classical mechanics - Physical meaning of the Lagrangian function

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…

   
   

Answer

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.