Classical Mechanics

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Exercise 12.5. A projection operator is an operator which is idempotent, that is, it is its own square. Show that P is a projection operator by showing that P P = P and show that u P = 0 and P u = 0. Now we have 0 = 1 du + 2 c d P ∂ ∂x This projection operator is necessary because the acceleration term is orthogonal to u . Dividing by , we see that 1 du c2 d = P ∂ ln ∂x If we identify V = exp mc2 then we arrive at the desired equation of motion m du d = P which now is seen to follow as the extremum of the functional ∫ V a 2 1 2 [ ] = 1 / S x e mc ( u u ) d c C ∂V ∂x (12.3) See the exercises for other ways of arriving at this result. It is suggestive to notice that the integrand is simply the usual line element multiplied by a scale factor. 2V 2V 2 2 2 = 1 2 2 mc ( ) = 1 d mc 2 c e u u d e dx dx c This is called a conformal line element because it is formed from a metric which V is related to the at space metric by a conformal factor, e mc2 , 2V mc 2 g = e 192
Conformal transformations will appear again when we study Hamiltonian mechanics. We can generalize the action further by observing that the potential is the integral of the force along a curve, ∫ V = F dx The potential is dened only when this integral is single valued. By Stokes theorem, this occurs if and only if the force is curl-free. But even for general forces we can write the action as ∫ 1 ∫ 2 1 2 [ ] = 1 mc C a F dx / S x e ( u u ) d c C In this case, variation leads to ∫ 0 = 1 ( ) ( ) c ∫ ∫ F dx d ∫ = 1 1 e F c x d c C 1 ∫ 1 2 mc2 C Fdx / e u u uu d C 1 1 ∫ 1 2 1 2 mc C F dx 2 ( ) / ∂ e u u x c C mc ∂x C d u 1 ∫ ∫ mc 2 C F dx mc C Fdx 2 C d c e mc 1 2 1 2 The equation of motion is ∫ du 1 1 0 = 2 d mc u u ∂ F dx ∂x m F e and therefore, C ∫ mc C F dx m du = P d F This time the equation holds for an arbitrary force. Finally, consider the non-relativistic limit of the action. If v
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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
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Notice that we have chosen to write
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forms and vectors, respectively, ev
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so that the linear map on curves is
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There is a natural duality between
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An inner product on the space of ve
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as a mapping from to by leaving one
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index is repeated, once up and once
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Exercise 4.12. The examples of Sect
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the metric in polar coordinates is
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4.4. Group representations One impo
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so the form of g is determined by i
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i j Of course, the most general 33
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use it elsewhere. Second, we use ea
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Part II Motion: Lagrangian mechanic
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i respect to q ( t) : 0 = S∫ = L
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again to nd k k dx˙ x = dt k k d
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Theorem 6.3. (Noethers Theorem) Sup
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6.2. Conserved quantities in restri
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or innitesimally, 1 2 S ∫ i i i
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Now consider the (vanishing) variat
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̸ Proof: Let x0 and v0 be the init
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Bringing both terms to the same sid
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for constants ( , , , ) . The Lagr
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so the variation gives 0 = S ) ∑
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th This time we must keep all of th
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so dL dt ∑n ∑k 1 ∑ m km m n
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7. The physical Lagrangian We have
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since the Newtonian assumption of u
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With this in mind, we dene Denition
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where we have used the symmetry, g
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and the equations for geodesics bec
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i m n ds d ds dx i ds dx ds dx 0 =
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We identify the rst term on the rig
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and the Lagrangian is m 2 2 2 2
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2. Power law potentials, n V ar . =
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and because ϕ is cyclic, the magni
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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
Page 169 and 170: of x parallel to n while the motion
Page 171 and 172: and this is not purely quadratic in
Page 173 and 174: This solution is well-dened on any
Page 175 and 176: With √ 2 v2 f( v) = mc 1 , we ha
Page 177 and 178: The remarkable fact is that the Ham
Page 179 and 180: Setting this to zero, we have two t
Page 181 and 182: an arbitrary, time-independent cons
Page 183 and 184: where We may also write in any of t
Page 185 and 186: 2 2 where v is the usual squared ma
Page 187 and 188: Since u is a 4-vector, its magnitu
Page 189 and 190: 12.3. Acceleration Next we consider
Page 191: Then the path of extremal proper ti
Page 195 and 196: 3. Substitute back into the equati
Page 197 and 198: i i valid. Using these constants, F
Page 199 and 200: so that 2 0 = ∂ ∂ ε + ∂ ∂
Page 201 and 202: Once again cycling the indices, add
Page 203 and 204: Exercise 13.3. An innitesimal dilat
Page 205 and 206: for x i and i i x = i i i x = x +
Page 207 and 208: The linear transformations that pre
Page 209 and 210: the connection transforms with tw
Page 211 and 212: 14.2. Consequences of the covariant
Page 213 and 214: equations which determine the form
Page 215 and 216: 14.4. Motion in biconformal space A
Page 217 and 218: In this form we may identify conjug
Page 219 and 220: which implies that each p n 0 is a
Page 221 and 222: = = ⎛ ∂H ∂pj ∂H 0 j + ∂x
Page 223 and 224: i A A p . This independence is gua
Page 225 and 226: Then there exists a function f (a f
Page 227 and 228: 14.8. Phase space and the symplecti
Page 229 and 230: Therefore, multiplying eq.(14.18) o
Page 231 and 232: Its occurrence in Hamiltons equatio
Page 233 and 234: then we have ′ AB ∂f ∂g { f,
Page 235 and 236: i for x with pj and nally { p , p }
Page 237 and 238: since ∂qj ∂pm = 0. For the seco
Page 239 and 240: For (a), let q i 1 1 i = = p i i x
Page 241 and 242: 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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