The Principle of Least Action
Cylinder rolling down an inclined plane
The principle of least action provides a different framework to summarize classical mechanics.
The action, denoted by S, is defined as the integral of the Lagrangian \mathcal{L} over a time interval from t_1 to t_2:
S = \int_{t_1}^{t_2} \mathcal{L}(q_1(t), \dots, q_n(t), \dot{q}_1(t), \dots, \dot{q}_n(t), t) \, \mathrm{\mathrm d}t
Here, q_k(t) represent the generalized coordinates, \dot{q}_k(t) are the generalized velocities, and the Lagrangian \mathcal{L} can also have an explicit time dependence.
A result is that Lagrange’s equations of motion imply an extremum of the action S, and conversely, an extremum of S leads to Lagrange’s equations. To demonstrate this, consider a simplified system with a single generalized coordinate q(t). We investigate what it means for S to be an extremum.
Suppose q(t) is the path for which S is an extremum.
If we introduce a small variation \delta q(t) to this path, with the constraint that the endpoints are fixed, i.e., \delta q(t_1) = 0 and \delta q(t_2) = 0, then the first-order change in S, denoted \delta S, must be zero.
The variation \delta S is calculated as:
\delta S = \int_{t_1}^{t_2} \left( \frac{\partial \mathcal{L}}{\partial q} \delta q + \frac{\partial \mathcal{L}}{\partial \dot{q}} \delta \dot{q} \right) \mathrm{\mathrm d}t
Recognizing that \delta \dot{q} = \frac{\mathrm{\mathrm d}}{\mathrm{\mathrm d}t}(\delta q), we apply integration by parts to the second term in the integrand:
\int_{t_1}^{t_2} \frac{\partial \mathcal{L}}{\partial \dot{q}} \delta \dot{q} \, \mathrm{\mathrm d}t = \frac{\partial \mathcal{L}}{\partial \dot{q}} \delta q \bigg|_{t_1}^{t_2} - \int_{t_1}^{t_2} \frac{\mathrm{\mathrm d}}{\mathrm{\mathrm d}t}\left(\frac{\partial \mathcal{L}}{\partial \dot{q}}\right) \delta q \, \mathrm{\mathrm d}t
Given the boundary conditions \delta q(t_1) = \delta q(t_2) = 0, the term \frac{\partial \mathcal{L}}{\partial \dot{q}} \delta q \bigg|_{t_1}^{t_2} vanishes.
The variation of the action then simplifies to:
\delta S = \int_{t_1}^{t_2} \left[ \frac{\partial \mathcal{L}}{\partial q} - \frac{\mathrm{\mathrm d}}{\mathrm{\mathrm d}t}\left(\frac{\partial \mathcal{L}}{\partial \dot{q}}\right) \right] \delta q \, \mathrm{\mathrm d}t
For S to be an extremum, \delta S must be zero for any arbitrary infinitesimal variation \delta q(t) (that respects the boundary conditions). This condition holds if and only if the expression multiplying \delta q(t) inside the integral is zero:
\frac{\partial \mathcal{L}}{\partial q} - \frac{\mathrm{\mathrm d}}{\mathrm{\mathrm d}t}\left(\frac{\partial \mathcal{L}}{\partial \dot{q}}\right) = 0
This is precisely Lagrange’s equation. The reasoning can be reversed to show that Lagrange’s equation implies an extremum for the action.
This principle provides an alternative viewpoint to Newtonian mechanics and is particularly useful for handling systems with constraints.
Instead of eliminating dependent variables due to constraints, the principle of stationary action can be formulated to include the constraint forces.
This is done by introducing “Lagrange multipliers”, which allows the independent variation of all coordinates while incorporating the constraints directly into the variational problem.
Lagrange multipliers
The method of Lagrange multipliers is a powerful mathematical technique used to find the local maxima and minima of a function subject to one or more equality constraints. The core idea is to convert a constrained optimization problem into an unconstrained one by introducing new variables called Lagrange multipliers.
