Derivation of the Lagrange equation

Question

The physical Lagrange equation is the same as the Euler-Lagrange equation. My question is why do we use the $$L=T-V$$ as the Lagrangian?

Answer 1

The Euler-Lagrange equation minimize (or maximize) the integral $$S=\int_{t=a}^{t=b}L(t,q,\dot q) dt$$ The function $L$ then must obey $$\frac d{dt}\frac{\partial L}{\partial \dot q}=\frac{\partial L}{\partial q}$$ The expressions above are for a single variable $q$, but it's easy to extend to multiple variables.

So how is this related to physics? You need to choose a function $L$ that has some physical meaning. Note that the function is not unique. If you have a function $L$ that is good, the function $\alpha L+\beta$ will give you the same equations of motion. So what would make a good function? In mechanics, we want $L$ to describe the same physics as Newton's laws of motion. We write the second law as $$\frac{d}{dt}(mv)=F$$ Newton's first law can be derived from this, if we set $m=const.$ and $F=0$. If we assume conservative forces (gravity, electromagnetism), then the force can be represented as a gradient of a potential:$$F=-\frac{dV}{dx}$$ So then we write $$\frac d{dt}(mv)=-\frac{dV}{dx}$$ Now we identify $q=x$ and $\dot q=v$. Then $$\frac{\partial L}{\partial v}=mv$$ and $$\frac{\partial L}{\partial x}=-\frac{dV}{dx}$$ One possible solution (if we assume $L$ is a sum of two functions, one that depends only on $x$, the other only on $v$) is $$L=\frac{mv^2}2-V(x)=T-V$$ Once again, this function is a choice.

After thinking about some of the comments, I realized that there are some changes to the form of the terms that have some definite physical meaning. For example, I can add a term linear in $v$ to $T$, so it becomes $$T=\frac m2 v^2+\alpha v$$ If you complete the square, one realizes that this is just the kinetic energy in a reference frame moving with a constant velocity. Then obviously one can change $x$ to some $x-x_0$, and the equation of motion does not change. Finally, one can add a constant. But since we have partial derivatives with respect to $x$ and $v$, constant means that it does not explicitly depend on those values. But it can depend explicitly on time.