3/30/2016

Outlines CHAPTER 10 POTENTIALS AND FIELDS 1. The potential formulation 3/21/2016

2. Continuous Distribution Fields and Potentials 10 Chapter 2.1 Retarded potentials 2.2 Jefimenko’s equations

3. Point Charges 3.1 Lienard-Wiechert Potentials 3.2 The field of a moving point charge

2

The potential Formulism And we can see in Chapter 9, we deal with fields exclusively. There, we started with Maxwell equations and then end up In electrostatics, it is more convenient to use potential V, with Wave equations for field and for field. rather that the electric field 3/21/2016 3/21/2016 where Now we want to ask a general question regarding the

Fields and Potentials 10 Chapter Fields and Potentials 10 Chapter presentation of materials in electrodynamics, “Is there any And the field satisfy advantages to use potentials instead of the fields?” · & The answer is YES. In magnetostatics, we have Here we will try to use potentials to describe the Maxwell · & equations. We start with Faraday’s law of induction: From the fact that ·, we can define a vector Since · potential , but we don’t use that much, because is 3 4 a vector. The Faraday’s law can be written as:

The Ampere’s law on the other hand can be re-written in terms of the scalar and vector potentials as follow: 3/21/2016 3/21/2016 So we can define a potential such that, Chapter 10 Potentials and Fields and Potentials 10 Chapter Fields and Potentials 10 Chapter · Take divergence of Eq. (1) and let · ⁄, we end up with Equations (2) and (3) contain all the information of the · four Maxwell equations. However, they look quite messy and may not be easy to solve. No advantages.

5 6 However in special conditions, (2)and (3) can be simplified.

1 3/30/2016

Example 10.1 Given potentials

3/21/2016 3/21/2016 & Chapter 10 Potentials and Fields and Potentials 10 Chapter Fields and Potentials 10 Chapter Find the source distributions, , , , , & , We can show that eqs. (4) and (5) satisfy the Maxwell equations. The fields associated with the potentials are · & · ∓ , ∓ And 7 8 , & ,

As we can see from page 8, the magnetic field has a For example, in electrostatics the fields are defined through discontinuity at x=0, this leads to a surface current at x=0. the scalar potential V and vector potential ∥ ∥ 3/21/2016 3/21/2016 & Fields and Potentials 10 Chapter If we add a constant to V or add a gradient to Fields and Potentials 10 Chapter & The fields remain the same.

Gauge Transformation (Choice of Gauge) & ′

Since and field are physical observables, they are This is a remarkable result, different potentials will yield uniquely defined, once the source terms and boundary the same results for the fields. (Potential is not unique). conditions are fixed. For a set of field, there exist many We will try to use this concept to simplify the two sets of potentials that will yield the same field. We will equations (2) & (3) we obtained on page 5 and 6. find ways to choose a convenient potentials. 9 10

Now we will show that, in electrodynamics we have Substitute (9) and (10) into equation (8)

& 3/21/2016 3/21/2016 ′ We want to choose a new set of and in such way that the fields still remain the same. We start out with the vector Fields and Potentials 10 Chapter Fields and Potentials 10 Chapter potential because the way we define the magnetic field does not change when we go from electrostatics to electrodynamics. ( See equation (6)) Equations (9) and (10) describe a transformation of one set of potentials , to another set of potentials ,′ . We want to choose a new set of potentials, ′ and ′ such This is called gauge transformation. that the fields and still remain the same. Let Next, we will try to use gauge transformation to simplify ′ equations (2) and (3). 11 12

2 3/30/2016

Coulomb Gauge We notice that in the Coulomb gauge, the scalar potential is still the same as in the static case: In Coulomb gauge, we choose 3/21/2016 3/21/2016 · ′, , ′

Eq. (2) becomes Fields and Potentials 10 Chapter Fields and Potentials 10 Chapter This potential has a peculiar property, namely, the charge distribution determine the potential instantaneously. This Eq. (3) is unusual in light of special relativity. Fortunately, the electric field contains the vector potential (see eq. (1)) We can see that in the Coulomb Gauge, the scalar potential is very easy to solve, while the vector potential still remains to be a difficult problem to solve.