Suppose we want to find the extremum (maximum or minimum) of a function f(x_1, x_2, \dots, x_n) subject to a set of m constraint equations:
\begin{aligned} g_1(x_1, x_2, \dots, x_n) &= c_1 \\ g_2(x_1, x_2, \dots, x_n) &= c_2 \\ &\vdots \\ g_m(x_1, x_2, \dots, x_n) &= c_m \end{aligned}
At an extremum point, the gradient of f must be a linear combination of the gradients of the constraint functions g_k. That is:
\nabla f = \sum_{k=1}^{m} \lambda_k \nabla g_k
where \lambda_k are the Lagrange multipliers.
This can be formulated by defining an auxiliary function, denoted \mathcal {L}^\prime:
\mathcal {L}^\prime(x_1, \dots, x_n, \lambda_1, \dots, \lambda_m) = f(x_1, \dots, x_n) - \sum_{k=1}^{m} \lambda_k (g_k(x_1, \dots, x_n) - c_k)
The extrema are then found by setting the partial derivatives of \mathcal {L}^\prime with respect to all x_i and all \lambda_k to zero:
\begin{aligned} \frac{\partial \mathcal {L}^\prime}{\partial x_i} &= 0 \quad \text{for } i=1, \dots, n \\ \frac{\partial \mathcal {L}^\prime}{\partial \lambda_k} &= 0 \quad \text{for } k=1, \dots, m \end{aligned}
The second set of equations simply recovers the original constraint equations g_k = c_k.
The method of Lagrange multipliers can be extended from functions of variables to functionals, which are typically integrals that depend on an unknown function and its derivatives. This is particularly relevant in the context of the Principle of Least Action.
Suppose we want to find the function y(x) that extremizes an integral:
I[y] = \int_a^b F(x, y, y^\prime) dx
where y^\prime = dy/dx, subject to an integral constraint:
J[y] = \int_a^b G(x, y, y^\prime) dx = \text{constant}
We define a new functional H[y] by introducing a Lagrange multiplier \lambda (which is a constant in this type of problem):
H[y] = I[y] + \lambda J[y] = \int_a^b (F(x, y, y^\prime) + \lambda G(x, y, y^\prime)) dx
The extremum of H[y] is found by treating the new integrand \mathcal{L}_{new}(x, y, y^\prime) = F(x, y, y^\prime) + \lambda G(x, y, y^\prime) as a new “Lagrangian” and applying the standard Euler-Lagrange equation to it:
\frac{\partial \mathcal{L}_{new}}{\partial y} - \frac{\mathrm d}{\mathrm dx} \left( \frac{\partial \mathcal{L}_{new}}{\partial y^\prime} \right) = 0
Solving this differential equation, along with the constraint J[y] = \text{constant} and boundary conditions, yields the function y(x) and the value of \lambda.
In Lagrangian mechanics, we often encounter systems where the motion of particles is restricted by holonomic constraints. These are constraints that can be expressed as algebraic equations relating the generalized coordinates \mathbf{q} = (q_1, q_2, \dots, q_N) and possibly time t:
g_k(\mathbf{q}, t) = 0 \quad k=1, \dots, m
These constraints mean that the N generalized coordinates are not all independent. The method of Lagrange multipliers allows us to incorporate these constraints directly into the Lagrangian formalism without needing to explicitly solve for a reduced set of independent coordinates.
The standard Lagrangian of the system is \mathcal{L}(\mathbf{q}, \dot{\mathbf{q}}, t) = T - V. We form an augmented Lagrangian (or modified Lagrangian), often denoted \mathcal{L}^\prime:
\mathcal{L}^\prime(\mathbf{q}, \dot{\mathbf{q}}, \mathbf{\lambda}, t) = \mathcal{L}(\mathbf{q}, \dot{\mathbf{q}}, t) - \sum_{k=1}^{m} \lambda_k(t) g_k(\mathbf{q}, t)
Here, \lambda_k(t) are the Lagrange multipliers, which are now generally functions of time. They are treated as additional generalized coordinates.
