Metamodel or Surrogated-model definition

=0

∂L

∂Φ^ −∂μ

 ∂L

∂(∂_μΦ^)



=0, we find the corresponding equations of motion

(∂μ−iqA_μ)(∂^μ+iqA^μ)Φ^−m²Φ^ =0 (7.42)

(∂μ−iqA_μ)(∂^μ+iqA^μ)Φ−m²Φ=0, (7.43) which describe each a charged spin 0 field coupled to a massless spin 1 field.

7.1.8 Interaction of Massive Spin 1 Fields

The interaction of a massive spin 1 field with a massless spin 1 field is dictated by symmetry, too. The Lagrangian for massless spin 1 fields is given by

L_Maxwell= ¹

4F^μνF_μν.

To distinguish between a massless and a massive spin 1 field, we name the massive field B^μand define

G^μν:=∂^μB^ν−∂^νB^μ.

The Lagrangian for this massive spin 1 field reads (Eq. 6.19) L_Proca= ¹

2G^μνG_μν+^m2BμBμ.

Lorentz symmetry dictates the interaction term in the Lagrangian to be of the form

LProca-interaction=^CGμνF^μν,

with the coupling constant C we need to measure in experiments. If you’re interested you can derive yourself the corresponding equa-tions of motion, by using the Euler-Lagrange equaequa-tions.

7.2 SU ( ² ) Interactions

Motivated by the success with U(1)symmetry we want to answer the question: Is U(1)the only internal symmetry of our Lagrangians?

It turns out that we can find an internal symmetry for two mass-less spin ¹₂ fields. We get the Lagrangian for two spin ¹₂ fields by addition of two copies of the Lagrangian we derived in Sec. 6.3. The final Lagrangian we derived can be found in Eq. 6.16:

L_Dirac=ψ^¯(iγ_μ∂^μ−^m)ψ.

Here we neglect mass terms, which means m = 0, because other-wise the Lagrangian isn’t invariant as we will see in a moment. We will see later how we can include mass terms, without spoiling the symmetry. The addition yields

L_D1+D2=^{i ¯}ψ₁γ_μ∂^μψ₁+^{i ¯}ψ2γ_μ∂^μψ2 (7.44) and this can be rewritten, if we define

Ψ :=

ψ₁ ψ2



→Ψ :^¯ =^ψ^¯₁ ψ¯₂

where the newly defined objectΨ is called a doublet. Thus

L_D1+D2=^{i ¯}Ψγ_μ∂^μΨ. (7.45)

This Lagrangian is invariant under global SU(2)transformations Ψ→Ψ^=e^iaiσi2Ψ (7.46)

⇒Ψ^¯ →Ψ^¯ =Ψe^{¯ −iaiσi2}, (7.47) where a sum over the index "i" is implicitly assumed, a_idenotes arbitrary real constants and ^σ₂ⁱ are the usual generators of SU(2), with the Pauli matricesσi.

To see the invariance we take a look at the transformed Lagrangian²³

23We neglect mass terms here, be-cause these would be−m1Ψ¯1Ψ1and

−m2Ψ¯2Ψ2and we could write them, using the two component definition for Ψ and defining

m :=^{m1 0} 0 m2



as

LD1+D2= −ΨmΨ.^¯

Unfortunately, this term is not invariant under SU(2)transformations, because LD1+D2= −Ψ^¯mΨ^=Ψ e^¯^−iaiσi2 me^iaiσi2

=m

Ψ.

For equal masses m1 =m2it would be invariant again, but we are going to see how we can include arbitrary mass terms without violating this symmetry.

We know from experiments that the two fields in the doublet do not create particles of equal mass, i.e. m1 =m2. This will be discussed later in detail.

L_D1+D2^ =^{i ¯}Ψ^γ_μ∂^μΨ^

=i ¯Ψe^−iaiσi2γ_μ∂^μe^iaiσi2Ψ

=^{i ¯}Ψγ_μ∂^μΨ=L_D1+D2,  (7.48) where we got to the last line because our transformation e^iaiσi2 acts on our newly defined two-component objectΨ, whereas γμacts on the objects in our doublet, i.e. the Dirac spinors. We can express this using indices

e^−iaiσi2

abδ_αβ 

δ_bcγ^βδ_μ  

e^iaiσi2

cdδ_δ

=δ_adγ^α_μ

This symmetry should be a local symmetry, too. The SU(2) trans-formations mix the two components of the doublet. Later we will give these two fields names like electron and electron-neutrino field.

