# Why is the Lagrangian defined as $L=T - V$? [duplicate]

Please try to provide a sufficient answer, and when it is just „because it satisfies Newton‘s equations“, please try to give an example or explain it. If you know it, I would be very happy if you could tell me how Lagrange himself came up with this.

• Does this answer your question? Deeper Meaning to the Nature of Lagrangian – Davide Morgante Jul 1 at 19:30
• Possible duplicate: physics.stackexchange.com/q/78138/2451 – Qmechanic Jul 1 at 22:34
• Here is how I understand your question: we have the two forms: Newton's second law on one hand and variational calculus with the Lagrangian $T - V$ on the other hand, and they are mathematically equivalent; proofs are abundant, and you accept these proofs. The nagging question is: how can two formulations that look so different be mathematically equivalent? How is that possible? The following visual demonstration is designed specifically to explain that: physics.stackexchange.com/a/556160/17198 – Cleonis Jul 3 at 13:39

The Lagrangian isn't defined as $$T-V$$, it just turns out that way in a lot of situations. The electromagnetic Lagrangian, for example, is not $$T-V$$. The Shankar quantum mechanics book states (pg 84):
$$L$$ is
that function $$L(q, \dot{q}, t)$$ which, when fed into the Euler-Lagrange equations, reproduces the correct dynamics. The rule $$L=T-U$$ becomes just a useful mnemonic for the case of conservative forces.
A very simple example for when it is $$L=T-U$$ is whenever an object is falling. $$T=\frac{1}{2}m\dot{y}^2$$ and $$U=mgy$$, so the Lagrangian is $$L=T-U=\frac{1}{2}m\dot{y}^ 2 - mgy$$. Feeding these into $$\frac{d}{dt}\frac{\partial L}{\partial \dot{y}}=\frac{\partial L}{\partial y} \ \ \ \text{(Euler-Lagrange equation)}$$ comes out to $$\ddot{y}=g$$ telling you that the acceleration is just $$g$$. You could ge this form Newton's second law also $$F=ma=m\ddot{y}$$ setting this equal to the force of gravity (since we're talking about a falling object) $$m\ddot{y} = mg$$ $$\ddot{y} = g$$ So I got the same thing from Newton and Lagrange