Electromagnetic Field Equation 4: UNIZOR.COM - Physics4Teens - Waves - F...

Notes to a video lecture on http://www.unizor.com

Electromagnetic Field Equations 4

This lecture is about the Fourth Maxwell equation, expressing in differential form the Ampere-Maxwell's Law - relationship between a changing in time electric field and produced by this process changing in time magnetic field.

But, before diving into mathematical concepts of the Fourth Maxwell's Equation, some general discussion is appropriate.

We mentioned previously that a changing magnetic field induces a changing electric field (Faraday's Law), which was expressed in the Third Maxwell's Equation.

Ampere's Law describes the magnetic field generated by an electric current, actual movement of electric charges.
Constant electric current generates constant magnetic field, variable current generates variable magnetic field.

But constant magnetic field will not generate an electric field. For propagation of electromagnetic field the initial electric current must be variable to produce a variable electric field, which will generate a variable magnetic field, which, in turn, generate a new variable electric field etc.

The addition of an impact of the variable electric field onto a generated magnetic field was introduced by James Maxwell in his Fourth Equation.

If we consider only transformations between variable electric and magnetic fields, it looks like the electric and magnetic components of an electromagnetic field are symmetrical, and the equation, expressing how variable electric field generates variable magnetic field (the Fourth Maxwell equation), should be symmetrical to the Third Maxwell equation describing how changing magnetic field generates a changing electric field.
But it's not exactly so.

We do need some initial source of energy, a variable electric current, to start this chain of transformations between electric and magnetic fields.

While changing electric field generates changing magnetic field and thus propagates through oscillations of a combined electromagnetic field, the first originating changing electric field in this chain (before the first changing magnetic field generated) should appear from somewhere, and that "somewhere" is the flow of electrons - a movement of material objects in our everyday's understanding of the word "material", which is different from "field".

The generation of a magnetic field must include two sources: electric current component capable of generation of magnetic field (constant or changing) and a variable electric field component generating variable magnetic field.

We will address both components separately and then present their sum as the combined source that generates a variable magnetic field.

The Fourth Maxwell Equation encompasses both effect on a generated magnetic field by electric current (Ampere's Law) and by a variable electric field (Maxwell's addition).

We start with magnetic field generated by an electric current.

The existence of a magnetic field around an electric current was introduced in the "Electromagnetism" part of this course, in the topic "Magnetism of Electric Current", which we strongly recommend to review.

Recall the familiar experiment of iron shavings positioned around a straight line electric wire with a current running through the wire that results in creation of visible magnetic field lines in circular formation around a wire.



There is no other source of magnetism that can cause this formation, so we must conclude that an electric current creates a magnetic field that moves iron shavings into circular positions.

In the same topic "Magnetism of Electric Current" mentioned above we suggested the logical basis for a formula for an intensity of the magnetic field around an infinite straight line electric current as dependent on the amperage of the electric current and a distance from a wire carrying this current:
B = μI/(2πR)
where
B is an intensity of the magnetic field,
I is an electric current,
R is a distance of a point of measurement of the magnetic field intensity from the wire with a current,
μ is a magnetic permeability of the media.

Notice that from this formula follows
μ·I = B·(2πR)
That is, an electric current I going through a circular loop times the magnetic permeability of the media (μ₀ for vacuum) equals to a circulation of a magnetic field generated by this current along its loop.

This description represents a particular case of a more general Ampere's Circuital Law that states that for any closed loop L around an electric current I the circulation of the magnetic field produced by this current equals to the electric current enclosed in the loop times a magnetic permeability constant:
μ·I = ∫_[L]B·dL

Indeed, if the path L is a circle of radius R in the plane perpendicular to a straight line conductor with an electric current I flowing through it, vectors B and dL are collinear with the length of B constant. Therefore, the integral above would be equal to a product of a constant magnitude of a magnetic field intensity by the length of a circumference of radius R:
μ·I = B·(2πR).

To transform this equation into point-based differential one, let's perform two operations:
(1) we will squeeze the loop around point {x,y} on the plane perpendicular to a direction of an electric current and
(2) we will divide both sides of an equation above, the electric current on the left and the circulation on the right, by an area of the loop of a magnetic field line around point {x,y}

As a result of these steps, as presented in the lecture about circulation, the ratio of a circulation to an area of the loop will tend to a curl of a magnetic field at point {x,y}.
On the other hand, division of the electric current by the same area produces average density of electricity flowing through the loop, which in limit, when the loop is squeezed to a point, will produce the density of electricity at that point {x,y}.

This leads us to an equation for point {x,y} in the magnetic field
curl B(x,y) = μ·J(x,y)
where J(x,y) is a density of electric current at point {x,y}.

As we know,
curl B(x,y) = ∇⨯B(x,y), which allows to right the equation above as
∇⨯B(x,y) = μ·J(x,y)

We might want to engage the electric field intensity E instead of current density J into our equation. To accomplish this, we can make one more variation of the above equation based on the empirical Ohm's Law.

Recall that Ohm's Law states
U = I·R
where U is difference in potential,
I - electric current,
R - resistance.
Resistance R is proportional to length of a conductor L and inversely proportional to the area A of its cross-section:
R = ρ·L/A
where ρ is resistivity of the material of a conductor, that is the resistance of the unit of area per unit of length of a conductor.
The conductivity σ is an inverse of ρ.
Therefore,
R = L/(σ·A)
where A is the area of a conductor,
L is the length of a conductor.
Therefore, the Ohm's Law can be written as
U = I·L/(σ·A)
or
U/L = (I/A)/σ
The expression I/A is the electric current density J.
Since U is the difference in potential, it's the work to transfer a unit of electric charge from one end of a conductor to another.
Since work is the force times distance, and the force acting on a unit charge in the electric field is the field's intensity E,
E = J/σ
σ·E = J

Now we can replace the current density J with a product of electric field intensity E and material conductivity σ: J=σ·E, which transforms the equation above into
∇⨯B(x,y) = μ·σ·E(x,y)

This expression for a curl of a magnetic field produced by electric current constitutes the half of the Fourth Maxwell's Equation.
The second half depends on changing electric field generating a changing magnetic field.

First of all, let's support by an experiment that a changing electric field generates a changing magnetic field.
Consider a capacitor.

The alternating current (AC) in the circuit goes through a capacitor connected to some generator of AC power.
The amount of electricity Q(t) accumulated on each plate of a capacitor is proportional to a variable voltage U(t) applied to its plates, and capacity of a capacitor C (see lecture "Electric Fields" - "Capacitors" in this course) is the constant proportionality factor that depends on a type of a capacitor C = Q(t)/U(t)
Q(t) = C·U(t)

Knowing the amount of electricity Q(t) accumulated in a capacitor as a function of time t, we can determine the electric current I(t) in a circuit, which is a rate of change (that is, derivative by time) of the amount of electricity
I(t) = dQ(t)/dt = C·dU(t)/dt

This I(t) is not as "material" as a current in a wire, no electrons are exchanged directly between the plates, but, nevertheless, it's real and, as the "material" current in a wire, it generates a magnetic field around it.
There is a special name for this electric current - displacement current, and we will use the symbol I_D(t) for it to differentiate it from the electric current caused by actual movement of charges.

