Electrically and Magnetically Charged Solitons in Gauge Field Th

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1 Electrically and Magnetically Charged Solitons in Gauge Field Theory Polytechnic Institute of New York University Talk at the conference Differential and Topological Problems in Modern Theoretical Physics, SISSA, Trieste, Italy, April 26 30, 2010

2 Outline

3 Magnetism, monopoles, and P. Curie P. Curie Scientific giant in magnetism Curie s law (for magnetization of paramagnetic materials): M = C T B Curie constant, Curie point. Magnetic monopoles? P. Curie: On the possible existence of magnetic conductivity and free magnetism, Séances Soc. Phys. pp (Paris), Scholar.google citation count: 4 P. Dirac: Born on August 8, 1902

4 Mathematics and formulation of monopoles by Dirac New mathematics with the birth of Quantum Mechanics is needed P. Dirac: Quantized singularities in the electromagnetic field, Proc. Roy. Soc. A 133, pp , Scholar.google citation count: 1396 Quoted from the first paragraph of the Dirac paper: The steady progress of physics requires for its theoretical formulation a mathematics that gets continually more advanced. This is only natural and to be expected. Dirac s toy: The Maxwell equations

5 Maxwell equations Variational structure Spacetime coordinates t = x 0, (x, y, z) = (x i ), x µ = (x 0, x i ) Gauge field A µ, µ = 0, 1, 2, 3 Electromagnetic field F µν = µ A ν ν A µ 4-Current density J µ = (ρ, J) Charge density J 0 = ρ Current density J k = J Action density (the Lagrangian density) L = 1 4 F µνf µν + A µ J µ Maxwell equations ν F µν = J µ

6 Static situation 0 = 0 A 0 electric potential A i = A magnetic potential E = A 0 electric field B = A magnetic field Maxwell equations E = ρ, B = J, B = 0, E = 0 or A 0 = ρ A = J

7 The Maxwell equations are the Euler Lagrange equations of the action functional I (A 0, A) = { A A 2 A 0 ρ + A J} R 3 which is indefinite. Finite-energy condition E = {E 2 + B 2 } R 3 = { A A 2 } < R 3

8 The point charge problems Point electric charge With ρ = 4πeδ(x), J = 0 we can solve the Maxwell equations to get E = A 0 = e x 3 x B = 0 (Coulomb s law) Infinite energy R 3 E 2 = 0 4πe 2 r 2 dr =

9 Point magnetic charge (a monopole ) As before, we solve B = 4πmδ(x) to obtain B = m x 3 x ( a magnetic Coulomb s law ) We arrive at infinite energy problem again. The real problem is more severe than this inconsistency with the existence of a magnetic potential A: B = A which requires B = 0 in all R 3 ( magnetic field is solenoidal ).

10 A deeper insight is needed to resolve the puzzle. Topological structure We hope to see what is needed to represent the magnetic field B of a monopole smoothly over the punctured space M = R 3 \ {0} by a vector potential field A. In fact, if we insist to achieve a smooth relation B = A in M, we run into trouble with the Stokes theorem: 4πm = B ds = ( A) ds S 2 R = lim C 0 C S 2 R A dr = 0 where C is any closed curve over the sphere S 2 R of radius R.

11 A way out of this puzzle is to admit that the process of shrinking a closed curve on any surface around the origin will inevitably encounter a singularity. The locus of singularities is a string connecting the monopole, called the Dirac string.

12 Local existence of magnetic potentials Let S be a Dirac string connecting the monopole at the origin with the infinity of R 3. Consider the open set U = R 3 \ S Then there is nothing preventing us from solving the equation B = A over U so that B is represented over U (at least). However, we are not satisfied since B should only be singular at the origin but not on the entire string S which could well be an artifact.

