My Physics teacher was reluctant to define Lagrangian as Kinetic Energy minus Potential Energy because he said that there were cases where a system's Lagrangian did not take this form. Are you are aware of any such examples?

Update: Here I'm of course assuming that $T$ and $U$ stands for the kinetic and the potential energy, respectively. Also:

  1. adding a total time derivative term to the Lagrangian, or

  2. scaling the Lagrangian with a non-zero multiplicative constant

do not change the Euler-Lagrange equations, as Dilaton and dmckee points out in the comments. Needless to say that I'm not interested in such trivial modifications (1&2).

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    $\begingroup$ Different kinds of terms can be added to a $T-U$ Lagrangian, as long as they do not change the corresponding equations of motion. $\endgroup$
    – Dilaton
    Commented Jan 13, 2013 at 10:38
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    $\begingroup$ Here is another example. $\endgroup$
    – Qmechanic
    Commented Jan 13, 2013 at 12:41
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    $\begingroup$ In a classical mechanics course, the nonconstraint forces are conservative and $L$ is definitely always $T-V$. The whole point of Lagrangian mechanics is to get at a minimal set of differential equations for the system as easily as possible. Definitions like "it is some function that gets the right equations blah blah blah" are useless, because you can never solve any problems with them. And if a method doesn't make predictions, it's not physics. $\endgroup$
    – user10851
    Commented Jan 13, 2013 at 17:39
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    $\begingroup$ It should be clear that you can multiply the Lagrangian by any non-zero constant (even a dimensional one) and obtain another Lagrangian, but I don't really consider that an enlightening case. $\endgroup$ Commented Jan 14, 2013 at 12:54
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    $\begingroup$ @ChrisWhite Once you leave the intro classical mechanics course, there are Lagrangians that are constructed because they give the right answer. And they do lead to new predictions. The Standard Model Lagrangian is one example. $\endgroup$
    – garyp
    Commented Dec 12, 2015 at 3:15

15 Answers 15


For a relativistic free particle you would think that the Lagrangian would be like $$ \tag{1} L ~=~ T ~=~ E-E_0~=~(\gamma -1)m_0c^2. \qquad(\leftarrow\text{Turns out to be wrong!}) $$ This is not the case! Instead it is $$ \tag{2} L ~=~ -\gamma^{-1}m_0c^2. $$ These two functions look like

$\gamma - 1$ and $-\gamma^{-1}$

and are not the same. This choice (2) of kinetic term gives a canonical momentum

$$p~:=~\frac{\partial L}{\partial v}~=~\gamma m_0v,$$

as it should be.


Just a pair of remarks. The second is the most interesting, in my view.

(1) The Lagrangian of a charged particle in an assigned electromagnetic field still has a Lagrangian ${\cal L}= T-U$, but here $U$ is not a standard position-dependent function, since it generally depends also on $\dot{q}$ and $t$ as is well known (see Jackson's textbook, for instance).

The difference between the structure of $T$ and $U$ is now that the dependence of $U$ on $\dot{q}$ is of the first order instead of the second one as in $T$. Otherwise the determinism (“normality” of Euler-Lagrange equations) could be violated. However one cannot think of $U$ as a potential energy. The same structure of $U=U(t,q, \dot{q})$ arises if one includes in ${\cal L}$ inertial forces when working in a generic non-inertial reference frame.

(2) Consider a classical particle on the real line immersed in a liquid giving rise to a friction force $-\gamma v$, with $\gamma>0$ constant. We can also suppose that there is a positional force with potential energy $U=U(x)$. $m>0$ is the mass of the particle and we use its coordinate $x$ as Lagrangian coordinate. This system is not invariant under time reversal, however there is a Lagrangian for this system:

$${\cal L}(t,x, \dot{x}) = e^{\gamma t/m} \left(\frac{1}{2}m \dot{x}^2 -U(x)\right)\:.$$