The equations of motion are obtained by applying the Euler-Lagrange equations to this augmented Lagrangian \mathcal{L}^\prime with respect to each generalized coordinate q_i (as if they were all independent) and also with respect to each Lagrange multiplier \lambda_k:
For each generalized coordinate q_i (i=1, \dots, N):
\frac{\mathrm d}{\mathrm dt} \left( \frac{\partial \mathcal{L}^\prime}{\partial \dot{q_i}} \right) - \frac{\partial \mathcal{L}^\prime}{\partial q_i} = 0
Since g_k are assumed to be holonomic constraints of the form g_k(\mathbf{q}, t), they do not depend on \dot{q_i} (i.e., \frac{\partial g_k}{\partial \dot{q_i}} = 0). Thus, \frac{\partial \mathcal{L}^\prime}{\partial \dot{q_i}} = \frac{\partial \mathcal{L}}{\partial \dot{q_i}}. The equation becomes:
\frac{\mathrm d}{\mathrm dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{q_i}} \right) - \left( \frac{\partial \mathcal{L}}{\partial q_i} - \sum_{k=1}^{m} \lambda_k \frac{\partial g_k}{\partial q_i} \right) = 0
Rearranging, we get the modified Euler-Lagrange equations:
\frac{\mathrm d}{\mathrm dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{q_i}} \right) - \frac{\partial \mathcal{L}}{\partial q_i} = -\sum_{k=1}^{m} \lambda_k \frac{\partial g_k}{\partial q_i}
The term on the right-hand side, Q_{i,c} = -\sum_{k=1}^{m} \lambda_k \frac{\partial g_k}{\partial q_i}, represents the generalized force of constraint corresponding to the coordinate q_i. The Lagrange multipliers \lambda_k(t) are directly related to the physical forces required to maintain the constraints.
For each Lagrange multiplier \lambda_k (k=1, \dots, m):
\frac{\partial \mathcal{L}^\prime}{\partial \lambda_k} = 0 \implies -g_k(\mathbf{q}, t) = 0 \implies g_k(\mathbf{q}, t) = 0
These equations simply recover the original constraint equations.
The full set of equations consists of N modified Euler-Lagrange equations for the q_i and m constraint equations g_k=0. Together, these form a system of N+m differential-algebraic equations that can be solved for the N functions q_i(t) and the m functions \lambda_k(t).
To solve a mechanics problem with holonomic constraints using Lagrange multipliers:
- Identify Coordinates and Constraints:
Choose a suitable set of N generalized coordinates q_i that describe the configuration of the system (initially, you can treat them as if they were independent).
Write down the m holonomic constraint equations in the form g_k(q_1, \dots, q_N, t) = 0.
- Write the Standard Lagrangian:
Determine the kinetic energy T and potential energy V of the system.
Form the standard Lagrangian \mathcal{L} = T - V.
- Introduce Lagrange Multipliers:
For each constraint equation g_k = 0, introduce a Lagrange multiplier \lambda_k(t).
- Form the Augmented Lagrangian \mathcal{L}^\prime:
Construct the augmented Lagrangian:
\mathcal{L}^\prime = \mathcal{L} - \sum_{k=1}^{m} \lambda_k g_k
(Note: Some texts use \mathcal{L}^\prime = \mathcal{L} + \sum \lambda_k g_k. The choice of sign for \lambda_k g_k affects the interpretation of \lambda_k as a force, but the resulting equations of motion for q_i are equivalent. The convention used here, with -\sum \lambda_k g_k, leads to Q_{i,c} = -\sum \lambda_k \frac{\partial g_k}{\partial q_i}.)
