Magnetic monopoles. Arjen Baarsma. January 14, 2009

Transcription

1 Magnetic s Arjen Baarsma January 14, 2009

2 GUT s GUT s 2

3 GUT s GUT s 3

4 (very briefly) Thoroughly discussed in previous talks Spontaneous symmetry occurs when a system with some symmetry (described by a symmetry group G) possesses vacuum states that are not invariant under this symmetry. Perturbations are made around one such a solution. Best explained through an example. GUT s 4

5 : An example We can for instance consider the Lagrangian: (in (2 + 1)D, so x µ = (t, x 1, x 2 )) L = µ φ µ φ V(φ), with V(φ) = λ(1 φ 2 ) 2 where φ is a complex scalar field and λ > 0. We have a U(1) symmetry under φ e iα φ. GUT s 5

6 : An example We can for instance consider the Lagrangian: (in (2 + 1)D, so x µ = (t, x 1, x 2 )) L = µ φ µ φ V(φ), with V(φ) = λ(1 φ 2 ) 2 where φ is a complex scalar field and λ > 0. We have a U(1) symmetry under φ e iα φ. The energy for this system is ( E = d 3 x φ ) 2 + φ 2 + V(φ) V(φ) is minimal on the vacuum manifold M = S 1, so this is minimised by the constant solution GUT s φ(x, t) = φ M 5

7 : Phase transitions V(φ) should actually be replaced by an effective potential V eff (φ) even at T = 0 due to loop diagrams. At finite temperatures this changes into Veff (φ, T) GUT s 6

8 : Phase transitions V(φ) should actually be replaced by an effective potential V eff (φ) even at T = 0 due to loop diagrams. At finite temperatures this changes into Veff (φ, T) At large temperatures the broken symmetry may be restored. V( φ). T>Tc T=T c T<T c 67 GUT s T=0 φ µ µ FIG. 47: Toy model potential for a real scalar field with two degenerate local minima used to illustrate a radiatively induced phase transition. Crossover or a second order transition. 6

9 GUT s GUT s 7

10 A topological soliton is a solution that cannot be continuously deformed into the vacuum solution due to some topological constraint (the exact definition varies). The constraint we put on our solutions is that their total energy is finite GUT s 8

11 We had energy density (still in dimensions) E = φ 2 + φ 2 + V(φ), so we need r φ 0 and r V(φ) 0 as r. GUT s 9

12 We had energy density (still in dimensions) E = φ 2 + φ 2 + V(φ), so we need r φ 0 and r V(φ) 0 as r. This tells us that φ(r, θ) φ (θ) M as r, which defines a function φ : S 1 M, θ φ(, θ). GUT s 9

13 We had energy density (still in dimensions) E = φ 2 + φ 2 + V(φ), so we need r φ 0 and r V(φ) 0 as r. This tells us that φ(r, θ) φ (θ) M as r, which defines a function φ : S 1 M, θ φ(, θ). We also need that r eθ φ = θ φ 0 as r, so φ (θ) = φ actually has to be constant GUT s 9

14 We had energy density (still in dimensions) E = φ 2 + φ 2 + V(φ), so we need r φ 0 and r V(φ) 0 as r. This tells us that φ(r, θ) φ (θ) M as r, which defines a function φ : S 1 M, θ φ(, θ). We also need that r eθ φ = θ φ 0 as r, so φ (θ) = φ actually has to be constant We can continuously deform such a solution to φ(r, θ) = φ everywhere, so there are no topological (according to this definition). GUT s 9

15 : Gauge theory The same definitions also apply to gauge theories, so suppose we add some gauge field A µ (in the usual manner) to make the symmetry local: L = D µ φd µ φ V(φ) 1 4 F µνf µν If we choose a gauge such that A0 = 0 and A r = 0 for r 1, then the energy density is E = ( 0 A i ) φ 2 + D i φ 2 + V(φ) (F ij) 2, GUT s 10

