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Statement of the problem

I need to solve the equation

\begin{align} 0 = \frac{1}{\phi} \partial_{\sigma}\partial_{\sigma} \phi \hspace{20mm} (1) \end{align}

where $\phi$ is a scalar field and we're working in Euclidean 4-space. More specific questions are asked below. Basic advice on this would also be appreciated.

Context

I'm trying to work through 't Hooft's Euclidean Yang-Mills N-instanton solution. The idea is that we make the ansatz for the gauge potentials

\begin{align} A_{\mu} = i \overline{\sigma}_{\mu \nu} \partial^{\nu}(\ln \phi(x)) \end{align}

where $\overline{\sigma}_{\mu \nu}$ can be thought of as a constant 4x4 matrix that is unimportant to my question. We use the self-duality of the field strength tensor that arises from this ansatz to obtain the equation (1) for $\phi$

The Difficulties

One generally good source has been Rajaraman's "Solitons and Instantons." The relevant section is displayed

Relevant section in Rajaraman

My difficulties are:

  1. I am assuming by "singular" Rajaraman means that the function contains singularities. Why, explicitly, does $\phi$ being non-singular imply that we we need only solve $\Box \phi = 0$? Surely if $\phi$ contains any zeros then $1/\phi$ raises issues?
  2. Why does $\Box \phi = 0$ only permit the constant solution and not say $\phi(x) = a_{\mu} x^{\mu} + b$ where $a_{\mu}$ and $b$ are constants? Is this because of the form of the ansatz, or is it something I'm missing?
  3. In the case where $\phi$ is allowed to be singular, why does $\phi = 1/|x|^2$ solve the equation (1) for $x \neq 0$? Surely we have

\begin{align} \frac{1}{\phi} \partial_{\sigma}\partial_{\sigma} \phi &= |x|^2 \partial_{\sigma}\partial_{\sigma} \frac{1}{|x|^2}\\ &= |x|^2 \partial_{\sigma}\bigg(-\frac{2 x_{\sigma}}{|x|^4}\bigg)\\ &=|x|^2\partial_{\sigma} (\frac{-2 x_{\sigma}}{|x|^4})\\ &= |x|^2\partial_{\sigma} (\frac{-2 x_{\sigma}}{(\sum_{i} x_i^2)^2})\\ &= |x|^2\frac{-2}{(\sum_i x_i^2)^2} + |x|^2\frac{8 x_{\sigma}^2}{(\sum_i x_i^2)^3}\\ &=\frac{-2}{|x|^2} + \frac{8 x_{\sigma}^2}{|x|^4}\\ &\neq 0 \end{align}

(edit) I see where I went wrong above, the last three lines should be

\begin{align} &= \sum_{\sigma}(|x|^2\frac{-2}{(\sum_i x_i^2)^2}) + |x|^2\frac{8 x_{\sigma}^2}{(\sum_i x_i^2)^3}\\ &=\frac{-8}{|x|^2} + \frac{8 |x|^2}{|x|^4}\\ &= 0 \end{align}

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  • $\begingroup$ For your question $3$, it is explained in $(4.64)$, you have $|x|^2 \delta^4(x)=0^2 \delta^4(x)=0$ $\endgroup$ – Trimok Sep 5 '14 at 8:22
  • $\begingroup$ 4.64 applies to the case where $x=0$. My question is referring to the $x\neq 0$ sector. That is they claim that the equation holds when $x$ is not 0 and I don't understand why. $\endgroup$ – Jonathan Rayner Sep 5 '14 at 12:28
  • $\begingroup$ No, in $4$ euclidean dimensions, the equation $\square \frac{1}{|x|^2} = -4 \pi^2 \delta^4(x)$ is true " for all $x$". Strictly speaking, it is an equation between distributions, meaning that, for any smooth function $f(x)$, you have $\int d^4x f(x)\, \square \frac{1}{|x|^2} = \int d^4x f(x) \,(-4 \pi^2 \delta^4(x)) = -4 \pi^2 f(0)$. $\endgroup$ – Trimok Sep 6 '14 at 10:13
  • $\begingroup$ Yes, it's true for all $x$, that is correct. When I said "4.64 applies to the case where $x=0$," what I'm implying is that 4.64 sheds no light on why 4.62 is true. 3 asks why 4.62 is true, which I've since seen (see edit). Sure, 4.64 implies 4.62, but unless you can prove 4.64 without knowledge of 4.62 (which would be interesting!), your suggestion isn't helpful. That is, my view is that when you integrate over all space and obtain a non-zero answer, you use the knowledge that the integrand is zero "almost everywhere" and so conclude that the contributions come from delta functions. $\endgroup$ – Jonathan Rayner Sep 6 '14 at 14:43
  • $\begingroup$ More concisely, what I'm asking is is it fair to conclude $\int\limits_{\text{all}} d^4 x\, g(x) = -4\pi^2 \implies g(x) = -4 \pi^2 \delta^4(x)$? It sounds like you're saying that it is. $\endgroup$ – Jonathan Rayner Sep 6 '14 at 14:46
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1) If $\phi$ is non-singular, you can safely multiply both sides by $\phi$ and get $\square \phi = \phi*0 = 0 $. If $\phi$ is singular you can't do this because $\phi * 0$ is undefined. Equivalently, where $1/\phi=0$ you can have $\square\phi$ nonzero, as shown in (4.64). The singularities in $\phi$ will map to zeroes in $1/\phi$.

2) I believe only the linear solution you mentioned is allowed, which results in unbounded field values at infinity. I assume that's reason enough to reject it.

3) I believe your math is incorrect. $\partial_\sigma (x_\sigma / |x|^4)$ is zero. In 4 dimensions, you should get two equal terms with opposite signs.

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  • $\begingroup$ For 1) This seems reasonable (I've never really formally learned this). Can you elaborate a little, maybe in a group theoretical sense why $0 * \phi$ is undefined for singular $\phi$? 2) Seems reasonable also. Not really clear from Rajaraman, so I wondered if there was another reason... 3) Sorry, I don't see this. Can you show explicitly? I have edited what I have in my question above. $\endgroup$ – Jonathan Rayner Sep 5 '14 at 13:16
  • $\begingroup$ Nevermind on 3, have seen my mistake and corrected above. Thanks! $\endgroup$ – Jonathan Rayner Sep 5 '14 at 13:32
  • $\begingroup$ 2) However, I'm not sure that 2) is correct. From the ansatz I've mentioned, the gauge fields corresponding to the linear solution would be $A_{\mu} = i \overline{\sigma}_{\mu\nu}\partial^{\nu} \ln (a_{\sigma} x^{\sigma} +b) = i \overline{\sigma}_{\mu\nu} a_{\nu}/(a_{\sigma} x^{\sigma} + b)$ which vanishes at infinity. $\endgroup$ – Jonathan Rayner Sep 5 '14 at 15:08
  • $\begingroup$ In regards to (1), what I'm saying is that $\infty * 0$ is a meaningless quantity, which is what $\phi*0$ will be at the singular points of $\phi$. We're concerned with $\square\phi/\phi=0$. If you know that $\phi$ contains no infinities (singular points) then at every point the denominator is finite and so $\square\phi$ must be zero everywhere - so you can drop the $1/\phi$. If $\phi$ does contain infinities, then at those points $\square\phi$ can be finite - or even infinite so long as it's a "lesser infinity" like in $x/x^2$. $\endgroup$ – adipy Sep 5 '14 at 15:41

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