- Derive Equations of Motion:
For each generalized coordinate q_i: Apply the Euler-Lagrange equation using \mathcal{L}^\prime:
\frac{\mathrm d}{\mathrm dt} \left( \frac{\partial \mathcal{L}^\prime}{\partial \dot{q_i}} \right) - \frac{\partial \mathcal{L}^\prime}{\partial q_i} = 0
This will result in N equations involving q_i, \dot{q_i}, \ddot{q_i}, and \lambda_k. As noted before, this often simplifies to:
\frac{\mathrm d}{\mathrm dt} \left( \frac{\partial \mathcal{L}}{\partial \dot{q_i}} \right) - \frac{\partial \mathcal{L}}{\partial q_i} = -\sum_{k=1}^{m} \lambda_k \frac{\partial g_k}{\partial q_i}
- Include Constraint Equations:
The m constraint equations g_k(\mathbf{q}, t) = 0 must also be satisfied. (These are formally obtained by \frac{\partial \mathcal{L}^\prime}{\partial \lambda_k} = 0).
- Solve the System:
You now have a system of N second-order differential equations (for q_i) and m algebraic equations (the constraints g_k=0). These N+m equations are solved simultaneously to find the time evolution of the coordinates q_i(t) and the Lagrange multipliers \lambda_k(t).
- Interpret Multipliers (Optional but useful):
The Lagrange multipliers \lambda_k(t) can be interpreted in terms of the forces of constraint. The generalized force of constraint associated with coordinate q_i due to all constraints is Q_{i,c} = -\sum_{k=1}^{m} \lambda_k \frac{\partial g_k}{\partial q_i}.
This method provides a systematic way to handle constrained motion in Lagrangian mechanics, and the multipliers themselves provide valuable information about the forces maintaining these constraints.
Cylinder rolling down an inclined plane
Let’s consider the example of a cylinder rolling down an inclined plane without slipping which was analyzed here.
The Lagrangian \mathcal{L} is expressed in terms of the coordinate x (position of the center of mass along the incline) and \theta (angle of rotation of the cylinder):
\mathcal{L} = \frac{1}{2} m \dot{x}^2 + \frac{1}{2} I \dot{\theta}^2 + mgx \sin\varphi
where m is the mass, I is the moment of inertia, g is the gravitational acceleration, and \varphi is the angle of the incline.
The constraint of rolling without slipping is given by:
x = R\theta
where R is the radius of the cylinder.
To incorporate this constraint, we use the method of Lagrange multipliers. We define an augmented Lagrangian \mathcal{L}^{\prime}:
\mathcal{L}^{\prime} = \mathcal{L} - F(x - R\theta) = \frac{1}{2} m \dot{x}^2 + \frac{1}{2} I \dot{\theta}^2 + mgx \sin\varphi - \lambda (x - R\theta)
Here, \lambda is the Lagrange multiplier, which will be identified as the force of constraint (static friction in this case). We now treat x and \theta as independent variables in \mathcal{L}^{\prime}.
The Euler-Lagrange equation for a coordinate q using \mathcal{L}^{\prime} is:
\frac{\mathrm{\mathrm d}}{\mathrm{\mathrm d}t}\left(\frac{\partial \mathcal{L}^{\prime}}{\partial \dot{q}}\right) - \frac{\partial \mathcal{L}^{\prime}}{\partial q} = 0
For the coordinate \theta:
\begin{aligned} \frac{\partial \mathcal{L}^{\prime}}{\partial \dot{\theta}} & = I \dot{\theta} \\ \frac{\partial \mathcal{L}^{\prime}}{\partial \theta} & = -\lambda (-R) = \lambda R \end{aligned}
The equation of motion for \theta is:
\begin{aligned} & I \ddot{\theta} - \lambda R = 0 \\ & I \ddot{\theta} = \lambda R \end{aligned}
This equation represents the rotational dynamics: the net torque (\lambda R) equals the moment of inertia times the angular acceleration.