13 From local to global representation In order to find a way to globally represent the magnetic field B of the monopole over the entire space M = R 3 \ {0} we pick two strings S + and S, which are the positive and negative x 3 -axises, respectively, form U + = R 3 \ S +, U = R 3 \ S and solve B = A over U + and U, separately, to get A + and A. Since M = U + U, we are able to represent B globally from its local representations over U + and U. Puzzle solved?

14 Puzzle not yet solved Consistency with physics Motion of a test particle of mass µ and electric charge e in B is quantum mechanically governed by the gauged Schrödinger equation i ψ t = 1 2µ ( ea)2 ψ which enjoys the gauge invariance ψ e iω ψ, A A + 1 e ω In other words, physical consistency requires that we must obey the gauge invariance in the Schrödinger equation when considering transition between the two magnetic potentials A + and A over the interaction U + U.

15 Fiber bundle terminology Fiber bundle (principal U(1)-bundle) P M: Ω = e iω U(1), ψ Ωψ, A A + i e Ω d Ω 1 The relation B = A may be translated as B = d A (exactness) which is locally valid over the coordinate patches U + and U, respectively, but can never be valid globally over the entire M. The quantity that measures the distance from such a global exactness is an integer N called the first Chern class which lies in the second cohomology of M: N = c 1 (P) H 2 (M, Z) = H 2 (S 2, Z) = Z = Topology

16 The Dirac quantization formula 2em = N, N = ±1, ±2, which explains why all electric charges are integer multiples of a basic unit charge.

17 Schwinger s dyons The motion of a dyon of electric and magnetic charges, e 1, m 1, in the electromagnetic field of another dyon of electric and magnetic charges, e 2, m 2, results in the Schwinger quantization formula 2(m 1 e 2 m 2 e 1 ) = N, N = ±1, ±2,

18 Modern physics Monopoles and dyons are inevitable predictions rather than conceptual constructs T. T. Wu, C. N. Yang, t Hooft, Polyakov, Julia, Zee, Bogomol nyi, Prasad, Sommerfield, Jackiw, Witten

19 Typical variational structure (once again) I (A 0, A) = { A A 2 + other terms} Monopoles positive definite case (A 0 0) Dyons indefinite case (when both A 0 and A are present) Constraints: finite energy; nontrivial topology

20 Abelian Higgs model (simplest gauge theory) φ complex scalar field D A φ = φ + i Aφ gauge-covariant derivative Action functional is I (φ, A 0, A) = { A A 2 A 2 0 φ 2 + D A φ 2 + λ R 3 2 ( φ 2 1) 2 } Energy is E(φ, A 0, A) = { A A 2 +A 2 0 φ 2 + D A φ 2 + λ R 3 2 ( φ 2 1) 2 }

21 Equations of motion are D 2 A φ = λ( φ 2 1)φ A 2 0φ, A = i 2 (φd Aφ φd A φ), A 0 = φ 2 A 0 Theorem, Trivial! Proof. π 2 (S 1 ) = 0. Open Problem # 1. Obtain an analytic proof that any finite energy solution must be gauge-equivalent to the trivial solution φ = 1, A 0 = 0, A = 0.

22 Modification (the advent of non-abelian theory) We need to enlarge the range of φ so that the set of images of φ at infinity become S 2. Thus, from π 2 (S 2 ) = Z we may expect to get a big zoo of monopoles and dyons. Hence, we modify φ so that it takes values in R 3, with the natural symmetry group SO(3). For sake of topological simplicity, we further replace SO(3) by SU(2) since the latter is a universal double covering of SO(3), which is simply connected, and the Lie algebras of SO(3) and SU(2) are identical so that we do not lose any gauge fields. In doing so, φ lies in the adjoint representation of SU(2) instead.

