January 2021 – biology, math, and physics

Analytical mechanics was developed in the 18th century as a way of simplifying the vector mechanics of Newton to quantities involving only scalars, i.e. single numbers. The central notion of Lagrangian mechanics is the Principle of Least Action which states that Newton’s laws can be equivalently expressed as a problem in the calculus of variations. We write the Lagrangian of a system $L$ as the kinetic energy $T$ minus the potential energy $V$

$$L=T-V$$

The principle of least action states that the configurations over time that the system adopts are a function which is the minimum of the action $S$ , where

$$S=\int L(x(t), \dot x(t)) dt$$

We write $L$ as a function of the position $x$ and the velocity $\dot x=\frac{dx}{dt}$ , since that is usually how the kinetic and potential energies are defined. They are both time dependent, because the position of a particle depends on the time.

To minimize $S$ , we draw inspiration from how functions are minimized: we take a derivative and set it to $0$ . However, $S$ is not a function, but rather a functional since it is an integral of $L$ , and this sort of problem is covered in great mathematical rigor in functional analysis. We’ll take the physicist’s approach and declare there is a new type of derivative, the functional derivative $\frac\delta{\delta \epsilon}$ so that it works like the ordinary derivative. In particular, let’s say that there is some path $x(t)$ for which $S$ is at a minimum. If we write with $f(t)$ an arbitrary function,

$$x_\epsilon(t)=x(t)+\epsilon f(t)$$

then the derivative with respect to $\epsilon$ of $F(x_\epsilon)$ should give $0$ at the minimum when $\epsilon=0$ .

Then by the multidimensional chain rule, we can write

$$\frac{\delta S}{\delta \epsilon}=\int\frac{\delta}{\delta\epsilon} L(x_\epsilon(t), \dot x_\epsilon(t))dt\\=\int\left[\frac{\partial L}{\partial x_\epsilon}\frac{\partial x_\epsilon}{\partial\epsilon}+\frac{\partial L}{\partial \dot x_\epsilon}\frac{\partial\dot x_\epsilon}{\partial\epsilon}\right]dt$$

Taking the second term and writing out $\dot x_\epsilon=\frac{dx_\epsilon}{dt}$

$$\frac{\delta S}{\delta \epsilon}=\int \left[\frac{\partial L}{\partial x}\frac{\partial x}{\partial\epsilon}+\frac{\partial L}{\partial \dot x}\frac{\partial}{\partial\epsilon}\left(\frac{d}{dt}x\right)\right]dt$$

We can switch the order of the derivatives in the second term

$$\frac{\partial L}{\partial \dot x_\epsilon}\frac{\partial}{\partial\epsilon}\left(\frac{d}{dt}x_\epsilon\right)=\frac{\partial L}{\partial \dot x_\epsilon}\frac{d}{dt}\left(\frac{\partial}{\partial\epsilon} x_\epsilon\right)$$

Now integrate the second term by parts ( $\int_b^a udv=uv\big\vert_b^a-\int_b^a vdu$ ), so the time derivative acts on the $\frac{\partial L}{\partial \dot x_\epsilon}$ rather than the $\frac{\partial}{\partial\epsilon} x_\epsilon$ ,

$$\frac{\delta S}{\delta\epsilon}=\int\left[\frac{\partial L}{\partial x}\frac{\partial x_\epsilon}{\partial\epsilon}-\frac{d}{dt}\left(\frac{\partial L}{\partial \dot x_\epsilon}\right)\frac{\partial\dot x_\epsilon}{\partial\epsilon}\right]dt$$

Here, the $uv\big\vert_b^a$ is ignored since it is just a constant, and adding a constant to a function doesn’t change where the minimum is. Finally, we see that we can factor out

$$\frac{\partial x_\epsilon}{\partial \epsilon}=f(t)$$

and

$$\frac{\partial L(x_\epsilon, \dot x_\epsilon)}{\partial x_\epsilon}=\frac{\partial L(x, \dot x)}{\partial x} \text{ and } \frac{\partial L(x_\epsilon, \dot x_\epsilon)}{\partial \dot x_\epsilon}=\frac{\partial L(x, \dot x)}{\partial \dot x}$$

$$\frac{\partial S}{\partial \epsilon}=\int \left[\frac{\partial L}{\partial x}-\frac{d}{dt}\frac{\partial L}{\partial x}\right]f(t)dt=0$$

For this to be true for arbitrary functions $f(t)$ , it must be that the term inside the brackets is 0, giving the Euler-Lagrange equations

$$\frac{\partial L}{\partial x}-\frac{d}{dt}\frac{\partial L}{\partial x}=0$$

After all that, lets write down the Lagrangian in 1D for a point particle moving in a potential $V(x)$

$$L=T-V=\frac{1}{2}m\dot x^2-V(x)$$

The Euler-Lagrange equations give us

$$-\frac{\partial V}{\partial x}-\frac{d}{dt}\left(m\dot x\right)=0$$

Now with the definition of force $F=-\frac{\partial V}{\partial x}$ and $a=\ddot x$

$$F-ma=0$$

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