Special Relativity and Classical Field Theory

RésuméShort Summary

Prof. Dr. Haye HinrichsenFaculty for Physics and Astronomy

University of WürzburgSummer term 2021

Version: 16. April 2021

Section 1.1

Groups, Vector Space, and Linear Algebra

Groups

A group G is a set of elements g ∈ G (typically transformations)with the properties

I Associativity: ∀a,b,c∈G :

(a ◦ b) ◦ c = a ◦ (b ◦ c)

I Identity element: There exists an element e ∈ G such that

e ◦ a = a ◦ e = a.

I Inverse element: ∀a ∈ G , there exists an a−1 such that

a ◦ a−1 = a−1 ◦ a = e.

A group is called commutative or Abelian ⇔ a ◦ b = b ◦ a

Linear vector spaces

A linear vector space V over R is a set of elements v ∈ Vequipped with the operations ’+’ and ’·’:

I V is a commutative group under addition.I Identity element of scalar multiplication: 1v = v.

Compatibility laws:I Compatibility with scalar multiplication: a(bv) = (ab)v.I Distributivity with respect to vectors: a(u + v) = au + av.I Distributivity with respect to scalars: (a + b)v = av + bv.

Special element: The neutral element for ’+’, called zero vector ~0.

So far we have no information about the length of a vector.

Representation of vectorsLinear dependence:

{v1, . . . , vn} linear dependent ⇔ ∃{ai} :∑n

i=1 aivi = 0.

Basis:

{e1, . . . , ed} ⊂ V is called basis if every v ∈ V can berepresented as a linear combination

v =d∑

i=1v iei

I d is the dimension.I v1, v2, . . . , vd ∈ R are the components (with upper indices)

which depend on the basis.I Einstein sum convention: v = v iei .

Covector spaceLine vectors versus column vectors

A linear form α is a linear map α : V → R.

I Linearity under vector addition: α(u + v) = α(u) + α(v)I Linearity under scalar multiplication: α(au) = aα(u)

In order to specify a linear form we only have to know how it mapsthe basis vectors:

α(u) = α(uiei ) = ui α(ei )︸ ︷︷ ︸=:αi

= αiui

α1, . . . , αd are the components of the linear form (lower indices).

Covector space – Dual basisThe set of all possible linear forms constitute a vector space V ∗,called dual space or covector space.

I Addition of linear forms

α + β : [α + β](u) := α(u) + β(u)

I Scalar multiplication of linear forms

aα : [aα](u) := aα(u)

For each basis {e1, . . . , ed} of V there exists a corresponding dualbasis {e1, . . . , ed} of V ∗ defined by the ’holy relation’

ei (ej) = δij ={1 if i = j0 if i 6= j

Covector spaceRepresentation of linear forms by components

ei (ej) = δij ={1 if i = j0 if i 6= j

Using this associated co-basis of V ∗, we can represent α by

α = αiei ,

where αi are the components of the linear form.

I Vector components with upper indices are calledcontravariant.

I Components of linear forms with lower indices are calledcovariant.

Basis and co-basis: An example

Recall

SUMMARY

I Vectors u ∈ V have contravariant components ui withupper indices.

I Linear forms α ∈ V ∗ have covariant components αi withlower indices.

I Einstein sum convention: Always add over pairs ofoppositely oriented indices.

I Contraction: α(u) = αiui

Basis transformationsTwo basis systems {e1, . . . , ed} and {ẽ1, . . . , ẽd} in the same vector space V

Since each basis vector can be represented in the respective otherbasis, we can write

ei = ẽjAj i ẽj = ekBkj .

where B is the inverse of A.

I An abstract vector v ∈ V remains invariant, but itscomponents transform as

ṽ j = Aj ivi vk = Bkj ṽ j .

I An abstract linear form α ∈ V ∗ remains invariant, but itscomponents transform as

αj = α̃iAi j α̃l = αkBkl

TensorsLinear machines with many connections

A tensor T of rank (p, q) is a multilinear map that maps p linearforms α(1), . . . , α(p) and q vectors v(1), . . . , v(q) onto a number:

α(1), . . . , α(p), v(1), . . . , v(q) 7→ T(α(1), . . . , α(p), v(1), . . . , v(q))

Special cases:I Tensors of rank (0,0) are scalars.I Tensors of rank (1,0) are vectors.I Tensors of rank (0,1) are linear forms.I Tensors of rank (1,1) are linear maps.

Tensor representation and transformation

Representation in a basis:

T(α(1), . . . , α(p), v(1), . . . , v(q)) = Ti1,...,ip

j1,...,jqα(1)i1 · · ·α

(p)ip v

j1(1) · · · v

jq(q)

whereT i1,...,ipj1,...,jq = T(e

i1 , . . . , eip , ej1 , . . . , ejq )

Behavior under basis transformations:

T̃ k1,...,kpl1,...,lq = Ak1

i1 · · ·Akp

ipTi1,...,ip

j1,...,jqBj1

l1 · · ·Bjq

lq

Folklore: A tensor is something which transforms in each upperindex like a vector and in each lower index like a linear form.

Scalar productAn additional structure that is not part of the vector space axioms.

A scalar product is a map g : V ⊗ V → R with the properties:

I Bilinearity (= linear in both arguments)I Symmetry: g(u, v) = g(v,u).I Positive definite: g(v, v) ≥ 0 and g(v, v) = 0⇔ v = 0

Representation of the scalar product:

gij := g(ei , ej). g(u, v) = gijuiv j .

The matrix gij isI realI symmetricI positive eigenvalues.

Scalar product – PropertiesTransformation behavior: g is a (0,2)-tensor, i.e., it is invariantunder basis transformations, while its representation changes as

g̃kl = gijB i kBjl ,

Induced norm:||u|| =

√g(u,u)

Induced distance measure:

d(u, v) = ||u− v||

Example: Euclidean metric:

gij = g(ei , ej) = δij ⇒ g(u, v) = gijuiv j =d∑

i=1uiv i

PseudometricA scalar product that is NOT positive definite

I g(v, v) can be positive or negative.I g(v, v) can be zero for v 6= ~0.

Simplest example in 2d:

g(u, v) = u1v1−u2v2

gij =(1−1

)

The set of vectors v for which g(v, v) = 0 is called light cone.

The sorted signs of the eigenvalues of g is called the signature

Mapping vectors to linear forms and vice versaMusical isomorphism

Flat: The scalar product can turn vectors into linear forms:

[ : u ∈ V 7→ α ∈ V ∗ : α(v) := g(v,u)

In components this means to lower the index: ui := αi = gij uj

Induced scalar product on V ∗

g∗(α, β) := g(u, v) g ij := g∗(ei , ej) ⇒ g ijgjl = δil

Sharp: The induced scalar product turns forms back into vectors:

] : α ∈ V ∗ 7→ u ∈ V : u(β) = β(u) := g∗(α, β)

In components this means to raise the index:uj = g jk uk = g jk αk

Why upper and lower indices?

I It helps us to distinguish vectors and linear forms.I We can switch freely from contravariant to covariant indices

and vice versa by raising or lowering the index:

uj = g jk uk , ui = gij uj

I The Einstein sum convention allows us to write the scalarproduct as

g(u, v) = uivi = uiv i .

There is no need to use upper and lower indices.It is just a matter of convenience.

Summary

Object FormulaBasis of V {e1, e2, . . .}Dual basis of V ∗ {e1, e2, . . .} ei (ej ) = δijVector representation v = v ieiLinear form representation α = αjej

Metric tensor gij = g(ei , ej ) gij g jk = δkiInverse metric tensor g jk = g∗(ej , ek )Lowering indices ui = gij uj Tij = gikgjl T kl

Raising indices uj = g jkuk T ij = g ikg jl TklScalar product 〈u, v〉 = g(u, v) = gij ui v j = g ij ui vj = ui vj = ui v j

Basis transformation ẽj = ekBk j , ei = ẽj Aj

i , Bk

j Aj

i = δki

Transformation of vectors ṽ j = Aj i vi , vk = Bk j ṽ j

Transformation of forms α̃l = αkBk l , αk = α̃i Ai

j

Transformation of metric g̃kl = gij Bi kBj

l , g̃kl = Ak i Al j g ij

WARNING

What you should never do:

Never raise or lower the index of a basis vector of form:

�����

��XXXXXXXei = g ijej

The reason is that ej ∈ V while ei ∈ V ∗.

This does not make sense.

It is like comparing apples with bananas.

Section 1.2

Symmetries of Metric Spaces

Coordinate SystemsNot to be confused with vector components!

Definition:

I A coordinate system is a set of d functions

x i : V → R : v 7→ x i (v)

labeled by i = 1, . . . , d , where d is the dimension.

I The coordinates describe the vector uniquely, that is, the setof coordinate functions has to be invertible.

I The coordinate functions have to be smooth and differentiablebut they can be nonlinear.

Examples of Coordinate Systems

Grid lines are curves where one of the coordinates is constant.

In Cartesian curvilinear coordinate systems the coordinates areidentical with the components of the position vector:

v(x1, . . . , xd ) = x iei ⇒ ei :=∂v∂x i

TransformationsActive and passive transformations

I Active transformations move the object physically:

I Passive transformations move the coordinate system:

I In both cases the coordinate changes as x 7→ x̃ = x + aIn this sense active and passive transformations are equivalent.

Transformations of functions

Active transformation Passive transformation

In both cases:f (x) 7→ f̃ (x) = f (x − a)

Applying an active transformation to a functionmeans to apply the inverse transformation to its

argument.

Coordinate transformationsNot to be confused with basis transformations

A coordinate transformation is a set of d functions

x̃ i : {x1, . . . , xd} 7→ x̃ i (x1, . . . , xd )

with i = 1, . . . , d . These maps may be nonlinear.

Example:x : {r , φ} 7→ x(r , φ) = r cosφ

y : {r , φ} 7→ y(r , φ) = r sinφ

Special case: Linear coordinate transformations:

x̃ i = Λi jx j + s i ⇒ Λi j =∂x̃ i∂x j

What is an Algebra?

An algebra consists of

I a set of symbols, say A,B,C ,I words formed from symbols, like ABABCA,I a space of linear combinations of words, such as 2AB + 7BA,I certain algebraic rules (commutation relations), restricting the

size of the word space.

Example: The Heisenberg algebra:

Symbols: 1, x i , and ∂/∂x j

Algebraic relations: ∂∂x j x

i = δij

The algebraic relations should hold in any coordinate system.

Transformation behavior of partial derivativesStarting point: The Heisenberg algebra holds in both coordinatesystems:

∂

∂xk xi = δik ,

∂

∂x̃k x̃i = δik .

Conclusion: This requires derivatives to transform inverselycompared to the coordinates.

Notation:

∂i :=∂

∂x i ∂i := ∂

∂xiThen we can write:

∂ix j = ∂̃i x̃ j = δji ∂i = g ij∂j

The partial derivative with respect to a component with upperindex transforms like a component with lower index, and vice versa.

Lie groups

Lie groups are continuous groups with infinitely many groupelements, parameterized by continuous parameters, which can beTaylor-expanded around the neutral element.

Example: Group of translations in Rn :

Ta : x 7→ x + a

Example: Special orthogonal group SO(2)(x̃1x̃2

)=(

cosφ − sinφsinφ cosφ

)(x1x2

)

Example: Lie group of translations in 1dConsider an infinitesimaltranslation of a function bythe distance � to the right:

f (x) 7→ f̃ (x) = f (x − �)

This may be Taylor-expanded:

f̃ (x) ≈(1− � ddx

)f (x)

Consider a finite translation:

f̃ (x) = f (x − a) = limN→∞

(1− aN

ddx)N

f (x) = exp(−a ddx

)f (x)

or in higher dimensions:

f̃ (x) = e−a·∇f (x) = e−ai∂i f (x).

