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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
{Vx(x,y,z),Vy(x,y,z),Vz(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
{Vx(x,y,z),Vy(x,y,z),Vz(x,y,z)}
Let curl V(x,y,z) be a vector K(x,y,z) with components
{Kx(x,y,z),Ky(x,y,z),Kz(x,y,z)}
Intuitively, the X-component Kx(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
{Vy(x,y,z),Vz(x,y,z)}
Therefore, Kx(x,y,z) equals to ∂Vz(x,y,z)/∂y − ∂Vy(x,y,z)/∂z
Analogously, the Y-component Ky(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
{Vz(x,y,z),Vx(x,y,z)}
Therefore, Ky(x,y,z) equals to ∂Vx(x,y,z)/∂z − ∂Vz(x,y,z)/∂x
Similarly, the Z-component Kz(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
{Vx(x,y,z),Vy(x,y,z)}
Therefore, Kz(x,y,z) equals to ∂Vy(x,y,z)/∂x − ∂Vx(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) = Kx(x,y,z)·i +
+ Ky(x,y,z)·j + Kz(x,y,z)·k
where the components of the curl vector were calculated above: Kx(x,y,z) =
= ∂Vz(x,y,z)/∂y − ∂Vy(x,y,z)/∂z Ky(x,y,z) =
= ∂Vx(x,y,z)/∂z − ∂Vz(x,y,z)/∂x Kz(x,y,z) =
= ∂Vy(x,y,z)/∂x − ∂Vx(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 = Px·i + Py·j + Pz·k Q = Qx·i + Qy·j + Qz·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 =
= Px·Qy·i⨯j + Px·Qz·i⨯k +
+ Py·Qx·j⨯i + Py·Qz·j⨯k +
+ Pz·Qx·k⨯i + Pz·Qy·k⨯j =
= Px·Qy·k + Px·Qz·(−j) +
+ Py·Qx·(−k) + Py·Qz·i +
+ Pz·Qx·j + Pz·Qy·(−i) =
= (PyQz−PzQy)·i +
+ (PzQx−PxQz)·j +
+ (PxQy−PyQx)·k
Now consider a pseudo-vector ∇ with components
{∂/∂x, ∂/∂y, ∂/∂z}
and a vector field V(x,y,z) with components
{Vx(x,y,z),Vy(x,y,z),Vz(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 =
= (∂Vz/∂y − ∂Vy/∂z)·i +
+ (∂Vx/∂z − ∂Vz/∂x)·j +
+ (∂Vy/∂x − ∂Vx/∂y)·k
which exactly the same as an expression for curlV(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}:
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 fieldF(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 fieldF(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 Vx(x,y,z), Vy(x,y,z) and Vz(x,y,z) so that the following vector equality is held: V(x,y,z) = Vx(x,y,z)·i +
+ Vy(x,y,z)·j + Vz(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 fieldV(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 Vx(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 Vx(x−½Δx,y,z)·i.
In the middle of the right side this projection is Vx(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 Vx(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 Vx(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 Ax(x,y,z) = [Vx(x+½Δx,y,z) −
− Vx(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 Vx(x+½Δx,y,z) −
− Vx(x−½Δx,y,z) ≅
≅ ∂Vx(x,y,z)/∂x · Δx
Now the expression for an amount of air accumulated in the vicinity of point {x,y,z} is Ax(x,y,z) ≅
≅ ∂Vx(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. Dx(x,y,z) ≅ ∂Vx(x,y,z)/∂x
Analogously, contribution to air density by Y-component of the vector field V(x,y,z) is Dy(x,y,z) ≅ ∂Vy(x,y,z)/∂y
And Z-component of this vector field adds Dz(x,y,z) ≅ ∂Vz(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):
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
{Vx(x,y,z),Vy(x,y,z),Vz(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.
Scientists involved in the mathematical representation of the laws of nature, starting with Newton and Leibniz, used differentiation to reflect the changing character of the world. Sometimes the mathematical representation of the results of research becomes too lengthy to deal with. To make the life easier and formulae shorter, different shorthands were invented.
In this lecture we will talk about operator ∇ called nabla - a convenient way to represent certain properties and characteristics of vectors.
First of all, ∇ is a set of three operators of partial differentiation (if needed, review the topic of partial derivatives in the "Calculus" part of the course "Math 4 Teens" on UNIZOR.COM), with partial differentiation performed for each of the three dimensions of space we live in, as represented by Cartesian coordinates.
More precisely, we use the symbol ∇ as a substitution for a triplet of operators of partial differentiation ∇ = {∂/∂x, ∂/∂y, ∂/∂z}
Consider a function F(x,y,z) defined in three dimensional space, our three operators on it result in three new functions: ∂F(x,y,z)/∂x, ∂F(x,y,z)/∂y, ∂F(x,y,z)/∂z
The result of each operator above is another function of three arguments and all three together can be interpreted as a vector of three functional components.
