Problem 20 - Classical Mechanics

Question

Another way of deriving the equation of motion (5.16) is to use Lagrange's equations. Express the kinetic energy \(\frac{1}{2} m(\mathrm{~d} \boldsymbol{r} / \mathrm{d} t)^{2}\) in terms of \((x, y, z)\), and show that Lagrange's equations (3.44) reproduce (5.16) for the case where the force is conservative.

Step-by-Step Solution

Answer

To summarize, we derived the equation of motion using Lagrange's equations by performing the following steps: 1. We expressed the kinetic energy in terms of (x, y, z): \(T = \frac{1}{2}m\left(\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}z}{\mathrm{d}t}\right)^2\right)\) 2. We wrote out Lagrange's equations for each coordinate (x, y, z): - \(m\ddot{x} = Q_x\) - \(m\ddot{y} = Q_y\) - \(m\ddot{z} = Q_z\) 3. We showed that these equations reproduce equation (5.16) when the force is conservative, resulting in the following equations: - \(m\ddot{x} = -\frac{\partial U}{\partial x}\) - \(m\ddot{y} = -\frac{\partial U}{\partial y}\) - \(m\ddot{z} = -\frac{\partial U}{\partial z}\) Hence, Lagrange's equations can be used to derive the equation of motion for conservative forces.

Step 1: Express the kinetic energy in terms of (x, y, z)

First, we'll express the given kinetic energy \(\frac{1}{2}m(\frac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t})^2\) in terms of \((x, y, z)\). We know that \(\boldsymbol{r} = x\boldsymbol{\hat{i}} + y\boldsymbol{\hat{j}} + z\boldsymbol{\hat{k}}\). Then, we can differentiate \(\boldsymbol{r}\) with respect to time and get the velocity vector \(\boldsymbol{v} = \frac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t} = \frac{\mathrm{d}x}{\mathrm{d}t}\boldsymbol{\hat{i}} + \frac{\mathrm{d}y}{\mathrm{d}t}\boldsymbol{\hat{j}} + \frac{\mathrm{d}z}{\mathrm{d}t}\boldsymbol{\hat{k}}\). Now, we can find the square of the velocity vector's magnitude: \((\boldsymbol{v})^2 = \left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}z}{\mathrm{d}t}\right)^2\) Then, the kinetic energy in terms of \((x, y, z)\) is given by \(T = \frac{1}{2}m\left(\left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}z}{\mathrm{d}t}\right)^2\right)\)

Step 2: Write out Lagrange's equations

Lagrange's equations are given by \(\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial T}{\partial \dot{q}_i}\right) - \frac{\partial T}{\partial q_i} = Q_i\), where \(q_i\) represents the generalized coordinates (in this case, \(x, y, z\)) and \(Q_i\) are the generalized forces. So, we need to calculate \(\frac{\partial T}{\partial \dot{x}}, \frac{\partial T}{\partial \dot{y}}, \frac{\partial T}{\partial \dot{z}}, \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y}, \frac{\partial T}{\partial z}\). Since \(T\) is dependent only on the velocities, we have \(\frac{\partial T}{\partial x} = \frac{\partial T}{\partial y} = \frac{\partial T}{\partial z} = 0\). And we can determine the partial derivatives with respect to the velocities as \(\frac{\partial T}{\partial \dot{x}} = m\dot{x}, \frac{\partial T}{\partial \dot{y}} = m\dot{y}, \frac{\partial T}{\partial \dot{z}} = m\dot{z}\). Now, let's write Lagrange's equations for each coordinate: 1. For \(x\): \(\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial T}{\partial \dot{x}}\right) - \frac{\partial T}{\partial x} = m\ddot{x} = Q_x\) 2. For \(y\): \(\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial T}{\partial \dot{y}}\right) - \frac{\partial T}{\partial y} = m\ddot{y} = Q_y\) 3. For \(z\): \(\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial T}{\partial \dot{z}}\right) - \frac{\partial T}{\partial z} = m\ddot{z} = Q_z\)

Step 3: Show that the equations reproduce (5.16) for conservative forces

When the force is conservative, we can express the force as the negative gradient of a potential energy function \(U(x, y, z)\), i.e., \(\boldsymbol{F} = -\boldsymbol{\nabla}U\). Thus for each component of the force, we have \(Q_x = -\frac{\partial U}{\partial x}\), \(Q_y = -\frac{\partial U}{\partial y}\), and \(Q_z = -\frac{\partial U}{\partial z}\). Let's substitute these into the Lagrange's equations from Step 2: 1. \(m\ddot{x} = -\frac{\partial U}{\partial x}\) 2. \(m\ddot{y} = -\frac{\partial U}{\partial y}\) 3. \(m\ddot{z} = -\frac{\partial U}{\partial z}\) These equations are equivalent to equation (5.16) for conservative forces. Therefore, we have shown that Lagrange's equations can be used to derive the equation of motion.

Key Concepts

Kinetic Energy

Kinetic energy, often abbreviated as \( T \), is an essential concept in understanding motion. It represents the energy an object possesses due to its motion. In the context of Lagrange's equations, it plays a pivotal role because Lagrangian mechanics often starts by defining the kinetic energy of the system.To calculate kinetic energy in our problem, we consider an object moving in three-dimensional space. The position of the object is captured by the vector \( \boldsymbol{r} \), which is composed of its components \( x, y, z \). The velocity vector \( \boldsymbol{v} = \frac{\mathrm{d}\boldsymbol{r}}{\mathrm{d}t} \) is the time derivative of this position vector. The magnitude of this velocity vector, squared, gives the sum of the squares of the rate of change of each coordinate.Thus, the expression for kinetic energy in terms of \( x,y,z \) becomes:

• \( T = \frac{1}{2}m\left( \left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}y}{\mathrm{d}t}\right)^2 + \left(\frac{\mathrm{d}z}{\mathrm{d}t}\right)^2 \right) \)

This formula quantifies how energy changes as the object accelerates through space.

Generalized Coordinates

To solve complex mechanical problems, physicists often use a powerful framework called generalized coordinates. Generalized coordinates allow the creation of a simpler representation of a system's configuration, especially when constraints are involved.In our exercise, generalized coordinates are directly applied as the Cartesian coordinates \( x, y, z \). They are generalized in the sense that they can describe any configuration of the system, even if it involves complex constraints or motion along paths that are not straightforward.Lagrange's equations harness these generalized coordinates to describe the dynamics of the system:

• They allow the transformation of kinetic information into simplified forms.
• These equations integrate the kinetic and potential energy, focused solely on the generalized coordinates.
• The elimination of direct constraints simplifies problem-solving, as they inherently handle the constraints.

Thus, using generalized coordinates, Lagrange's approach often resolves motion in a more straightforward manner than classical Newtonian mechanics.

Conservative Forces

Conservative forces are forces for which the work done is independent of the path taken. The work done by conservative forces such as gravity, for example, depends only on the initial and final positions. This characteristic allows us to define a potential energy \( U \) that is the same at any point of an isolated system when moving between those points.In the context of Lagrange's equations, conservative forces are crucial because they allow the use of potential energy to fully describe the force impacts without explicitly solving for the forces themselves. Here, the forces associated with \( x, y, z \) become:

• \( Q_x = -\frac{\partial U}{\partial x} \)
• \( Q_y = -\frac{\partial U}{\partial y} \)
• \( Q_z = -\frac{\partial U}{\partial z} \)

This means instead of directly dealing with forces, we can simply compute derivatives of the potential energy, making Lagrange's equation more manageable and insightful in systems subjected to conservative forces.