Our symmetry here tells us that it does not matter what we call electron and what electron-neutrino field, because by using SU(2) transformations we can mix them as we like. If this is only a global symmetry, as soon as we fix one choice²⁴, which means we decide

24This is known as choosing a gauge.

what we call electron and what electron-neutrino field, this choice would be fixed immediately for the complete universe. Therefore we investigate if this is a local symmetry. Again we find that it isn’t, but as for local U(1)symmetry we will do everything we can to make the Lagrangian locally SU(2)invariant.

The problem here is again the derivative, which produces an extra term:

L_D1+D2^ =i ¯Ψ^γ_μ∂^μΨ^

=i ¯Ψe^{−iai(x)σi2}γμ∂^μe^iai(x)σi2Ψ

 =

product rule

i ¯Ψγμ∂^μΨ−Ψγ^¯ μ(∂μa_i(x)^σi

2)Ψ=LD1+D2. (7.49) We already know what to do next from the experience with U(1), which we discussed in Sec. 7.1. We will, step by step, derive a locally SU(2)invariant Lagrangian, which will take some time. To avoid confusion, the steps that follow are summarized in the following list:

• The final result will be that we get a locally SU(2)invariant La-grangian, by adding the extra term i ¯ΨγμW_i^μσiΨ to the Lagrangian, which describes interactions between two spin ¹₂ fields

ψ1

ψ2

 and three massive spin 1 fields W_i^μ.

• In addition, this is only invariant if we transform

(W_μ)_i→ (W_μ^)_i = (W_μ)_i+∂_μa_i(x) +_ijka_j(x)(W_μ)_k instead of

(W_μ)_i→ (W_μ^)_i = (W_μ)_i+∂μa_i(x), which we used to fix local U(1)symmetry.

• This is only a symmetry for the free(W_μ)_iLagrangian L= ¹

4(W_μν)_i(W^μν)_i if we define

(W_μν)i=∂_μ(W_ν)i−∂_ν(W_μ)i+_ijk(W_μ)j(W_ν)k

instead of

(W_μν)_i =∂_μ(W_ν)_i−∂_ν(W_μ)_i.

In other words: In order to get a locally SU(2)invariant La-grangian, we must redefine the free Lagrangian for our three W_i^μ fields, such that it is invariant under

(W_μ)i→ (W_μ^)i = (W_μ)i+∂μa_i(x) +_ijka_j(x)(W_μ)_k.

Now we will go through these steps in detail. We know the inter-nal symmetry of spin 1 fields from Eq. 7.11

A_μ→A^_μ= A_μ+∂_μa(x)

and here we are going to use three spin 1 fields, W₁^μ, W₂^μand W₃^μ, because we have three generators ^σ₂ⁱ for SU(2)and need one field for each extra term, produced by the sum a_i(^x)^σ₂ⁱ in the exponent. These possess the local internal symmetry

W_i^μ→W_i^μ=W_i^μ+∂^μa_i(x) and therefore, we try adding the term

Ψγ¯ _μσj

2W_j^μΨ

to the Lagrangian. Unfortunately, this does not lead to a locally SU(2)invariant Lagrangian:

LD1+D2+Extra^ =i ¯Ψ^γ_μ∂^μΨ^+Ψ^¯γ_μσ_j 2W_j^μΨ^

=i ¯Ψe^{−iai(x)σi2}γμ∂^μe^iai(x)σi2Ψ +Ψe^{¯ −iai(x)σi2}γ_μσ_j

2(^W_j^μ+∂_μa_j(^x))e^iai(x)σi2Ψ

=i ¯Ψγ_μ∂^μΨ−Ψγ^¯ _μ(∂_μa_i(x)^σi 2)Ψ +Ψ e^{¯ −iai(x)σi2}γμσ_j

2W_j^μe^iai(x)σi2

 

=γμσj

2W_j^μbecause[^σi₂,^σj₂]=ijkσk 2=0

Ψ

+Ψe^{¯ −iai(x)σi2}γ_μσ_j

2(∂_μa_j(^x))e^{−iai(x)σi2}Ψ

=^{i ¯}Ψγ_μ∂^μΨ−₍₍₍₍Ψγ^¯ _μ(∂_μa_i⁽⁽⁽⁽(^x)σ_i)Ψ+Ψγ^¯ _μσ_iW_i^μΨ +₍₍₍₍Ψγ^¯ μσi(∂μ⁽⁽⁽⁽a_i(x))Ψ

