Saturday, July 18, 2015

quantum field theory - Dimension analysis in Derrick theorem


The following image is taken from p. 85 in the textbook Topological Solitons by N. Manton and P.M. Sutcliffe:


enter image description here enter image description here



What I don't understand from the above statement:



  • why e(μ) has minimum for d=2,3, whereas when d=4, e(μ) is scale independent and stationary points and vacuum solutions are possible?

  • How e(μ) is a continuous function bounded by zero?





Answer



Generalized versions of Derrick's No-Go Theorem compare spatially scaled, non-trivial, time-independent, finite-energy, classical, field-configurations to exclude the existence of static solitons. (Since we are only considering time-independent field-configurations, there is no kinetic energy T=0. Therefore the stationary action principle δS=0 amounts to minimize the (potential) energy T+V. Here S:=dt L, and L:=TV. See also this Phys.SE post.)


Let d be the number of spatial dimensions. The (potential) energy of the μ-scaled configuration is


e(μ) = n{0,2,4}μndEn0,


where we assume that the scale parameter


μ  ]0,[


is strictly positive, and that the energies


En  0,n{0,2,4}



are non-negative. From this it already follows that the (μ-scaled potential) energy e:]0,[[0,[ is a non-negative and continuous function, and in particular that it is bounded from below, cf. some of OP's subquestions (v1).




  1. Case E0,E4>0 and d3: Then e(0) := limμ0+e(μ) =  = limμe(μ) =: e(), so that there must exist an interior1 (relative) minimum, and therefore Derrick's No-Go conclusion does not apply.




  2. Case d4: The function e is monotonically weakly decreasing. [The word weakly means here that it could be (locally) constant.] The only possibility to have an interior minimum (and therefore evade Derrick's No-Go conclusion) is if E0=0=E2 and if moreover either (i) d=4 or (ii) E4=0. The former case corresponds to pure 4+1 gauge theory, which indeed has non-trivial static soliton solutions with E4>0. The latter case corresponds to vacuum solutions e0. The function e is in both cases a constant function, i.e. independent of the scale μ.




  3. Case E0=0=E2: The function e is monotonic. The only possibility to have an interior minimum (and therefore evade Derrick's No-Go conclusion) is, if (i) d=4, or if (ii) E4=0, i.e. we are back in the previous case (2).





References:




  1. N. Manton and P.M. Sutcliffe, Topological Solitons, 2004, Section 4.2.




  2. S. Coleman, Aspects of symmetry, 1985. Note that Sidney calls solitons for lumps.





  3. R. Rajaraman, Solitons and Instantons: An Introduction to Solitons and Instantons in Quantum Field Theory, 1987.




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1 An interior minimum point μ means that μ is different from the boundary 0 and .


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