Indeed, it immediately produces the correct Newtonian equation:

$$ m \frac{d^2x}{dt^2} = -\frac{dU}{dx} - \gamma \frac{dx}{dt}\:.$$

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    $\begingroup$ +1 for example 2 of a Lagrangian description of a friction force in 1D (even though it introduces explicit time dependence in the process). As for example 1 with velocity-dependent $U(q, \dot{q})$, one may argue that a velocity-dependent $U(q, \dot{q})$ still has an interpretation as potential energy on the tangent bundle $TM$ rather that the position manifold $M$ itself. See also e.g. this Phys.SE answer. $\endgroup$
    – Qmechanic
    Commented Dec 17, 2013 at 9:31
  • $\begingroup$ For the example with the classical particle immersed in a liquid, why is it justified to ignore the partial time derivative of the Lagrangian as time explicitly enters the Lagrangian through $e^{\gamma t/m}$. $\endgroup$
    – Dhuality
    Commented Oct 4, 2018 at 12:00

In one of his classical mechanics lectures (I believe the latest set), Leonard Susskind answered a similar question by saying (and I cannot directly quote because I don't have the video in front of me) that Lagrangians are simply functions that lead to the correct equations of motion. I will add that those equations of motion can be solved and the resulting behavior compared to Nature as a test of correctness. Susskind went on to day that there is no rule that a system's Lagrangian must be T - U and that there can be "cross terms" that describe certain interactions. He went further to say something that really stuck with me, and that is when we're learning calculus, we never ask, "From where do we get the functions that we're learning to analyze?" We basically make them up or guess them or deduce them from observed behaviors (in physics, anyway). That statement seemed rather profound to me.


The point is rather subtle for a non-mathematically literate physicist, since the distinction is quite technical. According to Arnold (see References), we give the following definitions.

Definition. Let $M$ be a differentiable manifold, $TM$ its tangent bundle and $L : TM \to M$ a differentiable application. An application $\gamma : \mathbb R \to TM$ is a motion in a lagrangian system with configuration manifold TM and lagrangian function $L$ if and only if $\gamma$ is extremal for the functional

\begin{equation} \Phi(\gamma) = \int_{t_0}^{t_1} \mbox{dt} \, L(\dot{\gamma}). \end{equation} $\dot{\gamma}$ is said velocity vector, $\dot{\gamma}(t) \in TM_{\gamma(t)}.$

Local coordinates $q_1, \dots, q_n$ of the point $\gamma(t)$ evolve according to the Euler-Lagrange equation

\begin{equation} \frac{\partial L}{\partial q} = \frac{\mbox{d}}{\mbox{d}{t}}\frac{\partial L}{\partial \dot{q}} . \end{equation}

Now, suppose $M$ is a riemannian manifold, i.e. a couple $(M, g)$, with $M$ differentiable manifold and $g$ a positive-definite quadratic form, usually indicated as $\langle \cdot , \cdot \rangle$. In this case, and only in this case, we can define a kinetic energy as is usually meant:

Definition Let $M$ be a riemannian manifold. A quadratic form $K = \frac{1}{2}\langle v, v \rangle$, where $v \in TM_x$, defined on all tangent bundles is called kinetic energy. We say that $U$ is a potential energy if and only if $U : M \to \mathbb R$ is a differentiable function.

Definition. A lagrangian system on a riemannian manifold is said natural if and only if $L = K - U$, for some $K$ and $U$ previously defined.

In classical mechanics, one deals with riemannian manifolds all the time (apart from "pathological" situations), so he does not care about the distinction. If fact, in basic courses that a problem never arises. But it should be pointed out (by teachers, I mean) that Minkowski space $\mathcal M^4$ of special relativity is not a riemannian manifold, actually it is a pseudo-riemannian one (the metric is not positive-definite), so the definition of lagrangian must be taken with care. It is clear that the situation in general relativity is even more "dramatic" and defining a lagrangian is a non-trival problem.

