Particles Lagrangian Formulation
Let’s introduce a different method than Newton \mathbf F=m \mathbf a for deriving the equations of motion, known as the Lagrangian method.
The resulting equations—called the Lagrange equations—form the basis of analytical mechanics, which shows underlying structures of classical mechanics, offering insights into its symmetries and invariants.
Let’s consider a system of N particles. The position of the i^{th} particle is given by the vector \mathbf r_i, where i = 1, \ldots, N. Suppose that the system is subject to a set of k geometric constraints which can be written in the form:
\begin{aligned} & f_1(\mathbf r_1,\ \mathbf r_2,\,\ldots,\,\mathbf r_N,\, t) = 0\\ & f_2(\mathbf r_1,\ \mathbf r_2,\,\ldots,\,\mathbf r_N,\, t) = 0\\ & \vdots \\ & f_k(\mathbf r_1,\ \mathbf r_2,\,\ldots,\,\mathbf r_N,\, t) = 0 \end{aligned}
Such constraints, expressed as functional relations involving only the positions (and possibly time), are called holonomic constraints. Note that while the positions \mathbf r_i generally depend on time (implicit time dependence), the constraints may also exhibit explicit time dependence through t.
Non-holonomic constraints would, for instance, involve velocities in a manner that cannot be reduced to positional relations of the above type. Here, however, we focus on holonomic constraints expressed by equations of the form above.
We now introduce a fundamental concept in Lagrangian mechanics: the notion of degrees of freedom. With N particles, the system has 3N coordinates specifying their positions in three-dimensional space. If there are k holonomic constraints then, it is possible to eliminate k of these coordinates, expressing them in terms of the remaining ones. The number of independent coordinates left is:
n = 3N - k
This value n is called the number of degrees of freedom of the system: it is the minimal number of generalized coordinates q_1,\,\ldots,\,q_n required to uniquely specify the positions of all the particles in the system.
Suppose there is a single particle constrained to move on the surface of a sphere. In this case, two angles (for instance, spherical coordinates \theta and \phi) are required to determine its position, so the system has two degrees of freedom (n = 2).
Now, constrain the particle further by attaching it to a massless, fixed-length arm anchored at one of the poles of the sphere. The particle still moves on the sphere’s two-dimensional surface, but now only a single coordinate (an angle along a circle of constant latitude) is needed to specify its position. The system has only one degree of freedom (n = 1).
The standard notation is to denote these generalized coordinates as q_1, \ldots, q_n. They are called generalized because they can represent either distances, angles, or any other variables sufficient to describe the configuration.
The positions of the N particles can then be expressed as functions of the generalized coordinates and possibly time:
\begin{aligned} & \mathbf r_1 = \mathbf r_1(q_1, \ldots, q_n, t) \\ & \mathbf r_2 = \mathbf r_2(q_1, \ldots, q_n, t) \\ & \vdots \\ & \mathbf r_N = \mathbf r_N(q_1, \ldots, q_n, t) \end{aligned}
Here, the time t may appear explicitly in these functions, in addition to the implicit time dependence through q_1, \ldots, q_n.
To define the virtual displacement compatible with the constraints in complete generality, let’s consider the position vector of particle i when the generalized coordinates q_1, \ldots, q_n undergo infinitesimal variations \delta q_1, \ldots, \delta q_n.
The virtual displacement is the difference between the position at the perturbed coordinates and the position at the original coordinates. For infinitesimal variations, the virtual displacement of particle i is:
\delta \mathbf r_i = \sum_{j=1}^n \frac{\partial \mathbf r_i}{\partial q_j}\, \delta q_j
This gives the component of the displacement of particle i along each generalized coordinate, in accordance with the system constraints.
We now focus on describing the dynamics in a way that eliminates the constraint forces. Forces are decomposed into constraint forces and all other (applied or active) forces, which we denote simply as \mathbf F. The total force on a particle is thus the sum of the constraint force and \mathbf F.