16 : Gauge theory The same definitions also apply to gauge theories, so suppose we add some gauge field A µ (in the usual manner) to make the symmetry local: L = D µ φd µ φ V(φ) 1 4 F µνf µν If we choose a gauge such that A0 = 0 and A r = 0 for r 1, then the energy density is E = ( 0 A i ) φ 2 + D i φ 2 + V(φ) (F ij) 2, GUT s We still need limr r V(φ) = 0, but we now require lim r e θ Dφ = lim θ φ r i e A θ φ 0. r r It is possible to choose an A i such that this holds. 10

17 : Gauge theory A goes like r 1, so F goes like r 2 and F 2 like r 4. The first time derivatives actually forms a separate boundary value problem, so we can find a solution with finite energy: d 2 x { ( 0 A i ) φ 2 + D i φ 2 + V(φ) (F ij) 2} < The same argument also works in three spatial dimensions (but not four!) and for other fields. GUT s 11

18 : Gauge theory A goes like r 1, so F goes like r 2 and F 2 like r 4. The first time derivatives actually forms a separate boundary value problem, so we can find a solution with finite energy: d 2 x { ( 0 A i ) φ 2 + D i φ 2 + V(φ) (F ij) 2} < The same argument also works in three spatial dimensions (but not four!) and for other fields. We haven t specified φ, only that limr r V(φ) = 0 Any two functions with the same behaviour at infinity can be continuously transformed into each other, so only φ is important to classify solutions. GUT s 11

19 : Gauge theory We re interested in classes of solutions that cannot be continuously deformed into a vacuum solution. This comes down to classes of functions φ : S d 1 M that cannot be deformed into a constant function. But when is this possible? GUT s 12

20 : Gauge theory We re interested in classes of solutions that cannot be continuously deformed into a vacuum solution. This comes down to classes of functions φ : S d 1 M that cannot be deformed into a constant function. But when is this possible? There is a word for continuous deformation: homotopy. We have the so-called homotopy groups π n (M) = {f : S n M}/ which exactly describe our classification. GUT s 12

21 : Gauge theory We re interested in classes of solutions that cannot be continuously deformed into a vacuum solution. This comes down to classes of functions φ : S d 1 M that cannot be deformed into a constant function. But when is this possible? There is a word for continuous deformation: homotopy. We have the so-called homotopy groups π n (M) = {f : S n M}/ which exactly describe our classification. GUT s Our theory admits topological of this type if and only if π d 1 (M) {1}. 12

22 : Gauge theory If d = 3 then such solutions are called s and we can see why from the 2-dimensional case: Our example had M = S 1 and π 1 (S 1 ) = Z {1}, so it admits topological of this type. GUT s 13

23 : Gauge theory If d = 3 then such solutions are called s and we can see why from the 2-dimensional case: Our example had M = S 1 and π 1 (S 1 ) = Z {1}, so it admits topological of this type. We can for instance find a non-trivial solution that points radially outwards: GUT s 13

24 GUT s GUT s 14

25 GUT s GUTs embed GSM = SU(2) Iw SU(3) c U(1) Y in a larger, more pleasing, compact connected gauge group G GUT. (e.g. G GUT = SU(5) or G GUT = SO(10), etc) The standard model is recovered after spontaneous symmetry. This happens after a phase transition at the GUT scale, so at around T = T GUT GeV. in stages also possible: G GUT... G SM SU(3) c U(1) em GUT s 15

26 GUT s GUTs embed GSM = SU(2) Iw SU(3) c U(1) Y in a larger, more pleasing, compact connected gauge group G GUT. (e.g. G GUT = SU(5) or G GUT = SO(10), etc) The standard model is recovered after spontaneous symmetry. This happens after a phase transition at the GUT scale, so at around T = T GUT GeV. in stages also possible: G GUT... G SM SU(3) c U(1) em Do GUTs predict the existence of s? GUT s 15

27 GUT s: Homotopy theorems Assume a symmetry G is broken to H. Since M = G/H we have π2 (M) = π 2 (G/H) GUT s 16

28 GUT s: Homotopy theorems Assume a symmetry G is broken to H. Since M = G/H we have π2 (M) = π 2 (G/H) There exists a canonical map ψ : π2 (G/H) π 1 (H), which is bijective if π 2 (G) = π 1 (G) = {1} Most GUTs have (a covering group with) π 2 (G) = π 1 (G) = {1}, so π 2 (G/H) π 1 (H). GUT s 16