Now we will follow the logic presented above for wire-based electric current generating a magnetic field to calculate the effect of the displacement current, which is a product of the changing electric field without actual movement of the charges.

First of all, we will evaluate the displacement current as follows
I_D(t) = C·dU(t)/dt
Capacity C, as described in the lecture "Electric Fields" - "Capacitors" in this course is
C = A·ε/d
where A is the area of a plate of a capacitor (that is, the area, through which displacement current is flowing),
ε is the permittivity of the media between the plates,
d is the distance between the plates.

Therefore,
I_D(t) = (A·ε/d)·dU(t)/dt =
= A·ε·d[U(t)/d]/dt

If E(t) is the intensity of the electric field between the plates with voltage U(t) between them,
E(t) = U(t)/d
From this follows
I_D(t) = A·ε·dE(t)/dt

In terms of density of the displacement current
J_D(t)=I_D(t)/A
this looks like
J_D(t) = ε·dE(t)/dt

Let's consider now any changing in time electric field of intensity E(t,x,y) in two-dimensional XY-coordinate space.
We can imagine a tiny capacitor with plates parallel to XY-plane at point {x,y}, for which the equation above, which is independent on the size of the plates and the distance between them, represents a displacement current density.
Now the expression above for a displacement current density can be stated for any point {x,y} as
J_D(t,x,y) = ε·∂E(t,x,y)/∂t

Using the equation above for a curl of a magnetic field intensity as a function of electric current density
curl B(x,y) = μ·J(x,y),
we evaluate another part of a total magnetic field intensity, which depends on variability of the electric field and displacement current density:
curl B(x,y) = μ·J_D(t,x,y) =
= μ·ε·∂E(t,x,y)/∂t

Combining both components of the magnetic field intensity, one that depends on the moving charges and another that depends on variability of the electric field, we obtain the Fourth Maxwell Equation
curl B(t,x,y) =
= μ·ε·∂E(t,x,y)/∂t + μ·σ·E(t,x,y)

Using our favorite symbol ∇ for curl, the Fourth Maxwell Equation looks like

∇⨯B(t,x,y) =
=μ·ε·∂E(t,x,y)/∂t + μ·σ·E(t,x,y)

More Field Problems: UNIZOR.COM - Physics4Teens - Waves - Field Waves

Notes to a video lecture on http://www.unizor.com

More Field Problems

Problem 1

Prove that a curl of two-dimensional conservative vector field equals to zero.

Solution

(a) From the definition of the conservative field (the integral of the vector field along any path depends only on the end points) follows that an integral around any closed path equals to zero.
(b) A curl is defined as a limit of the ratio of an integral around a closed path (which is zero for conservative vector field, as stated above) to the area enclosed by this path. The result of this division is zero.


Problem 2

Prove that a two-dimensional conservative field
F(x,y) = {P(x,y), Q(x,y)} =
= P(x,y)·i + Q(x,y)·j
is a gradient of some two-dimensional scalar field f(x,y) called its potential.
In other word, prove that there exists function f(x,y) such that
P(x,y) = ∂f(x,y)/∂x and
Q(x,y) = ∂f(x,y)/∂y.

Solution

Before offering a solution, let's try to use some knowledge about conservative fields that we have learned before.

Recall a concept of potential energy of a gravitation or an electrostatic field.
By definition, it's the work performed by or against a field by moving a unit of mass or a unit of electrical charge from infinitely remote location to a specific location of interest in space, thus making this potential energy a function of a point in space where gravitational or electrostatic forces act.
Validity of a concept of potential energy is based on its independence on the path taken from infinity to a point of interest. Does not it remind the conservative field? It sure does.

That means, a potential energy is a scalar field, a scalar function of a specific location in space. It does not depend on trajectory of movement.

The work participating in the definition of a potential energy is, roughly speaking, a product of a force by distance. More precisely, it's an integral of this product along a trajectory we move a unit of mass or a unit of electric charge, because the force is, generally speaking, variable in magnitude and direction. The force itself constitutes a vector field in space, and the integral of a product of this force by distance along a trajectory (that is, a work performed by or against this vector field by moving a unit of mass or a unit of electric charge) corresponds to a definition of a circulation of this vector field of force along some trajectory of movement.

As an example, let's find the work of moving a unit of mass in the gravitational field. For simplicity, we'll analyze this in one-dimensional case. Let's choose a point {x,0,0} as a point of interest and calculate the work to move a unit of mass along the X-axis from infinity to this point. Let the source of gravity be a mass M at the origin of coordinates.

According to the Newton's Law of Gravity, the force of gravity for a unit mass at distance s from the origin of coordinates is
F(s) = G·M/s²
Therefore, the potential energy at point {x,0,0} is
∫_[∞,x]F(s)·ds = ∫_[∞,x]G·M/s²·ds
The indefinite integral of 1/s² is −1/s.
Therefore, our potential energy at point {x,0,0} is
U(x) = −G·M/s|^x_∞ = −G·M/x

The potential energy U(x) is a scalar field.
What's interesting is that
dU(x)/dx = G·M/x² = F(x).
So, a scalar field of potential energy for a gravitational field plays the role of potential, gradient of which is the vector field of gravitational force.

Our guess then is, that in case of a given conservative vector field we can find its potential, gradient of which is our vector field, and this potential is just amount of work performed by or against the vector field of force by moving along some (actually, any) trajectory from some fixed location to a point of interest.
We can choose any trajectory because our vector field is conservative.

Consider a two-dimensional case and point of interest {x,y}.
The given conservative vector field is defined at all points as
F(x,y) = {P(x,y); Q(x,y)}


Let's calculate a work needed to move an object in this field from origin of coordinate to point {x,y}.
Since the field is conservative, we choose an easy path:
A{0,0} → B{x,0} → C{x,y}
and calculate the work separately on an interval AB, then on BC and then add them together.

W_AB = ∫_[0,x]P(u,0)du
W_BC = ∫_[0,y]Q(x,v)dv
Total amount work is W(x,y)=W_AB+W_BC
Can this scalar field W(x,y) be the needed potential, gradient of which is our vector field F(x,y)?