Examples: SO(2) and SO(3)

I SO(2) has only one generatorwith the Lie algebra L2 = −1:

L =(0 −11 0

)⇒ Rφ = exp

(φL)

=(

cosφ − sinφsinφ cosφ

)

I SO(3) has three generators with Lie algebra [Li , Lj ] = �ijkLk :

L1 =

0 0 00 0 −10 1 0

, L2 = 0 0 10 0 0−1 0 0

, L3 =0 −1 01 0 00 0 0

R~φ = exp

(φiLi

)

Isometries

I An isometry is a coordinate transformation that, wheninterpreted as an active transformation, preserves angles andlengths, i.e., a metric-preserving map.

I In practice, an isometry leaves the components of the metrictensor invariant:

gkl = g̃kl , gkl = g̃kl .

I Proof: see Lecture Notes.

Section 2.1

Lorentz Transformations

From Galilei to Lorentz InvarianceThe Galilei transformation reads:

x→ x̃ = x− vtt → t̃ = t

In matrix form: t̃x̃ỹz̃

=

1 0 0 0−vx 1 0 0−vy 0 1 0−vz 0 0 1

txyz

or (

t̃x̃

)=(

1 0−v 1

)(tx

)

I The transformation is linear.I The time t = t̃ is not modified.

Broken Galilei invariance in Electrodynamics

Maxwell equations → Wave equation in vacuum:

�B = 0 , �E = 0

Here ’quabla’ is the d’Alembert operator:

� := ∇2 − 1c2∂2

∂t2 with c =1

√µ0ε0

≈ 3× 108 m/s

� is not invariant under Galilei transformations

�̃ = � +2∂t(v · ∇)− (v · ∇)2︸ ︷︷ ︸symmetry-breaking terms

Broken Galilei invariance in Electrodynamics

The broken Galilei invariance comes as no surprise:I Velocities u = ẋ transform as

u 7→ ũ = u− v

where v is the relative velocity between the inertial systems.I Electrodynamics singles out a particular velocity, namely c.I Galilei invariance forbids to single out any velocity

(since c 7→ c + v)

Obvious solution:I There must be a medium, called ether.I The wave equation is only valid in the rest frame of the ether.

But the Michelson-Morley shows that there is no ether.

Hendrik Antoon LorentzIf electrodynamics is not invariant under Galilei, is it invariant under something else?

I Galilei transformation:(t̃x̃

)=(

1 0−v 1

)(tx

)

I Try instead a general linear transformation:(t̃x̃

)=(A BC D

)(tx

)

I Find A,B,C ,D in such a way that � = �̃.

Solution (see lecture notes):

⇒ A = D , C = Bc2 , AD − BC = 1.

The Lorentz transformationThe origin x̃ = 0 of S̃, when seen from S, moves with velocity

v = xt = −CD .

So we have four equations with four unknowns A,B,C ,D:

A = D = 1√1− v2/c2

, B = − vc2√1− v2/c2

, C = − v√1− v2/c2

Define the dimensionless Lorentz factor

γ := 1√1− β2

, β := vc

Final form of the Lorentz transformation

⇒(t̃x̃

)= γ

(1 −v/c2−v 1

)(tx

)

The Lorentz transformation

Use ct instead of t:(ct̃x̃

)=(

γ −βγ−βγ γ

)(ctx

);

(ctx

)=(

γ +βγ+βγ γ

)(ct̃x̃

)

4-vectors: x =

ctxyz

, A point in R1+3 is called event.

Remark:The Lorentz transformation can also be derived in the traditionalway by analyzing a sequence of events (see lecture notes).

The Lorentz transformation as a Lie groupConsider infinitesimal Lorentz transformation with β � 1.Expand Lorentz transformation to linear order in β:(

ctx

)≈(

1 +β+β 1

)(ct̃x̃

)=[1 + β

(1

1

)︸ ︷︷ ︸

=:λ

](ct̃x̃

)

⇒ x = (1 + βλ) x̃ + O(β2)

Make transformation finite using the exponential function:

x = exp(θλ) x̃

This leads to (exercise)(ctx

)=(

cosh θ sinh θsinh θ cosh θ

)︸ ︷︷ ︸

=Λ(θ)

(ct̃x̃

)

RapidityThe ’angle’ of the Lorentz transformation

Compare the two representation:(ctx

)=(

γ +βγ+βγ γ

)(ct̃x̃

)(ctx

)=(

cosh θ sinh θsinh θ cosh θ

)(ct̃x̃

)One can show that the rapidity θ is given by

θ = tanh−1(

vc

).

Additivity of the Rapidityand the Relativistic Addition Theorem of Velocities

Subsequent Lorentz transformations are additive in the rapidity(just in the same way as 2d rotations are additive in the angle)

Λ(θ) = Λ(θ1)Λ(θ2) ⇒ θ = θ1 + θ2

Proof: see lecture notes.

Because of θ = tanh−1( v

c)we can rewrite this as

vc = tanh

[tanh−1

( v1c)

+ tanh−1( v2

c)]

This boils down to the relativistic addition theorem for velocities:

β = β1 + β21 + β1β2⇒ v = v1 + v21 + v1v2c2

.

Lorentz boosts in 3+1 dimensions

A Lorentz boost does not affect the degrees of freedomperpendicular to the boost direction.

For example, a Lorentz boost in x direction does not modify y andz components:

ctxyz

=

cosh θ sinh θsinh θ cosh θ

11

︸ ︷︷ ︸

=Λx (θ)

ct̃x̃ỹz̃

Section 2.2

Minkowski Spacetime

Minkowski Space R1+3

Position vectors have components with upper indicesctxyz

=x0x1x2x3

using Greek indices running from 0 to 3:

x = xµeµ = cte0 + xe1 + ye2 + ze3.

Covectors have components with lower indices, for example thegradient:

∇ = eν∂ν(∂0, ∂1, ∂2, ∂3

)=(

1c∂∂t ,

∂∂x ,

∂∂y ,

∂∂z

)

Scalar product: The Minkowski metricThe metric for zero gravity

The Minkowski metric is so special that we use ηµν instead of gµν :

ηµν =

−1

+1+1

+1

This metric is self-inverse:

ηµν = ηµν

Raising and lowering indices in Special Relativityamounts simply to changing the sign in the zeroth

component.

Light cone

I xµxµ > 0: Spacelike distance: The events cannot be causally connected.I xµxµ = 0: Lightlike distance: The events can be connected by a light ray.I xµxµ < 0: Timelike distance: The events can be connected causally.

Wave equation in Minkowski spaceWave equation

�Φ = 0 ⇒ ∂µ∂µΦ = (∇2 − 1c2∂2t )Φ = 0

Plane-wave ansatz:

Φ(x) = eik·x = eikµxµ where k =(ω/c~k

)

Inserting the ansatz into the wave equation:

�Φ = −kµkµ︸ ︷︷ ︸=0

Φ

Dispersion relation of light:

kµkµ = 0 ⇒ ω = ±kc

Adding mass: The Klein-Gordon equation

Klein-Gordon equation:(�−M2

)ψ = 0

Insert plane wave:Ψ(x) = eik·x

⇒ The wave vector k has to lie on the mass shell:

kµkµ = −M2 ,

Dispersion relation:ω

c = ±√M2 − k2.

Mass shell

Isometries of the Minkowski spaceThe isometries are transformations between intertial systems

Recall: An isometry is a coordinate transformation x 7→ x̃ (activeor passive) which leaves the metric tensor invariant:

gµν = g̃µν , gµν = g̃µν

In general the transformation is nonlinear, the metric tensor isposition-dependent (like in polar coordinates), and thetransformation matrix reads:

Λρµ(x) =∂x̃ρ∂xµ = ∂µx̃

ρ

The isometry condition then reads:

g̃µν(x) = gρτ (x) Λρµ(x) Λτν(x) , g̃µν(x) = Λµρ(x) Λντ (x) gρτ (x)

In Special Relativity (zero gravity), where gµν = ηµν , theisometries are elements of the Poincaré group.

Zero gravity isometries: The Poincaré group and itssubgroups

Exercise: The Lie group SO(3)

Consider the isometry condition for the Euclidean metric gij = δij :

1 = Λ1ΛT ⇒ ΛΛT = 1.

I These are the orthogonal transformations O(3).I Adding the condition det(Λ) = 1 we get the SO(3)

Consider infinitesimal SO(3) transformations:

Λ = 1 + �λ ⇒ λ+ λT = 0

I The generators λ have to be antisymmetric.

Exercise: The Lie group SO(3)

Any antisymmetric 3×3 matrix can be written as a linearcombination of the three generators:

λ(12) =

0 1 0−1 0 00 0 0

, λ(13) = 0 0 10 0 0−1 0 0

, λ(23) =0 0 00 0 10 −1 0

For example the generator λ(12) generates a rotation in thex1-x2-plane around the z-axis :

exp(ϕλ(12)

)=

cosϕ sinϕ− sinϕ cosϕ1

= R(12)(ϕ)Similarly λ(13) and λ(23) generate rotations around the x , y axes.

SO(3) Angular momentum algebra

λ(12) =

0 1 0−1 0 00 0 0

, λ(13) = 0 0 10 0 0−1 0 0

, λ(23) =0 0 00 0 10 −1 0

In quantum physics, generators are renamed and multiplied by −i :

L1 =

0 0 00 0 −i0 i 0

, L2 = 0 0 i0 0 0−i 0 0

, L3 = 0 −i 0i 0 0

0 0 0

These three generators form the angular momentum operator ~Lobeying the angular momentum algebra

[Li , Lj ] = i�ijkLk .

Finite rotations:R(~φ) = ei~L·~φ.

~φ is the vector defining the rotational axis.

The SO+(3, 1) and its Lie algebracombining rotations and Lorentz boosts

Definition of the generators:

[λ(αβ)]µν = δµαηβν − δµβηαν

Commutation relations:

[λ(αβ), λ(γδ)] = −(ηαγλ(βδ) − ηαδλ(βγ) + ηβδλ(αγ) − ηβγλ(αδ)

)Matrix representation:

λ(10) =

11 λ(20) = 1

1

λ(30) = 1

1

λ(12) =

1−1

λ(13) = 1

−1

λ(23) =

1−1

The SO+(3, 1) and its Lie algebra

I Rotations:

exp(φλ(12)

)=

1

cosφ sinφ− sinφ cosφ

1

I Lorentz boosts:

exp(θλ(01)

)=

cosh θ sinh θsinh θ cosh θ

11

Rotationsmove things on circles

Lorentz boostsmove things on hyperbolas

Lorentz boosts

I Straight lines remain straight.I The light cone is mapped onto itself.

Redefining the presenceThere is no notion of simultaneity any more

I The present cannotchange my past.

I I cannot change thepresent in the future.

Section 3.1

Relativistic Mechanics – Equations of Motion

Mass points and world lines

In Relativity, we have the same notion of pointlike particles as inNewtonian Mechanics, but

I In Minkowski space, even if the particle is at rest, it proceedsin time, hence the mass point moves on a line, the so-calledworld line.

I The effective mass of the particle depends on its velocity anddiverges as the particle velocity approaches the speed of light.

I Therefore, m always denotes the rest mass of the particle.

Worldlines are timelike

Events along a worldline have tobe causally connected.

⇒ Tangent vector has to be timelike.

⇒ Worldlines are confined to lightcone.

