Retarded Potential, Radiation Field and Power for a Short Dipole Antenna

Module 1- Antenna: Retarded potential, radiation ﬁeld and power for a short dipole antenna

ELL 212 Instructor: Debanjan Bhowmik Department of Electrical Engineering Indian Institute of Technology Delhi

Abstract An antenna is a device that acts as interface between electromagnetic waves propagating in free space and electric current flowing in metal conductor. It is one of the most beautiful devices that we study in electrical engineering since it combines the concepts of flow of electricity in circuits and propagation of waves in free space. The governing physics behind antenna, e.g. how and why antenna radiates power, can be confusing to learn. It is only after a careful study of the Maxwell’s equations that we can start understanding the physics of antenna. In this module we shall discuss the physics of radiation of an antenna in details. We will first learn Green’s functions because that will help us in understanding the concept of retarded vector potential, without which we will not be able to derive the radiation field for time varying charge and current and show its ”1/r” dependence. We will then derive the expressions for radiated field and power for time varying current flowing through a short dipole antenna. (Reference: a) Classical electrodynamics- J.D. Jackson b) Electromagnetics for Engineers- T. Ulaby)

1 We need the Maxwell’s equations throughout the module. So let’s list them here ﬁrst (for vacuum): ρ ∇~ .E~ = (1) 0

∇~ .B~ = 0 (2)

∂B~ ∇~ × E~ = − (3) ∂t

1 ∂E~ ∇~ × B~ = µ J~ + (4) 0 c2 ∂t Also scalar potential φ and vector potential A~ are deﬁned as follows:

∂A~ E~ = −∇~ φ − (5) ∂t

B~ = ∇~ × A~ (6) Note that equation (1)-(4) are independent equations but equation (5) is dependent on equation (3) and equation (6) is dependent on equation (4).

1 Green’s function and retarded potential

Let L be an operator such that Lψ(~r) = f(~r) (7) L is a function of position vector ~r) If f(~r) is known and ψ(~r) needs to be evaluated, using Z f(~r) = δ(~r − r~0)f(r~0)d3r~0 (8) and LG(r, r0) = δ(~r − r~0) (9) we can show that Z ψ(~r) = G(r, r0)f(r0)d3r~0 (10) is a solution of equation (7) as below: Z Z Z Lψ(~r) = L G(r, r0)f(r0)d3r~0 = LG(r, r0)f(r0)d3r~0 = δ(~r − r~0)f(r0)d3r~0 = f(~r) (11)

2 Note that this Green’s function method is a multi variable extension of the formalism in continuous time unit impulse response and convolutional integral representation of Linear Time Invariant (LTI) systems (Reference: Signals and Systems- Oppenheim). In that formalism, if the input signal is given by x(t) and output signal is y(t) then we can write: Z ∞ Z ∞ x(t) = x(τ)δ(t − τ)dτ; y(t) = x(τ)h(t − τ)dτ (12) −∞ −∞ where h(t) is the impulse response to δ(t). Now we can think of the LTI system as solution to a ordinary differential equation, with initial condition equal to 0. Without stating an initial condition, multiple outputs are possible for a given input, then the system is non- deterministic. By making the initial/boundary condition 0 we make the system linear. Let the differential operator corresponding to the differential equation be D and it is only a function of independent variable t. We can write Dy(t) = x(t) (13) Also Dh(t − τ) = δ(t − τ) (14) We can show Z ∞ y(t) = x(τ)h(t − τ)dτ (15) −∞ is solution to equation (13) as follows: Z ∞ Z ∞ Dy(t) = D x(τ)h(t − τ)dτ = x(τ)Dh(t − τ)dτ −∞ −∞ Z ∞ = x(τ)δ(t − τ)dτ = x(t) (16) −∞ What we showed for Green’s function above is exactly same as this, just that there is more than one variable and the equation is a partially differential equation.

