Calculation of transversal fluctuations

We will work in the D3/D7-setup described in section 2.4 at vanishing quark mass, i.e. with flat D7-brane embeddings. The coordinates we use are those introduced in equation (3.12). In order to find the effective action which suffices to describe small gauge field fluctuations we start from the Dirac-Born-Infeld action (2.21) for a D7-brane, constant dilaton field e^Φ = gs

and vanishing transversal scalarsφi ≡0so that we get SD7 =−TD7

Z

d⁸ξStr q

det{g+ (2πα^′) ˆF} (4.15) where g is the pull-back of the originally ten-dimensional metric to the eight-dimensional brane and Fˆ is the non-Abelian field strength on the brane for a field A. Making use of theˆ determinant expansion formula for small values of|M|

pdet(1 +M) =e^{12tr(M−12M2+13M3+...)} = 1 + 1

2trM − 1

4trM²+1

8(trM)²+. . . , (4.16) we expand the action in gauge field fluctuations A up to quadratic order in A, which are contained in Fˆ. The non-Abelian field strength tensor F consists of flavor components F^a and representation matricesT^aas follows

Fˆ_µν = ˆF_µν^a T^a= 2∂_[µAˆ^a_ν]T^a+f^abcAˆ^b_µAˆ^c_νT^a, (4.17) and the fieldAˆis comprised of a background gauge field and fluctuationsAin the context of the background field method of quantum field theory

Aˆ^a_ν =δν0δ^a3µ+A^a_ν, (4.18) whereµis the constant time-component which is interpreted as the chemical potential at the AdS-boundary. Using (4.16), and noting thatM =g⁻¹Fˆ so thattr(g⁻¹Fˆ) = 0by tracing the

76 Chapter 4. Holographic thermo- and hydrodynamics

symmetricgtogether with the antisymmetricFˆ, we obtain SD7 = −TD7

Z

d⁸ξStr{√

−g q

det[1 +g⁻¹(2πα^′) ˆF]}

= −TD7

Z

d⁸ξStr{√

−g[1 + 1

2(2πα^′)tr(g^ΣΣ′Fˆ_Σ^′_Ω)

−1

4(2πα^′)²tr(g^ΣΣ′FˆΣ^′Ωg^ΩΩ′FˆΩ^′∆) +. . .]} (4.19)

= −TD7

Z

d⁸ξStr{√

−g[1−(2πα^′)²

4 g^ΣΣ′g^ΩΩ′FˆΣΩFˆΣ^′Ω^′+. . .]} (4.20)

= −TD7

Z

d⁸ξStr{√

−g[¹Nf×Nf −(2πα^′)²

4 g^ΣΣ′g^ΩΩ′Fˆ_ΣΩ^a Fˆ_Σ^b′_Ω^′T^aT^b+. . .]}(4.21). The symmetrized flavor trace Str{. . .}applied to the first two terms in the action merely gives a factor ofNf for the trace over unity while in the second term it gives

Str{T^aT^b} = trfund(Nf){T^aT^b+T^bT^a}

= trfund(Nf)[T^a, T^b]

| {z }

if^abcT^c

+2trfund(Nf)(T^bT^a)

= 2trfund(Nf)(T^bT^a), (4.22) where we have used that our flavor group generatorsT^c are traceless. Furthermore the gen-erators are HermitianT^a† = T^a and they live in the fundamental representation of the fla-vor gauge groupSU(Nf). It is in general possible to choose linear combinations of a given set{T^a}such that the trace property

trfund(Nf)(T^aT^b) =kaδ^ab(no sum), (4.23)

is satisfied [113, equation (II.7)]. The standard conventions [114] fix the factorka = TR for alla= 1,2, . . . ,(N_f²−1), where the Dynkin indexTRonly depends on the representation. For the fundamental representation we haveTR = 1/2as we can check explicitly in the example with Pauli matricesσ^afor theSU(2)isospin generatorsT_Iso^a =σ^a/2

tr_fund(Nf)(T_Iso^a T_Iso^b ) = 1

4tr_fund(Nf)(δ^ab1+iǫ^abcσ^c) = 2δ^ab 4 = 1

2δ^ab. (4.24) In the hypothetical case that our flavor generatorsT^awould live in the adjoint representation the Dynkin indexTRwould equal the value of the Casimir operatorCA =Tadj =Nf.

