Question

For a system specified by the coordinates \(q\) and \(t\), show that the equation of motion is unchanged if the Lagrangian \(L(q, \dot{q}, t)\) is replaced by $$ L_{1}=L+\frac{d \phi(q, t)}{d t} $$ where \(\phi\) is an arbitrary function. Deduce that the equation of motion of a particle that moves in one dimension subject to a force \(-d V(x) / d x\) ( \(x\) being measured from a point \(O\) ) is unchanged if \(O\) is forced to move with a constant velocity \(v\) \((x\) still being measured from \(O)\).

Short answer

The equation of motion is unchanged if \( L \) is replaced by \( L + \frac{d \phi}{dt} \). The equation of motion for a particle is also unchanged if the origin moves with a constant velocity.

Step-by-step solution

- Understanding the given Lagrangian

Begin by considering the original Lagrangian, given by: \[ L(q, \dot{q}, t) \]. The equation of motion can be derived from this Lagrangian using the Euler-Lagrange equation:

- Applying the Euler-Lagrange equation

The Euler-Lagrange equation for the coordinate \(q\) is: \[\frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \right) - \frac{\partial L}{\partial q} = 0.\]

- Incorporating the new Lagrangian

The new Lagrangian is given by: \[ L_1 = L + \frac{d \phi(q, t)}{dt} \] where \( \phi(q, t) \) is an arbitrary function of \( q \) and \( t \).

- Calculating the derivative terms

Compute the partial derivatives for the new Lagrangian: \[ \frac{\partial L_1}{\partial \dot{q}} = \frac{\partial L}{\partial \dot{q}} + \frac{d}{dt} \left( \frac{\partial \phi(q, t)}{\partial \dot{q}} \right) \] Since \( \phi \) does not depend on \( \dot{q} \), \( \frac{\partial \phi}{\partial \dot{q}} = 0 \).

- Substituting into Euler-Lagrange

Substitute \( \frac{\partial L_1}{\partial \dot{q}} \) and \( \frac{\partial L_1}{\partial q} \) into the Euler-Lagrange equation:\[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \right) + \frac{d}{dt} \left( \frac{d \phi}{d t} \right) - \left( \frac{\partial L}{\partial q} + \frac{\partial \phi}{\partial q} \right) = 0. \]

- Simplifying the equation

Noting that the total derivative of \( \phi \) with respect to time is: \[ \frac{d \phi}{dt} = \frac{\partial \phi}{\partial t} + \frac{\partial \phi}{\partial q} \dot{q}, \]and simplifying, we find that the additional terms involving \( \phi \) cancel out. Thus, the equation of motion remains:\[ \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \right) - \frac{\partial L}{\partial q} = 0. \]

- Deduction for a particle in one dimension

Consider a particle moving in one dimension with a Lagrangian: \[ L = \frac{1}{2} m \dot{x}^2 - V(x). \] If the origin \(O\) moves with a constant velocity \(v\), the coordinate transformation is \( x' = x - vt \). The new Lagrangian becomes: \[ L' = \frac{1}{2} m (\dot{x} - v)^2 - V(x). \]

- Transforming the Lagrangian and introducing \( \phi \)

Write the transformed Lagrangian as a modified Lagrangian of the old system: \[ L' = L + \frac{d \phi}{dt}, \] where \( \phi = -mvx \). Since \( \phi \) adds a total derivative with respect to time, the equation of motion remains unchanged.

Key concepts

Euler-Lagrange Equation

The Euler-Lagrange equation is a fundamental tool in Lagrangian mechanics. It helps us find the equations of motion for a given system. It's derived from the principle of least action. The general form of the Euler-Lagrange equation for a coordinate, say 𝑞, is:
\[\frac{d}{dt} \bigg( \frac{\partial L}{\partial \dot{q}} \bigg) - \frac{\partial L}{\partial q} = 0.\]
This equation is a set of differential equations that describe how the system evolves over time. It's crucial to respect the dependencies of the Lagrangian, typically on the coordinates, their time derivatives, and time.

Coordinate Transformation

Coordinate transformation allows us to change our perspective on a problem. If we move our frame of reference or change the variables we use to describe a system, we perform a coordinate transformation. For instance, consider a particle moving under a force in one dimension, where the coordinate is given by 𝑥. If our point of measurement, O, moves with a constant velocity 𝑣, we can introduce a new coordinate \(x' = x - vt\). This transformation helps us understand how the system behaves relative to the moving point.
When we transform coordinates, the form of the Lagrangian may change. However, as shown in the original exercise, adding a total time derivative of some function \(\phi(q, t)\) doesn't change the equations of motion. This property simplifies solving problems using different coordinate systems.

Equation of Motion

The equation of motion derived from the Euler-Lagrange equation indicates the dynamics of the system. For a particle in one dimension influenced by a potential \V(x)\, the Lagrangian is usually \[ L = \frac{1}{2} m \dot{x}^{2} - V(x).\]
Using the Euler-Lagrange equation for the Lagrangian above, we get:
\[\frac{d}{dt} \bigg( m \dot{x} \bigg) = -\frac{d V}{dx}.\]
That simplifies into:
\[m \ddot{x} + \frac{d V}{dx} = 0,\]
giving the familiar Newton's second law in disguise: \F = -d V(x)/dx\. If the origin O moves with a constant velocity v, our derivation showed that this doesn't change the motion equation, ensuring the invariance in a moving frame.

Total Derivative

Total derivative represents how a function changes as all of its variables change over time. When we take the total derivative of a function \(\phi(q, t)\), we mean looking at its rate of change considering both explicit time dependence and the dependence on the changing variable q. Formally, it is written as:
\[ \frac{d \phi}{dt} = \frac{\partial \phi}{\partial t} + \frac{\partial \phi}{\partial q} \dot{q}. \]
This form shows the sum of how \(\phi\) changes directly with time and how it changes indirectly via its dependence on q. In Lagrangian mechanics, adding the total derivative of \(\phi(q, t)\) to the Lagrangian creates a new Lagrangian that produces the same equations of motion. This keeps the physical predictions invariant under such transformations, simplifying many problems.