1.1 Electrostatics In the case of electrostatics, from equation (1) and (5), ρ(~r) ∇2φ(~r) = − (17) (Vector potential A~ and ﬁeld B~ are not time varying). Using the method of Green’s function above, we evaluate scalar potential φ(~r) for a given charge distribution ρ(~r). Now. 1 ∇2(− ) = δ(~r − r~0) (18) 4π(|~r − r~0|

3 (proof in Appendix A) Using equation (10), (17) and (18)

Z 1 ρ(r~0) Z ρ(r~0) φ(~r) = (− )(− )d3r~0 = d3r~0 (19) 0 0 4π(|~r − r~ | 0 4π0(|~r − r~ |

1.2 Magnetostatics The vector potential A~(~r) can be evaluated using Green’s function for the case of magnetostatics, similar to the case of electrostatics above. Using equation (4) and (6),

∇~ × ∇~ × A~ = µ0J~ (20)

(Field E~ does not vary with time)

2 ⇒ ∇~ (∇~ .A~) − ∇ A~ = µ0J~ (21) Different values of potentials can lead to the same fields. Moving between these different potentials is called gauge transformation. Choosing vector potential A~ such that

∇~ .A~ = 0 (22)

, which is called Coulomb gauge, we get

2 ∇ A~ = −µ0J~ (23)

2 2 2 ⇒ ∇ Ax = −µ0Jx; ∇ Ay = −µ0Jy; ∇ Az = −µ0Jz (24) Using the same method of Green’s function as in the case of electrostatics, we get

Z ~0 Z ~0 Z ~0 µ0Jx(r ) 3 0 µ0Jy(r ) 3 0 µ0Jz(r ) 3 0 Ax(~r) = d r~ ; Ay(~r) = d r~ ; Az(~r) = d r~ (25) 4π(|~r − r~0|) 4π(|~r − r~0|) 4π(|~r − r~0|)

Or, Z µ J~(r~0) A~(~r) = 0 d3r~0 (26) 4π(|~r − r~0|)

1.3 Electrodynamics For time varying charge ρ(~r, t) and and current J~(~r, t),

∂A~ ∂ ρ(~r, t) ∇~ .E~ (= ∇~ .(−∇~ φ − ) = −∇2φ − (∇~ .A~) = (27) ∂t ∂t 0

4 1 ∂E~ 1 ∂ ∂A~ ∇~ × B~ = µJ~ + ⇒ ∇~ × ∇~ × A~ = µJ~ + (−∇~ φ − ) c2 ∂t c2 ∂t ∂t 1 ∂ 1 ∂2 ⇒ ∇~ (∇~ .A~) − ∇2A~ = µJ~ − (∇~ φ) − A~ c2 ∂t c2 ∂t2 1 ∂φ 1 ∂2 ⇒ ∇~ (∇~ .A~ + ) − µJ~ = ∇2A~ − A~ (28) c2 ∂t c2 ∂t2

Vector potential A~ can be chosen such that 1 ∂φ ∇~ .A~ + = 0 (29) c2 ∂t This choice of gauge is known as the Lorentz gauge. Using equation (29) in equation (27) we get 2 2 1 ∂ ρ(~r, t) ∇ φ(~r, t) − 2 2 φ(~r, t) = − (30) c ∂t 0 and using equation (29) in equation (28) we get

1 ∂2 ∇2A~(~r, t) − A~(~r, t) = −µ J~(~r, t) (31) c2 ∂t2 0 It is to be noted that equation (28) can be broken down into three scalar equations identical to equation (25) with scalars Ax,Ay and Az. All these equations ﬁt into the following form: 1 ∂2 (∇2 − )ψ(~r, t) = f(~r, t) ⇒ ψ(~r, t) = f(~r, t) (32) c2 ∂t2 which is diﬀerent from the equation for scalar potential in electrostatics (12) and equation for vector potential in magnetostatics (18) that were of the form:

∇2ψ(~r, t) = f(~r, t) (33)

1 ∂2 The c2 ∂t2 factor modiﬁes the Green’s function we obtained in the cases of electrostatics and magnetostatics and results in a time delay factor in the solution of the equation. That’s why the potential functions ψ and A~ obtained as solution of equations (31) and (32) are called retarded scalar and vector potentials respectively. We show how equation (33) is solved to obtain the retarded potential term in details in the next subsection. Note that 2 1 ∂2 ∇ − c2 ∂t2 operator is known as the d’Alembert operator () in literature.