For the coordinate x:
\begin{aligned} \frac{\partial \mathcal{L}^{\prime}}{\partial \dot{x}} & = m \dot{x} \\ \frac{\partial \mathcal{L}^{\prime}}{\partial x} & = mg \sin\varphi - \lambda \end{aligned}
The equation of motion for x is:
\begin{aligned} & m \ddot{x} - (mg \sin\varphi - \lambda) = 0 \\ & m \ddot{x} = mg \sin\varphi - \lambda \end{aligned}
This is Newton’s second law for the translation of the center of mass along the incline: mass times acceleration equals the sum of the gravitational component (mg \sin\varphi) and the friction force -\lambda = -F.
These two equations, along with the constraint equation x = R\theta (and its time derivatives, e.g., \ddot{x} = R\ddot{\theta}), allow us to solve for the accelerations \ddot{x}, \ddot{\theta}, and the constraint force \lambda. This approach recovers the equations derived from Newtonian mechanics, including the theorem of angular momentum and the theorem of the center of mass.
Naturally, getting \lambda from the equation for \theta and using the no-slip condition:
m \ddot{x} = mg \sin\varphi - \lambda = mg \sin\varphi - \frac{I}{R} \ddot{\theta} = mg \sin\varphi - \frac{I}{R^2} \ddot{x}
we retrieve the previous result:
\ddot x = \frac{g \sin\varphi}{1 + \frac{I}{mR^2}}
Hanging chain (catenary)
A flexible chain or cable hanging freely under its own weight between two supports takes on a characteristic curve known as a catenary. This shape is not a parabola, as one might intuitively guess, but a hyperbolic cosine function. We can derive this shape by minimizing the potential energy of the chain, subject to the constraint that its length is fixed. This is a problem in the calculus of variations, solvable using the method of Lagrange multipliers for functionals.
Consider a uniform chain with constant linear mass density \rho. It hangs in the xy-plane between two points x = -a and x = a, with the attachment points at the same height. The total length of the chain is 2L_{total}. The chain will settle into a shape y(x) that minimizes its gravitational potential energy.
The potential energy I is given by:
I[y] = \int_{-a}^{a} \rho g y \, ds = \int_{-a}^{a} \rho g y \sqrt{1 + (y^\prime)^2} \, dx
where ds = \sqrt{1 + (y^\prime)^2} \, dx is the element of arc length, and g is the acceleration due to gravity.
The constraint is the fixed total length of the chain, 2L_{total}:
J[y] = \int_{-a}^{a} \sqrt{1 + (y^\prime)^2} \, dx = 2L_{total}
We form the new functional H = I + \lambda J:
H[y] = \int_{-a}^{a} (\rho g y \sqrt{1 + (y^\prime)^2} + \lambda \sqrt{1 + (y^\prime)^2}) \, dx = \int_{-a}^{a} (\rho g y + \lambda) \sqrt{1 + (y^\prime)^2} \, dx
The new Lagrangian \mathcal{L} is:
\mathcal{L}(y, y^\prime) = (\rho g y + \lambda) \sqrt{1 + (y^\prime)^2}
Since \mathcal{L} does not depend explicitly on x, we can use the Beltrami identity (a first integral of the Euler-Lagrange equation):
\mathcal{L} - y^\prime \frac{\partial \mathcal{L}}{\partial y^\prime} = C_1
where C_1 is a constant of integration. Calculating the partial derivative \frac{\partial \mathcal{L}}{\partial y^\prime}:
\frac{\partial \mathcal{L}}{\partial y^\prime} = (\rho g y + \lambda) \frac{y^\prime}{\sqrt{1 + (y^\prime)^2}}
Substituting into the Beltrami identity:
\begin{aligned} & (\rho g y + \lambda) \sqrt{1 + (y^\prime)^2} - y^\prime \left( (\rho g y + \lambda) \frac{y^\prime}{\sqrt{1 + (y^\prime)^2}} \right) = C_1 \\ & (\rho g y + \lambda) \left( \frac{1 + (y^\prime)^2 - (y^\prime)^2}{\sqrt{1 + (y^\prime)^2}} \right) = C_1 \end{aligned}