The most widely known example of such a lagrangian is, I think, that of a free particle in special relativity: $L = -mc^2 \sqrt{ 1 - \frac{v^2}{c^2} }$. (See Goldstein)

References. V.I. Arnold, Mathematical methods of classical and celestial mechanics, chapter IV.ù H. Goldstein, C. Poole, J. Safko, Classical mechanics, 3d edition, Par. 7.9.


Definition of the Lagrangian function

According to Landau-Lifchitz course, the definition of the least action principle contains two substantial points.

  • First it tells us that any mechanical system is fully characterized by one function that depends on generalized coordinates, on the first time derivative of the generalized coordinates and on the time. Such a function is called the Lagrangian.

  • The second point deals with the minimization problem itself. The motion of the system satisfies the following. Consider two distinct instants and the associated generalized coordinates that describe the position of the system at those two instants. Between those two points the motion is made such that the integral of the Lagrangian function between those two instants is minimized.

From there you can get the Lagrange's equation. Nothing is said about $L = T-U$.

Expression of the Lagrangian for a free particle

Considering a free material point we choose to describe the motion in a specific kind of frame. A frame where the space can be considered homogeneous, isotropic and where the time is uniform seems to be the wisest choice. Assuming such a frame exists (it is called a Galilean frame of reference) what would be the form of the Lagrangian?

Because the space is homogeneous the Lagrangian can't contain any term involving the generalized coordinates. In other words the laws of motion cannot depend on where the system actually is. Because the time is also homogeneous, we get the same conclusion, the time cannot explicitly appear in the Lagrangian.

The space is also isotropic, it means that the laws of motion cannot depend on the direction of the motion in the space. Then the Lagrangian only depends on the norm of the velocity and thus not on the direction of the velocity vector. Then the Lagrangian function only depends on the absolute value of the velocity or on the square of the velocity vector. $L = a v^2$.

If you put this form in Lagrange's equation you'll get that $v^2$ is a constant independent of time. Then you'll obtain Newton's first law. Pursuing this reasoning with the study of two Galilean frames moving from one to another will end on L proportional to the square of speed.

General expression of the Lagrangian

Consider an isolated system constituted of several particles. You can describe the interactions between all particles with a function that depends on the position of each particle only. You can call this function $-U$.

It's important to see why this function cannot depend on time. In classical mechanics we consider that the interaction propagates itself instantaneously from one particle to another. Then the time cannot explicitly appear int this -U function.

Hence the general form of the Lagrangian function is $L = T-U$. Using the uniformity of time and Lagrange's equations you will be able to find that a certain quantity doesn't depend on time : $$ E=\sum_i \dot{q_i} \frac{\partial L}{\partial \dot{q_i}} - L $$ Using the form $T-U$ of the Lagrangian, the above relation and the Euler's homogeneous function theorem you will get : $$ E = T + U $$ Now, and only now, you can say that the total energy of the motion is the sum of two distinct terms. The first one only depends on velocity and is called kinetic energy. The second term only depends on position and it is called potential energy.

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    $\begingroup$ Good review, but not directly answer the question. I am not quite sure the last sentence "Now, and only now". For Euler's homogeneous, we definitely have E=T+U, but I am not sure for the reverse direction $\endgroup$
    – unsym
    Commented Jan 14, 2013 at 1:40
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    $\begingroup$ The point was to show the way to obtain $L=T-U$ as an expression of the lagrangian. This last form highly depends on what assumptions we made, and one in particular : we only consider conservative forces, i.e. the total energy is always conserved. $\endgroup$
    – ChocoPouce
    Commented Jan 14, 2013 at 22:15

The action integral can be in "Jacobi action" form, which looks like:

$$ S = 2\int^{B}_{A}\sqrt{(E-V)T}\,\mathrm{d}t $$

where usually $E$ is constant, $V=V(x)$ is the potential energy, and $T=2m$ is the kinetic energy.

For more on this, see:

  1. Brown, J. D. and J. W. York (1989). "Jacobi’s action and the recovery of time in general relativity". Physical Review D40, 3312–3318. doi:10.1103/PhysRevD.40.3312.
  2. Lanczos, C. (1970). The Variational Principles of Mechanics. University of Toronto Press, Toronto.