As mentioned before, we work in the fundamental representation of the flavor groupSU(Nf), therefore we find the D7-brane action in quadratic order of gauge field fluctuationsA

S_D7⁽²⁾ =TD7

(2πα^′)²

4 TR(2π²R³) Z

dud⁴x√

−gg^µµ′g^νν′Fˆ_µν^a Fˆ_µ^a′_ν^′, (4.25) where we have already integrated over the three angular directions5,6and7(on which none of the fields depends) giving the factor(2π²R³). With the help of equation (2.22) the factor in

4.2. Analytical Hydrodymamics at finite isospin potential 77

front of the action integral in (4.25) can be written as TD7

(2πα^′)²

4 TR(2π²R³) = 2⁻⁷π⁻³gs−1(α^′)⁻²R³. (4.26) Note, that equation (4.25) still contains cubic and quartic terms in the fluctuations but we have deliberately chosen this covariant form since it is more compact. We will neglect cubic and quartic contributions in a later step. The fluctuations A^a_µ(t, x = 0, y = 0, z, u)without loss of generality are assumed to depend on time t, the third directionx3 = z and the radial AdS coordinate u only while we choose a gauge such that the field has components in the Minkowski directions only, i.e.ν = 0, 1, 2, 3.

Fˆ_µν^a Fˆ^{a µν} = 4∂[µAˆ^a_ν]∂^[µAˆ^ν]a+ 4f^abc∂[µAˆ^a_ν]Aˆ^{µ b}Aˆ^{ν c}+f^abcf^ab′c′Aˆ^b_µAˆ^b_νAˆ^{µ b′}Aˆ^{ν c′}

= 4∂[µ(δ^a3δν]0µ+A^a_ν])∂^[µ(δ^a3δ^ν]0µ+A^ν]a)

+4f^abc∂[µ(δ^a3δν]0µ+A^a_ν])(δ^b3δ^µ]0µ+A^µ]b)(δ^c3δ^ν]0µ+A^ν]c)

+f^abcf^ab′c′(δ^b3δµ]0µ+A^b_µ])(δ^c3δν]0µ+A^c_ν])(δ^b′3δ^µ]0µ+A^µ]b′)(δ^c′3δ^ν]0µ+A(4.27)^ν]c′). This expression simplifies considerably by noting that derivatives acting on the constant µ vanish. Furthermore the terms including more than two background fields µvanish because of the antisymmetrization. For example

f^abc∂[µA^a_ν]δ^µ0δ^b3µδ^ν0δ^c3µ= 0. (4.28) The mathematical reason for this to vanish is that more than one background gauge field term is contracted with one single structure constant. Since every term including the background gauge fieldµby our choice always has to contain the factorδ^3a, it is clear that more than one such factor forces two of the flavor indices inf^abcto be equal to 3:f^33c = 0. Since there are at most two different structure constants in one single term such asf f µµAA(schematically), we can have at most two background gauge fields in one term. One of the two µ has to be contracted with the first structure constantf the other has to be contracted with the other one.

Since we are interested in two-point functions we are also free to neglect all terms that are cu-bic or higher order in the field fluctuationsO(AAA, AAAA, . . .). After these considerations the action factor (4.27) becomes

Fˆ_µν^a Fˆ^{a µν} = 4∂[µA^a_ν]∂^[µA^ν]a+ 4f^a3cg⁰⁰∂[0A^a_ν]A^{ν c}µ+ 4f^ab3g⁰⁰∂[µA^a_0]A^{µ b}µ +f^a3cf^a3c′µ²A^c_νA^{ν c′}+f^a3cf^ab′3µ²g⁰⁰A^c₀A^b₀^′

+f^ab3f^a3c′µ²g⁰⁰A^b₀A^c₀^′ +f^ab3f^ab′3µ²A^b_µA^{µ b′} (4.29)

= 4∂_[µA^a_ν]∂^[µA^ν]a+ 8f^ab3µg⁰⁰∂_[νA^a_0]A^{ν b}

+2µ²(g^µµ′A¹_µA¹_µ′ +g^µµ′A²_µA²_µ′ −g⁰⁰A¹₀A¹₀−g⁰⁰A²₀A²₀). (4.30) Using this simplified factor (4.30) in the quadratic action (4.25) we derive the equations of

78 Chapter 4. Holographic thermo- and hydrodynamics

motion for the gauge field componentsA^a_µusing the Euler-Lagrange equation 0 = ∂κ

"

δS_D7⁽²⁾ δ(∂_κA^d_σ)

