Solving SHO with Time Translation Invariance

The SHO equation states that:

$$m\frac{d^{2}x}{d{t}^2}=-kx$$

With constant coefficients, physically there is no reference to a particular moment of time. This means that if $x(t)$ is a solution, then $x(t+t_0)$ should also be a solution, or in other words the system is time translation invariant.

Main assumption

Suppose that $x(t)$ and $x(t+t_0)$ are the same solutions related by some coefficient dependent on $t_0$:

$$x(t+t_0)=f(t_0)\cdot x(t)$$

There are two things to note. If $t_0=0$, then $f(0)=1$. If we perform two time translations $t_1,t_2$, then $x(t+t_1+t_2)=f(t_1+t_2)\cdot x(t)$, and $x(t+t_1+t_2)=f(t_2)\cdot x(t+t_1)=f(t_2)\cdot f(t_1)\cdot x(t)$, which implies that $f(t_1+t_2)=f(t_1)\cdot f(t_2)$.

Being tricky (smart)

From the expression above we find that

$$f(t+\delta t)=f(\delta t)f(t)$$

Perform another $\delta t$ translation to get

$$f(t+2\delta t)=f^2(\delta t)f(t)$$

It’s not hard to see that after N translation we get

$$\begin{array}{rl}f(t+N\delta t)=f^N(\delta t)f(t)& \text{set t=0 and use f(0)=1}\\ f(N\delta t)=f^N(\delta t)& \text{now use the Taylor expansion}\\ f(\delta t)\approx f(0)+\delta t\cdot df/dt& \\ f(N\delta t)\approx [1+\delta t\cdot \alpha {]}^N& \text{ with α≡df/dt}\\ f(t)\approx [1+\frac{t\cdot \alpha }{N}{]}^N& \text{as N→∞}\\ f(t)=e^{\alpha t}& \end{array}$$

and being tricky we arrive at the final result $x(t)=e^{\alpha t}\cdot x(0)$

Simple Harmonic Oscillator

But what’s $\alpha$? We have to use the SHO equation:

$$\begin{array}{c}m\frac{d^{2}x}{d{t}^2}=-kx\\ m\cdot \cancel{x(0)}\cdot \cancel{e^{\alpha t}}\cdot {\alpha }^2=-k\cdot \cancel{x(0)}\cdot \cancel{e^{\alpha t}}\\ {\alpha }^2=-{\omega }^2\\ \alpha =\pm iw\end{array}$$

where the imaginary arises naturally! So, the solution is well-known $x(t)=x(0)\cdot e^{\pm i\omega t}$. The only way to make sense of it is to have a linear combination

$$x(t)=a{e}^{i\omega t}+b{e}^{-i\omega t}$$

such that the imaginary parts vanish:

$$\begin{array}{c}a{e}^{i\omega t}=[b{e}^{-i\omega t}{]}^{\ast }=b^{\ast }{e}^{i\omega t}\\ a=b^{\ast }\\ \text{So we get the final result by plugging into the initial eq.}\\ x(t)=a{e}^{i\omega t}+a^{\ast }{e}^{-i\omega t}=z+z^{\ast }=2\text{Re}[z]\\ x(t)=2\text{Re}[a{e}^{i\omega t}]\\ x(t)=\text{Re}[a\cdot e^{i\omega t}]\text{Very Important Line}\end{array}$$

The latter line is the solution that we later always use in coupled oscillator, travelling waves on string, and actual EM waves in medium.