# Create bremsstrahlung feynman diagram with feyngraph

I'm using Latexit to create feynman diagrams and I'd like to create a bremsstrahlung diagram similar to this one but I don't know how to go about getting the nucleus to be vertically below one vertex, or how to have the A' in the right position.

So far all I've got is this:

 \begin{fmffile}{feyngraph}
\begin{fmfgraph*}(60,50)
\fmfleft{ie}
\fmflabel{$e$}{ie}
\fmfright{o1,o2}
\fmf{plain, tension=0.5}{ie,b}
\fmf{plain, tension=0.5}{b,o2}
\fmf{photon, tension=0.5}{b,o1}
\end{fmfgraph*}
\end{fmffile}


Any help will be appreciated

feynmf uses some algorithms to determine a reasonable layout for the diagram. You can still create horizontal lines by adding some auxiliary (phantom) paths and vertices.

\documentclass{article}
\usepackage[force]{feynmp-auto}
\begin{document}
\begin{fmffile}{feyngraph}
\begin{fmfgraph*}(60,60)
\fmfleft{l0,l1,ie,l2}
\fmfright{o0,o1,o2,o3}
\fmf{plain}{ie,v1}
\fmf{phantom}{v1,a1,o2}
\fmf{phantom}{l1,a2,a3,o1}
\fmf{phantom}{l2,a4,a5,o3}
\fmf{phantom}{l0,a6,a7,o0}
\fmffreeze
\fmf{plain}{v1,a3,o1}
\fmf{photon}{v1,a5}
\fmf{photon}{a3,a7}
\fmfv{label={$e^\pm$},label.angle=90}{ie}
\fmfv{label={$e^\pm$},label.angle=90}{o1}
\fmflabel{$A'$}{a5}
\fmfv{decor.shape=cross}{a7}
\end{fmfgraph*}
\end{fmffile}
\end{document} • That's extremely useful, I added an extra phantom line between the outgoing e and the nucleus just to get that photon line longer and I changed the cross to be a blob \fmfblob{.1w}{a7}. The only thing I'd now like to change is to move the label for the nucleus so that it doesn't overlap with the blob. I'm currently using \fmfv{label={N},label.angle=10}{a7} but I can't find an angle that prevents the overlap May 15 at 13:40
• @BethLong You can also play with label.distance.
– user242026
May 15 at 15:54
• \fmfv{label={N},label.distance=100}{a7} changing the distance isn't moving the label... May 16 at 9:18
• for future reference: it's label.dist not label.distance eg \fmfv{label={N},label.angle=0,label.dist=15}{a7} May 16 at 9:32