For a single particle, Newton’s second law reads:
\mathbf F_{\text{constraint}} + \mathbf F = m\mathbf a
or, rearranged,
(\mathbf F - m\mathbf a) + \mathbf F_{\text{constraint}} = 0
We introduce a virtual displacement \delta\mathbf r compatible with the constraints and take the scalar product of the above equation with \delta\mathbf r:
\left(\mathbf F - m\mathbf a\right) \cdot \delta\mathbf r + \mathbf F_{\text{constraint}} \cdot \delta\mathbf r = 0
Constraint forces are always normal to the surface or curve defining the constraint, while virtual displacements are tangential to it. Therefore,
\mathbf F_{\text{constraint}} \cdot \delta\mathbf r = 0
This leaves:
\left(\mathbf F - m\mathbf a\right) \cdot \delta\mathbf r = 0
For a system of multiple particles, sum over all i:
\sum_i \left(\mathbf F_i - m_i \mathbf a_i\right) \cdot \delta\mathbf r_i = 0
This gives a general dynamical description with all constraint forces eliminated.
This equation is known as d’Alembert’s principle. Let us focus on the term involving the non-constraint forces:
\sum_i \mathbf F_i \cdot \delta \mathbf r_i
where \delta \mathbf r_i is the virtual displacement of particle i, compatible with the constraints. Substituting, the work of the applied forces along virtual displacements is:
\sum_i \mathbf F_i \cdot \delta \mathbf r_i = \sum_{j=1}^n \sum_i \mathbf F_i \cdot \frac{\partial \mathbf r_i}{\partial q_j} \delta q_j
This expresses the contribution of the forces to d’Alembert’s principle in terms of the generalized coordinates q_j and their virtual displacements \delta q_j.
This formula, collecting terms by \delta q_j can be written as:
\sum_i \mathbf F_i \cdot \delta \mathbf r_i = \sum_{j=1}^n \sum_i \mathbf F_i \cdot \frac{\partial \mathbf r_i}{\partial q_j} \delta q_j = \sum_{j=1}^n Q_j\,\delta q_j
where the generalized force Q_j is defined as
Q_j = \sum_{i=1}^N \mathbf F_i \cdot \frac{\partial \mathbf r_i}{\partial q_j}
Here, Q_j is the generalized force conjugate to the coordinate q_j; it gathers the effect of all the applied (non-constraint) forces projected along the directions defined by the generalized coordinates.
We now examine the kinetic term in d’Alembert’s principle, \sum_i m_i \mathbf a_i \cdot \delta \mathbf r_i. Consider a single particle first (dropping the index i for simplicity):
m \mathbf a \cdot \delta \mathbf r = m \ddot{\mathbf r} \cdot \delta \mathbf r
Substitute the expression for the virtual displacement \delta \mathbf r = \sum_{j=1}^n \frac{\partial \mathbf r}{\partial q_j} \delta q_j:
m \ddot{\mathbf r} \cdot \delta \mathbf r = \sum_{j=1}^n \left( m \ddot{\mathbf r} \cdot \frac{\partial \mathbf r}{\partial q_j} \right) \delta q_j
We focus on the term m \ddot{\mathbf r} \cdot \frac{\partial \mathbf r}{\partial q_j}. Using the product rule for differentiation:
\frac{\mathrm d}{\mathrm d t} \left( m \dot{\mathbf r} \cdot \frac{\partial \mathbf r}{\partial q_j} \right) = m \ddot{\mathbf r} \cdot \frac{\partial \mathbf r}{\partial q_j} + m \dot{\mathbf r} \cdot \frac{\mathrm d}{\mathrm d t} \left( \frac{\partial \mathbf r}{\partial q_j} \right)
Therefore,
m \ddot{\mathbf r} \cdot \frac{\partial \mathbf r}{\partial q_j} = \frac{\mathrm d}{\mathrm d t} \left( m \dot{\mathbf r} \cdot \frac{\partial \mathbf r}{\partial q_j} \right) - m \dot{\mathbf r} \cdot \frac{\mathrm d}{\mathrm d t} \left( \frac{\partial \mathbf r}{\partial q_j} \right)
We need to evaluate the terms \frac{\partial \mathbf r}{\partial q_j} and \frac{\mathrm d}{\mathrm d t} (\frac{\partial \mathbf r}{\partial q_j}). Recall the particle velocity \mathbf v = \dot{\mathbf r}:
