# MSD of Sum and Difference Trajectories

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Given two trajectories $r_A$ and $r_B$, the sum trajectory is given by $r_A(t) + r_B(t) \over \sqrt 2$, while the difference trajectory is given by $r_A(t) - r_B(t) \over \sqrt 2$
Here, we consider trajectory A, with its independent displacement $\epsilon_A$, and trajectory B, with its independent displacement $\epsilon_B$. Since these two trajectories come from particles that are glued well to the same slide (the slide you used for your stability measurements), they share the same mechanical noise that comes from vibrations and movements of the microscope stage. We will denote this mechanical noise as $\epsilon_M$. Thus, the total error, or calculated displacement, from trajectory A would be $\epsilon_A + \epsilon_M$, while that of trajectory B would similarly be $\epsilon_B + \epsilon_M$. Note that we define displacement as the change in position over a time interval $\tau$, as when we are calculating the MSD $\left \langle {\left | r(t+\tau)-r(t) \right \vert}^2 \right \rangle$ . With this in mind, we can show that the difference between the two types of MSDs turns out to be two times the square of the mechanical noise as demonstrated below.