We used to think about three-dimensional vector as a set of three real numbers. In this case a triplet of operators also can be formally viewed as a vector (or, to distinguish, a pseudo-vector of operators or vector-operator).
Same with three functions that can be considered as a pseudo-vector of functions or vector-function.
The result of the application of a pseudo-vector ∇ to a function F(x,y,z) can be expressed as ∇F(x,y,z) or, even simpler, ∇F, assuming F stays for F(x,y,z) (there is nothing in between ∇ and F).
So, the expression ∇F(x,y,z) is a shorthand for a set of three functions ∂F(x,y,z)/∂x, ∂F(x,y,z)/∂y and ∂F(x,y,z)/∂z, as if a pseudo-vector ∇ is "multiplied" by a scalar F(x,y,z).
Let's emphasize that ∇F just helps to shorten the triplet of partial differentiation above, it's just a notation, a syntax, nothing more.
Now let's exemplify the use of a symbol ∇ in a particular physical case.
Consider a function F(x,y,z) that represents some physical characteristic at point (x,y,z) in space. It can be an air pressure or electric potential, or density of dust particle, or level of radiation etc.
Very often this function is called a scalar field. Scalar because for each point (x,y,z) this function equals to a single number and field because, traditionally, when a function is defined in some area of a three-dimensional space, this area is called a field.
We are interested in how this characteristic changes in space. In particular, which direction from point (x,y,z) the change is the most significant.
Let's set a point P(x,y,z) as our base and move from it by an infinitesimal vector Δr={Δx,Δy,Δz} to point Q(x+Δx,y+Δy,z+Δz).
The value of our function will change during this move from F(x,y,z) at point P to F(x+Δx,y+Δy,z+Δz) at point Q.
Increment of a function, when all three of its arguments are changing, can be represented as three increments caused by a move along each coordinate axis: F(x+Δx,y+Δy,z+Δz) −
− F(x,y,z) =
= [F(x+Δx,y,z) −
− F(x,y,z)] +
+ [F(x+Δx,y+Δy,z) −
− F(x+Δx,y,z)] +
+ F[(x+Δx,y+Δy,z+Δz) −
− F(x+Δx,y+Δy,z)]
A change in a value of a function, when only one argument is infinitesimally changing, can be represented as follows: F(x+Δx,y,z) − F(x,y,z) ≅
≅ (∂F(x,y,z)/∂x)·Δx F(x+Δx,y+Δy,z) −
− F(x+Δx,y,z) ≅
≅ (∂F(x,y,z)/∂y)·Δy F(x+Δx,y+Δy,z+Δz) −
− F(x+Δx,y+Δy,z) ≅
≅ (∂F(x,y,z)/∂z)·Δz
Now is the time to apply ∇ syntax.
Notice that the expression (∂F(x,y,z)/∂x)·Δx +
+ (∂F(x,y,z)/∂y)·Δy +
+ (∂F(x,y,z)/∂z)·Δz
can be viewed as a scalar (dot) product of two vectors ∇F(x,y,z) = {∂F(x,y,z)/∂x, ∂F(x,y,z)/∂y, ∂F(x,y,z)/∂z}
and
Δr = {Δx, Δy, Δz}
Therefore, F(x+Δx,y+Δy,z+Δz)−F(x,y,z)≅
≅ (∇F(x,y,z) · Δr)
where (a·b) denotes the scalar product of two vectors.
From the properties of a scalar product of two vectors we know that |(a,b)| = |a|·|b|·cos(φ)
where φ is an angle between these two vectors.
If lengths of these vectors are fixed, the maximum absolute value of their scalar product will be if cos(φ)=1.
From this we conclude that function F(x,y,z) will have the largest increment F(x+Δx,y+Δy,z+Δz)−F(x,y,z)
if vectors ∇F(x,y,z) and Δr are collinear.
The vector ∇F(x,y,z) = {∂F(x,y,z)/∂x, ∂F(x,y,z)/∂y, ∂F(x,y,z)/∂z}
is called a gradient of a scalar fieldF(x,y,z) and it points to a direction of the largest by absolute value change of the values of this field.
This lecture is about the third Maxwell equation, expressing in differential form the Faraday's Law of magnetic induction - relationship between magnetic and induced by it electric fields.
This law was introduced in the "Electromagnetism" part of this course, in the topic "The Laws of Induction", which we strongly recommend to refresh.
Recall the familiar experiment of a magnet moving through a wire loop that results in an electric current running through a wire loop.
There is no source of electricity that drives the electric current. so we must conclude that moving magnet creates an electric field that moves electrons inside a wire loop.
This electric field exists around a moving magnet whether a wire is there or not, and our task is to quantify this field based on information about changing magnetic field created by a moving magnet.
