where r ≡ r ( t ) , r ′ ≡ r ( t $^{′}$) and G ( r − r $^{′}$, t − t $^{′}$) is the Green’s function for the diffusion equation in free space with a sink at infinity (valid for comparison to our MC simulations when the MC boundary position L is large): where D$_{2}$ is the diffusion constant for the solute. We now define F ( r − r $^{′}$, t − t $^{′}$) = ∇ G ( r − r $^{′}$, t − t $^{′}$) . Since in our Langevin dynamics algorithm the particle moves in dis- crete steps, we can replace the integral in Eq.(3) by a sum of F ( r − r $^{′}$, t − t $^{′}$) over the previous trajectory of the particle, giving us a history-dependent expression for the force on the particle: where Δ t is the timestep. The sum in Eq.(5) is carried out only over the previous M steps, where M Δ t should be longer than the timescale over which the solute diffuses away. The equation of motion for the point chucker is then: where D$_{1}$ is the diffusion constant of the chucker and ξ represents Gaussian white noise with zero mean and variance 2 D$_{1}$ Δ t . Care is needed when selecting the parameters D$_{1}$ , D$_{2}$ and k$_{c}$ for comparability of the results of the Langevin and MC simulations, because of the different time units of the two algorithms; however, dimensionless combinations of parameters can be defined which are equivalent between the two simulation schemes. In our MC simulations, we expect that at high chucking rates, the solute particles will tend to form a cage around the chucker, restricting its diffusion. This effect is not included in the Langevin dynamics simulations where we consider a point chucker and a continuous solute field. The Langevin Dynamics simulations are therefore only valid for low chucking rates, where the solute concentration is small. In Sections III A and III B, we first discuss the behaviour of a chucker in free space, with no external applied forces. We then determine the response of the chucker to a dragging III. RESULTS G ( r − r $^{′}$, t − t $^{′}$) = 1 8( πD$_{2}$ ( t − t $^{′}$)) 3 / 2 exp [ − ( x − x $^{′}$) 2 + ( y − y $^{′}$) 2 + ( z − z $^{′}$) 2 4 D$_{2}$ ( t − t $^{′}$) ] (4) f ( t ) = − Ak$_{c}$ M ∑ i =1 Δ tF ( r ( t ) − r ( t − i Δ t ) , i Δ t ) (5) r ( t + Δ t ) = r ( t ) + D$_{1}$ k$_{B}$T Δ t f ( t ) + ξ (6) 8