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Chapter 2
Brief Review of Electromagnetics
2.1 Introduction
The specific equations on which the finite-difference time-domain (FDTD) method is based will be considered in some detail later. The goal here is to remind you of the physical significance of the equations to which you have been exposed in previous courses on electromagnetics. In some sense there are just a few simple premises which underlie all electromagnetics. One could argue that electromagnetics is simply based on the following:
- Charge exerts force on other charge.
- Charge in motion exerts a force on other charge in motion.
- All material is made up of charged particles.
Of course translating these premises into a corresponding mathematical framework is not trivial. However one should not lose sight of the fact that the math is trying to describe principles that are conceptually rather simple.
2.2 Coulomb’s Law and Electric Field
Coulomb studied the electric force on charged particles. As depicted in Fig. 2.1, given two discrete particles carrying charge Q 1 and Q 2 , the force experienced by Q 2 due to Q 1 is along the line joining Q 1 and Q 2. The force is proportional to the charges and inversely proportional to the square of the distance between the charges. A proportionality constant is needed to obtain Coulomb’s law which gives the equation of the force on Q 2 due to Q 1 :
F 12 = ˆa 12
4 Ãϵ 0
Q 1 Q 2
R^212
where ˆa 12 is a unit vector pointing from Q 1 to Q 2 , R 12 is the distance between the charges, and 1 / 4 Ãϵ 0 is the proportionality constant. The constant ϵ 0 is known as the permittivity of free space
Lecture notes by John Schneider. em-review.tex
14 CHAPTER 2. BRIEF REVIEW OF ELECTROMAGNETICS
Q 1
Q 2
F 12 = Q 2 E
Figure 2.1: The force experienced by charge Q 2 due to charge Q 1 is along the line which pass through both charges. The direction of the force is dictate by the signs of the charges. Electric field is assumed to point radially away from positive charges as is indicated by the lines pointing away from Q 1 (which is assumed here to be positive).
and equals approximately 8. 854 × 10 −^12 F/m. Charge is expressed in units of Coulombs (C) and can be either negative or positive. When the two charges have like signs, the force will be repulsive: F 12 will be parallel to aˆ 12. When the charges are of opposite sign, the force will be attractive so that F 12 will be anti-parallel to aˆ 12. There is a shortcoming with (2.1) in that it implies action at a distance. It appears from this equation that the force F 12 is established instantly. From this equation one could assume that a change in the distance R 12 results in an instantaneous change in the force F 12 , but this is not the case. A finite amount of time is required to communicate the change in location of one charge to the other charge (similarly, it takes a finite amount of time to communicate a change in the quantity of one charge to the other charge). To overcome this shortcoming it is convenient to employ the concept of fields. Instead of Q 1 producing a force directly on Q 2 , Q 1 is said to produce a field. This field then produces a force on Q 2. The field produced by Q 1 is independent of Q 2 —it exists whether or not Q 2 is there to experience it. In the static case, the field approach does not appear to have any advantage over the direct use of Coulomb’s law. This is because for static charges Coulomb’s law is correct. Fields must be time-varying for the distinction to arise. Nevertheless, to be consistent with the time-varying case, fields are used in the static case as well. The electric field produced by the point charge Q 1 is
E 1 = ˆar
Q 1
4 Ãϵ 0 r^2
where ˆar is a unit vector which points radially away from the charge and r is the distance from the charge. The electric field has units of volts per meter (V/m). To find the force on Q 2 , one merely takes the charge times the electric field: F 12 = Q 2 E 1. In general, the force on any charge Q is the product of the charge and the electric field at which the charge is present, i.e., F = QE.
16 CHAPTER 2. BRIEF REVIEW OF ELECTROMAGNETICS
D = ε 0 E 0
E 0
Figure 2.2: Charged parallel plates in free space. The dashed line represents the integration surface S.
− − − − − − − − − − − − − − − − − − −
E 0
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
− − − − −
bound surface charge
E m
Figure 2.3: Charged parallel plates with a polarizable material present between the plates. The elongated objects represent molecules whose charge orientation serves to produce a net bound negative charge layer at the top plate and a bound positive charge layer at the bottom plate. In the interior, the positive and negative bound charges cancel each other. It is only at the surface of the material where one must account for the bound charge. Thus, the molecules are not drawn throughout the figure. Instead, as shown toward the right side of the figure, merely the bound charge layer is shown. The free charge on the plates creates the electric field E 0. The bound charge creates the electric field Em which opposes E 0 and hence diminishes the total electric field. The dashed line again represents the integration surface S.