Worldline parameterizationA curve can be parameterized freely (gauge)

From the mathematical point of view, a worldline is aparameterized curve of the form

C : [a, b]→ R3+1 : λ 7→ x(λ) =

x0(λ)x1(λ)x2(λ)x3(λ)

constrained by the condition that the tangent vector is timelike:

dxµ(λ)dλ

dxµ(λ)dλ < 0

ReparameterizationChanging the gauge

Two parameterizations C(λ) and C̃(τ) describe the same curveif there is a function f : R→ R such that

τ = f (λ) , C(λ) = C̃(f (λ)).

The corresponding tangent vectors differ by a factor of df / dλ:

dxµ(λ)dλ =

dx̃µ(τ)dτ

df (λ)dλ .

⇒ The tangent vectors differ in their squared lengths:( dxµ(λ)dλ

dxµ(λ)dλ

)=( dx̃µ(τ)

dτdx̃µ(τ)

dτ)( df (λ)

dλ)2.

Eigenzeit (proper time) parameterizationThe eigenzeit is the time the particle experiences by itself

I We can always choose a special parameterization such that

dxµ(τ)dτ

dxµ(τ)dτ = −c

2.

τ is called proper time (eigenzeit).I The proper time τ of an object is defined as the time

displayed by a clock attached to the object.I τ is a scalar since it is invariant under basis transformations.

I Remark: Choosing the eigenzeit parameterization is like“fixing a gauge”.

Relativistic 4-velocityEven when sleeping, we move in time...

The relativistic velocity or 4-velocity of a mass point is defined by

u := ẋ = dxdτ

• τ is a scalar, x is a 4-vector, so ẋ has to be a 4-vector as well.

The squared norm of the velocity always equals the speed of light:

u2 = uµuµ = −c2

If we apply a Lorentz transformation we get the components

uµ =(γcγ~v

)= 1√

1− v2c2

cvxvyvz

Relativistic 4-momentumequals rest mass times relativistic velocity

The relativistic 4-momentum of a free particle is defined by

p := mu = mẋ

Since m is Lorentz-invariant, p transforms like a 4-vector.

I Components:

pµ =(E/c~p

)where E = γmc2 , ~p = γm~v

I Squared norm:

pµpµ = m2uµuµ = −m2c2

E = mc2

The squared-norm condition

pµpµ = −m2c2

translates into the energy-momentum relation

E 2 = m2c4 + ~p2c2.

Resting particle:E = ±mc2

Taylor expansion:

E = ±(mc2 + p

2

2m −p4

8m3c2 +O(p6/c4)

)

4-accelerationVery simple definition:

a := u̇ = ẍ = d2x

dτ2

Components (see exercise)

a =(

γγ̇cγ2~a + γγ̇~v

).

Remarks:I In an instantaneously co-moving inertial frame we have

a =(0~a

)

I The 4-acceleration is the curvature vector of a worldline.

Principle of Least ActionRecall lecture on Classical Mechanics

Action of a particle in a potential on trajectory q(t):

S[q] =∫ t2

t1L(q, q̇) dt where L = T − V = 12mq̇

2 − V (q)

Realized trajectory in Nature ⇔ Minimal action.

0 = δS =∫ t2

t1

(∂L(q, q̇)∂q δq +

∂L(q, q̇)∂q̇ δq̇

)Here δq and δq̇ are not independent, perform partial integration:

0 = δS =∫ t2

t1

(∂L(q, q̇)∂q −

ddt∂L(q, q̇)∂q̇

)δq ⇒ ∂L

∂q−ddt∂L∂q̇ = 0.

Relativistic action of a free point particleThe action is simply the length of the worldline

Heuristic Principle of Simplicity:

Nature always selects the simplest solution compatible withthe required symmetries.

The relativistic action should be invariant under frame changes:

Simplest solution:The action is defined as the relativistic length of the world line:

S = −mc∫

cds

where ds =√−dxµ dxµ .

!! Never fix the gauge before carrying out the variation !!Proposed action of a free particle:

S = −mc∫

cds

We know that ds = c dτ , so the action issimply the elapsed eigenzeit:

S = −mc2∫ τBτA

dτ = −mc2(τB − τA).

This is wrong!

I The ending points (ctA, xA) and (ctB, xB), but the eigenzeitdifference varies for each of the trajectories over which wewant to minimize.

I We won’t get meaningful Lagrange equations of motion.

Solution:

For each of the trajectories, choose a pa-rameterization such that the parameters atthe ending points coincide, say λA and λB:

S = −mc∫

cds = −mc

∫ λBλA

dsdλ dλ

=∫ λBλA−mc

√− dx

µ

dλdxµdλ︸ ︷︷ ︸

=L(x,ẋ)

dλ

This gives us directly the Lagrange function for a free particle

L(x, ẋ) = −mc√−ηµν ẋµẋν

Derivation of the equations of motionThe variation of the action is given by

δS =∫ λBλA

(∂L(x, ẋ)∂xµ δx

µ + ∂L(x, ẋ)∂ẋµ δẋ

µ)

dλ

Principle of least action ⇒ δS = 0.δxµ and δẋµ are not independent ⇒ Partial integration:

0 = δS =[∂L(x, ẋ)

∂ẋµ δxµ]λBλA︸ ︷︷ ︸

=0

+∫ λBλA

(∂L(x, ẋ)∂xµ −

ddλ

∂L(x, ẋ)∂ẋµ

)︸ ︷︷ ︸

=0

δxµ dλ

Variation δxµ is arbitrary, hence we obtain the relativisticEuler-Lagrange equations

∂L(x, ẋ)∂xµ −

ddλ

∂L(x, ẋ)∂ẋµ = 0 .

Canonically conjugated momentumis defined as the derivative of the Lagrange function

Define relativistic canoncially conjugated momentum:

pµ = ∂L∂ẋµ

, pµ =∂L∂ẋµ

This allows us to rewrite the EOM’s as

ṗµ = ∂L∂xµ

Free particle:

pρ = −mc∂√−ηµν ẋµẋν∂ẋρ =

mc ẋρ√−ηµν ẋµẋν

Fixing the gaugeThe momentum does not depend on the parameterization

Consider reparameterization:

λ 7→ λ̃ = f (λ) , ddλ 7→d

dλ̃= 1f ′(λ)

ddλ .

The Lagrange function is not invariant under reparameterization:

L 7→ L̃ = −mc√−ηµν ˜̇xµ ˜̇xν =

−mc√−ηµν ẋµẋν|f ′(λ)| .

But the conjugated momentum is invariant (f ′(λ) cancels out):

pρ = ∂L∂ẋρ

= mc ẋρ√

−ηµν ẋµẋν

At this point we can choose eigenzeit parameterization!

p = mu .

Relativistic 4-forceNatural definition:

f = ddτ p = ṗ = ma .

Components:

f =

γc dEdtγ~F

Something is strange here:

p(τ = 0) =

mc000

, f =a000

⇒ p(τ) =mc + aτ

000

⇒ The object is resting, but gaining weight!

“Non-weight-gaining” forceshave to be “perpendicular” on the world line.

Mass conservation:

||p||2 = pµpµ = −m2c2 ⇒ddt pµp

µ = 2ṗµpµ = 0

This leads to:f · u = fµuµ = 0

Such a mass-conserving force is of the form

f =(γc~F · ~vγ~F

).

Charged particle in a given electromagnetic fieldThe electromagnetic 4-potential

Recall: The electromagnetic fields ~E and ~B can be derived from ascalar potential φ and a vector potential ~A as follows:

~E (~x , t) = −∇φ(~x , t)− ∂~A(~x , t)∂t ,

~B(~x , t) = ∇× ~A(~x , t) .

In Special Relativity, we combine the potentials in a 4-potential A:

Aµ(x) =

1cφ(~x , t)~A(~x , t)

The SI unit is:

[A] = Vsm

Charged particle in a given electromagnetic fieldThe Action of a charged particle in an electromagnetic field

Heuristic principle of simplicity:Add a term to the Lagrange function L that is as simple as possible:

Lfree → Lem = Lfree + e A(x) · ẋ︸ ︷︷ ︸scalar

I The extra term has to be translation-invariant and scalar.I e is a coupling constant.

I L has the dimension of an energy: [L] = J = VAs.I e has the dimension of an electric charge:

[e] = [L][ẋ][A] = As = C

Charged particle in a given electromagnetic fieldThe Action of a charged particle in an electromagnetic field

L(x, ẋ) = −mc√−ηµν ẋµẋν︸ ︷︷ ︸

free particle

+ eAν ẋν︸ ︷︷ ︸electromagnetic

Put this into the equations of motion:

∂L(x, ẋ)∂xµ −

ddτ

∂L(x, ẋ)∂ẋµ = 0

I First term:∂L(x, ẋ)∂xµ = e

∂Aν∂xµ ẋ

ν = e(∂µAν)ẋν

I Second term:ddτ

∂L(x, ẋ)∂ẋµ =

ddτ(mẋµ + eAµ(x(τ))

)= mẍµ + e(∂νAµ)ẋν

Charged particle in a given electromagnetic fieldThe field tensor

⇒ Equations of motion:

mẍµ = e (∂µAν − ∂νAµ) ẋν

We define the electromagnetic tensor:

Fµν = ∂µAν − ∂νAµ

Equations of motion:

mẍµ = e Fµν ẋν

Tensor of the electromagnetic fieldDetermine the matrix elements

1. Take equations of motion

mẍµ = e Fµν ẋν

2. Consider non-relativistic limit v � c, where ddτ ≈ddt :

mẍi = e(Fi0c + Fijv j

)3. Compare this expression with formula for Lorentz force:

m~a = e(~E + ~v × ~B

)4. ⇒ Determine components of the antisymmetric tensor Fµν .

Tensor of the electromagnetic field

Result:

With lower indices:

Fµν =

0 −Ex/c −Ey/c −Ez/c

Ex/c 0 Bz −ByEy/c −Bz 0 BxEz/c By −Bx 0

With upper indices:

Fµν =

0 Ex/c Ey/c Ez/c

−Ex/c 0 Bz −By−Ey/c −Bz 0 Bx−Ez/c By −Bx 0

Remark: You may find different conventions in various textbooks.

Electromagnetic Gauge Invariance

A gauge is an non-physical redundancy in ourdescription of physical reality.

Claim:Electrodynamics is gauge-invariant under gauge transformations

Aµ → õ = Aµ + ∂µf .

where f (x) is an arbitrary scalar function.

Consistency check:The equations of motion mẍµ = e Fµν ẋν are in fact invariant:

Fµν → F̃µν = ∂µAν + ∂µ∂ν f − ∂νAµ − ∂ν∂µf = Fµν

Electromagnetic Gauge Invarianceworks also nicely on the level of the Lagrangian:

Every Lagrangian has a redundancy by adding

L(x, ẋ) → L̃(x, ẋ) = L(x, ẋ)+ dfdλ = L(x, ẋ)+(∂µf )ẋµ

because

S → S̃ =∫ λBλA

(L(x(λ), ẋ(λ)) + dfdλ

)= S + f (λB)− f (λA)︸ ︷︷ ︸

const

.

⇒ δS = δS̃

Compare this with:

L(x, ẋ) = −mc√−ηµν ẋµẋν + eAµẋµ

The redundancy is equivalent to replacing Aµ → Aµ + 1e∂µf .

Section 3.2

Noether Theorem

Noether TheoremRecall nonrelativistic Noether Theorem

I Let q(t) be a solution of the Lagrange equations.I Let qs(t) be a continuous transformation of the solution with

no change at s = 0.

I This transformation is called symmetry transformation if allqs(t) minimize the action (to first order in s).

