It is written in standard textbooks on classical mechanics that, advantage of Lagrangian equations is that nowhere do enter statement regarding (Newtonian) force.

e.g. To find Lagrangian $L = T - U$, for simple pendulum, we have

$$T=(1/2)ml^2\dot{\theta}^2 $$ and $$U=mgl(1-\cos\theta).$$

But definitely, we must know the force acting on the bob of simple pendulum to find the potential energy $U$. That force is $F_g = mg$, where symbols have their usual meanings. The potential energy is derived from force, as $U = -\int_{y=0}^{y=h} \vec{F}. \vec{ds} $. And we get $U = mgh$.

Then we make argument that there is one degree of freedom, and we need one generalized co-ordinate, namely, $\theta$, and then we will replace $h$ by $\theta$.

If we replace gravitational force acting on the bob of pendulum by some other force which is a function of space co-ordinates only (& not time), obviously, we will get different potential energy function.

So without knowledge of force, how can we find Lagrangian $L$?

  • $\begingroup$ You just do an experiment and determine it. Easy. $\endgroup$
    – user41826
    Sep 13, 2017 at 5:55

2 Answers 2

  1. OP asks (v2):

    So without knowledge of force, how can we find Lagrangian $L$?

    If the force is truly unknown, then we have no model to discuss. Go make an experiment or something, and come back with at least a model!

  2. It seems OP is essentially asking a chicken & egg question:

    What comes first: The (conservative$^1$) force $\vec{F}$ or the potential (energy) $U$?

    That is often a less relevant question that one might think: In practice, physicists have already worked out a long list of pairs $(\vec{F}, U)$, say, Lorentz (force, potential), centrifugal (force, potential), etc, which can be taken off the shelf and used as needed.

    One way requires differentiation; the other integration.

    From an experimental side, the force often comes first; while the potential often comes first from theoretical symmetry considerations.

  3. OP essentially wrote (v2):

    It is written in standard textbooks on classical mechanics that the advantage of Lagrange equations is that nowhere do enter statement regarding (conservative) forces.

    For non-conservative forces in Lagrangian mechanics, see e.g. this Phys.SE post.

    The advantages of Lagrangian mechanics over Newtonian mechanics are discussed in e.g. this & this related Phys.SE posts.


$^1$ For a discussion of the notion of a conservative force, see e.g. my Phys.SE answer here.


Both the kinetic energy $T$ and the potential energy $U$ are taken as inputs to the Lagrangian formalism. Lagrangian mechanics does not tell you what to write down for $T$ and $U$ - it only tells you the laws of motion that would result from the $T$ and $U$ that you choose.

That begin said, what makes you say that the gravitational potential energy requires us to first think about forces? The physical observation that we make is that if we drop an object of mass $m$, it accelerates downward at a rate $\ddot z = -g$.

In the Newtonian viewpoint, we would use our observation to infer that there was a gravitational force

$$ F_g = -mg$$ which we could identify with a potential energy $$ U = -\int F_g dz = mgz + \text{constant}$$

On the other hand, letting $L = \frac{1}{2}m \dot z^2 - U$, the Lagrangian equation of motion tells us that

$$ \ddot z = - \frac{1}{m} \frac{\partial U}{\partial z}$$

Upon comparison with our observation, we see that $$ \frac{1}{m}\frac{\partial U}{\partial z} = g$$ so $$U = mgz + \text{constant}$$

with not a word said about forces.

  • 1
    $\begingroup$ But that doesnt build a lagrangian $\endgroup$
    – user41826
    Sep 13, 2017 at 5:56
  • 1
    $\begingroup$ @user41826: I agree with you. I think Newtonian force is essential for building Lagrangian L; whether you find that force by experiment or if it is known function of co-ordinates does not matter. Lagrange equations avoids constraint forces and makes the mathematics, of finding space co-ordinates as function of time, simpler. Force is involved in Lagrangian approach but not directly. $\endgroup$
    – atom
    Sep 13, 2017 at 6:38

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