\mathbf v = \frac{\mathrm d \mathbf r}{\mathrm d t} = \sum_{k=1}^n \frac{\partial \mathbf r}{\partial q_k} \dot q_k + \frac{\partial \mathbf r}{\partial t}
Since \mathbf r depends on (q_1, \ldots, q_n, t), but not on \dot q_k, we have:
\frac{\partial \mathbf v}{\partial \dot q_j} = \frac{\partial}{\partial \dot q_j} \left( \sum_{k=1}^n \frac{\partial \mathbf r}{\partial q_k} \dot q_k + \frac{\partial \mathbf r}{\partial t} \right) = \sum_{k=1}^n \frac{\partial \mathbf r}{\partial q_k} \frac{\partial \dot q_k}{\partial \dot q_j} = \frac{\partial \mathbf r}{\partial q_j}
Next, consider \frac{\mathrm d}{\mathrm d t} (\frac{\partial \mathbf r}{\partial q_j}). Since \frac{\partial \mathbf r}{\partial q_j} is a function of (q_1, \ldots, q_n, t), its total time derivative is:
\frac{\mathrm d}{\mathrm d t} \left( \frac{\partial \mathbf r}{\partial q_j} \right) = \sum_{k=1}^n \frac{\partial}{\partial q_k} \left( \frac{\partial \mathbf r}{\partial q_j} \right) \dot q_k + \frac{\partial}{\partial t} \left( \frac{\partial \mathbf r}{\partial q_j} \right)
Using the commutativity of partial derivatives:
\frac{\mathrm d}{\mathrm d t} \left( \frac{\partial \mathbf r}{\partial q_j} \right) = \sum_{k=1}^n \frac{\partial^2 \mathbf r}{\partial q_j \partial q_k} \dot q_k + \frac{\partial^2 \mathbf r}{\partial q_j \partial t} = \frac{\partial}{\partial q_j} \left( \sum_{k=1}^n \frac{\partial \mathbf r}{\partial q_k} \dot q_k + \frac{\partial \mathbf r}{\partial t} \right) = \frac{\partial \mathbf v}{\partial q_j}
Substitute these results (\frac{\partial \mathbf r}{\partial q_j} = \frac{\partial \mathbf v}{\partial \dot q_j} and \frac{\mathrm d}{\mathrm d t} (\frac{\partial \mathbf r}{\partial q_j}) = \frac{\partial \mathbf v}{\partial q_j}) into the expression for m \ddot{\mathbf r} \cdot \frac{\partial \mathbf r}{\partial q_j}:
m \ddot{\mathbf r} \cdot \frac{\partial \mathbf r}{\partial q_j} = \frac{\mathrm d}{\mathrm d t} \left( m \mathbf v \cdot \frac{\partial \mathbf v}{\partial \dot q_j} \right) - m \mathbf v \cdot \frac{\partial \mathbf v}{\partial q_j}
Now, introduce the kinetic energy T = \frac{1}{2} m \mathbf v^2 = \frac{1}{2} m (\mathbf v \cdot \mathbf v). We calculate its partial derivatives:
\frac{\partial T}{\partial \dot q_j} = \frac{1}{2} m \frac{\partial}{\partial \dot q_j} (\mathbf v \cdot \mathbf v) = m \mathbf v \cdot \frac{\partial \mathbf v}{\partial \dot q_j}
\frac{\partial T}{\partial q_j} = \frac{1}{2} m \frac{\partial}{\partial q_j} (\mathbf v \cdot \mathbf v) = m \mathbf v \cdot \frac{\partial \mathbf v}{\partial q_j}
Substituting these into the previous expression:
m \ddot{\mathbf r} \cdot \frac{\partial \mathbf r}{\partial q_j} = \frac{\mathrm d}{\mathrm d t} \left( \frac{\partial T}{\partial \dot q_j} \right) - \frac{\partial T}{\partial q_j}
Finally, insert this back into the expression for m \mathbf a \cdot \delta \mathbf r:
m \mathbf a \cdot \delta \mathbf r = \sum_{j=1}^n \left( \frac{\mathrm d}{\mathrm d t} \left( \frac{\partial T}{\partial \dot q_j} \right) - \frac{\partial T}{\partial q_j} \right) \delta q_j
Summing over all particles i, and defining the total kinetic energy T = \sum_i T_i:
\sum_i m_i \mathbf a_i \cdot \delta \mathbf r_i = \sum_i \sum_{j=1}^n \left( \frac{\mathrm d}{\mathrm d t} \left( \frac{\partial T_i}{\partial \dot q_j} \right) - \frac{\partial T_i}{\partial q_j} \right) \delta q_j = \sum_{j=1}^n \left( \frac{\mathrm d}{\mathrm d t} \left( \frac{\partial T}{\partial \dot q_j} \right) - \frac{\partial T}{\partial q_j} \right) \delta q_j
This concludes the transformation of the kinetic term.