Consider a changing magnetic field in three-dimensional space. At each point {x,y,z} at each moment in time t it's characterized by a vector B(t,x,y,z) of magnetic field intensity.
Each such magnetic field intensity vector has three components - its projections on three coordinate axes: Bx(t,x,y,z), By(t,x,y,z) and Bz(t,x,y,z).
If i, j and k are unit vectors along X, Y and Z axes, we can represent any vector of magnetic field intensity as B(t,x,y,z) = Bx(t,x,y,z)·i +
+ By(t,x,y,z)·j + Bz(t,x,y,z)·k
In this lecture we will concentrate on the expression of the intensity of the electric field induced by a variable magnetic field, but initially we will consider only the magnetic field directed along the Z-axis. So, vectors Bx(t,x,y,z) and By(t,x,y,z) equal to zero and only Z-component Bz(t,x,y,z) is not equal to zero and is changing with time.
All magnetic lines of such a magnetic field are parallel to Z-axis.
Let's see what electric field is generated by such unidirectional field Bz(t,x,y,z).
Generally speaking, such a magnetic field will induce an electric field E(t,x,y,z) with some X-, Y- and Z-components Ex(t,x,y,z), Ey(t,x,y,z) and Ez(t,x,y,z), however Ez(t,x,y,z) should be equal to zero since electric field induced by Bz(t,x,y,z) must be perpendicular to it.
At any point in space {x,y,z} at any moment in time t the vector Bz(t,x,y,z) is parallel to a Z-axis. The lines of an electric field generated by this magnetic field lie in the plane perpendicular to the vector Bz(t,x,y,z), that is they are parallel to XY-coordinate plane.
To quantify the electric field intensity vector E(t,x,y,z) at point {x,y,z} at time t, consider a closed infinitesimal rectangular circuit ABCD positioned parallel to XY-plane, centered at point {x,y,z}, with sides Δx parallel to X-axis and Δy parallel to Y-axis and coordinates A(x−½Δx, y−½Δy, z), B(x+½Δx, y−½Δy, z), C(x+½Δx, y+½Δy, z), D(x−½Δx, y+½Δy, z).
This circuit lies in the plane perpendicular to magnetic field intensity vector Bz(t,x,y,z), so the change in this vector's magnitude will induce electric field with some intensity E(t,x,y,z), resulting in electromotive force in a circuit and electric current in it.
As we know from the above referenced topic "The Laws of Induction" of this course, changing magnetic field flux Φ of the magnetic field intensity Bz(t,x,y,z) through this circuit generates the electromotive force U in it that equals to U = −dΦ/dt
(also know as Lenz's Law)
Since our circuit ABCD is infinitesimal, the magnetic field flux through it can be approximated as Φ = Bz(t,x,y,z)·Δx·Δy
where the area of a circuit Δx·Δy does not depend on time, but the magnitude of the magnetic field intensity in the Z-axis direction Bz(t,x,y,z) does.
Therefore, the above equation for electromotive force induced in our circuit by changing Z-component of a vector of magnetic field intensity Bz(t,x,y,z) can be written as U = −[∂Bz(t,x,y,z)/∂t]·Δx·Δy
Note that electromotive force U depends on time t, location {x,y,z} and dimensions of our rectangular circuit Δx and Δy.
The dimensions are infinitesimal variables, that's why we approximated the magnetic field flux by a product of the magnetic field intensity at its center and an area of a circuit to obtain the electromotive force equation.
On the other hand, electromotive force (or voltage) U between two electric field points is a work needed (or performed by a field) to transfer a unit charge from one point to another.
If electric field intensity E (the force on a unit charge) is constant along a trajectory between these two points, the same work is a product of this intensity by the distance between points L: U = E·L
In case of variable intensity vector E(x,y,z), the differential of voltage dU on an infinitesimal trajectory of the length dL is a product of a projection of a field intensity vector on this trajectory EL(x,y,z) and an infinitesimal distance dL along this trajectory: dU(x,y,z) = EL(x,y,z)·dL
Let's calculate the electromotive force in our infinitesimal rectangular circuit as a function of an unknown yet electric field intensity vector E(t,x,y,z).
In this circuit the total electromotive force can be represented as a sum of four components along four sides of a circuit, as the current goes around. Along each line we will use a corresponding component of vector E(t,x,y,z): Ex(t,x,y,z) for a component along X-axis and Ey(t,x,y,z) for a component along Y-axis
(there is no Z-component as the electric field is perpendicular to Z-component of the magnetic field Bz(t,x,y,z), which is the only one we are considering now).
Since our circuit is infinitesimal, we can use the electric field intensity in the middle of each side as a constant along an entire side.
So, for side AB we will use Ex(t,x,y−½Δy,z),
for side BC we will use Ey(t,x+½Δx,y,z),
for side CD we will use Ex(t,x,y+½Δy,z),
for side DA we will use Ey(t,x−½Δx,y,z).