13 14

The Lorentz Gauge In the 4-vector notation we define In the Lorentz gauge, we choose: 3/21/2016 3/21/2016 · ,,, ,,,

Chapter 10 Potentials and Fields and Potentials 10 Chapter Fields and Potentials 10 Chapter Apply the above equation into (2) and (3), we obtain ⊡ Define a new operator, ⊡ , d’Alembertian ⊡≡ Skip 10.1.4 Lorentz force law in potential form

⊡ , ⊡ 15 16

Retarded Potentials In the electrostatic case, eqs. (17) (18) reduce to Poisson From now on we will use Lorentz Gauge only, namely equations

3/21/2016 3/21/2016 · &

So eq. (2) and (3) become Fields and Potentials 10 Chapter Fields and Potentials 10 Chapter The solutions to the above Poisson equations are already , known: z P , ′ r r’ y These equations are called inhomogeneous wave equations. They are not easy to solve. We will use a “handwaving” argument to guess a solution, and then x

show that the “guessed” solution is a reasonable 17 18 solution.

3 3/30/2016

Now, let’s start our “handwaving” argument. Define a “retarded time” First examine the following event of a charge moving

3/21/2016 3/21/2016 ≡ along a trajectory and try to find the potential at position P. B C Fields and Potentials 10 Chapter And assume the time-dependent potentials as giving by Fields and Potentials 10 Chapter A ′, V(B) at P V(C) at P , ′ V(A) at P

′, ′, Is it possible that , ′ , ′ NO! In general, this is not true! Since EM waves travel at a finite velocity. So the information of the

charge will take time to reach to point P, at a time 19 These two potentials are called retarded potentials 20 delay of

Next we want to prove that equations (21) and (22) Now if we want to solve an arbitrary inhomogeneous indeed are solutions to the inhomogeneous wave differential equation, such as the following equations (17) and (18). 3/21/2016 3/21/2016 Green function method , A point charge q at , the Poisson Equation is given by Chapter 10 Potentials and Fields and Potentials 10 Chapter We first solve Fields and Potentials 10 Chapter ′ And the potential at , due to charge q at ′ is given by , ′ The solution to eq. (25) is called the Green function, and the solution to eq. (24) is given by If we have a charge distribution (′), the potential at , ,′ ,,, ′ ′ 21 22 ′

This is superposition theorem!!!

Now we want to show that solving eq. (17) and (18) leads to Here we’ll solve a special case, assume that , · the solutions (21) and (22). We will use Green function

method. 3/21/2016 3/21/2016 , ′ · ′ Solve ,

Fields and Potentials 10 Chapter Fields and Potentials 10 Chapter The solution is given by (Let ) ′ · ,,, ′ ′ Substitute eq. (27) into eq. (26)

, , ′ , , ′ 23 24

If the above holds, then any arbitrary function of time also will be true, because we can use Fourier series.

4 3/30/2016

Next, we want to show that the retarded potentials [Eqs. Take divergence of equation (28), 1 (21), (22)] satisfy the inhomogeneous wave equations (17) 1 1 and (18). We start with equation (21) and take gradient: 3/21/2016 3/21/2016 · · , ′ Fields and Potentials 10 Chapter Fields and Potentials 10 Chapter 1 4 [ ] · · ′ [ ≡ ] Re-arrange the above expression, we end up with Substitute and ′ 25 26

Example 10.2 An infinite straight wire carries the current Eq. (30) describes the abruptness of the turn-on of the current from the point of view of the source. At point P, , 3/21/2016 3/21/2016 the appearance of the current is given by ,

Chapter 10 Potentials and Fields and Potentials 10 Chapter Z’ Fields and Potentials 10 Chapter , ⁄ P Assume the wire is electrically neutral, so . r , ⁄ , , ′ So only for , I is not zero, otherwise I = 0. Eq. (33) becomes , (32) Substitute into (31) , ′ ′ , ′ 27 28

Jefimenko’s equations To find the fields generated by these potentials When we derive the retarded potentials, Eq. (21), (22),

3/21/2016 we just used some hand-waving arguments and 3/21/2016 fortunately it works. Fields and Potentials 10 Chapter In general, we can not expect such simple argument Fields and Potentials 10 Chapter would work. For example, we can not write down the field using the retarded time argument, namely For →∞, eq.(35) (36) approach the statics case. , , ′ , Do they satisfy the Maxwell equations? YES , , ′

Do they satisfy the wave equations? YES 29 30

5 3/30/2016

The exact form of the and field can be obtained through ′ , 3/21/2016 3/21/2016 , , ′ , Chapter 10 Potentials and Fields and Potentials 10 Chapter , Fields and Potentials 10 Chapter , ′ As you can see , this is a complicated equation with no , , utility at all. So it is not very useful. Similarly can be found (see page 450 of Griffiths). , , , ′ , , 31 32

Lienard-Wiechert Potentials (Point Charge) For one single point charge moving in a trajectory

