1
$\begingroup$

I am slightly doubting something I have always thought obvious at the moment. Consider two vectors $\vec{a},\vec{b}$ in $\mathbb{R}^3$. We know that $\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\gamma$, where $\gamma$ is the angle between the two vectors. I use $\gamma$ to distinguish it from spherical coordinates. Clearly, in general $\gamma$ has nothing to do with polar or azimuthal angles $\theta,\phi$ in spherical coordinates, since in general $\gamma$ will depend on $\theta_a,\theta_b,\phi_a,\phi_b$, the angular coordinates of each vector.

Now suppose I wish to do integrals of the form \begin{align} I = \int d^3x\, f((\vec{x}-\vec{a})^2)\,. \end{align} One simple case would be e.g. when $f$ is Gaussian. In spherical coordinates, my volume element will be $d^3x = r^2\sin\theta\,dr\,d\theta\,d\phi$. Now, I always thought that I could now do the following: \begin{align} (\vec{x}-\vec{a})^2 = |\vec{x}|^2+|\vec{a}|^2-2|\vec{x}||\vec{a}|\cos\gamma \end{align} and the integral is straightforward. However, now this looks like I will have to need $\vec{a}$ to be along the $z$-axis in order for $\gamma$ to be the same as the $\theta$ in the volume element, which then can be straightforwardly integrated (crucially, it does not depend on $\phi$). Does that mean that if $\vec{a}$ is not along the $z$-direction, I cannot do this (i.e. $\phi$-independent integral)? The fact that the integral depends on the distance $|\vec{x}-\vec{a}|$ seems to me implying that I could always set $\gamma=\theta$ even if $\vec{a}$ is not along the $z$-axis.

$\endgroup$
1
  • 1
    $\begingroup$ If you are prepared to specify the domain of integration in a new coordinate system you can essentially re-coordinatize however you want. For example, you could use spherical coordinates in which the azimuthal angle measures declination relative to the vector $\vec{a}$, and define the polar angle accordingly. Or even better, you could use spherical coordinates centered around $\vec{a}$ so that $f((\vec{x}-\vec{a})^2)$ becomes $f(r^2)$. Both of these coordinatizations will use essentially the same volume element as spherical coordinates. $\endgroup$
    – subrosar
    Aug 11, 2019 at 5:33

1 Answer 1

1
$\begingroup$

$\gamma$ does depends on $(\theta,\phi)$ and, in the general case where $\vec{a}$ is not in the $z$-direction, you may not replace $\gamma$ with $\theta$.

To see so, suppose the orientation of $\vec{a}$ is $(\theta_a$, $\phi_a)$, you should decompose the vectors along three perpendicular directions and express the squared distance $d^2$ as,

$$(\vec{x}-\vec{a})^2=(r\sin\theta\cos\phi-a\sin\theta_a\cos\phi_a)^2 + (r\sin\theta\sin\phi-a\sin\theta_a\sin\phi_a)^2 + (r\cos\theta-a\cos\theta_a)^2 =d^2(r,\theta,\phi) $$

As a result, $\gamma$ depends on $\theta$ and $\phi$ according to, $$ \cos\gamma= \frac{d^2(r,\theta,\phi)-r^2-a^2}{2ar} $$

As seen from the expression, $\gamma$ depends on both $\theta$ and $\phi$ in a complex way. You may not simply replace $\gamma$ with $\theta$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .