classical mechanics fall 2011 chapter 7: lagrange’s...
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Classical Mechanics Fall 2011
Chapter 7: Lagrange’s Equations
1. Introduction to Lagrange’s Equations
Let us consider a single particle that moves under the influence of conservative forces. There are no constraints on the motion of the particle; however, its path must be consistent with Newton’s second law. The kinetic energy of the particle is
T = 1
2m x2 + y2 + z2( ), (7.1)
and the potential energy is
U =U(r ) =U(x, y, z). (7.2)
The Lagrangian for this system is defined as
L = T −U. (7.3)
Let us relate the Lagrangian to Newton’s second law. Since the forces are conservative,
Fx = − ∂U
∂x= ∂L∂x. (7.4)
Further,
∂T∂ x
= ∂L∂ x
= mx = px . (7.5)
By Newton’s second law,
Fx = px . (7.6)
In terms of L, Newton’s second law is given by
∂L∂x
= ddt
∂L∂ x. (7.7)
This equation is called Lagrange’s equation. Similar equations can be derived for the y and z components. Clearly, Lagrange’s equations (for all Cartesian coordinates) are equivalent to Newton’s second law.
Eq. (7.7) has exactly the same form as the Euler-‐Lagrange equation that we have previously seen. It follows that
δS = δ Ldt
t1
t2
∫ = 0, (7.8)
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i.e., the integral S (called the action integral) is stationary for any path followed by the particle within the time interval t1 to t2. Eq. (7.8) is called Hamilton’s principle. Hamilton’s principle is usually regarded as fundamental in mechanics and Lagrange’s equations are derived from it.
Generalized Coordinates
It is often much easier to use coordinate systems other than the Cartesian one to solve problems. Do Lagrange’s equations change when we move from the Cartesian coordinate system to another one? Let us assume that we can fully describe the configuration (position, orientation, etc.) of the system in terms of another set of coordinates. A set of quantities that can completely describe the configuration or state of a system is called a set of generalized coordinates. These could be other common coordinate systems such as the spherical polar system, or they may be other quantities that are not geometric in nature. For an unconstrained particle moving in three-‐dimensional space, three generalized coordinates are necessary to describe its configuration, which is just the position in the case of a single particle. The position of the particle can be uniquely specified in terms these three generalized coordinates. In other words, each Cartesian component of the position can be uniquely expressed in terms of these coordinates:
x = x(q1,q2,q3); y = y(q1,q2,q3); z = z(q1,q2,q3). (7.9)
In these coordinates, the action integral is
S = L(q1,q2,q3, q1, q2, q3)dt
t1
t2
∫ . (7.10)
Clearly, the value of the definite integral will not change with a change of coordinates. Thus, the action integral is still stationary and Lagrange’s equations have exactly the same form in the new (and any) coordinate system:
∂L∂qi
= ddt
∂L∂ qi
. [i = 1,2,3] (7.11)
In analogy to Newton’s second law in Cartesian coordinates, the quantity ∂L ∂qi is defined as the ith component of the generalized force and ∂L ∂ qi is the ith component of the generalized momentum. To make these ideas more concrete, let us use the Lagrangian formalism for a particle moving in a plane. We will specify the particle’s position using polar coordinates.
Prelude: In-‐class Problem: Taylor, Problem 7.5
Two-‐Dimensional Motion of a Single Particle Using Polar Coordinates
The kinetic energy of the particle is
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T = 1
2m r2 + r2 φ 2( ). (7.12)
The potential energy is
U =U(r,φ). (7.13)
The Lagrangian is therefore
L = 1
2m r2 + r2 φ 2( )−U(r,φ). (7.14)
Applying Lagrange’s equation for the radial coordinate gives
∂L∂r
= ddt
∂L∂ r
, which gives
mr φ 2 − ∂U∂r
= mr, or, − ∂U∂r
= mr −mr φ 2. (7.15)
Since Fr = −∂U ∂r, we see that the generalized force for the radial coordinate is the sum of the negative of the centripetal force (i.e., the centrifugal force) and the radial component of the conservative force. Of course, we obtained exactly the same equation using Newton’s second law in the form of the second equation in Eq. (7.15).
