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When you describe vectors in spherical or cylindric coordinates, that is, write vectors as sums of multiples of unit vectors in the directions defined by these coordinates, you encounter a problem in computing derivatives.
The unit vectors themselves change as you change coordinates, so that the change in your vector consists of terms arising from the changes of the multiples and also those from changes in the unit vectors.
We can find neat expressions for the divergence in these coordinate systems by finding vectors pointing in the directions of these unit vectors that have 0 divergence.
Then we write our vector field as a linear combination of these instead of as linear combinations of unit vectors.
By the product rule, the expression for the divergence we seek will be a sum over the three directions of the dot product of one of these vectors with the gradient of its coefficient. The second terms in the product rule will all be 0.
You may very well encounter a need to express divergence in these coordinates in your future life, so we will carry this approach out with spherical coordinates.
First please notice: whenever you differentiate functions in polar coordinates you must treat the origin in them separately and carefully. The coordinates themselves are singular there! The safest thing is to add "except perhaps at r = 0" or some similar statement, unless you are sure that your conclusion is sensible there. Often it will NOT hold at the origin of your coordinates.
Such caviats are omitted below but you should assume that they are present whenever differentiation by a polar parameter is involved.
The only non-trivial step in doing this is finding vectors in the various required directions that have 0 divergence. This can be done by finding the divergence of any vectors in these directions and figuring out what multiple you need apply in each case to cancel its divergence out, again using the product theorem for divergence.
The vector (x, y, z) points in the radial direction in spherical coordinates, which we call the 
direction. Its divergence is 3.
It can also be written as or as
A multiplier which will convert its divergence to 0 must therefore have, by the product theorem, a gradient that is multiplied by itself.
The function does this very thing, so the 0-divergence function in the direction is
Exercise 17.2 Notice that the divergence of (x, y, 0) otherwise known as r or as rur is 2. What function of r should you multiply it by to get a vector with divergence 0?
The vector (-y, x) points in the direction and has 0 divergence already. It can be written as
The direction is normal to both of these and you can get a vector in it by taking the cross product of (-y, x, 0) and (x, y, z), with result (xz, yz, -r2).
This vector has divergence 2z, and the form rzur - r2uz.
It is the first of these two terms, rzur (which is (xz, yz)) that has the non-vanishing divergence and it is the x and y which lead to it, not the z factor. We can invoke the result of the last exercise to introduce a multiplier that will get rid of that divergence.
The resulting vector has the form whose length is It can therefore be written as
To summarize, the vectors have 0 divergence.
If we define these combinations to be respectively, a vector of the form is also writable as
and its divergence is
or
or
and this is the form of the divergence in spherical coordinates.
In the last line here we used the form of the gradient in spherical coordinates: recall that is a polar variable with radius r and is a polar variable with radius . Also recall that r is sin.
This last expression is not very pretty but it is quite important in physical applications.
It appears in particular in the combination which is called the Laplacian of f.
In rectangular coordinates and spherical coordinates the Laplacian takes the following forms, which follow from the expressions for the gradient and divergence
Exercises:
17.3 Find the divergence of
17.4 Deduce the form of the divergence in cylindric coordinates using the logic used above for spherical coordinates. Apply it to find the Laplacian in cylindric coordinates.
FAQs
Divergence in Spherical Coordinates Derivation
Firstly, the partial derivatives with respect to x, y and z would be converted into the ones with respect to r, φ and θ. Then, the x, y and z components of the vector i.e. Ax, Ay and Az are equivalently written in terms of Ar, Aφ and Aθ.
What is the formula for spherical coordinates? ›
To convert a point from spherical coordinates to Cartesian coordinates, use equations x=ρsinφcosθ,y=ρsinφsinθ, and z=ρcosφ. To convert a point from Cartesian coordinates to spherical coordinates, use equations ρ2=x2+y2+z2,tanθ=yx, and φ=arccos(z√x2+y2+z2).
What does dV equal in spherical coordinates? ›
What is dV is Spherical Coordinates? Consider the following diagram: We can see that the small volume ∆V is approximated by ∆V ≈ ρ2 sinφ∆ρ∆φ∆θ. This brings us to the conclusion about the volume element dV in spherical coordinates: Page 5 5 When computing integrals in spherical coordinates, put dV = ρ2 sinφ dρ dφ dθ.
How do you find the divergence of a point? ›
The divergence of a vector field is a scalar field. The divergence is generally denoted by “div”. The divergence of a vector field can be calculated by taking the scalar product of the vector operator applied to the vector field. I.e., ∇ .
What is the 16.9 divergence theorem? ›
Theorem 16.9. 1 (Divergence Theorem) Under suitable conditions, if E is a region of three dimensional space and D is its boundary surface, oriented outward, then ∫∫DF⋅NdS=∫∫∫E∇⋅FdV.
