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Direct Energy, 2018a

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11 CALCULUS OF VARIATIONS 255<br />

The path x 1 (t) describes a case where the mass travels at a constant speed.<br />

The path x 2 (t) describes a case where the mass accelerates whenthe restraint<br />

is removed, and the path x 3 (t) describes a case where the mass rst<br />

accelerates thenslows. The actionof each path canbe calculated using Eq.<br />

11.36. For example purposes, the values of m =1kg and K = π 2 J m 2 are<br />

used. The path x 1 (t) has S =0.355, the path x 2 (t) has S =0.364, and the<br />

physical path x 3 (t) has zero action S =0.<br />

We can derive the path that minimizes the action and that is found in<br />

nature using the Euler-Lagrange equation.<br />

∂L<br />

∂x − d dt<br />

∂L<br />

) =0 (11.37)<br />

∂ ( dx<br />

dt<br />

The rst term is the generalized potential. The second term is the time<br />

derivative of the generalized momentum. The equationof motionis found<br />

by putting these pieces together.<br />

Kx + m d2 x<br />

=0 (11.38)<br />

dt2 The rst term of the equationof motionis − ∣ −→ ∣ ∣∣.<br />

F spring The second term<br />

represents the accelerationof the mass. We have just found the equationof<br />

motion, and it is a statement of Newton's second law, force is mass times<br />

acceleration. It is also a statement of conservation of force on the mass.<br />

Equation11.38 is a second order linear dierential equationwith constant<br />

coecients. It is the famous wave equation, and its solution is well<br />

known<br />

(√ ) (√ )<br />

K K<br />

x(t) =c 0 cos<br />

m t + c 1 sin<br />

m t (11.39)<br />

where c 0 and c 1 are constants determined by the initial conditions. If we<br />

securely attach the mass to the spring, as opposed to letting the mass get<br />

kicked away, it will oscillate as described by the path x(t).<br />

<strong>Energy</strong> is conserved in this system. To verify conservation of energy,<br />

we can show that the total energy does not vary with time. The total<br />

energy is given by the Hamiltonian of Eq. 11.31. In this example, both<br />

the Hamiltonian and the Lagrangian do not explicitly depend on time,<br />

∂H<br />

=0and ∂L =0. Instead, they only depend on changes intime. For<br />

∂t<br />

∂t<br />

this reason, we say both the total energy and the Lagrangian have time<br />

translation symmetry, or we say they are time invariant. The spring and<br />

mass behave the same today, a week from today, and a year from today.

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