Classical Mechanics

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Exercise 6.5. Consider the 3-dimensional action ∫ 1 2 S = mẋ mgz dt 2 where x = ( x, y, z ) . 1. Show that there are four symmetries of S. 2. Find the four conserved quantities. Exercise 6.6. Consider the 2-dimensional action functional ∫ 2 2 2 S = 1 ml ϕ˙ + mglϕ dt 2 Find all rescalings of the parameters and coordinates ( m, g, l, ϕ, t) which leave S changed by no more than an overall constant. √Use these rescalings to show that the period of the motion is proportional to . l g Exercise 6.7. The action functional ∫ t 2 1 2 K S = mẋ √ dt 2 x2 t1 Use a scaling argument to derive Keplers law relating the period and a characteristic length of the orbit. 118
7. The physical Lagrangian We have seen that the extrema of functionals give preferred paths of motion among the class of curves. These extremal paths appear to be smooth curves. Indeed, when the action is taken to be path length, the extremal curves are straight lines. Now we return to the problem of physical motion stated in Chapter III: can we specify a functional for specic physical problems in such a way that the extremals are the physical paths of motion? We have seen examples where this is the case. For example, the action functional 2 4 2 2 2 4 L = 1 m x˙ + 1 1 kmx˙ x k x (7.1) 12 2 4 describes the simple harmonic oscillator in 1-dimension. Now we take a more systematic approach, based on the principles of gauge theory. Our gauging of Newtons second law begins with the observation that the symmetry which preserves Newtons law is different than the symmetry of the Euler-Lagrange equation. Specically, Newtons second law is invariant under a special set of constant transformations called Galilean transformations. By contrast, the Euler-Lagrange equation is invariant under the diffeomorphism group, that is, the group of all coordinate transformations. It follows that, if we are to develop a systematic way of writing physical equations of motion using the Euler-Lagrange equation, then it must involve a generalization of Newtons second law to arbitrary coordinate systems. This is exactly what gauging accomplishes the systematic extension of a symmetry. There is nothing surprising in such a generalization. In fact, we would nd it odd not to be able to work out a form of the second law valid in any coordinate system of our choosing. The key point here is to do it systematically, rather than just substituting a specic transformation of the coordinates. There are several steps to our generalization. First, we derive the most general set of transformations under which Newtons law is covariant the Galilean group. This group has important subgroups. In particular, we are interested in the subgroup that also preserves Newtonian measurement theory, the Euclidean group. Next, we derive the geodesic equation. Geodesics are the straightest possible lines in arbitrary, even curved, spaces. In our Euclidean 3-space, the geodesic equation simply provides a diffeomorphism-invariant way of describing a straight line the condition of vanishing acceleration. As a result, we will have expressed the acceleration in arbitrary coordinates. 119
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Contents I Preliminaries 5 1 Physic
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11.2.1 Are inequivalent Lagrangians
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Part I Preliminaries Geometry is ph
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Several important features are impl
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and physicists have returned again
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and partition the decimal expansion
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2.1. Symmetry and groups Mathematic
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For the nal case, we have two subca
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one of the elements must be the ide
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2.2. Lie groups While groups having
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Denition 2.12. A topological space,
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for some xed dimension n. Here is
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1 which shows that TaTb = Ta+ b and
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We now have R R x x = x x This exp
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and may therefore be written as ⎛
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such that ′ ′ ′ ′′ p ( R
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3.2. Curves, lengths and extrema So
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Perhaps more important for our disc
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2 Most functionals that arise in ph
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′ where we omit terms of order (
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But the point 0 was arbitrary, so w
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such that at extrema, Let f be give
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sequences of functions. Therefore,
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∫ ( n) 3. Evaluate f( x) ( x) dx
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Since we have chosen hm ( ) to ap
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Remark 3. To dene higher order deri
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= ∫ ∫ + 1 2 ∂ d1 d2 ( 1 2
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for some tensors of each type, then
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1 mi or, multiplying both sides by
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then use the identity of eq.(4.2) t
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and we have, simply, dLi Ni = dt We
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Exercise 4.4. Find the metric tenso
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After the geometric description of
Page 67 and 68: Notice that we have chosen to write
Page 69 and 70: forms and vectors, respectively, ev
Page 71 and 72: so that the linear map on curves is
Page 73 and 74: There is a natural duality between
Page 75 and 76: An inner product on the space of ve
Page 77 and 78: as a mapping from to by leaving one
Page 79 and 80: index is repeated, once up and once
Page 81 and 82: Exercise 4.12. The examples of Sect
Page 83 and 84: the metric in polar coordinates is
Page 85 and 86: 4.4. Group representations One impo
Page 87 and 88: so the form of g is determined by i
Page 89 and 90: i j Of course, the most general 33
Page 91 and 92: use it elsewhere. Second, we use ea
Page 93 and 94: Part II Motion: Lagrangian mechanic
Page 95 and 96: i respect to q ( t) : 0 = S∫ = L
Page 97 and 98: again to nd k k dx˙ x = dt k k d
Page 99 and 100: Theorem 6.3. (Noethers Theorem) Sup
Page 101 and 102: 6.2. Conserved quantities in restri
Page 103 and 104: or innitesimally, 1 2 S ∫ i i i
Page 105 and 106: Now consider the (vanishing) variat
Page 107 and 108: ̸ Proof: Let x0 and v0 be the init
Page 109 and 110: Bringing both terms to the same sid
Page 111 and 112: for constants ( , , , ) . The Lagr
Page 113 and 114: so the variation gives 0 = S ) ∑
Page 115 and 116: th This time we must keep all of th
Page 117: so dL dt ∑n ∑k 1 ∑ m km m n
Page 121 and 122: since the Newtonian assumption of u
Page 123 and 124: With this in mind, we dene Denition
