Distribution change of variables I'm working on distributions for a physics problem and I am quite stumped on how to proceed properly.
The problem is as follows: I start at the point $(0,0)$ and go a distance $\eta$ in x direction $(\eta, 0)$, from there I go in a random direction $\phi$ (uniformly distributed) a distance $x$ that is gaussian distributed. I need the distribution of $y$, with $y$ being the distance of the resulting point from $(0,0)$
To formalize that:
$f(x)=\frac{1}{b\sqrt{\pi}}\exp(-x^2/b^2)$
(is the distribution of the random x direction) but I also know that
$y = \sqrt{2 x \eta  \cos (\phi )+\eta
   ^2 + x^2}$
with $\phi$ being uniformly randomly distributed between $0, 2\pi$
Is it correct to say (with performing change of variables between x to y, and phi:
$g(y, \phi)\propto f(x)|\frac{dx}{dy}||\frac{dx}{d\phi}|$?
I'm pretty sure this is wrong (I suppose I need the Jacobian in some way), but I can't find good references for it

Let $(X,\Phi,Y)$ be the random variables in question.  Let's suppose $X$ and $\Phi$ are independent.  $Y$ is a function of those two,
$$Y = f_\eta(X,\Phi)=\sqrt{2X\eta\cos(\Phi) + \eta^2+X^2}$$
for some number $\eta \ge 0.$  (If $\eta \lt 0,$ negating both it and $X$ does not change the distribution of $X$ and places us into the $\eta \gt 0$ situation.)  That is, when $X\ge 0,$ $Y$ is the length of the third side of a triangle of side lengths $\eta$ and $X$ with included angle $\Phi;$ and when $X\le 0,$ $Y$ is the length of the third side of a triangle of side lengths $\eta$ and $-X.$  This shows the formula makes sense (we're not trying to find the root of a negative number) and that it is almost everywhere a two-to-one mapping, because
$$f_{\eta}(X,\Phi) = f_{\eta}(-X,\Phi+\pi).$$
The question appears to ask for the joint density function of $(Y,\Phi) = (f_\eta(X,\Phi), \Phi).$
We may easily write the joint density of $(X,\Phi),$ because independence implies the joint density is the product of the univariate densities.  The joint probability element therefore is (up to sign)
$$f_{X,\Phi}(x,\phi) = \frac{1}{b\sqrt{\pi}} e^{-x^2/b^2}\,\mathrm{d}x \frac{1}{2\pi}\,\mathrm{d}\phi.\tag{*}$$
To change the coordinates, notice the definition of $Y$ implies
$$y^2 - x^2 - \eta^2 - 2x\eta\cos(\phi)=0.$$
From this we obtain $x$ and $x^2$ (sort of, up to a choice of two solutions) as
$$x = -B(\phi,\eta) \pm \sqrt{\Delta(\phi,\eta,y)}\tag{1a}$$
and
$$x^2 = B(\phi,\eta)^2 + \Delta(\phi,\eta, y) \mp 2B(\phi,\eta)\sqrt{\Delta(\phi,\eta, y)}\tag{1b}$$
where $B(\phi,\eta) = \eta\cos(\phi)$ and $\Delta(\phi,\eta, y) = \eta^2\cos^2(\phi) + (y^2-\eta^2).$
Differentiating (with respect to the variables $(y,x,\phi)$) we also find
$$0 = \mathrm{d}\left(y^2 - x^2 - \eta^2 - 2x\eta\cos(\phi)\right) = 2y\mathrm{d}y - 2x\mathrm{d}x - 2\eta\cos(\phi)\mathrm{d}x + 2x\eta\sin(\phi)\mathrm{d}\phi.$$
Consequently
$$0 = 0\wedge \mathrm{d}\phi = 2y\mathrm{d}y\wedge\mathrm{d}\phi - (2x + 2\eta\cos(\phi))\mathrm{d}x\wedge\mathrm{d}\phi,$$
which we can solve (almost everywhere) to convert the original differential element $\mathrm{d}x\mathrm{d}\phi$ to the new variables,
$$\mathrm{d}x\mathrm{d}\phi = \frac{y}{x + \eta\cos(\phi)}\,\mathrm{d}y\mathrm{d}\phi.\tag{2}$$
Plugging  $(2)$ into $(*)$ gives an expression for the joint probability element $f_{Y,\Phi})(y,\phi)\mathrm{d}y\mathrm{d}\phi$ in which "$x$" and "$x^2$" can be replaced by expressions $(1a)$ and $(1b),$ respectively.
Finally, because the map from $(X,\Phi)$ to $(Y,\Phi)$ is two-to-one, we must double the preceding result: that's the answer.

I'm not going to work this out further for several reasons.  One is that the many-to-one transformation suggests the problem hasn't been stated quite as clearly as it ought to be.  Another is that in the question, the expression given for the density of $X$ is incorrect.  I have silently fixed it above, assuming the intention is that $X$ have a Normal distribution: but perhaps that's the wrong assumption.  A change in the density of $X$ would not change how the answer is worked out, but it would change the details of the final answer.  Finally, I suspect that this is just one step in the solution of a problem that, if it were disclosed, might be (much) simpler to solve some other way.  Analyzing the density of $(Y,\Phi)$ is not an appetizing prospect.
