Is there a proof from the first principle that for the Lagrangian $L$,

$$L = T\text{(kinetic energy)} - V\text{(potential energy)}$$

in classical mechanics? Assume that Cartesian coordinates are used. Among the combinations, $L = T - nV$, only $n=1$ works. Is there a fundamental reason for it?

On the other hand, the variational principle used in deriving the equations of motion, Euler-Lagrange equation, is general enough (can be used to to find the optimum of any parametrized integral) and does not specify the form of Lagrangian. I appreciate for anyone who gives the answer, and if possible, the primary source (who published the answer first in the literature).

Notes added on Sept 22:

  • Both answers are correct as far as I can find. Both answerers were not sure about what I meant by the term I used: 'first principle'. I like to elaborate what I was thinking, not meant to be condescending or anything near to that. Please have a little understanding if the words I use are not well-thought of.
  • We do science by collecting facts, forming empirical laws, building a theory which generalizes the laws, then we go back to the lab and find if the generalization part can stand up to the verification. Newton's laws are close to the end of empirical laws, meaning that they are easily verified in the lab. These laws are not limited to gravity, but are used mostly under the condition of gravity. When we generalize and express them in Lagrangian or Hamiltonian, they can be used where Newton's laws cannot, for example, on electromagnetism, or any other forces unknown to us. Lagrangian or Hamiltonian and the derived equations of motion are generalizations and more on the theory side, relatively speaking; at least those are a little more theoretical than Newton's laws. We still go to lab to verify these generalizations, but it's somewhat harder to do so, like we have to use Large Hadron Collider.
  • But here is a new problem, as @Jerry Schirmer pointed out in his comment and I agreed. Lagrangian is great tool if we know its expression. If we don't, then we are at lost. Lagrangian is almost as useless as Newton's laws for a new mysterious force. It's almost as useless but not quite, because we can try and error. We have much better luck to try and error on Lagrangian than on equations of motion.
  • Oh, variational principle is a 'first principle' in my mind and is used to derive Euler-Lagrange equation. But variational principle does not give a clue about the explicit expression of Lagrangian. This is the point I'm driving at. This is why I'm looking for help, say, in Physics SE. If someone knew the reason why $n=1$ in $L=T-nV$, then we could use this reasoning to find out about a mysterious force. It looks like that someone is in the future.
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    $\begingroup$ I think your question kind of gets this backward... saying $L = T - V$ is a nice way to give a general Lagrangian that gives you Newtonian mechanics, but it isn't the most general Lagrangian, and this form, in fact doesn't work for, say Electromagnetism. The modern view would say that the Lagrangian (density) is an object that is defined prior to a notion of energy. $\endgroup$ Commented Sep 21, 2013 at 22:18
  • $\begingroup$ Related: physics.stackexchange.com/q/9/2451 , physics.stackexchange.com/q/15899/2451 and links therein. $\endgroup$
    – Qmechanic
    Commented Sep 21, 2013 at 22:22
  • 1
    $\begingroup$ @ChinYeh There is a proof that builds from Newton's Laws and D'Alembert's principle! You want something more basic than that? $\endgroup$
    – Cheeku
    Commented Sep 21, 2013 at 23:28
  • 4
    $\begingroup$ @ChinYeh: sure--you define the type sof fields/particles you need, define the symmetry of the problem, and then write down the most general lagrangian that respects these two things . Then you make predictions, and refine your lagrangian. $\endgroup$ Commented Sep 22, 2013 at 0:18

7 Answers 7


We assume that OP by the term first principle in this context means Newton's laws rather than the principle of stationary action$^1$. It is indeed possible to derive Lagrange equations from Newton's laws, cf. this Phys.SE answer.

Sketched proof: Let us consider a non-relativistic$^2$ Newtonian problem of $N$ point particles with positions ${\bf r}_1, \ldots, {\bf r}_N$, with generalized coordinates $q^1, \ldots, q^n$, and $m=3N-n$ holonomic constraints.

Let us for simplicity assume that the applied force of the system has generalized (possibly velocity-dependent) potential $U$. (This e.g. rules out velocity-dependent friction forces.)

It is then possible to derive the following key identity

$$\tag{1} \sum_{i=1}^N \left(\dot{\bf p}_i-{\bf F}_i\right)\cdot \delta {\bf r}_i ~=~ \sum_{j=1}^n \left(\frac{d}{dt} \frac{\partial (T-U)}{\partial \dot{q}^j} -\frac{\partial (T-U)}{\partial q^j}\right) \delta q^j. $$

Here $\delta$ denotes an infinitesimal virtual displacement consistent with the constraints. Moreover, ${\bf F}_i$ is the applied force (i.e. the total force minus the constraint forces) on the $i$'th particle. The Lagrangian $L:=T-U$ is here defined as the difference$^3$ between the kinetic and the potential energy. Note that the rhs. of eq. (1) precisely contains the Euler-Lagrange operator.

