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Lagrange's Theorem


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  • 1. Physics 430: Lecture 15 Lagrange’s Equations Dale E. Gary NJIT Physics Department
  • 2.
    • Statement of the problem:
      • A surface of revolution is generated as follows: Two fixed
      • points ( x 1 , y 1 ) and ( x 2 , y 2 ) in the x-y plane are joined by a
      • curve y = y ( x ) . [Actually, you’ll make life easier if you start
      • out writing this as x = x ( y ) .] The whole curve is now rotated
      • about the x axis to generate a surface. Show that the curve
      • for which the area of the surface is stationary has the form
      • where x o and y o are constants. (This is often called the soap-bubble problem, since the resulting surface is usually the shape of a soap bubble held by two coaxial rings of radii y 1 and y 2 .)
    • Solution:
      • What we have to minimize now is not a line, but rather a surface, dA = y d  ds .
      • We could write this
      • but then both Euler-Lagrange terms will be non-zero. Instead, follow the hint:
    Problem 6.19 October 21, 2010 (x 1 ,y 1 ) (x 2 ,y 2 ) y(x) y x ds
  • 3.
    • Solution, cont’d:
      • We identify
      • so and
      • Squaring, we have
      • Solving for
      • Integrating
      • The statement of the problem wants y ( x ) , so we have to solve for y , to get
      • where we identify C = y o .
    Problem 6.19, cont’d October 21, 2010
  • 4. 6.4 More Than Two Variables
    • Aside: Maximum and minimum vs. stationary—geodetics on surfaces.
    • When we discussed the shortest path between two points, we examined the situation in the figure below.
    • However, this is not the most general path. Here is one that cannot be expressed as a function y ( x ) .
    • To handle this case, we need to consider a path specified by an independent variable u along the path, i.e. x = x ( u ), y = y ( u ) .
    • Now the length of the path is then
    • In general, the problem is
    • This may seem more difficult than the previous case,
    • but in fact we just proceed in the same manner with
    • the “wrong” paths
    October 21, 2010 x y x 1 x 2 y 2 y 1 y = y(x) (right) 2 1
  • 5. More Than Two Variables-2
    • The stationary path is the one for which
    • Following the same procedure as before, we end up with two Euler-Lagrange equations that the function f must satisfy:
    • In Chapter 7, we will meet problems with several variables, with time t as the independent variable. In these cases, there will be two or more Euler-Lagrange equations to satisfy (for example, equations for x , y and z ).
    • One of the great strengths of Lagrangian mechanics is its ability to deal with cartesian, cylindrical, spherical, and any other coordinate systems with ease.
    • In order to generalize our discussion, we write ( x , y , z ), (  ,  , z ) or ( r ,  ,  ) as generalized coordinates ( q 1 , q 2 , …, q N ) .
    October 21, 2010
  • 6. The Lagrangian
    • In the next chapter, we will use slightly different notation, and refer to our integrand function
    • as the Lagrangian
    • Note that because the independent variable is t , we can use rather than .
    • We will then make the action integral stationary
    • which requires that L satisfy
    • We will develop this more next time.
    October 21, 2010
  • 7. Advantages Over Newtonian Mechanics
    • We are now going to use the ideas of the previous lecture to develop a new formalism for mechanics, called Lagrangian mechanics, invented by Lagrange (1736-1813).
    • There are two important advantages of the Lagrange formalism over that of Newton.
    • First, as we have seen, Lagrange’s equations take the same form for any coordinate system, so that the method of solution proceeds in the same way for any problem.
    • Second, the Lagrangian approach eliminates the forces of constraint, which we talked about in Chapter 4. This makes the Lagrangian formalism easier to solve in constrained problems.
    • This chapter is the heart of advanced classical mechanics, but it introduces some new methods that will take getting used to. Once you master it, you will find it an extraordinarily powerful way to solve mechanics problems.
    • To help you master it, we will spend the next three lectures on it, and you will be solving a number of problems. However, there are many more at the end of the chapter, and you are advised to try a few more.
    October 21, 2010
  • 8. 7.1 Lagrange’s Equations for Unconstrained Motion
    • Recall our general function f , that played such a large role in the Euler-Lagrange equation
    • To make the connection to mechanics, we are now going to show that a function called the Lagrangian, L = T – U , is the function that, when used in the Euler-Lagrange equation, leads to the equation of motion for a particle.
    • Here, T is the particle’s kinetic energy
    • and U is the potential energy
    • Note that L is NOT the total energy, E = T + U . One can ask why the quantity T – U should give rise to the equation of motion, but there seems to be no good answer.
    • I do note, however, that the problem can be cast in terms of the total energy, which gives rise to Hamiltonian mechanics (sec. 7.8 and chap. 13).
