I understand that in Lagrangian mechanics, a Lagrangian can be written as $L=T-V$ where $T$ and $V$ are the kinetic and potential energy of the system, respectively. However, in this paper, it proposes a Lagrangian which is not related to those energies.

Basically, it considers the paraxial wave equation, $$(2ik \partial_z+\nabla ^2_{\perp})a=0\tag{1} $$ where $a$ is the envelope of Gaussian beam $A(x,y,z)=a(x,y,z)e^{-ik(t-z)}$ and $\nabla ^2_{\perp}=\partial^2_x+\partial^2_y$. Then the author proposes a Lagrangian $$L=ik(a\partial_z a^*-a^* \partial_z a)+\nabla_\perp a^* \cdot \nabla_\perp a .\tag{2}$$ The corresponding equations of motion $$ \partial_z \left [\frac {\partial L}{ \partial(\partial a^*/\partial z)} \right ]+\nabla_\perp \cdot \left [\frac {\partial L}{ \partial(\nabla_\perp a^*)} \right ]-\frac {\partial L }{\partial a^*}=0\tag{3} $$ do give the above paraxial wave equation. But I wonder if this is a valid Lagrangian because this is not related to energies. Also, if I somehow derive its corresponding Hamitonian, will it be valid too?


Historically, the Lagrangian approach may have been first introduced for mechanics, but the concept of the Lagrangian has been generalized across many different areas of physics. A key property across these formulations is that the Lagrangian is a function such that the stationary trajectories of the action give the equations of motion. In mechanics, we know that the right Lagrangian is equal to the difference in kinetic and potential energies because the action principle gives us Newton's laws. In geometric optics, the Lagrangian is valid because it leads to the eikonal equation. In electromagnetic field theory, the Lagrangian takes its form because it leads to Maxwell's equations. In short, verifying the validity of a Lagrangian amounts to verifying that the equations of motion are correct.

  1. user318039's answer is exactly right: An action principle only has to reproduce the given equations of motion (EOMs) as its Euler-Lagrange (EL) equations.

  2. The Lagrangian does not have to have dimension of energy. For instance, if one scales a Lagrangian with a dimensionfull (non-zero) constant, the EOMs will stay the same. See also e.g. this & this Phys.SE posts.

  3. By the way: OP's paraxial wave equation (1) has the same form as the TDSE, cf. e.g. this Phys.SE post.

  • $\begingroup$ Thanks. The similarities between the paraxial wave equation and the Schrodinger equation have been noticed for a long time. I will dig in that, as well. $\endgroup$
    – Haorong Wu
    Nov 2 '21 at 8:28

As pointed out in an earlier answer: in the history of mathematical physics there have been multiple instances of putting calculus of variations to use. Those instances aren't necessarily connected, and the specific Lagrangian may not contain energy terms.

My overall aim in this answer is to discuss the relation between differential calculus and calculus of variations.

I will first discuss a specific case (catenary problem), and from there I will generalize.

To my knowledge: calculus of variation was initially created to address problems in statics.

As an example of application of calculus of variations in statics I take the problem of the shape of the catenary.

(Catenary: the shape of a chain suspended betweeen two points)

The problem of the shape of the catenary is comparitively simple; it can be solved without involvement of calculus of variations. The comparison gives insight into how calculus of variations works.

Differential calculus approach

In the wikipedia article, in the section 'Analysis', it is demonstrated how to set up an equation for the catenary curve.

At each point along the curve the force can be decomposed in horizontal component and vertical component. The horizontal force-component is the same value along the entire length of the chain. The vertical force component at some point 'r' along the chain is equal to the weight of the section of chain below point 'r'

In the wikipedia article it is demonstrated how the above constraints are sufficient to state the problem in the form of a differential equation.

Calculus of variations approach

As we know: if you have a system such that motion is subject to friction, and you start it in some non-equilibrium state, then the constituent parts of the system will move, and when all energy that can be dissipated has been dissipated the end state is an equilibrium state.

Any state other than the equilibrium state has a higher potential energy than the equilibrium state.

In statics we are only interested in the equilibrium state, not in how the system gets there. Therefore only one form of energy enters the evaluation: potential energy.

