area element in spherical coordinates

\underbrace {r \, d\theta}_{\text{longitude component}} *\underbrace {r \, \color{blue}{\sin{\theta}} \,d \phi}_{\text{latitude component}}}^{\text{area of an infinitesimal rectangle}} {\displaystyle (r,\theta ,-\varphi )} 32.4: Spherical Coordinates is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts. as a function of $\phi$ and $\theta$, resp., the absolute value of this product, and then you have to integrate over the desired parameter domain $B$. In baby physics books one encounters this expression. Another application is ergonomic design, where r is the arm length of a stationary person and the angles describe the direction of the arm as it reaches out. $$y=r\sin(\phi)\sin(\theta)$$ Coming back to coordinates in two dimensions, it is intuitive to understand why the area element in cartesian coordinates is \(dA=dx\;dy\) independently of the values of \(x\) and \(y\). the spherical coordinates. In the case of a constant or else = /2, this reduces to vector calculus in polar coordinates. Find \(A\). Connect and share knowledge within a single location that is structured and easy to search. , $$, So let's finish your sphere example. We already know that often the symmetry of a problem makes it natural (and easier!) That is, \(\theta\) and \(\phi\) may appear interchanged. The relationship between the cartesian coordinates and the spherical coordinates can be summarized as: (26.4.5) x = r sin cos . {\displaystyle (r,\theta ,\varphi )} The polar angle, which is 90 minus the latitude and ranges from 0 to 180, is called colatitude in geography. , . We see that the latitude component has the $\color{blue}{\sin{\theta}}$ adjustment to it. The radial distance r can be computed from the altitude by adding the radius of Earth, which is approximately 6,36011km (3,9527 miles). To apply this to the present case, one needs to calculate how Coming back to coordinates in two dimensions, it is intuitive to understand why the area element in cartesian coordinates is dA = dx dy independently of the values of x and y. The answer is no, because the volume element in spherical coordinates depends also on the actual position of the point. These coordinates are known as cartesian coordinates or rectangular coordinates, and you are already familiar with their two-dimensional and three-dimensional representation. If it is necessary to define a unique set of spherical coordinates for each point, one must restrict their ranges. We'll find our tangent vectors via the usual parametrization which you gave, namely, In each infinitesimal rectangle the longitude component is its vertical side. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. A spherical coordinate system is represented as follows: Here, represents the distance between point P and the origin. , There are a number of celestial coordinate systems based on different fundamental planes and with different terms for the various coordinates. A common choice is. ) Because \(dr<<0\), we can neglect the term \((dr)^2\), and \(dA= r\; dr\;d\theta\) (see Figure \(10.2.3\)). The volume element spanning from r to r + dr, to + d, and to + d is specified by the determinant of the Jacobian matrix of partial derivatives, Thus, for example, a function f(r, , ) can be integrated over every point in R3 by the triple integral. The brown line on the right is the next longitude to the east. where \(a>0\) and \(n\) is a positive integer. $$ , This will make more sense in a minute. As the spherical coordinate system is only one of many three-dimensional coordinate systems, there exist equations for converting coordinates between the spherical coordinate system and others. I'm just wondering is there an "easier" way to do this (eg. Define to be the azimuthal angle in the -plane from the x -axis with (denoted when referred to as the longitude), + The result is a product of three integrals in one variable: \[\int\limits_{0}^{2\pi}d\phi=2\pi \nonumber\], \[\int\limits_{0}^{\pi}\sin\theta \;d\theta=-\cos\theta|_{0}^{\pi}=2 \nonumber\], \[\int\limits_{0}^{\infty}e^{-2r/a_0}\,r^2\;dr=? These choices determine a reference plane that contains