D'Alembert's principle says that the lhs. of eq. (1) is zero. Then Lagrange equations follows from the fact that the virtual displacement $\delta q^j$ in the generalized coordinates is un-constrained and arbitrary.

D'Alembert's principle in turn follows from Newton's laws using some assumptions about the form of the constraint forces. (E.g. we assume that there is no sliding friction.) See Ref. 1 and this Phys.SE post for further details.


  1. H. Goldstein, Classical Mechanics, Chapter 1.


$^1$ One should always keep in mind that, at the classical level (meaning $\hbar=0$), the Lagrangian $L$ is far from unique, in the sense that many different Lagrangians may yield the same eqs. of motion. E.g. it is always possible to add a total time derivative to the Lagrangian, or to scale the Lagrangian with a constant. See also this Phys.SE post.

$^2$ It is possible to extend to a special relativistic version of Newtonian mechanics by (among other things) replacing the non-relativistic formula $T=\frac{1}{2}\sum_{i=1}^N m_i v^2_i $ with $T=-\sum_{i=1}^N \frac{m_{0i}c^2}{\gamma(v_i)}$ rather than the kinetic energy $\sum_{i=1}^N [\gamma(v_i)-1]m_{0i}c^2$. See also this Phys.SE post.

$^3$ OP is pondering why the Lagrangian $L$ is not of the form $T-\alpha U$ for some constant $\alpha\neq 1$? In fact, the key identity (1) may be generalized as follows

$$\tag{1'} \sum_{i=1}^N \left(\dot{\bf p}_i-\alpha{\bf F}_i\right)\cdot \delta {\bf r}_i ~=~ \sum_{j=1}^n \left(\frac{d}{dt} \frac{\partial (T-\alpha U)}{\partial \dot{q}^j} -\frac{\partial (T-\alpha U)}{\partial q^j}\right) \delta q^j. $$

So the fact that the Lagrangian $L$ is not of the form $T-\alpha U$ for $\alpha\neq 1$ is directly related to that Newton's 2nd law is not of the form $\dot{\bf p}_i=\alpha {\bf F}_i$ for $\alpha\neq 1$.

  • $\begingroup$ This is a fantastically interesting response, but as I’m reading it I’m not sure how you derive the “key identity.” $\endgroup$ Commented Jan 7 at 18:20
  • 3
    $\begingroup$ Hi @Matt Hanson. Thanks for the feedback. See e.g. Goldstein. $\endgroup$
    – Qmechanic
    Commented Jan 8 at 7:47

Lagrangian mechanics can be derived directly from Newton's second law using only algebraic manipulation and a some calculus. This includes both the general form of the Euler-Lagrange equation and the specific form Langangian $L = T - V$. No assumptions of stationarity, use of the calculus of variations, or even any reference to the concept of action are needed.

This is shown in Brian Lee Beers: Geometric Nature of Lagrange's Equations. Similar derivation is also in James Casey: Geometrical derivation of Lagrange’s equations for a system of particles. Casey also wrote a series of follow on papers extending the idea to rigid bodies, fluid dynamics, ...

