where we assume a Maxwell-Boltzmann distribution of relative velocities given by with v 2 r = v 2 x + v 2 $_{y}$. A factor of v 2 z in both the numerator and denominator of Eqn. (7) accounts for the fact that the m$_{L}$ = 0 p -wave inelastic collision rate scales as collision energy ∝ v 2 $_{z}$. Similarly, for m$_{L}$ = ± 1 collisions, the collision rate scales as v 2 r and we find the average relative energy for two colliding particles using Adding this to the k$_{B}$T$_{i}$ from the center-of-mass energies, we get Putting this into the expression for the heating rate (Eqn. (6)), we find that In the main text, we defined the loss rate coefficient as f ( v$_{z}$, v$_{r}$ ) ∝ exp ( − µv 2 r 2 k$_{B}$T$_{r}$ ) exp ( − µv 2 z 2 k$_{B}$T$_{z}$ ) (9) E 1 z,rel = 1 2 µ ∫ ∞ $_{−∞}$f ( v$_{z}$, v$_{r}$ ) v 2 $_{z}$dv$_{z}$ ∞ $_{−∞}$f ( v$_{z}$, v$_{r}$ ) dv$_{z}$ = 1 2 k$_{B}$T$_{z}$ (10) E 1 x,rel = 1 4 µ ∫ ∞ 0 f ( v$_{z}$, v$_{r}$ ) v 4 $_{r}$(2 πv$_{r}$ ) dv$_{r}$ ∞ 0 f ( v$_{z}$, v$_{r}$ ) v 2 $_{r}$(2 πv$_{r}$ ) dv$_{r}$ = k$_{B}$T$_{x}$. (11) Δ E 0 z = 5 2 k$_{B}$T$_{z}$ (12) Δ E 0 x = 3 2 k$_{B}$T$_{x}$ (13) Δ E 1 z = 3 2 k$_{B}$T$_{z}$ (14) Δ E 1 x = 2 k$_{B}$T$_{x}$. (15) dT$_{z}$ dt = − 1 2 T$_{z}$ Γ 0 coll + 1 2 T$_{z}$ Γ 1 coll (16) dT$_{x}$ dt = + 1 2 T$_{x}$ Γ 0 $_{coll}$. (17) β = K$_{0}$T$_{z}$ + 2 K$_{1}$T$_{x}$ (18) = 2 n ( Γ 0 coll + Γ 1 coll ) , Γ 0 coll = K$_{0}$T$_{z}$n 2 (19) Γ 1 coll = 2 K$_{1}$T$_{x}$n 2 . (20) 3 where K$_{0}$ and K$_{1}$ depend on the induced dipole moment. The inelastic collision rate per particle is related to β through