the origin and is perpendicular to the zenith. because this orbital is a real function, \(\psi^*(r,\theta,\phi)\psi(r,\theta,\phi)=\psi^2(r,\theta,\phi)\). \[\int\limits_{all\; space} |\psi|^2\;dV=\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}\int\limits_{0}^{\infty}\psi^*(r,\theta,\phi)\psi(r,\theta,\phi)\,r^2\sin\theta\,dr d\theta d\phi=1 \nonumber\]. Computing the elements of the first fundamental form, we find that X_{\phi} = (-r\sin(\phi)\sin(\theta),r\cos(\phi)\sin(\theta),0), \\ , $$ atoms). The geometrical derivation of the volume is a little bit more complicated, but from Figure \(\PageIndex{4}\) you should be able to see that \(dV\) depends on \(r\) and \(\theta\), but not on \(\phi\). The Schrdinger equation is a partial differential equation in three dimensions, and the solutions will be wave functions that are functions of \(r, \theta\) and \(\phi\). r Here's a picture in the case of the sphere: This means that our area element is given by Calculating Infinitesimal Distance in Cylindrical and Spherical Coordinates Calculating \(d\rr\)in Curvilinear Coordinates Scalar Surface Elements Triple Integrals in Cylindrical and Spherical Coordinates Using \(d\rr\)on More General Paths Use What You Know 9Integration Scalar Line Integrals Vector Line Integrals These markings represent equal angles for $\theta \, \text{and} \, \phi$. Spherical coordinates are the natural coordinates for physical situations where there is spherical symmetry (e.g. These formulae assume that the two systems have the same origin and same reference plane, measure the azimuth angle in the same senses from the same axis, and that the spherical angle is inclination from the cylindrical z axis. The radial distance is also called the radius or radial coordinate. In spherical coordinates, all space means \(0\leq r\leq \infty\), \(0\leq \phi\leq 2\pi\) and \(0\leq \theta\leq \pi\). The standard convention The same value is of course obtained by integrating in cartesian coordinates. Polar plots help to show that many loudspeakers tend toward omnidirectionality at lower frequencies. The angle $\theta$ runs from the North pole to South pole in radians. The wave function of the ground state of a two dimensional harmonic oscillator is: \(\psi(x,y)=A e^{-a(x^2+y^2)}\). The relationship between the cartesian coordinates and the spherical coordinates can be summarized as: (25.4.5) x = r sin cos . We will see that \(p\) and \(d\) orbitals depend on the angles as well. Coming back to coordinates in two dimensions, it is intuitive to understand why the area element in cartesian coordinates is d A = d x d y independently of the values of x and y. When using spherical coordinates, it is important that you see how these two angles are defined so you can identify which is which. Use your result to find for spherical coordinates, the scale factors, the vector d s, the volume element, and the unit basis vectors e r , e , e in terms of the unit vectors i, j, k. Write the g ij matrix. You have explicitly asked for an explanation in terms of "Jacobians". $$h_1=r\sin(\theta),h_2=r$$ r As we saw in the case of the particle in the box (Section 5.4), the solution of the Schrdinger equation has an arbitrary multiplicative constant. Theoretically Correct vs Practical Notation. Spherical coordinates (r, . The straightforward way to do this is just the Jacobian. We assume the radius = 1. This can be very confusing, so you will have to be careful. Spherical coordinates are useful in analyzing systems that are symmetrical about a point. The corresponding angular momentum operator then follows from the phase-space reformulation of the above, Integration and differentiation in spherical coordinates, Pages displaying short descriptions of redirect targets, List of common coordinate transformations To spherical coordinates, Del in cylindrical and spherical coordinates, List of canonical coordinate transformations, Vector fields in cylindrical and spherical coordinates, "ISO 80000-2:2019 Quantities and units Part 2: Mathematics", "Video Game Math: Polar and Spherical Notation", "Line element (dl) in spherical coordinates derivation/diagram", MathWorld