Let's check by calculating the gradient
grad W(x,y) = ∇W(x,y).
If we are correct in the assumption that work by or against the conservative vector field along some trajectory to a point of interest gives a potential of this field, the gradient of work W(x,y) (that is, partial derivatives by each argument) should deliver the vectors of the field F(x,y).
In other words, we should check that
∂W(x,y)/∂x = P(x,y) and
∂W(x,y)/∂y = Q(x,y)

Here are the calculations.
∂W(x,y)/∂x = ∂W_AB(x,y)/∂x +
+ ∂W_BC(x,y)/∂x =
= (∂/∂x)∫_[0,x]P(u,0)du +
+ (∂/∂x)∫_[0,y]Q(x,v)dv

The first expression is a derivative by an upper limit of integration, which reverses the integration and results in
(∂/∂x)∫_[0,x]P(u,0)du = P(x,0)

The second expression allows to bring differentiation under the integral, since differentiation and integration are for different arguments
(∂/∂x)∫_[0,y]Q(x,v)dv =
= ∫_[0,y][(∂Q(x,v)/∂x]dv

From the definition of the conservative field (the integral of the vector field along any path depends only on the end points) follows that an integral around any closed path equals to zero.
A curl is defined as a limit of the ratio of an integral around a closed path (which is zero for conservative vector field, as stated above) to the area enclosed by this path. The result of this division is zero.
The curl in two dimensional case was calculated as
curl F(x,y) =
= ∂Q(x,y)/∂x − ∂P(x,y)/∂y
Since curl is zero for a conservative field,
∂Q(x,y)/∂x = ∂P(x,y)/∂y
and the expression for the second integral above can be changed to
(∂/∂x)∫_[0,y]Q(x,v)dv =
= ∫_[0,y][(∂Q(x,v)/∂x]dv =
= ∫_[0,y][(∂P(x,v)/∂v]dv
Integral by v and differentiation by v reverse each other and the result is the original function
∫_[0,y][(∂P(x,v)/∂v]dv =
= P(x,v)|^y₀ = P(x,y) − P(x,0)
Adding this to the first of two integrals above, that we found to be equal to P(x,0), the result will be P(x,y), which corresponds to our expectations.

Absolutely analogous calculations for ∂W(x,y)/∂x will yield Q(x,y) as a result.
These results prove that
grad W(x,y) = F(x,y)
and that proves the existence and construction of the scalar field W(x,y), gradient of which is our conservative vector field F(x,y).


Problem 3

Prove that a curl of a gradient of a three-dimensional scalar field is zero.

Solution

The solution in three-dimensional space is based on the formula for a curl and an expression of a gradient of a scalar field.
Let vector field F(x,y,z) be a gradient of a scalar field f(x,y,z):
F_x(x,y,z) = ∂f(x,y,z)/∂x
F_y(x,y,z) = ∂f(x,y,z)/∂y
F_z(x,y,z) = ∂f(x,y,z)/∂z

A curl of F(x,y,z) is a vector with components
curl F(x,y,z) = ∇⨯F(x,y,z) =
= {∂F_z(x,y,z)/∂y−∂F_y(x,y,z)/∂z;
∂F_x(x,y,z)/∂z−∂F_z(x,y,z)/∂x;
∂F_y(x,y,z)/∂x−∂F_x(x,y,z)/∂y}

Substituting F_x, F_y, F_z with their expressions in terms of partial derivatives of f(x,y,z), we obtain
∂F_z(x,y,z)/∂y−∂F_y(x,y,z)/∂z =
= ∂²f(x,y,z)/∂z∂y −
− ∂²f(x,y,z)/∂y∂z = 0
because a mixed partial second derivative by two different arguments does not depend on the order of differentiation (assuming the participating function is twice continuously differentiable)

Analogously,
∂F_x(x,y,z)/∂z−∂F_z(x,y,z)/∂x =
= ∂²f(x,y,z)/∂x∂z −
− ∂²f(x,y,z)/∂z∂x = 0

Finally,
∂F_y(x,y,z)/∂x−∂F_x(x,y,z)/∂y =
= ∂²f(x,y,z)/∂y∂x −
− ∂²f(x,y,z)/∂x∂y = 0

Since a conservative vector field can be represented as a gradient of some scalar field (its potential), we have proven that
curl grad f(...) = 0 or
∇⨯(∇f(...)) = 0
for any scalar vector field f(...).

Circulation: UNIZOR.COM - Physics4Teens - Waves - Field Waves

Notes to a video lecture on http://www.unizor.com

Circulation of Vector Field

Consider a two-dimensional vector field F(x,y) with components
{F_x(x,y),F_y(x,y)}
For definitiveness, assume that vectors F(x,y) represent some force that can act on objects positioned at any point {x,y} on the coordinate XY-plane. For example, our plane is a surface of water and the wind blows on paper ships on it. So, at any given moment of time the vectors at any point of this two-dimensional vector field are defined by the magnitude and direction of the wind at that point, as represented by arrows on a picture below.


Consider now that the wind does not change and our task is to move a paper ship along some trajectory from point A to point B.
How much work should we perform?
We will consider different cases, increasing the complexity.

Uniform Vector Field and Straight Path

So, our field F(x,y) consists of vectors equal to each other and equal to the same vector F for all points {x,y} on a plane, and our path AB is a straight line.
This is a simple problem from Mechanics.
First, we have to find a projection F_AB of vector F onto line AB and then multiply the magnitude of this projection by the length from A to B:
W = |F_AB|·|AB|

Using scalar (dot) product of vectors, the same can be expressed as
W = (F·AB)
where AB is a vector from A to B.

If vector F is perpendicular to AB, the scalar product is zero and the work is zero, since the wind does not prevent us to move the ships from A to B nor helps us.

If vector F is collinear with AB, the scalar product is a product of a magnitude of F and a length of AB, taken with a positive sign if the direction of vector F is from B to A and we actually perform work against the wind or taken with a negative sign if the direction of vector F is from A to B and it's the field instead, not we, moves the ship.

If our segment AB is positioned along the X-axis and vector F has components {F_x,F_y}, the work is a product of F_x by a length of AB, taken with appropriate sign, as above.

Non-uniform Vector Field and Straight Path

Assume again that points A and B are on the X-axis.
Now vectors F(x,y) are not the same, their magnitude and directions are functions of point {x,y}.

Since F(x,y)=F_x(x,y)+F_y(x,y), the work against vector F(x,y) can be expressed as a sum of work against its component F_x(x,y), that is collinear with our path AB, and work against its component F_y(x,y), that is perpendicular to our path AB.

Obviously, the latter is zero since F_y(x,y) is perpendicular to path AB.

To calculate the work against F_x(x,y) along path AB (that is, for x from x=A to x=B, while y=0), we break segment AB into many tiny intervals of the same length Δx by points
x₀=A, x₁, x₂,...,x_n=B
Assuming that each interval is small enough, so F_x(x,0) does not significantly change within it, the work, approximately, equals to
W = Σ_i∈[1,n]F_x(x_i,0)·Δx

All we need now to have exact value for work is to go to a limit with Δx→0 (assuming, the limit exists), getting
W = ∫_[A,B]F_x(x,0)·dx

Non-uniform Vector Field and Curved Path

Our task is to move a paper ship from point P(x_p,y_p) to point Q(x_q,y_q) along a defined curved path in the two-dimensional vector force field F(x,y) and calculate the work performed on the way.

This case leads us to integration along a curved path.
This process is not much different from regular definite integral of some function f(x) on a segment x∈[a,b].

First, we break a curved path from point P(x_p,y_p) to point Q(x_q,y_q) into many tiny pieces, so small that each one is practically straight and vector field F(x,y) within each piece can be considered constant.

Now our problem to find a total work needed to move a paper ship against the wind from P to Q can be expressed approximately as a sum of works on each almost straight tiny segment against almost constant vector force field.
As we break the path into smaller and smaller pieces, our calculation will be more and more precise, and in the limit case we will come up with exact value of work performed.