2.4. STATIC ELECTRIC FIELDS 17
With the material present the electric field due to the charge on the plates is still E 0 , i.e., the same field as existed in Fig. (2.2). However, there is another field present due to the displacement of the bound charge in the polarizable material between the plates. The polarized material effectively acts to establish a layer of positive charge adjacent to the bottom plate and a layer of negative charge adjacent to the top plate. The field due to these layers of charge is also uniform but it is in the opposite direction of the field caused by the “free charge” on the plates. The field due to bound charge is labeled Em in Fig. (2.3). The total field is the sum of the fields due to the bound and free charges, i.e., E = E 0 + Em. Because E 0 and Em are anti-parallel, the magnitude of the total electric field E will be less than E 0. Since the material is neutral, we would like the integral of the electric flux over the surface S to yield just the enclosed charge on the bottom plate—not the bound charge due to the material. In some sense this implies that the integration surface cannot separate the positive and negative bound charge of any single molecule. Each molecule is either entirely inside or outside the integration surface. Since each molecule is neutral, the only contribution to the integral will be from the free charge on the plate. With the material present, the integral of
S ϵ^0 E^ ·^ ds^ yields too little charge. This is because, as stated above, the total electric field E is less than it would be if only free space were present. To correct for the reduced field and to obtain the desired result, the electric flux density is redefined so that it accounts for the presence of the material. The more general expression for the electric flux density is D = ϵrϵ 0 E = ϵE (2.6)
where ϵr is the relative permittivity and ϵ is called simply the permittivity. By accounting for the permittivity of a material, Gauss’s law is always satisfied. In (2.6), D and E are related by a scalar constant. This implies that the D and E fields are related by a simple proportionality constant for all frequencies, all orientations, and all field strengths. Unfortunately the real world is not so simple. Clearly if the electric field is strong enough, it would be possible to tear apart the bound positive and negative charges. Since charges have some mass, they do not react the same way at all frequencies. Additionally, many materials may have some structure, such as crystals, where the response in one direction is not the same in other directions. Nevertheless, Gauss’s law is the law and thus always holds. When things get more complicated one must abandon a simple scalar for the permittivity and use an appropriate form to ensure Gauss’s law is satisfied. So, for example, it may be necessary to use a tensor for permittiv- ity that is directionally dependent. However, with the exception of frequency-dependent behavior (i.e., dispersive materials), we will not be pursuing those complications. A scalar permittivity will suffice.
2.4 Static Electric Fields
Ignoring possible nonlinear behavior of material, superposition holds for electromagnetic fields. Therefore we can think of any distribution of charges as a collection of point charges. We can get the total field by summing the contributions from all the charges (and this summing will have to be in the form of an integration if the charge is continuously distributed). Note from (2.2) that the field associated with a point charge merely points radially away from the charge. There is no “swirling” of the field. If we have more than a single charge, the total
2.5. GRADIENT, DIVERGENCE, AND CURL 19
Dy ( x , y +∆ y /2)
Dx ( x −∆ x /2, y ) (^) Dx ( x +∆ x /2, y )
Dy ( x , y −∆ y /2)
x
y
Figure 2.4: Discrete approximation to the divergence taken in the xy-plane.