Substituting these expressions back into d’Alembert’s principle gives:
\sum_{j=1}^n Q_j \delta q_j - \sum_{j=1}^n \left( \frac{\mathrm d}{\mathrm d t} \left( \frac{\partial T}{\partial \dot q_j} \right) - \frac{\partial T}{\partial q_j} \right) \delta q_j = 0
Combining the sums:
\sum_{j=1}^n \left[ Q_j - \left( \frac{\mathrm d}{\mathrm d t} \left( \frac{\partial T}{\partial \dot q_j} \right) - \frac{\partial T}{\partial q_j} \right) \right] \delta q_j = 0
Since the generalized coordinates q_j are independent, the virtual displacements \delta q_j are also independent and arbitrary. For this sum to be zero for any choice of \delta q_j, the coefficient of each \delta q_j must vanish:
\frac{\mathrm d}{\mathrm d t} \left( \frac{\partial T}{\partial \dot q_j} \right) - \frac{\partial T}{\partial q_j} = Q_j \quad \text{for } j = 1, \ldots, n
These are the Lagrange equations of motion in their first form.
We now specialize to the case where all applied forces \mathbf F_i derive from a potential energy function V(\mathbf r_1, \ldots, \mathbf r_N, t). In Cartesian coordinates, the force on particle i is given by \mathbf F_i = -\nabla_i V, where \nabla_i denotes the gradient with respect to the coordinates of particle i.
The generalized force Q_j becomes:
Q_j = \sum_{i=1}^N \mathbf F_i \cdot \frac{\partial \mathbf r_i}{\partial q_j} = \sum_{i=1}^N (-\nabla_i V) \cdot \frac{\partial \mathbf r_i}{\partial q_j}
Recognizing the chain rule for the partial derivative of V (which depends on q_k through the \mathbf r_i) with respect to q_j:
\frac{\partial V}{\partial q_j} = \sum_{i=1}^N (\nabla_i V) \cdot \frac{\partial \mathbf r_i}{\partial q_j}
For conservative forces, the generalized force is simply:
Q_j = -\frac{\partial V}{\partial q_j}
Substituting this into the first form of Lagrange’s equations:
\frac{\mathrm d}{\mathrm d t} \left( \frac{\partial T}{\partial \dot q_j} \right) - \frac{\partial T}{\partial q_j} = Q_j
gives:
\frac{\mathrm d}{\mathrm d t} \left( \frac{\partial T}{\partial \dot q_j} \right) - \frac{\partial T}{\partial q_j} = -\frac{\partial V}{\partial q_j}
Rearranging:
\frac{\mathrm d}{\mathrm d t} \left( \frac{\partial T}{\partial \dot q_j} \right) - \left( \frac{\partial T}{\partial q_j} - \frac{\partial V}{\partial q_j} \right) = 0
Since the potential energy V typically depends only on positions and possibly time, but not on generalized velocities, we have \frac{\partial V}{\partial \dot q_j} = 0. Therefore, we can write:
\frac{\mathrm d}{\mathrm d t} \left( \frac{\partial T}{\partial \dot q_j} \right) - \frac{\mathrm d}{\mathrm d t} \left( \frac{\partial V}{\partial \dot q_j} \right) - \left( \frac{\partial T}{\partial q_j} - \frac{\partial V}{\partial q_j} \right) = 0
Combining the terms:
\frac{\mathrm d}{\mathrm d t} \left( \frac{\partial (T-V)}{\partial \dot q_j} \right) - \frac{\partial (T-V)}{\partial q_j} = 0
Define the Lagrangian \mathcal L as the difference between the kinetic and potential energies:
\mathcal L = T - V
Lagrange’s equations for conservative systems take the simple form:
\frac{\mathrm d}{\mathrm d t} \left( \frac{\partial \mathcal L}{\partial \dot q_j} \right) - \frac{\partial \mathcal L}{\partial q_j} = 0 \quad \text{for } j = 1, \ldots, n
If some forces derive from a potential V (conservative forces) while others do not (non-conservative forces, denoted \mathbf F_i^{(nc)}), the total generalized force is Q_j = Q_j^{(c)} + Q_j^{(nc)}, where Q_j^{(c)} = -\frac{\partial V}{\partial q_j} and Q_j^{(nc)} = \sum_i \mathbf F_i^{(nc)} \cdot \frac{\partial \mathbf r_i}{\partial q_j}. The equations then become:
\frac{\mathrm d}{\mathrm d t} \left( \frac{\partial \mathcal L}{\partial \dot q_j} \right) - \frac{\partial \mathcal L}{\partial q_j} = Q_j^{(nc)}
Here, \mathcal L = T - V includes only the potential V corresponding to the conservative forces, and Q_j^{(nc)} represents the generalized forces associated with the non-conservative interactions.