29 GUT s: Homotopy theorems Assume a symmetry G is broken to H. Since M = G/H we have π2 (M) = π 2 (G/H) There exists a canonical map ψ : π2 (G/H) π 1 (H), which is bijective if π 2 (G) = π 1 (G) = {1} Most GUTs have (a covering group with) π 2 (G) = π 1 (G) = {1}, so π 2 (G/H) π 1 (H). The fundamental group of GSM is π 1 (G SM ) = π 1 (SU(3) SU(2) U(1)) = π 1 (SU(3)) π 1 (SU(2)) π 1 (U(1)) = π 1 (U(1)) = π 1 (S 1 ) = Z GUT s 16

30 GUT s: Homotopy theorems Assume a symmetry G is broken to H. Since M = G/H we have π2 (M) = π 2 (G/H) There exists a canonical map ψ : π2 (G/H) π 1 (H), which is bijective if π 2 (G) = π 1 (G) = {1} Most GUTs have (a covering group with) π 2 (G) = π 1 (G) = {1}, so π 2 (G/H) π 1 (H). The fundamental group of GSM is π 1 (G SM ) = π 1 (SU(3) SU(2) U(1)) = π 1 (SU(3)) π 1 (SU(2)) π 1 (U(1)) = π 1 (U(1)) = π 1 (S 1 ) = Z GUT s Therefore π2 (M) π 2 (G GUT /G SM ) π 2 (G SM ) Z 16

31 GUT s GUT theories allow for the existence of s. If we assume a single phase transition at the GUT scale, then s with a mass of GeV would form. GUT s 17

32 GUT s GUT theories allow for the existence of s. If we assume a single phase transition at the GUT scale, then s with a mass of GeV would form. At time of the phase transition, the Higgs field has a correlation length ξ, so domains of size ξ 3 form. At the intersection point of domains there is some probability (p 0.1) that s will form. Monopole density can be estimated to be nmm pξ 3 GUT s 17

33 GUT s GUT theories allow for the existence of s. If we assume a single phase transition at the GUT scale, then s with a mass of GeV would form. At time of the phase transition, the Higgs field has a correlation length ξ, so domains of size ξ 3 form. At the intersection point of domains there is some probability (p 0.1) that s will form. Monopole density can be estimated to be nmm pξ 3 By causality ξ < lgut cm This gives n cm 3 at the phase transition GUT s 17

34 GUT s Annihilation can reduce this density, but not significantly, so n MM a 3 (and d dt n MM = 3 H n MM ). The Entropy density s scales in the same way, so n/s is approximately conserved (without inflation). Monopole density today would therefore be n MM,now = n MM, GUT (s now )/(s GUT ), with s g T 3 GUT s 18

35 GUT s Annihilation can reduce this density, but not significantly, so n MM a 3 (and d dt n MM = 3 H n MM ). The Entropy density s scales in the same way, so n/s is approximately conserved (without inflation). Monopole density today would therefore be n MM,now = n MM, GUT (s now )/(s GUT ), with s g T 3 This gives us approximately nnow 10 7 cm 3, which is absurd (comparable to the baryon density). This is the problem. Inflation solves this problem: As long as the temperature after preheating is below the GUT scale the density will be too low to oberve. GUT s 18

36 GUT s GUT s 19

37 Classical Magnetic s with Bmm = g 4 π r 2 ˆx is allowed by Maxwell s equations (after extension). E = 4 π ρ e B = 4 π ρ m E = B t 4 πj m B = E t + 4 πj e Charge density: Bmm (x) = g δ 3 (x) = 4 π ρ m (x) Net magnetic flux: S B mm ds = g (for any surface S around the ) GUT s 20

38 Classical Magnetic s with Bmm = g 4 π r 2 ˆx is allowed by Maxwell s equations (after extension). To formulate quantum mechanics, we need B = A. No such potential can be defined for the magnetic, even if the origin is excluded from its domain since by Stokes theorem: B ds = A ds = A dl = 0 S S S GUT s 20