Now we can construct an expression for voltage U in terms of electric field intensity U = UAB+UBC+UCD+UDA =
= Ex(t,x,y−½Δy,z)·Δx +
+ Ey(t,x+½Δx,y,z)·Δy −
− Ex(t,x,y+½Δy,z)·Δx −
− Ey(t,x−½Δx,y,z)·Δy
(plus or minus signs in front of each member depend on the direction of a circuit's current to positive or negative direction of the corresponding axis)
Grouping members with the same multiplier Δx or Δy, we obtain U = [Ex(t,x,y−½Δy,z) −
− Ex(t,x,y+½Δy,z)]·Δx +
+ [Ey(t,x+½Δx,y,z) −
− Ey(t,x−½Δx,y,z)]·Δy
From the Calculus we know that for infinitesimal α we can use the approximation f(x+α) − f(x) ≅ (df(x)/dx)·α
or f(x+½α) − f(x−½α) ≅
≅ (df(x)/dx)·α
and the approximation tends to equality as α→0.
Considering the sides of our wire rectangle Δx and Δy are infinitesimal, we can substitute Ex(t,x,y−½Δy,z) −
− Ex(t,x,y+½Δy,z)
with −[∂Ex(t,x,y,z)/∂y]·Δy
Analogously, Ey(t,x+½Δx,y,z) −
− Ey(t,x−½Δx,y,z)
can be replaced with
[∂Ey(t,x,y,z)/∂x]·Δx
Now the expression for an electromotive force U would look like U = [∂Ey(t,x,y,z)/∂x]·Δx·Δy −
− [∂Ex(t,x,y,z)/∂y]·Δy·Δx
Recall the above expression of U in terms of magnetic flux: U = −[dBz(t,x,y,z)/dt]·Δx·Δy
Comparing these two expressions for U we conclude that in a limit case, when the size of our rectangular wire loop tends to zero, U = [∂Ey(t,x,y,z)/∂x]·Δx·Δy −
− [∂Ex(t,x,y,z)/∂y]·Δy·Δx =
= −[∂Bz(t,x,y,z)/∂t]·Δx·Δy
or ∂Ey(t,x,y,z)/∂x−∂Ex(t,x,y,z)/∂y=
= −∂Bz(t,x,y,z)/∂t
or, shorter, considering that all members are functions of a quartet (t,x,y,z), ∂Ey/∂x − ∂Ex/∂y = −∂Bz/∂t
At this point it's easy to come up with analogous equations for two other important cases:
a case of a unidirectional magnetic field with all magnetic field lines parallel to Y-axis with a direction of the vector of magnetic field intensity always parallel to Y-axis at any point in space, but its magnitude can change with time;
a case of a unidirectional magnetic field with all magnetic field lines parallel to X-axis with a direction of the vector of magnetic field intensity always parallel to X-axis at any point in space, but its magnitude can change with time.
For B(t,x,y,z) = By(t,x,y,z) the corresponding equation will be
∂Ex/∂z − ∂Ez/∂x = −∂By/∂t
For B(t,x,y,z) = Bx(t,x,y,z) the corresponding equation will be
∂Ez/∂y − ∂Ey/∂z = −∂Bx/∂t
There is one more step to combine all the obtained results into one short formula.
Recall the triplet of operators of partial differentiation ∇ = {∂/∂x, ∂/∂y, ∂/∂z}
that we used as a pseudo-vector in the "E-M Equations 1" lecture to express the first Maxwell's equation as a scalar (dot) product of this pseudo-vector and a vector of electric field intensity E={Ex,Ey,Ez}.
Using this operator, the first Maxwell's equation looks like ∇·E = ρ/ε
In this case of the third Maxwell equation we will use the same symbol ∇ as a pseudo-vector, but will use it in a vector (cross) product between it and an electric field intensity vector E={Ex,Ey,Ez}.
For this we need to express the result of such a vector product ∇⨯E in coordinate form.
