This question (in Lagrangian mechanics) might be silly, but why is the Lagrangian $L$ defined as: $L = T - V$?

I understand that the total mechanical energy of an isolated system is conserved, and that energy is: $E = T + V$. This makes good sense to me, but why would we subtract $V$ from $T$ in the expression of the Lagrangian? What does the Lagrangian represent, after all? Every text or reference I consult just glosses over that and presents $L = T - V$ without explaining anything...

Background: In classical mechanics class, there are often multiple ways to solve the problem at hand. For example, if the problem at hand is the motion of a physical pendulum, or some Atwood machine, you can solve by figuring out the forces, moments, and applying Newton's law. But you can also solve by considering that the total mechanical energy is conserved. Usually, it's faster that way. Then there is a third way, which would be to write the Lagrangian of the system and use Euler-Lagrange's equation. I am trying to understand how the third method should be interpreted and compared to the second one.


7 Answers 7


You can think of the Euler-Lagrange equations $$ \frac{\rm d}{\rm dt} \frac{\partial L}{\partial v} = \frac{\partial L}{\partial x} $$ as a generalization of Newton's equations $$ \frac{\rm d}{\rm dt} p = F $$ where momentum and force are functions on (velocity) phase space derived from a potential $L$ via $$ {\rm d}L = p\,{\rm d}v + F\,{\rm d}x $$ Now, if $\frac{\partial^2 L}{\partial v\,\partial x} = 0$, ie $p=p(v)$ and $F=F(x)$, this integrates to a function $$ L(x,v) = T(v) - V(x) $$ where the minus sign occurs due to the fact that by convention, $F = -\frac{\partial V}{\partial x}$.


We have to point out some simple remark of capital importance. First of all, the lagrangian is not defined as $L = T - V$. This turns out to be true only on a riemannian manifold; the fact this is almost always true in classical mechanics is an "accident" due to the postulates of classical (non relativistic) mechanics. For the most precise definition of the lagrangian, see my answer to this question.

Here we are interested to the fact the lagrangian is by definition the function that satisfies the Euler-Lagrange equations along the curves of motions and so reproduce the equations of motion.

In fact, if you follow, for example, Landau's treatment (see References), you can see how the form of lagrangian is established from a physical point of view. Assume the action principle and denote with $S$ the action for our system and with $q$ the set of generalized coordinates we are using. Then the lagrangian is, by definition, the function such that the integral

$$S = \int_{t_0}^{t_1} L( \dot q, q, t ) \mbox{d}t$$

get an extremal value along the trajectories of the motion. By varying $S$, we find Euler-Langrange equations and the fact that $L$ is defined up to a total derivative with respect to time of an arbitrary differentiable function of time and coordinates, $f(q,t)$. Why is $L$ a function of $\dot q, q$ and $t$ but not, for example, of $\ddot q$? Because newtonian mechanics postulates state that the state of a system is completely determined by the knowledge of all positions and velocities of its constituents at a given instant. Now we would establish the functional form of $L$. In order to do this, we consider the galilean postulate of relativity. We can rephrase it in a more suitable form for our goals:

Postulate(Galilean relativity) There exists a class of frame of reference respect to which the space is homogeneus and isotropic. In those frames, equations of motion retain the same form.

Consider a free-moving particle in an inertial frame of reference. Let's indicate with $\mathbf{r}$ its position and with $\mathbf{v}$ its velocity in such a frame. Since there are no preferred points in the space (nor in time: time flows in the same manner for all observers - i.e. in all frames - in newtonian mechanics), $L$ can't depend on $\mathbf{r}$ or $t$. So it is a function of $\mathbf{v}$ only. Moreover, in virtues of absence of preferred directions, it depends only on $v^2$. Hence $L = L(v^2)$. It follows that

$$ \frac{\partial L}{\partial{\mathbf{r}}} = 0 \implies \frac{\mbox{d}}{\mbox{d}t} \frac{\partial L}{\partial{\mathbf v}} = 0$$.

This implies that $\mathbf{v} = \mbox{constant}$ also. From here we obtain Galilei's transformations. Take two inertial frames of reference, $K$ and $K'$, moving one respect to other with a small velocity $\mathbf{a}$. In $K'$ lagrangian is $L'(v'^2)$, while in $K$ is $L(v^2)$. Since the form of equation of motion must be the same,

$$L'(v'^2) = L(v^2) = L(v^2 + 2\mathbf{a \cdot v} + a^2)$$

Expanding the last member in series of powers of $\mathbf{a}$ and retaining only the first order term, we get

$$L'(v'^2) = L(v^2) + \frac{\partial L}{\partial{v^2}} 2 \mathbf{a \cdot v}.$$

The last term in RHS is a total derivative with respect to time. From the above remark, this means that $\frac{\partial L}{\partial{v^2}}$ can't depend on $v^2$ and hence

$$L = a v^2.$$

We reproduce known equation of motion for a free moving particle if we set $a=m/2$.

