# Physical meaning of the Lagrangian function [duplicate]

1. In Lagrangian mechanics, the function $L=T-V$, called Lagrangian, is introduced, where $T$ is the kinetic energy and $V$ the potential one. I was wondering: is there any reason for this quantity to be introduced? Does it have any physical meaning?

2. A professor of mine once said its differential $\delta L=dT-dV$ is "the quantity of energy you must pump into the system to move it from a point $(q,\dot q)$ in the phase space to a point $(q+dq,\dot q+d\dot q)$". Is this interpretation correct? I'm not too convinced…

## marked as duplicate by John Rennie, ACuriousMind♦, Sebastian Riese, Qmechanic♦Sep 19 '15 at 15:43

• The way I've seen it until know only makes sense if you know the derivation of Euler's Equation in the context of calculus of variations. I recommend you see that if you havent yet. – DLV Sep 18 '15 at 18:37
• Okay. Possibly related: physics.stackexchange.com/q/9686 and physics.stackexchange.com/q/41138 – Kyle Kanos Sep 18 '15 at 18:38
• Derivation: on a physical path $\vec r(t),$ created by a potential force and constraint force $m\vec a=-\nabla U+\vec F_c,$ we vary the path a little within the bounds of its constraints with a "path-perturbation operator" $\delta$ so $\vec r(t)\mapsto\vec r(t)+\delta\vec r(t),$ etc. Then $\delta U=U(\vec r+\delta\vec r,t)=\nabla U\cdot\delta\vec r$ and $m\vec a\cdot\delta r=-\delta U$ (since $F_c\cdot\delta r=0$ since $\delta$ satisfies the constraints.) Then we can take a time integral and integrate by parts to get $\int dt (-m v\delta v+\delta U)=0,$ or $\delta\int dt(\frac12 mv^2 - U)=0.$ – CR Drost Sep 28 '15 at 14:26

is there any reason for this quantity to be introduced?

It is a quantity for which the actual dynamics makes the integral of the thing be stationary with respect to changes of paths when you consider alternate, but nearby changed paths.

Does it have any physical meaning?

One problem is that many Lagrangians give the same equations of motion, so it is like trying to give a physical meaning to a vector potential. It isn't super straightforward. The linked questions might address your question.

But I will point out that the Lagrangian isn't always $L(\vec q,\dot{\vec q})=T-V.$ It is what you need it to be so that the stationary paths are the physically correct paths.

A professor of mine once said its differential $\delta L=dT-dV$ is "the quantity of energy you must pump into the system to move it from a point $(q,\dot q)$ in the phase space to a point $(q+dq,\dot q+d\dot q)$". Is this interpretation correct?

It is not correct. The function could change for instance without you having to do anything. You could consider two equal masses attached by a spring, moving in a line, with the center of mass in chancing. As it oscillates; the Lagrangian, as a function of time, changes. But you do not have to put any energy into it. The Lagrangian is a function of $(x_1,\dot x_1,x_2,\dot x_2).$ And the path through the coordinates $(x_1,\dot x_1)$ moves in a circle as do the coordinates $(x_2,\dot x_2).$ The energy of the system doesn't change, but the Lagrangian is largest when the kinetic energy is high and lowest when the kinetic energy is zero. You can pump energy in and make them move on larger circles or have the circles both translate in some direction over time. So you can put energy in and change the Lagrangian. But the Lagrangian can change without you putting any energy into the system.

I think my professor would reply «The spring is pumping in energy!» :).

I had two particles (instead of a single particle attached by a spring to an immobile point) exactly so there would be no room to say that. The energy is just sloshing back and forth between kinetic and potential, there is no pumping into the system.

If the energy is sloshing back and forth, I would expect $dT=dU$ and so $\delta\mathcal{L}=0$, so indeed no pumping of energy :). But anyway I got the message: that interpretation is incorrect.

The Lagrangian does change, even when you don't pump energy into the system. That was my entire point.

If $L(\vec q,\dot{\vec q})=T-V$ then $L(\vec q,\dot{\vec q})$=$T+(T-T)-V$=$2T-(T+V)$ so during the actual dynamics $\Delta L=2\Delta T-0.$ So maybe your professor has confused you on other issues too, since it seems like you have the literal same confusion as your professor is claimed. Or maybe you misunderstood all along.

A Lagrangian is different than a Hamiltonian. It increases when kinetic energy increases it is not a constant, it isn't energy. It's a thing, that when stationary under variation, gives the true dynamics.

Now, someone tried to mark this question as a duplicate of this other one. From my point of view, having derived the Euler-Lagrange equations before even mentioning the least action principle or the action, it didn't seem too related. The point was, I wanted to have a physical interpretation of the Lagrangian, and leave the action and the principle as abstract constructions done for who knows what reason, probably because the principle is equivalent to the EL equations. Because that is basically the way things were presented to me: we start from Newton, introduce general coordinates, make use of the D'alembert principle (What did the teacher call it? Smooth constraint principle, or the likes) to derive some equations, and see they are written in term of this function, $L=T-V$, which we call Lagrangian; oh but then we know those equations are equivalent to this strange abstract principle regarding the integral of the Lagrangian; then transform the Lagrangian and we simplify the EL equations, getting into Hamilton, which is more symmetric in that no total time derivatives of partial derivatives of the Hamiltonian appear, as opposed to EL. So I was a bit biased towards interpreting $L$ physically and keeping the action on another planet.