What is spherical divergence? ›
1. n. [Geophysics]
The apparent loss of energy from a wave as it spreads during travel. Spherical divergence decreases energy with the square of the distance.
What is the format for spherical coordinates? ›
In mathematics, a spherical coordinate system is a coordinate system for three-dimensional space where the position of a given point in space is specified by three numbers, (r, θ, φ): the radial distance of the radial line r connecting the point to the fixed point of origin (which is located on a fixed polar axis, or ...
What is the spherical coordinates ϕ? ›
The coordinate ρ is the distance from P to the origin. If the point Q is the projection of P to the xy-plane, then θ is the angle between the positive x-axis and the line segment from the origin to Q. Lastly, ϕ is the angle between the positive z-axis and the line segment from the origin to P.
How to find limits in spherical coordinates? ›
As the circle is rotated around the z-axis, the relationship stays the same, so ρ = 2 sinφ is the equation of the whole surface. To determine the limits of integration, when φ and θ are fixed, the corresponding ray enters the region where ρ = 0 and leaves where ρ = 2 sinφ.
What is the Jacobian for spherical coordinates? ›
Now, since the Jacobian is the absolute value of this, we can conclude that the Jacobian in spherical coordinates is r²sinθ. Therefore, when performing a change of variables from Cartesian coordinates to spherical we need to make the following changes: Thank you for reading.
volume element d V for a spherical shell of radius r and thickness d r is equal to 4 π r 2 d r .
How to convert spherical coordinates to Cartesian coordinates? ›
To convert a point from spherical coordinates to Cartesian coordinates, use equations x=ρsinφcosθ,y=ρsinφsinθ, and z=ρcosφ. To convert a point from Cartesian coordinates to spherical coordinates, use equations ρ2=x2+y2+z2,tanθ=yx, and φ=arccos(z√x2+y2+z2).
How to derive divergence in spherical coordinates? ›
Idea: In the Cartesian gradient formula ∇F(x,y,z)=∂F∂xi+∂F∂yj+∂F∂zk, put the Cartesian basis vectors i, j, k in terms of the spherical coordinate basis vectors eρ,eθ,eφ and functions of ρ,θ and φ. Then put the partial derivatives ∂F∂x,∂F∂y,∂F∂z in terms of ∂F∂ρ,∂F∂θ,∂F∂φ and functions of ρ,θ and φ.
How to calculate a divergence? ›
The divergence of a vector field can be calculated using the del operator, also known as the nabla operator (∇). It involves taking the dot product of the del operator with the vector field. For a three-dimensional vector field (F=Fi+Fj+Fk), divergence is calculated as ∇•F = ∂F/∂x + ∂F/∂y + ∂F/∂z.
What is the rule for divergence? ›
The divergence theorem states that the volume integral of the divergence of a vector field over a volume V bounded by a surface S is equal to the surface integral of the vector field projected on the outward facing normal of the surface S.
Can you use divergence theorem on a sphere? ›
We cannot apply the divergence theorem to a sphere of radius a around the origin because our vector field is NOT continuous at the origin. Applying it to a region between two spheres, we see that Flux = . because div E = 0.
How do you rotate a point in spherical coordinates? ›
To plot any dot from its spherical coordinates (r, θ, φ), where θ is inclination, the user would: move r units from the origin in the zenith reference direction (z-axis); then rotate by the amount of the azimuth angle (φ) about the origin from the designated azimuth reference direction, (i.e., either the x– or y–axis, ...
How do you find dS in spherical coordinates? ›
On the surface of the sphere, ρ = a, so the coordinates are just the two angles φ and θ. The area element dS is most easily found using the volume element: dV = ρ2 sin φ dρ dφ dθ = dS · dρ = area · thickness so that dividing by the thickness dρ and setting ρ = a, we get (9) dS = a2 sin φ dφ dθ.
What is the formula for divergence in polar coordinates? ›
For a vector field X, the divergence in coordinates is given by ∇⋅X=∑nXi∂xi. In polar coordinates, the metric is [100r2], and so 1√g(∂∂r,∂∂r)∂∂r=∂∂r and 1√g(∂∂θ,∂∂θ)∂∂θ=1r∂∂θ are unit vectors.
direction. Its divergence is 3.
or as 
multiplied by itself. 
does this very thing, so the 0-divergence function in the 
direction and has 0 divergence already. It can be written as 
direction is normal to both of these and you can get a vector in it by taking the cross product of (-y, x, 0) and (x, y, z), with result (xz, yz, -r2).
whose length is 
It can therefore be written as 
have 0 divergence.
respectively, a vector of the form 
is also writable as
which is called the Laplacian of f.