Let's represent each participating vector (or pseudo-vector) in coordinate form using unit vectors i, j and k along X-, Y- and Z-axis correspondingly: ∇= (∂/∂x)·i + (∂/∂y)·j + (∂/∂z)·k E = Ex·i + Ey·j + Ez·k
Vector (cross) product of these vectors in coordinate form looks as follows. ∇⨯E = (∂Ex/∂x)·(i⨯i) +
+ (∂Ex/∂y)·(j⨯i) +
+ (∂Ex/∂z)·(k⨯i) +
+ (∂Ey/∂x)·(i⨯j) +
+ (∂Ey/∂y)·(j⨯j) +
+ (∂Ey/∂z)·(k⨯j) +
+ (∂Ez/∂x)·(i⨯k) +
+ (∂Ez/∂y)·(j⨯k) +
+ (∂Ez/∂z)·(k⨯k)
According to the rules of the vector product, (i⨯i) = 0 (j⨯i) = −k (k⨯i) = j (i⨯j) = k (j⨯j) = 0 (k⨯j) = −i (i⨯k) = −j (j⨯k) = i (k⨯k) = 0
Using the above equations, the general vector product ∇⨯E can be represented in coordinate form as follows ∇⨯E =
= (∂Ez/∂y − ∂Ey/∂z)·i +
+ (∂Ex/∂z − ∂Ez/∂x)·j +
+ (∂Ey/∂x − ∂Ex/∂y)·k
At the same time any general vector of magnetic field intensity can be represented as a vector sum of its coordinate components B = Bx·i + By·j + Bz·k
Above we derived three differential equations that relate each component of the magnetic field intensity with components of the induced electric field intensity
Combining an expression for ∇⨯E and the above three equations, we see that ∇⨯E =
= −(∂Bx/∂t)·i −
−(∂By/∂t)·j −
− (∂Bz/∂t)·k =
= −∂(Bx·i+By·j+Bz·k)/∂t =
= −∂B/∂t
Our final equation, which is the third Maxwell equation that relates induced electric field to a changing magnetic field - the Faraday's Law - is ∇⨯E = −∂B/∂t
where ∇ = {∂/∂x, ∂/∂y, ∂/∂z} is a triplet of operators of partial differentiation interpreted as a pseudo-vector for convenience, E = E(t,x,y,z) is the intensity of an induced electrical field and B = B(t,x,y,z) is the intensity of the magnetic field tha changes with the time.
We continue discussion of certain laws related to properties of electromagnetic field expressed in a format of equations.
We assume that the concepts of electric and magnetic fields are known by now from the "Electromagnetism" part of this course.
It's also important to go through the previous lectures of this topic "Field Waves", especially "E-M Equations 1".
One of the concepts introduced in the "Electromagnetism" part of this course is magnetic field intensityB - a vector defined at each point of space where magnetic field exists, that is a vector field, acting as a force on any moving charge proportional to its amount and its speed in the direction perpendicular to both the direction of magnetic field intensity vectors and to the direction of the moving charge.
As an example, consider a magnetic field produced by an infinitely long infinitesimally thin wire with constant electric current I flowing through it.
This wire is the source of a magnetic field with circular magnetic lines around the wire and tangential to these lines vectors of magnetic field intensityB=μI/(2πr).
Here the constant r is the distance from the wire and the constant μ is the permeability of the media around the wire. In many cases it's expressed as μ=μ0·μr, where μ0 is permeability of the vacuum and μr=μ/μ0 is relative permeability of a media.
In general, a magnetic field intensityB (a vector in three-dimensional space) produces a force Fm (also a vector) on any charge q moving with velocity V (also a vector): Fm = q·(V⨯B)
An important difference between electric field lines and magnetic field lines is that magnetic ones are always cyclical, they do not have a beginning or end.
The electric field lines are always initiated at some source and do not end where they are initiated. That's why we have a non-zero electric field flux through a closed surface around an electric charge, which is called monopole.
In case of magnetic field the situation is different in a sense that there is no such thing as magnetic monopole. Every magnet has two poles, which we call north and south, every magnet is dipole, and magnetic lines around it are always closed in some kind of a loop, as the picture above shows.
Here is another example of circular magnetic field lines around a solenoid. Each magnetic field line goes from one pole into another and loops back.
An important consequence of this property of magnetic field lines is that the magnetic field flux through any closed surface, whether there is a source of magnetic field inside it or not, is zero.
In other words, there is no magnetic monopole, and that is exactly the Gauss Law for magnetic field
Mathematically, it can be expressed as ΦM(S) = 0
where S is any closed surface.
The meaning of this equation is that magnetic field lines are entering and the same lines exiting the closed surface, there is no unbalanced source of magnetic lines (that is, unbalanced magnetic field source) inside this surface.
It can also be expressed differentially, as we did in the previous lecture "E-M Equations 1".
Consider any magnetic field in three-dimensional coordinate space with vector B(x,y,z) of its intensity. The X-, Y- and Z-components of this vector are Bx(x,y,z), By(x,y,z) and Bz(x,y,z).
Then the Gauss Law for this magnetic field states ∂Bx(x,y,z)/∂x +
+ ∂By(x,y,z)/∂y +
+ ∂Bz(x,y,z)/∂z = 0
To simplify a notation, mathematicians came up with a construction that looks like a vector with three components: ∇ = {∂/∂x, ∂/∂x, ∂/∂x}
Then a construction that resembles a scalar (dot) product of this pseudo-vector ∇ and a real vector B={Bx,By,Bz} would look like ∇·B=∂Bx/∂x+∂By/∂y+∂Bz/∂z
With all these assumptions the Gauss Law for magnetic field in this differential form looks really short and simple: ∇·B = 0
The above equation is the second of four Maxwell differential equations that describe the electromagnetic field.