Rectilinear motion
Let’s consider the simplest case: rectilinear motion. A particle of mass m is constrained to move along a straight line, which we take as the x-axis. The position is described by the single coordinate x.
The generalized coordinate is q_1 = x.
The kinetic energy is T = \frac{1}{2} m \dot x^2.
We assume a generic potential energy V(x).
The Lagrangian is \mathcal L = T - V = \frac{1}{2} m \dot x^2 - V(x).
Lagrange’s equation is:
\frac{\mathrm d}{\mathrm d t} \left( \frac{\partial \mathcal L}{\partial \dot x} \right) - \frac{\partial \mathcal L}{\partial x} = 0
Calculating the partial derivatives:
\frac{\partial \mathcal L}{\partial \dot x} = \frac{\partial}{\partial \dot x} \left(\frac{1}{2} m \dot x^2 - V(x)\right) = m \dot x
\frac{\partial \mathcal L}{\partial x} = \frac{\partial}{\partial x} \left(\frac{1}{2} m \dot x^2 - V(x)\right) = -\frac{\partial V}{\partial x}
Substituting into Lagrange’s equation:
\frac{\mathrm d}{\mathrm d t} (m \dot x) - \left(-\frac{\partial V}{\partial x}\right) = 0
\frac{\mathrm d}{\mathrm d t} (m \dot x) = -\frac{\partial V}{\partial x}
Recognizing that F_x = -\frac{\partial V}{\partial x} is the force acting on the particle, and \frac{\mathrm d}{\mathrm d t} (m \dot x) = m \ddot x is the mass times acceleration, we recover Newton’s second law:
m \ddot x = F_x
Pendulum
Let us analyze the simple mathematical pendulum using the Lagrangian formalism. The system consists of a point mass m attached to a massless rod of fixed length r, constrained to swing in a vertical plane under uniform gravity g.
The configuration of the pendulum is fully described by a single generalized coordinate, the angle \theta that the rod makes with the downward vertical. Thus, the system has one degree of freedom, n=1, with q_1 = \theta.
