Goldstein, 3rd ed

$$ \frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}_{j}}\right)-\frac{\partial L}{\partial q_{j}}=0\tag{1.57} $$ expressions referred to as "Lagrange's equations."

Note that for a particular set of equations of motion there is no unique choice of Lagrangian such that above equation lead to the equations of motion in the given generalized coordinates.

I'm not able to understand what does the highlighted statement mean. How can we have different Lagrangians. While deriving the above equation we went through the derivation and ended up defining $L=T-V$, so the Lagrangian is fixed and always $L=T-V$, so why talk about a different Lagrangian?

Here are the preceding steps of derivation

$$ \frac{d}{d t}\left(\frac{\partial(T-V)}{\partial \dot{q}_{j}}\right)-\frac{\partial(T-V)}{\partial q_{j}}=0 $$ Or, defining a new function, the Lagrangian $L$, as $$ L=T-V\tag{1.56} $$ the Eqs. (1.53) become $$ \frac{d}{d t}\left(\frac{\partial L}{\partial \dot{q}_{j}}\right)-\frac{\partial L}{\partial q_{j}}=0.\tag{1.57} $$

So we see that we have defined that $L=T-V$ so why talk of a different Lagrangian? Can anyone please help me.

  • $\begingroup$ bio-physics.at/wiki/… $\endgroup$
    – user65081
    Sep 1, 2021 at 15:03
  • $\begingroup$ @JEB, please see the edit. $\endgroup$
    – Kashmiri
    Sep 1, 2021 at 15:10
  • 2
    $\begingroup$ a physically meaningful example is a frame transformation $x\to x’=x-vt$, for constant $v$. Clearly we know from Galilean invariance that the EOM stay the same, yet the Lagrangian will certainly change. $\endgroup$ Sep 1, 2021 at 21:30
  • $\begingroup$ I edit the title : Non-unique is the Lagrangian not the Lagrange Equations. $\endgroup$
    – Frobenius
    Sep 2, 2021 at 10:28
  • $\begingroup$ I note that $L$ always appears differentiated, so adding a constant to $L$ will yield a different, equivalent Lagrangian. $\endgroup$ Sep 2, 2021 at 16:37

3 Answers 3


It may not be obvious that different Lagrangians can lead to the same equations of motion. Here is a simple example. Take theses two Lagrangians

$$L = \frac{m}{2}\dot{x}^2-\frac{k}{2}x ^2 \tag{1}$$ $$L' = \frac{m}{2}\dot{x}^2-\frac{k}{2}x ^2+ax\dot{x} \tag{2}$$ These two Lagrangians differ just by the extra term $ax\dot{x}$.

From the Lagrangian (1) you get the equation of motion $$\frac{d}{d t}\left(\frac{\partial L}{\partial \dot{x}}\right) = \frac{\partial L}{\partial x}$$ $$m\ddot{x}=-kx \tag{3}$$

From the Lagrangian (2) you get the equation of motion $$\frac{d}{d t}\left(\frac{\partial L'}{\partial \dot{x}}\right) = \frac{\partial L'}{\partial x}$$ $$m\ddot{x}+a\dot{x}=-kx+a\dot{x} \tag{4}$$ which is the same as (3).

The above was just an example. Actually you can add any function $F$ of the form $$F(x,\dot{x},t)=\frac{\partial G(x,t)}{\partial x}\dot{x}+\frac{\partial G(x,t)}{\partial t}$$ with an arbitrary function $G(x,t)$ to the Lagrangian, and the equation of motion will remain the same. This function $F$ can equivalently be written as a total time derivative $$F(x,\dot{x},t)=\frac{dG(x,t)}{dt}$$

By the way: The example above was constructed using $G(x,t)=\frac{a}{2}x^2$, thus giving $F(x,\dot{x},t)=ax\dot{x}$.

  • $\begingroup$ Thank you so much. Is there any other way to construct Lagrangians that lead to the same EOM apart from what you cited? $\endgroup$
    – Kashmiri
    Sep 6, 2021 at 16:12
  • $\begingroup$ @Kashmiri You can also multiply the Lagrangian by a constant factor (as mentioned in Qmechanics's answer) $\endgroup$ Sep 6, 2021 at 16:18
  • $\begingroup$ Yes but are there any other ways other than those. They seem to be just tinkering with $T-V$ . Do you have any Lagrangian in mind that isn't of the form $T-V$. Thank you $\endgroup$
    – Kashmiri
    Sep 6, 2021 at 16:28
  • $\begingroup$ @Kashmiri Maybe there are other ways which I haven't heard of yet. $\endgroup$ Sep 6, 2021 at 16:33
  • $\begingroup$ Thank you for your detailed answer and the time. I learnt a lot ,Take care :) $\endgroup$
    – Kashmiri
    Sep 6, 2021 at 16:35
  1. Given a set of EOMs, there is not a unique Lagrangian (and there might not exists a Lagrangian), whose Lagrange equations reproduce the EOMs.