This and a few subsequent lectures are about certain laws related to properties of electromagnetic field expressed in a format of equations.
These laws, in their general mathematical form, were formulated in 19th century by a Scottish mathematician and scientist James Maxwell. We will concentrate on their meaning and mathematical expression in some simple cases.
We assume that the concepts of electric and magnetic fields are known by now from the "Electromagnetism" part of this course.
One of the concepts introduced there was the electric field intensityE - a vector defined at each point of space where electric field exists, that is a vector field, acting as a force on any electric charge proportional to its amount of charge.
In particular, if the electric field is produced by a point charge q0 located at point O, all vectors of electric field intensity are radial, directed towards or away from point O depending on its charge (positive or negative). Then on the distance r from O at point P the magnitude of the electric field intensity equals to E=k·q0 /r², it's directed from O to P for positive charge q0 or from P to O for negative q0.
In this formula the constant k is often expressed as k=1/(4πε), where ε is the permittivity of the media between points O and P with ε0 used for permittivity of vacuum and εr=ε/ε0 defined as relative permittivity of the media.
In general, an electric field intensityE (a vector) produces a force Fe (also a vector) on any charge q: Fe = q·E
The Gauss Law - Electric
At any point of the electric field, where its intensity vector E is defined, we can position an infinitesimal flat piece of surface σ and calculate the electric field fluxΦE(σ) flowing through it.
This flux is defined as a scalar product of two vectors - the vector of electric field intensityE and a normal vector A to a piece of a surface σ proportional in its magnitude to an area of this piece of surface and directed perpendicularly to it: ΦE(σ) = E·A
where |A| = area(σ) and A⊥σ.
The picture below illustrates this:
If electric field intensity vector is already perpendicular to a piece of a surface, the flux is just a product of magnitude of this vector and an area of a piece of a surface.
Consider a simple case of a point charge q. Space around it constitutes the electric field and, as mentioned above, the vector of electric field intensityE is radial and its magnitude on a distance r equals to E=k·q/r² where constant k=1/(4πε).
Consider a sphere of radius r with a center at this point charge. Since the magnitude of a vector of electric field intensity is constant on this sphere and the direction of this vector is always radial (which implies that it's perpendicular to a sphere's surface), the total electric field flux through a sphere can be expressed as a product of the magnitude of the electric field intensity by an area of a sphere: ΦE=[1/(4πε)]·(q/r²)·4πr²=q/ε
We have received a wonderfully simple formula that states that the total flux of electric field through a sphere does not depend on a radius of this sphere, it depends only on amount of charge inside this sphere and the permittivity of a media.
Remarkably, the same result we would be able to obtain, if the surface around a point charge is not exactly spherical, but is any closed surface. The mathematics of it are a little more involved, and we just state this as a fact.
Moreover, the charge inside a closed surface does not have to be a point charge. Any distribution of charge inside a closed surface will produce the same result - the total flux through a surface is the total amount of charge divided by permittivity of media. It's easy to understand because multiple fields have a property of superposition - total intensity is a vector sum of individual component intensities.
The above is the Gauss Law for electric field.
As a trivial consequence, imagine any closed surface, like a cube or ellipsoid, in the electric field with no charge inside it.
According to Gauss Law, the flux of electric field through this closed surface must be zero, which is understandable because what comes in should get out without any addition or subtraction since there is no charge inside.
On the other hand, if there is a charge inside this surface, it will contribute its electric field intensity to the one that came from outside, which will distort the balance between incoming and outgoing electric field intensities. It the charge outside was positive and the one inside was positive as well, the resulting outgoing from a surface intensity will be greater than incoming. Same with negative outside and negative inside.
A different form of the same law can be obtained by making the following thought experiment.
Assume we have electric field in three-dimensional Cartesian coordinate space of intensity E(x,y,z) at point {x,y,z} produced by electric charge distributed in space with density ρ(x,y,z).
Let's consider a very small (of course, we intend to shrink it to infinitesimal) parallelepiped aligned along coordinate planes with one corner at point A{x,y,z} and dimensions Δx⨯Δy⨯Δz.
Inside this parallelepiped there is some charge, which we can consider to be uniform since our parallelepiped is very small: q=ρ(x,y,z)·Δx·Δy·Δz
This charge will emit electric field to all six sides of a parallelepiped.
Our plan is to calculate the total flux of the electric field through a surface of this parallelepiped and equate it to a total charge inside, which, according to the Gauss Law discussed above, should be equal to this flux (with a factor ε).
Let's consider X-, Y- and Z-components Ex(x,y,z), Ey(x,y,z) and Ez(x,y,z) of a vector of field intensity E(x,y,z).
Let's calculate the flux through the right side of a parallelepiped and its left side, separated by Δx.