Now, if we take a compound system and we separate its constituents in such a way that interactions turn out to be negligible, then for each constituent we can write a free-body lagrangian and the lagrangian for the whole system will the sum of all of them. This means that the free-body lagrangian (in an inertial frame of reference) enjoys the expected characteristics for a kinetic energy.

Finally, consider a system in which interactions between particles are not negligible. We make the following ansatz:

$$L = \sum_\alpha \frac{m_\alpha v_\alpha^2}{2} - V( \mathbf{r_1,r_2 \dots} ).$$

($V$ is a differentiable function of coordinate.) Using Euler-Langrange equations, we get

$$m_\alpha \frac{\mbox{d}\mathbf{v}_\alpha}{\mbox{d}t} = -\frac{\partial V}{\partial \mathbf{r_\alpha}},$$

that is Newton's Second Law. (at least when masses are fixed.) So our ansatz works.

We have to note that in order to derive the form of $L$, we have used the postulates of newtonian mechanics. So our result is no longer valid when those do not apply. In particular, $L = T - V$ is only a particular case and is not a good definition for $L$.

A side question is: does $L$ have a physical significance? No. We measure the equations of motion, not the lagrangian! So the form of $L$ is not important, provided that it reproduces the equations of motion.

Now, the energy. I think you would mean that the hamiltonian, often written as $H = T + V$, makes more sense. In effect, energy is measured quantity, but the hamiltonian is not defined as $H = T + V$. It is, by definition, the Legendre transform of $L$ with respect to $\mathbf{v}$ ($\dot q$ in general). It turns out to be the energy only in the case $L = T - V$ and, in addition, $T$ takes the form $T = \sum_{i,j} a_{ij}(q, t) \dot q_i \dot q_j$, i.e. is a symmetric quadratic form in $\dot q$ whose coefficients are function of $q$ and $t$ only and are symmetric in $i,j$.


L. Landau, E. Lifshitz, Mechanics

V. Arnold, Mathematical methods of classical mechanics

  • 1
    $\begingroup$ Summarizing, Lagrangian, Newtonian and Hamiltonian mechanics are different mathematical frameworks whose goal is to describe the same physics. The postulates of classical mechanics hold for all formalism, because they are facts of Nature, and we can use them to realize the passage from one to another scheme. Of course, one must treat mathematical objects with care in order to gain the equivalence. $\endgroup$
    – user91126
    Commented Nov 11, 2013 at 17:02
  • $\begingroup$ Amazing answer. $\endgroup$
    – Jeff
    Commented Jan 5, 2021 at 17:20
  • $\begingroup$ Kinetic energy comes naturally from general lagrangian when fundamentals of classical mechanics are taken into account. This being: homogeneity and isotropy of space and time in inertial frames as well as galilean invariance. However, when potential energy is taken into account, Landau textbook doesn't explain where does it come from in lagrangian. It just says: "It is found that the interaction between particles can be explained by adding -U term in Lagrangian for system of free particles.", which isn't a great explanation. Do you know how to explain this? $\endgroup$ Commented Jan 14, 2022 at 12:47

The Lagrangian is what it is. At root, it is a more fundamental concept than energy.

After all, not all Lagrangian formulations predict paths that have conserved energy, but all systems with a conserved energy can be formulated with a Lagrangian.

  • $\begingroup$ Is there some perturbation theory coupled with functional analysis at the heart of the lagrangian and the euler lagrange equations? $\endgroup$
    – user28823
    Commented Nov 10, 2013 at 3:18

The minus sign is there because the adjoint of the derivative operator is the negation of the derivative operator.

Let me get the adjoint part out of the way. Consider this inner product on your favourite set of functions on some 1-dimensional domain D:

$$\left<f,g\right> = \int_D f(t)g(t)dt$$

Now consider $\left<f,\frac{\partial g}{\partial t}\right>$. This is

$$\int_D f(t)\frac{\partial g}{\partial t}(t)dt = -\int_D \frac{\partial f}{\partial t}(t)g(t) dt = -\left<\frac{\partial f}{\partial t},g\right>$$

where I used integration by parts and assumed that we're working with a set of functions that goes to zero on the boundary of our domain so we don't need to deal with boundary conditions. (BTW You can redo what I've done without this restriction but I'm trying to save typing.)

