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?



\unitlength=1.00 mm






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





1 Answer 1


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


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 __.

  • 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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