the change in the field realized by moving an amount dℓ is
∇f · dℓ =
∂f ∂x
dx +
∂f ∂y
dy +
∂f ∂z
dz. (2.13)
Returning to (2.8), when the del operator is dotted with a vector field, one obtains the diver- gence of that field. Divergence can be thought of as a measure of “source” or “sink” strength of the field at a given point. The divergence of a vector field is a scalar field given by
∇ · D =
∂Dx ∂x
∂Dy ∂y
∂Dz ∂z
Let us consider a finite-difference approximation of this divergence in the xy-plane as shown in Fig. 2.4. Here the divergence is measured over a small box where the field is assumed to be constant over each edge of the box. The derivatives can be approximated by central differences:
∂Dx ∂x
∂Dy ∂y
Dx
x + ∆ 2 x , y
− Dx
x − ∆ 2 x , y
∆x
Dy
x, y + ∆ 2 y
− Dy
x, y − ∆ 2 y
∆y
where this is exact as ∆x and ∆y go to zero. Letting ∆x = ∆y = ¶, (2.15) can be written
∂Dx ∂x
∂Dy ∂y
Dx
x +
, y
− Dx
x −
, y
x, y +
− Dy
x, y −
Inspection of (2.16) reveals that the divergence is essentially a sum of the field over the faces with the appropriate sign changes. Positive signs are used if the field is assumed to point out of the box and negative signs are used when the field is assumed to point into the box. If the sum of these values is positive, that implies there is more flux out of the box than into it. Conversely, if the sum is negative, that means more flux is flowing into the box than out. If the sum is zero, there must
20 CHAPTER 2. BRIEF REVIEW OF ELECTROMAGNETICS
Ex ( x , y +∆ y /2)
Ey ( x −∆ x /2, y ) (^) Ey ( x +∆ x /2, y )
Ex ( x , y −∆ y /2)
x
y
( x , y )
Figure 2.5: Discrete approximation to the curl taken in the xy-plane.
be as much flux flowing into the box as out of it (that does not imply necessarily that, for instance, Dx (x + ¶/ 2 , y) is equal to Dx (x − ¶/ 2 , y), but rather that the sum of all four fluxes must be zero). Equation (2.8) tells us that the electric flux density has zero divergence except where there is charge present (as specified by the charge-density term Äv). If the charge density is zero, the total flux entering some small enclosure must also leave it. If the charge density is positive at some point, more flux will leave a small enclosure surrounding that point than will enter it. On the other hand, if the charge density is negative, more flux will enter the enclosure surrounding that point than will leave it. Finally, let us consider (2.7) which is the curl of the electric field. In Cartesian coordinates it is possible to treat this operation as simply the cross product between the vector operator ∇ and the vector field E:
∇ × E =
ˆax ˆay aˆz ∂ ∂x
∂ ∂y
∂ ∂z Ex Ey Ez
= ˆax
∂Ez ∂y
∂Ey ∂z
∂Ex ∂z
∂Ez ∂x
∂Ey ∂x
∂Ex ∂y
Let us consider the behavior of only the z component of this operator which is dictated by the field in the xy-plane as shown in Fig. 2.5. The z-component of ∇ × E can be written as
∂Ey ∂x
∂Ex ∂y
Ey
x + ∆ 2 x , y
− Ey
x − ∆ 2 x , y
∆x
Ex
x, y + ∆ 2 y
− Ex
x, y − ∆ 2 y
∆y
The finite-difference approximations of the derivatives are again based on the fields on the edges of a box surrounding the point of interest. However, in this case the relevant fields are tangential to the edges rather than normal to them. Again letting ∆x = ∆y = ¶, (2.18) can be written
∂Ey ∂x
∂Ex ∂y
Ey
x +
, y
− Ey
x −
, y
− Ex
x, y +
x, y −
22 CHAPTER 2. BRIEF REVIEW OF ELECTROMAGNETICS
The integrals in (2.22) and (2.23) can be equated. Since the equality holds for any two arbitrary points, the integrands must be equal and we are again left with E = −∇V. The electric flux density can be related to the electric field via D = ϵE and the behavior of the flux density D is dictated by ∇ · D = Äv. Combining these with (2.21) yields
E =
ϵ
D = −∇V. (2.24)
Taking the divergence of both sides yields
1 ϵ
∇ · D =
ϵ
Äv = −∇ · ∇V. (2.25)
Rearranging this yields Poisson’s equation given by
∇^2 V = −
Äv ϵ
where ∇^2 is the Laplacian operator
∇^2 ≡ ∇ · ∇ =
∂^2
∂x^2
∂^2
∂y^2
∂^2
∂z^2
Note that the Laplacian is a scalar operator. It can act on a scalar field (such as the potential V as shown above) or it can act on a vector field as we will see later. When it acts on a vector field, the Laplacian acts on each component of the field. In the case of zero charge density, (2.26) reduces to Laplace’s equation
∇^2 V = 0. (2.28)
We have a physical intuition about what gradient, divergence, and curl are telling us, but what about the Laplacian? To answer this, consider a function of a single variable. Given the function V (x), we can ask if the function at some point is greater than, equal to, or less than the average of its neighboring values. The answer can be expressed in terms of the value of the function at the point of interest and the average of samples to either side of that central point:
V (x + ¶) + V (x − ¶) 2
− V (x) =
