity: v(t) = v0 + 11

a0 (1 − e−λt ) λ

As t → ∞, the velocity approaches a constant, v → v0 +a0 /λ, but this final maximum velocity seems to depend on arbitrary constants of integration: we know this is not so! If we integrate v(t), we would obtain an expression for z(t) with three constants of integration z0 , v0 , a0 , but we know we should only have two, since we started with a second order differential equation for x. However, the equation for v tells us that v0 , a0 are not independent: 0 = mv˙ + kv − mg   a0 mg = ma0 e−λt + k v0 + (1 − e−λt ) λ mg = kv0 + ma0 a0 = g − λv0 and then the velocity is v(t) = v0 +

g − λv0 mg (1 − e−λt ) = v0 e−λt + (1 − e−λt ) λ k

and the final (maximum) terminal velocity is mg/k. At that point, gravity force mg is in equilibrium with the friction force −dF/dz˙ = −k z. ˙ 6. A pendulum hanging from a spring A spring of rest length La (no tension) is connected to a support at one end and has a mass M attached to the other. Assuming the motion is confined to a vertical plane, the position of the mass M is, using polar coordinates r, θ with the x-axis pointing down: r = rˆ er = r(cos θ, sin θ) Notice that the length of the spring is not constrained like in a rigid pendulum, we do have two independent generalized coordinates in this case. The kinetic energy is 1 1 T = mv 2 = (r˙ 2 + r2 θ˙2 ) 2 2 The potential energy is 1 1 V = −mg · r + k(r − La )2 = −mgr cos θ + k(r − La )2 2 2 and the Lagrangian is 1 1 L = T − V = m(r˙ 2 + r2 θ˙2 ) + mgr cos θ − k(r − La )2 2 2 In equilibrium, the spring will be stretched to provide a force compensating for gravity on the mass: L0 = La + ∆L = La + mg/k. We can choose a generalized coordinate 12

ξ, instead of r, representing the length the spring has stretched from its equilibrium length L0 : ξ = r − L0 . The Lagrangian is then 1 1 L = m(ξ˙2 + (ξ + L0 )2 θ˙2 ) + mg(ξ + L0 ) cos θ − k(ξ + mg/k)2 2 2 Lagrange’s equation for ξ is : ∂L d ∂L = ˙ dt ∂ ξ ∂ξ ¨ mξ = m(ξ + L0 )θ˙2 + mg cos θ − k(ξ + mg/k) mξ¨ = m(ξ + L0 )θ˙2 − mg(1 − cos θ) − kξ and Lagrange’s equation for θ is: ∂L d ∂L = dt ∂ θ˙ ∂θ 2¨ ˙ + L0 ) = −mgL0 sin θ m(ξ + L0 ) θ + 2ξ˙θ(ξ If we assume small departures from equilibrium: ξ  L0 and θ  1, we can keep leading oder terms in the equations, and obtain: mξ¨ = −kξ mL20 θ¨ = −mgL0 θ which have oscillatory solutions with frequencies ωk2 = k/m for the spring stretching and ωg2 = g/L0 for the angular displacement.

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