⇔ ∂∂s L(qs , q̇s)

∣∣∣s=0

= dΛdt

Claim: For a symmetry transformation, the Noether charge

Q(t) = p∂qs(t)∂s

∣∣∣s=0− Λ(t)

is conserved in time.

Relativistic Noether Theorem

I Let x(τ) be a solution of the Lagrange equations.I Let xs(τ) be a continuous transformation of the solution with

no change at s = 0.

I This transformation is called symmetry transformation if allxs(τ) minimize the action (to first order in s).

⇔ ∂∂s L

(xs(τ), ẋs(τ)

)∣∣∣s=0

= dΛ(τ)dτ

Claim: For a symmetry transformation, the Noether charge

Q(τ) = pµ(τ)∂xµs (τ)∂s

∣∣∣s=0− Λ(τ)

is conserved along the worldline of the particle.

Applications of the Noether TheoremTranslations

Move everything by a constant 4-vector f:

x → xs = x + sf ⇒ xµ → xµs = xµ + sf µ

I For a free particle the Lagrangian depends only on ẋ.I The Lagrangian does not depend on s, hence Λ = 0.

Now apply Noether theorem:I

Q = ∂L(xs , ẋs)∂ẋµs

∂xµs∂s

∣∣∣s=0

= pµf µ.

I For a translation in direction f = eν :

Qν = pµ(eν)µ = pν

Applications of the Noether TheoremRotations / Lorentz transformations

Consider a rotation / Lorentz-transformation in the (α, β)-plane:

x → xs = (1 + sλ(αβ))x

where s is the angle / rapidity.I The Lagrangian is scalar, i.e., invariant under SO+(1, 3).I Λ = 0.

xµ → xµs =(δµν + s [λ(αβ)]µν

)xν

ẋµ → ẋµs =(δµν + s [λ(αβ)]µν

)ẋν

⇒ ∂sL =∂L∂xµ

∂xµs∂s +

∂L∂ẋµ

∂ẋµs∂s =

( ∂L∂xµ x

ν + ∂L∂ẋµ ẋ

ν)

[λ(αβ)]µν = 0

⇒ Qαβ = pαxβ − pβxα

Section 4.1

Classical Scalar Field

Concept of classical fields

In Special Relativity, a classical field lives in space-time, e.g.

φ : R3+1 → R : x 7→ φ(x)

The field is the solution of a field equation, e.g., the Klein-Gordon equ.:(�−M2

)φ = 0

Lagrange formalism for Classical Fields

I Principle of Least Action: The solution of the fieldequations minimizes the action S[φ].

I The action is a functional, mapping a function onto R.

I The field lives not only on a line, but in the whole space-time.I The action can be written as an integral over a Lagrange

density:S =

∫d4x L

I The Lagrange density should only depend on the field and itsfour partial derivatives

S[φ] =∫

d4x L(φ(x),∇φ(x)

)

Variational calculus for fields

Consider an infinitesimal variation of the field

φ(x) → φ(x) + χ(x)

with |χ(x)| � |φ(x)| which vanishes on ∂Ω.

S[φ]→ S[φ+ χ] =∫

Ωd4x L

(φ(x) + χ(x), ∇φ(x) +∇χ(x)

)=∫

Ωd4x

[L(φ,∇φ) + ∂L(φ,∇φ)

∂φχ+ ∂L(φ, ∇φ)

∂(∂µφ)∂µχ

]δS = S[φ+ χ]− S[φ] ≈

∫Ω

d4x[∂L(φ,∇φ)

∂φχ+ ∂L(φ,∇φ)

∂(∂µφ)∂µχ

]How can we integrate by parts?

Variational calculus for fields

Recall Gauss divergence theorem in R3:∫V~∇ · ~F dV =

∮∂V

~F · ~n

A similar theorem holds in Minkowski spaceR1+3 for any differentiable 4-vector field F:∫

Ω∂µFµ d4x =

∮∂Ω

Fµ dnµ

Variational calculus for fieldsPartial integration

Gauss divergence theorem in R1+3.∫Ω∂µFµ d4x =

∮∂Ω

Fµ dnµ

Now replaceFµ := ∂L

∂(∂µφ)χ

and apply product rule:

∫Ω

([∂µ

∂L∂(∂µφ)

]χ+ ∂L

∂(∂µφ)[∂µχ

])=∮∂Ω

∂L∂(∂µφ)

χ dnµ︸ ︷︷ ︸zero since χ=0 on ∂Ω

This is like partial integration in the variational integral.

Variational calculus for fieldsField equations

δS ≈∫

Ωd4x

[∂L(φ,∇φ)∂φ

χ+ ∂L(φ,∇φ)∂(∂µφ)

∂µχ]

Partial integration:

δS ≈∫

Ωd4x

[∂L(φ,∇φ)∂φ

− ∂µ∂L(φ,∇φ)∂(∂µφ)

]︸ ︷︷ ︸

=0

χ

Field equations:

⇒ ∂L(φ,∇φ)∂φ

− ∂µ∂L(φ,∇φ)∂(∂µφ)

= 0

Action for the wave equationWhich action S gives us the wave equation

�φ = 0 ?

We have no potential, therefore:

�φ = ∂µ∂µφ ∝ ∂µ∂L(∇φ)∂(∂µφ)

⇒ ∂µφ ∝ ∂L∂(∂µφ)

The simplest action would be:

L(∇φ) = −12(∂νφ)(∂νφ).

Note that

L = 12c2 (∂tφ)2︸ ︷︷ ︸

Ekin

− 12(~∇φ)2︸ ︷︷ ︸Epot

= Ekin − Epot

Action for the Klein Gordon equationNow add another ’potential’ V to the action:

L(φ,∇φ) = −12(∂νφ)(∂νφ)−V (φ).

Explicitely:

L(φ,∇φ) = 12c2 (∂tφ)2︸ ︷︷ ︸

=Ekin

− 12(~∇φ)2−V (φ)︸ ︷︷ ︸

=Epot

,

⇒ Equations of motion:∂L∂φ− ∂µ

∂L∂(∂µφ)

= −V ′(φ) +�φ = 0.

Choose potential of the form V (φ) = 12M2φ2:

L(φ,∇φ) = −12(∂νφ)(∂νφ)− 12M

2φ2.

(�−M2

)φ = 0

Scalar field as a “wave machine”

(�−M2

)φ =

(∇2 − 1c2

∂2

∂t2 −m2c2~2

)φ = 0

Vector fieldsI Vector fields are vector-valued fields, for example Φ(x) ∈ RN .I Such vector fields have N components Φa(x) labelled by Latin

indices a = 1, . . . ,N.I Usually they come with a symmetry among the components.

Example:

I Free O(N) vector model:

L(Φ,∇Φ) = −12

N∑a=1

(∂νΦa)(∂νΦa)−12M

2N∑

a=1

ΦaΦa

I Interacting O(N) vector model with Φ4-coupling:

L(Φ,∇Φ) = −12

N∑a=1

(∂νΦa)(∂νΦa)−12M

2N∑

a=1

ΦaΦa−124λ

N∑a,b=1

ΦaΦaΦbΦb

Noether theorem for fieldsA coordinate transformation can be interpreted as a changing field

x→ x̃ = x− �f(x)⇒ xµ → x̃µ = xµ − �f µ(x)

φ(x) = φ̃(x− �f(x)

)' φ̃(x)− �(∂µφ̃)f µ(x)︸ ︷︷ ︸

=: �∆φ(x)

In short:

φ̃(x) ' φ(x) + �∆φ(x).

Noether theorem for fields

φ̃(x) ' φ(x) + �∆φ(x).

This transformation is called a symmetry transformation if theLagrangian changes only by a gauge transformation

δL ' �∂µΛµ

If we manage to determine Λ, the Noether current is given by

jµ := πµ∆φ − Λµ

where πµ = ∂L∂(∂µφ) is the conjugate field momentum.

This current is conserved via continuity equation

∂µjµ(x) = 0.

Noether theorem for fieldsGeneralization

Noether theorem for several fields ψ1, ψ2, . . . , ψN :

δψi (x) = �∆ψi (x)

jµ(x) =( N∑

i=1πµψi (x)∆ψi (x)

)− Λµ(x) with πµψi (x) =

∂L∂(∂µψi )

Noether theorem for vector fields A(x) with components Aµ(x):

δAµ(x) = �∆ µA (x)

jµ(x) = πµν (x)∆ νA (x)− Λµ(x) with πµν (x) =∂L

∂(∂µAν)

In both cases:∂µjµ(x) = 0

Noether theorem for fieldsTranslation

Move coordinates (space & time) infinitesimallyby a constant vector �a:

x̃ = x + �a

Equivalently, the fields change as

δφ(x) = �∆φ(x) = �(∂µφ(x)

)aµ.

One can show that in this case

Λ = La

The corresponding Noether current is

jµ := πµ∆φ − Λµ = πµ(∂ρφ)aρ − Laµ

and we have as usual the continuity equation

∂µjµ(x) = 0

The energy momentum tensorof a field is the Noether current related to translation

Choose translation in the direction of the basis vector a := eν :

aρ = δρν

Then the Noether current carries two indices:

jµν = πµ(∂ρφ)δρν − Lδµν = πµ∂νφ− δµνL .

We call it the energy-momentum tensor:

T µν = πµ∂νφ− δµνL

which obeys the continuity equation

∂µT µν = 0.

Energy momentum tensor

T µν = πµ∂νφ− δµνL ∂µT µν = 0.

Names used in the literature:I Stress-energy tensorI Stress-energy-momentum tensorI Energy-momentum tensor

Why do we need a tensor to describe energy and momentum?Why not a simple (position-dependent) 4-vector?

Answer:A field is not a single particle, it is a soup of many particles.

Energy momentum tensor - InterpretationExercise: Consider cloud of dust

I Density of particles ρ(~x , t)I Local average velocity ~v(~x , t)I ⇒ Particle current

~j(~x , t) = ρ(~x , t)~v(~x , t)

Now consider test volume V , containing mass M.Apply Gauss divergence theorem:

dMdt =

ddt

∫V

dV ρ(~x) = −∮∂V

d~n ·~j = −∫

VdV ~∇ ·~j(~x).

Compare integrands ⇒ 3D Continuity equation:

ddt ρ(

~x) = −~∇ ·~j(~x).

Energy momentum tensor - InterpretationDefine 4-current for dust:

ddt ρ(

~x) = −~∇ ·~j(~x).

I Define 4-current

jµ(x) =(cρ(~x , t)~j(~x , t)

).

I Rewrite continuity equation

∂µjµ = 0.

Energy momentum tensor - InterpretationInterpretation of the 4-current

jµ(x) =(cρ(~x , t)~j(~x , t)

).

I Zeroth component:The differential mass dM contained in infinitesimal volumedV is

dM = 1c j0 dx dy dz

I First component:The differential mass dM passing the area dy dz in time dt is

dM = j1 dy dz dt

The content of an infinitesimal 3-volume dA dt canbe interpreted as the flux of this content through

the surface dA within the time span dt.

Energy momentum tensor - InterpretationNormal vectors in momentum space

I A timelike vector dn ∝ e0 is normal on dx dy dz .I A spacelike vector dn ∝ e1 is normal on dy dz dt.I A spacelike vector dn ∝ e2 is normal on dx dz dt.I A spacelike vector dn ∝ e3 is normal on dx dy dt.

dM = j · dn = jµ dnµ

Energy momentum tensor - InterpretationThe matrix elements

Section 5.1

Electrodynamics - Geometric Content

ElectrodynamicsMaxwell equations

Name Integral equation Differential equation

Gauß law for ~E∮∂V

~E · d~A = 1�0

∫Vρ dV ~∇ · ~E = 1

�0ρ

Gauß law for ~B∮∂V

~B · d~A = 0 ~∇ · ~B = 0

Faraday’s law∮∂A

~E · d~l − ddt

∫A

~B · d~A = 0 ~∇× ~E + ∂~B∂t = 0

Ampère’s law∮∂A

~B · d~l = µ0∫

A

~J · d~A ~∇× ~B − µ0�0∂~E∂t = µ0

~J

+µ0�0ddt

∫A

~E · d~A

Electrodynamics is nothing but a ring

A ring is attached to each point in space-time.

