Symmetry and Conservation: Time Translation
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You should be comfortable with the Euler-Lagrange equations and the definition of the Lagrangian \( L(q, \dot{q}, t) \). Familiarity with the concept of generalized coordinates and momenta is assumed.
Why this matters
Time symmetry is the deepest reason energy is conserved.
In Newtonian mechanics, energy conservation is often treated as a separate postulate or a derived result specific to conservative forces. In analytical mechanics, we discover that energy conservation is not a special case but a direct, inevitable consequence of the fact that the laws of physics do not change with time. This lecture bridges the gap between the abstract variational principle and the tangible conservation of energy, setting the stage for the general Noether's theorem which links all continuous symmetries to conservation laws.
The Intuition: Why Time Matters
Imagine performing a physics experiment today and repeating it exactly the same way tomorrow. If the laws of nature are the same, the outcome must be identical. This is time-translation symmetry. In the Lagrangian formalism, this symmetry manifests in a specific way: the Lagrangian function itself does not explicitly depend on time. It may depend on positions and velocities, but if you shift the clock forward by an hour, the mathematical form of \( L \) remains unchanged. This lack of explicit time dependence is the key. It implies that the "cost" of a path through configuration space is independent of when you traverse it. This geometric property of the action functional forces a quantity associated with the time direction to remain constant along the physical trajectory. That quantity is what we identify as the Hamiltonian, or total energy.
Explicit vs Implicit Time Dependence
It is crucial to distinguish between explicit and implicit time dependence. A Lagrangian \( L(q, \dot{q}, t) \) has explicit time dependence if the variable \( t \) appears directly in the expression, such as in a time-dependent potential \( V(q, t) \). If \( t \) does not appear explicitly, we say \( \partial L / \partial t = 0 \). However, the generalized coordinates \( q(t) \) and velocities \( \dot{q}(t) \) are functions of time. Therefore, the total time derivative of \( L \) is generally non-zero even if \( \partial L / \partial t = 0 \). This distinction is the source of the conservation law. We use the chain rule to expand the total derivative, separating the explicit partial derivative from the implicit changes due to the motion of the system.
If the Lagrangian has no explicit time dependence, the first term vanishes. The remaining terms are related to the equations of motion.
Deriving the Conserved Quantity
Let us assume \( \partial L / \partial t = 0 \). We examine the total time derivative of the quantity \( \sum_i \dot{q}_i \frac{\partial L}{\partial \dot{q}_i} - L \). This quantity is defined as the Hamiltonian \( H \). We calculate its time derivative to see if it is constant.
Expanding this derivative using the product rule and the chain rule, we get:
Now, we substitute the expression for \( dL/dt \) from the previous section, noting that \( \partial L / \partial t = 0 \):
Substituting this back into the expression for \( dH/dt \), the terms involving \( \ddot{q}_i \) cancel out exactly. We are left with:
Recall the Euler-Lagrange equations: \( \frac{d}{dt} \frac{\partial L}{\partial \dot{q}_i} - \frac{\partial L}{\partial q_i} = 0 \), which implies \( \frac{d}{dt} \frac{\partial L}{\partial \dot{q}_i} = \frac{\partial L}{\partial q_i} \). Substituting this identity into the equation for \( dH/dt \), every term cancels, resulting in \( dH/dt = 0 \). Thus, \( H \) is a constant of motion.
Identification with Energy
For a standard mechanical system where the kinetic energy is quadratic in velocities and the potential energy depends only on positions, the Hamiltonian \( H \) takes the familiar form of total mechanical energy. The generalized momentum is defined as \( p_i = \frac{\partial L}{\partial \dot{q}_i} \). In Cartesian coordinates, \( p_i = m \dot{x}_i \), so \( \sum_i p_i \dot{q}_i = \sum_i m \dot{x}_i^2 = 2T \). If \( L = T - V \), then \( H = 2T - (T - V) = T + V \). This confirms that the conserved quantity associated with time-translation symmetry is indeed the total energy. This result holds generally for systems with time-independent constraints and potentials.
The derivation
Noether's Theorem for Time Translation
Derives that invariance of the action under time translation implies conservation of the energy function (Jacobi integral).
A common pitfall is assuming that \( H \) is always equal to the total energy \( T + V \). This is only true if the transformation from generalized to Cartesian coordinates is time-independent and the kinetic energy is quadratic in velocities. If the constraints are time-dependent, or if the Lagrangian has explicit time dependence, \( H \) is still conserved only if \( \partial L / \partial t = 0 \), but it may not equal \( T + V \). In such cases, \( H \) is a conserved quantity associated with time symmetry, but it is not necessarily the sum of kinetic and potential energies. C
Worked example
Problem. Consider a particle of mass \( m \) moving in one dimension under the influence of a potential \( V(x) = \frac{1}{2} k x^2 \). The Lagrangian is \( L = \frac{1}{2} m \dot{x}^2 - \frac{1}{2} k x^2 \). Determine the conserved quantity associated with time-translation symmetry and verify that it equals the total energy.
- Check for explicit time dependence: \( L \) contains no \( t \), so \( \partial L / \partial t = 0 \). Time-translation symmetry exists.
- Calculate the generalized momentum: \( p = \frac{\partial L}{\partial \dot{x}} = m \dot{x} \).
- Compute the Hamiltonian: \( H = p \dot{x} - L = (m \dot{x}) \dot{x} - \left( \frac{1}{2} m \dot{x}^2 - \frac{1}{2} k x^2 \right) \).
- Simplify: \( H = m \dot{x}^2 - \frac{1}{2} m \dot{x}^2 + \frac{1}{2} k x^2 = \frac{1}{2} m \dot{x}^2 + \frac{1}{2} k x^2 \).
- Identify the terms: \( \frac{1}{2} m \dot{x}^2 \) is kinetic energy \( T \), and \( \frac{1}{2} k x^2 \) is potential energy \( V \). Thus \( H = T + V \).
Units check. \( m \dot{x}^2 \) has units \( \mathrm{kg \cdot (m/s)^2} = \mathrm{kg \cdot m^2/s^2} = \mathrm{J} \). \( k x^2 \) has units \( \mathrm{N/m \cdot m^2} = \mathrm{N \cdot m} = \mathrm{J} \). Both terms are in Joules, consistent with energy.
Three quick tests
- Does the Lagrangian \( L = \frac{1}{2} m \dot{x}^2 - m g x \cos(\omega t) \) possess time-translation symmetry?
Answer
No, because \( L \) has explicit time dependence via \( \cos(\omega t) \), so \( \partial L / \partial t \neq 0 \). - If \( \partial L / \partial t = 0 \), what is the time derivative of the Hamiltonian \( H \)?
Answer
Zero. \( H \) is a constant of motion. - For a free particle in 3D Cartesian coordinates, what is the form of the conserved Hamiltonian?
Answer
\( H = \frac{1}{2} m (\dot{x}^2 + \dot{y}^2 + \dot{z}^2) \), which is the total kinetic energy.
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.