The distance between point P_{1}(\rho_{1},\ \theta_{1},\ \phi_{1}) and P_{2}(\rho_{2},\ \theta_{2},\

oliviayychengwh 2021-12-21 Answered
The distance between point \(\displaystyle{P}_{{{1}}}{\left(\rho_{{{1}}},\ \theta_{{{1}}},\ \phi_{{{1}}}\right)}\) and \(\displaystyle{P}_{{{2}}}{\left(\rho_{{{2}}},\ \theta_{{{2}}},\ \phi_{{{2}}}\right)}\) given in spherical coordinates extract the formula.

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Expert Answer

Cheryl King
Answered 2021-12-22 Author has 4767 answers

Step 1
Given point \(\displaystyle{P}_{{{1}}}{\left(\rho_{{{1}}},\ \theta_{{{1}}},\ \phi_{{{1}}}\right)},\ {P}_{{{2}}}{\left(\rho_{{{2}}},\ \theta_{{{2}}},\ \phi_{{{2}}}\right)}\)
Let corresponding coutesian cordinates are,
\(\displaystyle{P}_{{{1}}}{\left({x}_{{{1}}},\ {y}_{{{1}}},\ {z}_{{{1}}}\right)}\ \&\ {P}_{{{2}}}{\left({x}_{{{2}}},\ {y}_{{{2}}},\ {z}_{{{2}}}\right)}\)
Relation between coutesian cordinates and sherical coordinates are
\(\displaystyle{x}_{{{1}}}=\rho_{{{1}}},\ {{\sin{\theta}}_{{{1}}},}\ {{\cos{\phi}}_{{{1}}}}\)
\(\displaystyle{y}_{{{1}}}=\rho_{{{1}}},\ {{\sin{\phi}}_{{{1}}},}\ {{\sin{\theta}}_{{{1}}}}\)
\(\displaystyle{z}_{{{1}}}=\rho_{{{1}}},\ {{\cos{\theta}}_{{{1}}}}\)
\(\displaystyle{x}_{{{2}}}=\rho_{{{2}}},\ {{\sin{\theta}}_{{{2}}},}\ {{\cos{\phi}}_{{{2}}}}\)
\(\displaystyle{y}_{{{2}}}=\rho_{{{2}}}{{\sin{\theta}}_{{{2}}}\ }{{\sin{\phi}}_{{{2}}}}\)
\(\displaystyle{z}_{{{2}}}=\rho_{{{2}}}\ {{\cos{\theta}}_{{{2}}}}\)
Distance between paints \(\displaystyle{P}_{{{1}}}\ \&\ {P}_{{{2}}}\)
\(\displaystyle{d}=\sqrt{{{\left({x}_{{{1}}}-{x}_{{{2}}}\right)}^{{{2}}}+{\left({y}_{{{1}}}-{y}_{{{2}}}\right)}^{{{2}}}+{\left({z}_{{{1}}}-{z}_{{{2}}}\right)}^{{{2}}}}}\)
\(\displaystyle=\sqrt{{{{x}_{{{1}}}^{{{2}}}}+{x}_{{{2}}}-{2}{x}_{{{1}}}{x}_{{{2}}}+{{y}_{{{1}}}^{{{2}}}}+{{y}_{{{2}}}^{{{2}}}}-{2}{y}_{{{1}}}{y}_{{{2}}}+{{z}_{{{1}}}^{{{2}}}}+{{z}_{{{2}}}^{{{2}}}}-{2}{z}_{{{1}}}{z}_{{{2}}}}}\)
\(\displaystyle=\sqrt{{{\left({{x}_{{{1}}}^{{{2}}}}+{{y}_{{{1}}}^{{{2}}}}+{{z}_{{{1}}}^{{{2}}}}\right)}+{\left({{x}_{{{2}}}^{{{2}}}}+{{y}_{{{2}}}^{{{2}}}}+{{z}_{{{2}}}^{{{2}}}}\right)}-{2}{\left({x}_{{{1}}}{x}_{{{2}}}+{y}_{{{1}}}{y}_{{{2}}}+{z}_{{{1}}}{z}_{{{2}}}\right)}}}\)
But \(\displaystyle{{x}_{{{1}}}^{{{2}}}}+{{y}_{{{1}}}^{{{2}}}}+{{z}_{{{1}}}^{{{2}}}}={\rho_{{{1}}}^{{{2}}}}\)
and \(\displaystyle{{x}_{{{2}}}^{{{2}}}}+{{y}_{{{2}}}^{{{2}}}}+{{z}_{{{2}}}^{{{2}}}}={\rho_{{{2}}}^{{{2}}}}\)
and also put the values of \(\displaystyle{x}_{{{1}}},\ {x}_{{{2}}},\ {y}_{{{1}}},\ {y}_{{{2}}},\ {z}_{{{1}}},\ {z}_{{{2}}}\)
we get,
\(\displaystyle{d}=\sqrt{{{\rho_{{{1}}}^{{{2}}}}+{\rho_{{{2}}}^{{{2}}}}-{2}{x}{\left(\rho_{{{1}}}\ {{\sin{\theta}}_{{{1}}}\ }{\cos{\phi}}_{{{1}}}\right)}{\left(\rho_{{{2}}}\ {{\sin{\theta}}_{{{2}}}\ }{\cos{\phi}}_{{{2}}}\right)}}}\)
\(\displaystyle=\sqrt{{-{2}{\left(\rho_{{{1}}}{{\sin{\theta}}_{{{1}}}{\sin{\phi}}_{{{1}}}}\right)}{\left(\rho_{{{2}}}{{\sin{\theta}}_{{{2}}}{\sin{\phi}}_{{{2}}}}\right)}-{2}{\left(\rho_{{{1}}}{\cos{\theta}}_{{{1}}}\right)}{\left(\rho_{{{2}}}{\cos{\theta}}_{{{2}}}\right)}}}\)
\(\displaystyle{d}=\sqrt{{{\rho_{{{1}}}^{{{2}}}}+{\rho_{{{2}}}^{{{2}}}}-{2}\rho_{{{1}}}\rho_{{{2}}}{{\sin{\theta}}_{{{1}}}{{\sin{\theta}}_{{{2}}}{\left({{\cos{\phi}}_{{{1}}}{{\cos{\phi}}_{{{2}}}+}}{{\sin{\phi}}_{{{1}}}{\sin{\phi}}_{{{2}}}}\right)}}}-{2}\rho_{{{1}}}\rho_{{{2}}}{{\cos{\theta}}_{{{1}}}{\cos{\theta}}_{{{2}}}}}}\)
Step 2
\(\displaystyle\Rightarrow{d}=\sqrt{{{\rho_{{{1}}}^{{{2}}}}+{\rho_{{{2}}}^{{{2}}}}-{2}\rho_{{{1}}}\rho_{{{2}}}{{\sin{\theta}}_{{{1}}}{{\sin{\theta}}_{{{2}}}{\cos{{\left(\phi_{{{1}}}-\phi_{{{2}}}\right)}}}}}-{2}\rho_{{{1}}}\rho_{{{2}}}{{\cos{\theta}}_{{{1}}}\times}{\cos{\theta}}_{{{2}}}}}\)
\(\displaystyle{\left\lbrace∵{\cos{{A}}}{\cos{{B}}}+{\sin{{A}}}{\sin{{B}}}={\cos{{\left({A}-{B}\right)}}}\right\rbrace}\)