39 Classical Magnetic s with Bmm = g 4 π r 2 ˆx is allowed by Maxwell s equations (after extension). To formulate quantum mechanics, we need B = A. No such potential can be defined for the magnetic, even if the origin is excluded from its domain since by Stokes theorem: B ds = A ds = A dl = 0 S S S GUT s We can define an A such that B = A on any contractible region that does not contain the origin. 20

40 If we cut out a line, the magnetic field lines can disappear through it and there is no problem. We can define a potential A everywhere except in this line (so for 0 θ < π) A (1) = g 4 πr Annu. Rev. Nucl. Part. Sci :461 by Utrecht University 1 cos θ e ϕ sin θ such that B mm = A. discovered that the existence of magnetic monop electric charge quantization. Dirac envisaged a magnetic as a semi-i finitesimally thin solenoid (Figure 2). The end of such a GUT s Fioure 2 The end o 21

41 If we cut out a line, the magnetic field lines can disappear through it and there is no problem. end of a semi-infinite solenoid. We can define a potential A everywhere except in this line (so for 0 < θ π) A (2) = g 4 πr 2 The 1 + cos θ e ϕ sin θ such that B mm = A. Fioure Measured electric charges are always found to be integer multiples of the electron charge. This quantization of electric charge is a deep property of Nature crying out for an explanation. More than fifty years ago, Dirac (1) discovered that the existence of magnetic s could "explain" electric charge quantization. Dirac envisaged a magnetic as a semi-infinitely long, infinitesimally thin solenoid (Figure 2). The end of such a solenoid looks like 2.1 Monopoles.and Charge Quantization GUT s Annu. Rev. Nucl. Part. Sci : Download by Utrecht University on 01/01/09. Fo 22

42 We had found A (1) = g 1 cos θ e ϕ (0 θ < π) 4 πr sin θ and also A (2) = g 1 + cos θ e ϕ (0 < θ π). 4 πr sin θ These are related by the gauge transformation A (2) = A (1) α with α = g 2 π ϕ. Find a similar gauge transformation wherever you choose your Dirac string. GUT s 23

43 We had found A (1) = g 1 cos θ e ϕ (0 θ < π) 4 πr sin θ and also A (2) = g 1 + cos θ e ϕ (0 < θ π). 4 πr sin θ These are related by the gauge transformation A (2) = A (1) α with α = g 2 π ϕ. Find a similar gauge transformation wherever you choose your Dirac string. One problem: α(ϕ = 2π) α(ϕ = 0) = g, i.e. α is multiply-valued (up to multiples of g). GUT s 23

44 α itself is not observable, so is this really a problem? A field ψ with charge e couples to A via D j ψ = j ϕ i e A j ψ and it transforms under gauge transformations as ψ(x) e i e α(x) ψ(x) GUT s 24

45 α itself is not observable, so is this really a problem? A field ψ with charge e couples to A via D j ψ = j ϕ i e A j ψ and it transforms under gauge transformations as ψ(x) e i e α(x) ψ(x) Because α is multiply-defined we want e i e α(x) ψ(x) = e i e (α(x)+g) ψ(x), for which we need e g 2 π Z. GUT s 24

46 : Charge quantisation If e g 2 π Z then everything is consistent. The charge therefore needs to be a multiple of 2 π/e. GUT s 25

47 : Charge quantisation If e g 2 π Z then everything is consistent. The charge therefore needs to be a multiple of 2 π/e. Conversely, if a single magnetic with charge g exists, then the electric charge of any particle has to be an integer multiple of 2 π/g (charge quantisation). GUT s 25

48 : Charge quantisation If e g 2 π Z then everything is consistent. The charge therefore needs to be a multiple of 2 π/e. Conversely, if a single magnetic with charge g exists, then the electric charge of any particle has to be an integer multiple of 2 π/g (charge quantisation). Attempts to make a quantum field theory with Dirac s have failed. GUT s 25