5 1.4 Retarded potential Following the method of Green’s function described above, equation (33) can be solved as below: Z Z f(~r, t) = f(r~0, t0)δ(~r − r~0, t − t0)d3r~0dt0 (34) t0 r~0 If 1 ∂2 (∇2 − )G(~r, t, r~0, t0) = δ(~r − r~0, t − t0) (35) c2 ∂t2 then Z Z ψ(~r, t) = G(~r, t, r~0, t0)f(r~0, t0)d3r~0dt0 (36) t0 r~0 is the solution to equation (33). So basically we have to solve for G(~r, t, r~0, t0) in equation (35). Those solutions are called retarded and advanced Green’s functions as we see next. Using the Fourier transform formalism as below:

Z ∞ 1 Z ∞ F (ω) = f(τ)e−iωτ dτ; f(τ) = F (ω)eiωτ dω (37) −∞ 2π −∞

Let r = |~r − r~0|,τ = t − t0 Then equation (35) becomes

1 ∂2 (∇2 − )G(r, τ) = δ(~r − r~0)δ(τ) c2 ∂τ 2 1 ∂2 Z ∞ 1 1 Z ∞ 2 iωτ ~0 iωτ ⇒ (∇ − 2 2 ) G(r, ω)e dω = δ(~r − r ) e dω c ∂τ −∞ 2π 2π −∞ ω ⇒ (∇2 + ( )2)G(r, ω) = δ(~r − r~0) (38) c (using standard Fourier transform result for delta function and exponential function:Appendix ω 2 B) which is same as equation (18) solved in Appendix A but the ( c ) factor.

ω (∇2 + ( )2)G(r, ω) = δ(~r − r~0) c 1 d2 ω ⇒ (rG(r, ω)) + ( )2G(r, ω) = δ(~r − r~0) r dr2 c d2 ω ⇒ (rG(r, ω)) + ( )2(rG(r, ω)) = rδ(~r − r~0) (39) dr2 c When r 6= 0 the solution to equation (39) is:

i( ω )r −i( ω )r i( ω )r −i( ω )r e c e c rG(r, ω) = Ae c + Be c ⇒ G(r, ω) = A + B (40) r r

6 When ω = 0 (static case) equation (38) becomes equation (18) which we have solved in Appendix A. The solution is − 1 . So, 4π|~r−r~0| 1 A + B = − (41) 4π Now,

1 Z ∞ 1 G(r, τ) = G(r, ω)eiωτ dω 2π −∞ 2π Z ∞ Z ∞ 1 1 i( ω )r iωτ 1 i(− ω )r iωτ = (A( e c e ) + B( e c e ))dω r 2π −∞ 2π −∞ 1 r r = (Aδ(τ + ) + Bδ(τ − )) (42) r c c (using standard Fourier transform result for delta function and exponential function: Appendix B) or

1 |~r − r~0| |~r − r~0| G(~r, t, r~0, t0) = (Aδ(t − t0 + ) + Bδ(t − t0 − )) (43) |~r − r~0| c c Then from equation (36) Z Z ψ(~r, t) = G(~r, t, r~0, t0)f(r~0, t0)d3r~0dt0 t0 r~0 Z Z 1 |~r − r~0| |~r − r~0| = (Aδ(t − t0 + ) + Bδ(t − t0 − ))f(r~0, t0)d3r~0dt0 t0 r~0 |~r − r~0| c c Z 1 |~r − r~0| |~r − r~0| = (Af(r~0, t + ) + Bf(r~0, t − ))d3r~0 (44) r~0 |~r − r~0| c c

1 where A+B = − 4π . The function with coeﬃcient B is called retarded Green’s function and the function with coeﬃcient A is called advanced Green’s function. Such choice of names is obvious from the delay or advanced factors with respect to time in equation (44). For physical situations like the case of scalar potential φ and vector potential A~ the advanced Green’s function is discarded because it leads to unphysical result. So A=0 and 1 B= − 4π . Hence, solutions to equations (30) and (31) are as follows

Z 1 |~r − r~0| φ(~r, t) = ρ(r~0, t − )d3r~0; r~0 4π|~r − r~0| c Z 1 |~r − r~0| A~(~r, t) = J~(r~0, t − )d3r~0 (45) r~0 4π|~r − r~0| c

