Conservation of Energy

In principle, everything can be solved with only F = ma equations, but the conservation laws are very helpful. We’ll derive them from F = ma.

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1D movement

Imagine, we have a particle that can move only in the x direction, and there is a force $F(x)$ that acts on it.

$$\begin{array}{c}F(x)=mv\cdot \frac{dv}{dx}\\ F(x)\cdot dx=mv\cdot dv\end{array}$$

Then, let’s pick an arbitrary point $x_0$, where a particle has a velocity $v_0$

$$\begin{array}{c}{\int }_{x_0}^{x}F(x)\cdot dx={\int }_{v_0}^{v}mv\cdot dv\\ {\int }_{x_0}^{x}F(x)\cdot dx=\frac{m{v}^2}{2}-\frac{m{v}_0^2}{2}\\ E=\frac{m{v}^2}{2}-{\int }_{x_0}^{x}F(x)\cdot dx\\ E=\frac{m{v}^2}{2}+V(x)\end{array}$$

We indeed see that the sum of potential and kinetic energy is constant. However, I want to stress that potential energy is not real! We just made it up and defined it as integral of force over a displacement. It fully depends on our choice of $x_0$, and we always speak of it as a relative quantity. Here, we can only speak of the difference between E and V(x).

Relevant words about potential energy

Fugayzi, fugazi. It’s a whazy. It’s a woozie. It’s fairy dust. It’s never landed. It is no matter. It’s not on the elemental chart. It’s not fucking real.”>

Equivalently, given V(x): $F(x)=-dV(x)/dx$. One moment to notice is that in 3D, we would take a line integral between two points. Logically, then it makes sense to talk about $V(x,y,z)$ if and only if the line integral is the same for all paths. This is true (if you remember) when the vector field $\text{F}$ is conservative ($\nabla \times \text{F}=0$).

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Small oscillations

Imagine, we have a potential function $V(x)$ and an equilibrium point $x_0$. What is the motion around it? First, Taylor expand $V(x)$:

$$\begin{array}{c}V(x)=V(x_0)+V^{\prime }(x_0)(x-x_0)+\frac{V^{\prime \prime }}{2!}(x_0)(x-x_0{)}^2+\frac{V^{\prime \prime \prime }}{3!}(x_0)(x-x_0{)}^3+...\end{array}$$

Well, we don’t care about $V(x_0)$ because we are concerned with differences of potential energy; the constant is not real (the quote above). By definition of $x_0$, $V(x_0)=0$. Then, for small enough displacements, we can derive Hooke’s law!

$$\begin{array}{c}V(x)\approx \frac{V^{\prime \prime }}{2}(x_0)(x-x_0{)}^2\\ F(x)=-V^{\prime \prime }(x_0)(x-x_0)=-k(x-x_0)\end{array}$$