positive if center point less than average of neighbors zero if center point equals average of neighbors negative if center point greater than average of neighbors (2.29) Here the left-most term represents the average of the neighboring values and ¶ is some displace- ment from the central point. Equation (2.29) can be normalized by ¶^2 / 2 without changing the properties of this equation. Performing that normalization and rearranging yields
1 ¶^2 / 2
ø V (x + ¶) + V (x − ¶) 2
− V (x)
ø
¶^2
{(V (x + ¶) − V (x)) − (V (x) − V (x − ¶))}
V (x+δ)−V (x) δ −^
V (x)−V (x−δ) δ ¶
≈
∂V (x+δ/2) ∂x −^
∂V (x−δ/2) ∂x ¶
≈
∂^2 V (x) ∂x^2
2.6. LAPLACIAN 23
-4 -2 2 4
1
uniform sphere of charge
potential
distance from center
Figure 2.6: Potential along a path which passes through a uniform sphere of positive charge (arbi- trary units).
Thus the second partial derivative can be thought of as a way of measuring the field at a point relative to its neighboring points. You should already have in mind that if the second derivative is negative, a function is tending to curve downward. Second derivatives are usually discussed in the context of curvature. However, you should also think in terms of the field at a point and its neighbors. At points where the second derivative is negative those points are higher than the average of their neighboring points (at least if the neighbors are taken to be an infinitesimally small distance away). In lieu of these arguments, Poisson’s equation (2.26) should have physical significance. Where the charge density is zero, the potential cannot have a local minima or maxima. The potential is always equal to the average of the neighboring points. If one neighbor is higher, the other must be lower (and this concept easily generalizes to higher dimensions). Conversely, if the charge density is positive over some region, the potential should increase as one moves deeper into that region but the rate of increase must be such that at any point the average of the neighbors is less than the center point. This behavior is illustrated in Fig. 2.6 which depicts the potential along a path through the center of a uniform sphere of charge.
2.8. ELECTRIC FIELD BOUNDARY CONDITIONS 25
2.8 Electric Field Boundary Conditions
Consider an interface between two homogeneous regions. Because electric flux density only begins or ends on charge, the normal component of D can only change at the interface if there is charge on the interface, i.e., surface charge is present. This can be stated mathematically as
n ˆ · (D 1 − D 2 ) = Äs (2.35)
where Äs is a surface charge density (C/m^2 ), nˆ is a unit vector normal to the surface, and D 1 and D 2 are the field to either side of the interface. One should properly argue this boundary condition by an application of Gauss’s law for a small volume surrounding the surface, but such details are left to other classes (this is just a review!). If no charge is present, the normal components must be equal n^ ˆ · D 1 = ˆn · D 2. (2.36) The boundary conditions on the tangential component of the electric field can be determined by integrating the electric field over a closed loop which is essentially a rectangle which encloses a portion of the interface. By letting the sides shrink to zero and keeping the “top” and “bottom” of the rectangle small but finite (so that they are tangential to the surface), one essentially has that the field over the top must be the same as the field over the bottom (owing to the fact that total integral must be zero since the field is conservative). Stated mathematically, the boundary condition is
n ˆ × (E 1 − E 2 ) = 0. (2.37)
2.9 Conductivity and Perfect Electric Conductors
It is possible for the charge in materials to move under the influence of an electric field such that currents flow. If the material has a non-zero conductivity Ã, the current density is given by
J = ÃE. (2.38)
The current density has units of A/m^2 and the conductivity has units of S/m. If charge is building up or decaying in a particular region, the divergence of the current density must be non-zero. If the divergence is zero, that implies as much current leaves a point as enters it and there is no build-up or decay of charge. This can be stated as
∇ · J = −
∂Äv ∂t
If the divergence is positive, the charge density must be decreasing with time (so the negative sign will bring the two into agreement). This equation is a statement of charge conservation. Perfect electric conductors (PECs) are materials where it is assumed that the conductivity ap- proaches infinity. If the fields were non-zero in a PEC, that would imply the current was infinite. Since infinite currents are not allowed, the fields inside a PEC are required to be zero. This sub- sequently requires that the tangential electric field at the surface of a PEC is zero (since tangential fields are continuous across an interface and the fields inside the PEC are zero). Correspondingly, the normal component of the electric flux density D at the surface of a PEC must equal the charge density at the surface of the PEC. Since the fields inside a PEC are zero, all points of the PEC must be at the same potential.