Luckily, Kyle Kalos posted this link in his last comment. THe turning point was this answer. This tells me the leaast action principle is equivalent to a principle which is much more physical, if the amplitudes are physical enough. SO I looked for that principle, and voilà: I found this, but most importantly this lecture by Feynman, which clearly illustrates the physical relevance of action.

So I conclude that it is action that has a physical interpretation in terms of phases -- and in fact the formula $S=\int(T-V)dt$ need not hold! It is just a case that in classical mechanics, that is the action functional. Unfortunately, I can't go too deep, since I know practically nothing of Quantum Mechanics, and delving into that for a geometric thesis that deals with Hamiltonian fields is decidedly off-topic.

As for the professor's interpretation, I guess I can safely conclude it need not be correct, as $L$ might not be $T-V$. This answer will probably give more details about this point.

Potentially related threads: 1, 2 and 3.

• Even when L=T-V it still is not the case that changes in the Lagrangian equal energy pumped into the system from outside. Again as my example of two particles connected by a spring shows. – Timaeus Sep 28 '15 at 14:08

As you have already mentioned, L is NOT in general T-V. T-V only holds in classical mechanics. And I will try to motivate the construction of T-V in classical mechanics following "The Variational Principles of Mechanics" by Cornelius Lanczos.

To start out, let's talk about statics. It it well known that the condition for a physical system to be at equilibrium is that $F_{net}=0$.

However, when we consider dynamics, this law is replaced by the famous $F= ma$. While $F=ma$ is very useful in most everyday scenarios, it fails for two reasons: First, it is a vector equation and thus difficult to generalize to general curvilinear coordinates. Second, it does not work in arbitrary reference frames. In the 18th century, Physicist Jean D'Alembert thought about treating dynamical problems as statics problems. Since Newton's law no longer holds in accelerated reference frames, D'Alembert invented a new function called the "effective force":

$$F_{effective}=F_{impressed}+F_{inertial}.$$

Where impressed force is the usual force in Newtonian mechanics and inertial force is what we now refer to as fictitious forces. Then D'Alembert's principle states that the dynamical system follows this particular equation:

$$\sum_i(\mathbf{F}_i-m_i\mathbf{a}_i)\cdot\delta\mathbf{r}_i=0.$$

Where the function inside the bracket is the effective force and the delta Ri's are "virtual displacements". This equation basically says that if you consider the system in some configuration, and you do a virtual experiment where you move the system in some arbitrary directions, the sum of the work done by the effective forces must add up to 0. This is a generalization of the principle of virtual work from statics to dynamics.

But you may be wondering. This is still a vector equation! We haven't solved the problem yet. You are right! Now let's make the jump from vector to scalar through integration.

let us take the integral of this equation with respect to $t$:

$$\sum_i(\mathbf{F}_i-m_i\mathbf{a}_i)\cdot\delta\mathbf{r}_i=0.$$

What you get is this:

$$\int\limits_{t_1}^{t_2}\sum_i[\mathbf{F}_i-m_i\mathbf{a}_i]\cdot\delta\mathbf{r}_idt=0.$$

• The first part of the integral, $\int_{t_1}^{t_2}\sum_i\mathbf{F}_i\cdot\delta\mathbf{r}_idt$, can be interpreted as the potential energy for conservative systems. Let's rewrite it as $-\delta\int_{t_1}^{t_2}Vdt$, where $V$ is the potential energy.
• Now let's look at the second part. It is clear from integration by parts that:

$$-\int_{t_1}^{t_2}\frac{d}{dt}(m_i\mathbf{v_i})\cdot\delta\mathbf{r}_idt=-\int_{t_1}^{t_2}\frac{d}{dt}(m_i\mathbf{v}_i\cdot\delta\mathbf{r}_i)dt+\int_{t_1}^{t_2}m_i\mathbf{v}_i\cdot\frac{d}{dt}(\delta\mathbf{r}_i)dt.$$

The first term on the right hand side is trivially integral and yields a boundary term: $-(m_i\mathbf{v}_i\cdot\delta\mathbf{r}_i)|_{t_1}^{t_2}$. Since we have the freedom to fix the end points, let us set $\delta\mathbf{r}_i=0$ at $t_1$ and $t_2$. Then we are left with $\int_{t_1}^{t_2}m_i\mathbf{v}_i\cdot\frac{d}{dt}(\delta\mathbf{r}_i)dt$, which we can integrate to obtain $\int_{t_1}^{t_2}\frac12m_iv_i^2dt$. Now the picture is starting to surface. The integral we started with can be expressed as:

$$\delta\int_{t_1}^{t_2}(\frac12m_iv_i^2-V)dt.$$

If we make the substitution $T=m_iv_i^2\cdot\frac12$, the integral reduces to:

$$\delta\int_{t_1}^{t_2}(T-V)dt.$$

Magic!!! $L = T-V$ !! Finally obtained the Hamilton's Principle. Variation of the action is zero!

So the Lagrangian function really came from integrating D'Alembert's principle. It is not arbitrary at all. It is formally different, but mathematically equivalent to D'Alembert's Principle. The remarkable feature of Hamilton's Principle is that $L$ is a scalar quality (a frame-independent quantity). Taking Hamilton's Principle as a fundamental construction, physicists went far beyond classical mechanics. In Standard Model, the Lagrangian is literally a page long instead of just a simple $T-V$. These complicated lagrangians are outside the scope of my understanding. So I will leave it to experts to shed more insight on a general lagrangian formalism.

I hope my somewhat historical account helped to motivate the Lagrangian construction in classical mechanics.