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Hints: Real Gaussian integral: $$ \tag{1} \int_{\mathbb{R}^n} \! d^n x ~e^{- x^t M x} ~=~ \sqrt{\frac{(2\pi)^n}{\det (M+M^t)}}.$$ Determinant of block matrices: $$ \tag{2} \det\begin{pmatrix} A & B \\ C & D \end{pmatrix} ~=~\det(A) \det(D-CA^{-1}B)~=~ \det(D) \det(A-BD^{-1}C). $$


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Although I think the question is related more to algorithm (as a branch of programming), but since it incorporates aspects of physics, I will give it a go. OP, know that there is not one answer to your problem. It can be accomplished in several ways. Here I would introduce you to some important variables in your scenario: The kinetic energy of the body in ...


5

There's really two questions here, one about the definition of distribution functions and one about the derivation of the BBGKY hierarchy. I will address them in turn. Definitions Let's define, for convenience, $\mathbf{r}^n = \mathbf{r}_1,\dots,\mathbf{r}_n$ and $\mathbf{r}^{(N-n)} = \mathbf{r}_{n+1},\dots,\mathbf{r}_N$. Next, let's denote the ...


2

$$δ^3(q⃗ )=\frac{δ(q)δ(\theta)}{2\pi q^2\sin(\theta)}$$ is wrong. The delta function is spherically symmetric, and thus has no θ dependence. Simply use: $$d^3(q⃗ )=\frac{δ(q)}{2\pi q^2}$$ instead. Use the Jacobian when you switch coordinate systems (from Cartesian to spherical) ($r^2 \sin(\theta)$), and you should get the result.


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Hints: In mathematics, a distribution is usually only defined wrt. smooth testfunctions. However the function ${\bf q}\mapsto({\bf q}\cdot{\bf p})^2/q^2$ is not continuous at the origin ${\bf q}={\bf 0}$. Nevertheless, we can e.g. try to evaluate the triple integral using the following representation of the 3D Dirac delta distribution $$\tag{1} ...


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First go to spherical coordinates: $$ A(t) = \frac{1}{2\pi^3}\int_0^{\infty}\text{d} p\int_0^{2\pi}\text{d} \phi \int_0^{\pi}\text{d}\theta\text{ } p^2 \sin\theta e^{-it\sqrt{p^2+m^2}} e^{ip\cos\theta|r-r_0|}, $$ and perform the trivial integral over $\phi$. Subsequently substitute $y=\cos\theta$ such that $\text{d}y=\sin\theta\text{d}\theta$ and integrate ...


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I suppose your title would rather be ; "... if not, why ?". In momentum space (which I assume is the focus of your question), these do not appear; the integrals are made of Feynman propagators $(p^2+m^2+i\epsilon)$ in the denominator and in the numerator you find various tensor contractions of your loop momentum (or momenta) with the external momenta / ...



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