Page 125 and 126: where we have used the symmetry, g
Page 127 and 128: and the equations for geodesics bec
Page 129 and 130: i m n ds d ds dx i ds dx ds dx 0 =
Page 131 and 132: We identify the rst term on the rig
Page 133 and 134: and the Lagrangian is m 2 2 2 2
Page 135 and 136: 2. Power law potentials, n V ar . =
Page 137 and 138: and because ϕ is cyclic, the magni
Page 139 and 140: n V = ax n. 1 m f 1 1 2 ′ d du d
Page 141 and 142: = = 2 ′ J f m f 2 J m f 2 3 2 3 d
Page 143 and 144: 8.2. General central potentials We
Page 145 and 146: which can always be satised by choo
Page 147 and 148: Now consider perturbations around c
Page 149 and 150: the energy and angular momentum giv
Page 151 and 152: When q = 1, each single orbital swe
Page 153 and 154: is Newtons gravitational force. By
Page 155 and 156: √ ) m 2 = 1 + 1 + 2 EL L2 m2 Deni
Page 157 and 158: Exercise 8.13. Show that the Laplac
Page 159 and 160: since g, being a test function, has
Page 161 and 162: 9. Constraints We are often interes
Page 163 and 164: Because y is cyclic we immediately
Page 165 and 166: Exercise 9.4. A ball moves friction
Page 167 and 168: the magnitude ω and angle ϕ may b
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of x parallel to n while the motion
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and this is not purely quadratic in
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This solution is well-dened on any
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With √ 2 v2 f( v) = mc 1 , we ha
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The remarkable fact is that the Ham
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Setting this to zero, we have two t
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an arbitrary, time-independent cons
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where We may also write in any of t
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2 2 where v is the usual squared ma
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Since u is a 4-vector, its magnitu
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12.3. Acceleration Next we consider
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Then the path of extremal proper ti
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Conformal transformations will appe
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3. Substitute back into the equati
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i i valid. Using these constants, F
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so that 2 0 = ∂ ∂ ε + ∂ ∂
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Once again cycling the indices, add
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Exercise 13.3. An innitesimal dilat
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for x i and i i x = i i i x = x +
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The linear transformations that pre
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the connection transforms with tw
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14.2. Consequences of the covariant
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equations which determine the form
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14.4. Motion in biconformal space A
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In this form we may identify conjug
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which implies that each p n 0 is a
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= = ⎛ ∂H ∂pj ∂H 0 j + ∂x
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i A A p . This independence is gua
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Then there exists a function f (a f
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14.8. Phase space and the symplecti
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Therefore, multiplying eq.(14.18) o
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Its occurrence in Hamiltons equatio
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then we have ′ AB ∂f ∂g { f,
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i for x with pj and nally { p , p }
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since ∂qj ∂pm = 0. For the seco
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For (a), let q i 1 1 i = = p i i x
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The rst equation j j ∂f( x , q ,
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Then cancelling the exponential, Co
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Therefore, solving the Hamilton-Jac
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′ E H, H , q Because = the new Ha
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The rst equation relates p to the e
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Substituting into the expression fo
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terms of X : m L = tr XX ˙ ˙ V
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each element of the Lie algebra gen
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Varying, we get two sets of equatio
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is clearly just dz ∧ dz and is t
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17. Gauge theory Recall our progres
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numbers required to specify the tra
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1 The last line is the statement th
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where if i, j = 1, 2 , . . . , n th
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= 1 k! Therefore ∑n k 1 0 1 k
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so O (3) preserves the Pythagorean
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Similarly, we nd that [ J 2, J3] =
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tell us which components of the mat
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so that to linear order, J x + J x
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To rst order, this simply implies a
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dx ∧ dy, dy ∧ dz, dz ∧ dx wit
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l ∧ V = ( A dx) ∧ ( B dx ∧ dy
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Lets sum up. We have dened forms, h
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By the same argument we used to get
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Exercise 17.16. Write the curl of A
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Exercise 17.19. Notice that the ide
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is also a p-form, while for any two
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Divergence The use of differential
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This combines to dω ∂ = e g +
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2 [ ] = 1 ∇ ω ∇ ω z ∂
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Remark 11. Start with the magnetic
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[23] Wolsky, A. M., Amer. Journ. Ph
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[53] May, R. and Oster, G. F., Bifu
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[83] P.A.M. Dirac, The Principles o
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[111] L. ORaifeartaigh, The Dawning
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Classical Mechanics

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