Problem Set II Solutions

Transcription

1 Physics R. Wal Classical Mechanics Autun, 2002 Proble Set II Solutions 1) Let L(q, q; t) be a Lagrangian [where, as in class, q stans for (q 1,..., q n )]. Suppose we introuce new coorinates (Q 1 (q),..., Q n (q)) on configuration space. Relate the new oenta, P, to the ol oenta p an show that i P i Q i = i p i q i. (For the purposes of this proble, it is convenient to view all quantities as functions of the inepenent variables (q, q).) The oentu has the for: Q j q i p i = L = L q i Q j + L Q j (1) Q j q i (There is an iplie su over ouble inices in the above equation an those to follow.) But the secon ter is zero because the Q i s are functions only of q i. Because of this we also have that: So: Q j = Q j q i (2) Q j q i = Q j (3) Thus the new oentu P i is relate to the ol by the forula: We also ieiately have: p i = L Q j Q j = P j Q j (4) Q j p i q i = P j q i = P j Q j (5) Recall that there is an iplie su over ouble inices. 2) (a) Show that the Euler-Lagrange equations for the Lagrangian L = 1 2 x 2 e eφ + A t c x t yiel the usual Lorentz force equations of otion of a charge particle in an electroagnetic fiel. 1

2 (b) Obtain the corresponing Hailtonian forulation of the proble. Write out Hailton s equation of otion an show explicitly that they also are equivalent to the usual Lorentz force law. (a) First calculate the conjugate oenta: L = x i + e x i c A i (6) The force ter is: L = e φ + e x i x i c x A (7) x i So the Euler-Langrange equations are: t (x i + e c A i) = e φ + e x i c x A (8) x i Subtracting the tie erivative of A fro both sies an using the chain rule we have: ẍ i = [ e φ e A i x i c t ] + e c x A e x i c ( x )A i (9) The first ter in brackets is ee i, where E is the electric fiel. The secon ter can be calculate as follows: x j i A j x j j A i = x j l A k (δ li δ kj δ lj δ ki ) = x j l A k (ɛ lk ɛ ij ) (10) But l A k ɛ lk = B an x j B ɛ ij = ( x B) i, so: which is the Lorentz Force Law. (b) Fro equation 6 above we know that: Thus, H = p x L = p p e c A which gives x = e E + e c x B (11) p = x + e c A (12) [ 2 ( p e c A )2 eφ + e A c ( p e A c )] (13) H = (1 1 A 2 )p2 + ( )e p + ( 1 c 2 + 1)(eA)2 c + eφ 2 = p2 e p A 2 c 2 + (ea)2 + eφ (14) 2c2

3 Hailton s equations are: x i = H = p i p i ea i c (15) p i = H = + e p x i c A e2 A x i c A e φ (16) 2 x i x i Using the equation for ẋ we can write the ṗ equation as: t (x i + ea i c ) = x e c A e φ (17) x i x i This is the sae as equation 8 above; thus Hailton s equations also reuce to the Lorentz Force Law. 3) In the context of special relativity, it is uch ore in keeping with the covariant nature of the theory to treat all four spacetie coorinates (t, x, y, z) on an equal footing, an thus to escribe particle otion as a path t(λ), x(λ), y(λ), z(λ) in a 4-iensional configuration space (with λ an arbitrary paraeter along the path) rather than as a curve x(t), y(t), z(t) in a 3-iensional configuration space (with t the tie coorinate of a particular global inertial coorinate syste). (a) Show that the Lagrangian [( ) t 2 x L = λ λ yiels the correct equations of otion for a free particle. (In keeping with the above reark, treat (t, x, y, z) as the egrees of freeo an λ as tie.) 2 ] 1 2 (b) Show that the conjugate oenta satisfy the relation p 2 t (p 2 x + p 2 y + p 2 z) = 2 an, thus, are not inepenent, i.e. one cannot eliinate the q s in favor of p s. (c) Nevertheless, obtain a (constraine) Hailtonian forulation for the free relativistic particle by the proceure escribe in class, with α = t/λ. (a) First, let s efine soe sipler notation: q µ = u µ = ( t λ, x λ, y λ, z λ ) (18) q µ = u µ = ( t λ, x λ, y λ, z λ ) (19) 3

4 where q µ are the coorinates an u µ are the first erivatives with respect to λ. Then the Lagrangian is now expresse as: L = [u µ u µ ] 1 2 (20) The oenta are: p µ = L u = u µ[u ν u µ ν ] 1 2 (21) Since the coorinates o not explicitly appear in L, the equation of otion is: λ ( u µ[u ν u ν ] 1 2 ) = 0 (22) which says the ter in parenthesis is a constant, which we will call P µ. P µ = u µ [u ν u ν ] 1 2 (23) Note that this expression is inepenent of the paraetrization λ along the worl-line. If I choose a new λ = f(λ) then P µ is: P µ = βu µ [βu ν βu ν ] 1 2 = uµ [u ν u ν ] 1 2 (24) where β = λ. So we can choose λ equal to the proper tie τ an equation λ 23 becoes: P µ = q µ (25) τ which is the failiar expression for the (constant) oentu of a relativistic particle. The paraeter λ is affine if the function f(λ) = 0 in the following equation: u µ µ u ν = f(λ)u ν (26) Carrying out the erivative in equation 22, we will get (after iviing by ): u µ [u ν u ν ] 1 2 uµ [u ν u ν ] 3 2 uσ u σ = 0 (27) This leas to: u µ = [ u σu σ ]u u ν µ (28) u ν After noticing that the left-han sie of equation 24 is u ν, we see that the function in brackets is f(λ). Thus, choosing λ affine gives u µ = 0, which allows us to set u µ u µ to a constant (which is 1 if λ is the proper tie). It is easy to show that if λ is not affine, one can choose a new λ that is affine. (b) Using equation 21, we see that: p µ p µ = 2 u µ u µ [u ν u ν ] 1 2 [u σ u σ ] 1 2 = 2 u µ u µ [u ν u ν ] 1 = 2 (29) 4

5 (c) Since the p µ are not inepenent, we cannot eliinate all of the u µ in favor of the. We choose one of the (α = t ) to serve as a non-ynaical λ constraining variable. Thus we have the Hailtonian: H = p 0 α + p i u i L(α, p i ) (30) where i runs over the spatial variables. Define γ = α[u µ u µ ] 1 2 has no α epenence.) Thus: (note that this H = p 0 α p ip i Note that p i p i = p 2 an that: [uµ u µ ] [u µ u µ ] 1 2 = p0 α p ip i α γ + α γ (31) Thus: (γ) 2 = 2 1 v = 1 v2 + v = (γ v) 2 = 2 + p 2 (32) 1 v 2 So Hailton s equations rea: H = p 0 α + α p (33) q i = H = p i p i p2 + 2 = p i γ (34) p i = H = 0 (35) H α = p 0 + p = 0 (36) 5