We have to consider only Ex component of the electric field intensity vector since other components are parallel to surfaces under consideration.
The flux on the right side is Φright = Ex(x+Δx,y,z)·Δy·Δz
The flux on the left side is Φleft = −Ex(x,y,z)·Δy·Δz
(minus sign in front is used because the direction of a normal vector on the left side of the parallelepiped is opposite to the one on the right side)
Sum of both is the flux going through parallelepiped in the X-direction of the space: Φx = [Ex(x+Δx,y,z)−Ex(x,y,z)]·
·Δy·Δz
Considering we intend to make all Δ dimensions infinitesimal, the difference between two values of Ex can be expressed as Ex(x+Δx,y,z)−Ex(x,y,z) =
= [∂Ex(x,y,z)/∂x]·Δx
Now the flux along the X-axis is Φx = [∂Ex(x,y,z)/∂x]·Δx·Δy·Δz
Analogously, flux along the Y-axis is Φy = [∂Ey(x,y,z)/∂y]·Δx·Δy·Δz
And along Z-axis: Φz = [∂Ez(x,y,z)/∂z]·Δx·Δy·Δz
The total electric field flux through and the total charge inside a parallelepiped, according to the Gauss Law are related as Φx + Φy + Φz = q/ε
As we noted above, the total charge inside a parallelepiped is q=ρ(x,y,z)·Δx·Δy·Δz
Therefore, ∂Ex(x,y,z)/∂x +
+ ∂Ey(x,y,z)/∂y +
+ ∂Ez(x,y,z)/∂z = ρ/ε
To simplify a notation, mathematicians came up with a construction that looks like a vector with three components: ∇ = {∂/∂x, ∂/∂x, ∂/∂x}
Then a construction that resembles a scalar (dot) product of this pseudo-vector ∇ and a real vector E={Ex,Ey,Ez} would look like ∇·E=∂Ex/∂x+∂Ey/∂y+∂Ez/∂z
With all these assumptions the Gauss Law in this differential form looks really short and simple: ∇·E = ρ/ε
The above equation is the first of four Maxwell differential equations that describe the electromagnetic field.
Assume that we have a rope stretched along the X-axis (horizontally).
Its end (x=0) is forcefully oscillated up and down according to a formula D(t)=A·cos(ω·t).
This causes a rope's piece positioned at distance x from the forcefully oscillated end (x=0) to perform harmonic up and down oscillations along the Y-axis (vertically) according to a formula y(x,t) = A·cos(ω·t−k·x)
where y(x,t) is vertical deviation of the rope's piece at distance x from the source of oscillations as a function of time t, A is an amplitude of oscillations along the Y-axis, ω is an angular speed of oscillations, t is time, k is a rope's characteristic.
The physical meaning of k will be clear, if the formula for oscillations at distance x from the rope's end is presented as y=A·cos(ω·(t−x/v)), where v is a speed of wave propagation along a rope and, therefore, x/v is a time delay between oscillations at their origin x=0 and oscillations at distance x from the origin.
Then we can construct different relations between
parameter k,
speed of wave propagation v,
period of oscillation τ,
frequency of oscillations f and
wavelength λ
as follows: k = ω/v v = λ/τ τ = 1/f v = λ·f 2π·f = ω k = ω/v = ω/(λ·f) = 2π/λ
Previous lecture was about modeling the oscillations of a horizontally stretched rope with oscillations of an infinite number of vertically oriented tiny springs, each attached to an independent infinitesimal piece of a rope.
The characteristics of this approach to model rope's oscillations are as follows:
1. Mass attached to each tiny spring equals to dm=μ·dx.
2. Location of each tiny spring is x - its distance from the rope's end.
3. All springs have the same elasticity ke=ω²·μ·dx
4. All springs are initially stretched by A, so their amplitude will be the same as that of the rope.
5. All springs will have the same angular speed ω, the same as waves on a rope.
6. After initial stretch the springs will be let go, but not simultaneously; the time delay of a spring at distance x from the beginning will correspond to time needed for a wave on a rope to reach that distance, that is the time delay will be equal to x/v=x/(λ·f)=2π·x/(λ·ω).
7. As a result, a spring at distance x from the beginning oscillates according to a formula y(x,t) = A·cos(ω·t−k·x) =
= A·cos[ω·(t−2π·x/(λ·ω))]
where k = 2π/λ
Modeled as described above, rope oscillations are identical to oscillations of all those tiny springs. The point-masses at the top of these springs will oscillate exactly the same as if they are attached to a rope.
The above characteristics of the springs, that we have used to model the propagating waves on a rope, are sufficient to analyze the distribution of energy, using what we have already analyzed about springs.
Our task is to find what energy is carried by one single wave of a rope. For this we will calculate the energy carried by a single tiny spring that models oscillations of an infinitesimal piece of a rope dx and integrate it along the wavelength of oscillations λ.