23 Non-Abelian Gauge Field Theory For convenience, we use iso-vectors for elements in su(2). Thus we have A µ R 3 (µ = 0, 1, 2, 3) gauge field F µν = µ A ν ν A µ + ea µ A ν curvature tensor D µ φ = µ + ea µ φ gauge-covariant derivative Non-Abelian Yang Mills Higgs action density is L = 1 4 F µν F µν D µφ D µ φ λ 4 ( φ 2 1) 2

24 Static case 0 = 0 Lagrangian density Energy density Topology L = 1 2 ia 0 + ea i A e2 A 0 φ F ij D iφ 2 + λ 4 ( φ 2 1) 2 H = 1 2 ia 0 + ea i A e2 A 0 φ F ij D iφ 2 + λ 4 ( φ 2 1) 2 deg(φ) deg ( φ S 2 ) = N or [φ S 2 ] π 2 (S 2 )

25 Equations of motion We have D i D i φ = e 2 A 0 (A 0 φ) + λφ( φ 2 1) D j F ij = e(a 0 F i0 + φ D i φ) A 0 = e i (A i A 0 ) +e( i A 0 + ea i A 0 ) A i + e 2 φ (A 0 φ) The A 0 sector looks hard Q: Is A 0 component important for electricity?

26 Electromagnetism The t Hooft electromagnetic tensor is proposed to be F µν = φ φ F µν 1 e φ 3 φ (D µφ D ν φ) The induced electric field E = (E i ) is given by E i = F 0i = φ φ ( ia 0 + ea 0 A i ) 1 φ 3 φ ([A 0 φ] D i φ) which dictates the condition A 0 0 in order to have a nontrivial electric sector.

27 Magnetic charge The magnetic charge Q m can be obtained through computing the total flux of the magnetic field induced to yield the quantization formula eq m = N (= deg(φ)) (Arafune, Freund, and Goebel) Facts The magnetic charge Q m has a topological origin Although the above quantization formula generalizes that of Dirac, the electric charge e is put in by hand externally which may not be the electric charge induced from the t Hooft tensor

28 Electric charge The electric charge Q e induced from the t Hooft tensor can be computed as the total flux generated from the electric field which gives us Q e = 1 i F 0i 4π R 3 which is not topological and requires as observed already A 0 0 In other words, we need to obtain solutions with A 0 0 in order that both magnetic and electric charges are present (dyons)

29 Existence of Dyons For any N Z, obtain a finite-energy N-dyon solution with nonvanishing magnetic and electric charges Q m and Q e. Remark: Too ambitious a problem!

30 Special cases A 0 = 0 ( temporal gauge = monopole case) The equations become D i D i φ = λφ( φ 2 1) D j F ij = eφ D i φ N = 1, radially symmetric solution due to Belavin, Polyakov, Schwartz, and Tyupkin N = 1, λ = 0 (BPS), reduction to the self-dual or BPS equations B i = ±D i φ, B i = 1 2 ɛ ijkf jk Prasad Sommerfield, Weinberg, Taubes, Ward, Atiyah, Manton

31 Theorem 1 (forthcoming 2010) For N = 1, there is a one-parameter family of dyon solutions with Q m = 1 e (topological) Q e = any value in an open interval (non-topological) Remark: In particular, Q e is not quantized

32 Special case λ = 0 Julia and Zee found an explicit family of dyon solutions for which the electric and magnetic charge relation is given by Q e = sinh α Q m, < α < where the parameter α determines the asymptotic value of the modulus of the Higgs field, lim φ(x) = cosh α x

33 Open problems (#2 #4) Any dyon solution for N 2 and λ > 0? What mechanism fixes Q e? How to compute dyons?

34 Progress in other dyon problems Dyons in the electroweak theory of Weinberg Salam (Yang, Proc. Royal Soc. 1998) Dyons in gauged Skyrme model (joint work with F. H. Lin, 2010) Q e is still not fixed

35 Dimension reduction Can we lower the dimension so that electric charge is fixed by topology? Motivation: Vortex-lines vs monopoles Answer: Yes (?) and No (?) About the No answer this is the Julia Zee theorem which is a fundamental property of classical gauge field theory About the Yes answer this is the existence theory for Chern Simons vortices

