2

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 $e(\mu)$ has minimum for $d=2,3$, whereas when $d=4$, $e(\mu)$ is scale independent and stationary points and vacuum solutions are possible?
  • How $e(\mu)$ is a continuous function bounded by zero?
Qmechanic
  • 220,844
Raisa
  • 661

1 Answers1

2

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 $\delta S=0$ amounts to minimize the (potential) energy $T+V$. Here $S:=\int \! dt ~L$, and $L:=T-V$. See also this Phys.SE post.)

Let $d$ be the number of spatial dimensions. Assume there exists a stationary field configuration $(\Phi({\bf x}), A_i({\bf x}))$ in temporal gauge $A_0({\bf x})=0$.

Define a 1-parameter family of field configurations is $$ \Phi^{(\mu)}({\bf x})~=~\Phi(\mu{\bf x}), \tag{4.11} $$ $$ A^{(\mu)a}_i({\bf x})~=~\mu A_i^a(\mu{\bf x}), \qquad i~\in\{1,\ldots,d\}. \tag{4.13} $$ The (potential) energy of the $\mu$-scaled configuration is

$$ e(\mu)~=~\sum_{n\in\{0,2,4\}} \mu^{n-d} E_n\geq 0, \tag{4.22} $$

where we assume that the scale parameter

$$ \tag{A}\mu~\in~]0,\infty[ $$

is strictly positive, and that the energies

$$ E_n~\geq ~0, \qquad n~\in~\{0,2,4\},\tag{B}$$

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

  1. Case $E_0,E_4>0$ and $d<4$: Then $$ e(0)~:=~\lim_{\mu\to 0^{+}}e(\mu)~=~\infty~=~\lim_{\mu\to \infty}e(\mu)~=:~e(\infty),\tag{C}$$ so that there must exist an interior$^1$ (relative) minimum, and therefore Derrick's No-Go conclusion does not apply.

  2. Case $d\geq 4$: 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 $E_0=0=E_2$ and if moreover either

    • (i) $d=4$ or
    • (ii) $E_4=0$.

    The case (i) corresponds to pure 4+1 gauge theory, which indeed has non-trivial static soliton solutions with $E_4>0$. The case (ii) corresponds to vacuum solutions $e\equiv 0$. In both cases (i) and (ii) the function $e$ is a constant function, i.e. independent of the scale parameter $\mu$.

  3. Case $E_0=0=E_2$: 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) $E_4=0$,

    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.

$^1$ An interior minimum point $\mu$ means that $\mu$ is different from the boundary $0$ and $\infty$.

Qmechanic
  • 220,844