There are many other versions for deriving the equations of motion from variational calculus, see:

  1. Spivak, M. (2010). Physics for Mathematicians, Mechanics I. Publish or Perish.
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    $\begingroup$ but this just begs the question: why does the action integral have this form? $\endgroup$ Commented Dec 4, 2021 at 2:52

I) It is interesting to observe that if the Hamiltonian $H=\frac{p^2}{2m}+U$ is of the form kinetic plus potential energy, then the so-called Hamiltonian Lagrangian

$$\tag{A} L_H~:=~p \dot{q}-H ~=~ \underbrace{(p \dot{q}-\frac{p^2}{2m})}_{\approx ~\frac{m}{2}\dot{q}^2} - U $$

is also of the form kinetic minus potential energy if we use one of Hamilton's equations $p\approx m \dot{q}$. Off-shell, such interpretation is more challenging. (Here the words on-shell and off-shell refer to whether the equations of motion (eom) are satisfied or not.)

II) A more general Hamiltonian Lagrangian is of the form

$$\tag{B} L_H~=~ \theta_I \dot{z}^I - H - \lambda^a \chi_a, $$

where $z^I$ are the fundamental variables in the theory, $\theta=\theta_I(z) \mathrm{d}z^I$ is a (pre)symplectic potential one-form, $H=H(z)$ is the Hamiltonian, $\lambda^a$ are Lagrange multipliers, and $\chi_a=\chi_a(z)$ are constraints. There are several mechanisms in the Hamiltonian formulation that could complicate or even obstruct an interpretation as kinetic minus potential energy for the Hamiltonian Lagrangian $L_H$:

  • a) The Hamiltonian $H$ is not of the form kinetic plus potential energy.

  • b) Constraints $\chi_a$ are only satisfied on-shell. Off-shell, the term $\lambda^a \chi_a$ does not have an interpretation as a kinetic nor potential energy.

  • c) The two-form $\omega= \mathrm{d}\theta$ may be degenerated, i.e. the phase space may be presymplectic rather than symplectic. In such cases, there is no Darboux' theorem to ensure that $\theta$ is locally of the form $p_i \mathrm{d}q^i$.

III) If OP just want a simple example, here is an example of a free point particle in two dimensions [1]

$$\tag{C} L~=~m\dot{x}\dot{y}.$$

This Lagrangian (C) is different from the kinetic energy & standard Lagrangian

$$\tag{D} L_0~=~T~=~\frac{m}{2}(\dot{x}^2+\dot{y}^2).$$

Yet the Euler-Lagrange equations are the same:

$$\tag{E} \ddot{x}~=~0~=~\ddot{y}. $$

It is a straightforward exercise to check that Lagrangian (C) is non-trivial in OP's sense 1 & 2, i.e. that the difference between $L$ and $L_0$ (where the latter is multiplied with a constant $\alpha$) is never a total time derivative:

$$\tag{F} L-\alpha L_0~\neq~ \frac{dF}{dt}. $$

Hint to prove eq. (F): It is enough to check that the functional derivative of $\int \! \mathrm{d}t~(L-\alpha L_0)$ is non-zero. Why?

IV) For another elementary example, see this Phys.SE post.


  1. M. Henneaux, Equations of motion, commutation relations and ambiguities in the Lagrangian formalism, Ann. Phys. 140 (1982) 45.

To derive the field equations of general relativity (in vacuum), the Lagrangian density is simply the Ricci scalar, which measures deviations from flat space-time. This is a good example of a Lagrangian that has no real "energy" interpretation: in vacuum there is clearly no energy in classical mechanics!


As far as I know, in classical mechanics $L$ is defined exactly as difference between kinetic and potential energy. Conversely, it's the Hamiltonian that not always equals $T+U$, and should be defined as Legendre transform of Lagrangian.