36 For the Abelian Higgs model over R 2, Julia and Zee proved, assuming radial symmetric and sufficient decay rate at infinity, that the field configuration must stay in the temporal gauge A 0 = 0 In other words, the Abelian Higgs model over R 2 must be electrically neutral and must be the same as the classical Ginzburg Landau model for superconductivity A generally accepted principle in classical gauge field theory

37 Theorem 2 (with Joel Spruck, Comm. Math. Phys., 2009) In classical gauge field theory, Abelian or non-abelian, finite-energy static solutions must stay in the temporal gauge A 0 = 0 Thus, the Julia Zee theorem is rigorously established.

38 PDE problems Abelian case Energy density over R 2 is D i D i φ = 2V ( φ 2 )φ A 2 0φ j F ij = i 2 (φd iφ φd i φ) A 0 = φ 2 A 0 H = 1 2 ia A2 0 φ F 2 ij D iφ 2 + V ( φ 2 ) Remark: It seems quite trivial that A 0 = 0 should follow readily from the finite-energy condition. However, the real situation is trickier because the method is expected to work over R 2 only

39 Non-Abelian case The A 0 -equation is A 0 + i (A i A 0 ) + A i i A 0 + A i (A i A 0 ) = φ (A 0 φ) The part of the energy density over R 2 that involves A 0 is 1 2 ia 0 + (A i A 0 ) A 0 φ 2 Remark It may be interesting to recall similar results regarding entire solutions of PDEs such as the Liouville theorem and the Bernstein theorem

40 The Chern Simons vortices over R 2 Extra topological term added the Chern Simons term In 3-dimensional spacetime and in terms of differential forms, the term reads A da Existence of electrically and magnetically charged vortices (Dyons in R 2 ) Two areas of research: Self-dual case (extensive work) (Italian school, in particular) Non-self-dual case (hard situation) progress report

41 Abelian Case (the simplest Chern Simons Higgs model) The Lagrangian density is L = 1 4 F µνf µν + κ 4 ɛµνα A µ F να D µφd µ φ λ 8 ( φ 2 1) 2 φ complex scalar D µ φ = µ φ + ia µ φ κ > 0 (the Chern Simons coupling constant)

42 Static equations D i D i φ = λ 2 ( φ 2 1)φ A 2 0φ j F ij = κɛ ij j A 0 + i 2 (φd iφ φd i φ) A 0 = κf 12 + φ 2 A 0 F 12 magnetic field We now formally compute the magnetic and electric charges

43 Magnetic charge We have Q m = 1 F 12 dx 2π R 2 = c 1 = the first Chern class = N = the winding number of φ near infinity = topological invariant

44 Electric charge In order to compute the electric charge, recall that the current density is given by J µ = i 2 (φdµ φ φd µ φ) Hence the electric charge density is given by ρ = J 0 = i 2 (φd0 φ φd 0 φ) = A 0 φ 2 So Q e = 1 ρ dx 2π R 2 = 1 A 0 φ 2 dx = topological? 2π R 2

45 Theorem 3 (with R. M. Chen, Y. Guo, and D. Spirn, Proc. Royal Soc., 2009) For any N Z, the static Chern Simons Higgs equations on R 2 have a finite-energy smooth solution of topological class N. The solution decays fast enough at infinity so that R 2 A 0 dx = 0 As a consequence, the magnetic and electric charges are all quantized topologically and obey the formulas Q m = N, Q e = κn Remark: This concludes our Yes answer discussion

46 Proof of Theorem 3 The action functional is of the form I = I + I ( 1 = R 2 2 F D iφ 2 + λ ) 8 ( φ 2 1) 2 dx ( 1 2 A κa 0 F ) 2 φ 2 A 2 0 dx R 2 The energy is E = I + + R 2 ( 1 2 A ) 2 φ 2 A 2 0 dx The admissible class is { } C = (φ, A 0, A i ) E(φ, A 0, A i ) <, δi = 0

47 Proof continued From δi, we get R 2 ( A0 η + κf 12 η + φ 2 A 0 η ) dx = 0, Taking η = A 0 as a test function, we have ( A φ 2 A 2 0) dx = κ F 12 A 0 dx R 2 R 2 Inserting this into the action functional I, we find I = E = positive definite η Q.E.D.