We first determine the kinetic energy T. The speed of the mass is v = r |\dot\theta|, so the kinetic energy is given by
T = \frac{1}{2} m v^2 = \frac{1}{2} m (r \dot\theta)^2 = \frac{1}{2} m r^2 \dot\theta^2
Next, we determine the potential energy V. Choosing the reference level V=0 at the lowest point of the swing (\theta=0), the height of the mass at angle \theta is h = r - r \cos \theta. The gravitational potential energy is then:
V = mgh = mgr(1 - \cos \theta)
The Lagrangian \mathcal L is defined as \mathcal L = T - V. Substituting the expressions for T and V, we get:
\mathcal L = \frac{1}{2} m r^2 \dot\theta^2 - mgr(1 - \cos \theta)
Since the equations of motion involve derivatives of \mathcal L, any constant term in \mathcal L (like -mgr) will vanish. We can therefore work with a simplified Lagrangian:
\mathcal L = \frac{1}{2} m r^2 \dot\theta^2 + mgr \cos \theta
Lagrange’s equation for the generalized coordinate \theta is:
\frac{\mathrm d}{\mathrm d t} \left( \frac{\partial \mathcal L}{\partial \dot\theta} \right) - \frac{\partial \mathcal L}{\partial \theta} = 0
We compute the required partial derivatives. The derivative with respect to the generalized velocity \dot\theta is:
\frac{\partial \mathcal L}{\partial \dot\theta} = \frac{\partial}{\partial \dot\theta} \left( \frac{1}{2} m r^2 \dot\theta^2 + mgr \cos \theta \right) = m r^2 \dot\theta
This term represents the generalized momentum conjugate to \theta, which in this case is the angular momentum about the pivot. The derivative with respect to the generalized coordinate \theta is:
\frac{\partial \mathcal L}{\partial \theta} = \frac{\partial}{\partial \theta} \left( \frac{1}{2} m r^2 \dot\theta^2 + mgr \cos \theta \right) = -mgr \sin \theta
This term is related to the generalized force, which corresponds to the torque exerted by gravity. Substituting these derivatives into Lagrange’s equation yields:
\frac{\mathrm d}{\mathrm d t} (m r^2 \dot\theta) - (-mgr \sin \theta) = 0
Performing the time differentiation gives:
m r^2 \ddot\theta + mgr \sin \theta = 0
Finally, dividing by the constant factor mr^2, we obtain the equation of motion for the simple pendulum:
\ddot\theta + \frac{g}{r} \sin \theta = 0
This derivation showcases the efficiency of the Lagrangian method. By calculating the kinetic and potential energies and applying the Lagrange equation, we systematically arrive at the equation of motion without explicitly dealing with constraint forces or decomposing vector forces and accelerations.
Cycloidal pendulum
Christiaan Huygens (1629-1695) sought to improve the accuracy of pendulum clocks. He realized that a simple pendulum is only approximately isochronous (its period is nearly independent of amplitude) for small oscillations. Huygens discovered that if the bob of a pendulum were constrained to move along a cycloidal path, its oscillations would be truly isochronous, regardless of the amplitude.
A cycloid is the curve traced by a point on the circumference of a circle as it rolls without slipping along a straight line. For our pendulum, we’ll consider an inverted cycloid, with the cusp pointing upwards.
Let R be the radius of the generating circle. The parametric equations for a point (x, z) on an inverted cycloid, where z is measured downwards from the cusp (highest point, \theta=0) and x=0 at the cusp, is given by:
\begin{aligned} x & = R(\theta - \sin\theta) \\ z & = R(1 - \cos\theta) \end{aligned}
Here, \theta is the angle through which the generating circle has rolled. The lowest point of the path occurs at \theta = \pi, where z = 2R.
Consider a bob of mass m moving along this cycloidal path.
If we set the gravitational potential energy to be zero at the lowest point of the path (z=2R), then at any point z along the path:
V = mg(2R - z)
Substituting z = R(1 - \cos\theta):
V = mg(2R - R(1 - \cos\theta)) = mgR(1 + \cos\theta)