=i ¯Ψγ_μ∂^μΨ+Ψγ^¯ _μσ_iW_i^μΨ=LD1+D2+Extra  (7.50)

We can see that the reason here is that the generators of SU(2)do not commute²⁵. Let us take a closer look at the difficulty. We will look

25Mathematicians call a group with non commuting elements non-abelian. In contrast, U(1)is abelian and therefore everything was easier.

at, as usual for Lie groups, infinitesimal transformations, ignoring higher order terms:

e^{−iai(x)σi2} σ_j

2e^iai(x)σi2 ≈^1−ia_i(x)^σi 2

 σj

2



1+ia_i(x)^σi 2



= ^σj

2 −a_i(x)^σi 2

σ_j

2 +a_i(x)^σj 2

σ_i

2 + O(a²)

= ^σj

2 +a_i(x)^{ σj} 2,σ_i

2



+ O(a²)

= ^σj

2 −a_i(x)_ijkσ_k

2 + O(a²) = ^σj

2 (7.51)

Therefore, the infinitesimal transformed, which means we can ignore

terms of orderO(a²)and higher, Lagrangian is LD1+D2+Extra^ =i ¯Ψ^γ_μ∂^μΨ^+Ψ^¯γ_μσ_j

2W_j^μΨ^

 =

Eq. 7.50

i ¯Ψγμ∂^μΨ−Ψγ^¯ μ(∂μa_i(x)^σi

2)Ψ+Ψe^{¯ −iai(x)σi2}γμσ_j

2W_j^μe^iai(x)σi2Ψ +Ψe^{¯ −iai(x)σi2}γ_μσj

2(∂_μa_j(x))e^{−iai(x)σi2}Ψ

 ≈

Eq. 7.51

i ¯Ψγ_μ∂^μΨ−Ψγ^¯ _μ(∂_μa_i(x)^σi

2)Ψ+Ψγ^¯ _μ(^σj

2 −a_i(x)_ijkσ_k 2)W_j^μΨ +Ψγ^¯ μ(^σj

2 −a_i(x)_ijkσ_k

2)(∂μa_j(x))Ψ

=i ¯Ψγμ∂^μΨ−

Ψγ¯ μ(∂μa_i(x)^σⁱ

2)Ψ+Ψγ^¯ μσ_j

2W_j^μΨ−Ψγ^¯ μa_i(x)_ijkσ_k

2 W_j^μΨ+_ Ψγ¯ μσ_j

2(∂μa_j(x))Ψ

−Ψγ^¯ _μa_i(x)_ijkσ_k

2 (∂_μa_j(x))Ψ

 

O(a²)

=i ¯Ψγμ∂^μΨ+Ψγ^¯ μσ_j

2W_j^μΨ−Ψγ^¯ μa_i(x)_ijkσ_k

2W_j^μΨ (7.52) The internal symmetry of(W_μ)ithat would fix local SU(2)invariance is

(W_μ)i→ (W_μ^)i = (W_μ)i+∂_μa_i(x) +_ijka_j(x)(W_μ)k

because this gives an extra term in the Lagrangian that cancels ex-actly the symmetry-destroying term

LD1+D2+Extra^ =i ¯Ψ^γ_μ∂^μΨ^+Ψ^¯γ_μσ_j 2W_j^μΨ^

 =

Eq. 7.52

i ¯Ψe^{−iai(x)σi2}γ_μ∂^μe^iai(x)σi2Ψ+Ψe^{¯ −iai(x)σi2}γ_μσ_j

2(W_j^μ+∂_μa_j(x) +_jlma_l(x)W_m^μ)e^iai(x)σi2Ψ

=i ¯Ψγ_μ∂^μΨ+Ψγ^¯ _μσ_j

2W_j^μΨ−

 Ψγ¯ _μa_i(x)_ijkσ_k

2W_j^μΨ+

Ψγ¯ _μσ_j

2_jlma_l(x)W_m^μΨ

=LD1+D2+Extra  (7.53)

We therefore have to take a step back now and have a look if this is an internal symmetry of the three free spin 1 field Lagrangian. The Lagrangian for the three massless spin 1 fields is

L³_xMaxwell = ¹

4(W_μν)₁(W^μν)₁+¹

4(W_μν)2(W^μν)2+¹

4(W_μν)3(W^μν)3

= ¹

4(W_μν)_i(W^μν)_i (7.54)

with

(W_μν)_i =∂_μ(W_ν)_i−∂_ν(W_μ)_i. We want to see if this Lagrangian is invariant under