    For starters, one can add total time derivative terms to the Lagrangian, or scale the Lagrangian with a non-zero constant.

    See also this example from Goldstein.

  2. On the other hand, it is true that starting from Newton's Laws and d'Alembert's principle Goldstein derives a Lagrangian in chapter 1. However, it is not unique, cf. point 1.

  • 1
    $\begingroup$ Thank you. So there may be other exotic functions which if I use in lagrange equations I'll get the correct EOM. However Goldstein just proves one of the function namely L=T-V. Am i correct sir? $\endgroup$
    – Kashmiri
    Sep 1, 2021 at 16:41
  • $\begingroup$ Yes, except that some of the other Lagrangians may in principle also be of the form $L=T-V$. $\endgroup$
    – Qmechanic
    Sep 1, 2021 at 17:14
  • 3
    $\begingroup$ @Kashmiri: Note that the changes to the Lagrangian do not have to be "exotic". As pointed out in the answer, a Lagrangian $L = T - V + C$ for any constant C will give you the exact same EOM. So for a given $T$ and $V$, there are an infinite set of Lagrangians, parametrized by C, that give the same EOM. And the total time derivative doesn't really count as "exotic" to me, but that's personal flavor I suppose. $\endgroup$
    – levitopher
    Sep 1, 2021 at 17:44
  • $\begingroup$ @Qmechanic The aim that the gauge-transformed lagrangian can also have a L=T-V form is nontrivial. Is there a clean example? $\endgroup$ Sep 1, 2021 at 18:41
  • $\begingroup$ @levitopher, yes I agree with you. What I had in mind was something more complex than those examples but which give us the same EOM. $\endgroup$
    – Kashmiri
    Sep 6, 2021 at 16:14

There is indeed no unique Lagrangian that gives particular set of equations of motion. Here's an example taken from David Tong's lectures on the subject (Problem Set 1 of Classical Dynamics, Problem 2). We're given the Lagrangian $$L=\frac{1}{12}m^2\dot{x}^4 + m\dot{x}^2V -V^2$$ and we're told that this is classically equivalent to Newton's Lagrangian. Hence, by applying Euler-Lagrange equations we expect to obtain Newton's equations. We have $$\frac{d}{dt}\frac{\partial L}{\partial \dot{x}} - \frac{\partial L}{\partial x}=0$$ We have

$$\cfrac{\partial L}{\partial x} = m\dot{x}^2V'(x) -2VV'(x)$$ $$\cfrac{d}{dt}\cfrac{\partial L}{\partial \dot{x}} = m^2\dot{x}^2\ddot{x} + 2m\ddot{x}V+2m\dot{x}^2V'(x)$$ By standard calculations, where we have used the fact that $\cfrac{dV}{dt} = \cfrac{\partial V}{\partial x}\cfrac{dx}{dt}$, and primes denote derivative with respect to $x$. Now, the equation becomes $$m^2\dot{x}^2\ddot{x} + 2m\ddot{x}V+m\dot{x}^2V'(x) +2VV'(x)=0\Rightarrow m\ddot{x}(m\dot{x}^2+2V) + V'(x)(m\dot{x}^2+2V)=0$$ Now since $m\dot{x}^2+2V=2E \neq 0$ (cannot be zero regardless of $x$ and $\dot{x}$ or else the particle does not have energy to move) we have $$m\ddot{x} + V'(x)=0$$ which is Newton's equation. In this particular case Newton's Lagrangian is not scaled by a number neither is a total derivative added (at least one i could see) but gives the same equations of motion.

There are theories like general relativity Einstein-Hilbert action where the lagrangian does not satisfy the definition $T-V$ but gives the correct equations of motion.

  • $\begingroup$ How do you come to the conclusion that energy is conserved so that $m\dot{x}^2/2 + V =E$ in the middle of your proof? Normally I would start from $L = m\dot{x}^2/2 - V$, then get the standard N2L equation of motion, and then integrate this once to get energy conservation. So if you start from a different EoM and then use conservation of energy to get back to N2L, the logic seems circular to me $\endgroup$
    – Joe
    Sep 3, 2021 at 12:05
  • $\begingroup$ I just used the fact that E= T+V in the particular setting and the fact that energy cannot be zero everywhere or else the particle does not move. $\endgroup$
    – Noone
    Sep 3, 2021 at 12:34
  • $\begingroup$ You can find the exercise here damtp.cam.ac.uk/user/tong/dynamics.html $\endgroup$
    – Noone
    Sep 3, 2021 at 12:35
  • $\begingroup$ Yes my question is; How do you know that $E = T+V$ in this setting? This relationship is usually derived from $L=T-V$, but you've started from a different $L$. I looked on your link but I couldn't see which pdf the exercise is in $\endgroup$
    – Joe
    Sep 3, 2021 at 12:42
  • 1
    $\begingroup$ Your argument that the particle doesn't move when $E=0$ is not correct; the particle will move unless $T=0$. Thanks for a nice answer though it's been interesting to think about $\endgroup$
    – Joe
    Sep 3, 2021 at 13:10

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