content_image/360238.jpg content_image/360237.jpg content_image/360236.jpg content_image/360235.jpg
The result is 

<img src='content_image/360238.jpg'>

which is Eq. (20). 

Since the summation over k and k ′ is dominated by the imaginary part of the Green’s functions, we use g r k ,σ = − i 2 τ$_{σ}$ | g r k ,σ | $^{2}$+ o ( ℏ ϵ$_{F}$ $_{τ}$) and g r k $^{′}$, − $_{σ}$( g a k $^{′}$, − $_{σ}$) 2 = i 2 τ $^{−}$σ | g r k $^{′}$, − σ | $^{4}$+ o ( ℏ ϵ$_{F}$ $_{τ}$), and obtain the leading contribution as 

<img src='content_image/360237.jpg'>

Here spin-dependent density of states and the Fermi wavelength are represented by ν$_{σ}$ and k$_{F σ}$ , respectively. 

Another contribution to the relaxation torque, T A,α so of Eq. (19), represented by the third diagram of Fig. 1, reads 

<img src='content_image/360236.jpg'>

As is seen, T A,α so is an odd function of k or k $^{′}$, and thus it vanishes. α 

The second diagram in Fig .1 (we denote it as δ T $_{so}$) reads 

<img src='content_image/360235.jpg'>

Here, the wave vectors of the two Green’s function, k and k $^{′}$, are independent, and thus the magnitude of δ T α so turns out to be smaller than T z so by the order of 1 / ( ϵ$_{F}$ τ ). 

We therefore see that the dominant contribution to the relaxation torque arises from the first process in Fig. 1, which is estimated to be Eq. (B9) (Eq. (20)). 