(W_μ)i → (W_μ^)i=^(W_μ)i+∂_μa_i(x) +_ijka_j(x)(W_μ)k



(7.55)

The transformed Lagrangian is

L3^xMaxwell= ¹

4(W_μν^ )i(W^μν)i

 =

See Eq. 6.25

(∂^μ(W^ν)i∂_μ(W_ν^)i−∂^μ(W^ν)i∂_ν(W_μ^)i)

= (∂^μ

(W^ν)i+∂^νa_i(x) +_ijka_j(x)(W^ν)_k^∂μ

(W_ν)i+∂νa_i(x) +_ijka_j(x)(W_ν)_k^

−∂^μ

(W^ν)i+∂^νa_i(x) +_ijka_j(x)(W^ν)_k^∂ν

(W_μ)i+∂μa_i(x) +_ijka_j(x)(W_μ)_k^)

 =

Ignoring terms of order O(a²)

∂^μ(W^ν)i∂μ(W_ν)i+₍₍₍₍(∂^μ(W^ν)i⁽⁽⁽⁽⁽)∂μ∂νa_i(x) + (∂^μ(W^ν)i)∂μ_ijka_j(x)(W_ν)_k

+₍₍₍₍∂^μ∂^νa_i(x)⁽⁽⁽⁽∂_μ(W_ν)_i+∂^μ_ijka_j(x)(W^ν)_k∂_μ(W_ν)_i

−∂^μ(W^ν)i∂_ν(W_μ)i−₍₍₍₍∂^μ(W^ν)i∂⁽⁽⁽⁽_ν∂_μa_i(x) −∂^μ(W^ν)i∂_νijka_j(x)(W_μ)k

−₍₍₍₍∂^μ∂^νa_i(x)⁽⁽⁽⁽∂_ν(W_μ)_i−∂^μ_ijka_j(x)(W^ν)_k∂_ν(W_μ)_i

=∂^μ(W^ν)i∂_μ(W_ν)i −∂^μ(W^ν)i∂_ν(W_μ)i

=L³xMaxwell

+(∂^μ(W^ν)i)∂_μijka_j(x)(W_ν)k

+∂^μ_ijka_j(x)(W^ν)_k∂μ(W_ν)_i−∂^μ(W^ν)_i∂ν_ijka_j(x)(W_μ)_k−∂^μ_ijka_j(x)(W^ν)_k∂ν(W_μ)_i,

 =

product rule

L³_xSpin1− (∂^μ(^Wν)i)_ijk(∂_μa_j(^x))(^Wν)k− (∂^μ(^Wν)i)_ijka_j(^x)(∂_μ(^Wν)k)

−_ijk(∂^μa_j(x))(W^ν)_k∂μ(W_ν)i−_ijka_j(x)(∂^μ(W^ν)_k)∂μ(W_ν)i

+∂^μ(W^ν)_iijk(∂_νa_j(x))(W_μ)_k+∂^μ(W^ν)_iijka_j(x)(∂_ν(W_μ)_k)

+_ijk(∂^μa_j(x))(W^ν)k∂_ν(W_μ)i+_ijka_j(x)(∂^μ(W^ν)k)∂_ν(W_μ)i (7.56) which show that this is no internal symmetry of the Lagrangian.

However we can make this an internal symmetry by adding the term

−_ijk(W_μ)_j(W_ν)_k to the field tensor Wμν. We have then

(W_μν)i=∂_μ(W_ν)i−∂_ν(W_μ)i−_ijk(W_μ)j(W_ν)k, which gives us a new Lagrangian for the three free spin 1 fields

L³_xMaxwell= ¹

4(^Wμν)i(^Wμν)i

=^∂_μ(^Wν)i−∂_ν(^Wμ)i−_ijk(^Wμ)j(^Wν)k

×

×^∂^μ(W^ν)i−∂^ν(W^μ)i−_ijk(W^μ)j(W^ν)k



(7.57) You can check the invariance of this Lagrangian by yourself, but be warned the computation is quite long and tedious. We have therefore found the locally SU(2)invariant Lagrangian, which we write LD1+D2+Interaction+ 3xMaxwell =i ¯Ψγμ∂^μΨ+Ψγ^¯ μσ_j

2W_j^μΨ−¹

4(W_μν)i(W^μν)i

(7.58)

In document Nose Shape of a High-speed Train (página 59-65)