#

− δS_D7⁽²⁾

δA^d_σ (4.31)

= ∂κ

δ δ(∂κA^d_σ)(√

−gg^µµ′g^νν′Fˆ_µν^a Fˆ_µ^a′_ν^′)

− δ δA^d_σ(√

−gg^µµ′g^νν′Fˆ_µν^a Fˆ_µ^a′_ν^′) (4.32) (4.33) After a few simplifications by interchanging indices the equations of motion can be written as

0 = 2∂κ

h√

−gg^κκ′g^σσ′ ∂[κ^′A^d_σ^′_]i +µf^db3h

δσ0∂κ(√

−gg⁰⁰g^κκ′A^b_κ^′) +√

−gg⁰⁰g^σµ∂µA^b₀−2√

−gg⁰⁰g^σµ∂0A^b_µi

−µ²√

−gg⁰⁰g^σσ′

δ^d1(A¹_σ^′ −A¹₀δ0σ^′) +δ^d2(A²_σ^′ −A²₀δ0σ^′)

. (4.34)

There is one free space-time index σ which can take values in the four Minkowski direc-tions (x0 = t, x1 = x, x2 = y, x3 = z) and in the radial AdS-direction x4 = u as well.

Therefore we can split equation (4.34) into five distinct differential equations which are cou-pled with each other. There is also one free flavor indexd which we will consider in detail shortly. Let us start choosing the free indexσ= 1

0 = 2∂κ

h√

−gg^κκ′g^1σ′ ∂[κ^′A^d_σ^′_]i +µf^db3h

δ10∂κ(√

−gg⁰⁰g^κκ′A^b_κ^′) +√

−gg⁰⁰g^1µ∂µA^b₀−2√

−gg⁰⁰g^1µ∂0A^b_µi

−µ²√

−gg⁰⁰g^1σ′

δ^d1(A¹_σ^′ −A¹₀δ0σ^′) +δ^d2(A²_σ^′ −A²₀δ0σ^′)

. (4.35)

This equation only involves the gauge field components in thex1-direction and writing down the other four equations we will see later, that this equation decouples from all of them and is therefore the simplest one to solve. We note here that the inverse metric is diagonal such thatg^µµ′ =g^µ′µand it vanishes forµ6=µ^′, so we get

0 = 2∂κ

h√

−gg^κκ′g¹¹ ∂_[κ^′A^d_1]i +µf^db3√

−gg⁰⁰g¹¹∂1A^b₀−2√

−gg⁰⁰g¹¹∂0A^b₁

−µ²√

−gg⁰⁰g¹¹

δ^d1A¹₁+δ^d2A²₁

. (4.36)

Now recall that we have chosen the geometry such thatA^a_µ(x₀, x₁ = 0, x₂ = 0, x₃, x₄), which implies that the derivatives of fluctuations in all other thanx0, x3, x4-directions vanish

∂1,2A^a_µ ≡0, ∂5,6,7A^a_µ ≡0. (4.37)

Considering this gives 0 = ∂κ

h√

−gg^κκ′g¹¹ ∂κ^′A^d₁i

−2µf^db3√

−gg⁰⁰g¹¹∂0A^b₁

−µ²√

−gg⁰⁰g¹¹

δ^d1A¹₁+δ^d2A²₁

. (4.38)

4.2. Analytical Hydrodymamics at finite isospin potential 79

Now we transform to Fourier space with conventions given in equation (3.15) 0 = −iω√

−gg⁰⁰g¹¹ −iωA^d₁

+iq√

−gg³³g¹¹ iqA^d₁

+∂u√

−gg⁴⁴g¹¹ ∂uA^d₁

−2µf^db3√

−gg⁰⁰g¹¹∂0A^b₁−µ²√

−gg⁰⁰g¹¹

δ^d1A¹₁+δ^d2A²₁

. (4.39)

We abbreviate derivatives in radial AdS-direction∂uA = A^′ and sort the equation in deriva-tives of the gauge field fluctuationsA, A^′, A^′′and normalize it such that the second derivative has the coefficient one

0 =A^d₁^′′+∂u(√

−gg¹¹g⁴⁴)

√−gg¹¹g⁴⁴ A^d₁^′−g⁰⁰

ω²A^d₁ −2if^db3ωµA^b₁+µ²(δ^d1A¹₁+δ^d2A²₁)

+g³³q²A^d₁

g⁴⁴ .

(4.40) Turning to the free flavor indexdwe recall that it can take the values1,2,3corresponding to the three flavor directions we introduced by assuming anSU(Nf = 2)-isospin flavor symme-try. We split (4.40) into three equations

0 = A¹₁^′′+∂_u(√

−gg¹¹g⁴⁴)

√−gg¹¹g⁴⁴ A¹₁^′− g⁰⁰

ω²A¹₁−2if^1b3ωµA^b₁+µ²(δ¹¹A¹₁+δ¹²A²₁)

+g³³q²A¹₁

g⁴⁴ ,

0 = A²₁^′′+∂u(√

−gg¹¹g⁴⁴)

√−gg¹¹g⁴⁴ A²₁^′− g⁰⁰

ω²A²₂−2if^2b3ωµA^b₁+µ²(δ²¹A¹₁+δ²²A²₁)

+g³³q²A²₁

g⁴⁴ ,

0 = A³₁^′′+∂u(√

−gg¹¹g⁴⁴)

√−gg¹¹g⁴⁴ A³₁^′− g⁰⁰ω²+g³³q² g⁴⁴ A³₁.

By using the antisymmetry of the structure constantsf^3b3 = 0, f^abc =−f^bacwe arrive at 0 = A¹₁^′′+∂_u(√

−gg¹¹g⁴⁴)

√−gg¹¹g⁴⁴ A¹₁^′− g⁰⁰[ω²A¹₁−2iωµA²₁+µ²A¹₁] +g³³q²A¹₁

g⁴⁴ , (4.41)

0 = A²₁^′′+∂u(√

−gg¹¹g⁴⁴)

√−gg¹¹g⁴⁴ A²₁^′− g⁰⁰[ω²A²₂+ 2iωµA¹₁+µ²A²₁] +g³³q²A²₁

g⁴⁴ , (4.42)

0 = A³₁^′′+∂u(√

−gg¹¹g⁴⁴)

√−gg¹¹g⁴⁴ A³₁^′− g⁰⁰ω²+g³³q²

g⁴⁴ A³₁. (4.43)

Decoupling transformation These three differential equations for flavor components of the gauge field in x₁-direction are coupled in the first two flavor directions while the third equation for the componentA³₁ decouples from all others. We decouple the first two equations as well by a field transformation

X1 =A¹₁+iA²₁, Y1 =A¹₁−iA²₁. (4.44) After this transformation the equations of motion for the three new fieldsX₁, Y₁, A³₁are given by

0 = X₁^′′+ ∂_u(√

−gg¹¹g⁴⁴)

√−gg¹¹g⁴⁴ X₁^′ −g⁰⁰(µ−ω)²+g³³q² g⁴⁴ X₁, 0 = Y1′′+ ∂u(√

−gg¹¹g⁴⁴)

√−gg¹¹g⁴⁴ Y1′− g⁰⁰(µ+ω)²+g³³q² g⁴⁴ Y1, 0 = A³₁^′′+∂u(√

−gg¹¹g⁴⁴)

√−gg¹¹g⁴⁴ A³₁^′− g⁰⁰ω²+g³³q²

g⁴⁴ A³₁. (4.45)

80 Chapter 4. Holographic thermo- and hydrodynamics

We are working in the background given by the metric (3.12) with the inverse components and determinant

g⁰⁰ = − u

b²R²f(u), g¹¹=g²² =g³³ = u

b²R² , g⁴⁴=g^uu= 4u²f(u) R² ,

√−g = b⁴R⁵

2u³ , b=πT , (4.46)

so that the coefficients can be evaluated to

∂u(√

−gg¹¹g⁴⁴)

√−gg¹¹g⁴⁴ = f^′(u)

f(u) , −g⁰⁰(µ∓ω)²+g³³q²

g⁴⁴ = (m∓w)²−q²f(u)

uf(u)² , (4.47) where we used the dimensionless frequency, momentum and chemical potential

w=ω/(2πT), q=q/(2πT), m=µ/(2πT), (4.48) respectively, which have already been introduced at the beginning of section 3.1.2. These coefficients (4.47) are singular at the horizonu= 1and at the boundaryu= 0just like in the example given in section 3.1.2. Therefore we apply exactly the same steps in order to gain the indices at the horizon

β =∓i

2(w∓m), (4.49)

where the upper sign inside the bracket belongs to the index for the fieldX1and the lower one gives the index forY1. The indices at the boundary for both fields are given by

α1 = 0, α2 = 1. (4.50)

Now the question which index produces the solution that satisfies the incoming wave condi-tion (which tells us to choose only those solucondi-tions which propagate into the black hole horizon, see section 3.1.2 for a detailed discussion) is a bit more subtle than in the previous example.