This simplifies to:
\frac{\rho g y + \lambda}{\sqrt{1 + (y^\prime)^2}} = C_1
Rearranging the equation for y^\prime:
\begin{aligned} & \sqrt{1 + (y^\prime)^2} = \frac{\rho g y + \lambda}{C_1} \\ & 1 + (y^\prime)^2 = \left( \frac{\rho g y + \lambda}{C_1} \right)^2 \\ & (y^\prime)^2 = \left( \frac{\rho g y + \lambda}{C_1} \right)^2 - 1 \end{aligned}
So, \frac{\mathrm dy}{\mathrm dx} = \sqrt{\left( \frac{\rho g y + \lambda}{C_1} \right)^2 - 1}. This is a separable differential equation. Let u = \frac{\rho g y + \lambda}{C_1}, then \mathrm du = \frac{\rho g}{C_1} \mathrm dy. The equation becomes:
\int \frac{\mathrm du}{\sqrt{u^2 - 1}} = \int \frac{\rho g}{C_1} \mathrm dx
Integrating gives:
\text{arccosh}(u) = \frac{\rho g}{C_1} x + D_1
where D_1 is another constant of integration. Substituting back u:
\begin{aligned} & \text{arccosh}\left( \frac{\rho g y + \lambda}{C_1} \right) = \frac{\rho g x}{C_1} + D_1 \\ & \frac{\rho g y + \lambda}{C_1} = \cosh\left( \frac{\rho g x}{C_1} + D_1 \right) \end{aligned}
Solving for y(x):
y(x) = \frac{C_1}{\rho g} \cosh\left( \frac{\rho g x}{C_1} + D_1 \right) - \frac{\lambda}{\rho g}
The chain is suspended symmetrically between x=-a and x=a. For symmetry y(-a) = y(a). This implies that the lowest point of the chain must be at x=0 if the support heights are equal.
\cosh\left( \frac{-\rho g a}{C_1} + D_1 \right) = \cosh\left( \frac{\rho g a}{C_1} + D_1 \right)
Since \cosh(A) = \cosh(B) implies A = \pm B, for a \neq 0, we must have D_1 = 0. So:
y(x) = \frac{C_1}{\rho g} \cosh\left( \frac{\rho g x}{C_1} \right) - \frac{\lambda}{\rho g}
We can choose the origin of the y-axis such that the lowest point of the chain is at y=0. So, y(0)=0.
0 = \frac{C_1}{\rho g} \cosh(0) - \frac{\lambda}{\rho g} = \frac{C_1}{\rho g} - \frac{\lambda}{\rho g}
This implies \lambda = C_1. The equation for the catenary simplifies to:
y(x) = \frac{C_1}{\rho g} \left( \cosh\left( \frac{\rho g x}{C_1} \right) - 1 \right)
The total length is 2L_{total}.
y^\prime = \sinh\left( \frac{\rho g x}{C_1} \right)
So:
\sqrt{1 + (y^\prime)^2} = \sqrt{1 + \sinh^2\left( \frac{\rho g x}{C_1} \right)} = \cosh\left( \frac{\rho g x}{C_1} \right)
The length integral is:
\begin{aligned} 2L_{total} & = \int_{-a}^{a} \cosh\left( \frac{\rho g x}{C_1} \right) dx = \left[ \frac{C_1}{\rho g} \sinh\left( \frac{\rho g x}{C_1} \right) \right]_{-a}^{a} \\ & = \frac{C_1}{\rho g} \left( \sinh\left( \frac{\rho g a}{C_1} \right) - \sinh\left( \frac{-\rho g a}{C_1} \right) \right) = \frac{2C_1}{\rho g} \sinh\left( \frac{\rho g a}{C_1} \right) \end{aligned}
The constant C_1 is determined by the transcendental equation:
L_{total} = \frac{C_1}{\rho g} \sinh\left( \frac{\rho g a}{C_1} \right)
Or, rearranging to match a common form:
\frac{L_{total} \rho g}{C_1} = \sinh\left( \frac{\rho g a}{C_1} \right)
The shape of the hanging chain is given by:
y(x) = \frac{C_1}{\rho g} \left( \cosh\left( \frac{\rho g x}{C_1} \right) - 1 \right)
where the constant C_1 is found by solving:
L_{total} = \frac{C_1}{\rho g} \sinh\left( \frac{\rho g a}{C_1} \right)
with 2L_{total} being the total length of the chain and 2a being the horizontal distance between the suspension points. The parameter C_1 has units of force and is related to the horizontal tension in the chain.