Beers starts with Newton's second law and projects it onto the coordinate basis vectors. For a single particle this is $$\mathbf{F} \cdot \frac{\partial \mathbf{r}}{\partial q^i} = m \mathbf{\ddot{r}} \cdot \frac{\partial \mathbf{r}}{\partial q^i}$$ From this a few simple algebraic steps produce $$\frac{d}{dt} \frac{\partial T}{\partial \dot{q}_i} - \frac{\partial T}{\partial q_i} = F_i = \mathbf{F} \cdot \frac{\partial \mathbf{r}}{\partial q^i}$$ This is the more general form the Lagrange equation that covers dissipative systems. The conservative case is obtained by setting $\mathbf{F} = -\nabla V$. Substituting that in the above equation gives $$ \begin{align} \frac{d}{dt} \frac{\partial T}{\partial \dot{q}_i} - \frac{\partial T}{\partial q_i} &= - \frac{\partial V}{\partial q_i} \text {, since } \frac{\partial V}{\partial q_i} = \nabla V \cdot \frac{\partial \mathbf{r}}{\partial q^i} \\ \therefore \frac{d}{dt} \frac{\partial T}{\partial \dot{q}_i} - \frac{\partial (T - V) }{\partial q_i} &= 0 \\ \frac{d}{dt} \frac{\partial T}{\partial \dot{q}_i} - \frac{\partial L }{\partial q_i} &= 0 \end{align} $$ Now $\frac{\partial V}{\partial \dot{q}_i} = 0$ since by definition $V$ is a function of only the $q_i$ and independent of $\dot{q}_i$, so: $$ \begin{align} \frac{d}{dt} \left( \frac{\partial T}{\partial \dot{q}_i} - \frac{\partial V}{\partial \dot{q}_i} \right) - \frac{\partial L }{\partial q_i} = 0 \\ \frac{d}{dt} \frac{\partial (T - V)}{\partial \dot{q}_i} - \frac{\partial L }{\partial q_i} = 0 \\ \frac{d}{dt} \frac{\partial L}{\partial \dot{q}_i} - \frac{\partial L }{\partial q_i} = 0 \\ \end{align} $$ Nowhere in this it is assumed that $T - V$ is stationary or is even special in any way. Seen this way defining $L = T - V$ looks like a bit of a hack to tidy up the equations for a conservative system rather than something fundamental. One can use $T$ as the Lagrangian at least for classical mechanics. This is actually necessary for dealing with dissipative systems.

The above derivation goes through for general systems, like multiparticle systems, rigid bodies, etc. The main change is that the mass scalar must be replaced by the inertia tensor of the system. This is covered in Caseys papers mentioned above as well as Synge: On the geometry of dynamics and Crouch: Geometric structures in systems theory.

  • 3
    $\begingroup$ I like this derivation, as it is direct, clear and simple (even though I really like variational principles). However it is not immediately clear how you handle the case with velocity dependent conservative forces (e.g. magnetic fields) as you explicitly require $V$ to be velocity-independent. Can you handle that by redefining $T$? $\endgroup$ Commented May 27, 2016 at 21:23
  • $\begingroup$ @SebastianRiese Interesting question. Let me think about it. $\endgroup$ Commented May 28, 2016 at 15:17
  • $\begingroup$ Man, I hate it when people subtract zero :-) +1 $\endgroup$
    – user541686
    Commented May 1, 2019 at 0:54

Let me assume that "first principles" means Newton's laws but in the somewhat more encompassing formulation of Hamilton's equations, which say that given a Hamiltonian function $H$, then the canonical momentum (I'll display a single one, just for ease of notation) is related to the velocities by

$$ \dot q = \frac{\partial H}{\partial p} $$

and that the dynamical equation of motion (generalizing $F = m a$) is

$$ \dot p = -\frac{\partial H}{\partial q} \,. $$

So in an infinitesimal time span $\epsilon$ the coordinates and momenta evolve as

$$ q_\epsilon = q + \frac{\partial H}{\partial p} \epsilon $$


$$ p_\epsilon = p - \frac{\partial H}{\partial q} \epsilon \,. $$

At the same time, the change of canonical coordinates/canonical momenta is related to the Lagrangian $L$ by ("generating functions for canonical transformations")

$$ p_\epsilon \mathbf{d}q_{\epsilon} - p \mathbf{d}q = \epsilon \mathbf{d}L \,. $$

Now we compute:

$$ \begin{aligned} p_\epsilon \, \mathbf{d} q \epsilon - p \mathbf{d} q & = \left(p - \frac{\partial H}{\partial q} \epsilon \right) \mathbf{d} \left( q + \frac{\partial H}{\partial p} \epsilon \right) - p \mathbf{d}q \\ & = \epsilon \left( p \mathbf{d}\frac{\partial H}{\partial p} - \frac{\partial H}{\partial q} \mathbf{d}q \right) \\ & = \epsilon \left( \mathbf{d}\left( p \frac{\partial H}{\partial p}\right) - \frac{\partial H}{\partial p} \mathbf{d} p - \frac{\partial H}{\partial q} \mathbf{d}q \right) \\ & = \epsilon \mathbf{d} \left( p \frac{\partial H}{\partial p} - H \right) \end{aligned} \,. $$

Hence in general the Lagrangian is

$$ L := p \frac{\partial H}{\partial p} - H \,. $$

Now if $H$ has the standard form (setting $m = 1$ for simplicity)

$$ H = H_{kin} + H_{pot} = \tfrac{1}{2}p^2 + V(q) $$


$$ L = H_{kin} - H_{pot} \,. $$

By the way, anyone enjoying a more general abstract perspective on what's going on here might enjoy to learn this story translated to the language of "prequantized Lagrangian correspondences", for more on this see on the nLab here,


I found a wikilink, Lagrange_multiplier, that answers my question:

"Thus, the force on a particle due to a scalar potential, $F=-\nabla V$, can be interpreted as a Lagrange multiplier determining the change in action (transfer of potential to kinetic energy) following a variation in the particle's constrained trajectory."