The calculation of work on each straight path with constant vector field along it was described above in the part Uniform Vector Field and Straight Path.
All we need is to construct a scalar (dot) product of the vector of force F(x,y), considered constant in the tiny interval around each point {x_i,y_i}, by the length of i^th interval, considered as a vector from {x_i-1,y_i-1} to {x_i,y_i}. Then we will sum up all the results from all pieces of a path.

To accomplish this, assume that we actually move along a curved path from P to Q in real time t. At t=0 we are at point P and at time t=T we arrive at point Q.
Our journey along the path on a two-dimensional plane can be described as a function of time: {x(t),y(t)}.

Let's break the total time from 0 to T into many tiny intervals, each equal to Δt:
{t₀=0, t₁, t₂,...t_n=T}
Our position along a path at time t_i will be {x(t_i),y(t_i)} with {x(t₀),y(t₀)}=P and {x(t_n),y(t_n)}=Q.

Let the velocity of our movement along a path be
V(x,y) = {dx(t)/dt, dy(t)/dt}
This vector is always tangential to a path of motion, that is directed along each tiny piece of a curved path.
The length of the i^th piece of a path equals to |V(x_i,y_i)|·Δt.
Therefore, the work performed on the i^th piece of a path equals
ΔW_i = (F(x_i,y_i)·V(x_i,y_i))·Δt
(here the first multiplication is a scalar product of two vectors, the second one is a product of two scalars)

Now we are ready to calculate approximate work performed on the whole path:
W_[PQ] ≅ Σ_i∈[0,n]ΔW_i =
= Σ_i∈[0,n](F(x_i,y_i)·V(x_i,y_i))·Δt

Familiarity with a definition of the definite integral would lead to expression for the limit of approximate work above to its exact value
W_[PQ] =
= ∫(F(x(t),y(t))·V(x(t),y(t)))·dt
where integration is performed in the time interval t∈[0,T]

As you see, an integration along a curved path is reduced to familiar process of integration on a time interval [0,T], as long as we know how we move along a path:
position r(t)={x(t),y(t)} and
velocity, which is a derivative of a position by time
V(x(t),y(t)) = dr(t)/dt =
= {dx(t)/dt,dy(t)/dt}.

Using a position vector r(t)={x(t),y(t)}, the formula about total work performed on the path from P to Q can be written as W_[PQ] =
=∫_[0,T](F(x(t),y(t))·dr(t)/dt)·dt=
= ∫_[0,T](F(x(t),y(t))·dr(t))
or, simpler,
W_[PQ] = ∫_[PQ](F·dr),
assuming the integration is performed along a path defined by vector
r(t)={x(t),y(t)}.

All the above analysis of a circulation of two-dimensional vector fields is easily transformed onto three-dimensional case, including the latest formula for a path integral with the only difference that vectors are dependent on three coordinates:
W_[PQ] = ∫_[PQ](F·dr),
where F = F(x,y,z) and
r = {x, y, z}.

Conservative Vector Fields

There are a few equivalent definitions of conservative vector field.
Here we will define a vector field as conservative if the above path integral (presented as a work done by moving a paper ship along some path from P to Q) depends only on the starting and ending points P and Q and is independent of which path from one to another we choose.

An immediate consequence of this is that a path integral along any closed loop is zero.
Indeed, choose two points on any closed loop, P and Q. Now we have two different paths from P to Q. Since integrals along both paths are the same, integral from P to Q along one of the paths is equal to integral from Q to P along the second path with an opposite sign:
∫_[PQ](F·dr) = −∫_[QP](F·dr)
The sum of these two integrals is a total integral on a closed loop, and it's equal to zero.

An equivalent definition of a conservative vector field F(x,y) is that it is a gradient of some scalar field f(x,y):
F(x,y) = grad f(x) = ∇f(x,y)
It can be proven that this definition is equivalent to the one above about path integral being dependent only on starting and ending points, each definition can be derived from the other.
The participating scalar function f(x,y) is called a potential for a given vector field F(x,y).

Curl of a Vector Field

By definition, a curl of a two-dimensional vector field at any point equals to a limit of this vector field's circulation along a closed path around that point, divided by the area enclosed in this path, assuming that this path's length and area it encloses tend to zero.

We leave open an issue about existence of this limit. Under general conditions of smoothness of the vector field this limit exists and, therefore, independent on how exactly we shrink the path around a point on a plane to reach the limit.

Let's find the curl analytically.
Choose a rectangular path ABCD around a point {x,y} of sizes Δx⨯Δy.


The choice of a rectangular form is related to ability to analyze separately X- and Y-components of the vector field with only one of them playing a role for each side of this rectangle.

Considering infinitesimal character of the sizes of this rectangle, we can assume that the X-component of the vector field F_x(x,y) is constant along the horizontal sides of a rectangle and equals to its value in the middle of a corresponding side, that is F_x(x,y−½Δy) for side AB and F_x(x,y+½Δy) for side CD.
This X-component should not participate in the calculation along vertical sides BC and DA, as it is perpendicular to them.

Similarly, we can assume that the Y-component of the vector field F_yy(x,y) is constant along the vertical sides of a rectangle and equals to its value in the middle of a corresponding side, that is F_y(x+½Δx,y) for side BC and F_y(x−½Δx,y) for side DA.
This Y-component should not participate in the calculation along horizontal sides AB and CD, as it is perpendicular to them.

Now the total circulation along our rectangular path is
W_[ABCD] =
= F_x(x,y−½Δy)·Δx +
+ F_y(x+½Δx,y)·Δy −
− F_x(x,y+½Δy)·Δx −
− F_y(x−½Δx,y)·Δy
Signs of the members of the above sum depend on which vertex is a start of a path and which direction we choose. The formula above is related to point A as a start and counterclockwise movement along the path.

From the classical definition of a partial derivative for a function of two argument f(x,y)
∂f(x,y)/∂x =
= lim[f(x+Δx,y)−f(x,y)]/Δx
where lim is assumed as Δx→0.
Obviously, the same result is from a similar limit of
[f(x+½Δx,y)−f(x−½Δx,y)]/Δx

Therefore,
F_x(x,y+½Δy)−F_x(x,y−½Δy) ≅
≅ [∂F_x(x,y)/∂y]·Δy

Similarly,
F_y(x+½Δx,y)−F_y(x−½Δx,y) ≅
≅ [∂F_y(x,y)/∂x]·Δx

Using these, the circulation above equal to
[∂F_y(x,y)/∂x]·Δx·Δy −
− [∂F_x(x,y)/∂y]·Δx·Δy

Notice the area of a rectangle Δx·Δy as a multiplier in both component.
Therefore, dividing the circulation around an infinitesimal path around our point by the area encompassed by this path, we obtain the curl of the vector field
curl F(x,y) =
= ∂F_y(x,y)/∂x − ∂F_x(x,y)/∂y
We will use this definition in the next lecture that discusses curl of a two-dimensional vector field in more details.

Field Problems: UNIZOR.COM - Physics4Teens - Waves - Field Waves

Notes to a video lecture on http://www.unizor.com

Problems on Field Characteristics

Problem 1

The gravitation field is characterized by a vector of force of gravity per unit of mass of a probe object (a point-mass) as a function of its distance from the source of the field (like Earth).
The direction of this force is always towards a center of mass that is the source of gravity.

