Assume the relation $(ab)^6=1$, for $a$ and $b$ Dehn twists about the meridian and the longitude of a torus. Now if we glue the two ends of $T\times I$ together by either the diffeomorphism $(ab)^6$ or the identity map we should get the same three manifold $T^3$. I know there are ways to get from one surgery diagram to another,when we are getting from one Dehn twist presentation of an element of mapping class group to another presentation, by an algebraic process.(for instance: http://www.ams.org/journals/tran/1992-331-01/S0002-9947-1992-1065603-2/S0002-9947-1992-1065603-2.pdf).
My guess is for relations like the above there is no algebraic process (we can check that the two maps are isotopic but not by an algebraic process), so I was wondering if there is a "canonical way" that the isotopy between the two words as above (as elements of mapping class group not $\pi_1(T) $) tells us how to change between their corresponding surgery diagrams.