I am trying to understand how to use the Euler-Lagrange formulation when my system is subject to external forces. Consider the system pictured below:

my system

Let's define the lagrangian, as always, as $L = K - V$, where the external forces play no roll at all.

If $F_x \equiv F_\theta \equiv 0$, the standard Euler-Lagrange formulation for the system would be:

$$\frac{d}{dt} \left ( \frac{\partial L}{\partial \dot x} \right ) - \frac{\partial L}{\partial x} = 0$$

$$\frac{d}{dt} \left ( \frac{\partial L}{\partial \dot \theta} \right ) - \frac{\partial L}{\partial \theta} = 0$$

Now, according to a paper I am reading, if we include the force $F_x(t)$ (i.e. $F_x(t) \not \equiv 0$), the first equation should now be replaced by

$$\frac{d}{dt} \left ( \frac{\partial L}{\partial \dot x} \right ) - \frac{\partial L}{\partial x} = F_x(t)$$

This makes sense, of course, but I'm trying to understand how to extend this procedure to different forces, and I am a bit lost. So, for example, let's now include the force $F_\theta(t)$. How would the Euler Lagrange equations change to account for it?

  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/153302/2451 , physics.stackexchange.com/q/283238/2451 and links therein. $\endgroup$
    – Qmechanic
    Jun 30, 2017 at 17:46
  • $\begingroup$ @Qmechanic I had stumbled on that question before, and I didn't find the answer to be clear enough. Moreover, that is a general question, whilst mine is asking about a specific example. $\endgroup$
    – LGenzelis
    Jun 30, 2017 at 18:03
  • $\begingroup$ Hi @LGenzelis: The question (v4) is unclear. Are you talking about external forces (main text) or non-conservative forces (title)? What variables does the force $F_i$ depend on? The title (v4) is changed. $\endgroup$
    – Qmechanic
    Jun 30, 2017 at 18:59

1 Answer 1


If the force is not derived from a potential, then the system is said to be polygenic and the Principle of Least Action does not apply. However, the Euler-Lagrange equations can be derived from d'Alembert Principle.

If we decompose the applied (or specified) forces acting on particle $\alpha$ into monogenic (derived from a potential), $\vec F_\alpha^m$ and polygenic forces, $\vec F_\alpha^p$, then d'Alembert Principle reads, $$\sum_\alpha(\vec F_\alpha^m+\vec F_\alpha^p-\dot{\vec p}_\alpha)\cdot\delta\vec r_\alpha=0.$$ The next step is to write this equation in terms of generalized coordinates $q_i$. The result is the following equation of motion $$\frac{d}{dt}\frac{\partial T}{\partial \dot q_i}-\frac{\partial T}{\partial q_i}=Q_i^m+Q_i^p,$$ where $$Q_i^p\equiv\sum_\alpha\vec F_\alpha\cdot\frac{\partial \vec r_\alpha}{\partial q_i}.\tag1$$

The monogenic force can be obtained from a potential $V$, $$Q_i^m=-\frac{\partial V}{\partial q_i},$$ hence the equation of motion $$\frac{d}{dt}\frac{\partial T}{\partial \dot q_i}-\frac{\partial T}{\partial q_i}+\frac{\partial V}{\partial q_i}=Q_i^p.$$ If the potential does not depend on velocities, then this equation can also be written as $$\frac{d}{dt}\frac{\partial L}{\partial \dot q_i}-\frac{\partial L}{\partial q_i}=Q_i^p,\tag2$$ where $L=T-V$ is the Lagrange function. Equation (2) is the one you shall use, together with Eqn. (1) to obtain the generalized force $Q_i^p$.


Let's now apply this approach to the example posed in the question. There are two external forces, which can be written as $\vec F_1 = [F_x(t) \; , 0]^T$ and $\vec F_2 = [0 \; , -F_\theta(t)]^T$. The position of each body (regarded as a point mass) is $\vec r_1 = [x \; , 0]^T$ and $\vec r_2 = [x + l \sin \theta \; , -l \cos \theta]^T$. Therefore, we calculate $$Q_1^p = \vec F_1 \cdot\frac{\partial \vec r_1}{\partial x} + \vec F_2 \cdot\frac{\partial \vec r_2}{\partial x} = F_x(t)$$ and $$Q_2^p = \vec F_1 \cdot\frac{\partial \vec r_1}{\partial \theta} + \vec F_2 \cdot\frac{\partial \vec r_2}{\partial \theta} = -F_\theta(t) l \sin \theta .$$

Finally, the corresponding Euler-Lagrange equations are $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot x}\right )-\frac{\partial L}{\partial x}= F_x(t)$$ $$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot \theta}\right )-\frac{\partial L}{\partial \theta}=-F_\theta(t) l \sin \theta,$$ where $$L = T - V = \frac{M}{2} \left\lVert\dot {\vec r_1}\right\rVert^2 + \frac{m}{2} \left\lVert\dot {\vec r_2}\right\rVert^2 + m g l \cos \theta .$$

  • $\begingroup$ Thank you. Could please expand your answer to apply your procedure to my example? Note that $F_\theta(t)$ is an arbitrary function of time. If it makes it easier, replace it by any know function (e.g. $F_\theta(t) = t \sin t$) $\endgroup$
    – LGenzelis
    Jun 30, 2017 at 20:23
  • $\begingroup$ @LGenzelis It's quite straightforward. I leave you to do that. Write the force vector and the position vector of the bob and use Eqn (1). The sum contains just one term and the generalized coordinate is $\theta$. $\endgroup$
    – Diracology
    Jun 30, 2017 at 20:33
  • $\begingroup$ thanks a lot. I edited your answer so as to apply the methodology you suggested to my example problem. Once you've accepted the edits, I'll mark this as the accepted answer. Just a couple more quick questions about your answer. 1) Will this approach be applicable to any kind of force? Let's say that I've got forces which depend on velocities (such as friction), is it as simple as pluggin these forces into Eq. (1)? And the last one, 2) you wrote in your answer "If the potential does not depend on velocities" : can a potential ever depend on velocities? I've never seen such a thing. $\endgroup$
    – LGenzelis
    Jul 4, 2017 at 21:09
  • 1
    $\begingroup$ 1) Yes it is a general procedure, valid even for dissipative forces. For instance, you can google about Rayleigh's dissipation function, which is a particular example. 2) Those potentials are normally called generalized potential. A particular example is a particle in the presence of an electromagnetic field. You can learn more about both questions in section 1.5 of Goldstein, third ed. $\endgroup$
    – Diracology
    Jul 4, 2017 at 22:25
  • $\begingroup$ The term $\alpha$ is associated to the degrees of freedom of the system, right? $\endgroup$
    – Vajra
    May 6, 2021 at 8:06

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