how do I formulate the action-functional of the Lagrangian together with constraints in form of Lagrange-Multipliers in the sense of the calculus of variations and secondly how does one derive the Euler-Lagrange-Equation from that? The normal way is like here https://de.universaldenker.org/argumentationen/298 or https://en.wikipedia.org/wiki/Euler%E2%80%93Lagrange_equation#Statement

I also found another thread but there was a "Lagrangian" system of motion formulated such that it only consists of actualized energy (kinetic energy) without potential energy (how is this even possible? ^^) So this here isn't a dup. Just saiyan.


  • $\begingroup$ I don't think you need to derive the Euler-Lagrange equations per se, you can find them on Wikipedia. If you already have a Lagrangian, you can add a Lagrange multiplier directly to it to turn it into a constrained optimization problem, and then plug your Lagrangian into the equation from Wikipedia. Now, how to build the original Lagrangian is a completely different question, and that depends on the problem. $\endgroup$
    – Szgoger
    Commented Jul 14, 2022 at 8:36

1 Answer 1


Suppose you have $n$ coordinates $\boldsymbol{q}=(q_1,...,q_n)$ and one constraint equation $f(\boldsymbol{q})=0$, then you introduce one lagrange multipler $\lambda$ and modify the action functional accordingly:

$$S[\boldsymbol{q},\lambda] = \int \big( L(\boldsymbol{q}(t),\boldsymbol{\dot{q}}(t),t)-\lambda(t) f(\boldsymbol{q}(t)) \big)dt$$

Now you vary with respect to each $q_i$ separately as well as for $\lambda$ separately. This may seem confusing at first since at first sight $q_i$'s are not independent since they are related to each other via a constraint but remember that's the whole point of Lagrange multiplier method: You treat the coordinates as if they're independent and then impose the constraint. You can, on the other hand, just impose the constraint directly from the start, this can be done for instance by introducing a convenient coordinate system where the constraints are already built in (for example, in the planar pendulum, by going to polar coordinates you immediately incorporate the constraint since the coordinate $r$ will be constant). In any case you do the standard variation procedure and get the following Euler Lagrange equations:

$$\dfrac{d}{dt}\dfrac{\partial L}{\partial \dot{q_i}}-\dfrac{\partial L}{\partial {q_i}}=\lambda\dfrac{\partial f}{\partial {q_i}} $$

If there are $k$ constraints ($f_1,...,f_k$) instead of just $1$ you would introduce $k$ lagrange multipliers $\lambda_1,...,\lambda_k$ and get the following EL equations:

$$\dfrac{d}{dt}\dfrac{\partial L}{\partial \dot{q_i}}-\dfrac{\partial L}{\partial {q_i}}=\sum_{j=1}^k \lambda_j\dfrac{\partial f_j}{\partial {q_i}}$$

If you want a good reference for all this see "Classical Mechanics" by Goldstein section 2.4. Note that the physics textbooks are not completely rigorous but essentially the only key ingredient their arguments are missing is the fundamental lemma of the calculus of variations; once you know that, everything can be put in a rigorous footing.

As to your second question, Whenever you see a Lagrangian with just kinetic energy this means that the system in question is just a free particle (or several free particles). More concretely, this means that there are no external forces on the particles and they don't interact with one another, they are just moving freely in space.

  • $\begingroup$ Thank you very much! $\endgroup$
    – von spotz
    Commented Jul 16, 2022 at 8:42

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