7 Thus unlike the case of electrostatics (equation 19) and magnetostatics (equation 26) in the dynamic case, potential at a given point in space at a given time t is due to the charge/ current distribution at the source at a time t’ where t-t’ is equal to the distance between the source and the point in space divided by c (speed of light). This potential term is called the retarded potential. A physical interpretation of this is that t-t’ is the time taken for the wave to propagate from source to the point in space where its is measured, so the delay factor has to be included. However merely using this physical intuition to put the retardation factor in the expressions obtained for the case of electrostatics and magnetostatics and obtaining equation (45) for electrodynamics case without doing the mathematical analysis is incomplete. We will use these expressions for retarded scalar and vector potentials (equation 45) for rest of the antenna analysis.

2 Deriving radiation ﬁeld patterns and power for time varying current: A short dipole antenna

Let us consider a short dipole antenna. It is assumed to be very short such that the length of the antenna is much shorter than the wavelength of the electromagnetic wave corresponding to the ac current ﬂowing through the antenna. Hence we can consider the current density through the antenna to be uniform. Let us consider the antenna be excited by an ac source of frequency ω. Current ﬂows through the antenna in the z direction. We can write current density through the antenna as:

I I J~ = 0 zˆ = Re( 0 eiωt)ˆz (46) s s where s is the cross-sectional area of the conductor. From equation (45) under the assumption the antenna is small and close to origin so that |~r − r~0| ≈ |~r| = r The vector potential is given by:

L iω(t− r ) µ I Z 2 e c A~(~r, t) = Re( 0 0 sdzzˆ) = 4π s L r − 2 µ ei(ωt−kr) µ ei(ωt−kr) Re( 0 I L zˆ = Re( 0 I L (cos(θ)ˆr − sin(θ)θˆ) (47) 4π 0 r 4π 0 r (Converting from cartesian coordinates to spherical coordinates). The expression for magnetic ﬁeld can be obtained as follows:

8 µ 1 ∂ ∂ 1 B~ (~r, t) = ∇ × A~(~r, t) = Re[ 0 I L( [ (ei(ωt−kr)(−sinθ)) − ( ei(ωt−kr)cosθ)])]φˆ 4π 0 r ∂r ∂θ r µ ik 1 = Re[ 0 I L( + )sinθei(ωt−kr)]φˆ (48) 4π 0 r r2 ω where k = c At any point in space other than the antenna, current density is 0 (J~ = 0). Also the time dependent term of electric ﬁeld has to be eiωt So, the expression for electric ﬁeld can be obtained as follows: iωt E~ (~r, t) = Re(E~0(~r)e ) (49)

1 ∂ E~ (~r, t) = ∇ × B~ (~r, t) ⇒ Re(iωE~ (~r)eiωt) = c2(∇ × B~ (~r, t)) c2 ∂t 0 µ 1 ∂ ik 1 1 ∂ ik 1 c2Re( 0 I L[( (( + )sin2θ)ei(ωt−kr))ˆr + ( (− (r( + )sinθe−ikr)eiωt))θˆ]) 4π 0 rsinθ ∂θ r r2 r ∂r r r2 µ 1 ik 1 1 1 1 = c2Re( 0 I L[( ( + )(2cosθ)ei(ωt−kr))ˆr+( (( sinθei(ωt−kr))+(iksin(θ)(ik+ )ei(ωt−kr)))θˆ] 4π 0 r r r2 r r2 r (50)

1 Only the r component of the magnetic field and electric field is considered for the 1 1 purpose of radiation. The other components ( r2 and r3 ) of the field are neglected because 1 they decay much faster than the r component as we move away from the source. From equation (50)

µ −k2 µ kI L Re(iωE~ (~r)eiωt) = c2Re( 0 I L( sin(θ)ei(ωt−kr)))θˆ ⇒ Re(iE (~r)eiωt) = −c 0 0 sin(θ)cos(ωt−kr)θˆ 0 4π 0 r 0 4πr µ kI L ⇒ E~ (~r) = ic 0 0 sin(θ)e−ikrθˆ (51) 0 4πr Hence µ kI L µ kI L E~ (~r, t) = Re(E~ (~r)eiωt) = Re(ic 0 0 sin(θ)e−ikreiωt)θˆ = −c 0 0 sin(θ)sin(ωt − kr)θˆ 0 4πr 4πr (52) From equation (48)