    October 21, 2010
  • 9. Lagrangian
    • Obviously, with the dependences of T on the x, y, z velocities, and U on the x, y, z positions, the Lagrangian depends on both, i.e.
    • Let’s look at the first two derivatives
    • But note that if we differentiate the second equation with respect to time we get
    • So you can see that the Lagrange equation is manifestly true for a free particle:
    • In Cartesian coordinates (so far) in three dimensions, we have:
    October 21, 2010 Note: this last equality is only true in an inertial frame.
  • 10. Connection to Euler-Lagrange
    • If you compare the Lagrange equations with the Euler-Lagrange equation we developed in the previous chapter, you see that they are identical.
    • Since the Euler-Lagrange equation is the solution to the problem of stationarity of a path integral, we see that
    • is stationary for the path followed by the particle.
    • This integral has a special name in Physics—it is called the action integral , and when it is a minimum it is called the principle of least action , although that is a misnomer in the sense that this could be a maximum, or even an inflection point.
    • The principle is also called Hamilton’s Principle:
    October 21, 2010 The actual path that a particle follows between two points 1 and 2 in a given time interval, t 1 to t 2 , is such that the action integral is stationary when taken along the actual path.
  • 11. Generalized Coordinates
    • We have now seen that the following three statements are completely equivalent:
      • A particle’s path is determined by Newton’s second law F = m a .
      • The path is determined by the three Lagrangian equations (at least in Cartesian coordinates).
      • The path is determined by Hamilton’s Principle.
    • Again, the great advantage is that we can prove that Lagrange’s equations hold for any coordinate system such that, for any value r = ( x, y, z ) there is a unique set of generalized coordinates ( q 1 , q 2 , q 3 ) and vice versa.
    • Using these, we can write the Lagrangian in terms of generalized coordinates as
    • In terms of these, we then have
    • These are true in any coordinate system, so long as the coordinates are measured in an inertial frame.
    October 21, 2010
  • 12. Example 7.1
    • For a single particle in two dimensions, in Cartesian coordinates, under some arbitrary potential energy U ( x, y ) , the Lagrangian is
    • In this case, there are two Lagrange equations
    • The left side of each equation is just
    • The right side of each equation, in turn, is just
    • Equating these, we have Newton’s second law
    October 21, 2010
  • 13. Generalized Force and Momentum
    • Notice that the left side term in Cartesian coordinates gives the force
    • When this is expressed in terms of generalized coordinates (which, for example, could be angular coordinates) this term is not necessarily a force component, but it plays a role very like a force, and indeed is called the generalized force .
    • Likewise, the right side in terms of generalized coordinates plays the role of a momentum, and is called the generalized momentum .
    • Thus, the Lagrange equation can be read as
    • generalized force = rate of change of generalized momentum .
    October 21, 2010
  • 14. Example 7.2
    • Let’s repeat example 7.1, but in polar coordinates. In this case, the potential energy is U ( r,  ) , and you should be able to write down the kinetic energy directly without much thought as
    • so the Lagrangian is just
    • In this case, there are two Lagrange equations
    • Inserting the above Lagrangian into these equations gives
    • The first of these equations says
    • The second equation is best interpreted by recalling the gradient in polar coordinates
    October 21, 2010 centripetal acceleration r  2 . angular momentum mr 2  . torque.
  • 15. N Free Particles
    • It should be obvious how to extend these ideas to a larger number of particles. In the case of two particles, at positions r 1 and r 2 , for example, Newton’s second law says each component of each particle obeys the equations
    • In the Lagrangian formalism, we have the equivalent Lagrange equations
    • We could likewise write these in generalized coordinates as
    • An important example we will use repeatedly in Chapter 8 is to replace r 1 and r 2 with R CM = ( m 1 r 1 + m 2 r 2 )/( m 1 + m 2 ) , and r = r 1  r 2 .
    • For N particles, then, there are 3 N Lagrange equations.
    October 21, 2010
  • 16. 7.2 Constrained Systems Example
    • We have said that the great power of the Lagrange formalism is that constraint forces disappear from the problem.
    • Before getting into the general case, let’s do a familiar problem—instead of a free particle, let’s consider one that is tied to the ceiling, i.e. the simple pendulum.
    • The pendulum bob moves in both x and y , but it moves under the constraint
    • A perfectly valid way to proceed might be to eliminate the y variable,
    • and express everything in terms of x , e.g. replace y with
    • However, it is much simpler to express the problem in terms of the
    • natural coordinate  .
    • The first task is to write down the Lagrangian L = T – U in terms of  . Clearly, for this problem U = mgh = mgl (1  cos  ) . Likewise, the kinetic energy is
    • The relevant Lagrange equation is and you can basically write down the solution
    October 21, 2010 Equivalent to  = I   h