2

I have a Feynman diagram like the following:

The red lines on the right connecting the final states overlap with the black lines of the Feynman diagram. If the lowermost and uppermost black lines of the t quarks were extended beyond the extent of the rightmost black lines of the b quarks, then the overlapping would be lessened and the diagram made possibly clearer.

How could this be done?

\documentclass{article}

\usepackage{feynmp}
\usepackage{feynmp-auto}

\unitlength=1.00 mm

\begin{document}

\begin{figure}

\begin{fmffile}{gg}
\begin{fmfchar*}(100,70)

\fmfleftn{i}{2}
\fmfrightn{o}{4}

\fmf{curly}{i1,v1}
\fmf{curly}{i2,v2}
\fmf{fermion}{o1,v1}
\fmf{fermion}{v1,v3}
\fmf{fermion}{v3,v2}
\fmf{fermion}{v2,o4}

\fmf{curly, label=\(g\)}{v3,v4}
\fmf{fermion}{o2,v4}
\fmf{fermion}{v4,o3}

\fmf{double,fore=red}{o1,o2}
\fmf{double,fore=red}{o1,o3}
\fmf{double,fore=red}{o1,o4}
\fmf{double,fore=red}{o2,o3}
\fmf{double,fore=red}{o2,o4}
\fmf{double,fore=red}{o3,o4}

\fmflabel{\(g\)}{i1}
\fmflabel{\(g\)}{i2}
\fmflabel{\(\bar{t}\)}{o1}
\fmflabel{\(\bar{b}\)}{o2}
\fmflabel{\(b\)}{o3}
\fmflabel{\(t\)}{o4}

\end{fmfchar*}
\end{fmffile}
\end{figure}

\end{document}

1 Answer 1

4

You can shift the top quark anchor points to the right using \fmfforce. Adding

\fmfforce{xpart(vloc __o4)+50,ypart(vloc __o4)}{o4} % upper top quark
\fmfforce{xpart(vloc __o1)+50,ypart(vloc __o1)}{o1} % lower top antiquark

right after \fmfrightn{o}{4} results in

feynman

Note that xpart(...) returns the x coordinate of a point, and vloc switches from „vertex mode“ to „immediate mode“. Also note that vloc requires the point to be prefixed with __.

2
  • Just in case you wonder, the 50 are in pt units, but you can use others as well, e.g. 20mm. Sep 15, 2016 at 11:52
  • You can also specify the new x-coordinate using the built-in w-variable (which is the width of the original figure), e.g. \fmfforce{1.1*w,ypart(vloc __o4)}{o4} Sep 15, 2016 at 11:53

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