description of spherical coordinates, Coordinate Converter converts between polar, Cartesian and spherical coordinates, https://en.wikipedia.org/w/index.php?title=Spherical_coordinate_system&oldid=1142703172, This page was last edited on 3 March 2023, at 22:51. The relationship between the cartesian and polar coordinates in two dimensions can be summarized as: \[\label{eq:coordinates_1} x=r\cos\theta\], \[\label{eq:coordinates_2} y=r\sin\theta\], \[\label{eq:coordinates_4} \tan \theta=y/x\]. (b) Note that every point on the sphere is uniquely determined by its z-coordinate and its counterclockwise angle phi, $0 \leq\phi\leq 2\pi$, from the half-plane y = 0, How to match a specific column position till the end of line? Let P be an ellipsoid specified by the level set, The modified spherical coordinates of a point in P in the ISO convention (i.e. the orbitals of the atom). The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Then the area element has a particularly simple form: Stack Exchange network consists of 181 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Would we just replace \(dx\;dy\;dz\) by \(dr\; d\theta\; d\phi\)? Here is the picture. The simplest coordinate system consists of coordinate axes oriented perpendicularly to each other. (g_{i j}) = \left(\begin{array}{cc} For positions on the Earth or other solid celestial body, the reference plane is usually taken to be the plane perpendicular to the axis of rotation. The spherical system uses r, the distance measured from the origin; , the angle measured from the + z axis toward the z = 0 plane; and , the angle measured in a plane of constant z, identical to in the cylindrical system. the area element and the volume element The Jacobian is The position vector is Spherical Coordinates -- from MathWorld Page 2 of 11 . F & G \end{array} \right), (25.4.6) y = r sin sin . Would we just replace \(dx\;dy\;dz\) by \(dr\; d\theta\; d\phi\)? When solving the Schrdinger equation for the hydrogen atom, we obtain \(\psi_{1s}=Ae^{-r/a_0}\), where \(A\) is an arbitrary constant that needs to be determined by normalization. In the cylindrical coordinate system, the location of a point in space is described using two distances (r and z) and an angle measure (). Blue triangles, one at each pole and two at the equator, have markings on them. $$z=r\cos(\theta)$$ It is because rectangles that we integrate look like ordinary rectangles only at equator! Then the integral of a function f (phi,z) over the spherical surface is just $$\int_ {-1 \leq z \leq 1, 0 \leq \phi \leq 2\pi} f (\phi,z) d\phi dz$$. where we do not need to adjust the latitude component. These relationships are not hard to derive if one considers the triangles shown in Figure 26.4. Now this is the general setup. This will make more sense in a minute. Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site. Learn more about Stack Overflow the company, and our products. Angle $\theta$ equals zero at North pole and $\pi$ at South pole. The difference between the phonemes /p/ and /b/ in Japanese. The geometrical derivation of the volume is a little bit more complicated, but from Figure \(\PageIndex{4}\) you should be able to see that \(dV\) depends on \(r\) and \(\theta\), but not on \(\phi\). In linear algebra, the vector from the origin O to the point P is often called the position vector of P. Several different conventions exist for representing the three coordinates, and for the order in which they should be written. In three dimensions, this vector can be expressed in terms of the coordinate values as \(\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\), where \(\hat{i}=(1,0,0)\), \(\hat{j}=(0,1,0)\) and \(\hat{z}=(0,0,1)\) are the so-called unit vectors. If measures elevation from the reference plane instead of inclination from the zenith the arccos above becomes an arcsin, and the cos and sin below become switched. Visit http://ilectureonline.com for more math and science lectures!To donate:http://www.ilectureonline.com/donatehttps://www.patreon.com/user?u=3236071We wil. $$. The result is a product of three integrals in one variable: \[\int\limits_{0}^{2\pi}d\phi=2\pi \nonumber\], \[\int\limits_{0}^{\pi}\sin\theta \;d\theta=-\cos\theta|_{0}^{\pi}=2 \nonumber\], \[\int\limits_{0}^{\infty}e^{-2r/a_0}\,r^2\;dr=? A bit of googling and I found this one for you! Degrees are most common in geography, astronomy, and engineering, whereas radians are commonly used in mathematics and theoretical physics. By contrast, in many mathematics books, + So to compute each partial you hold the other variables constant and just differentiate with respect to the variable in the denominator, e.g. The cylindrical system is defined with respect to the Cartesian system in Figure 4.3. Case B: drop the sine adjustment for the latitude, In this case all integration rectangles will be regular undistorted rectangles. The answer is no, because the volume element in spherical coordinates depends also on the actual position of the point. ) "After the incident", I started to be more careful not to trip over things. The volume of the shaded region is, \[\label{eq:dv} dV=r^2\sin\theta\,d\theta\,d\phi\,dr\]. 167-168). Recall that this is the metric tensor, whose components are obtained by taking the inner product of two tangent vectors on your space, i.e. \[\int\limits_{all\; space} |\psi|^2\;dV=\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}\int\limits_{0}^{\infty}\psi^*(r,\theta,\phi)\psi(r,\theta,\phi)\,r^2\sin\theta\,dr d\theta d\phi=1 \nonumber\]. But what if we had to integrate a function that is expressed in spherical coordinates? But what if we had to integrate a function that is expressed in spherical coordinates? How do you explain the appearance of a sine in the integral for calculating the surface area of a sphere? Solution We integrate over the entire sphere by letting [0,] and [0, 2] while using the spherical coordinate area element R2 0 2 0 R22(2)(2) = 4 R2 (8) as desired! If the inclination is zero or 180 degrees ( radians), the azimuth is arbitrary. If you preorder a special airline meal (e.g. In order to calculate the area of a sphere we cover its surface with small RECTANGLES and sum up their total area. Spherical coordinates are somewhat more difficult to understand. Use the volume element and the given charge density to calculate the total charge of the sphere (triple integral). ) The volume of the shaded region is, \[\label{eq:dv} dV=r^2\sin\theta\,d\theta\,d\phi\,dr\]. $$ In any coordinate system it is useful to define a differential area and a differential volume element. {\displaystyle (r,\theta ,\varphi )} {\displaystyle \mathbf {r} } for physics: radius r, inclination , azimuth ) can be obtained from its Cartesian coordinates (x, y, z) by the formulae. However, in polar coordinates, we see that the areas of the gray sections, which are both constructed by increasing \(r\) by \(dr\), and by increasing \(\theta\) by \(d\theta\), depend on the actual value of \(r\). \[\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}\int\limits_{0}^{\infty}\psi^*(r,\theta,\phi)\psi(r,\theta,\phi) \, r^2 \sin\theta \, dr d\theta d\phi=\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}\int\limits_{0}^{\infty}A^2e^{-2r/a_0}\,r^2\sin\theta\,dr d\theta d\phi=1 \nonumber\], \[\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}\int\limits_{0}^{\infty}A^2e^{-2r/a_0}\,r^2\sin\theta\,dr d\theta d\phi=A^2\int\limits_{0}^{2\pi}d\phi\int\limits_{0}^{\pi}\sin\theta \;d\theta\int\limits_{0}^{\infty}e^{-2r/a_0}\,r^2\;dr \nonumber\]. Lines on a sphere that connect the North and the South poles I will call longitudes. Find \(A\). To define a spherical coordinate system, one must choose two orthogonal directions, the zenith and the azimuth reference, and an origin point in space. $$ The spherical coordinate systems used in mathematics normally use radians rather than degrees and measure the azimuthal angle counterclockwise from the x-axis to the y-axis rather than clockwise from north (0) to east (+90) like the horizontal coordinate system. The lowest energy state, which in chemistry we call the 1s orbital, turns out to be: This particular orbital depends on \(r\) only, which should not surprise a chemist given that the electron density in all \(s\)-orbitals is spherically symmetric.