For the polar-‐angle coordinate, the Lagrange equation is
∂L∂φ
= ddt
∂L∂ φ. (7.16)
This gives
− ∂U∂φ
= ddt
mr2 φ( ). (7.17)
Now, in polar coordinates
∇U = ∂U
∂rr̂ + 1
r∂U∂φ
φ̂ (7.18)
Thus, Fφ = − 1r ∂U ∂φ( ) and so Eq. (7.17) becomes
rFφ =dldt, (7.19)
where l = mr2 φ is the angular momentum. Thus, the generalized momentum in this case is
the angular momentum and the generalized force is the torque. One could say that the
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generalized force and momentum are the quantities that arise naturally from the generalized coordinates used to describe the system.
The Lagrangian formalism can be easily extended to systems consisting of more than one unconstrained particles. Thus, if we have a system of N particles, we would need three generalized coordinates per particle, and so a total of 3N generalized coordinates would be needed. The corresponding Lagrange’s equations are
∂L∂qi
= ddt
∂L∂ qi
. [i = 1,2,...., 3N ] (7.20)
The kinetic and potential energies in the Lagrangian are the total values for the system.
2. Lagrange’s Equations: Constrained Motion
A particle moving on a horizontal table is constrained to move in two dimensions because of the action of the normal force. The normal force is called a constraint force. Constraints reduce the number of generalized coordinates needed to describe the motion of the system. For example, to describe the motion of a single particle moving on a horizontal table, only two coordinates are needed; the number of coordinates have been reduced from three to two. In general, for a system of N particles, the x-‐component of the position of the ith particle is given by
xi = xi q1,q2,....qn ,t( ), [i = 1,2,...,N ] (7.21)
where the number of generalized coordinates n < 3N in a constrained system. The other Cartesian coordinates can be similarly expressed in terms of the generalized coordinates. The number of degrees of freedom of a system is the number of independent coordinates necessary to specify an arbitrary displacement. Put another way, a system has p degrees of freedom if exactly p independent parameters are needed to specify the configuration of the system at any point in time. If the number of degrees of freedom equals the number of generalized coordinates needed to describe the system, the system is said to be holonomic. A holonomic system has holonomic constraints, which can be described by equations relating the generalized coordinates of the unconstrained system:
gj q1,q2,...,q3N ,t( ) = 0, [j = 1,2,..,m] (7.22)
where gj is the jth functional relationship among the generalized coordinates and there are m constraints. The number of degrees of freedom (and the number of generalized coordinates for the constrained system) is n = 3N – m. Though it seems natural for the number of generalized coordinates to be equal to the number of degrees of freedom, it does not necessarily have to be so. However, we will not deal with non-‐holonomic systems in this course.
Using the Lagrangian formalism for holonomic systems, one finds that one can determine the equations of motion without considering the constraint forces. This is not possible with Newtonian dynamics. The equations of motion will involve the independent generalized
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coordinates (and their time derivatives) used to describe the system in the presence of constraints and so will be consistent with the constraints.
Consider Eq. (7.21) for a single particle. The n generalized coordinates are consistent with the constraints. The x -‐component of the velocity can be expressed in terms of the generalized coordinates via the chain rule:
x = ∂x
∂q1
dq1dt
+ ∂x∂q2
dq2dt
+ ...+ ∂x∂qn
dqndt
+ ∂x∂t
= ∂x∂qi
dqidt
+ ∂x∂ti=1
n
∑ . (7.23)
If we assume natural coordinates, x is not an explicit function of time and so ∂x ∂t = 0 in Eq. (7.23). We can define the generalized velocity corresponding to the ith generalized coordinate as
qi =
dqidt. (7.24)
Thus, x is a function of both the generalized coordinates and velocities:
x = x q1,q2,...qn , q1, q1,..qn( ). (7.25)
One can obtain similar expression for y and z . One can then evaluate the kinetic energy
T = 1
2m x2 + y2 + z2( ) (7.26)
in terms of the generalized coordinates and velocities using Eq. (7.25) and its analogs for y and z . One can take the derivative of T with respect to the generalized velocities, and identifying the ith component of the generalized force as
Qi = Fx∂x∂qi
+ Fy∂y∂qi
+ Fz∂z∂qi
= − ∂U∂qi, (7.27)
one finds, upon using Newton’s second law, that
ddt
∂L∂ qi
⎛⎝⎜
⎞⎠⎟= ∂L∂qi. (7.28)
Thus, Lagrange’s equations have exactly the same form if the motion is constrained. Note that Eq. (7.28) is valid for any choice of generalized coordinates (the number of coordinates must be equal to the number of degrees of freedom; i.e., the system must be holonomic) because the Lagrangian is a scalar quantity, which is invariant under coordinate transformations. Again, Lagrange’s equations can be straightforwardly extended to systems of more than one particles. In this case, the kinetic energy and potential energy in the Lagrangian are the total values for the system.
