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Good day everyone,

I am reading a paper and trying to reproduce some of the results in there. I do have to say this is a physics paper, so I am essentially sorry if the question I'm asking might seem a bit weird, but I feel that it belongs here.

Essentially, I have derived such an expression:

$$\frac{\mathrm{d}L^{\mu\nu}(q,k_1,k_2)}{\mathrm{d}^4k_1\mathrm{d}^4k_2}=\left[k_1^{\mu}k_2^{\nu}+k_1^{\nu}k_2^{\mu}-g^{\mu\nu}(k_1\cdot k_2)\right]\delta^4(q-k_1-k_2)\frac{\delta((k_1^0)-|\vec{k_1}|)}{2k_1^0}\frac{\delta((k_2^0)-|\vec{k_2}|)}{2k_2^0}$$

Where $k_1,k_2, q$ are four-vectors, $\vec{q}=0$, i.e. the $q^1,q^2,q^3$ components of q are zero, $g^{\mu\nu}=\mathrm{diag}(1,-1,-1,-1)$ is the metric tensor, $k_1\cdot k_2$ denotes the scalar product of the two 4-vectors, i.e. $g^{\mu\nu}k_{\mu}k_{\nu}$ and $k^{\mu}_{1,2}=\frac{Q}{2}(1,\pm\sin\theta\cos\phi,\pm\sin\theta\sin\phi,\pm\cos\theta)$, where $Q=\sqrt{q^2}$.

My goal is to get to $$\frac{\mathrm{d}L^{\mu\nu}(q,k_1,k_2)}{\mathrm{d}(\cos\theta)\mathrm{d}\phi}=?,$$

however, I am not sure how to approach this. I need to integrate over the 6 deltas that I have, obviously, but how does integration over deltas work with the index structure of my original equation? I think, what I should be left with is a very similar expression with some overal factor in front. But I am not sure how do I get there. This shouldn't be hard, assuming the deltas allow me to evaluate the integrals rather trivially. Thank you for any input.

Henrikas
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  • Is $\delta$ the delta distribution? And what does the $\delta^4$ mean exactly? – Botond Jul 31 '18 at 14:22
  • Yes $\delta$ is the Dirac delta. As for $\delta^4$, sorry for not making it clear, it's used in physics as a notation, meaning the delta works in all 4 dimensions (of time and space). Namely $\delta^4(x-y)=\delta(x^0-y^0)\delta(x^1-y^1)...\delta(x^3-y^3)$. I'm sorry if the notation seems a bit clunky, but that's common to use in QFT. – Henrikas Jul 31 '18 at 14:26
  • Most of your notations are familiar from special relativity, but not everything. Also, what does the $\frac{\mathrm{d}L^{\mu\nu}(q,k_1,k_2)}{\mathrm{d}^4k_1\mathrm{d}^4k_2}$ mean? Mostly the $d^4$; and are you doing the derivation with respect to 2 vectors? – Botond Jul 31 '18 at 14:36
  • Yeah, the $d^4x$ notation is similar to delta I explained in previous comment; if you have $d^4y$ that's like saying $dy^0dy^1dy^2dy^3$. A quick google found me this: https://math.stackexchange.com/questions/426263/meaning-of-int-mathop-mathrmd4x so maybe that can help you understand the idea better – Henrikas Jul 31 '18 at 14:43
  • So it is the $\frac{\partial^8}{\partial k_1^0 \partial k_1^1 \dots \partial k_2^3 \partial k_2^4}=\partial_{k_1^0} \partial_{k_1^1} \dots \partial_{k_2^3} \partial_{k_2^4}$ operator, right? (If I didn't mess up the position of any index) – Botond Jul 31 '18 at 14:45
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    The expression $\frac{\mathrm{d}L^{\mu\nu}(q,k_1,k_2)}{\mathrm{d}^4k_1\mathrm{d}^4k_2}$ is not that wrong as it can be seen as a Radon-Nikodym derivative and the derivative fraction notation is used for such. – md2perpe Jul 31 '18 at 14:56
  • @Botond. No, $\frac{\mathrm{d}L^{\mu\nu}(q,k_1,k_2)}{\mathrm{d}^4k_1\mathrm{d}^4k_2}$ is a density with respect to two volume measures $d^4 k_1$ and $d^4 k_2,$ which actually are just Lebesgue (or Jordan) measures at $k_1$ and $k_2$ respectively. That is, $$L_{\Omega_1\times\Omega_2}^{\mu\nu}(q) = \iint_{\Omega_1\times\Omega_2} \frac{\mathrm{d}L^{\mu\nu}(q,k_1,k_2)}{\mathrm{d}^4k_1\mathrm{d}^4k_2} , d^4 k_1 , d^4 k_2.$$ – md2perpe Jul 31 '18 at 15:03
  • @md2perpe I believe you, it was just really strange to see $\mathrm{d}$ instead of $\partial$, and "a different amount of $\mathrm{d}$s (1 vs 8)". – Botond Jul 31 '18 at 15:06
  • @Botond. I can understand that. Maybe the best would have been to write $\mathrm{d}^4\mathrm{d}^4 L,$ but that also looks a bit silly. – md2perpe Jul 31 '18 at 15:12
  • @Henrikas. I'm not sure if I get one thing: on the one hand you have a density with respect to two four-space variables, but you want it expressed as a density with respect to only two real variables ($\theta$ and $\phi$). I understand that the integrand might depend only on the distance between $\vec k_1$ and $\vec k_2,$ but still I want some more explanation of the parametrization $k^{\mu}_{1,2}=\frac{Q}{2}(1,\pm\sin\theta\cos\phi,\pm\sin\theta\sin\phi,\pm\cos\theta).$ – md2perpe Jul 31 '18 at 15:22
  • The idea is that an intermediate particle with virtual 4-momentum q decays into the two final state particles with 4-momenta $k_1$ and $k_2$ (specifically - leptons) in their center-of-mass frame. Essentially, it is enough to know the two angles $\phi, \theta$ in such a frame to be able to fully understand where the particles are going. Does that make things clearer? – Henrikas Jul 31 '18 at 15:30

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