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

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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?
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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.) –  Michael Brown Aug 28 '13 at 13:44
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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? –  Raisa Aug 28 '13 at 17:37

1 Answer 1

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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).

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 $\infty$.

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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? –  Raisa Aug 30 '13 at 17:15
    
@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. –  Qmechanic Aug 30 '13 at 17:21
    
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!!! –  Raisa Aug 30 '13 at 17:42
    
@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 $2^{-3}=\frac{1}{8}$. –  Qmechanic Aug 30 '13 at 17:47
    
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"? –  Raisa Aug 30 '13 at 18:25

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