Being a physic student I first heard the term: "Lagrangian" during a course about Lagrangian mechanics; at that time this term was defined to me in the following way:
For a classic, non relativistic, mechanical system, the Lagrangian ($\mathcal{L}$) is a function, defined as the difference between kinetic energy (K) and potential energy (V):
$$\mathcal{L} := K-V. \tag{1}$$
This definition is consistent with what we find in the related section of the Wikipedia page regarding Lagrangian mechanics.
Soon after I learned that the Lagrangian can also be defined in contexts that aren't strictly classical or mechanical, one good example of such context is the Lagrangian in presence of an EM field; but here definition (1) doesn't hold anymore. We need to redefine the Lagrangian in the following way:
The Lagrangian is any function which generates the correct equations of motion once plugged into Euler-Lagrange equation:
$$\frac{\mathrm{d}}{\mathrm{d}t} \left ( \frac {\partial L}{\partial \dot{x}} \right ) = \frac {\partial L}{\partial x}.\tag{2}$$
This, way more cloudy, definition is also reported in the related Wikipedia page. This new way of defining the Lagrangian came as a shock to me, since it seems to me as a non-definition. But fair enough. This second way of defining the term Lagrangian is of course consistent with our first definition; it's a generalization of our previous definition (1).
But then I stumbled upon a third definition for the term Lagrangian: in the context of the method of Lagrange multipliers (a field strongly related with Lagrangian mechanics) the Lagrangian is defined in the following way:
Given the function $f(x)$ that we want to maximize, and given the equation $g(x)=0$ that represents our constraints; the Lagrangian is defined as the following function:
$$\mathcal{L}(x,\lambda) := f(x)-\lambda g(x).\tag{3}$$
My question is: Why is this last function called Lagrangian? Is it just a coincidence? Is there a link between our first two definitions of Lagrangian and this last third definition? Is the Lagrangian in the context of Lagrangian mechanics in some way the same thing as the Lagrangian in the context of the Lagrange multipliers?
Best Answer
The first two definitions of the Lagrangian are not equivalent but very closely related in the sense that one is a generalization of the other.
The second one may be a non-definition, but that can be easily helped with. I would argue that it is more logically sound to define a Lagrangian in the context of variation problems rather than in physics.
The connection with physics is that you want to specify the dynamical equation of your system as a variational problem and specifically one which can be formulated in terms of a Lagrangian (there are more general variation problems where the "Lagrangian" is nonlocal/not of finite order).
The third definition is different. Of course it is also related in the sense that in Lagrangian formulations of dynamics or variational problems, Lagrange multiplier methods often appear. But this third definition of a Lagrangian gets its name because of Lagrange multipliers, rather than because they are in any ways related to Lagrangians of variational problems.
I am not the best person to talk about historical aspects but probably both get the name Lagrangian because Lagrange played an important part in their discovery or popularization through application.
Just like in differential geometry half of the things that appear are named after Cartan but not all of those are related or the same.