July 13, 2020

Lagrangian mechanics is an alternative formulation of Newtonian mechanics. Newtonian mechanics solves movements using successive approximations. Lagrangian mechanics figures out the whole path all at once. Lagrangian mechanics is useful for solving problems subject to certain constraints. It is also a prerequisite to understanding quantum field theory and the path integral formulation of quantum mechanics.

In the classical formulation of Lagrangian mechanics, the Lagrangian $L$ is the difference between a particle's kinetic energy $T$ and potential energy $V$.

$$ L=T-V $$The action $S$ of a particle is the path integral of its Lagrangian.

$$ S=\int_{t_i}^{t_f}Ldt $$The path of a particle extremizes $S$.

$$\delta S=0$$We therefore derive the Euler–Lagrange equations.

$$ \frac d{dt}\left(\frac{\partial L}{\partial\dot q_j}\right)=\frac{\partial L}{\partial q_j} $$Does this make sense to you? Because **subtracting** potential energy from kinetic energy does **not** make conceptual sense to **me**. Besides, what, conceptually, is "action" and why should it be extremized? Lagrangian mechanics make better sense in special relativity.

To make things more intuitive, let's look at Lagrangian mechanics in the context of special relativity, from which classical Lagrangian mechanics is an approximation. I will set $c=1$ because space and time ought to have the same units. I will set $m_0=1$ for convenience because the rest mass of our particle does not change.

The most important number in special relativity is the Lorentz factor $\gamma$, the instantaneous ratio of coordinate time $t$ to proper time $\tau$.

$$ \gamma=\frac{dt}{d\tau}=\frac1{\sqrt{1-v^2}} $$We can^{[1]} take the action $S$ to be the length of the particle's world line between proper times $\tau_1$ and $\tau_2$. (Note that $\gamma$ is a function of $t$.)

Locally maximizing $S$ equals locally maximizing the proper time experienced by a particle. The Lagrangian $L$ is the expression inside of the coordinate time integral.

$$ L=\gamma^{-1}=\sqrt{1-v^2}=\alpha $$If you think of all matter particles as moving through spacetime at a speed of 1 along the hypotenuse of a right triangle with one spatial leg and one temporal leg then the particle's spatial velocity is $v=\dot q$ and its temporal velocity is $\alpha$.

$$ \alpha^2+v^2=1 $$ $$ S=\int_{t_1}^{t_2}\alpha dt $$The Laplacian $L$ is simply $\alpha$, the particle's temporal velocity. The action $S$ is the integral of the temporal velocity. Therefore, extremizing (maximizing^{[2]}) the action $S$ equals maximizing proper time.

Hamilton's principle, that the evolution of a system is a stationary point of the action functional $S$, naturally follows.

$$\delta S=0$$ $$ \frac d{dt}\frac{\partial\alpha}{\partial v}=\frac{\partial\alpha}{\partial q} $$Under general relativity, time slows down for a particle in a gravity well. In other words, a gravity well decreases the particle's temporal velocity. (This generalizes to the other fundamental forces.) Flipping this around, increases to a particle's potential increase its temporal velocity. It therefore makes intuitive sense to add potential $V$ to special relativistic temporal velocity $\alpha$.

$$ L=\alpha+V $$Thus we arrive at the Euler–Lagrange equations.

$$ \frac d{dt}\frac{\partial L}{\partial v}=\frac{\partial L}{\partial q} $$Classical Lagrangian mechanics follows as an approximation.

[1] The action $S$ also includes a factor of $-m_0c^2$. We have already declared $m_0=c=1$. I have removed the factor of $-1$ too because mechanics in general and $\delta S=0$ in particular are symmetric with respect to time parity. It does not matter to the Euler–Lagrange equations whether $S$ is flipped by a minus sign. ↩

[2] beepConventionally, $S$ is minimized. Since I removed the minus sign from $S$ we maximize it instead. ↩