This says that you can slide the $\frac{\partial}{\partial t}$ operator back and forth between the two factors in the integral of a product, as long as you flip sign each time. This is just another way of saying that the adjoint of the derivative is minus the derivative because the definition of the adjoint of the operator $A$ is the operator $A^\ast$ such that $\left<f,Ag\right>=\left<A^\ast f,g\right>$ for all $f$ and $g$.

At the end of the day, Newtonian dynamics is largely about solving linear systems of equations. As we're solving for functions these are infinite-dimensional systems. Let's think about this in a finite-dimensional context first. Suppose we have the vector equation

$$Bx = 0$$

where B is some matrix.

Now suppose we want to minimise $\frac{1}{2}|Ax|^2$. Using elementary calculus we can do this by solving:


So if we ever spot that $B$ above factors as $B=A^TA$, we can recast $Bx = 0$ as minimising $|Ax|^2$.

Something similar happens when we're solving the equations $$(B_1+B_2)x=0$$ and each $B_i$ factors as $A_i^\ast A_i$. The equation is then equivalent to minimising $|A_1x|^2+|A_2x|^2$.

Note that the matrix transpose functions like the adjoint because $x\cdot Ay=(A^Tx)\cdot y$ and that the infinite dimensional analogue of $|Ax|^2$ is $\left<Ax,Ax\right>=\left<x,A^\ast Ax\right>$.

Back to Newtonian dynamics. Consider the harmonic oscillator defined by

$$\frac{\partial^2 x}{\partial t^2} + \omega^2 x = 0\label{a}\tag{1}$$

Now watch out - I'm about to do a sign flip. We can rewrite this as:

$$\frac{\partial}{\partial t}\frac{\partial}{\partial t}x+\omega^2x = -(\frac{\partial}{\partial t})^\ast\frac{\partial}{\partial t}x +\omega^2 x = 0$$


$$(-(\frac{\partial}{\partial t})^\ast\frac{\partial}{\partial t}+\omega^2)x=0$$

As discussed above, this means we are doing something equivalent to minimising

$$\frac{1}{2}\left<\frac{\partial x}{\partial t},\frac{\partial x}{\partial t}\right>-\frac{1}{2}\left<\omega x,\omega x\right> = \int(T(t)-V(t))dt\label{b}\tag{2}$$

That minus sign on going from \eqref{a} to \eqref{b} is coming straight from the "anti-self-adjointness" of the derivative.

To make this short I've pared things down to what I think is the minimum heuristic argument. As you add back in the missing details (e.g. a more general force field, boundary conditions at the edge of the domain) the source of the minus sign should remain the same. For example, if you follow through the standard derivation of the Euler-Lagrange equations you'll notice a sign flip when integration by parts is perfomed.

  • $\begingroup$ interesting point of view. There is a book that expand this in detail in a more general setting, and not just applied to the harmonic oscillator? $\endgroup$
    – Masacroso
    Commented May 21 at 7:20
  • $\begingroup$ I don't know of a specific text. This is just me noticing the same sign flip in the same place in mechanics and the kinds of optimization problems that arise in image processing. $\endgroup$
    – Dan Piponi
    Commented May 21 at 17:49

The Lagrangian for the system is whatever happens to give the right equations of motion when you apply the Euler-Lagrange equations, and for systems with conservative forces that is $T-V$. Though this isn't always the case, not even in classical physics. Even when dealing with electrodynamics the Lagrangian takes a different form.


Because it works. In a Newtonian system, if you take $\mathcal L=T-V$, and apply the Euler-Lagrange equations when your generalized coordinates are just the coordinates of the particle, you get $$\mathcal L=\sum \frac12 m\dot x_i^2- V(x_1,x_2,x_3,...))$$ $$\therefore 0=\frac{\mathrm d}{\mathrm dt}\left(\frac{\partial L}{\partial \dot x_i}\right)-\frac{\partial L}{\partial x_i}= \frac{\mathrm d}{\mathrm dt}(m\dot x_i)+\frac{\partial V}{\partial x_i}$$ $$\implies 0=m\ddot{x_i}-F_i\implies F_i=m\ddot{x_i}\implies F=ma$$

So we get $F=ma$ out of this, and from there the rest of Newtonian mechanics can be worked out. Since Newton's laws can be derived from $\mathcal L=T-V$, this is a choice for the Lagrangian of the system. When we choose a Lagrangian, we basically look at the differential equation caused by the system and backtrack from there. For example, for a velocity dependant field (eg the electromagnetic field), a term of the form $q\phi -q\mathbf v\cdot\mathbf A$ is added to the Lagrangian because the final differential equation turns out to be consistent with reality. Indeed, we can derive this by noting that the electromagnetic force is $q\mathbf E + q\mathbf v\times\mathbf B$, and finding out a function $V'$ such that $\frac{\partial V'}{\partial x_i}=\mathbf{\hat{e_i}}\cdot(q\mathbf E + q\mathbf v\times\mathbf B)$ and subtracting it from the Lagrangian.