Every subsection of the chain influences the shape of the chain as a whole. The sides tend to move outward, lifting the center, the center tends to move down, pulling the sides inward.

Now: we know that the catenary problem can be solved with a differential equation.

A differential equation operates at infinitisimal level: a differential equation doesn't evaluate the entire chain as whole. That raises a question: how is is possible that a differential equation can solve the problem?

The key is:
Take two points along the curve, marking a sub-interval. Evaluate the shape of the curve. Along that sub-interval the potential energy is minimal.

Proof by contradiction: if it would not be minimal then the curve as a whole would not be at minimal potential energy.

The sub-interval can be made arbitrarily small; the same reasoning still applies.

So we see that the constraint of being a minimal state is actually extremely constraining. It constrains all the way down to arbitrarily small intervals. This constraint property generalizes to all of calculus of variations.

That is why the Euler-Lagrange equation is a differential equation.

For a treatment of solving the catenary with calculus of variations I recommend the following article by Preetum Nakkiran: 'Geometric derivation of the Euler-Lagrange equation'

The relation between differential and variational calculus

Here I discuss difference between solving the catenary problem with differential calculus, and solving with calculus of variations.

Why is it that in order to solve the catenary problem with calculus of variations we need to use the quantity Energy instead of the quantity Force?

As we know: the solution of the catenary problem is the hyperbolic cosine:

$$ f(x) = \cosh(x) $$

That is: by convention we express the solution as a function of the x-coordinate. (For the conventional coordinate system with the horizontal axis as the x-axis, and the vertical axis as the y-axis.)

We have that the variation of the calculus of variations is in the vertical direction, and thus the Euler-Lagrange equation expresses derivation with respect to the y-coordinate

$$ \frac {\partial f}{\partial y} - \frac{d}{dx}\frac{\partial f}{\partial y'} = 0 $$

The first term $\frac{\partial f}{\partial y}$ is straightup differentiation with respect to the y-coordinate. The second term is actually also differentiation with respect to the y-coordinate. (One way to see that it must be so: it has to be dimensionally the same as the $\frac{\partial f}{\partial y}$ term.)

We take a coordinate system and use the convention of designating horizontal axis the x-axis and vertical axis the y-axis.

Differential calculus is about derivation with respect to the x-coordinate: $\frac{dy}{dx}$, to obtain a solution that is a function of the x-coordinate

Calculus of variations is about derivation with respect to the y-coordinate, but for the purpose of obtaining a solution that is a function of the x-coordinate. That is what the Euler-Lagrange equation does.

Statics and Mechanics

The reason that in the case of statics and mechanics the variational approach uses the quantity Energy:
Potential energy is the integral of force over distance. (More precisely: potential energy is defined as the negative of Work Done, and work done is the integral of force over distance.)

So: You construct the energy by integrating with respect to the position coordinate. In statics/mechanics: the Euler-Lagrange equation takes the derivative with respect to the position coordinate, so you end up recovering the force.

What we see is that in many branches of physics it has been demonstrated that it is possible to construct a bespoke Lagrangian that when inserted in the Euler-Lagrange equation produces the appropriate equation(s) of motion.

We have that any problem that is stated in terms of calculus of variations can be restated in differential calculus form. That is what the Euler-Lagrange equation does: the Euler-Lagrange equation is a differential equation.

This suggests the following mathematical conjecture:
Any problem that can be expressed in differential equation form can also be expressed in calculus of variations form.

There is no general method for converting from differential form to variational form. Converting from differential form to variational form is more of a reverse engineering problem.

This reverse engineering involves integration, since the Euler-Lagrange equation does a derivation.

Generalization in general

To my knowledge calculus of variations was initially developed to address problems in statics.

At some later stage Calculus of variations was ported to Mechanics.

In subsequent stages Calculus of variations was ported to more abstract form of mathematical physics.

The core characteristic is the pivot from x-coordinate to y-coordinate.

We take a coordinate system and use the convention of designating horizontal axis the x-axis and vertical axis the y-axis. Given that convention:

Differential calculus is about derivation with respect to the x-coordinate: $\frac{dy}{dx}$

Calculus of variations is about derivation with respect to the y-coordinate.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.