John Magorrian-Comprehensive but very consize note.
S7 and BT VII: Classical mechanicsJohn Magorrian, HT firstname.lastname@example.org
These notes accompany the lectures for the Physics Short Option S7 / Physics & Philosophy BT VII course on classical mechanics. You can nd the most up-to-date version at http://www-thphys.physics.ox.ac.uk/user/JohnMagorrian/cm.pdf . Last updated: 13 Feb 2012.
are not covered in the HT2012 course)
Calculus of variations: EulerLagrange equation, (variation subject to constraints) . Lagrangian mechanics: principle of least action; generalized co-ordinates; conguration space. Application to motion in strange co-ordinate systems, particle in an electromagnetic eld, normal modes , rigid bodies . Noethers theorem and conservation laws. Hamiltonian mechanics: Legendre transform; Hamiltons equations; examples; principle of least action again ; Liouvilles theorem ; Poisson brackets; symmetries and conservation laws; canonical transformations. HamiltonJacobi equation .
Recommended reading T. W. B. Kibble & F. H. Berkshire, Classical mechanics, 5th ed. About 19. The single most suitable book for this course. L. D. Landau & E. M. Lifshitz, Mechanics. About 30. First volume of the celebrated Course of Theoretical Physics. Succinct. H. Goldstein, C. Poole & J. Safko, Classical mechanics, 3rd ed. About 50. Covers more advanced topics too. Verbose. Supplementary reading The following books are more dicult, but some might nd them inspiring for a second pass at the subject. V. I. Arnold, Mathematical methods of classical mechanics Adopts a more elegant, more mathematically sophisticated approach than the other books listed here, but develops the maths along with the mechanics. G. J. Sussman & J. Wisdom, Structure and interpretation of classical mechanics. About 45, but also freely available online. Uses a modern, explicit functional notation and breaks everything down into baby steps suitable for a computer.
S7 & BT VII: Classical mechanics
0 Some maths0.0 NotationVectors:x = xi + y j + z k
(in 3d) orx = x1 x 1 + x2 x 2 + + xn x n
for the general case. Gradients of function f (x, x ): f f x 1 + x x1 f f x 1 + x x 1 So,p
f x 2 + , x2 f x 2 + . x 2
f f f = p1 + p2 + . x x1 x2
0.1 An introduction to the calculus of variationsRecall that a function is simply a rule for mapping elements of one set (the functions domain) to elements of another set (its range). A functional is a mapping from the set of all functions that satisfy some specied conditions (e.g., the set of all smooth maps from the real line to three-dimensional space) to the real numbers. Examples include: the nth momentx1
In [y ] =
xn y (x) dx
of a one-dimensional function y (x) dened for x0 < x < x1 and having y (x0 ) = y (x1 ) = 0; the length of a curve x(t) joining two xed points x(t0 ) and x(t1 ) in n-dimensional space,t1
|x | dt;
the gravitational potential energy of a mass distribution (x),1 V  = 2 G
(x)(x ) d3 xd3 x . |x x |
For this course we need only consider functionals of the formt1
F [ x] =t0
L(x, x , t) dt
that eat smooth one-dimensional curves x(t) with xed endpoints x(t0 ) = x0 , x(t1 ) = x1 in an n-dimensional space. The rst two examples above are of this form. The third is not. Internally, the functional runs over the curve, feeding the local values of (x, x , t) to a function L and accumulating the results. Note that this L treats x, x and t as independent variables; it does not know that x = dx/dt!
S7 & BT VII: Classical mechanics Now lets look at how the output of the functional changes when we distort the curve slightly from x(t) to x(t) + h(t). The variation of the curve h(t) must be smooth and vanish at the endpoints in order that x + h be admissible to F , but is otherwise arbitrary. The variation or dierential of the functional F [x; h] lim0
F [ x + h] F [ x ]
An extremal is a curve x(t) for which F [x; h] = 0 for all admissible h(t). Finding these extremals (if they exist) is the business of the calculus of variations. Fundamental lemma of the calculus of variations t0 < t < t1 , satisest1 t0
If a smooth curve f (t), dened on the range (0.10)
f (t) h(t) dt = 0
for all continuous h(t) having h(t0 ) = h(t1 ) = 0, then f (t) 0. Proof by contradiction: we show that if f = 0 then equation (0.10) would not be true for all h(t). Suppose that there were some tblip between t0 and t1 for which f (tblip ) = 0. Then, because f has no discontinuous jumps we can always nd a small interval (tleft , tright ) around this tblip where f = 0. Now consider the functionh(t) = f (t)
(tright t)(t tleft ), for tleft < t < tright , 0, otherwise.