${\ \ }$In other words, the potential energy $V$ becomes a set of constraints for the Lagrangian $L=T-nV$ where $n$ is the Lagrange multiplier that needs to be determined. The variation

$$\delta \int_{t_1}^{t_2}L(\dot q_1,...,\dot q_N,q_1, ..., q_N)dt=0 $$
is turned into $2N$ equations, $N$ of which are equations of motion

$$\frac{d}{dt} ( \nabla_{\dot q}T)+n \nabla{_q}V=0$$

and the other $N$ equations are constraints. It turns out $n=1$.

The Lagrange multiplier method makes sense because $V$ is path-independent, therefore, its variation along different paths is always zero:
$$\delta \int_{t_1}^{t_2}Vdt≡0 $$
When we apply variational principle to $\delta \int_{t_1}^{t_2}Ldt≡0$, only the $T$ term varies.
When we add $n \int_{t_1}^{t_2}Vdt $ with arbitrary $n$, nothing changes.
But if we think the $V$ term as constraints on which the particle moves, then we get the right equations of motion.

  • $\begingroup$ Comment to the answer (v1): Note that the two actions $S[q] := \int \! dt~(\frac{m}{2}\dot{q}^2-V)$ and $S[q,\lambda]:= \int \! dt~(\frac{m}{2}\dot{q}^2-\lambda V)$ describe two different physical theories. The latter model constrains the particle to move on the equipotential surface $V=0$, while the former doesn't. For instance, if $V=\frac{m}{2}(\omega q)^2$ is a harmonic potential in 1D, then the solution is $q=q_0\cos(\omega (t-t_0))$ in the former case, while $q=0$ in the latter case. $\endgroup$
    – Qmechanic
    Commented Sep 24, 2013 at 11:25
  • $\begingroup$ It looks like the resulting equations of motion are the same for the 2 settings. I've to scratch my head more. My point is that potential energy plays no role in variation even though it has role in trajectory. $\endgroup$
    – Chin Yeh
    Commented Sep 24, 2013 at 16:47

Proof 1: I have one of my own which is less intensive:enter image description here

As a footnote to that proof: Because what we found as the total energy of the system is conserved, the equation in line 3 has new a new meaning: at every point along the real trajectory aka the solution of the Euler-Lagrange equation, the particle will move in the direction which keeps its total energy constant. In other words, the Lagrangian path is the path which minimizes the change of total energy from point to point (WHICH SHOULD BE ZERO).

Proof 2: (Functional Calculus) There is another proof in the textbook (Quantum Field Theory For The Gifted Amateur): Essentially if T and U are both functionals and their functional derivatives are taken:

The functional derivative of dT/d(x(t)) = -ma, and the functional derivative of U = dU/d(x(t))

If compared to Newtons Equation: (-dU/dx = ma) aka (dU/dx = -ma) we find that Newtons equation states that the functional derivative of T is equal to the functional derivative of U.

d/d(x(t))(T)=d/d(x(t))(U) (Where these are functional derivatives with respect to a change in functional)

Which when factored turns into: d/d(x(t))(T-U)=0 Which is the principle of least action: The stationary integral of the functional (T-U).

  • 5
    $\begingroup$ Please type out your answer instead of using images. Images reduce the discoverability and hamper the visibility of the post. Also, use MathJax for typesetting mathematical expressions. $\endgroup$
    – user258881
    Commented Apr 12, 2020 at 11:38

The $n$ in $L=T-nV$ can be seen as a rescaling factor of potential. $n$ does not change the physics. For example for gravity, $n$ can be absorbed in gravitational constant. See also this.