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3. Applications of Lagrange’s Equations
Taylor, Problem 7.23
The motion of the larger cart is specified, so the coordinate X is specified and represents a constraint. We need only the coordinate x to describe the motion of the smaller cart. We must remember to write down the Lagrangian with respect to an inertial reference frame. This is because Newton’s second law has the usual form only in inertial frames. Thus, even though the coordinates themselves may be defined with respect to non-‐inertial frames (as is x in this problem), the Lagrangian itself must be constructed in an inertial frame.
The velocity of the smaller cart (relative to the inertial frame attached to the floor) is
v = d dt X + x( ) = X + x. As both carts move, the extension of the spring relative to the floor is simply x. Thus, the Lagrangian for the smaller cart is
L = 1
2m x + X( )2 − 12 kx
2.
Lagrange’s equation for the coordinate x is
ddt
∂L∂ x
⎛⎝⎜
⎞⎠⎟ −
∂L∂x
= 0,
i.e.,
m ddtx + X( ) + kx = 0.
m x + X( ) +mω 02x = 0.
x +ω 02x = − X =ω 2Acosωt,
which is the required form, with B =ω 2A . The equation of motion represents the forced harmonic oscillator without damping. What would happen to the cart eventually in this ideal case?
Taylor, Problem 7.38
(a) In spherical polar coordinates, the velocity is given by
v = r = ddt
rr̂( ) = rr̂ + r ̂r. (7.29)
Now, any infinitesimal change in r̂ must be perpendicular to it (the magnitude cannot change!) and so the tip of the r̂ unit vector must move over the surface of a sphere. It follows that
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dr̂ = dθθ̂ + sinθdφφ̂.
Thus,
dr̂dt
= dθdt
θ̂ + sinθ dφdt
φ̂,
or,
̂r = θθ̂ + sinθ φφ̂. (7.30)
Thus, the velocity in spherical coordinates is
v = rr̂ + r θθ̂ + r sinθ φφ̂. (7.31)
(a) The Lagrangian is
L = 1
2m r2 + r2 θ 2 + r2 sin2θ φ 2( )−mgr cosθ .
(Remember that z = r cosθ .) For a cone, the polar angle θ is fixed: θ =α . Thus, θ = 0 . Therefore, the Lagrangian becomes
L = 1
2m r2 + r2 sin2α φ 2( )−mgr cosα . ----- (i)
(b) The equations of motion can be found from Lagrange’s equations. For the radial coordinate, we have
ddt
∂L∂ r
⎛⎝⎜
⎞⎠⎟ =
∂L∂r.
This yields
mr = mr sin2α φ 2 −mgcosα . -‐-‐-‐-‐-‐ (ii)
For the φ coordinate, we have
ddt
∂L∂ φ
⎛⎝⎜
⎞⎠⎟= ∂L∂φ.
This gives
ddt
mr2 sin2α φ( ) = 0.
We can rewrite the above equation as
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ddt
mρ2 φ( ) = 0,
where ρ = r sinα . Now,
lz = mρ2 φ, ----- (iii)
so the equation of motion becomes
dlzdt
= 0,
or,
lz = constant.
Using Eq. (iii) to substitute for φ in Eq. (ii), we obtain
mr = mr sin2α lz
mr2 sin2α⎛⎝⎜
⎞⎠⎟2
−mgcosα ,
or,
mr = lz
2
mr3 sin2α−mgcosα . ---- (iv)
If lz = 0, Eq. (iv) becomes
mr = −mgcosα .
This tells us that the radial force (away from the apex of the cone) equals the radial component of the weight (up is positive, so the component pointing away from the apex of the cone is negative). This makes perfect sense if the z-‐component of the angular momentum is zero.