This clearly satises the conditions of the lemma, butt1 t0 tright
f (t) h(t) dt =
f 2 (t)(tright t)(t tleft ) dt > 0,
since the integrand is positive between tleft and tright . So, weve shown that if f = 0 anywhere then we can always nd some h(t) that makes f h dt = 0. Turning this around, if there is no h for which the integral is non-zero, then we must have f = 0 between t0 and t1 . Now we come to the key result of this section. Let F [x] be a functional of the formt1
L(x, x , t) dt,t0
dened on the set of smooth functions x(t) satisfying boundary conditions x(t0 ) = x0 and x(t1 ) = x1 . Then a curve x(t) is an extremal of F if and only if it satises the EulerLagrange equation, d dt L x L = 0. x (0.14)
Proof: If x is an extremal of F then for any variation h we have 0 = F = lim =t0 t1
, t) L(x, x L(x + h, x + h , t) dt (0.15) L h x t1
L L h+ h dt x x L d x dt L x h dt + ,t0
where the last line follows from the previous one using integration by parts. The nal term on the last line vanishes because the boundary conditions mean that h(t0 ) = h(t1 ) = 0. Thus (0.15) becomest1
L d x dt
S7 & BT VII: Classical mechanics for any smooth h. Applying the fundamental lemma, our extremal curve x(t) must satisfy the Euler Lagrange equation (0.14). Conversely, if a curve x(t) satises (0.14) then it is clear from (0.15) that it is an extremal of the functional F . Example: the shortest path between two points Consider the set of smooth curves in the (t, x) plane that pass between two xed points x (t0 ) = x0 and x(t1 ) = x1 . The path length of any such curve x(t) 2 . The EL equation for this L is is given by the functional (0.13) with L = 1 + x d dt x 1+x 2 = 0, (0.17)
since L/x = 0 and L/ x = x/ 1+x 2 . Therefore extremals satisfy x = A, a constant. Integrating, x = At + B , with the constants A and B completely determined by the condition that the curve pass through the two xed points. Important: When writing down the EL equation, remember that x and x are independent arguments of L. Use the fact that along extremals x(t) satisies x = dx/dt only when solving (i.e., integrating) for x(t). Easy rst integral when L does not depend explicitly on t (Beltrami identity) Solving the EL equation often leads to lots of messy algebra. But if L = L(x, x ), the EL equation can be reduced to the rst-order dierential equation x (L/ x ) L = constant. To see this, note that on solution paths f dx f dx f f df = + = x + x dt x dt x dt x x for any function f (x, x ). So, d x dt L x L = x L d L L L +x x x x dt x x x d L L =x = 0. dt x x (0.18)
Look out for a more physical way of deriving this result later in the course! Example: minimal surface of revolution Among all the curves that pass through the points (t0 , x0 ) and (t1 , x1 ), nd the one that generates the surface of minimum area when rotated about the t-axis. A physical example is a soap bubble drawn between two coaxial circular hoops. The surface area generated by a curve x(t) ist1
x 1+x 2 dt.
Comparing to equation (0.13), we see that L(x, x ) = 2x 1 + x 2 , independent of t. Using the result above for general L(x, x ), the extremals of (0.20) satisfy xx x 1+x 2 1+x 2 = A. (0.21)
This is easily rearranged, via x = A 1 + x 2 , to give Ax =
x 2 A2 . t+B A
Integrating, the curve that extremizes the surface area of revolution is x(t) = A cosh , (0.23)
where the constants A and B are chosen to satisfy the boundary conditions x(t0 ) = x0 and x(t1 ) = x1 . Depending on the choice of x0 and x1 , there can be zero, one or two solutions for (A, B ). Of course, in the zero-solution case an extremal does exist, but it is not smooth and therefore lies beyond the remit of the machinery developed above.
S7 & BT VII: Classical mechanics
0.2 Variation subject to constraintsSometimes it is necessary to nd extremals of a functional F [x] (equation (0.13)) subject to a constraint of the form g (x, t) = 0 (0.24) among the co-ordinates. From (0.15), the condition for a curve x(t) to be an extremal is thent1
F [x, h] =t0
L d x dt
h dt = 0
for any smooth h(t) that satises g g + + hn = 0, (0.26) x1 xn since we must have g (x + h, t) = 0. This last condition means that we cannot use the fundamental lemma directly. Instead let us multiply (0.26) by an arbitrary function (t) and insert it into the integrand of (0.25). This combines the two conditions (0.25) and (0.26) into one:h g