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?
  • $\begingroup$ To clarify: do you not understand how eq. 4.22 is derived, or how eqs. 4.23-4.24 are gotten from 4.22, or how the conclusions are derived given eqs. 4.22-4.24? If it's the latter try plotting the individual terms. (Hint: you don't need to know anything about $E_0,E_2,E_4$ except that they are positive real numbers. Pick any convenient values if you want to use a plotting package - the argument doesn't depend on the actual values.) $\endgroup$ – Michael Brown Aug 28 '13 at 13:44
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    $\begingroup$ Related: physics.stackexchange.com/q/62858 and physics.stackexchange.com/q/62924 $\endgroup$ – Michael Brown Aug 28 '13 at 13:46
  • $\begingroup$ I understood how they got equation 4.22 and 4.23 but don't understand the limit they saying. May be I did get the equation but failed to understand the physics. Equation (4.23) doesn't allow d=4 but why? $\endgroup$ – Raisa Aug 28 '13 at 17:37

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. The (potential) energy of the $\mu$-scaled configuration is

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

where we assume that the scale parameter

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

is strictly positive, and that the energies

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

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 subquestions (v1).

  1. Case $E_0,E_4>0$ and $d\leq 3$: Then $$\tag{C} e(0)~:=~\lim_{\mu\to 0^{+}}e(\mu)~=~\infty~=~\lim_{\mu\to \infty}e(\mu)~=:~e(\infty),$$ 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 former case corresponds to pure 4+1 gauge theory, which indeed has non-trivial static soliton solutions with $E_4>0$. The latter case corresponds to vacuum solutions $e\equiv 0$. The function $e$ is in both cases a constant function, i.e. independent of the scale $\mu$.

  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).


  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 $\infty$.

  • $\begingroup$ If we apply, $D \leq 3$, I mean $d=3$ in equation(4.22) then we get $$e(\mu)= \mu^{1} E_4+\mu ^{-1} E_2 + \mu^{-3} E_0 $$> Because $\mu$ has to be positive then, $d=3$ makes then potential negative I mean $\mu$ is negative. This is why we don't apply the Derricks theorem because we get negative potential here? What is actually monotonically here? $\endgroup$ – Raisa Aug 30 '13 at 17:15
  • $\begingroup$ @Raisa: To be able to uniquely answer your comment, you should specify which $E_n$'s (if any) are zero, cf. the various cases in the answer. $\endgroup$ – Qmechanic Aug 30 '13 at 17:21
  • $\begingroup$ There are three potential energy in the equation 4.22. I thought, i needed to keep the algebraic sum of these potential energy positive. So when I consider any value of d, i get negative. Am i doing wrong in calculating the value of $\mu$ by inserting different values of $d$? I guessed, because the power of $\mu$ is negative so the $\mu$, therefore we get negative potential energy and eventually some of the potential energy appears negative. But according to are consideration, the potential energy can not be negative!!! $\endgroup$ – Raisa Aug 30 '13 at 17:42
  • $\begingroup$ @Raisa: Note that a positive number raised to any real power is a positive number. In symbols: $\forall \mu>0, \forall p\in \mathbb{R} : \mu^p>0$ positive. For instance: $1^{-1}=1$ and $2^{-3}=\frac{1}{8}$. $\endgroup$ – Qmechanic Aug 30 '13 at 17:47
  • $\begingroup$ Got it, then any value of $d$ makes the potential energy positive. But $e(0)=\infty=e(\infty)$ indicates you inserted $\mu=0$ but why it is equal to infinity? Can you please elaborate the term "intermediate minimum"? $\endgroup$ – Raisa Aug 30 '13 at 18:25

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