26 CHAPTER 2. BRIEF REVIEW OF ELECTROMAGNETICS
2.10 Magnetic Fields
Magnetic fields circulate around, but they do not terminate on anything—there is no (known) magnetic charge. Nevertheless, it is often convenient to define magnetic charge and magnetic current. These fictions allow one to simplify various problems such as integral formulations of scattering problems. However for now we will stick to reality and say they do not exist. The magnetic flux density B is somewhat akin to the electric field in that the force on a charge in motion is related to B. If a charge Q is moving with velocity u in a field B, it experiences a force F = Qu × B. (2.40)
Because B determines the force on a charge, it must account for all sources of magnetic field. When material is present, the charge in the material can have motion (or rotation) which influences the magnetic flux density. Alternatively, similar to the electric flux density, we define the magnetic field H which ignores the local effects of material. These fields are related by
B = μrμ 0 H = μH (2.41)
where μr is the relatively permeability, μ 0 is the permeability of free space equal to 4 Ã × 10 −^7 H/m, and μ is simply the permeability. Typically the relative permeability is greater than unity (although usually only by a small amount) which implies that when a material is present the magnetic flux density is larger than when there is only free space. Charge in motion is the source of magnetic fields. If a current I flows over an incremental distance dℓ, it will produce a magnetic field given by:
H =
Idℓ × ar 4 Ãr^2
where ar points from the location of the filament of current to the observation point and r is the distance between the filament and the observation point. Equation (2.42) is known as the Biot- Savart equation. Of course, because of the conservation of charge, a current cannot flow over just a filament and then disappear. It must flow along some path. Thus, the magnetic field due to a loop of current would be given by
H =
L
Idℓ × ar 4 Ãr^2
If the current was flowing throughout a volume or over a surface, the integral would be correspond- ingly changed to account for the current wherever it flowed. From (2.43) one sees that currents (which are just another way of saying charge in motion) are the source of magnetic fields. Because of the cross-product in (2.42) and (2.43), the magnetic field essentially swirls around the current. If one integrates the magnetic field over a closed path, the result is the current enclosed by that path
L
H · dℓ = Ienc. (2.44)
28 CHAPTER 2. BRIEF REVIEW OF ELECTROMAGNETICS
If a loop carries a current but is otherwise neutral, it will produce a magnetic field and only a magnetic field. If a charge is stationary, it will produce an electric field and only an electric field. The charge will not “feel” the loop current and the current loop will not feel the stationary charge (at least approximately). The magnetic field and electric field are decoupled. If a charge Q moves with velocity u in the presence of both an electric field and a magnetic field, the force on the charge is the sum of the forces due to the electric and magnetic fields
F = Q(E + u × B). (2.54)
2.13 Time Varying Fields
What happens when a point charge moves? We know that charge in motion gives rise to a magnetic field, but if the charge is moving, its associated electric field must also be changing. Thus, when a system is time-varying the electric and magnetic fields must be coupled. There is a vector identity that the divergence of the curl of any vector field is identically zero. Taking the divergence of both sides of (2.53) yields
∇ · ∇ × H = ∇ · J = −
∂Äv ∂t
