122
15. QCD II: Asymptotic Freedom, the Renormalization Group
where β em (α μ ) is defined by
∂α μ
β em (α μ ) ≡ μ
2
.
(15.35)
∂μ 2
|
|
|
|
e0
From (15.35) and (15.26) we deduce that, to the one-loop order to which we
are working,
α
2
β
[2]
e μ β
[2]
μ
(α μ ) =
(e μ ) =
.
(15.36)
em
em
4π
3π
Now introduce the important variable
t = ln(|q
2
|/μ
2 ).
(15.37)
Equation (15.34) then becomes
[
∂
∂
] (
)
− + β em (α μ )
S e
t , α μ = 0.
(15.38)
∂t
∂α μ
This is a first-order differential equation which can be solved by implicitly
defining a new function – the running coupling α(|q
2
|) – as follows (compare
(15.30):
∫ α(|q
2 |)
dα
t =
.
(15.39)
αμ
β em (α)
To see how this helps, we have to recall how to differentiate an integral with
respect to one of its limits – or, more generally, the formulae
∫ f (a)
∂
∂f
g(x)dx = g (f (a))
.
(15.40)
∂a
∂a
First, let us differentiate (15.39) with respect to t at fixed α μ ; we obtain
1
∂α(|q
2
|)
1 =
.
(15.41)
β em (α(|q 2 |))
∂t
Next, differentiate (15.39) with respect to α μ at fixed t (note that α(|q
2
|) will
depend on μ and hence on α μ ); we obtain
∂α(|q
2
|)
1
1
0 =
−
(15.42)
∂α μ β em (α(|q 2 |)) β em (α μ )
the minus sign coming from the fact that α μ is the lower limit in (15.39).
From (15.41) and (15.42) we find
[
]
∂
∂
− + β em (α μ )
α(|q
2
|) = 0.
(15.43)
∂t
∂α μ
It follows that S(1, α(|q
2
|)) is a solution of (15.38).