Electrodynamics is nothing but a ring

The ring is not attached in a single point.Rather each point of space-time is associated with a ring.

Electrodynamics is nothing but a ring

Think of yourself as having a position on the ring.

Electrodynamics is nothing but a ring

For characterizing the position, we have to define coordinates.

Electrodynamics is nothing but a ringThe Circe Group U(1)

I The circle group U(1) is the

group of translations on a circle.

I The U(1) is an Abelian Lie group.

I Representation:• Angular: ϕ→ ϕ̃ = (ϕ+ ∆ϕ)mod2π

• Complex: z → z̃ = ze i∆ϕ

Electrodynamics is U(1)

If space-time was one-dimensional,the U(1) rings would form a tube.

Electrodynamics is U(1)

Living on the tube, you cannot say where you are,but there is a clear notion of “moving straight”.

Electrodynamics is U(1)

We can ’twist’ the tube physically.Electromagnetism is the story of tube twists.

Electrodynamics is U(1)

IMORTANT:If the tube is twisted by force, the trajectories of

“moving straight” would follow the twist.

Electrodynamics is U(1)

In terms of the coordinate ϕ, one cannot decidewhether the tube is physically twisted or whether the

origins φ = 0 are just chosen in a fancy way.

Local gauge transformations

A gauge transformation is nothing butthe change of the local coordinate sys-tems on the U(1) circles:

ϕ(x) 7→ ϕ̃(x) = ϕ(x) + θ(x)

Equivalently with z = eiϕ:

z(x) 7→ z̃(x) = z(x)eiθ(x)

Dreaming of a unified electrodynamics......would be to include ϕ as a fifth dimension!

Consider φ as an additional dimension:

~x =

ctxyzϕ

, gµν =−1 ∗

+1 ∗+1 ∗

+1 ∗∗ ∗ ∗ ∗ ∗

This is known as Kaluza-Klein theory.

I Very cool, but difficult.I Requires knowledge of differential geometry.

Most people consider the electromagnetic field as a separate thing.

How electrodynamics is usually done...Slice 5d space into a 4+1-dimensional fiber bundle

I We have to install a map that relates the fibers.I We pay a price: The map will be gauge-dependent.

How electrodynamics is usually done...

I Choose a gauge (coordinate system) in the U(1) circles.I Choose a position x in space-time.I Walk “straight” in direction eµ and see at which rate ϕ(x)

changes. Call this quantitity Aµ(x) (up to a proportionalityfactor).

⇒ dϕ ∝ A(x) · dx = Aµ(x) dxµ

The proportionality constant turns out to be the electric charge:

dϕ = e~A(x) · dx = e

~Aµ(x) dxµ

The Aµ(x) are the connection coefficients which tell us how to “go straight”. By construction, they depend on the chosen gauge.

Behavior of A(x) under gauge transformationsConnection of two circles at infinitesimal distance dx:

dϕ = e~A(x) · dx

Gauge transformation:ϕ(x) 7→ ϕ̃(x) = ϕ(x) + θ(x)

Transformation behavior of the infinitesimal angle difference:

dϕ(x) 7→ dϕ̃(x) = dϕ(x) +∇θ(x) · dx

Gauge transformation of the connection:e~A(x) · dx 7→ e

~Ã(x) · dx = e

~A(x) · dx +∇θ(x) · dx

⇒ A 7→ Ã = A + ~e∇θ.

Electromagnetism - Gauge transformations

A 7→ Ã = A + ~e∇θ.

In coordinates:

Aµ(x) → õ(x) = Aµ(x) +~e ∂µθ(x)

Usually, the factor ~e is suppressed by setting θ(x) =e~ f (x):

Aµ 7→ Aµ + ∂µf

In a gauge transformation, the electromagnetic 4-vectorpotential A changes by a gradient of a scalar function.

Detection of twists / curvature

Intrinsic curvature:

In two dimesions, curvaturemanifests itself in the factthat the angular sum in atriangle is larger than π.

Idea: Study mismatcheson closed contours

Detection of twists in U(1) tubes

Walking on a closed contour means to walk on a torus.

Detection of twists in U(1) tubesFind mismatch along a closed tube

Generally, a twist leads to a mismatch if we “go straight”all the way along a closed contour.

Compute the twist along a closed contour

I For simplicity set

e/~ = 1.

I Compute the fourcontributions via∫

dϕ =∫Aµ dxµ.

I Add them up.

Quantitative computation of the twist

Consider connection A with components Ax ,Ay .

For simplicity let coordinate origin be in the center.

Compute contribution on A→ B:

∆ϕA→B =∫ +�−�

Ax (x ,−�) dx

Insert Taylor expansion:

A(x , y) = A(0, 0) + x∂xA(0, 0) + y∂yA(0, 0) +O(x2, y2)

⇒ ∆ϕA→B '∫ +�−�

(Ax (0, 0) + x∂xAx (0, 0)− �∂yAx (0, 0)

)dx

⇒ ∆ϕA→B ' 2�Ax (0, 0)− 2�2∂yAx (0, 0) + O(�3)

Quantitative computation of the twist

Add up the four contributions:

∆ϕA→B = 2�Ax (0, 0)− 2�2∂yAx (0, 0) + O(�3)∆ϕB→C = 2�Ay (0, 0) + 2�2∂xAy (0, 0) + O(�3)∆ϕC→D = −2�Ax (0, 0)− 2�2∂yAx (0, 0) + O(�3)∆ϕD→A = −2�Ay (0, 0) + 2�2∂xAy (0, 0) + O(�3) .

Altogether the resulting change is

∆ϕ = ∆ϕA→B + ∆ϕB→C + ∆ϕC→D + ∆ϕD→A= 4�2

(∂xAy (0, 0)− ∂yAx (0, 0)

)+ O(�3) .

The electromagnetic field tensor

∆ϕ = 4�2(∂xAy (0, 0)− ∂yAx (0, 0)

)+ O(�3) .

⇒ Fxy := lim�→0

∆ϕ4�2 = ∂xAy − ∂yAx .

⇒ Fµν := lim�→0

∆ϕ4�2 = ∂µAν − ∂νAµ .

Electromagnetic field tensor in 3+1 dimensions

Fµν := ∂µAν − ∂νAµ .

Fµν =

0 −Ex/c −Ey/c −Ez/c

Ex/c 0 Bz −ByEy/c −Bz 0 BxEz/c By −Bx 0

Electrodynamics in a 2+1-dimensional world

Fµν =

0 −Ex/c −Ey/cEx/c 0 BEy/c −B 0

The homogeneous Maxwell equationsare purely geometric

Homogeneous Maxwell equations

I Each face of the cube contributeswith 4�2Fµν , evaluted at thecenter of the respective face.

I Consider e.g. the front face (0,1):

4�2F01(x + �e2) ' 4�2(1 + �∂2)F01(x).

front: F01(x + �e2) ' (1 + �∂2)F01(x)rear: − F01(x− �e2) ' (−1 + �∂2)F01(x)top: F12(x + �e0) ' (1 + �∂0)F12(x)

bottom: − F12(x− �e0) ' (−1 + �∂0)F12(x)right: − F02(x + �e1) ' (−1− �∂1)F02(x)left: F02(x− �e1) ' (1− �∂1)F02(x)

⇒ 2�(∂0F12(x)− ∂1F02(x) + ∂2F01(x)

)= 0

Homogeneous Maxwell equationsin 1+3 dimenions

Orientation of the cube Differential equation

txy ∂0F12(x)− ∂1F02(x) + ∂2F01(x) = 0

txz ∂0F13(x)− ∂1F03(x) + ∂3F01(x) = 0

tyz ∂0F23(x)− ∂2F03(x) + ∂3F02(x) = 0

xyz ∂1F23(x)− ∂2F13(x) + ∂3F12(x) = 0

Homogeneous Maxwell equations

Shortcut: Cyclic bracket notations:I Round parenthesis where all permutations are added

symmetrically with a positive sign:

T(µν...) =1n!∑σ∈Sn

Tσ{µν...}

I Square brackets where all permuations are addedantisymmetrically with the sign of the permutation:

T[µν...] =1n!∑σ∈Sn

sign(σ)Tσ{µν...}

Shortest form of the homogeneous Maxwell equations:

∂[µFνρ] = 0.

There are three spatio-temporal ones:

∂[0F12] = 0 ⇒ ∂tBz + ∂xEy − ∂yEx = 0∂[0F13] = 0 ⇒ −∂tBy + ∂xEz − ∂zEx = 0∂[0F23] = 0 ⇒ ∂tBx + ∂yEz − ∂zEy = 0

⇒ ∂t~B + ~∇× ~E = 0

and a single equation among the three spatial indices:

∂[1F23] = 0 ⇒ ∂xBx + ∂yBy + ∂zBz = 0

⇒ ~∇ · ~B = 0

The dual field tensor ?F

I Define the fully antisymmetric Levy-Civita symbols in 1+3dimenions:

�µνρτ :=

1 if (µνρτ) is an even permutation of (0123)−1 if (µνρτ) is an odd permutation of (0123)0 if (µνρτ) is not a permutation of (0123)

together with�µνρτ := −�µνρτ

I Define the dual field tensor ?F with the components

?Fµν = �µνρτF ρτ , ?Fµν = �µνρτFρτ

The dual field tensor ?FThe dual field tensor has the components

?Fµν =

0 Bx By Bz−Bx 0 Ez/c −Ey/c−By −Ez/c 0 Ex/c−Bz Ey/c −Ex/c 0

and

?Fµν =

0 −Bx −By −BzBx 0 Ez/c −Ey/cBy −Ez/c 0 Ex/cBz Ey/c −Ex/c 0

With this tensor, the homogeneous Maxwell equations can bewritten in a continuity-equation-like form as

∂µ?Fµν = 0

Section 5.2

Electrodynamics - Physical Content

Action of the electromagnetic field

Apply heuristic principle of simplicity.

Search for the simplest action that is compatible with symmetries

I Poincaré invariance (Lorentz + rotation + translation):

⇒ Action should be a Lorentz scalar

I Gauge invariance:

⇒ Action should depend on F, not on A

Fµµ: too simple FµνFµν : this works!

Action of the electromagnetic field

SEM =∫

d4x L(A,∇A)

L(A,∇A) = − 14µ0FµνFµν

Rewrite Lagrangian in A and ∇A:

L(A,∇A) = − 12µ0

((∂µAν)(∂µAν)− (∂µAν)(∂νAµ)

)Rewrite Lagrangian in the physical field ~E and ~B:

L = 14µ0

( 2c2~E 2 − 2~B2

)= 12

(�0~E 2 −

1µ0~B2)

Action of the electromagnetic field

L(A,∇A) = − 14µ0FµνFµν

The Lagrangian is just the sum of the squares of the twist in eachpossible orientation of the loop.

