Building Lagrangians: L = T - V and Newton Recovered
~6 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You should be comfortable with the Euler-Lagrange equation and the definition of kinetic energy \(T\) for a single particle in Cartesian coordinates.
Why this matters
The form \(L = T - V\) is not a guess: it is the simplest scalar function that reproduces Newtonian mechanics (not the only one; see the Tier 3 note).
In the previous lecture, we established that the Euler-Lagrange equations provide the equations of motion for any system described by a Lagrangian \(L\). However, the Euler-Lagrange equation is a differential operator that acts on an arbitrary function. It does not tell us what that function must be. To connect this abstract variational framework to the physical world we know, we must demonstrate that the specific combination of kinetic and potential energy, \(L = T - V\), yields the familiar Newton's second law. This lecture bridges the gap between the variational principle and classical dynamics, confirming that the Lagrangian formalism is not merely a mathematical curiosity but a rigorous reconstruction of standard mechanics.
The Physical Intuition
Think of the Lagrangian as a measure of the "efficiency" of a system's motion. Kinetic energy \(T\) represents the energy associated with motion, while potential energy \(V\) represents the energy stored in the configuration of the system. The difference \(T - V\) captures the trade-off between these two forms. When a particle moves quickly, \(T\) is large. When it is in a region of high potential, \(V\) is large. The principle of stationary action suggests that nature chooses the path that balances these contributions in a specific way. By constructing \(L\) as the difference rather than the sum, we ensure that the resulting equations of motion describe inertial motion against conservative forces. This construction is universal for systems where forces are derivable from a potential, making it the standard starting point for analytical mechanics.
Constructing the Lagrangian
For a single particle of mass \(m\) moving in one dimension under a conservative force, the kinetic energy is defined as \(T = \frac{1}{2}m\dot{x}^2\). The potential energy \(V(x)\) depends only on the position \(x\), not on the velocity \(\dot{x}\). We propose the Lagrangian:
This form is the simplest scalar function that depends on position and velocity. It is quadratic in the velocity and arbitrary in the position. The key assumption is that the force is conservative, meaning it can be written as the negative gradient of a potential: \(F = -\frac{\partial V}{\partial x}\). This assumption is crucial because it allows us to define a potential energy function. If the forces were non-conservative, such as friction, this simple form would not suffice, and we would need to introduce additional terms or use a different formalism.
Applying the Euler-Lagrange Equation
We now apply the Euler-Lagrange equation to our proposed Lagrangian. The equation is:
First, we calculate the partial derivative of \(L\) with respect to the velocity \(\dot{x}\). Since \(V(x)\) does not depend on \(\dot{x}\), it vanishes in this derivative:
Next, we take the total time derivative of this result. Using the chain rule, \(\frac{d}{dt}(m\dot{x}) = m\ddot{x}\), assuming the mass \(m\) is constant:
Now, we calculate the partial derivative of \(L\) with respect to the position \(x\). Here, the kinetic energy term \(\frac{1}{2}m\dot{x}^2\) does not depend on \(x\), so it vanishes:
Substituting these two results back into the Euler-Lagrange equation:
Rearranging the terms gives:
Recovering Newton's Second Law
The result \(m\ddot{x} = -\frac{\partial V}{\partial x}\) is precisely Newton's second law for a conservative force. We define the force \(F\) as the negative gradient of the potential energy: \(F = -\frac{\partial V}{\partial x}\). Therefore, the equation becomes \(m\ddot{x} = F\), or equivalently \(F = ma\). This demonstrates that the Lagrangian \(L = T - V\) is not an arbitrary choice but the specific function that encodes the dynamics of conservative systems. The derivation confirms that the variational principle, when applied to this Lagrangian, is mathematically equivalent to Newtonian mechanics. This recovery is a powerful validation of the Lagrangian approach, showing that it can reproduce the known laws of motion while offering a more general and flexible framework for complex systems.
The derivation
Recovery of Newton's Second Law from L = T - V
For a system of particles subject only to conservative forces derivable from a scalar potential V(q), inserting the Lagrangian L = T − V into the Euler-Lagrange equations d/dt(∂L/∂q̇i) − ∂L/∂qi = 0 returns exactly Newton's second law mᷤi = −∂V/∂xi. This fixes the otherwise-undetermined form of the physical Lagrangian.
A common misconception is that \(L = T - V\) is the only possible Lagrangian. In fact, any function of the form \(L' = L + \frac{d}{dt}f(x,t)\) yields the same equations of motion because the total derivative term vanishes in the Euler-Lagrange equation. This gauge freedom means that the Lagrangian is not unique, but its physical content is. C
Worked example
Problem. A particle of mass \(m = 2.0\ \mathrm{kg}\) moves along the \(x\)-axis under the influence of a conservative force derived from the potential \(V(x) = 5x^2\ \mathrm{J}\). At \(t=0\), the particle is at \(x=0\) with velocity \(\dot{x}=3.0\ \mathrm{m/s}\). Determine the acceleration of the particle at \(t=0\).
- Identify the Lagrangian: \(L = T - V = \frac{1}{2}m\dot{x}^2 - 5x^2\).
- Calculate \(\frac{\partial L}{\partial \dot{x}} = m\dot{x}\).
- Calculate \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right) = m\ddot{x}\).
- Calculate \(\frac{\partial L}{\partial x} = -10x\).
- Apply the Euler-Lagrange equation: \(m\ddot{x} - (-10x) = 0 \implies m\ddot{x} = -10x\).
- Substitute the initial conditions: \(x=0\), so \(m\ddot{x} = 0\).
- Solve for acceleration: \(\ddot{x} = 0\ \mathrm{m/s^2}\).
Units check. The term \(10x\) has units of \(\mathrm{J/m} = \mathrm{N}\). Dividing by mass \(m\) in \(\mathrm{kg}\) gives \(\mathrm{N/kg} = \mathrm{m/s^2}\), which is consistent with acceleration.
Three quick tests
- Why does the kinetic energy term not depend on position \(x\)?
Answer
Because kinetic energy is defined purely in terms of velocity, \(T = \frac{1}{2}m\dot{x}^2\), and is independent of the particle's location in space. - If the potential energy is constant, what is the equation of motion?
Answer
If \(V\) is constant, \(\frac{\partial V}{\partial x} = 0\), so \(m\ddot{x} = 0\). The particle moves with constant velocity, consistent with Newton's first law. - Can the Lagrangian \(L = T + V\) produce Newton's second law?
Answer
No. Using \(L = T + V\) would yield \(m\ddot{x} = +\frac{\partial V}{\partial x}\), which implies a force in the direction of increasing potential, contradicting the definition of conservative forces.
Problem set for PU-201 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.