48 Self-dual Chern Simons equations Two prototype equations The Liouville equation (integral, non-relativistic, doable ) Extension: Toda systems u = ±e u u a = K ab e u b, a, b = 1, r, r = rank of gauge group The Jaffe Taubes equation (non-integrable, relativistic ) u = e u 1 Full of challenges and many still intractable (recent focus) (contributors from Italian school include: G. Tarantello, M. Nolasco, T. Ricciardi, D. Bartolucci)

49 Example 1 (solved with C. S. Lin and A. Ponce, J. Funct. Anal., 2007) The governing equations over R 2 are u = e v (e u 1) v = e u (e v 1) How to approach this problem? (Lack of apriori estimates or compactness; no maximum principle) Variational structure?

50 Example 1 continued With u + v = f, we transform the equations into u v = g f = 2e f e 1 2 (f +g) e 1 (f g) 2 g = e 1 2 (f +g) e 1 (f g) 2 which are the Euler Lagrange equations of the action functional ( 1 I (f, g) = 2 f 2 1 ) 2 g 2 + 2e f 2e 1 2 (f +g) 2e 1 (f g) 2 dx R 2 which is indefinite

51 Example 2 (unsolved; non-relativistic; Open Problem # 5) The governing equations over R 2 are u = e v 1 v = e u 1 With the same change of variables, f = u + v and g = u v, we have the new equations f = e 1 2 (f +g) + e 1 2 (f g) 2 g = e 1 2 (f +g) + e 1 (f g) 2 which are the Euler Lagrange equations of the action functional ( 1 I (f, g) = 2 f 2 1 ) 2 g 2 + 2e 1 2 (f +g) 2e 1 2 (f g) 2f dx which is again indefinite ( most off-diagonal )

52 Example 3 (non-abelian; relativistic ; Yang, Comm. Math. Phys., 1997) The governing equations over R 2 are u a = λ ( K ab K bc e u b+u c K ab e u b ), a, b, c = 1,, r (K ab ) is the Cartan matrix and r is the rank of the gauge group G Suppose (a very general condition for standard vacua to exist) r (K 1 ) ab > 0, a = 1,, r b=1 Existence of topological solutions satisfying ( r ) lim u a(x) = ln (K 1 ) ab, a = 1,, r x b=1 Method: Calculus of variations; positive-definite functional; suitable transformations

53 Example 3 continued For G = SU(N), N 3, r = N 1, and the matrix K is K = When K is not symmetric, it is a curious inquiry whether there is a solution theory when K is replaced by K τ. Indeed, in the original physics literature, K is wrongly given as K τ which defied solution for a while. This leads to our Open Problem # 6

54 Example 3 continued It will be interesting to develop an existence theory when the domain R 2 is replaced by a closed 2-surface such as T 2. There are some results due to M. Nolasco and G. Tarantello when G = SU(3) A general existence study is yet to be fully carried out Open Problem # 7

55 Summary There are locally concentrated static solutions carrying both electric and magnetic charges called dyons Dyons are topologically originated Dyons are realized as critical points of indefinite action functionals The magnetic charge of a dyon is quantized topologically The electric charge of a dyon takes continuous values and is non-topological Dyons cannot exist in two spatial dimensions in classical theory Dyons can be generated in two spatial dimensions if the Chern Simons terms are added in classical models so that both electric and magnetic charges are topologically quantized

56 Quotation from P. Dirac (again) I played with a few equations. Words of Paul Dirac when answering questions from an interviewer Encouraging? End of Talk