µ kI L B~ (~r, t) = − 0 0 sin(θ)sin(ωt − kr)φˆ (53) 4πr Thus equation (52) and (53) determine the electric and magnetic ﬁelds as a function of position ~r and time t for a short dipole antenna. We can make the following observations

9 about the field patterns which can give us insights about the nature of radiation from a short dipole antenna, or even antennas in general: i. The phase of the field is given by ωt − kr. Thus the wave is propagating outward from source (approximately at origin) in the radial direction. ii. From the direction cosine of the magnetic field and electric field, we can see that magnetic field and electric field are orthogonal to each other and also orthogonal to direction of propagation of the wave, like a TEM wave. Again it is to be remembered that we are only looking at the 1/r component here, there are other components of the fields which do not follow this rule. iii. The fields are very similar to that for plane wave, but the amplitude decays as 1/r where r is distance from the source (antenna) unlike a plane wave. At very large distance from the source, amplitude kind of becomes constant and it can be thought of as plane wave. iv. The propagation is symmetric about the azimuthal angle (φ) with respect to the antenna but not with respect about the polar angle (θ). This is because the current in the antenna flows in the ”z” direction and the symmetry of the fields is generated accordingly. This directionality with respect to the radiation pattern is signature of the short dipole antenna, or any general antenna.

Appendix

Appendix A: Green’s function for Laplacian operator in three dimensions Equation (18) is essentially the Green’s function for Laplacian operator in 3 dimensions. Let us show here how we get that Green’s function.

1 ∂2 ∂2 ∂2 1 ∂2 ∂2 ∂2 1 ∇2( ) = ( + + )( ) = ( + + )( ) |~r − r~0| ∂x2 ∂y2 ∂z2 |~r − r~0| ∂(x − x0)2 ∂(y − y0)2 ∂(z − z0)2 |~r − r~0| 1 ∂ ∂ 1 3a2 (r2 √ ) = − (54) r2 ∂r ∂r r2 + a2 (r2 + a2)5/2

Since there is a singularity at r=0 if we take 1 we take √ 1 with a → 0. r r2+a2

ZZZ 3a2 Z R 3a2r2 R (− )dV = −4π dr = −4π(√ )3 = −4π (55) 2 2 5/2 2 2 5/2 2 2 (r + a ) 0 (r + a ) R + a

2 since a → 0. − 3a is 0 when r is not equal to 0 and diverges when r=0. So it can be (r2+a2)5/2 thought of a delta function with a coeﬃcient. Volume intergral of delta function (in 3D) is 1, but volume intergral of this function is −4π.

10 So we can write, 1 1 ∇2( ) = −4πδ(~r − r~0) ⇒ ∇2( ) = δ(~r − r~0) (56) |~r − r~0| −4π|~r − r~0|

The same method can be followed to obtain Green’s function in one and two dimensions.

Appendix B: Fourier transform of complex exponential and delta function i( ω )r −i( ω )r Evaluating Fourier transform of a function like e c or e c (complex exponential function) is fairly involved since it needs complex analysis. On the other hand, evaluating Fourier transform of delta functions is straightforward. So we will evaluate Fourier trans- ω ω i( c )r −i( c )r r r form of e and e assuming delta function solutions δ(τ + c ) and δ(τ − c ), and i( ω )r taking Fourier transform of them to show that they indeed are Fourier transform of e c ω ∞ −i( c )r R −iωτ and e respectively. We show it next: Using F (ω) = −∞ f(τ)e dτ, we get Z ∞ r −iωτ i( ω )r δ(τ + )e dτ = e c (57) −∞ c Z ∞ r −iωτ −i( ω )r δ(τ − )e dτ = e c (58) −∞ c Similarly, Z ∞ δ(τ)e−iωτ dτ = 1 (59) −∞ 1 R ∞ iωτ Thus Fourier transform of delta function is 1. So, using that and f(τ) = 2π −∞ F (ω)e dω we get:

1 Z ∞ δ(τ) = eiωτ dω (60) 2π −∞