Partial differentiation with respect to 4 vector

Question

I am trying to write for the following differentiation $$ \frac{\partial}{\partial q^2}\frac{1}{k^2(k-q)^2} \quad\text{where $q^2=q^{\mu}q_{\mu}$}\tag{1} $$ This can be written as $$ \frac{q^{\mu}}{2q^2}\frac{\partial}{\partial q^{\mu}}\frac{1}{k^2(k^{\mu}k_{\mu}-2q^{\mu}k_{\mu}+q^{\mu}q_{\mu})} $$ And the differentiation leads to $$ \frac{q^{\mu}}{2q^2} \frac{2(k_{\mu}-q_{\mu})}{k^2(k-q)^4}=\frac{1}{q^2}\frac{(k\cdot q-q^2)}{k^2(k-q)^4} $$ I tried to use the partial differentiation command

\[PartialD]_q^2 F(q^2)

But that obviously gives wrong result.

Answer 1

I'll be using mostly minus signature here. If you are using mostly plus, just adapt the functions accordingly. First, $\frac{1}{k^2(k-q)^2}$ is not a function of $q^2$ only, it is really a function of the whole four-vector $q$ (because of the mixed term $k_\mu q^\mu$).

If you don't really need to use complex tensor structures in your computations, you don't need to learn a new package for now (it would be useful for the future though), you can just define the Minkowskian product by yourself as

MinkProd[q1_List, q2_List] := 2 q1[[1]] q2[[1]] - q1.q2;

Now you can define kvec = {k[0], k[1], k[2], k[3]};, then the function you want to take the derivative of is

F[q_List] := 1/(MinkProd[kvec, kvec] MinkProd[kvec - q, kvec - q]);

If qvec={q[0], q[1], q[2], q[3]};, the derivative $\partial_{q^i}$ is just D[F[qvec],q[i]], then the result you want in terms of its components is just

qvec.Table[D[F[qvec], q[i]], {i, 0, 3}]/(2 MinkProd[qvec, qvec])

Notice that the scalar product in the expression above is just the dot product between the two lists since the Table gives you a covariant vector as output. MinkProd should only be used with two covariant or two contravariant vectors.

A little extra: if you want to convert a contravariant list into covariant or vice-versa, use

ConvertIndex[q_List] := Flatten@{q[[1]], Table[-q[[i]], {i, 2, Length[q]}]}