49 GUT s GUT s 26

50 Consider the Georgi-Glashow SU(2) theory with the following Lagrangian density: L = 1 2 Tr(D µφd µ Φ) 1 4 λ(η2 Φ 2 ) Tr(F µνf µν ), with Φ 2 = (Φ 1 ) 2 + (Φ 2 ) 2 + (Φ 3 ) 2 = 2Tr(Φ 2 ) GUT s 27

51 Consider the Georgi-Glashow SU(2) theory with the following Lagrangian density: L = 1 2 Tr(D µφd µ Φ) 1 4 λ(η2 Φ 2 ) Tr(F µνf µν ), with Φ 2 = (Φ 1 ) 2 + (Φ 2 ) 2 + (Φ 3 ) 2 = 2Tr(Φ 2 ) Ingredients: A Higgs field Φ = Φ a t a A gauge field Aµ = A a µt a Fµν = µ A ν ν A µ i e [A µ, A ν ] Dµ Φ = µ Φ i e [A µ, Φ] Here t a = 1 2 τ a with [t a, t b ] = i ε abc t c generate of SU(2) (τ 1, τ 2 and τ 3 are the Pauli matrices.) The field Φ transforms in the adjoint representation: Φ g Φ g 1 or Φ [ξ a t a, Φ] (infinitesimally) GUT s 27

52 V(Φ) = 1 4 λ(η2 Φ 2 ) 2 is minimised at Φ = η, so we have a spontaneously broken symmetry with vacuum manifold S 2. GUT s 28

53 V(Φ) = 1 4 λ(η2 Φ 2 ) 2 is minimised at Φ = η, so we have a spontaneously broken symmetry with vacuum manifold S 2. What we usually do: In the unitary gauge we can write Φ = (η + φ)t 3 and A µ = W 1 µt 1 + W 2 µt 2 + a µ t 3 The Higgs field (φ) gets a mass 2 λ η, the W-bosons get mass 2 η and the photon a µ is massless. The unbroken U(1) is generated by t 3. GUT s 28

54 A static solution to the field equations of the form Φ(x) = η h(r) xa r ta, A i = 1 e (1 k(r))ε x j ija r 2 ta, A 0 = 0 with h(0) = k( ) = 0 and h( ) = k(0) = 1 (A 0 = 0) exists and has been computed numerically. GUT s 29

55 A static solution to the field equations of the form Φ(x) = η h(r) xa r ta, A i = 1 e (1 k(r))ε x j ija r 2 ta, A 0 = 0 with h(0) = k( ) = 0 and h( ) = k(0) = 1 (A 0 = 0) exists and has been computed numerically. Core size is of order (e η) π η 1 due to exponential convergence. For r (e η) 1 we get Φ(x) η xa r ta S 2 and A i 1 e ε ija x j r 2 ta GUT s The solution is stable and the mass (energy) can be calculated to be of order 8 πη/e η. 29

56 GUT s h (h(0) = 0) and k (k(0) = 1) for λ 0 (solid), λ = e 2 /40 (dashed) and λ = e 2 /4 (dotted). The horizontal axis is scaled by 2/(e η). 30

57 We can choose fµν = 2 Tr(F µν Φ/ Φ ) = a Fa µν Φa Φ and define the magnetic field as B i 1 2 ε ijkf jk 1 e x i r 3 = r e r 2 (r (e η) 1 ) This describes a magnetic monople of charge 4 π/e GUT s 31

58 We can choose fµν = 2 Tr(F µν Φ/ Φ ) = a Fa µν Φa Φ and define the magnetic field as B i 1 2 ε ijkf jk 1 e x i r 3 = r e r 2 (r (e η) 1 ) This describes a magnetic monople of charge 4 π/e We can choose the unitary gauge (Φ = (1 + φ)t 3 ) on a contractible region that does not contain the origin. In this gauge we can write Fµν = F 1 µνt 1 + F 2 µνt 2 + f µν t 3 and A µ = W 1 µ + W 2 µt 2 + a µ t 3 and obtain GUT s f µν = µ a ν ν a µ The field equations tell us that now µ f µν = 0. 31