The particle can remain in a horizontal circular path if r = 0 . Substituting this in Eq. (iv) and solving for r yields
r = r0 =lz2
m2gsin2α cosα⎛⎝⎜
⎞⎠⎟
1/3
.
(c) We substitute r = r0 + ε(t) into Eq. (iv) yielding
mε = lz
2
m r0 + ε( )3 sin2α−mgcosα . ---- (v)
Now,
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r0 + ε( )−3 = r0−3 1+ εr0( )−3 ≈ r0−3 1− 3ε
r0( ). Substituting this approximate result into Eq. (v) gives
mε = lz
2
mr03 sin2α
1− 3 εr0
⎛⎝⎜
⎞⎠⎟−mgcosα . ------ (vi)
But, as established above,
lz2
mr03 sin2α
= mgcosα ----- (vii)
is the condition for a horizontal circular orbit. Thus Eq. (vi) becomes
ε = −
3lz2
m2r04 sin2α
ε = 0. ----(viii)
Eq. (viii) is the harmonic oscillator equation in ε ; therefore, the particle oscillates about a stable circular path with angular frequency
Ω =3lz2
m2r04 sin2α
= 3gcosαr0
.
[The last expression is obtained by using Eq. (vii).]
In-‐class Problem: Taylor, Problem 7.29.
4. Ignorable Coordinates and Conservation Laws
Consider a Lagrangian expressed in terms of generalized coordinates q1,q2,....,qn :
L = L(q1,q2,...,qn , q1, q2,..., qn ,t). (7.32)
The Lagrange equation for the ith coordinate is
ddt
∂L∂ qi
⎛⎝⎜
⎞⎠⎟= ∂L∂qi. (7.33)
If L is not an explicit function of qi, then the coordinate qi is said to be ignorable or cyclic. If a coordinate qi is ignorable, then
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∂L∂qi
= 0. (Ignorable coordinate) (7.34)
Substituting this in Eq. (7.33) gives
ddt
∂L∂ qi
= 0. (7.35)
This indicates that the generalized momentum pi ≡
∂L∂ qi
is conserved. (Unfortunately, we
are using the same symbol for momentum and generalized momentum. This is not the best state of affairs, but it is the norm.) Thus, if the Lagrangian is invariant under changes in a coordinate, then the corresponding generalized momentum is conserved. This very important relationship between the invariance of the Lagrangian and conservation laws is called Noether’s theorem. For example, if the Lagrangian is invariant under spatial translations in a given direction, then the component of the total momentum in that direction is conserved. Noether’s theorem expresses the beautiful relationship between symmetry and conservation laws in physics.
It is also useful to note (though I will not derive it in these notes) that if the Lagrangian does not depend explicitly on time, a quantity called the Hamiltonian is conserved. The Hamiltonian is defined as
H = pi
i=1
n
∑ qi − L. (7.36)
Further, if the relationship between the Cartesian coordinates and the generalized coordinates is time independent, i.e.,
xi = xi (q1,q2,...,qn ); yi = yi (q1,q2,...,qn ); zi = zi (q1,q2,...,qn ), (7.37)
then the Hamiltonian is simply the total energy:
H = T +U. (7.38)
Thus, if Eq. (7.37) is satisfied, and the Lagrangian is independent of time, the total energy is conserved.
5. Lagrange Multipliers and Constraint Forces
One of the benefits of the Lagrangian formalism is that one can obtain equations of motion without needing to deal with constraint forces. However, if one needs to calculate the constraint forces, the Lagrangian method can also accommodate this.
As indicated before, holonomic constraints can be expressed as
gj q1,q2,...,qN ,t( ) = 0, [j = 1,2,..,m] (7.39)
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where the gj are algebraic relationships among the generalized coordinates. In Eq. (7.39), there are N generalized coordinates in the unconstrained system and m constraints. Eq. (7.39) may be alternatively expressed as
f j q1,q2,...,qN ,t( ) = constant. [j = 1,2,..,m] (7.40)
An example of a holonomic constraint is a particle moving on the surface of a sphere. Using Cartesian coordinates, the equation of constraint is
x2 + y2 + z2 = a2. (7.41)
Since one coordinate can be expressed in terms of the other two via Eq. (7.41), there are only two independent coordinates.