Pendulums
Consider a simple pendulum of mass m and length l. We use polar coordinates (r, \theta).
The kinetic energy is:
T = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2)
and the potential energy (with zero at pivot height) is:
V = -mgr\cos\theta
The Lagrangian is:
\mathcal{L} = T - V = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2) + mgr\cos\theta
The constraint for this system is that the length is fixed at l, which can be written as:
f_1(r) = r - l = 0
We form the modified Lagrangian \mathcal{L}^{\prime} by introducing a Lagrange multiplier \lambda_1:
\mathcal{L}^{\prime} = \mathcal{L} + \lambda_1 f_1(r) = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2) + mgr\cos\theta + \lambda_1(r-l)
Now, we derive the equations of motion from \mathcal{L}^{\prime}. For the \theta coordinate, the Euler-Lagrange equation:
\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial \mathcal{L}^{\prime}}{\partial \dot{\theta}}\right) - \frac{\partial \mathcal{L}^{\prime}}{\partial \theta} = 0
yields:
\frac{\mathrm d}{\mathrm dt}(mr^2\dot{\theta}) - (-mgr\sin\theta) = 0
which simplifies to:
mr^2\ddot{\theta} + 2mr\dot{r}\dot{\theta} + mgr\sin\theta = 0
Applying the constraint r=l (which implies \dot{r}=0 and \ddot{r}=0), this becomes:
ml^2\ddot{\theta} + mgl\sin\theta = 0
or:
\ddot{\theta} + \frac{g}{l}\sin\theta = 0
This is the standard equation of motion for a simple pendulum. Since \frac{\partial f_1}{\partial \theta} = 0, the generalized constraint force Q_\theta^{NC} = \lambda_1 \frac{\partial f_1}{\partial \theta} is zero, consistent with no \lambda_1 term appearing in this equation.
For the r coordinate, the Euler-Lagrange equation:
\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial \mathcal{L}^{\prime}}{\partial \dot{r}}\right) - \frac{\partial \mathcal{L}^{\prime}}{\partial r} = 0
gives:
\frac{\mathrm d}{\mathrm dt}(m\dot{r}) - (mr\dot{\theta}^2 + mg\cos\theta + \lambda_1) = 0
simplifying to:
m\ddot{r} - mr\dot{\theta}^2 - mg\cos\theta - \lambda_1 = 0
Applying the constraint conditions r=l, \dot{r}=0, \ddot{r}=0, we get
0 - ml\dot{\theta}^2 - mg\cos\theta - \lambda_1 = 0
Solving for \lambda_1:
\lambda_1 = -(ml\dot{\theta}^2 + mg\cos\theta)
To interpret \lambda_1 as the non-conservative force of constraint Q_r^{NC}, we use Q_r^{NC} = \lambda_1 \frac{\partial f_1}{\partial r}. Since f_1(r) = r-l, we have \frac{\partial f_1}{\partial r} = 1. Therefore,
Q_r^{NC} = \lambda_1 = -(ml\dot{\theta}^2 + mg\cos\theta)
This Q_r^{NC} represents the radial force of constraint. The tension T_{rod} in the pendulum rod is the reaction to this force, acting inwards on the mass, so:
T_{rod} = -Q_r^{NC} = ml\dot{\theta}^2 + mg\cos\theta
This is the expression for the tension in a pendulum string.