59 GUT s GUT s 32

60 : Overclosure The Friedmann equation H 2 = 8 π G N ρ + Λ 3 c 2 3 c2 k can be a 2 rewritten as ρ cr = ρ b + ρ γ ρ Λ + ρ MM c 2 k/h 2 0, so we need ρ MM < ρ cr = 3 c2 8 π G N H GeV/cm 3 to prevent overclosure of our universe. Thus nmm < (m/10 17 GeV) 1 cm 3 Magnetic s expected to have velocity 10 3 c Resulting flux limit: F < F u (m MM /10 17 GeV) 1 cm 2 s 1 sr 1. GUT s 33

61 : The Parker bound Our Galaxy has a magnetic field of the order of 3µG 10 9 T varying over typical distances of L cm 300 pc. This is generated by the dynamo effect and refreshed over a typical timescale of τ 10 8 yrs. Magnetic s are accelerated by these magnetic fields over a distance L and gain energy. The energy of the magnetic field should not be trained faster than τ Parker bound F < FP cm 2 s 1 sr 1 obtained Result was later improved: GUT s F < F PE (m/10 17 GeV) cm 2 s 1 sr 1 34

62 : Induction D. E. Groom, In search ofthe supermassive magnetic ji~ asa goes throughasuperconducting loop. Although lines never cut the conductor, theydistort andthen pinch radial lines from the after it goes through the loop, as well as those due to the induced current in the loop. A magnetic travelling through a superconducting loop leaves behind a flux 2 Φ 0 (flux quantum). Flux can be measured using SQUIDs Independent of velocity, mass etc. Shielding is a problem No s found but upper bound obtained: cm 2 s 1 sr 1 GUT s duced current. More complicated loop structures can be constructed, in two or three 35

63 its of cm 2 s 1 sr 1 ) as a function of β for an isotropic flux of g = g D he limits discussed in Section : 2 are given inenergy columns twoloss to six; the global the last column. the spatial ant analysis. y comparing fined as the overed solar he temporal t 1; otherwise, tector, there equal to 1. ependence of he difference obtained by ing isotropic subdetector nd N 1,2 are th analyses, Fast (v > 10 2 c) magnetic s are strongly ionizing Due to their large mass they are nevertheless very penetrating Exact energy loss rate depends on the velocity and the material used GUT s then: of magnetic al MACRO 36

64 : Catalysed nucleon decay A GUT may allow for the following M + p M + e + + π 0, M + p M + µ + + K 0, M + n M + e + + π, M + n M + µ + + K. Monopoles might be able to catalyse such reactions (Rubakov-Callan mechanism) Cross-section may be of order /(v/10 3 ) cm 2 If this is the case then this would be measurable. No catalysed nucleon decay has been been measured, which limits the flux (under the above assumption) F < cm 2 s 1 sr 1 (10 4 < v < ), F < cm 2 s 1 sr 1 (10 5 < v < 10 1 ), F < cm 2 s 1 sr 1 (10 2 < v < 1) GUT s 37

65 : Catalysed nucleon decay Neutron stars and white dwarfs capture all s that hit them and these accumulate in the core Under the conditions from the previous slide s will catalyse nucleon decay and cause these objects to heat up. Comparison with measured luminosity flux: NETIC MONOPOLES... PHYSICAL REVIEW D urements of the luminosity hat the radius is R 3.9 oling models. First we use from the calculations of Seated by Chabrier. The Segfor gravitational energy rerentiation at crystallization. yields significantly longer h in turn imply an older age These white dwarf models f initially unstratified white s carbon and oxygen, with e models, the age of white comparison we also use the ich do not include chemical ation provides an additional way; thus models that lack it ng models, the ages of white Hence these models give GUT s 38

66 A in linear gauge theory can be defined as a topological soliton and the existence of s is a topological property of the vacuum manifold. Grand Unified Theories generally predict the allow the existence of massive magnetic s. The problem doesn t need to be a problem. We ve studied at an example of a magnetic People are still searching, but no s have thus far been observed. If s exist they must therefore be very rare. GUT s 39

67 Questions Any questions? GUT s 40