Let $\;σ:[0, 2\pi) \rightarrow \mathbb R^2\;$ be a curve such that $\;σ(t)=(\cos t,\sin t)\;$. It is known $\;σ\;$ represents the unit circle. However if I have $\;σ(t-1)=(\cos (t-1),\sin (t-1))\;$ then $\;σ\;$ continues to represent the unit circle.
I'm having hard time to understand where is the difference between these two curves. To be more specific, what is happening when the angle changes? I believe a plot would make it easier but I don't know how to make it...
Generally if $\;σ:\mathbb R \supset \mathcal I \rightarrow \mathbb R^n\;$ be a curve, then what exactly is the difference between $\;σ(t)\;$ and $\;σ(t-t_0)\;$? Could somebody claim that $\;σ(t-t_0)\;$ is a "translation" of $\;σ(t)\;$?
Thanks in advance!
If $$\sigma:\quad[a,b]\to{\mathbb R}^2,\qquad t\mapsto\bigl(x(t),y(t)\bigr)$$ is a parametric representation of a curve in the plane then $\tilde\sigma(\tau):=\sigma(\tau-t_0)$ is defined not on the $t$-interval $[a,b]$, but on the $\tau$-interval $[a+t_0,b+t_0]$. The two maps $\sigma$ and $\tilde\sigma$ represent the same curve. Physically one may interpret the difference between the the two representations as a time shift.
Admissible reparametrizations are strictly increasing continuous (preferentially $C^1$) "rescaling of time" maps $$\psi:\quad[\alpha,\beta]\to[a,b],\qquad \tau\mapsto t=\psi(\tau)\ .$$ In particular beginning and end point of a curve, as well as "unit tangent vectors in the forward direction" stay the same.