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I'm trying to derive the D-dimensional laplacian over Euclidean space for a function $f$ that is invariant under D-dimensional Euclidean rotations.

Specifically i'm trying to go from the equation

$$
Delta_D phi = U'(phi)
$$

to the equation

$$
fracd^2phidr^2+fracD-1rfracdphidr = U'(phi)
$$

where $r = (x_1^2+x_2^2+x_3^2+dots+x_D^2)^1/2$. I'm a physicist and currently I don't have much knowledge about differential geometry and operators over manifolds, but still i wanted to know how, in a rigorous manner, to derive that equation under that change of coordinates.

Searching on the internet i found that the general form for the laplacian is given by the Laplace-Beltrami operator

$$
Delta_D phi= frac1sqrtdet gpartial_ileft(sqrtdet gg^ijpartial_jphiright)
$$

but i don't know how the metric tensor is in the coordinates specified. I know that in Euclidian space is just a Kroneker delta, but what about spherical coordinates?
On the Wikipedia article gives the result right away without any explanation.

I'm really curious to see how to derive it. I didn't find any explanation with computation in my google search.

Since you're a physicist, I thought a variational derivation might be appealing. Consider the functional $$F[phi]=int left[frac12(nabla phi)^2+U(phi)right] dV.$$ Then the functional derivative is $$fracdeltadelta phiF[rho]=fracpartial Upartial phi-nabla^2 phi =U'(phi)-Delta_D phi,$$ so your first equation is equivalent to $delta F[phi]/delta phi=0$. But $phi$ is a function of $r$ alone, so we can integrate out the irrelevant angular degrees of freedom to obtain $$F[phi]=C_Dint left[frac12(partial_r phi)^2+U(phi)right] r^D-1dr$$ where $C_D$ is some $D$-dependent multiplicative constant. We may then compute the functional derivative once again as
$$dfracdeltadelta phiF[phi]=C_n left[U'(phi)r^D-1-partial_r(r^D-1 partial_rphi)right]=C_n r^D-1left[U'(phi)-fracD-1rfracdphidr-fracd^2phidr^2 right].$$
For this variational derivative to vanish again, we must therefore have $$Delta_D phi = U'(phi) = fracd^2phidr^2+fracD-1rfracdphidr$$ which was the result to be derived.

Here is a longer derivation that requires mostly basic calculus.

The Laplacian is invariant under rotations. Specifically, if $f:mathbb R^nrightarrow mathbb R^n$ and $nabla^2 f(x);=alpha$, then $nabla^2 g;
(R^-1 x)=alpha$ where $R$ is any rotation matrix ($R^TR=I$ and $det(R)=1$) and $g(x) := f(Rx)$.

Now suppose that $f:mathbb R^nrightarrow mathbb R^n$ has the property that $f(x)= h(||x||)$ where $h:mathbb [0,infty) rightarrow mathbb R$, $h$ is twice differentiable, and $||x||=sqrtsum_i x_i^2.$

At $x$, we can form an orthonormal basis $x$. Now
$$
nabla^2 f(x) = sum_i=1^n fracpartial^2 fpartial x_i^2(x)
$$
and by invariance under rotation, we can use the basis $x$ to get
beginequation
(1)quadquadnabla^2 f(x)= h''(||x||) + sum_i=2^n p_i
endequation
where $p_i:= k_i''(0)$ and $k_i(epsilon) := f(x+epsilon v_i)$.

The fact that $f$ only depends on the radius implies that all of the $k_i$ are the same and for $i=2, 3, ldots n$,
$$f(x+epsilon v_i) = k_i(epsilon) = k(epsilon) = f(x+ epsilon v_2)=h(sqrt).$$

Let $r=||x||$. Using calculus,
$$k'(epsilon) = h'(sqrtr^2+epsilon^2)fracepsilonsqrtr^2+epsilon^2, quad mathrmand $$
$$k''(epsilon) = h''(sqrtr^2+epsilon^2)fracepsilon^2r^2+epsilon^2 + h'(sqrtr^2+epsilon^2)fracsqrtr^2+epsilon^2 -epsilon fracepsilonsqrtr^2+epsilon^2 r^2+epsilon^2.$$
Thus $p_i:= k''(0) = h''(r)cdot 0 + h'(r)fracrr^2=h'(r)/r$. Applying this to equation (1) gives
$$
nabla^2f(x)= h''(||x||) + sum_i=2^n p_i = h''(r) + (n-1)frach'(r)r.
$$