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ambarakaq8
Answered 2021-12-23 Author has 1239 answers
Assuming that \(\displaystyle\theta\) is the latitude and \(\displaystyle\phi\) is the longitude we have that the cartesian coordinates of the first point are:
\(\displaystyle{\left(\rho_{{{1}}}{{\cos{\theta}}_{{{1}}}{{\cos{\phi}}_{{{1}}},}}\ \rho_{{{1}}}{{\cos{\theta}}_{{{1}}}{{\sin{\phi}}_{{{1}}},}}\ \rho_{{{1}}}{\sin{\theta}}_{{{1}}}\right)}\)
so the distance between the two points is given by:
\(\displaystyle\sqrt{{{\rho_{{{1}}}^{{{2}}}}+{\rho_{{{2}}}^{{{2}}}}-{2}\rho_{{{1}}}\rho_{{{2}}}{\left({{\cos{\theta}}_{{{1}}}{{\cos{\theta}}_{{{2}}}{\cos{{\left(\phi_{{{1}}}-\phi_{{{2}}}\right)}}}}}+{{\sin{\theta}}_{{{1}}}{\sin{\theta}}_{{{2}}}}\right\rbrace}}}\)
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nick1337
Answered 2021-12-28 Author has 9467 answers

Step 1
The expression of the distance between two vectors in spherical coordinates provided in the other response is usually expressed in a more compact form that is not only easier to remember but is also ideal for capitalizing on certain symmetries when solving problems.
\(||r-r'||=\sqrt{(x-x')^{2}+(y-y')^{2}+(z-z')^{2}}\)
\(=\sqrt{r^{2}+r'^{2}-2rr'[\sin(\theta)\sin(\theta')\cos(\phi)\cos(\phi')+\sin(\theta)\sin(\theta')\sin(\phi)\sin(\phi')+\cos(\theta)\cos(\theta')]}\)
\(=\sqrt{r^{2}+r'^{2}-2rr'[\sin(\theta)\sin(\theta')(\cos(\phi)\cos(\phi')+\sin(\phi)\sin(\phi'))+\cos(\theta)\cos(\theta')]}\)
\(=\sqrt{r^{2}+r'^{2}-2rr'[\sin(\theta)\sin(\theta')\cos(\phi-\phi')+\cos(\theta)\cos(\theta')]}\)
This form makes it fairly transparent how azimuthal symmetry allows you to automatically eliminate some of the angular dependencies in certain integration problems
Another advantage of this form is that you now have at least two variables, namely \(\phi\ and\ \phi'\),  that appear in the equation only once, which can make finding series expansions w.r.t. these variables a little less of a pain than the others.

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