Second set of Maxwell equations without sources

Lagrangian of the electromagnetic field without sources:

L(A,∇A) = − 12µ0

((∂µAν)(∂µAν)− (∂µAν)(∂νAµ)

)Euler-Lagrange equations of motion:

−∂ρ∂L

∂(∂ρAσ)= 0

⇒ 1µ0∂ρ(∂ρAσ − ∂σAρ

)= 0

Short form:∂ρF ρσ = 0

Second set of Maxwell equations without sources

∂ρF ρσ = 0 with Fµν =

0 Ex/c Ey/c Ez/c

−Ex/c 0 Bz −By−Ey/c −Bz 0 Bx−Ez/c By −Bx 0

For σ = 0:

−1c (∂xEx + ∂yEy + ∂zEz) = 0 ⇒ div~E = 0

For σ = 1, 2, 3:

1c2∂tEx − ∂yBz + ∂zBy = 01c2∂tEy + ∂xBz − ∂zBx = 01c2∂tEz − ∂xBy + ∂yBx = 0

⇒ rot~B − µ0�0 ∂t~E = 0

Electromagnetic waves in vacuum

−∂ρ∂L

∂(∂ρAσ)= 0 ⇒ 1

µ0∂ρ(∂ρAσ − ∂σAρ

)= 0

Wave equation, valid in any gauge:

�Aµ − ∂µ ∂ρAρ = 0

We can simplify the wave equation by choosing a particular gauge:

Chose Lorenz gauge: ∂µAµ = 0

⇒ �Aµ = 0

The wave equation �A = 0 is only validin the Lorenz gauge ∂µAµ = 0.

Electromagnetic field with charged sourcesCharges are twist generators

jµ(x) =

cρ(~x , t)j1(~x , t)j2(~x , t)j3(~x , t)

.

Electromagnetic field with charged sourcesInduced twists in the vicinity

Easiest way to couple the charged source to the EM field

L(A,∇A) = − 14µ0FµνFµν + Aµjµ.

Attention: This term should be gauge-invariant!

Inhomogeneous Maxwell equations:

∂ρF ρσ = −µ0jσ

Inhomogeneous Maxwell equations old-fashioned:

div~E = 1�0ρ0 , rot~B − µ0�0 ∂t~E = µ0~J

Wave equation in Lorenz gauge:

�Aµ = −µ0jµ

Gauge invariance of the coupling term

L(A,∇A) = − 14µ0FµνFµν + Aµjµ.

Aµ → õ = Aµ + ∂µf ⇒ L → L̃ = L+ (∂µf )jµ

The solution must not change under gauge transformations!

⇒ S̃ − S = −∫

d4x (∂µf )jµ = 0

Partial integration

S̃ − S =∫

d4x f ∂µjµ = 0

Charge conservation: ∂µjµ = 0 ⇒ ∂tρ(~x , t) +∇~j(~x , t) = 0

Charge conservation follows from gauge invariance.

Energy-momentum tensor of the electromagnetic field

L = − 14µ0FµνFµν + Aµjµ.

The conjugate momentum can be shown to be given by:

∂L∂(∂ρAσ)

= Fσρ.

The energy-momentum tensor is defined as

T µν =∂L

∂(∂µAρ)∂νAρ − δµνL

⇒ T µν = F ρµ∂νAρ +14FρσF

ρσδµν

Section 5.3

Electrodynamics using differential forms

Extra section, see lecture notes for further details

Section 5.4

Electromagnetically charged fields

Real-valued scalar fieldMassless spring chain → Wave equation

L = −12(∂νφ)(∂νφ) = 12c2 (

∂

∂t φ)2︸ ︷︷ ︸

Ekin

− 12(∇φ)2︸ ︷︷ ︸

Epot

Real-valued scalar fieldMassive spring chain → Klein-Gordon equation

L = −12(∂νφ)(∂νφ)− 12M

2φ2

Attaching U(1)-circles: Complex-valued scalar fieldMassless spring chain

Epot =12(∇Re[φ]

)2 + 12(∇Im[φ])2

Complex-valued scalar field

L = 12c2( ∂∂tRe[φ]

)2 − 12(∇Re[φ])2 − 12M2Re[φ]2+ 12c2

( ∂∂t Im[φ]

)2 − 12(∇Im[φ])2 − 12M2Im[φ]2

Simplify expression:

L = 1c2( ∂∂t φ

∗)( ∂∂t φ

)−(∇φ∗

)(∇φ)−M2φ∗φ

Final form:L = −(∂νφ

)∗(∂νφ

)−M2φ∗φ

Global and local gauge invariance

Massive complex scalar field:

L = −(∂νφ)∗(∂νφ

)−M2φ∗φ

This Lagrangian is invariant under global gauge transformations:

φ(x)→ eiθφ(x)(θ = const

)

It is not invariant under local gauge transformations

φ(x)→ eiθ(x)φ(x)(θ(x) 6= const

)⇒ We have to modify the Lagrangian to make in gauge-invariant!

Restore local gauge invarianceModify the derivative

The ordinary derivativegenerates translations:

exp(a∂x )f (x) = f (x + a)

exp(aµ∂µ)f (x) = f (x + a)

⇒ We should modify thederivative in such a waythat it also goes straightinside the tube!

Covariant derivative

Begin with a real-valued field and perform the following gaugetransformation:

φ(x)→ eiϕ(x)φ(x)

Modify the partial derivative by the following steps:

I First revert the gauge transformation.I Then apply the usual partial derivative.I Finally restore the gauge transformation.

⇒ Dµ := eiϕ(x)∂µe−iϕ(x)

⇒ Dµ = ∂µ − i(∂µϕ(x)

)

Covariant derivative

Dµ = ∂µ − i(∂µϕ(x)

)Recall that

dϕ(x) = e~Aµ(x) dxµ

where e is the elementary charge. This means that

⇒ ∂µϕ(x) =e~Aµ(x)

Hence the covariant derivative reads:

Dµ = ∂µ −ie~

Aµ .

Minimal coupling by covariant derivativeCooking recipe

I Take the Lagrangian:

L = −12(∂νφ)(∂νφ)− 12M

2φ2

I Complexify it:

L = −(∂νφ)∗(∂νφ

)−M2φ∗φ

I Replace partial by covariant derivative:

L = −(Dνφ)∗(Dνφ)−M2φ∗φ.

I Add the Lagrangian of the free electromagnetic field:

L = − 14µ0FµνFµν − (Dνφ

)∗(Dνφ)−M2φ∗φ.

Minimal coupling: Solving the equations of motion

L = − 14µ0FµνFµν − (Dνφ

)∗(Dνφ)−M2φ∗φ.Rewrite:

(Dνφ)∗(Dνφ) = (∂νφ∗ + ie~ Aνφ∗

)(∂νφ− ie

~Aνφ

)= (∂νφ∗)(∂νφ) +

ie~

(Aνφ∗(∂µφ)− Aνφ(∂νφ∗)

)+ e

2

~2AνAνφ∗φ

Simply solve the Lagrange equations of motion:

∂L∂φ− ∂µ

∂L∂(∂µφ)

= 0

∂L∂φ∗− ∂µ

∂L∂(∂µφ∗)

= 0

∂L∂Aν − ∂µ

∂L∂(∂µAν)

= 0

Section 6.1

Relativistic quantum mechanics - without spin

Quantum mechanics of spinless particlesRecall the non-relativistic Schrödinger equation

Starting point for “deriving” the free Schrödinger equation is

E = p2

2m .

Then substitute according to the correspondence rules:

E → i~∂t ~p → −i~~∇

Let the relation act on a complex-valued wave function ψ(~x , t):

i~∂tψ(x, t) = −~2~∇2

2m ψ(x, t) .

Interpretation: |ψ(x, t)|2 = ψ∗(x, t)ψ(x, t) is the probabilitydensity to find a particle at x at time t. This requires normalization∫

ddx ψ∗(x, t)ψ(x, t) = 1.

Galilei invariance of the Schrödinger equationUnder Galilei transformation x→ x̃ = x− vt a function T (x)transforms as

T̃ (x̃) = T (x)

Galilei invariance of the Schrödinger equation

I Recall Galilei transformation:

x→ x̃ = x− vt , t̃ = t , ∂̃t = ∂t + v · ∇ , ∇̃ = ∇ .

I Assume “naive” transformation behavior of the wave function:

ψ̃(x̃, t̃) = ψ(x, t) .

I Insert both into the Schrödinger equation:

−i~∂̃t ψ̃(x̃, t̃) = −~2∇̃2

2m ψ̃(x̃, t̃)

⇒ −i~(∂t + v · ∇)ψ(x, t) 6= −~2∇2

2m ψ(x, t)

...violates form invariance!

Galilei invariance of the Schrödinger equation

This is not so surprising:I A resting particle has a constant wave function.I A moving particle has a wave function with oscillations in

space and time.ψ̃(x̃, t̃) = ψ(x, t) cannot work .

Possible solution: We have to add some kind of waviness

ψ̃(x̃, t̃) = ei θ(x ,t)ψ(x, t)

with a real function θ(x, t).I This preserves the transformation of the probability density

ψ̃∗(x̃, t̃)ψ̃(x̃, t̃) = ψ∗(x, t)ψ(x, t)

Galilei invariance of the Schrödinger equationBoost of the wave function

ψ̃(x̃, t̃) = ei θ(x ,t)ψ(x, t)

Choose phase function θ(x, t) such that the Schrödinger equationis form-invariant under Galilei transformations:

Exercise:

ψ̃(x̃, t̃) = exp(− imv · x

~+ imv

2t2~

)ψ(x, t)

= exp(− imv · x̃

~− imv

2t̃2~

)ψ(x̃ + vt̃, t̃)

This boosted wave function obeys the transformed wave equation

−i~∂̃t ψ̃(x̃, t̃) = −~2∇̃2

2m ψ̃(x̃, t̃) .

Klein-Gordon equation

Relativistic replacement rule:

p→ −i~∂ ⇔ pµ → −i~∂µ .

Recall that

pµ =

E/cpxpypz

, pµ =−E/cpxpy

pz

T

, ∂µ =

1c ∂t∂x∂y∂z

T

This is compatible with E → i~∂t , ~p → −i~~∇ .

Insert into energy-momentum relation pµpµ = −m2c2:

⇒ −~2∂µ∂µφ(x) = −m2c2φ(x) ⇒ (�−M2)φ(x) = 0

where M = mc~ is the mass parameter.

Klein-Gordon equationThe problem of negative probability currents

φ∗(�−M2)φ− φ(�−M2)φ∗ = 0.⇒ φ∗∂µ∂µφ− φ∂µ∂µφ∗ = 0.

Because of∂µ(φ∗∂µφ) = (∂µφ∗)(∂µφ) + φ∗(∂µ∂µφ)∂µ(φ∂µφ∗) = (∂µφ)(∂µφ∗) + φ(∂µ∂µφ∗)

this can be rewritten as

∂µ(φ∗∂µφ− φ∂µφ∗︸ ︷︷ ︸4-current jµ

) = ∂µjµ = 0 .

This is a nothing but a relativistic continuity equation of the form

− 1c2 ∂t( φ∗∂tφ− φ∂tφ∗︸ ︷︷ ︸

probability density ρc2

) +∇ · ( φ∗∇φ− φ∇φ∗︸ ︷︷ ︸probability current −~j

) = 0

or in short ∂tρ+∇~j = 0.

Klein-Gordon equationThe problem of negative probability currents

− 1c2 ∂t( φ∗∂tφ− φ∂tφ∗︸ ︷︷ ︸

probability density ρc2

) +∇ · ( φ∗∇φ− φ∇φ∗︸ ︷︷ ︸probability current −~j

) = 0

I The probability density in the KG equation is no longerpositive definite!

I This reflects the existence of antimatter, but needs a deeperunderstanding.

