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I am reading this research paper authored by NS Manton on the Force between 't Hooft-Polyakov monopoles. I have a doubt in equation 3.6 and 3.7. We assume the gauge field for a slowly accelerating monopole to be $A_0 = \epsilon^2 a_i t A_1$, where $\epsilon^2$ is an infinitesimal. Also, we write $\partial_0 \phi = -\epsilon^2 a_i t \partial_i \phi$. Using this he writes $D_0\phi=-\epsilon^2 a_i t D_i \phi$, where $D_i\phi=\partial_i \phi + [A_i,\phi]$. Isnt the sign of the second term wrong?

Secondly, he says differentiation wrt t gives us, $D^0 D_0 \phi = \epsilon^2 a_i D_i \phi$. Shouldnt it be $\partial^0D_0 \phi$? Cause we are taking the actual derivative wrt t rather than the covariant derivative, WE should get some extra terms, do they cancel out? How does the minus sign disappear?

Does the covariant derivative behave like a normal derivative in any case?

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  • $\begingroup$ OK. I got how the minus sign disappears, he has taken the metric signature as -+++ apparently $\endgroup$
    – user7757
    Commented Jul 17, 2012 at 5:37
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    $\begingroup$ I like how 't Hooft's sirname becomes two Hooft in the question(v1). $\endgroup$
    – Qmechanic ♦
    Commented Jul 17, 2012 at 9:12

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The sign of the gauge part of the covariant derivative is a convention, you can choose it any way you want, it just defines the sign of A. This sign has nothing to do with the metric convention, mostly + or mostly -. Its arbitrary in either convention.

The second part is just differentiating both sides of the previous equation for $D_0\phi$, there's a t in the right hand side. So it's $\partial_0 (t D_0 \phi)$, since A_0 is infinitesimal and gives a higher order correction, and he keeps the first part of this, where you differentiate t with respect to t, and ignores the second part, since time derivatives of $\phi$ are small by assumption that the monopole is stationary at t=0 and slowly accelerating.

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  • $\begingroup$ Hey, thanks a lot for the answer, but in the definition of covariant derivative he has taken a plus sign, so his covariant derivative is $D_\mu \phi = \partial_\mu \phi+ [A_\mu,\phi]$. See equation 2.3. As regards my comment, I meant I have resolved the sign problem for $D^0 D_0$ where the negative sign disappears because we are taking the contravariant indices. $\endgroup$
    – user7757
    Commented Jul 17, 2012 at 5:50
  • $\begingroup$ @ramanujan_dirac: I couldn't follow which sign was giving you trouble. I can't access the paper (it's paywalled). It's unfortunate, because it's a classic, and I haven't read it. $\endgroup$
    – Ron Maimon
    Commented Jul 17, 2012 at 6:07
  • $\begingroup$ Manton defines $D_\mu \phi = \partial_\mu \phi + e[A_\mu,\phi]$, and $A_0=\epsilon^2 a_i t A_i$, and $\partial_0 \phi = \epsilon^2 a_i t \partial_i \phi $, then how is $D_0 \phi = -\epsilon^2 a_i t D_i \phi$? Is there some resolution to this doubt or is there any typo in the paper (This is the first research paper I have read, so I don't know how common it is to have a typo). $\endgroup$
    – user7757
    Commented Jul 17, 2012 at 12:22
  • $\begingroup$ @ramanujan_dirac: In classic papers, it is vanishingly rare that there are typos that substantially affect the result in, but there are occasional minor typos (maybe 1 eqn in 500). But to understand it, you can just pretend that it is full of typos and you are proofreading it. A reseach paper leaves easoning steps that are obvious to the author and referee to the reader, here, it's going back and forth between the time covariant derivative and the time ordinary derivative. This is higher order term in the acceleration, since $A_0$ is infinitesimal for infinitesimal a. $\endgroup$
    – Ron Maimon
    Commented Jul 17, 2012 at 14:42
  • $\begingroup$ apparently there is some flaw in that paper, I have not been completely able to understand it(its related to the smoothness of elliptical functions requiring some functions for the acceleration and gauge fields something like that ) which was pointed out here: prd.aps.org/abstract/PRD/v18/i2/p542_1 $\endgroup$
    – user7757
    Commented Jul 24, 2012 at 6:55

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