In more complicated models, as in field theory, the Lagrangian could be more complicated. This is because Lagrangians, as Hamiltonian operators in quantum mechanics, are not determined by a universal rule or by a theorem. They are chosen only because they work, i.e. because of an analogy with classical mechanics, or because they lead to physically verified Euler equations. In this case, there is no special reason for which a lagrangian should be separable in two distinct $U$ and $T$ terms.


I always considered the canonical example to be the lagrangian for a point charge (with charge $q$ and mass $m$) in an external EM field: $L(\vec{x},\vec{v}) = \frac{1}{2} m v^2 -q \phi + q \vec{A}(\vec{x}) \cdot \vec{v}$, where $\phi$ is the scalar potential for the electric field, and $\vec{A}$ is the vector potential for the magnetic field.

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    $\begingroup$ It is part of the example (1) I mentioned in my answer. I did not write the explicit expression as you did instead, thanks: I always have problems with signs and coefficients when I try to write down that Lagrangian function! $\endgroup$ Commented Dec 7, 2013 at 23:49

Also note that there are many examples in fluid mechanics where $L\neq T-V$. Particularly when the Eulerian reference frame is used. For instance, for irrotational deep water surface gravity waves, the Lagrangian is written as

$L = \int \left(\int_{-\infty}^{\eta} \phi_t +\frac{1}{2}(\nabla \phi)^2 \ dz \right)+ \frac{g}{2}\eta^2 \ dx$

where $\phi$ is the velocity potential, $g$ is the acceleration due to gravity and $\eta$ is the surface height. In this case, $L\neq T-V$, rather we can recognize it as (minus) the pressure at the free surface. This is the case because the transformation from Lagrangian (in the sense of following particles in the fluid) to Eulerian variables is not canonical.

Also, note that the Lagrangian density yields unique dynamics up to multiplication by constants and the addition of perfect gradients.


I assume the case that you can write $L=T-U$ has the structure $$ L=T(\dot{q})-U(q) $$ with $T(\dot{q})$ as kinetic energy depending on momentum/velocity $\dot{q}$, and $U({q})$ as potential energy depending on coordinates ${q}$.

2+1D Chern-Simons theory is an example which cannot be written in this form.

For non-Abelian Chern-Simons has the action $$ S=\int L dt=\int \frac{k}{4\pi}\big( a \wedge d a + (2/3) a \wedge a \wedge a \big) $$

Even for Abelian Chern-Simons theory, has the action, $$ S=\int L dt=\int \frac{k}{4\pi} \big( a \wedge d a \big) $$ which does the job.

The Abelian 1-form gauge field has $a=a_0 dt+a_1 dx_1+a_2 dx_2$ If you choose temporal gauge $a_0=0$, you will see the Abelian Chern-Simons theory has the form: $$ S=\int L dt=\int\frac{k}{4\pi} \big( a_2 \frac{\partial}{\partial t} a_1 -a_1 \frac{\partial}{\partial t} a_2 \big) \;dt\, dx_1\, dx_2 $$

By identifying $a_1 \sim x$ and $a_2 \sim y$, so effectively Lagrangian is like: $$ \boxed{L=\frac{k}{4\pi} \big( \dot{x}\;y -\dot{y}\;x\big) = \dot{\vec{q}} \cdot \vec{A}(x,y)} $$

where $\vec{q}=(x,y)$ and $\vec{A}=(A_x,A_y)=\frac{k}{4\pi}(y,-x)$.

Effectively it is like a quantum mechanical problem - a particle with displacement $\vec{q}$ moving in a uniform magnetic field: $B=\nabla \times A=-\frac{k}{2\pi} \hat{z}$.

You see that Chern-Simons theory derives $L=\frac{k}{4\pi} \big( \dot{x}\;y -\dot{y}\;x\big) = \dot{\vec{q}} \cdot \vec{A}(x,y)$ does not obey this structure $L=T(\dot{q})-U(q)$.