Now, let’s add a second constraint: the pendulum is forced to rotate such that \theta = at, where a is a constant angular velocity. The Lagrangian
\mathcal{L} = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2) + mgr\cos\theta
and the first constraint:
f_1(r) = r-l=0
remain the same as in the previous example.
The second constraint is:
f_2(\theta, t) = \theta - at = 0
We form the modified Lagrangian \mathcal{L}^{\prime \prime} with two Lagrange multipliers, \lambda_1 and \lambda_2:
\begin{aligned} \mathcal{L}^{\prime \prime} & = \mathcal{L} + \lambda_1 f_1(r) + \lambda_2 f_2(\theta, t) \\ & = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2) + mgr\cos\theta + \lambda_1(r-l) + \lambda_2(\theta - at) \end{aligned}
The equations of motion from \mathcal{L}^{\prime \prime} can be derived. For the r coordinate, the equation of motion:
m\ddot{r} - mr\dot{\theta}^2 - mg\cos\theta - \lambda_1 = 0
is obtained similarly to the first example. Applying the constraints r=l, \dot{r}=0, \ddot{r}=0 and also \theta=at, \dot{\theta}=a, this equation becomes:
0 - mla^2 - mg\cos(at) - \lambda_1 = 0
Solving for \lambda_1:
\lambda_1 = -(mla^2 + mg\cos(at))
For the \theta coordinate, the Euler-Lagrange equation is:
\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial \mathcal{L}^{\prime \prime}}{\partial \dot{\theta}}\right) - \frac{\partial \mathcal{L}^{\prime \prime}}{\partial \theta} = 0
yields:
\frac{\mathrm d}{\mathrm dt}(mr^2\dot{\theta}) - (-mgr\sin\theta + \lambda_2) = 0
which is:
mr^2\ddot{\theta} + 2mr\dot{r}\dot{\theta} + mgr\sin\theta - \lambda_2 = 0
Applying all constraints (r=l, \dot{r}=0, \ddot{r}=0 and \theta=at, \dot{\theta}=a, \ddot{\theta}=0), this simplifies to:
0 + 0 + mgl\sin(at) - \lambda_2 = 0
Solving for \lambda_2:
\lambda_2 = mgl\sin(at)
Now we interpret these multipliers as generalized non-conservative forces of constraint. For Q_r^{NC}, we have:
Q_r^{NC} = \lambda_1 \frac{\partial f_1}{\partial r} + \lambda_2 \frac{\partial f_2}{\partial r}
Since \frac{\partial f_1}{\partial r} = 1 and \frac{\partial f_2}{\partial r} = 0, we get:
Q_r^{NC} = \lambda_1 = -(mla^2 + mg\cos(at))
This is the radial force of constraint.
For Q_\theta^{NC}, we have:
Q_\theta^{NC} = \lambda_1 \frac{\partial f_1}{\partial \theta} + \lambda_2 \frac{\partial f_2}{\partial \theta}
Since \frac{\partial f_1}{\partial \theta} = 0 and \frac{\partial f_2}{\partial \theta} = 1, we find:
Q_\theta^{NC} = \lambda_2 = mgl\sin(at)
This Q_\theta^{NC} is the generalized force (a torque in this case) required to maintain the constraint f_2=0, i.e., to drive the pendulum at the specified angular velocity \dot{\theta}=a. The physical torque \tau^{NC} is equal to Q_\theta^{NC} in this case.
The total non-conservative constraint force vector can be expressed as:
\mathbf{F}^{NC} = Q_r^{NC} \mathbf{e}_r + \frac{Q_\theta^{NC}}{r} \mathbf{e}_\theta
Substituting the values and r=l:
\mathbf{F}^{NC} = (-(mla^2 + mg\cos(at))) \mathbf{e}_r + (mg\sin(at)) \mathbf{e}_\theta
This represents the total external force (beyond gravity accounted for in V) that must be applied to the mass m to enforce this specific constrained trajectory.