Q:

Dehn twist and length of a curve.

Let $S$ be a closed hyperbolic surface and $\alpha$ be a simple closed curve. Let $T_{\alpha}$ be the left Dehn twist along $\alpha$. Let $l_x$ denote the length of the geodesic representative in the free homotopy class of $x$.  
Q) Prove that if $\beta$ is a closed curve such that the geometric intersection number between $\alpha$ and $\beta$ is zero then $l_{\beta}=l_{T_{\alpha}(\beta})$. 
A stronger statement (I guess!) ("Dehn twist fixed all geodesics not intersecting the simple closed curve.") was mentioned Here without any proof.
It seems true from definition but I don't know how to prove it. The reason of my confusion is if two curves have simple representatives then their geometric intersection number is realized by their geodesic representatives. Hence the proof of the above result follows from the definition. But if one of them is not simple then I don't know whether this result is true or not. 

A:

Perhaps the main point is that the definition of "the" Dehn twist $T_\alpha$ contains choices, and so it is not well-defined as a homeomorphism, although it is well-defined as a mapping class (meaning an isotopy class of homeomorphisms). One can take advantage of this non-well-definedness to answer the question.
To define $T_\alpha$, choose a closed collar neighborhood $N_\alpha$ of $\alpha$ and a homeomorphism $h : N_\alpha \to S^1 \times [0,1]$. First define $T_\alpha$ to be the identity map on $S - N_\alpha$. Next, given $x \in N_\alpha$, write $h(x) = (e^{i \theta},t)$ for $\theta \in [0,2\pi]$, and then define $T_\alpha(x) = h^{-1}(e^{i(\theta + 2 \pi t)},t)$.
The point is that when you "choose" a collar neighborhood, it can be chosen as narrowly as you like, and $T_\alpha$ is well-defined up to isotopy independent of this choice. 
Now take $\beta$ to be a closed curve having geometric intersection number zero with $\alpha$. This implies that $\beta$ may be homotoped to a closed curve $\beta'$ which is disjoint from $\alpha$. Next, the collar neighborhood $N_\alpha$ may be chosen to be disjoint from $\beta'$, and therefore $T_\alpha$ fixes $\beta'$. Altogether, $T_\alpha(\beta)$ is homotopic to $T_\alpha(\beta')$ which equals $\beta'$ which is homotopic to $\beta$, and therefore $T_\alpha(\beta)$ is homotopic to $\beta$. The geodesic representatives of $T_\alpha(\beta)$ and of $\beta$ are therefore the same.