Note that $\mathcal L=T-V$ is not a unique choice of the Lagrangian for a simple Newtonian system with non-velocity dependant forces. We can add any term of the form $\frac {\mathrm dG}{\mathrm dt}$, where $G$ is a function of the generalized coordinates and time only. So $L=T-V+\sum \dot q_i$ is also a valid choice for the Lagrangian, though it is not independent of the coordinates chosen. A term of the form $\mathbf v\cdot \nabla\phi$ could also be added with no effect.

Essentially, the choice for $\mathcal L$ is the main "axiom" of the system. Instead of having a system dependant on many laws, we have a rigorous mathematical framework that only assumes that the principle of least action holds for the form of the Lagrangian given. This basically takes the abstract notion of the "physical basis" of the system and condenses it into the Lagrangian, leaving the rest to be pure mathematics. New physical ideas can be expressed as modifications to the form of the Lagrangian, instead of having to change the entire framework of physics.


From D'Alembert virtual work principal, the virtual work is zero. So for a particle moving along a "surface" then it has some degree of freedom.(ex: if it moves on a sphere then the degree of freedom is the two angles $\theta,\phi$). So if it has n degree of freedom, then by the Newton's second law:

$$ \begin{split} \vec{F}\cdot\delta\vec{r}&=\sum^3_{i=1}F_i\delta x_i=\sum^3_{i=1}F_i\sum^n_{j=1}\frac{\partial x_i}{\partial q_j}\delta q_j\\ &=\sum^3_{i=1}\sum^n_{j=1}m\ddot{x_i}\frac{\partial x_i}{\partial q_j}\delta q_j\\ &=\sum^3_{i=1}\sum^n_{j=1}m\Big[\frac{d}{dt}\big(\dot{x_i}\frac{\partial x_i}{\partial q_j}\big)-\dot{x_i}\frac{d}{dt}\big(\frac{\partial x_i}{\partial q_j}\big)\Big]\delta q_j\\ &=\sum^3_{i=1}\sum^n_{j=1}m\Big[\frac{d}{dt}\big(\dot{x_i}\frac{\partial \dot{x_i}}{\partial \dot{q_j}}\big)\big)-\dot{x_i}\big(\frac{\partial \dot{x_i}}{\partial q_j}\big)\Big]\delta q_j\ \Bigg(\dot{x_i}=\sum^n_{j=1}\frac{\partial x_i}{\partial q_j}\delta\dot{q_j}\rightarrow\frac{\partial\dot{x_i}}{\partial\dot{q_k}}=\frac{\partial x_i}{\partial q_k}\Bigg)\\ &=\sum^n_{j=1}\Big[\frac{d}{dt}\frac{\partial}{\partial\dot{q_j}}\Big(\frac{1}{2}\sum^3_{i=1}m\dot{x_i}^2\Big)-\frac{\partial}{\partial q_j}\Big(\frac{1}{2}\sum^3_{i=1}m\dot{x_i}^2\Big)\Big]\delta q_j\\ \end{split} $$

Virtual work is:

$$ \vec{F}\cdot\delta\vec{r}=\sum^3_{i=1}F_i\delta x_i=\sum^3_{i=1}F_i\sum^n_{j=1}\frac{\partial x_i}{\partial q_j}\delta q_j=\sum^n_{j=1}\frac{\partial U}{\partial q_j}\delta q_j $$ So combine two terms and define $T=\frac{1}{2}\sum^3_{i=1}m\dot{x_i}^2$:

$$ \sum^n_{j=1}\Big[\frac{d}{dt}\frac{\partial}{\partial\dot{q_j}}(T)-\frac{\partial}{\partial q_j}(T-U)\Big]\delta q_j=0 $$

Because each $q_j$ is independent, every term in the squared bracket is zero. If $U$ is time independent then:

$$ \frac{d}{dt}\frac{\partial}{\partial\dot{q_j}}(T-U)-\frac{\partial}{\partial q_j}(T-U)=0 $$

Note: $\delta$ means virtual displacement. $q_j$ are degree of freedom of the particle. $U$ is potential energy of the particle.

Of course if $U$ is time dependent, the above result still works.


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