Section 6.2

Dirac equation

Dirac equationMotivation

Warmup exercise: Klein-Gordon equation in 0+1 dimensions:( 1c2∂

2t + M2

)φ(t) = 0.

Define χ(t) = iMc ∂tφ(t) and split into two equations:

i 1c ∂tχ(t)−Mφ(t) = 0i 1c ∂tφ(t)−Mχ(t) = 0 .

Define vectorψ(t) =

(φ(t)χ(t)

)Rewrite system of first-order differential equations:

i(0 11 0

)1c ∂t

(φ(t)χ(t)

)−M

(φ(t)χ(t)

)= 0

Dirac equationWarmup exercise

i(0 11 0

)︸ ︷︷ ︸

=:γ

1c ∂t

(φ(t)χ(t)

)−M

(φ(t)χ(t)

)= 0

In short: (iγ 1c ∂t −M

)ψ(t) = 0.

Iterate two times:

M2ψ(t) = − 1c2γ2∂2t ψ(t)

Each component separately should obey the Klein-Gordon equation( 1c

2∂2t + M2)ψ = 0

⇒ γ2 = 1

Dirac equationWarmup exercise

(iγ 1c ∂t −M

)ψ(t) = 0 with γ2 = 1

γ2 = 1 is the Dirac algebra in this simple case.

Representations:I γ = ±1

I γ = σx =(0 11 0

)

I γ = σy =(0 −ii 0

)

I γ = σz =(1 00 −1

)

Full Dirac equationin 1+3 dimensions

(iγµ∂µ −M

)ψ(x) = 0.

γ0, γ1, γ2, γ3 are four operators acting in anauxiliary k-dimensional space, called spinor space.

Shortcut notation: Fermi dagger:

/∂ = γµ∂µ = γµ∂µ.

⇒ (i/∂ −M)ψ = 0.

Dirac algebra

Iterate the Dirac equation i/∂ψ = Mψ two times:

M2ψ(x) = −/∂/∂ψ(x) = −γµγν∂µ∂νψ(x).

Each component obeys the Klein-Gordon equation

−γµγν∂µ∂νψ(x) = �ψ(x) = ηµν∂µ∂νψ(x)

I Comparing both sides it seems that γµγν = −ηµν .I However, the partial derivatives commute, leading to the

weaker condition γµγν + γνγµ = −2ηµν1k

Dirac algebra:

{γµ, γν} = −2ηµν1k

Representation of the γ-matricesWarmup exercise: The Pauli algebra

{σi , σj} = 2δij1k . (i , j = x , y)

Category Monomial NumberScalar S: 1 1Vector V: σx , σy 2Pseudoscalar P: σxσy = iσz 1

total number: M=4

Therefore, we need at least 2× 2 matrices. And in fact, this iswhat we get:

σx =(0 11 0

), σy =

(0 −ii 0

)

Representation of the γ-matricesNow consider the Pauli algebra

{γµ, γν} = −2ηµν1k

Category Monomial #Scalar S: 1 1Vector V: γ0, γ1, γ2, γ3 4Tensor T: γ0γ1, γ0γ2, γ0γ3, γ1γ2, γ1γ3, γ2γ3 6Pseudovector A: γ0γ1γ2, γ0γ1γ3, γ0γ2γ3, γ1γ2γ3 4Pseudoscalar P: γ0γ1γ2γ3 1

total number: M=16

⇒ So we need at least 4× 4 matrices.

The γ5 matrix

Define:γ5 := iγ0γ1γ2γ3.

This is analogous to σz in the Pauli case.

(Why not γ4?)

Properties:I γ5 is Hermitean, i.e., (γ5)† = γ5.I (γ5)2 = γ5γ5 = 1, therefore the eigenvalues of γ5 are ±1.I γ5 anticommutes with all other γ-matrices:{γ5, γµ} = γ5γµ + γµγ5 = 0.

Dirac representationThe standard choice

γ0 =(1

−1

), γj =

(σj

−σj

), γ5 =

(1

1

),

Explicitely:

γ0 =

1 1 −1−1

γ1 = 11−1−1

γ2 =

−iii− i

γ3 = 1 −1−1

1

and γ5 = iγ0γ1γ2γ3 =

1 111

.

Proof of the Dirac representation

We can write the γ-matrices in Dirac representation as tensorproducts of Pauli matrices:

γ0 = σz ⊗ 12 , γj = iσy ⊗ σj (j = 1, 2, 3)

This allows us to verify the Dirac algebra:

1. {γ0, γ0} = 2(γ0)2 = 2(σz)2 ⊗ 1 = 2122. {γ0, γj} = (σzi σy )⊗ σj + (iσyσz)⊗ σj = i{σz , σy} ⊗ σj = 03. {γj , γk} = (iσy )2 ⊗ {σj , σk} = −2δjk 14

⇒ {γµ, γν} = −2ηµν 14.

Weyl and Majorana representation

Weyl representation:

γ0 =(

1

1

), γj =

(σj

−σj

), γ5 =

(−1

1

),

Majorana representation:

γ0 =(

σ2

σ2

), γ1 =

(iσ3

iσ3

), γ2 =

(−σ2

σ2

)

γ3 =(−iσ1

−iσ1

), γ5 =

(σ2

−σ2

),

Spinor space

I All the representations introduced above are four-dimensional.I The purpose of this internal space is to provide a realization

of the anticommutation relations {γµ, γν} = −2ηµν .I This internal space is called spinor space.

I A spinor is a vector living in spinor space.I A spinor field ψ(x) is a vector field mapping the Minkowski

space to the spinor space.

Internal spinor space 6= Minkowski space

Form-invariance of the Dirac equation under Lorentztransformation

I The Dirac equation (iγµ∂µ −M

)ψ(x) = 0

is expected to be invariant under Lorentz transformations

xµ → x̃µ = Λµνxν , ∂µ → ∂̃µ = Λ νµ ∂νI Obviously this requires the Dirac γ-matrices to transform like

a 4-vectorγµ → γ̃µ = Λµνγν .

I This makes only sense if the transformed γ̃-matrices obeyagain the Dirac algebra, namely, {γ̃µ, γ̃ν} = −2ηµν .

can be proven easily!

Covariant spinor transformation

What we have shown so far:

If ψ(x) is a solution of the Dirac equation(iγµ∂µ −M

)ψ = 0

and ifx̃µ = Λµνxν and γ̃µ = Λµνγν ,

then ψ̃(x̃) = ψ(x) is also solution of(i γ̃µ∂̃µ −M

)ψ̃ = 0 .

But for form-invariance, we would like to have the same Diracequation with the same γ-matrices in the moving frame.

Covariant spinor transformation

I Idea: Instead of transforming γ̃µ = Λµνγν , we transform thewave function by

ψ̃(x̃) = Sψ(x) ,

where S is a 4× 4 matrix in spinor space.I This leads us to the condition(

iγµ∂̃µ −M)Sψ =

(iγµΛ νµ ∂ν −M

)Sψ = 0 .

I This is equivalent to (see lecture notes)

Λρνγν = S−1γρS.

Unfortunately, this is quadratic in S...

Covariant spinor transformationCalculation of the spinor transformation matrix S

If you have a nonlinear relation between elements of Liegroups, try it infinitesimally and get a linear relation.

Consider infinitesimal Lorentz transformation:

Λµν = δµν + �(αβ)λ(αβ)µν +O(�2)

Assume that spinor transformation has the same form:

S = 1 + i2 �(αβ)σ(αβ) +O(�2)

Then the inverse is super-trivial:

S−1 = 1− i2 �(αβ)σ(αβ) +O(�2)

Insert all this into the equation above:

Λµνγν = S−1γµS

Covariant spinor transformationCalculation of the spinor transformation matrix S

Insert this into Λµνγν = S−1γµS:

⇒ γµ + �(αβ)λ(αβ)µνγν +O(�2) =

=(1− i2 �

(αβ)σ(αβ) +O(�2))

︸ ︷︷ ︸S−1

γµ(1 + i2 �

(αβ)σ(αβ) +O(�2))

︸ ︷︷ ︸S

Compare to first order in �:

⇒ �(αβ)λ(αβ)µνγν =

[γµ ,

i2 �

(αβ)σ(αβ)]

This is now a linear system of equations!Linear equations are solvable.

Covariant spinor transformationCalculation of the spinor transformation matrix S

Intuitive approach: Recall the defintion of the generators[λ(αβ)]µν = δµαηβν − δ

µβηαν

and insert them into the linear system of equations:

γµσ(αβ) − σ(αβ)γµ = −2i(δµαηβν − δ

µβηαν

)γν

⇒ σ(αβ) =i2 [γα, γβ] =

{iγαγβ if α 6= β0 otherwise

Proof: Trivial for α = β. For α 6= β we have to show that

i(γµγαγβ − γαγβγµ

)= −2i

(δµαγβ − δµβγα

).

Now we use {γµ, γν} = −2ηµν1. If µ 6= α and µ 6= β we have {γα, γµ} ={γβ , γµ} = 0, hence both sides are zero. If µ = α then

γαγαγβ − γα γβγα︸︷︷︸=−γαγβ

= {γα, γα}︸ ︷︷ ︸=−2

γβ = −2γβ .

The proof for µ = β is analogous.

Covariant spinor transformationCalculation of the spinor transformation matrix S

Remark: We have done a very common trick in Theoretical Physics.Given a non-linear relation between transformations Λ and S, we studythem infinitesimally to first order in �. This gives a linear relationbetween the corresponding generators λ and σ.

To get the relation between the (non-infinitesimal) fulltransformations, we have to exponentiate the generators:

Λ = exp( ∑

0≤α

Covariant spinor transformation

Example: Rotation O(3):

Λ(12)(θ) =

1

cos θ sin θ− sin θ cos θ

1

goes together with

S(12)(θ) =

eiθ/2

e−iθ/2eiθ/2

e−iθ/2

For a rotation by 2π = 360◦ we get Λ = 1 but S = −1.

Covariant spinor transformation acting on fieldsRecall Lorentz transformation acting on a scalar field

Under a translation a scalar field φ : Rd → R changes as

x→ x− a⇒ φ(x)→ φ(x + a)

⇒ φ(xµeµ)→ φ((xµ + aµ)eµ) = eaµ∂µφ(xµeµ)

Similarly, for a Lorentz boost or a rotation we have

x→ Λ−1x⇒ φ(x)→ φ(Λx)

⇒ φ(xµeµ)→ φ(Λµνxνeµ) = e???φ(xµeµ)

What has to be put in the exponential?

Covariant spinor transformation acting on fieldsRecall SO+(1, 3) transformation acting on a scalar field

φ(xµeµ)→ φ(Λµνxνeµ) = e???φ(xµeµ)

Consider infinitesimal transformation Λ = 1 + �(αβ)λ(αβ):

φ(xµeµ)→ φ(xµeµ + �(αβ)λαβµνx

νeµ)

This is like a small (x-dependent) translation:

φ(xµeµ)→(1 + �(αβ)λ(αβ)µνx

ν∂µ)φ(xµeµ)

Exponentiate it:

φ(x)→ exp(θ(αβ)λ(αβ)

µνxν∂µ

)φ(x).

Covariant spinor transformation acting on fieldsExample: Rotation SO(3)

Consider for example λ(12):

φ(x)→ exp(φ(−x∂y + y∂x

))φ(x).