In differential form, the holonomic constraint Eq. (7.40) can be expressed as
∂ f j∂qii
N
∑ dqi = 0, (j = 1,2,...,m) (7.42)
where for simplicity, we assume that the constraint equations are not explicitly dependent on time. Hamilton’s principle tells us that
δ Ldt1
2
∫ = dt ∂L∂qi
− ddt
∂L∂ qi
⎛⎝⎜
⎞⎠⎟
⎡
⎣⎢
⎤
⎦⎥
i=1
N
∑ δ1
2
∫ qi = 0. (7.43)
Taking dqi ≈ δqi in Eq. (7.42), we can incorporate Eq. (7.42) into Eq. (7.43):
δ Ldt1
2
∫ = dt ∂L∂qi
− ddt
∂L∂ qi
⎛⎝⎜
⎞⎠⎟+ λ j
∂ f j∂qij=1
m
∑⎡
⎣⎢
⎤
⎦⎥
i=1
N
∑ δ1
2
∫ qi = 0, (7.44)
where λ j is an undetermined Lagrange multiplier. Because of the constraints, the displacements δqi are not all independent; they are related by the m constraint equations in Eq. (7.42). We can separate Eq. (7.44) into independent and non-‐independent parts. There are N –m independent coordinates; thus, we have
δ Ldt1
2
∫ = dt ∂L∂qi
− ddt
∂L∂ qi
⎛⎝⎜
⎞⎠⎟+ λ j
∂ f j∂qij=1
m
∑⎡
⎣⎢
⎤
⎦⎥
i=1
N−m
∑ δ1
2
∫ qi
+ dt ∂L∂qi
− ddt
∂L∂ qi
⎛⎝⎜
⎞⎠⎟+ λ j
∂ f j∂qij=1
m
∑⎡
⎣⎢
⎤
⎦⎥
i=N−m+1
N
∑ δ1
2
∫ qi = 0. (7.45)
We can now choose the Lagrange multipliers λ j such that the second integral is zero. Thus, the quantity in the square brackets is zero. Since the “independent” δqi displacements are arbitrary, it follows that the first integral must also vanish. Hence, we have
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ddt
∂L∂ qi
⎛⎝⎜
⎞⎠⎟= ∂L∂qi
+ λ j
∂ f j∂qij=1
m
∑ . (i = 1,2,...,N ); (j = 1,2,...,m) (7.46)
There are N equations for the generalized coordinates qi and including the multipliers, N +m unknowns. But there are m equations of constraint, so the system can be solved completely. The multipliers λ j are related to the constraint forces.
Example: A particle of mass m is placed at the top of a smooth hemisphere of radius a. Find the normal force exerted on the particle by the sphere. If the particle is disturbed and slides on the sphere, find the vertical distance below the top at which the particle leaves the sphere.
We will use spherical polar coordinates. The Lagrangian is
L = 1
2m r2 + r2 θ 2 + r2 sin2θ φ 2( )−mgr cosθ .
Note that L is independent of φ , which means the z-‐component of the angular momentum is conserved. Since its initial value is zero (because φinitial = 0 ), it is always zero. Thus, we can rewrite the Lagrangian as
L = 1
2m r2 + r2 θ 2( )−mgr cosθ .
The constraint equation is r = a. Thus, f (r,θ ) = r = a (a constant). Hence,
∂ f∂r
= 1 and ∂ f∂θ
= 0. Therefore, we need only one multiplier, which we will call λ .
The Lagrange equation for the radial coordinate is therefore (note that r = 0 and r = a)
ma θ2 −mgcosθ + λ = 0. ---- (i)
The Lagrange equation for the angular coordinate is
ma2 θ = mgasinθ . --- (ii)
We can integrate Eq. (ii) to obtain (fill in the details!)
θ 2 = 2g
a1− cosθ( ). -‐-‐-‐-‐ (iii)
Using Eq. (iii) to substitute for θ2 in Eq. (i) yields
λ = mg(3cosθ − 2).
The quantity λ is the normal force. Setting λ = 0 gives the angle of departure
N
mg
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θ = θ f = arccos23
⎛⎝⎜
⎞⎠⎟ . The required vertical distance is
h = r(1− cosθ f ) =r3.