Let us assume for definiteness that bothm, w ≥ 0. So in the rest of this thesis we assume that the chemical potentialµormis real and writingwin order relations we mean only the real part ofw. In this case there is only one index choice for the fieldY1 sincew+m ≥ 0 and we know that the negative indexβ=−i/2(w+m)corresponds to the incoming wave. In contrast to this we have to distinguish four cases for the index ofX1

β=











−₂ⁱ(w−m)forw≥m:incoming

−₂ⁱ(w−m)forw<m:outgoing +₂ⁱ(w−m)forw≥m:outgoing +₂ⁱ(w−m)forw<m:incoming

, (4.51)

so fixing mwe choose the incoming solution by choosing the first index if the frequency w is greater or equal to the chemical potential m, and we choose the last index if wis smaller.

Let us carry on considering X1 first. We also need to modify the hydrodynamic expansion Ansatz (3.24). Recall that our approach is to split the singular from the regular behavior in the solution according to

X1 = (1−u)^βF(u), (4.52)

4.2. Analytical Hydrodymamics at finite isospin potential 81

where F is a regular function of u. Our first choice is that the chemical potential is of the same order as the frequency w∼ mand therefore the small quantities to expand the solution in are (w−m)and q². In other words we expand in the spatial momentumq² around zero while we expand in the frequencywaround the fixed value of the chemical potentialm.

X1(u) = (1−u)^β F0+ (w−m)F1+q²G1+. . .

, (4.53)

X1′(u) = −β(1−u)^β−1 F0 + (w−m)F1+q²G1+. . . +(1−u)^β F0′+ (w−m)F1′+q²G1′+. . .

, (4.54)

X₁^′′(u) = β(β−1)(1−u)^β−2 F₀+ (w−m)F₁+q²G₁+. . .

−2β(1−u)^β−1 F₀^′+ (w−m)F₁^′+q²G₁^′ +. . . +(1−u)^β F₀^′′+ (w−m)F₁^′′+q²G₁^′′+. . .

. (4.55)

For definiteness let us consider only the caseβ =−i(w−m)/2wherew≥m. Plugging this expansion into the equation of motion (4.45) and seperating ordersO(1), O(q²)andO(w−m) from each other gives

O(1) : 0 = F0′′− 2u 1−u²F0′, O(w−m) : 0 = i

2(1−u)²F₀+ i

1−uF₀^′+F₁^′′− iu

(1−u²)(1−u)F₀− 2u

(1−u)²F₁^′, O(q²) : 0 = G1′′− 2u

1−u²G1′− 1

u(1−u²)F0. (4.56)

Alternative hydrodynamic expansion By choosing the hydrodynamic Ansatz (4.53) we assumed from the beginning that the frequency and chemical potential have to be treated at equal footing. We can check this assumption by taking a slightly more general Ansatz

X1(u) = (1−u)^β F0+wF1+mH1+q²G1+. . .

. (4.57)

The key point here is that we still assumew, m, q² to be of the same order but we allow an individual expansion coefficient H₁ for the chemical potential. Using this more general ex-pansion in the equation of motion (4.45) and seperating ordersO(1),O(q²),O(w)andO(m) from each other gives

O(1) : 0 = F0′′− 2u 1−u²F0′, O(w) : 0 = i

2(1−u)²F0+ i

1−uF0′+F1′′− iu

(1−u²)(1−u)F0− 2u

(1−u)²F1′, O(m) : 0 = − i

2(1−u)²F0− i

1−uF0′−H1′′+ iu

(1−u²)(1−u)F0 + 2u

(1−u)²H1′, O(q²) : 0 = G₁^′′− 2u

1−u²G₁^′− 1

u(1−u²)F₀. (4.58)

Here we see that the coefficientsH₁andF₁have to satisfy the same equation of motion. This is already clear from the start if we look at the differential equation (4.45) and the Ansatz so

82 Chapter 4. Holographic thermo- and hydrodynamics

that we note thatmandwalways appear as a sum(w−m), at least at linear order in w, m which we solely consider here. So there is no singlemor w, so both have identical factors in the equation of motion and thus their expansion coefficients have to be identical (provided both satisfy the same boundary conditions)

H1 =F1. (4.59)

We have now learned explicitly that our first Ansatz (4.53) is fully justified.

