In Taylor, read sections 7.7, 7.8, and 7.10 for today, and 10.1 for Monday.
Misplaced, we will conclude after the discussion of constraint forces.
As we've seen in some of the example problems, conservation of momentum and energy is built into the Lagrangian approach. It is insightful to see this explicitly.
Conservation of momentum is related to the uniformity of space. We expect the laws of physics to hold equally well in any inertial frame. So if we translate our reference frame by a small amount, say in the x direction, then we expect the motion of a system to be unchanged (except for the small change in x coordinate). Explicitly, if L(r1, ..., rn) is the Lagrangian in the original reference frame, then in the translated frame the Lagrangian is L' ( r1 + ε xhat, ..., rn + ε xhat ). Note that the first derivatives of the coordinates are unchanged. In order that the motion is unchanged, we require that δL = 0. Using the notation for differentials shown in a previous lecture, we can express δL in terms of derivatives of the generalized coordinates and the time derivatives of the generalized coordinates. For simplicity, let's assume that the generalized coordinates are cartesian coordinates, xi, yi, zi. The differential becomes
Using Lagrange's equations the partial derivatives can be equated to the time derivative of the x component of the momentum:
We are assuming that space is uniform and want the equality to hold for any ε, therefore we require that
Where P = ∑ pi is the total momentum of the system. This relation indicates that total momentum of the system is conserved if it is unchanged when displaced by a small amount. So we see that conservation of momentum is related to the uniformity of space.
Conservation of energy rests on the Lagrangian being independent of time; it can depend on qi and qidot, but must lack explicit t dependence. To see this, we need to evaluate the total time derivative of the Lagrangian. Recall that the Lagrangian depends on generalized coordinates qi, their time derivatives, qidot, and possibly t,
We need to use the chain rule to evaluate the total derivative:
This looks a little nicer when written with the sum notation
Using Lagrange's equation and our definition of generalized momentum, we can replace the first term in the sum by
the derivative of the generalized momentum. The derivative of the Lagrangian in the second term is just the generalized momentum, so
When the Lagrangian isn't explicitly dependent on time, then the final partial derivative is zero and we can write
or that the quantity ∑ ( pi qidot ) - L is a constant. This quantity is special, and is called the Hamiltonian
The Hamiltonian is important for a number of reasons: there's a formulation of the laws of mechanics based on the Hamiltonian, similar to the Lagrangian formulation we are using now; the Hamiltonian plays an important role in quantum mechanics; and, under certain restrictions, the Hamiltonian is the total energy of the system. We shall prove the last point.
The condition that is needed for the Hamiltonian to be the total energy of the system is that the relation between the generalized coordinates and Cartesian coordinates be time independent,
Generalized coordinates that satisfy this condition are called natural.
To show that the Hamiltonian is the total energy for natural coordinates, begin by writing the kinetic energy in terms of the generalized coordinates, starting with the velocities
and insert this into the expression for kinetic energy
Since the cartesian coordinates are assumed to be time independent, their partial derivatives are time independent. These can be combined with the mass and summed over $alpha; to form a quantity we'll call Ajk,
With this definition, the kinetic energy is written
Now, to prove that the Hamiltonian is the total energy, we need to evaluate the quantity ∑ ( pi qidot ). Begin by evaluating the generalized momentum
Using this we have
This gives us the result we want, since inserting this in the definition of the Hamiltonian yields
Recall that this entire argument began by consider how the Lagrangian would change by a translation in time. We've shown that if the Lagrangian is unaffected by a translation in time then energy is conserved. This complements the earlier proof that if the Lagrangian is unaffected by a translation in space then momentum is conserved. Likewise, one can show that if the Lagrangian is unchanged by a rotation along a particular axis, then angular momentum along that axis is conserved. Thus we arrive at an interesting connection between the fundamental conservation laws of energy, momentum, and angular momentum, and the uniformity of the laws of physics in time, space, and angle of rotation.
Interesting, but does involve the use of the vector potential, something beyond the present level. I recommend that students refer back to this when doing E&M at the appropriate level. This is an example of how topics in physics tie together; our divisions into standalone subjects is somewhat arbitrary.
This technique is discussed in Taylor 7.10. This material is optional.
Sometimes it is necessary to constrain the relationship between variables in a problem. For instance, to calculate the shortest path between two points on a cylinder one can use cylindrical coordinates. But how would one calculate the shortest path between two points on an arbitrary surface (with no particular symmetry, only known through some parameterization (x(t), y(t), z(t))? Here it is convenient to use an equation of constraint that relates x, y, and z, and incorporate that constraint into the Euler-Lagrange equations. Let's investigate how this is done.
For simplicity, begin with a problem involving 3 variables, x, y, and t. Say that we wish to minimize the integral
with the constraint g ( x, y ) = 0. (For convenience we assume f doesn't depend on t -- could be put back later without changing our result. If there were no constraint, we would begin by saying that the minimum occurs when δS = 0, or
We can apply integration by parts to the second and fourth terms:
therefore
and since the first term on the right is the integral of a derivative, the integral is simply the derivand evaluated at the limits. But by definition, δx = 0 at the endpoints (the path must go through the endpoints, so the deviation is zero at the endpoints), and this term evaluates to zero. The same result applies to the fourth term. The result is
Now, if there is no constraint, then δx and δy can vary independently, so both terms in square brackets must vanish separately, yielding the normal pair of Euler-Lagrange equations. With a constraint, δx and δy can't vary independently; in particular, from the constraint equation
Since the sum is zero, we can multiply it by any function (within reason) and it is still zero. In particular, we can multiply it by a function of t, λ(t), and add it to the integral. The functoin λ is called a Lagrange multiplier. Then
Since the function λ(t) is an arbitrary function, we can choose it such that the expression in square brackets is zero (and consequently, the second expression is also zero), that is, choose λ(t) such that
and then
Now we must solve for x(t), y(t), and λ(t). To accomplish this we need a third equation, which we have from the constraint,
In principle, these three equations can be solved for x(t), y(t), and λ(t).
In addition, there is more information we can extract from the Lagrange multiplier. Since we have used Cartesian coordinates explicitly, we know the form of the kinetic energy and the Lagrangian
We also know that the potential energy does not include the constraint force, that is contained in the constraint equation g ( x, y ) = 0. And, the Lagrange equations are
and
Since the derivatives of U cannot include the constraint force, the constraint force must be given by the product of the Lagrange multiplier and the derivative of the constraint equation