Gauge invariance of Euler-Lagrange equations

[andrewkirk]

I have been trying to teach myself Lagrangian mechanics from a textbook “Lagrangian and Hamiltonian Mechanics” by MC Calkin. It has covered virtual displacements, generalised coordinates, d’Alembert’s principle, the definition of the Lagrangian, the Euler-Lagrange differential equation and how it can be used to derive equations of motion for a system.

In discussing the motion of a charged particle in electromagnetic fields, the book introduces the concept of gauge transformations and how they make no change to the derived equations of motion. I can follow the logic of that, but it also claims that it is “easy to show” that the Euler-Lagrange equation $$\frac{d}{dt}(\frac{\partial L}{\partial\.q_a}) = \frac{\partial L}{\partial q_a}$$ is unchanged by the addition to the Lagrangian L of a term that is a total time derivative of a real scalar field $$\lambda(t,x,y,z)$$ defined in the space-time under consideration. I can't see why this should be the case, and I have been unable to turn up a proof through internet searching, although I have found another site that suggests it is true.

Can anyone point me to a place that proves this result?

Thanks very much for any help.

 

[eljose79]

Thank you for bringing up this interesting topic. The concept of gauge transformations and their effect on the Lagrangian mechanics is indeed a fundamental and important concept in physics. I would be glad to provide some clarification on this topic and point you towards a proof of the result you are looking for.

Firstly, let us briefly review the concept of gauge transformations. In physics, a gauge transformation is a mathematical operation that can be applied to a physical system without changing its observable properties. In other words, a gauge transformation does not change the physical behavior or dynamics of a system. Instead, it changes the mathematical description of the system, such as the Lagrangian or Hamiltonian, without altering the physical predictions of the theory.

Now, in the case of Lagrangian mechanics, the Euler-Lagrange equation is derived by taking the variations of the action integral with respect to the generalized coordinates. This is done by considering virtual displacements, which are infinitesimal changes in the coordinates of a system that do not affect its overall motion. In other words, these virtual displacements are equivalent to gauge transformations. This is why the addition of a total time derivative of a real scalar field to the Lagrangian does not change the Euler-Lagrange equation, as it represents a gauge transformation that does not alter the physical behavior of the system.

To prove this result, we can start by considering the action integral S, which is defined as the integral of the Lagrangian L over time t:

S = \int_{t_1}^{t_2} L(q,\dot{q},t) dt

Now, let us assume that the Lagrangian can be expressed as the sum of two terms, with the second term being a total time derivative of a real scalar field \lambda(t,x,y,z):

L(q,\dot{q},t) = L_1(q,\dot{q},t) + \frac{d}{dt}(\lambda(t,x,y,z))

Substituting this into the action integral, we get:

S = \int_{t_1}^{t_2} L_1(q,\dot{q},t) + \frac{d}{dt}(\lambda(t,x,y,z)) dt

Using the property of integration by parts, we can rewrite this as:

S = \int_{t_1}^{t_2} \left[L_1(q,\dot{q},t)