The Lienard-Wiechert potential is the retarded potential -- position of q at time tr due to one moving point charge. In particular when v ~ c. 3/21/2016 3/21/2016 It has its origin in the special theory of Relativity. ) Chapter 10 Potentials and Fields and Potentials 10 Chapter Fields and Potentials 10 Chapter Here we will use hand-waving argument to derive it. . We started with the retarded potentials and we assume It is important to note that at any time, there is only one point Lorentz gauge. on the trajectory contribute to the potential at point P

, V(, Naively, we may think that since , ′, , , 33 However, the above is not correct!!! 34

For a point charge, we have What happen is the following: ′, 3/21/2016 · 3/21/2016 , , ′

The denominator can come out of the integral Fields and Potentials 10 Chapter One way to verify equation (42) is shown on page 452 of Fields and Potentials 10 Chapter Griffiths. The time it takes the train to travel a distance without any problem, but the integral of charge is not so easy. Specifically: is the same as the time for light to travel a distance of ′. ′ , ′

Because, will be different for each different points. So we need to evaluate at different times for

different ′ and this leads to a distortion of the total 35 36 charge calculated.

6 3/30/2016

Notice that this effect does not distort the dimension Alternate approach to derive eqs. 44, 45 perpendicular to the velocity, such that the apparent

volume is related to the actual volume by 3/21/2016 3/21/2016

Fields and Potentials 10 Chapter Fields and Potentials 10 Chapter ·/ The formal way to solve equation (41) is to transform it into a new coordinate system, where time is the same It follows that the equations (21) (22) become such that the integration can take place. , ·/ ′, ′ ′ , , , ·/

These are the Lienard-Wiechert Potentials for a 37 ≡ 38 moving point charge.

⋯ 1 2! 3! ⋯ ⋯ ′ ′ We expand in terms of ′ , using Taylor expansion 3/21/2016 3/21/2016 ′ ′ / . . And Chapter 10 Potentials and Fields and Potentials 10 Chapter Fields and Potentials 10 Chapter ,, ≡ Then ; ; , , ′ ′ ′ where and it is the unit vector in the ̂ ̂ direction. ′ ′ ′ , , ,, · ⋯ ,,′ ′ ′ ′ ,,′ ′ ′ ′ ′ 39 40 ·

We first compute the retarded time, using (10.44) Example 10.3 · (10.48)

Find the potentials of a point charge moving with a 3/21/2016 3/21/2016 constant velocity. · · Since it is a constant velocity motion, the trajectory is a Chapter 10 Potentials and Fields and Potentials 10 Chapter Fields and Potentials 10 Chapter straight line. Let’s choose the line such that it pass Next through the origin at t = 0. · · . · Square above eq. and then solve for (See page 49). · Since is the retarded time, we choose = · (2) 41 42 Substitute the above into (10.46) and (10.47), we end up with:

7 3/30/2016

The fields due to Lienard-Wiechert potentials , · Once the Lienard-Wiechert potentials are known, the

3/21/2016 3/21/2016 , fields can be obtained through ·

Fields and Potentials 10 Chapter & Fields and Potentials 10 Chapter Now if we let ̂ and Conceptually, it is straight forward and simple. But in , reality, it is an exercise in vector calculus. See page 456-457 of Griffiths for details. In the end we have , · At t = 0 , and ≡ , 43 44

There are two terms in eq. (47) The static field satisfies the Gauss Law. No energy radiated out. Let 1st term 3/21/2016 3/21/2016 For the radiation field, it does not satisfy the Gauss law. · But it does agree with conservation of energy.

Fields and Potentials 10 Chapter Fields and Potentials 10 Chapter · This eq. does not depend on acceleration. It is called generalized Coulomb field, because if v=0, this part goes back to the static field. · 2nd term

The 2nd term depends on acceleration. This is called ∝ ∝ radiation field. ∝ 45 46

Example 10.4 The direction of the non-radiative field is in the direction of , namely from the present position of the charge. This is an extraordinary coincidence, since the “message” came from the retarded position. Calculate the and field of a point charge moving with 3/21/2016 3/21/2016 a constant velocity along the x-axis. y ·· · Fields and Potentials 10 Chapter Fields and Potentials 10 Chapter P But ; · Equation (10.72) becomes x · , / 47 48 ≡ ,

8 3/30/2016

New notes added From eq. (2) of slide 42, we can see that

3/21/2016 ·

Fields and Potentials 10 Chapter = · Re-arrange the above equation, and solve the quadratic equation of , we end up with

· · To fix the sign, consider the limit, v = 0

49

9