Spherical coordinates are also called spherical polar coordinates. The spherical polar coordinate system is denoted as (r, θ, Φ) which is mainly used in three dimensional systems. In three dimensional space, the spherical polar coordinate system is used for finding the surface area. Through these coordinates, three numbers are specified that is the radial distance, the polar angles, and the azimuthal angle. Also, these coordinates are determined by the help of Cartesian coordinates (x,y,z). The radius can be measured from a fixed origin, the polar angle can be measured from an imaginary point which is near to the exact location and lastly, the azimuthal angle passes through the origin on a reference plane. We can also call the radial distance as a radial coordinate apart from the radius. The polar angle can be mentioned as colatitude, zenith angle, normal angle, or inclination angle. ‘r’ is the radius of the system, ‘θ’ is an inclination angle and ‘Φ’ is azimuth angle.
The spherical coordinates with respect to the cartesian coordinates can be written as:
r = \[\sqrt{x^{2}+y^{2}+z^{2}}\]
Tan\[\theta\] = \[\frac{\sqrt{x^{2}+y^{2}}}{z}\]
Tan \[\varphi\] = \[\frac{y}{z}\]
The coordinates can be converted from cartesian to spherical as:
x = r sin θ cos Φ
y = r sin θ sin Φ
z = r cos θ
Example 1) Convert the point ( \[\sqrt{6}\], \[\frac{\pi}{4}\], \[\sqrt{2}\] )from cylindrical coordinates to spherical coordinates equations.
Solution 1) Now since θ is the same in both the coordinate systems, so we don’t have to do anything with that and directly move on to finding ρ.
\[\rho\] = \[\sqrt{r^{2}+z^{2}}\] = \[\sqrt{6 + 2}\] = \[\sqrt{8}\] = \[\sqrt[2]{2}\]
Finally, we will move on to finding φ and to do that we will have to use the conversion for either r or z. So let us use the conversion for z.
z = \[\rho\] cos\[\varphi\] => cos \[\varphi\] = \[\frac{z}{\rho }\] = \[\frac{\sqrt{2}}{\sqrt[2]{2}}\] => \[\varphi\] = cos-1\[\frac{1}{2}\] = \[\frac{\pi}{3}\]
Now, we can see that there can be many possible values of that will produce cos\[\varphi\] = \[\frac{1}{2}\]
However, φ is restricted to a range \[0\leq \varphi \leq \pi\] also because this seems to be the only possible value in that range.
Therefore, the spherical coordinates equations of this point are \[(\sqrt[2]{2}, \; \frac{\pi}{4}, \; \frac{\pi}{3})\]
Example 2) How can we find the kinetic energy in terms of (r, \[\theta\], \[\varphi\])?
Solution 2) According to the question, it says that we have to find kinetic energy in terms of spherical coordinates. The kinetic energy in terms of Cartesian coordinates is basically represented as:
T=1/2 m(x^{2}+y^{2}+z^{2}) (1)
Here, x, y, and z are the derivatives of z.
y and z with respect to time. Therefore, the cartesian coordinates x, y, and z in terms of r, \[\theta\], \[\varphi\] are:
x = r sin \[\theta\] cos\[\varphi\]
y = r \[\theta\] sin \[\varphi\]
z = r cos \[\theta\]
Now the derivatives of x, y, and z are:
x = \[\frac{dx}{dt}\] = r sin \[\theta\] cos \[\varphi\] + r cos \[\theta\] cos \[\varphi\] \[\theta\] - r sin \[\theta\] sin \[\varphi\] \[\varphi\]
y = \[\frac{dy}{dt}\] = r sin \[\theta\] sin \[\varphi\] + r cos \[\theta\] sin \[\varphi\] \[\theta\] + r sin \[\theta\] cos\[\varphi\]\[\varphi\]
z = \[\frac{dz}{dt}\] = r cos \[\theta\] - r sin \[\theta\]\[\theta\]
Where, r = \[\frac{dr}{dt}\], \[\theta\] = \[\frac{d\theta }{dt}\], \[\varphi\] = \[\frac{d\varphi }{dt}\]
This means that r, \[\theta\] and \[\varphi\] changes with time as the particle moves or changes its position with time. So now we will have to calculate for x2, y2 and z2 and then add them to get:
(x)2+(y)2+(z)2=r2\[\theta\]2+r2sin2\[\theta\] \[\varphi\]2
Now putting the equation (x)2+(y)2+(z)2 in equation 1, we will get the kinetic energy of the system in terms of r, \[\theta\] and\[\varphi\] .
Therefore, T = \[\frac{1}{2}\]m(r2 + r2\[\theta\]2 + r2sin2\[\theta\] \[\varphi\]2)
Example 3) Convert (3,4,7) from rectangular coordinates to spherical coordinates.
Solution 3) we will first use the formula for \[\varphi\]
\[\varphi\] = \[\sqrt{x^{2}+y^{2}+z^{2}}\] = \[\sqrt{74}\]
Next, we would find keeping in mind that cos \[\varphi\] = \[\frac{z}{p}\]
\[\varphi\] = cos-1\[\frac{z}{p}\] = cos-1\[\frac{7}{\sqrt{74}}\] = 0.62 radians
Finally, using the fact that cos\[\theta\] = \[\frac{x}{p sin\varphi }\] to find \[\theta\]
\[\theta\] = cos-1\[\frac{x}{p sin\varphi }\] = cos-1\[\frac{3}{\sqrt{74}sin0.62}\] = 0.93 radians
Therefore, the point in spherical coordinate is (\[\sqrt{74}\], 0.93, 0.62)
1. What Are The Applications Of Spherical Polar Coordinates?
The term spherical is drawn from the term sphere which means a geometrical object in 3-dimensional space. Therefore, spherical coordinates are generally easy and understandable when we deal with something that is somewhat spherical, for example, a ball or a planet, or maybe black holes, and even planetary objects. The basic example that we can give will be the use of spherical coordinates while dealing with the latitude and the longitude of our planet. There are other examples of the fields where Spherical coordinate or Polar coordinate systems are mostly used.
2. Who Brought The Term Polar Coordinate System To Light?
The polar coordinate system is a two-dimensional coordinate system that was invented in 1637 by a French Mathematician called René Descartes (1596–1650). Several decades later Descartes published his two-dimensional coordinate system. Sir Isaac Newton (1640–1727) worked and presented ten different coordinate systems. One of them was the polar coordinate system. Newton and others used the polar coordinate system for plotting a complex curve which is known as a spiral. It was the Swiss Mathematician known as Jakob Bernoulli (1654–1705) who first used a polar coordinate system for a very wider array of calculus problems. And also coined terms such as "pole" and "polar axis" that we still use today in polar coordinate systems.
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