Lecture 6 Conservation of Angular Momentum (talk about LL p. 9.3) In the previous lecture we saw that time invariance of the L leads to Energy Conservation, and translation invariance of L leads to Momentum Conservation. Angular Momentum and Roataion Invariance of L x = xcosφ + ysinφ y = ycosφ xsinφ for φ << 1, sinφ φ, cosφ 1 x x + yφ y = y xφ, z = z { = yφ, xφ, 0 1
L ({r, r{ ) = L({r + {, {ṙ + { ) L L L L = L({ r, { r) + yφ xφ + yφ x x y ẋ ẏ Rotational Invariance L L = 0 0 = φ( p x y + p x ẏ ( p y x + p y ẋ)) d 0 = φ for any small φ dt p x y p y x = constant L z = constant NB: L { = {r p{ L z = xp y yp x Invariance to rotation about z-axis leads to conservation of z-component of angular momentum! Of course there is nothing special about z; invariance to rotation about any axis implies conservation of AM about that axis. If the L is invariant to all rotations aobut a point (e.g. the origin), then L i is conserved. if U( {r ) central potential, then L { about origin is conserved Note that like E and P i, L i is additive and frame dependent. LL uses M { for AM, presumably to avoid confusion with Lagrangian. We will use L { as is customary. 1 Frame Dependence of L { { { { L = L i +R { cm P i:), AM in CoM frame
So like energy total AM may be non-zero in the CoM frame. Unlike energy, there exists a frame with L i = 0, though I don t know how it is used or called... Let s back up a moment and look at the big picture. For any isolated system of interacting particles we have 7 conserved quantities (E++). By isolated I mean the L doesn t depend on location, orientation or time. This is trivially true if the potential depend sonly on the relative positions of the particles (e.q. due to gravity or electric charge). Given particles A and B 1. if A has p{ A and B has p{ B, what is p{ AB?. if A has L { A and B has L { B, what is L { AB? 3. if A has E { A and B has E { B, what is E { AB? A and B are attracted to each other. They collide elastically. { Is P {, L, E conserved in the collision? They collide in-elastically, what is not conserved? (Where does it go?) Example: Diatomic Molecule 1 U(r{ 1, {r ) = l r{ 1 r{ 1 L = m(v 1 + v ) U(r{ 1, {r ) 3
e ave two equal mass particles connected by a spring (only 1 DOF - longitudinal). They have initial velocities v 1 and v. Find the EoM for the distance between the particles. First move to CoM frame in CoM frame 4
v CM { v{ 1 R CM r{ 1 E i {L i v{ 1 + v{ 1 = v{ 1 v{ =, v{ = {v r{ 1 + r{ = = r{ 1 R { CM = r{ {r 1 1 = (m)v + (4k)r = m{r {v fixed CoM In the CoM frame, both masses move symmetrically about the CoM which appears fixed. With the CoM at the origin, our problem has some easy symmetries we can take advantage of. The potential depends only on ir, so...??... is conserved: 5
L = m( ṙ + r θ + r sin θ φ ) kr L F r = r = mr(θ + sin θ phi ) = const f { f ff L i let s define λ = m λ F r = r 4kr 3 λ m r = r 4kr 3 EoM for r In the end, we have a fairly simple EoM. This is entirely due to: 1. conservation of momentum, which makes the CoM coordinate choice special (i.e. CoM is fixed point). conservation of AM, which removes all of the angles from the EoM for r To turn this into a trajectory, we need only add the initial radius and radial velocity (and solve the nd order ODE). Another way to find an EoM for r is to use energy conservation: E = E i = m(ṙ + r (θ + sin θφ ) + kr λ = mṙ + + kr r mṙ = E i ( λ + kr ) r Which is not a nice ODE, but it is first-order. This means that the initial velocity information is already contained in the constant E i, so you only need r @ t=0 to solve. in general, initial conditions are trivially constants of motion, ut exchanging them for conserved quantities can significantly simplify the EoM. 6
Mechanical Similarity This curious little chapter of LL turns out to be very interesting if only for the fact that the concept it explores led to the discovery of dark matter. The idea is that the form of the E-L equations can tell us something about motion in a particular potential even before we solve them to find the EoM. Mechanical Similarity U(αr 1, αr ) = α k U(r 1, r ) 1 e.g. U(r 1, r ) = k {r 1 {r U(α{r 1, α{r ) = α U({r 1, {r ) This just says If you double the size of the system, you quadruple the potential energy. What does this do to the EoM? 1 d α q = αq, t = βt q = = q β dt β L (q, q ) = L(αq, α q β L = 1 mq + U(q ) = AC 1 m( α ) + α k U(q) βq α = α k L iff = α β = α 1 k β Multiplying L by a constant doesn t change the EoM, so with condition on β (scaling of time), the EoM are equal. A few important examples of this come from the harmonic oscillator and orbits in a gravitational potential. First the Harmonic Oscillator: 7
Harmonic Oscillator Potential Period, Amplitude A, U(x) k = A = αa = β, k = β = α 1 = 1 is independent of A This is the founding principle of all kinds of tic-toc clocks. (Even digital clocks use crystals, which have a k= mechanical resonance). For gravity we ll take a different approach to scaling... we ll look at how the kinetic energy scales relative to potential energy given a special r k potential. There are tricks we need for this: d (pq) = pp + qp dt 1 1 U T = mv = qp, p = F = q d U (pq) = q + T = ku + T dt q k U U(q) for U(q) = aq = akq k 1 = k q q The dramatic part happens when we average this over any finite trajectory (think orbits or oscillations in any potential well). 1 f(t) = lim τ τ 0 τ f(t)dt d 1 (pq) = lim p t=τ τ t=0 = 0 dt τ 1 for U T = ku for U r T = U r (harmonic oscillator) T = U gravity Virial Theorem 8
What this means for astronomy is that if you can measure the velocities of stars in a galaxy (actually, the velocity dispersion), you can measure T. From that you can compute U. You can then compare that to the potential you would expect from the stars you see. And you would discover that they don t match at all! Dark matter! (factor 4). Simply put, stars in galaxies move too fast for the gravity of the stars. If there were nothing else, the galaxy would fly apart! For next time 1. read LL 11, 13, 14. do pset 15-? 9
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