Another characteristic of a gravitational field is its potential, which is a work (a scalar) needed to bring a unit of mass from infinity to a point of interest (see "Energy" part of this "Physics 4 Teens" course, chapter "Potential Energy", topic "Gravity").

Let the source of gravitational field be at location {0,0,0} and its mass is M.
Our point of interest is at location {x,y,z}.
The gravitational potential is inversely proportional to a distance from the center of the source of gravity and can be expressed as a scalar function of three coordinates
U(x,y,z) = −G·M·(x²+y²+z²)^−½
where G is a gravitational constant.

Determine the vector
grad U(x,y,z) = V(x,y,z) of gradient of this field at any point {x,y,z}, describe its direction and magnitude.

Solution

Gradient of a scalar field is a vector directed towards the highest rate of change of the value of a scalar field.
The components of a vector of gradient are partial derivatives of a scalar field by each coordinates
grad U(x,y,z) = V(x,y,z) =
= {∂U/∂x,∂F/∂y,∂F/∂z}
which can also be written using a symbol ∇ as
grad U(x,y,z) = V(x,y,z) =
= ∇U(x,y,z)
that looks like a multiplication of a pseudo-vector
∇ = {∂/∂x,∂/∂y,∂/∂z}
by a scalar U(x,y,z).

In our case of
U(x,y,z) = −G·M·(x²+y²+z²)^−½

Partial derivative by argument x of this function is
∂U(x,y,z)/∂x =
= G·M·x·(x²+y²+z²)^−3/2
Similarly,
∂U(x,y,z)/∂y =
= G·M·y·(x²+y²+z²)^−3/2
and
∂U(x,y,z)/∂z =
= G·M·z·(x²+y²+z²)^−3/2

The direction of vector grad U(x,y,z) is along the line from the origin of coordinates towards a location of a point of interest {x,y,z}.

The magnitude of this vector is
|grad U(x,y,z)| =
= G·M·(x²+y²+z²)⁻¹
which happened to be the magnitude of the gravitational force of mass M onto a unit mass (that is, the field intensity) at location {x,y,z}, according to the Newton's Law of Gravitation.
The only difference between the force of gravity and the gradient of a gravitational potential field is, the gradient directed from the center to point {x,y,z}, while the force of gravity is in the opposite direction.


Problem 2

Given a three-dimensional vector field
V(x,y,z) = {0,0,z}

(a) Describe this vector field verbally.
(b) Is there a front surface of this field, where all vectors at it are equal?
(c) Find a divergence of this field.
(d) Is there a part of space where divergence is zero?
(e) Find a curl of this field.

Solution

(a) For any point {x,y,z} the vector V(x,y,z)={0,0,z} has only Z-component not equal to zero. It means, all vectors of this vector field are parallel to Z-axis with zero projection on two other axes.
(b) All vectors at points that belong to the same plane parallel to XY-plane are of the same magnitude and direction since their Z-components are the same, while X- and Y-components are zero. So, any
(c) Divergence of V(x,y,z) is
div V(x,y,z) =
= div (0·i + 0·j + z·k) =
= ∂0/∂x + ∂0/∂y + ∂z/∂z =
= 0 + 0 + 1 = 1
(d) No. The divergence div V(x,y,z) is equal to 1 everywhere.
(e) Curl of V(x,y,z) is
curl V(x,y,z) =
= curl (0·i + 0·j + z·k) =
= (∂z/∂y−∂0/∂z)·i +
+ ∂0/∂z−∂z/∂x)·j +
+ ∂0/∂x−∂0/∂y)·k) =
= (0·i + 0·j + 0·k) = 0

Problem 3

Given a two-dimensional vector field
V(x,y) =
= {−y/√(x²+y²), x/√(x²+y²)}

(a) Describe this vector field verbally.
(b) Is there a front line of this field, where all vectors at it are equal?
(c) Find a divergence of this field.
(d) Is there a part of space where divergence is zero?
(e) Find a curl of this field.

Solution

(a) This vector field is circular, that is a vector V(x,y) at any point of interest is perpendicular to a radius-vector to this point from the origin of coordinates, since the scalar product of V(x,y) and r(x,y)={x,y} is zero.
(b) Yes. Any radial line from the origin of coordinates represents a front line because any field vector originated on it have the same direction (perpendicular to this radial line) and magnitude of 1.
(c) Divergence of V(x,y) is
div V(x,y) =
= div [−y/√(x²+y²)]·i +
+ [x/√(x²+y²)]·j =
= ∂[−y/√(x²+y²)]/∂x +
+ ∂[x/√(x²+y²)]/∂y =
= (−y)·(−x)·(x²+y²)^−3/2 +
+ (x)·(−y)·(x²+y²)^−3/2 =
= (−y·x+x·y)·(x²+y²)^−3/2 = 0
(d) Yes, everywhere the divergence is zero. The equality of the magnitude of the field vectors plays a major role.
(e) Curl of V(x,y) is
curl V(x,y) =
= curl [−y/√(x²+y²)]·i +
+ [x/√(x²+y²)]·j =
= {∂[x/√(x²+y²)]/∂x −
− ∂[−y/√(x²+y²)]/∂y}·k =
= x²·(x²+y²)^−3/2 +
+ y²·(x²+y²)^−3/2 =
= 1/√(x²+y²)
As we see, the curl is the same for all points of interest on the same distance from the origin of coordinates and the curl is diminishing as a point of interest is moving away from the origin of coordinates.

Curl 2D: UNIZOR.COM - Physics4Teens - Waves - Field Waves

Notes to a video lecture on http://www.unizor.com

Curl in 2D

Previous lectures addressed different usages of a pseudo-vector (a triad, a triplet, a set or simply a vector) of partial differentiation operators
∇ = {∂/∂x, ∂/∂y, ∂/∂z}.

We have discussed a gradient of the scalar field that looks like a multiplication of "vector" ∇ by scalar F(x,y,z)
as follows:
grad F(x,y,z) = ∇F =
= {∂F/∂x, ∂F/∂y, ∂F/∂z}
which is a vector.

We have also introduced a divergence of the vector field that looks like a scalar (dot) product of "vector" ∇ and vector V(x,y,z) with components
{V_x(x,y,z),V_y(x,y,z),V_z(x,y,z)}
as follows:
div V(x,y,z) = (∇·V) =
= ∂V_x/∂x + ∂V_y/∂y + ∂V_z/∂z
which is a scalar.

Our next goal is to introduce yet another usage of a symbol ("vector") ∇ for vector fields, which will describe an operation similar to a vector (cross) product of pseudo-vector ∇ and vector V(x,y,z) in three-dimensional space, that is
curl V(x,y,z) = ∇⨯V
which is a vector.

The concept of a curl in three-dimensional case, where it's the most practical, is easier to begin explaining in two-dimensional world.

Imagine a vector field on the XY-plane, assigning some vector V(x,y) (direction and magnitude) to each point {x,y} on this plane.