  • 1
    $\begingroup$ Comment to the answer (v1): What do you mean $n\neq 1$ does not change physics? Changing the gravitational constant $G$ does change physics. $\endgroup$
    – Qmechanic
    Commented Sep 30, 2013 at 20:12
  • $\begingroup$ @Qmechanic. I probably didn't phrase it right. Or, I might not have thought it right. What I thought was this. $V$ has no role in the variation. We can add a factor $n$ to $V$, the nature of physics does not change, even though the trajectory does. The example I gave is even more risque. If we double $V$ and reduce gravitational constant by half, we still get the same equations of motion. Disclosure: I wouldn't do any calculation with $n\neq 1$. I just wanted to explain away this lump in my mind. $\endgroup$
    – Chin Yeh
    Commented Oct 1, 2013 at 20:17

There is a great way to show that the Lagrangian (the thing you want to minimize) is actually equal to $T-V$. The "proof" comes from the book "Quantum Field Theory for Amateurs". To begin the proof we first of all should consider what the average kinetic and potential energy is as a functional $$T_{avg}[x(t)]=\frac{1}{t_2 -t_1}\int_{t_i}^{t_f}dt \frac{1}{2}m\dot{x}^2(t)$$ $$V_{avg}[x(t)]=\frac{1}{t_2 -t_1}\int_{t_i}^{t_f}dtV(x(t))$$ Next if we take functional derivatives of both sides we find that $$\frac{\delta T_{avg}}{\delta x(t)}=-\frac{1}{t_2 -t_1}m\ddot{x}$$ $$\frac{\delta V_{avg}}{\delta x(t)}=\frac{1}{t_2 -t_1}\frac{dV(x)}{dx}$$ The equations of motion for a object in newtonian mechanics is given by the following equation $$F=-\frac{dV(x)}{dx}$$ However we can also write this equation as $$m\ddot{x}=-\frac{dV(x)}{dx}$$ If we now impose that the equations of motion are satisfied we can substitute the expression we have above into our functional derivative $$\frac{\delta V_{avg}}{\delta x(t)}=-\frac{1}{t_2 -t_1}m\ddot{x}$$ Which is the same the functional derivative of the average kinetic energy. Which means that $$\frac{\delta T_{avg}}{\delta x(t)}=\frac{\delta V_{avg}}{\delta x(t)}$$ Moving the terms to one side gives us $$\frac{\delta T_{avg}}{\delta x(t)}-\frac{\delta V_{avg}}{\delta x(t)}=0$$ Since the functional derivative is linear we see that the following is also true $$\frac{\delta}{\delta x(t)}\left(T_{avg} - V_{avg}\right)=0$$ Now if we substitute the average kinetic energy and the average potential energy back into the equation we see that the following is true $$\frac{\delta}{\delta x(t)}\frac{1}{t_2 -t_1}\int_{t_i}^{t_f}dt \left( \frac{1}{2}m\dot{x}^2(t) - V(x)\right)=0$$ If we make this expression nicer by multiplying through by the constant term we see that $$\frac{\delta}{\delta x(t)}\int_{t_i}^{t_f}dt \left( \frac{1}{2}m\dot{x}^2(t) - V(x)\right)=0$$ Which is the same as $$\frac{\delta}{\delta x(t)}\int_{t_i}^{t_f}dt \left( T - V\right)=0$$ We can see here that we are minimising this functional and this functional leads to the equations of motions because we imposed it into our definition. This is what the action is and the term inside the action is the lagrangian. So we have shown that $$L = T - V$$

  • $\begingroup$ When deriving the kinetic energy, where did the negative sign come from? $\endgroup$
    – mohamed
    Commented Mar 1, 2022 at 15:54
  • $\begingroup$ @mohamed when taking the functional derivative you can show that a negative sign should be there. Try taking the functional derivative yourself and verifying this is the correct result. $\endgroup$ Commented Mar 1, 2022 at 23:15
  • $\begingroup$ i didn't make my point clear. I mean isn't this counter-intuitive? average kinetic energy increases while average potential energy decrease because energy is conserved (i,e, slope of kinetic energy has the opposite sign of potential energy slope) .your answer really makes things alot easier. i am beginner-intermediate level in physics can you recommend easy and clear books (like the one you refered to in the answer) preferably from springerlink or online wiley library. Note: I haven't studied functionals and their derivatives, yet. $\endgroup$
    – mohamed
    Commented Mar 2, 2022 at 15:54
  • $\begingroup$ Yeah it is a bit counter intuitive, but that's why you have to look at it from a different perspective. It makes it a lot more intuitive if you think about energy exchange. If you have potential energy, then naturally that potential energy will start to get converted into kinetic energy. Or if you have kinetic energy you can convert it into potential energy by moving away from the object causing the potential. So what you are trying to find is the path with either the most amount of potential energy or the most amount of kinetic energy, as it gets converted into eachother. $\endgroup$ Commented Mar 5, 2022 at 21:10
  • $\begingroup$ @mohamed What are you studying? I can give book recommendations based on that. $\endgroup$ Commented Mar 5, 2022 at 21:11

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