For this purpose we consider the tiny springs along the X-axis within any segment of the length λ, for example, from x=0 to x=λ.
Each individual tiny spring at fixed X-coordinate x carries potential and kinetic energies, and these energies are changing with the time, that is they are functions of time for each fixed spring at distance x from origin.
Consider a potential energy of an individual tiny single spring U(x,t) for any fixed x as a function of time t.
This potential energy depends only on its vertical deviation y(x,t) which determines the degree of a stretch.
Earlier we derived a formula for potential energy of a spring with elasticity ke stretched by d from a neutral position as ½ke·d².
In our case distance a spring is stretched is d=y(x,t).
The elasticity coefficient was derived above as a function of required angular speed ω of oscillations for a particular mass dm=μ·dx attached to a spring: ke=ω²·μ·dx.
So, the potential energy of a spring at distance x from the origin of oscillations at time t is U(x,t) = ½ke·y²(x,t) =
= ½ω²·μ·dx·A²·cos²(ω·t−k·x),
where k = 2π/λ.
For each moment in time t (that is, t is fixed and participates in the integration by x as a constant) the potential energy of all springs within a segment [0,λ] of a distance from the origin of oscillation is a definite integral of the above expression by x from x=0 to x=λ.
Let's find integral I = ∫0λcos²(ω·t−k·x)dx,
where k = 2π/λ, and multiply the result by a constant ½ω²·μ·A².
To calculate I, we substitute z = ω·t−k·x dz = −k·dx dx = −dz/k
We have to express the limits of integration in terms of z:
if x=0, z=ω·t
if x=λ, z=ω·t−k·λ=ω·t−2π
Now our integral looks like I = −∫0λcos²(z)dz/k
Calculating the indefinite integral ∫cos²(z)·dz = ½z + ¼sin(2z) + C
Using Newton-Leibnitz formula for definite integral, we have to substitute upper limit for z, then lower limit and subtract the latter from the former: ½(ω·t−2π)+¼sin(2·(ω·t−2π)) −
− ½ω·t−¼sin(2ω·t) = −π
Note, this integral is independent of time t. It means that combined potential energy of all tiny springs on a segment of the length λ does not change with time, as one spring is stretched or squeezed more, another is less, so the total sum remains constant. For a rope it means that a single rope's wave has potential energy in all its pieces not changing with time.
Therefore, I = π/k = π·λ/(2π) = ½λ
Consequently, the potential energy of all tiny springs on a segment of the length λ will be Uλ = ½ω²·μ·A²·I =
= ¼ω²·μ·A²·λ
Let's consider a kinetic energy of all tiny springs on a segment of the length λ with attached to them infinitesimal masses.
The general formula for kinetic energy is K = ½M·V²,
where M is mass and V is speed of an object.
For a single tiny spring with an infinitesimal mass attached to it we have M = μ·dx - product of linear mass density μ by the length of a piece of a rope attached to the top of a spring dx V = ∂y(x,t)/∂t - the first derivative of vertical position of the mass attached to a spring by time
Since we know the expression for vertical movements of the top of a spring y(x,t) = A·cos(ω·t−k·x),
where k = 2π/λ
we can find the speed of vertical movement at any moment of time: ∂y(x,t)/∂t = −ω·A·sin(ω·t−k·x)
The kinetic energy of a point-mass attached to this tiny spring is K(x,t) = ½M·V² =
= ½μ·dx·ω²·A²·sin²(ω·t−k·x)
As you see, the expression for K(x,t) resembles the expression for U(x,t) with the only difference that this time we have sin²() instead of cos²().
When we integrate this expression by x from 0 to λ (that is, across all tiny springs that cover a distance equaled to a wavelength λ), we will have a similar result - the total kinetic energy of all point-masses attached to springs on a segment of the length λ does not depend on time t and equals to the same value Kλ = ¼ω²·μ·A²·λ
It's quite remarkable that potential and kinetic energies of one wavelength of a rope are the same and independent of time.
Now we can calculate the combined energy carried by a single piece of a rope of the length equaled to a wavelength of oscillations: Eλ = Uλ + Kλ = ½ω²·μ·A²·λ
The above amount of energy is carried by one wavelength of a oscillations. That takes an amount of time equaled to a period of oscillations τ.
Therefore, the oscillations produce an average power Pave = Eλ/τ = ½ω²·μ·A²·λ/τ
Since λ/τ = v (speed of wave propagation), we conclude Pave = Eλ/τ = ½ω²·μ·A²·v
The following picture schematically represents a wave (red line) and the level of its kinetic energy (blue line), which is based on the shape of function sin²(ω·t−k·x)
The potential energy graph is based on cos²(ω·t−k·x) and its graph would be similar, just shifted to be in anti-phase with kinetic energy.
Their sum is constant because sin²()+cos²()=1.