Solving the hydrodynamic differential equations Our efforts have recast our problem into a set of differential equations (4.45) which are only coupled through the leading order function F0. Choosing F0 to be constant (with respect to the radial coordinate) F0 = C is compatible with all the equations of motion and decouples the system

O(1) : F0 = C , O(w−m) : 0 = iC

2(1−u)² +F1′′− iCu

(1−u²)(1−u) − 2u

(1−u)²F1′, O(q²) : 0 = G1′′− 2u

1−u²G1′− C

u(1−u²). (4.60)

These are effectively first order differential equations with an inhomogeneity and we can solve them with

F0 = C , F1 = iC

2 ln1 +u 2 , G1 = C

24

π² + 12 lnuln(1 +u) + 12Li2(1−u) + 12Li2(−u)

. (4.61) The functionLi2(u)is the double logarithm and the polylogarithm in general is defined as

Lin(u) =

n=∞X

n=1

u^k

kⁿ. (4.62)

Note, that we would not get these solutions (4.61) simply using Mathematica since the bound-ary conditions we have to satisfy here are a bit tricky. Just as described in section 3.1.2 the general solutions forF1 andG1 each come with two integration constants which have to be fixed by requiring that limu→1F1 = 0 andlimu→1F1 = 0. In this horzion limit two terms in each solution become divergent and one has to impose the condition that these cancel each other in order to get a regular solution. See also equation (3.29) and the discussion below it. The constantC can now be determined in terms of the boundary fields, momentum and frequency as described in section 3.1.2 and we get

C = 8X₁^bdy

8−4wln 2 +π²q² . (4.63)

4.2. Analytical Hydrodymamics at finite isospin potential 83

Now using the solutions (4.61) and the expression forC from (4.63) in the hydrodynamic Ansatz (4.53) we get the solution to the transversal field

X1,2 = 8X_1,2^bdy(1−u)^−iw−m2 8 +π²q²−4iln 2(w−m)

1 + (w−m)i

2ln1 +u

2 (4.64)

+q²

24 π²+ 12 lnuln(1 +u) + 12Li₂(1−u) + 12Li₂(−u)

forw≥m, while the derivative of its finite part turns out to be

X1,2′ =i(w−m)X_1,2^bdyforw≥m. (4.65) We have also included the Minkowski index 2 here because writing down the equations of motion for the component X2 we discover that it is identical to the equation for X1. Now recall that we have choosen w ≥ m. Finding the solution for smaller frequencies w < m amounts to redoing the above equation with replacing all the frequency potential brackets by absolute values(w−m)→ |w−m| = (m−w)and keeping all the signs as they are. So we only have to switch the order in the final solution to get the small frequency solution and we can write

X_1,2 = 8X_1,2^bdy(1−u)^−im−w2 8 +π²q²−4iln 2(m−w)

1 + (m−w)i

2ln1 +u

2 (4.66)

+q²

24 π²+ 12 lnuln(1 +u) + 12Li2(1−u) + 12Li2(−u)

forw<m, while the derivative of it’s finite part is given by

X1,2′ =i(m−w)X_1,2^bdyforw<m. (4.67) Similarly we get the solution for the other flavor combination fields Y_1,2 by an analogous computation replacing(w−m)→(w+m)

Y1,2 = 8Y_1,2^bdy(1−u)^−iw+m2 8 +π²q²−4iln 2(w+m)

1 + (w+m)i

2ln1 +u

2 (4.68)

+q²

24 π²+ 12 lnuln(1 +u) + 12Li2(1−u) + 12Li2(−u)

for anyw, and its derivative

Y1,2′ =i(w+m)X_1,2^bdyfor anyw. (4.69) Finally the third flavor direction components are obtained as in [28]

A³_1,2^′ =iwA³_1,2^bdyfor anyw. (4.70) Comparing our solutions with those at vanishing chemical potential µ ≡ 0[28] we learn that turning on a constant chemical potentialmresults in the substitution

w→(w−m) forw≥m

w→(m−w) forw<m . (4.71)

This is due to the fact that the wayAν →µδ0ν+Aνin which we introducemmakesµformally identical to a time derivative. The easiest way to understand this fact is to note the form of the covariant derivative appearing in the Lagrangian in time directionD0 =∂0+A0 =∂0+µ.

84 Chapter 4. Holographic thermo- and hydrodynamics

Im Dokument Holographic quark gluon plasma with flavor (Seite 81-90)