For example, if our plane is a surface of water (flat surface, no waves), the vector field V(x,y) can represent the velocity of each dust particle on this surface at some moment in time, as all these particles are chaotically sliding around a water surface.

As in other cases, we consider vector V(x,y) as represented by its two components along X- and Y-coordinates:
V(x,y) = {V_x(x,y), V_y(x,y)}

The intuitive understanding of a curl at some point of a two-dimensional field (a plane) is that, if we place a tiny paddle wheel with the axis perpendicular to our plane into some point, it will or will not rotate by the vectors of the field, interpreted as forces.
The direction and speed of rotation are expressed as a curl at that point on a plane at that time moment (positive for counterclockwise rotation, negative for clockwise and zero if there is no rotation at all).

The formal definition of a curl in a two-dimensional case is
curl V(x,y) =
= ∂V_y(x,y)/∂x − ∂V_x(x,y)/∂y

There is a logical base for this definition.

The term ∂V_y(x,y)/∂x characterizes how fast and in what direction (increases, pulling up or decreases pulling down) the Y-components of the vector field is changing if we step to the right by infinitesimal increment of argument x.

The term ∂V_x(x,y)/∂y characterizes how fast and in what direction (increases, pulling right or decreases pulling left) the X-components of the vector field is changing if we step up by infinitesimal increment of argument y.

Their difference characterizes whether the vector field curves towards X- or Y-axis at point {x,y}.
In other words, if we have a perpendicular to our XY-plane at point {x,y}, curl indicates whether there is a vector field's tendency to rotate around this point clockwise (towards X-axis) or counterclockwise (towards Y-axis).

Now let's analyze what this formula represents.

Unidirectional Vector Field

Let's start with constant direction and magnitude vector field
V(x,y) = {V_x(x,y), V_y(x,y)}
It can be described as
V_x(x,y)=a, V_y(x,y)=b
for all pairs {x,y}.

Intuitively, there is no curl in this field.
Indeed,
∂V_y(x,y)/∂x = 0
∂V_x(x,y)/∂y = 0
Therefore,
curl V(x,y) = 0

Consider now a unidirectional vector field (all vectors are parallel to some line, like y=k·x) with variable magnitude at different points.

If two infinitesimally closed to each other points, not lying on the same line as the direction of the vector field, have vectors of the same direction but different magnitude, the tiny pedal wheel positioned at that point would still rotate.

The absence of a rotation of a tiny pedal wheel that can be placed at any point on a plane necessitates that the magnitude of neighboring vectors along parallel lines must be the same (so, there is no momentum of rotation).

If we draw a line perpendicular to a graph of a function y=k·x, which is y=−x/k, the magnitude of parallel vectors on the same distance from that perpendicular line (in other words, on the same front line) must be the same.
This is illustrated on a picture below with the length of red lines corresponding to a magnitude of collinear vectors, equal for vectors on the same front line.


To prove that this field has curl equaled to zero, let's change the coordinate axes.
The fact of rotation or not rotation of our tiny pedal wheel does not depend on coordinates, so let's choose the new X-axis along the line y=k·x in old coordinates and Y-axis - along a line y=−x/k.

In this new system of coordinates all vectors are directed along the new X-axis, their new Y-coordinate is always zero.
Now our vector field looks like
V(x,y) = {V_x(x,y), 0}
The requirement of all vectors on the same front line to have the same magnitude means that all vectors with the same X-coordinate (x=a) have the same magnitude (equaled to V_x(a,y), regardless of Y-coordinate y.

Let's calculate the curl of this vector field at any point.
Again, the definition of a curl is
curl V(x,y) =
= ∂V_y(x,y)/∂x − ∂V_x(x,y)/∂y

Since V_y(x,y) = 0 at any point on a plane,
∂V_y(x,y)/∂x = 0

The situation with ∂V_x(x,y)/∂y is only slightly more complex.
Recall that the definition of a partial derivative of V_x(x,y) by argument y means
lim[V_x(x,y+Δy)−V_x(x,y)]/Δy
as Δy→0.
But the value of V_x(x,y), as we stated above, does not depend on argument y and is the same for all values of y for any fixed argument x.
Therefore, the difference V_x(x,y+Δy)−V_x(x,y) is always zero, which makes ∂V_x(x,y)/∂y to be zero as well.
So, curl is zero for this unidirectional vector field with the same magnitude of vectors lying on the same front line.

Circular Vector Field

Let's consider a circular vector field with direction of vectors changing from point to point, always perpendicular to a radius to this point from the origin of coordinates {0,0}.

The radius vector to a point with coordinates {x,y} is
r={x,y}.
The perpendicular vector must have a scalar product with r equal to zero.
There are many examples of a vector field that at any point is perpendicular to a radius-vector to this point from the origin of coordinates.
For example, p(x,y)={−y²·x, x²·y}
q(x,y)={−1/y², x/y³}
You can check that their scalar product with r={x,y} is zero.

We, as an example, choose a simpler case of vectors a(x,y)={−y,x} or b(x,y)={y,−x}. They are opposite to each other and both are perpendicular to a radius vector from the origin, therefore describe a circular vector field.
Indeed, the scalar (dot) product of r with any of them is
(r·a) = (r·b) = −x·y+x·y = 0.

The magnitude of the vector field a(x,y) is √x²+y² and is growing, as a point {x,y} moves farther from the origin of coordinates.

(click right mouse and open this picture in another tab to see it in full format)

Quick analysis shows that vector field a(x,y) defines counterclockwise circulation, while b(x,y) defines clockwise circulation of a vector field.

Let's calculate the curl of both fields.

Vector a(x,y)={−y,x} has components
a_x(x,y) = −y
a_y(x,y) = x
Therefore,
curl a(x,y) =
= ∂a_y(x,y)/∂x − ∂a_x(x,y)/∂y =
= ∂x/∂x − ∂(−y)/∂y =
= 1 + 1 = 2 (positive)

Vector b(x,y)={y,−x} has components
b_x(x,y) = y
b_y(x,y) = −x
Therefore,
curl b(x,y) =
= ∂b_y(x,y)/∂x − ∂b_x(x,y)/∂y =
= ∂(−x)/∂x − ∂y/∂y =
= −1 − 1 = −2 (negative)

The constant curl throughout an entire vector field a(x,y) or b(x,y) is an interesting phenomenon.
On one hand, the curvature of circles of the vector field is diminishes, as the distance from a center increases.
On the other hand, the strength (magnitude) of the vector field is increasing with the distance from the center.
These two factors balance each other and the "curliness" of the field stays the same.
As an analogy, going in a circle of a small radius with a lower speed and going in a circle of a large radius with a higher speed result in similar feelings of centripetal force.

Curl in 2D as Rotation

The last example of a circular vector field rotating around the origin of coordinates can be viewed as rotation within XY-plane around Z-axis.
In theory, any circulation of a vector field within XY-plane around any point can be viewed as rotation around a perpendicular to XY-plane at that point. Then the curl of the vector field at that point can be represented as a vector perpendicular to XY-plane (that is, parallel to Z-axis) with a magnitude equal to the value of a curl at that point.

