where Δ E$_{i}$ is the average E$_{i}$ for a pair of particles that is lost from the trap. Putting this together, we have To find the heating caused by inelastic collisional loss we need to calculate Δ E$_{z}$ and Δ E$_{x}$ . The total energy of any pair of particles is the sum of four terms: kinetic and potential energy of the center-of-mass motion of the two colliding particles, and their relative kinetic and potential energy. In any of the three directions, the ensemble average for the energy is 2 k$_{B}$T with equal contributions (1 / 2 k$_{B}$T ) coming from each of the four terms. However, the average energy for pairs of particles that undergo inelastic collisions is, in general, different from the ensemble average because the collision rate depends on the relative position and relative velocity. Because collisions are only sensitive to relative motion, the kinetic and potential energy of the center-of-mass motion of the two colliding particles will be on average the same as the ensemble average, (1 / 2 k$_{B}$T ). The relative potential energy is zero for colliding particles (since they must be at the same position to collide). This is the reason that collisional loss in a trap leads to“anti-evaporative” heating, in contrast to the elastic- collision-based evaporative cooling where energetic particles are preferentially removed from a trap. It remains then to consider the relative kinetic energy. For p -wave inelastic collisions, the collision rate increases linearly with the relative energy of the colliding particles. Further- more, as discussed in the main text, the spatial anisotropy of the dipole-dipole interaction can result in different rates for m$_{L}$ = 0 and m$_{L}$ = ± 1 collisions. Therefore, we calculate the heating for m$_{L}$ = 0 and m$_{L}$ = ± 1 collisions separately and then add the heating rates in the final result. For m$_{L}$ = 0 collisions, we find the average relative energy for two colliding particles using 2 E 0 z,rel = 1 2 µ ∫ ∞ $_{−∞}$f ( v$_{z}$, v$_{r}$ ) v 4 $_{z}$dv$_{z}$ ∞ $_{−∞}$f ( v$_{z}$, v$_{r}$ ) v 2 $_{z}$dv$_{z}$ = 3 2 k$_{B}$T$_{z}$ (7) E 0 x,rel = 1 4 µ ∫ ∞ 0 f ( v$_{z}$, v$_{r}$ ) v 2 $_{r}$(2 πv$_{r}$ ) dv$_{r}$ ∞ 0 f ( v$_{z}$, v$_{r}$ )(2 πv$_{r}$ ) dv$_{r}$ = 1 2 k$_{B}$T$_{x}$ (8) dT$_{i}$ dt = 2 T$_{i}$ Γ$_{coll}$ − Δ E$_{i}$ k$_{B}$ Γ$_{coll}$ . (6)