If you like quantum mechanics, rewrite it as

φ(x)→ exp(− i~φ[−i~(x∂y − y∂x︸ ︷︷ ︸

Lz

)])φ(x).

whith the orbital angular momentum

Lz = ~r × ~p = −i~(x∂y − y∂x )

Covariant spinor transformation acting on fieldsHow a Lorentz transformation acts on a spinor field

If the scalar field φ(x) is replaced by a 4-component spinor fieldψ(x), then the transformation of the internal degrees of freedomhas to be included as well:

ψ(x)→ exp(− i~θ(αβ)

[i~λ(αβ)µνx

ν∂µ︸ ︷︷ ︸orbital

−~2σ(αβ)︸ ︷︷ ︸spin

])ψ(x)

Example: Rotation around the z-axis:

ψ(x)→ exp(− i~

(Lz + Sz︸ ︷︷ ︸=Jz

))ψ(x)

operator acting on eigenvalues

Lz = i~λ(12)µνxν∂µ Minkowski space n~

Sz = −~2σ(12) Spinor space ±~2

Plane-wave solutions and their interpretation

ψ(x) = u(p)ei~ ε p·x

I u(p) is a 4-component “amplitude”, called spinor.I ε is a dimensionless constant to be determined later.

Insert into Dirac equation (iγµ∂µ −M)ψ = 0 with M = mc/~:(−1~εγµpµ −M

)u(p) = 0

This gives the eigenvalue problem

−εγµpµ u(p) = mc u(p)

Plane-wave solutions and their interpretationAnalyze eigenvalue problem −εγµpµ u(p) = mc u(p)

−γµpµ =

E/c 0 −pz −p−0 E/c −p+ pzpz p− −E/c 0p+ −pz 0 −E/c

.Eigenvalues:

±√E 2/c2 − p+p− − p2z = ±

√E 2/c2 − p2

Compare with E 2 = p2c2 + m2c4:

�1 = �2 = +1 (solutions with positive energy)�3 = �4 = −1 (solutions with negative energy)

Plane-wave solutions and their interpretationAnalyze eigenvalue problem −εγµpµ u(p) = mc u(p)

Eigenvectors for � = 1 in Dirac representation:

u1(p) =

E/c + mc

0pzp+

, u2(p) =

0E/c + mc

p−−pz

,Eigenvectors for � = −1:

u3(p) =

pzp+

E/c + mc0

, u4(p) =

p−−pz0

E/c + mc

.

Vecolity-momentum problem of negative-energy solutions

First consider positive energy � = +1:

ψ(x) = u(p) ei~pµx

µ.

For simplicity consider wave in x -direction with py = pz = 0:

I In which direction does the wave move?Determine velocity of a “wave crest” with phase zero:

0 = pµxµ = −Ec ct+pxx ⇒ x =Epx t ⇒ v = E/px > 0.

I What is the momentum P = −i~∇ ?

Pψ = −i~∇ei~pµx

µ = −i~∇ei~ (−Et+~p·~x) = +~pψ

For positive energy, ~v and ~p point in the same direction.

Vecolity-momentum problem of negative-energy solutions

Now consider negative energy � = −1:

ψ(x) = u(p) e−i~pµx

µ.

For simplicity consider wave in x -direction with py = pz = 0:

I In which direction does the wave move?Determine velocity of a “wave crest” with phase zero:

0 = pµxµ = −Ec ct+pxx ⇒ x =Epx t ⇒ v = E/px > 0.

I What is the momentum P = −i~∇ ?

Pψ = −i~∇e−i~pµx

µ = −i~∇e−i~ (−Et+~p·~x) = −~pψ

For negative energy, ~v and ~p point in opposite directions.

Vecolity-momentum problem of negative-energy solutions

Remember: Relativistic particles with negative energy have the pa-radoxical property that velocity and momentum point in oppositedirections.

Problems:I Particle transfers a momentum opposite to its own flight

direction, contradicting with our everyday experience.I Negative pressure.I Systems with an unlimited negative energy are

thermodynamically unstable.

Dirac hole theory

At T = 0 a physical system selects the state of lowest energy.Therefore, all modes with negative energy have to be occupied.I The Dirac sea is phy-

sically unobservable.I The total energy of

the sea is set to zero.I Absence of negative

energy and negativemomentum = presenceof positive energyand positive momentum.

⇒ Concept of an antiparticle

Dirac hole theory

I If the particles carry an additional electrical charge, thecorresponding anti-particles carry the opposite charge. Forexample, the antiparticles of electrons (e−) are calledpositrons (e+) since they carry a positive charge.

I The bare Dirac equation describes a two-particle theory(particle and antiparticle). However, Dirac’s hole theoryimplicitly uses the scenario of a multi-particle theory obeyingthe Pauli principle.

I Particles and antiparticles recombine to radiation, i.e.,e+e− → 2γ.

I The opposite reaction γ → e+e− is only possible in thepresence of an atom, for example a crystal lattice, providedthat the energy of the photon is high enough.

Feynman-Stückelberg Interpretation

Probability conservation in the Dirac equationLet ~∂ be a derivative acting to the right:(

iγµ~∂µ −M)ψ = 0.

Adjoint Dirac equation

ψ†(−i(γµ)† ~∂µ −M

)= 0 .

The adjoint Dirac matrices are given by

γ0 = (γ0)† , γk = −(γk)† (k = 1, 2, 3)

⇒ (γµ)† = γ0γµγ0

ψ†(−iγ0γµγ0 ~∂µ −M

)= 0

⇒ ψ†(iγ0γµ ~∂µγ0 + M

)= 0

∣∣∣ γ0γ0 = −η00 = 1⇒ ψ†

(iγ0γµ ~∂µγ0 + γ0Mγ0

)= 0

⇒ ψ†γ0(iγµ ~∂µ + M

)γ0 = 0

Probability conservation in the Dirac equationDefine the adjoint spinor:

ψ̄ := ψ†γ0

Then the Dirac equation and its adjoint read

(iγµ~∂µ −M

)ψ = 0 ψ̄

(iγµ ~∂µ + M

)= 0.

Probabilty current and continuity equation

ψ̄γµ( ~∂µ + ~∂µ)ψ = 0 ⇒ ∂µ(ψ̄γµψ︸ ︷︷ ︸

=jµ

)= 0.

jµ := ψ̄γµψ =(ρ1c~j

).

Probability conservation in the Dirac equation

∂µjµ = 0.

In terms of the probability density

ρ = ψ̄γ0ψ = ψ†ψ

and the probability 3-current ~j with

(~j)k = cψ̄γkψ = cψ†γ0γkψ

this continuity equation can be written in the usual form as

∂tρ = ~∇ ·~j .

Action for the Dirac equationfor electrically neutral fermions

We can guess the action:

S =∫

d4x ψ̄(x)(iγµ∂µ −M

)ψ(x).

ψ̄γµψ transforms as a 4-vector ⇒ S is a scalar.

S depends linearly on two (independent) fields ψ and ψ̄:

I Variation with respect to ψ̄:

δS =∫

d4x δψ̄(x)(iγµ∂µ −M

)ψ(x)︸ ︷︷ ︸

=0

⇒(iγµ∂µ −M

)ψ(x) = 0.

I Variation with respect to ψ (needs partial integration):

δS =∫

d4x ψ̄(x)(iγµ∂µ −M

)︸ ︷︷ ︸=0

δψ(x) ⇒ ψ̄(x)(iγµ ~∂µ+M

)= 0.

Dirac equation: Adding an electric field

Although the Dirac equation involves the imaginary unit, itdescribes electrically neutral fermions.Recall Majorana representation:

γ0 =(

σ2

σ2

), γ1 =

(iσ3

iσ3), γ2 =

(−σ2

σ2

)γ3 =

(−iσ1

−iσ1), γ5 =

(σ2

−σ2),

Then the Dirac equation (iγµ∂µ −M)ψ = 0 reads:−∂1 ∂3 0 ∂0 − ∂2∂3 ∂1 −∂0 + ∂2 00 ∂0 + ∂2 −∂1 ∂3

−∂0 − ∂2 0 ∂3 ∂1

ψ = Mψ.⇒ This equation is real. There is no U(1) circle.

Dirac equation: Adding an electric fieldComplexification

I Complexify ψ(x) in order to implementing a U(1) circle.I Replace derivative by covariant derivative:

∂µ → Dµ = ∂µ −ie~Aµ

⇒ Dirac equation of a charged fermion in an electromagnetic field:

(iγµ(∂µ −

ie~Aµ)−M

)ψ = 0.

Substitute pµ = −i~∂µ and M = mc/~:(γµ(pµ − eAµ

)︸ ︷︷ ︸Min. coupling

+mc)ψ = 0 .

Charged Dirac equation: Probability conservation

Dirac equation and adjoint Dirac equation:(iγµ(~∂µ − ie~ Aµ)−M

)ψ = 0

∣∣∣ ψ̄(iγµ( ~∂µ + ie~ Aµ)+ M)

= 0

Multiply from left and right and add them:

ψ̄(iγµ(~∂µ − ie~ Aµ)−M

)ψ = 0

ψ̄(iγµ(~∂µ +

ie~Aµ)

+ M)ψ = 0

—————————————-

iψ̄γµ( ~∂µ + ~∂µ)ψ = 0

∂µjµ = 0 , jµ = ψ̄γµψ

Charged Dirac equation: ActionSimply replace derivative by the covariant derivative and add thecontribution of the electromagnetic field:

S =∫

d4x[ψ̄(x)

(iγµ(∂µ −

ie~Aµ)−M

)ψ(x) − 14µ0

FµνFµν].

I Varying with respect to ψ̄ gives the charged Dirac equation:(iγµ(∂µ −

ie~Aµ)−M

)ψ = 0

I Varying with respect to A gives the inhomogeneous Maxwellequations

∂L∂Aµ

− ∂ν∂L

∂(∂νAµ)= 0

with∂νF νµ = −µ0

e~ψ̄γµψ︸ ︷︷ ︸

=jµ.

As we have shown above, this current is conserved, reflectingcharge conservation.

Charge conjugationReflections:

in space: parityin time: time reversalin the circle: charge conjugation

Charge conjugation: ψ(x)→ ψ∗(x)

ψ∗(x) obeys the complex-conjugated Dirac equation:

(−i(γµ)∗∂µ −M)ψ∗ = 0

Charge conjugation is a symmetry if and only if −(γµ)∗ is again avalid representation of the Dirac algebra. If so, it differs from γµ byunitary transformation (see exercise):

(−γµ)∗ = C†γµC

Chirality

Chirality refers to the eigenvalues of γ5 = iγ0γ1γ2γ3.

Define the projectorsP± :=

12(1± γ5

)with the usual properties P2± = P±, P+P− = 0, and P+ + P− = 1.

If the fermion is massless, obeying the massless Dirac equation

γµ∂µψ = 0

then the so-called chiral spinors

ψR = ψ+ := P+ψ , ψL = ψ− := P−ψ

are also solutions of the Dirac equation. ψR is the right-handedspinor, ψL is the left-handed spinor.

ChiralityThe most suitable representation is the Weyl representation with

γ5Weyl =(−1

1

)

Here the Dirac equation takes the form:

(iγµ∂µ −M)ψ =(

−M i(∂0 + ~σ · ~∇)i(∂0 − ~σ · ~∇) −M

)(ψLψR

)= 0

The mass terms −M mix the two components. For M = 0 theydecouple into two equations, the so-called Weyl equations:

i(∂0 − ~σ · ~∇)ψL = 0 ,i(∂0 + ~σ · ~∇)ψR = 0

This is also interesting if the mass is small (for example neutrino oscillations).

Section 7.1

Non-relativistic Hamilton formalism

Lagrange is fine.Why then do we need another formalism?

I Different viewpoint, first-order differential equations.I Much closer to quantum mechanics than Lagrange.I Quantization straight-forward by replacing momenta by

derivatives and replacing Poisson brackets with commutators.I Possibly a good investment for