Just as an association, remember the representation of the angular speed of rotation as a vector perpendicular to the plane of rotation, having a magnitude equal to the angular speed - angular velocity vector. This vector representation of angular speed of rotation was just for a convenience of representing other concepts related to rotation.
The similar vector representation of a curl in two-dimensional case as a vector perpendicular to a plane of rotation is also a convenience, that, in particular, will be used for three-dimensional case.

Curl 3D: UNIZOR.COM - Physics4Teens - Waves - Field Waves

Notes to a video lecture on http://www.unizor.com

Curl in 3D

This lecture continues analyzing a concept of curl of a vector field, this time in three-dimensional space.
We will extensively use the material of a previous lecture on curl in two-dimensional space.

Consider a three-dimensional vector field - the one that really exists in our three-dimensional world V(x,y,z) with components
{V_x(x,y,z),V_y(x,y,z),V_z(x,y,z)}

As a working model, we will use velocities of air molecules as a vector field, sometimes calling it a "wind", and we assume that time does not participate in our analysis, so everything is related either to one particular moment in time or the wind does not change with time.
Wind might steadily blow in one direction, which would be an example of unidirectional vector field or might form a tornado, which is a perfect example of a vector field with a non-zero curl.

As in the two-dimensional case, let's consider a tiny pedal wheel that can be placed at any point in the air without fixing its axis of rotation, as we did in the two-dimensional case, where this axis was always parallel to Z-axis.

After some turning around with the wind, provided the wind is not changing, our pedal wheel will establish some position and will or will not spin in some direction with some angular speed.
Its axis of rotation then will point to a direction of the three-dimensional curl of our vector field at that point and the speed of rotation gives the magnitude. These direction and magnitude define a curl as a vector in three-dimensional vector field.

The complication in three-dimensional case is that our tiny pedal wheel can rotate around any axis directed anywhere in space.
Moreover, if it rotates around some axis at some point in space because a vector field curls around that axis, a slight change in the direction of this pedal wheel might still result in its rotation, maybe with lower velocity.

It helps to simplify the problem by breaking it in three already solved ones.
Consider a vector field V(x,y,z) with components
{V_x(x,y,z),V_y(x,y,z),V_z(x,y,z)}

Let curl V(x,y,z) be a vector K(x,y,z) with components
{K_x(x,y,z),K_y(x,y,z),K_z(x,y,z)}

Intuitively, the X-component K_x(x,y,z) of the vector K(x,y,z) should represent the curl of the projection of the vector field V(x,y,z) onto the YZ-plane, that is a curl of a two-dimensional vector field with fixed X-component and two other components equal to
{V_y(x,y,z),V_z(x,y,z)}
Therefore, K_x(x,y,z) equals to
∂V_z(x,y,z)/∂y − ∂V_y(x,y,z)/∂z

Analogously, the Y-component K_y(x,y,z) of the vector K(x,y,z) should represent the curl of the projection of the vector field V(x,y,z) onto the ZX-plane, that is a curl of a two-dimensional vector field with fixed Y-component and two other components equal to
{V_z(x,y,z),V_x(x,y,z)}
Therefore, K_y(x,y,z) equals to
∂V_x(x,y,z)/∂z − ∂V_z(x,y,z)/∂x

Similarly, the Z-component K_z(x,y,z) of the vector K(x,y,z) should represent the curl of the projection of the vector field V(x,y,z) onto the XY-plane, that is a curl of a two-dimensional vector field with fixed Z-component and two other components equal to
{V_x(x,y,z),V_y(x,y,z)}
Therefore, K_z(x,y,z) equals to
∂V_y(x,y,z)/∂x − ∂V_x(x,y,z)/∂y

Now we know all three components of the vector K(x,y,z) that represents curl V(x,y,z).

This vector can be represented as a sum of its components using the unit vectors along the coordinate axis i along OX axis, j along OY axis and k along OZ axis:
K(x,y,z) = K_x(x,y,z)·i +
+ K_y(x,y,z)·j + K_z(x,y,z)·k
where the components of the curl vector were calculated above:
K_x(x,y,z) =
= ∂V_z(x,y,z)/∂y − ∂V_y(x,y,z)/∂z
K_y(x,y,z) =
= ∂V_x(x,y,z)/∂z − ∂V_z(x,y,z)/∂x
K_z(x,y,z) =
= ∂V_y(x,y,z)/∂x − ∂V_x(x,y,z)/∂y

Curl as a Vector Product

All we need now to make all the formulae more beautiful is to apply some vector algebra.

Let's take any two vectors in their coordinates form:
P = P_x·i + P_y·j + P_z·k
Q = Q_x·i + Q_y·j + Q_z·k

Now let's form a vector product of these two vectors and express it also in coordinate form, using trivial identities
i⨯i=0; j⨯j=0; k⨯k=0;
i⨯j=k; j⨯i=−k;
j⨯k=i; k⨯j=−i;
k⨯i=j; i⨯k=−j.

Skipping the terms equaled to zero, the vector (cross) product of P by Q equals to
P⨯Q =
= P_x·Q_y·i⨯j + P_x·Q_z·i⨯k +
+ P_y·Q_x·j⨯i + P_y·Q_z·j⨯k +
+ P_z·Q_x·k⨯i + P_z·Q_y·k⨯j =
= P_x·Q_y·k + P_x·Q_z·(−j) +
+ P_y·Q_x·(−k) + P_y·Q_z·i +
+ P_z·Q_x·j + P_z·Q_y·(−i) =
= (P_yQ_z−P_zQ_y)·i +
+ (P_zQ_x−P_xQ_z)·j +
+ (P_xQ_y−P_yQ_x)·k

Now consider a pseudo-vector ∇ with components
{∂/∂x, ∂/∂y, ∂/∂z}
and a vector field V(x,y,z) with components
{V_x(x,y,z),V_y(x,y,z),V_z(x,y,z)}
Use pseudo-vector ∇ as vector P in the above expression for a vector (cross) product and vector V(x,y,z) as vector Q.
Then, omitting (x,y,z) in vector field and its components for brevity, their vector product would look like
∇⨯V =
= (∂V_z/∂y − ∂V_y/∂z)·i +
+ (∂V_x/∂z − ∂V_z/∂x)·j +
+ (∂V_y/∂x − ∂V_x/∂y)·k
which exactly the same as an expression for curl V(x,y,z) derived earlier in this lecture.

We have come to a very short form of a vector K(x,y,z) representing a curl of a vector field V(x,y,z) at point {x,y,z}:

curl V(x,y,z) = ∇⨯V(x,y,z)

Divergence: UNIZOR.COM - Physics4Teens - Waves - Field Waves

Notes to a video lecture on http://www.unizor.com

Divergence

In this lecture we continue studying the characteristics of fields and, in particular, discuss another usage of symbol ∇ (nabla) that represents a set (or a triplet, or a pseudo-vector, or simply "vector") of three operators of partial differentiation, performed for each of the three dimensions of space we live in, in given Cartesian coordinates.
If needed, review the topic of partial derivatives in the "Calculus" part of the course "Math 4 Teens" on UNIZOR.COM.

