Why the Lagrangian is the difference of energies?

[MichPod]

The Lagrangian in classical mechanics is known to be a difference of the kinetic and potential energy. My first question is - why? I.e. are there any reasons (except for "because it works this way") to have it as this difference of energies?

The second question is why is it this very value which takes a minumum when integrated over time. But this, of course, is nearly the same question, as the first one. But really, what is the Action? What is the meaning of the value of Energy*Time?

I don't have a problem to understand that there must be some variational problem for which the Newton equations are the solution. Nor that the "minimal action" principle may follow from quantum mechanics. I'm rather interested to understand why the Lagrangian (or action) takes this form in classical mechanics.

 

[Dr.D]

I think that the best answer is "because this is what works."

 

[barek]

You might find this interesting:

http://ph.qmul.ac.uk/sites/default/files/Lagrange.pdf

There’s also something similar near the beginning of Goldstein’s Classical Mechanics

 

[hilbert2]

If potential energy $V$ was defined in the "opposite" way, having it increase when going to the direction of force, then the Lagrangian would be the sum of $K$ and $V$ instead of a difference. It's just a matter of convention. The Lagrange equation has the form that it does, because it produces the same equation of motion as Newton's 2nd law.

 

[muscaria]

The Lagrangian in classical mechanics is known to be a difference of the kinetic and potential energy. My first question is - why?

This is not strictly the case. For instance, the Lagrangian of a charged particle in an EM field takes the form $L=\frac{1}{2}mv^2+q\textbf{v}\cdot\textbf{A}-q\phi$. I think the point is not about it being related strictly to kinetic and potential energies, but rather simply to the fact that the Lagrangian is a function of the generalised coordinates $q_i$ and velocities $\dot{q}_i$. (See below)

why is it this very value which takes a minumum when integrated over time.

If you take the Lagrangian to have the functional dependence $L(q_i,\dot{q}_i)$, then the variation of the Lagrangian is $$\delta L=\sum_i\left(\frac{\partial L}{\partial q_i}\delta q_i+\frac{\partial L}{\partial \dot{q}_i}\delta \dot{q}_i\right).$$
Notice that $q_i$ and $\dot{q}_i$, although dynamically independent of each other (both need to be specified), are not unrelated, one being the time derivative of the other. However, $\dot{q}_i$ is obtained from $q_i$ by differentiation only after having solved the equations of motion. It would be nice though if the variation of $L$ due to $\delta\dot{q}_i$ could somehow be integrated out and written in terms of $\delta q_i$. This is achieved by taking the time integral of $\delta L$, since $$\sum_i\int dt\left[\frac{\partial L}{\partial q_i}\delta q_i+\frac{\partial L}{\partial \dot{q}_i}\delta \dot{q}_i\right]=\sum_i\int dt\left[\frac{\partial L}{\partial q_i}-\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}_i}\right)\right]\delta q_i,$$ through integration by parts.

 

[stevendaryl]

If potential energy $V$ was defined in the "opposite" way, having it increase when going to the direction of force, then the Lagrangian would be the sum of $K$ and $V$ instead of a difference. It's just a matter of convention. The Lagrange equation has the form that it does, because it produces the same equation of motion as Newton's 2nd law.

Well, it's not completely a matter of convention. There is a conserved quantity $E = K + V$, so the real fact about the Lagrangian is that the potential energy enters with the opposite sign as in the energy.