where the conservation of charge equation was used to write the last equality. Since the first term must be zero, this implies that ∂Äv/∂t must also be zero. However, that is overly restrictive. In general, for a time-varying system, the charge density will change with respect to time. Therefore something must be wrong with (2.53) as it pertains to time-varying fields. It was Maxwell who recognized that by adding the temporal derivative of the electric flux density to the right-hand side of (2.53) the equation would still be valid for the time-varying case. The correct equation is given by
∇ × H = J +
∂D
∂t
The term ∂D/∂t is known as the displacement current while J is typically called the conduction current. Equation (2.56) is known as Ampere’s law. Taking the divergence of the right-hand side of (2.56) yields
∇ · J +
∂∇ · D
∂t
∂Äv ∂t
∂∇ · D
∂t
∂Äv ∂t
∂Äv ∂t
where use was made of (2.8) and the conservation of charge equation (2.39). The electromotive force (EMF) is the change in potential over some path. It has been observed experimentally that when a magnetic field is time-varying there is a non-zero EMF over a closed path which encloses the varying field (i.e., the electric field is no longer conservative). The symbol ¼ is often use to represent total magnetic flux through a given surface, i.e.,
S
B · ds. (2.58)
2.14. SUMMARY OF TIME-VARYING FIELDS 29
For time-varying fields, the EMF over a closed path L can be written
Vemf =
d¼ dt
L
E · dℓ =
d dt
S
B · ds, (2.60)
S
∇ × E · ds =
S
∂B
∂t
· ds, (2.61)
where Stokes’ theorem was used to write the last equation. Since this equality holds over any surface, the integrands must be equal. This yields
∇ × E = −
∂B
∂t
which is known as Faraday’s law.
2.14 Summary of Time-Varying Fields
When a system is changing with respect to time, the governing equations are
∇ · D = Äv, (2.63) ∇ · B = 0 , (2.64)
∇ × E = −
∂B
∂t
∇ × H = J +
∂D
∂t
Note that the divergence equations are unchanged from the static case. The two curl equations have picked up terms which couple the electric and magnetic fields. Since the additional terms both involve temporal derivatives, they go to zero in the static case and the equations reduce to those which governed static fields. For time-varying fields the same boundary conditions hold as in the static case.
2.15 Wave Equation in a Source-Free Region
Equations (2.63)–(2.66) provide a set of coupled differential equations. In the FDTD method we will be dealing directly with the two curl equations. We will stick to the coupled equations and solve them directly. However, it is also possible to decouple these equations. As an example, taking the curl of both sides of (2.65) yields
∇ × ∇ × E = −∇ ×
∂B
∂t
= −μ∇ ×
∂H
∂t
There is a vector identity that the curl of the curl of any field is given by
∇ × ∇ × E = ∇(∇ · E) − ∇^2 E. (2.68)
2.16. ONE-DIMENSIONAL SOLUTIONS TO THE WAVE EQUATION 31
Employing the chain rule in a similar fashion, the second derivatives can be written as
∂ ∂t
∂f (À) ∂t
∂t
∂f (À) ∂À
∂^2 f (À) ∂À^2
∂À
∂t
∂^2 f (À) ∂À^2
∂x
∂f (À) ∂x
∂x
c
∂f (À) ∂À
c
∂^2 f (À) ∂À^2
∂À
∂x
c^2
∂^2 f (À) ∂À^2
Thus, (2.79) and (2.80) show that
∂^2 f ∂t^2
∂^2 f ∂À^2
∂^2 f ∂x^2
c^2
∂^2 f ∂À^2
Substituting these into (2.73) yields
1 c^2
∂^2 f ∂À^2
c^2
∂^2 f ∂À^2
The two terms on the left-hand side cancel, thus satisfying the equation. Consider a constant argument of f , say t − x/c = 0. Assume this argument is obtain by simultaneously having t = 0 and x = 0. Now, let time advance by one second, i.e., t = 1 s. How must the position x change to maintain an argument of zero? Solving for x yields x = c(1 s). In other words to move along with the field so as to maintain the value f (0) (whatever that value happens to be), at time zero, we would be at position zero. At time one second, we would have to have moved to the location x = c(1 s). Speed is change in position over change in time. Thus the speed with which we are moving is x/t = c(1 s)/(1 s) = c. In these notes c will typically be used to represent the speed of light in free space. Using the permittivity and permeability of free space, we obtain c = 1/
ϵ 0 μ 0 ≈ 3 × 108 m/s.
32 CHAPTER 2. BRIEF REVIEW OF ELECTROMAGNETICS
