What does Lagrangian actually represent?
$L=T-U$ Here, $L$ is Lagrangian. T is kinetic energy. U is potential energy. But, what Lagrangian actually is? I know what Holonomic and non-holonomic is. But, I was thinking what the Lagrangian represent. Like, F represent force applied on some body. To me, Lagrangian is just representing some kind energy. But, what type of energy?
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Lagrangian is no energy. It’s just the Lagrangian. It's perhaps more fundamental than energy in a certain sense. In general, you can think of it as a function that minimizes the action. That's the definition of Lagrangian. ~ said by Golam Ishtiak (You can find him right here also)
I don't have any opinion on it. Just posted what he told me.
There's not really a fundamental interpretation of the Lagrangian because the Lagrangian that describes the dynamics of a system isn't unique - more than one Lagrangian can yield the correct equations of motion. For instance, let's say we have a particle of mass $m$ experiencing a gravitational force $mg$ in the $-z$ direction. The Lagrangian for this system of the form $L=T-V$ is $$L(z,\dot{z})=\frac{1}{2}m\dot{z}^2-mgz$$ Plug this into the Euler-Lagrange equations and you should find that the equation of motion is $\ddot{z}=-g$, as expected. However, the following Lagrangian is just as valid: $$L'(z,\dot{z})=\frac{1}{2}m\dot{z}^2-mgz+\alpha z\dot{z}$$ for some constant $\alpha$. Go ahead and plug it into the Euler-Lagrange equations, and you'll find that it, too, predicts that $\ddot{z}=-g$. The physical interpretation of this function isn't clear, but it turns out to work just as well.
In general, we can add a total derivative to a system's Lagrangian and get a new Lagrangian that will still yield the same equations of motion. If we take our original Lagrangian $L(z,\dot{z})$ and any function $G(z,\dot{z},t)$, the new Lagrangian $$L''(z,\dot{z},\ddot{z},t)=L(z,\dot{z})+\frac{\partial G}{\partial t}+\frac{\partial G}{\partial z}\frac{\mathrm{d}z}{\mathrm{d}t}+\frac{\partial G}{\partial \dot{z}}\frac{\mathrm{d}\dot{z}}{\mathrm{d}t}$$ will give us the same equation of motion.
In fact, a Lagrangian in classical mechanics doesn't even need to have the units of energy. There are some absurd functions you can construct that are still valid Lagrangians, such as $$L=\frac{1}{3}T^2+2TV-V^2$$ All of this should hopefully convince you that it's not really possible to ascribe a physical meaning to a Lagrangian, even in classical mechanics. The quantity $T-V$ does have a (trivial) meaning - the difference between the kinetic and potential energies - but it is only one of many possible valid Lagrangians.
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