Form of hadronic tensor as constrained by current conservation While studying deep inelastic scattering of an electron off a proton, I need to determine the most general form of the hadronic tensor, such as in this presentation, at page 25: it is a tensor that describes the interaction of the proton with an off-shell virtual photon, resulting in a final (undetermined) state composed of multiple hadrons.
The linked source, and many others, state that such most general form is
\begin{equation}
W(p,q)^{\mu\nu}=
w_1\biggl(\eta^{\mu\nu}-\frac{q^\mu q^\nu}{q^2}\biggr)+
w_2\biggl(p^\mu-\frac{p\cdot q}{q^2}q^\mu\biggr)\biggl(p^\nu-\frac{p\cdot q}{q^2}q^\nu\biggr).
\tag{1}
\label{source}
\end{equation}
where $w_1$ and $w_2$ are some scalar functions of $p$ and $q$.
I tried to derive this formula, but I was not able to: my proof starts assuming that $W(p,q)$ is a rank-2 contravariant tensor, so that I can write it as
\begin{equation}
W(p,q)^{\mu\nu}=
a_1\eta^{\mu\nu}
+a_2p^\mu p^\nu
+a_3p^\mu q^\nu
+a_4q^\mu p^\nu
+a_5q^\mu q^\nu,
\end{equation}
the $a_i$ coefficients being scalar functions of $p$ and $q$, since $p$ and $q$ are the only vectors "available" to construct something with two upper indices, other than the metric tensor $\eta$.
Then current conservation implies $q_\mu W^{\mu\nu}=0$ and $q_\nu W^{\mu\nu}=0$ which in turn forces the coefficients to satisfy the equations
\begin{equation}
\begin{cases}
a_1+a_3(p\cdot q)+a_5 q^2=0,\\
a_2(p\cdot q)+a_4 q^2=0,\\
a_1+a_4(p\cdot q)+a_5 q^2=0,\\
a_2(p\cdot q)+a_3q^2=0.
\end{cases}
\end{equation}
Leaving $a_3$ and $a_5$ as free solutions, I get
\begin{equation}
W(p,q)^{\mu\nu}=
a_5(q^\mu q^\nu-q^2\eta^{\mu\nu})+a_3\Bigl[p^\mu q^\nu+q^\mu p^\nu-(p\cdot q)\eta^{\mu\nu}-\frac{q^2}{p\cdot q}p^\mu p^\nu\Bigr].
\tag{2}
\label{my_result}
\end{equation}
Comparing \eqref{source} and \eqref{my_result} I see that $a_5$ may be set equal to $w_1/q^2$ to get the first term, but according to me there's no way to put the second term as in \eqref{source}: in particular, I believe the metric tensor prevents such a factorisation, which would happen if $\eta^{\mu\nu}$ was replaced by $q^\mu q^\nu/q^2$.
My question is, is there a way to get \eqref{source} with my proof?
 A: Okay, I found the solution, it's just a clever rearrangement of the terms in (2).
Define $w_1$ and $w_2$ by
\begin{equation}
a_3=-\frac{p\cdot q}{q^2}w_2,
\quad
a_5=-\frac{1}{q^2}w_1+\frac{(p\cdot q)^2}{(q^2)^2}w_2
\end{equation}
to transform (2) in
\begin{equation}
-\frac{p\cdot q}{q^2}w_2\Bigl[p^\mu q^\nu+q^\mu p^\nu-(p\cdot q)\eta^{\mu\nu}-\frac{q^2}{p\cdot q}p^\mu p^\nu\Bigr]
+\Bigl[-\frac{1}{q^2}w_1+\frac{(p\cdot q)^2}{(q^2)^2}w_2\Bigr](q^\mu q^\nu-q^2\eta^{\mu\nu})=\\
w_2\Bigl[-\frac{p\cdot q}{q^2}(p^\mu q^\nu+q^\mu p^\nu)+p^\mu p^\nu+\frac{(p\cdot q)^2}{q^2}\eta^{\mu\nu}\Bigr]
+w_1\Bigl(\eta^{\mu\nu}-\frac{q^\mu q^\nu}{q^2}\Bigr)
+w_2\Bigl[\frac{(p\cdot q)^2}{(q^2)^2}q^\mu q^\nu-\frac{(p\cdot q)^2}{q^2}\eta^{\mu\nu}\Bigr]=\\
w_2\Bigl[-\frac{p\cdot q}{q^2}(p^\mu q^\nu+q^\mu p^\nu)+p^\mu p^\nu+\frac{(p\cdot q)^2}{(q^2)^2}q^\mu q^\nu\Bigr]
+w_1\Bigl(\eta^{\mu\nu}-\frac{q^\mu q^\nu}{q^2}\Bigr)=\\
w_1\Bigl(\eta^{\mu\nu}-\frac{q^\mu q^\nu}{q^2}\Bigr)
+w_2\Bigl(p^\mu-\frac{p\cdot q}{q^2}q^\mu\Bigr)\Bigl(p^\nu-\frac{p\cdot q}{q^2}q^\nu\Bigr).
\end{equation}
