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When doing calculations of Feynman diagrams in QFT I've seen a trick used that goes something like this $$\int \frac{d^dk}{(2\pi)^d} \frac{k_\mu k_\nu}{f(k^2)}\quad\longrightarrow\quad\int \frac{d^dk}{(2\pi)^d} \frac{ k^2 \eta_{\mu\nu}/d}{f(k^2)}.$$ Sometimes also linear terms in $k_\mu$ would get dropped. Unfortunately, I haven't really seen an explanation for this...

I think the part about dropping linear terms has something to do with integrating an uneven function over a symmetric interval. But I really have no clue where the $k_\mu k_\nu\to k^2 \eta_{\mu\nu}/d$ comes from..

Qmechanic
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Sito
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2 Answers2

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The reason is really symmetry. I think considering the problem with $d=2$ should be elucidating. Additionally let us take $f(k)=e^{k^2}$, in general it should be some polynomial on $k^2$ (spherical symm.) such that the integral converges. Let us call the integral $I_{\mu\nu}$ and consider w.l.g. $\mu=1,\nu=2$, so let us see $$I_{12} = \frac{1}{4\pi^2}\int {\rm d} x \,{\rm d} y\, x y\,e^{-x^2-y^2} = 0$$

So indeed you see that what you get is an odd function integrated over the full range so, different indices will give you always zero.

The computation also checks out when the indices are the same, that I will leave you to check. Hint: $k^2$ overcounts directions... therefore the $1/d$.

So at the end of the day what we have is that such replacement is true, provided we have a spherically symmetric situation, that is the function $f$ must depend only on the norm square of $k$ and the integration must be trough out the whole volume. (And as always in physics... assuming the integral actually converges)

ohneVal
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If one replaces $k^2=k_0 k^0+k_1 k^1+\cdots$ by $k^\mu k^\nu$ in the numerator of your integral, the only components that give anything but $0$ are for $\mu=\nu$ and there is(are) sign(s) if you are in Minkowski space-time (depending on your choice of signature). Then, we have something proportional to the Minkowski metric. For the factor $d$ you just have to see that integrating over $k^2$ gives just $d$ times the integral over $k^a k^a$ (without summation !). That's all, this is just maths.

Jeanbaptiste Roux
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