| May 18 2005 8-10 am |

- (as promised, one of 7 questions on a 3-hour exam your prof
had to pass as an undergraduate)

If the force on a particle moving in a plane is always directed toward or away from a fixed point in the plane, the magnitude of the force being a function only of the distance r from that fixed point, then:- Show that the forces are conservative and that the angular
momentum about the center of force is constant.

One way would be to show that Ñ×F = 0, which is true because in the expression for the curl the (only) radial component is differentiated only with respect to the other coordinates. But more correct and direct would be to point out that the work done along any path from**r**_{1}to**r**_{2}is just ò_{r1}^{r2}F dr, which contains no q, hence depends only on the endpoints.

We have d**L**/dt =**r**×**F**= 0 because**r**and**F**are parallel, so angular momentum is conserved.

- Show that the angular momentum is still constant, but that the force is not conservative.
The same angular momentum argument still holds; but now we have to show that the work definitely depends on the path. Thus, consider one path that is first purely radial, then in the angular (q) direction, and another between the same points first angular and then radial. The work is given by ò

_{r1}^{r2}F dr as before, but evaluated for the initial q for the first path, and the final q for the second. Since F depends on q, it must be possible to choose beginning and end points so the two integrals are definitely different.

- Show that the forces are conservative and that the angular
momentum about the center of force is constant.
- A rigid uniform bar of mass M, length 2
*l*, is supported in equilibrium in a horizontal position by two identical vertical springs of spring constant k, attached one at each end of the bar. The moment of inertia of the bar about its center is M*l*^{2}/3. We consider vertical motion only -- that is, the springs always remain vertical -- and small deviations from equilibrium.- Choose suitable coordinates and write the Lagrangian as well
as the Hamiltonian for this system.
If we have foresight we choose the position of the CM, x, and the angle q of the bar from the horizontal as the two generalized coordinates; otherwise we choose the positions of the two endpoints, call them y

_{+}and y_{-}. In the latter case we still have to express the KE of the bar as the contribution of the CM + that about the CM, so we need y_{±}= x ±*l*q. Thus

For the Hamiltonian we need pL = T - U = ^{1}/_{2}M×x2

+ ^{1}/_{2}(M*l*^{2}/3)×q2

- ^{1}/_{2}k (y_{+}^{2}+ y_{-}^{2}) =^{1}/_{2}M×x2

+ ^{1}/_{2}(M*l*^{2}/3)×q2

- ^{1}/_{2}k (2x^{2}+ 2*l*^{2}q^{2})_{x}= Mx^{·}and p_{q}= (M*l*^{2}/3)q^{·}, so H = T + U = (p_{x}^{2}/2M) + (3p_{q}^{2}/M*l*^{2}) +^{1}/_{2}k (2x^{2}+ 2*l*^{2}q^{2}). - Find the normal modes and normal frequencies of the system.
"By inspection" we see that the M- and k-matrices are already diagonal , we can write L = L

_{x}(x, x^{·}) + L_{q}(q, q^{·}) so the normal modes consist of an excitation of x alone (with q = 0), and one of q alone (with x = 0). From L_{x}we get the frequency (2k/M)^{1/2}, and from L_{q}we get (6k/M)^{1/2}.

- Choose suitable coordinates and write the Lagrangian as well
as the Hamiltonian for this system.
- Two particles having mass m and M attract each other
according to the law of gravitation. Initially they are at rest an
infinite distance apart.
- Show that their relative velocity of approach when they are
a distance X apart is

where G is the constant of gravitation.æ

Ö2G (M + m) XWe can assume that the CM is at rest, then the KE is the same as that of a single particle of reduced mass m at position X, and the potential energy is -GMm/X. The total energy is that at X = ¥, namely zero, so

Solving for X0 = 1 2æ

èmM m+Mö

ø×X2

- GMm X^{·}gives the required relation. - Find the momentum p
_{X}conjugate to X (in terms of m, M, [X\dot]). In the phase space with coordinates X, p_{X}sketch several orbits (including initial conditions different from those of part (a)). Restrict X to positive values, but allow p_{X}to be of either sign.p _{X}=¶T ¶ ×X= mM m+M×X. The graphs follow from H = T + U = m+M 2mMp ^{2}-GMm X= const.

- Show that their relative velocity of approach when they are
a distance X apart is
- In 2-dimensional space-time, (x, t) with c = 1, the
four-velocity of a mirror at rest is _u = (0, 1). A photon
incident on this mirror from the left has momentum-energy vector
_k = (k, k) where k = hn and n is the photon's
frequency. After reflection the photon has momentum-energy vector
_k¢ = (-k, k), that is, the space (momentum) component is
reversed, and the time (energy) component remains the same.
- Consider the space-time vector
^{1}equation

where the dot (·) denotes the space-time scalar product. Verify that this equation correctly describes the reflection in the mirror's rest frame, and argue that it must be valid in_k¢ = -_k - 2(_k·_u)_u *any*frame.It is valid in any Lorentz frame because it is a space-time vector equation.

- Show that the transformation _k ® _k¢ is a
Lorentz transformation for any vector _k (not just for null
vectors). It is enough to show _k¢
^{2}= _k^{2}.

The above uses the equation in (a) as intended. The question could also be interpreted in the mirror frame only, in which case you should show that the transformation of any vector__k__¢^{2}=__k__^{2}+ 2·(2__k__·__u__)(__u__·__k__) + 4 (__k__·__u__)^{2}__u__^{2}=__k__^{2}since__u__^{2}= -1.__k__= (a, b) into__k__' = (-a, b) is a Lorentz transformation, that is, that__k__'² =__k__². This was also counted as correct (though no one did it exactly right). - Let the mirror move with speed v, so that u
_{1}= v/Ö{1-v^{2}}, u_{4}= 1/Ö{1-v^{2}}. Use the equation in (a) to evaluate k¢ and hence find the reflected frequency n¢ in terms of the incident frequency n and the speed v.k¢ _{4}= hn¢ = -k_{4}- 2æ

èk _{1}v-k_{4}(1-v^{2})^{1/2}ö

øæ

è1 (1-v^{2})^{1/2}ö

ø= k æ

èv ^{2}-1 -2v + 21-v^{2}ö

ø= hn æ

è1-v 1+vö

ø.

- Consider the space-time vector

File translated from T

On 18 May 2005, 10:29.