We use the symbol ∇ as a shorthand for a triplet of operators of partial differentiation
∇ = {∂/∂x, ∂/∂y, ∂/∂z}

Previous lecture about gradient of a scalar field F(x,y,z) defined ∇F(x,y,z) as a "multiplication" of a "vector" ∇ by "scalar" F(x,y,z), resulting in a "vector", as traditional multiplication of a vector by a constant should:
∇F(x,y,z) =
= {∂F()/∂x, ∂F()/∂y, ∂F()/∂z}

The physical meaning of the gradient at any point {x,y,z} of a scalar field F(x,y,z) was a vector field (direction and magnitude at each point) of the greatest change of this field, thereby defining from a scalar field F(x,y,z) a vector field ∇F(x,y,z).

As an application of this concept, if F(x,y,z) represents an air density at point {x,y,z}, which is a scalar, the ∇F(x,y,z) represents the direction and magnitude of the most significant potential movement of the air, a vector, the main wind direction and its strength at each point.

The divergence is the result of a different operation with "vector" ∇.
Consider, again, an example of air distribution with different air density (and, therefore, pressure) in different areas of space. As a result, we can observe the direction and strength of a wind at any point - a vector field of air velocities (speed and direction of a wind) at each point.

The air goes from zones of higher density towards zones of lower density. While this process goes, the amount of air in a zone of higher density diminishes, while the amount of air in a zone of lower density increases until the density (and, therefore, pressure) equalizes. During this process the amount of air in a zone of average density in between the above two zones will remain steady, what air comes from a high density zone will go to a lower density.

The divergence of a vector field (for example, of a vector field of velocities of the wind at each point in space) quantitatively reflects the situation with air zones described above, showing the areas, from which air flows from and those it flows into.

Divergence is a scalar defined for each point of a vector field. It's positive for those areas, from which something (like air, or electrons, or energy) is flowing out, negative for areas receiving that something and zero if the amounts of coming into and going out are the same.

Let's approach this mathematically.
For each point {x,y,z} in space there is vector V(x,y,z) with projections on coordinate axes V_x(x,y,z), V_y(x,y,z) and V_z(x,y,z) so that the following vector equality is held:
V(x,y,z) = V_x(x,y,z)·i +
+ V_y(x,y,z)·j + V_z(x,y,z)·k
where i, j and k are unit vectors along X-, Y- and Z-axis.

Consider a parallelepiped ABCDEFGH centered at point P(x,y,z) and infinitesimal dimensions Δx⨯Δy⨯Δz.
Then coordinates of vertices are
A{x−½Δx,y−½Δy,z−½Δz}
B{x+½Δx,y−½Δy,z−½Δz}
C{x+½Δx,y+½Δy,z−½Δz}
D{x−½Δx,y+½Δy,z−½Δz}
E{x−½Δx,y−½Δy,z+½Δz}
F{x+½Δx,y−½Δy,z+½Δz}
G{x+½Δx,y+½Δy,z+½Δz}
H{x−½Δx,y+½Δy,z+½Δz}


To bring some flavor into mathematics, let's use our analogy of the vector field V(x,y,z) as a direction and magnitude of the wind at point P(x,y,z).

To evaluate the air contribution by this wind into point P(x,y,z), we will calculate how much air coming in and going out of the parallelepiped ABCDEFGH per unit of time and take its limit, as the dimensions of this parallelepiped are infinitesimal.
We will consider three major directions along the three axes of the coordinate system and coordinate representation of the vector V(x,y,z).

Projection of the vector V(x,y,z) on the X-axis V_x(x,y,z)·i brings air into our parallelepiped through its left side and takes the air out through its right side.
In the middle of the left side this projection is V_x(x−½Δx,y,z)·i.
In the middle of the right side this projection is V_x(x+½Δx,y,z)·i.

Since we assume that our parallelepiped has infinitesimal dimensions, we can assume that the projection of the vector V(x,y,z) on the X-axis at each point of the left side ADHE is the same as in the middle of this side at point {x−½Δx,y,z}.
Similarly, we assume that the projection of the vector V(x,y,z) on the X-axis at each point of the right side BCGF is the same as in the middle of this side at point {x+½Δx,y,z}.

The amount of air moving out through the right side of a parallelepiped is proportional to the average speed of wind at the right side V_x(x+½Δx,y,z) times the area of the right side Δy·Δz.
Similarly, the amount of air moving in through the left side of a parallelepiped is proportional to the average speed of wind at the left side V_x(x−½Δx,y,z) times the area of the left side Δy·Δz.
The net result of air moving in and out of the parallelepiped along the X-axis is proportional to
A_x(x,y,z) = [V_x(x+½Δx,y,z) −
− V_x(x−½Δx,y,z)]·Δy·Δz

Recall from the calculus the following property of differentiable function f(x):
f(b) − f(a) = df(c)/dx · (b−a)
where point c is somewhere between a and b.
If a→b (that is, b−a is an infinitesimal variable), c also tends to a (or b) and
df(c)/dx ≅ df(a)/dx ≅
≅ df(b)/dx ≅ df(½(a+b))/dx.

Using this property in our case with a=x−Δx and b=x+Δx, with Δx infinitesimal, we derive
V_x(x+½Δx,y,z) −
− V_x(x−½Δx,y,z) ≅
≅ ∂V_x(x,y,z)/∂x · Δx

Now the expression for an amount of air accumulated in the vicinity of point {x,y,z} is
A_x(x,y,z) ≅
≅ ∂V_x(x,y,z)/∂x · Δx·Δy·Δz

Since Δx·Δy·Δz is the volume of our parallelepiped, dividing both sides of the equation above by it will give a change (positive for increment or negative for decrement) of air density in the infinitesimal neighborhood of point {x,y,z}, contributed by a component of the vector field in the X-axis direction.
D_x(x,y,z) ≅ ∂V_x(x,y,z)/∂x

Analogously, contribution to air density by Y-component of the vector field V(x,y,z) is
D_y(x,y,z) ≅ ∂V_y(x,y,z)/∂y

And Z-component of this vector field adds
D_z(x,y,z) ≅ ∂V_z(x,y,z)/∂z

The combined contribution of air density will be a sum of these individual contributions, a divergence (div) of the vector field V(x,y,z):

divV(x,y,z) = ∂V_x(x,y,z)/∂x +
+ ∂V_y(x,y,z)/∂y + ∂V_z(x,y,z)/∂z

To simplify this expression, ∇ to the rescue.
Considering ∇ as a "vector"
∇ = {∂/∂x, ∂/∂y, ∂/∂z}
and V(x,y,z) being a vector in coordinate representation
{V_x(x,y,z),V_y(x,y,z),V_z(x,y,z)}
the expression for a divergence above can be represented as a scalar (dot) product of these two "vectors":

divV(x,y,z) = ∇ · V(x,y,z)

Other examples of the usage of a concept divergence are: divergence of the electric field (not zero where there are electric charges, see the First Maxwell equation),
divergence of the magnetic field (always zero, see the Second Maxwell equation),
divergence of the gravitational field (non-zero inside any material object, zero in vacuum),
and any others, where it makes sense to talk about vector field.