The velocity components are:
\begin{aligned} \dot{x} & = \frac{\mathrm dx}{\mathrm dt} = R\dot{\theta}(1 - \cos\theta) \\ \dot{z} & = \frac{\mathrm dz}{\mathrm dt} = R\dot{\theta}\sin\theta \end{aligned}
The kinetic energy T = \frac{1}{2}m(\dot{x}^2 + \dot{z}^2) is derived as follows:
\begin{aligned} T & = \frac{1}{2}m [R^2\dot{\theta}^2(1 - \cos\theta)^2 + R^2\dot{\theta}^2\sin^2\theta] \\ & = \frac{1}{2}m R^2\dot{\theta}^2 [(1 - 2\cos\theta + \cos^2\theta) + \sin^2\theta] \\ & = \frac{1}{2}m R^2\dot{\theta}^2 (2 - 2\cos\theta) \\ & = mR^2\dot{\theta}^2(1 - \cos\theta) \end{aligned}
The Lagrangian \mathcal L of the system is \mathcal L = T - V:
\mathcal L = mR^2\dot{\theta}^2(1 - \cos\theta) - mgR(1 + \cos\theta)
The equation of motion can be found using the Euler-Lagrange equation for the coordinate \theta:
\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial \mathcal L}{\partial \dot{\theta}}\right) - \frac{\partial \mathcal L}{\partial \theta} = 0
Let’s calculate the partial derivatives:
\begin{aligned} & \frac{\partial \mathcal L}{\partial \dot{\theta}} = 2mR^2\dot{\theta}(1 - \cos\theta) \\ & \frac{\mathrm d}{\mathrm dt}\left(\frac{\partial \mathcal L}{\partial \dot{\theta}}\right) = 2mR^2\ddot{\theta}(1 - \cos\theta) + 2mR^2\dot{\theta}(\sin\theta \dot{\theta}) = 2mR^2\ddot{\theta}(1 - \cos\theta) + 2mR^2\dot{\theta}^2\sin\theta \\ & \frac{\partial \mathcal L}{\partial \theta} = mR^2\dot{\theta}^2(\sin\theta) - mgR(-\sin\theta) = mR^2\dot{\theta}^2\sin\theta + mgR\sin\theta \end{aligned}
Substituting these into the Euler-Lagrange equation:
2mR^2\ddot{\theta}(1 - \cos\theta) + 2mR^2\dot{\theta}^2\sin\theta - (mR^2\dot{\theta}^2\sin\theta + mgR\sin\theta) = 0
This simplifies to:
2mR^2\ddot{\theta}(1 - \cos\theta) + mR^2\dot{\theta}^2\sin\theta - mgR\sin\theta = 0
Using the trigonometric identities 1 - \cos\theta = 2\sin^2(\theta/2) and \sin\theta = 2\sin(\theta/2)\cos(\theta/2), it becomes:
2mR^2\ddot{\theta}(2\sin^2(\theta/2)) + mR^2\dot{\theta}^2(2\sin(\theta/2)\cos(\theta/2)) - mgR(2\sin(\theta/2)\cos(\theta/2)) = 0
Assuming \sin(\theta/2) \neq 0 (i.e., not at the cusp \theta=0), we can divide by 2mR\sin(\theta/2):
2R\ddot{\theta}\sin(\theta/2) + R\dot{\theta}^2\cos(\theta/2) - g\cos(\theta/2) = 0
To simplify this equation, let’s introduce a change of variable: u = \cos(\theta/2). Then, the time derivatives are:
\begin{aligned} \dot{u} & = -\frac{1}{2}\sin(\theta/2)\dot{\theta} \\ & = -\frac{1}{4}\cos(\theta/2)\dot{\theta}^2 - \frac{1}{2}\sin(\theta/2)\ddot{\theta} \end{aligned}
We can write:
\sin(\theta/2)\ddot{\theta} = -\frac{1}{2}\dot{\theta}^2\cos(\theta/2) + \frac{g}{2R}\cos(\theta/2)
Substituting this into the expression for \ddot{u}:
\begin{aligned} \ddot{u} & = -\frac{1}{4}\cos(\theta/2)\dot{\theta}^2 - \frac{1}{2}\left(-\frac{1}{2}\dot{\theta}^2\cos(\theta/2) + \frac{g}{2R}\cos(\theta/2)\right) \\ & = -\frac{1}{4}\cos(\theta/2)\dot{\theta}^2 + \frac{1}{4}\dot{\theta}^2\cos(\theta/2) - \frac{g}{4R}\cos(\theta/2) \\ & = -\frac{g}{4R}\cos(\theta/2) \end{aligned}
Since u = \cos(\theta/2), we get the equation for a simple harmonic oscillator:
\ddot{u} + \frac{g}{4R}u = 0
The equation \ddot{u} + \omega^2 u = 0 describes simple harmonic motion with an angular frequency \omega. For the cycloidal pendulum, this angular frequency is:
\omega_u = \sqrt{\frac{g}{4R}}
The period of oscillation T_{osc} is given by T_{osc} = 2\pi/\omega_u:
T_{osc} = 2\pi \sqrt{\frac{4R}{g}} = 4\pi\sqrt{\frac{R}{g}} This period is independent of the amplitude of oscillation (i.e., the maximum value of \theta), making the cycloidal pendulum truly isochronous.
One way to construct a cycloidal pendulum is to suspend the bob by a string of length 4R and have the string wrap around two cycloidal “cheeks” or “jaws” as it swings. The shape of these cheeks ensures that the bob itself traces out a cycloidal path.