Hi all,

Thank you for all your comments! I just wanted to share the approach we took to address this question (sorry for the equation formatting, main conclusions are in bold). Any feedback is welcome!

We expect that the lateral translation is more affected by the viscosity and the interaction forces between the objective, immersion medium and the coverslip, compared to the axial translation which is dictated mostly by capillary forces. Axial translation used is also in steps of ~100 nm, three orders of magnitude smaller than the lateral translation of ~100 µm. Therefore we will focus on the lateral translation since this will have the stronger effect.

To model the system during lateral translation, we consider the immersion medium as a viscous, incompressible Newtonian fluid confined between two parallel planes separated by a distance *d*, representing the coverslip and the objective, one of which is moving at a velocity *U*.

Starting from the Navier Stokes equation:

*∂u/∂t + (u∙∇)u - ν∇^2u = -∇w + g*

Where *u* is the velocity profile of the fluid, *t* time, *ν* the kinetic viscosity of the fluid such that *ν=μ/ρ* with *μ* the dynamic viscosity and *ρ* the density, *w* an internal source and *g* and external source of the fluid. Since we are working with an incompressible Newtonian fluid, we ignore the convection and source terms and can rewrite the remaining variation and diffusion terms with respect to the forces experienced by the fluid:

*μ∇^2 u + f = 0*

To consider the lateral motion, we can introduce a force parallel to the two planes as *f_x*, and we write the equation explicitly along the *x* coordinate:

*μ((∂^2 u_x)/(∂x^2 ) + (∂^2 u_x)/(∂y^2 ) + (∂^2 u_x)/(∂z^2 )) = -f_x*

Since the gradient will only exist along the *y* direction, we are left with:

*μ (∂^2 u_x)/(∂y^2 ) = -f_x*

Solving this differential equation with the appropriate boundary conditions gives:

*u(y) = -f_x/2μ y^2*

At the boundary between the moving plane and the fluid, we get:

*u(d) = U = -f_x/2μ d^2*

Approximating the distance d as the focal length of the objective *d = 3* mm and the viscosity and density of the immersion oil (Olympus Type F immersion oil) as *ν = 450* mm^2/s and density of *ρ =* (28 g)/(30000 mm^3 ) ≈ 10^(-3) g mm^(-3), gives us a dynamic viscosity of *μ = νρ = 0.45* kg m^(-1) s^(-1).

If the translation stage applies a constant force to shift position, the velocity of the stage will therefore scale inversely with the viscosity of the fluid and proportionally to the square of the distance from the objective to the coverslip.

**Therefore, for certain applications it could be beneficial to increase the separation of the objective from the coverslip before stage translation. Similarly, less viscous immersion media would allow faster translation speed.**

This solution closely resembles a phenomenon known as the Couette flow (1,2), in which the time required for a fluid to form a velocity gradient is estimated as:

*t ~ d^2/ν ≈ (d^2 ρ)/μ ≈ 20* ms

In our case, given the maximal velocity of the stage specified by the manufacturer as 2 mm s^(-1)=2 μm ms^(-1) (Mad City Labs, MicroStage), changing a full field of view (FOV) by translating the stage by 100 µm will require approximately 100 ms (also considering rise and fall time). Since this is larger than the timescale required for the immersion medium to form a velocity gradient, we would expect the stage displacement to be sufficiently slow to avoid significant contribution of the drag force.

References:

- Batchelor, George Keith. An introduction to fluid dynamics. Cambridge university press, 2000.
- Acheson, David J. Elementary fluid dynamics. Oxford University Press, 1990.