  • $\begingroup$ Of course, the Lagrangian is in the form of $L(\vec{x},\vec{v}) = \frac{1}{2} m v^2 -e \phi + e \vec{A}(\vec{x}) \cdot \vec{v}$ of the earlier post. If we take our $ \vec{A}(\vec{x}) \cdot \vec{v}$ and regard it as $e \phi - e \vec{A}(\vec{x}) \cdot \vec{v} \to (\rho, \vec{J}) \cdot (\Phi, \vec{A})$, as a generalized covariant potential. Then it is another story. $\endgroup$
    – wonderich
    Commented Dec 17, 2013 at 3:34

Another example, expanding on Qmechanic answer, can be the 2D harmonic oscillator, with the Lagrangian:

$L = m\dot{q}_1\dot{q}_2 - m\omega^2q_1q_2$

this lagrangian has the same EoM as the usual standard harmonic oscillator, but it is quite different, the Noether theorem makes a big mess of the usual symmetries and the conserved quantities, for example the angular momentum $l = m(q_1\dot{q}_2-q_2\dot{q}_1)$ has a squeezing symmetry associated to it:

$$ \left\{\begin{array}{l} q_1\mapsto q'_1 = e^{-\eta}q_1\\ q_2\mapsto q'_2 = e^{\eta}q_2 \end{array} \right. $$

it's kind of weird, but a nice thing.

  • $\begingroup$ This Lagrangian isn't the standard 2-d isotropic harmonic oscillator , is it? The Lagrangian of the 2-d HO should be: $L = \frac{m}{2}(\dot q_1^2 + \dot q_2^2) - \frac{m}{2} \omega ^2 (q_1^2 + q_2^2)$. Has your Lagrangian a practical relevance? $\endgroup$
    – abu_bua
    Commented Jun 17, 2018 at 14:27

In cases of scalar fields, the Lagrangian no longer takes the form of kinetic minus potential but is generalized as kinetic energy minus gradient energy minus potential energy.

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    $\begingroup$ Note that the gradient energy term can in principle be viewed as a potential energy term. $\endgroup$
    – Qmechanic
    Commented Aug 27, 2019 at 19:12

What? Just come up with one? Okay, we can do that.

Let $k_{ab}$, $l_{ab}$ and $m_{ab}$ be a set of coefficients for $a,b = 1,2,⋯$ such that $k_{ab} = k_{ba}$, $l_{ab} = -l_{ba}$ and $m_{ab} = m_{ba}$. Set $U(q) = ½ \sum_{ab} k_{ab} q^a q^b$, $T(v) = ½ \sum_{ab} m_{ab} v^a v^b$. The Lagrangian will be $$L(q,v) = T(v) - U(q) + ½ \sum_{ab} l_{ab} \left(q^a v^b - v^a q^b\right).$$ where $v^a = dq^a/dt$. Then $$p_a ≡ \frac{∂L}{∂v^a} = \sum_b \left(m_{ab} v^b - l_{ab} q^b\right), \hspace 1em f_a ≡ \frac{∂L}{∂q^a} = \sum_b \left(l_{ab} v^b - k_{ab} q^b\right).$$ Therefore, for the Euler-Lagrange equations, we have: $$\frac{dp_a}{dt} = f_a ⇒ \sum_b \left(m_{ab} \frac{dv^b}{dt} - l_{ab} \frac{dq^b}{dt}\right) = \sum_b \left(l_{ab} v^b - k_{ab} q^b\right),$$ or, substituting in $v^a = dq^a/dt$: $$\sum_b \left(m_{ab} \frac{d^2q^b}{dt^2} - 2 l_{ab} \frac{dq^b}{dt} + k_{ab} q^b\right) = 0.$$ In matrix form: $$m \frac{d^2q}{dt^2} - 2l \frac{dq}{dt} + k q = 0.$$

What does it mean? I don't know. I just made it up. It's an example of what you're asking for.

So, this gives you an equation of the form $ma - 2lv + kq = 0$, while $L = T(v) - U(q)$ gives you $m(v) a = f(q)$, where $a = dv/dt$, $v = dq/dt$, $m(v)$ is the matrix of second derivatives of $T(v)$ and $f(q)$ is the gradient of $-U(q)$.


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