Retarded Potentials and Jefimenko’s Equations (1)

To calculate the scalar and vector potentials for continu- ous time-dependent charge-current distributions, we do the same integrals as for statics, except we evaluate the = − λ λ ′ densities at the retarded time tr t c where = r - r

The retarded potentials are

1 ρ(r ′, t ) µ J (r ′, t ) φ(r ,t) = r dv ′ A( r ) = 0 r dv ′ πε ∫ λ π ∫ λ 4 0 4

The electric and magnetic fields are

1  ρ(r ′ ,t ) ρ˙ (r ′, t ) J˙ (r ′ ,t ) E = ∫  r + r  λˆ − r dv ′ πε  λ2 λ  2λ  4 0  c c  µ  1  1  B = 0 ∫ (∇ × J ) − J × ∇ dv ′ 4π  λ  λ  Apparent Length and Charge Density

The integral is supposed to be done using the apparent of the charge sources that an observer at the measure- ment point would see through a telescope using light.

We are not supposed to make any attempt to correct for the speed of light for where the charges are “now” vs when the light was emitted.

This has the consequence that the apparent dimensions of an object change depending on if it is moving toward or away from us.

But the charge density we are supposed to use is the “real” charge density, not the total charge of the object divided by the apparent volume. Apparent Length of a Moving Body

Let’s call ∆t the time between light being emitted from the far face and the time that light passes the near face.

During time ∆t , light travels the apparent length of the cube a ′ . The near face of the cube had a head start by the length a, so the cube moves distance a ′ − a in the same time ∆t at velocity u.

a a’-a

u

a’

If we equate the two times

a ′ a ′ − a ac 1 ∆t = = ⇒ a ′ u = a ′c − ac ⇒ a ′ = = a c u c − u 1− u c

So the apparent length of a cube moving toward us with velocity u is increased by a factor 1 (1− u c). For u nearly the speed of light, this factor becomes infinite.

If the cube is moving away from us, we just reverse the sign of u. The cube looks shorter by a factor 1 (1+ u c). For velocity approaching light, this factor becomes 1/2, not infinitely small. Apparent Length with Coordinates

Let z1 and z2 be the coordinates of the ends of an object of length a moving with velocity v. At time t = 0, the far end of the object is at z0 , and a forward light ray z3 is emitted. = + z1 z0 vt = + + z2 z0 a vt = + z3 z0 ct

The time when the light ray reaches z2 is

a z = z ⇒ z + a + vt = z + ct ⇒ t = 2 3 0 0 c − v

The apparent length of the object is the distance bet- ween the point where the light ray left one end and the point where it arrives at the other end

a  v  1 z (t) − z = z + a + v − z = a 1+  = a 2 0 0 c − v 0  c − v 1− v c

a So the object appears to have length > a 1− v c

If the light ray is moving backwards, we reverse the sign a of c, and the length appears to be < a 1+ v c Apparent Volume

The dimensions of the object transverse to the motion are not distorted by the retardation. If the motion of the object is not directly toward or away from the observer, we break it into components. The component along the line to the observer is all that matters.

The apparent volume of a moving object is then

1 V = V apparent true 1− λˆ ⋅ v c

“True” volume means the volume of the object in the observer’s frame, after correcting for the distortion due to the finite speed of light. “Apparent” volume means the volume as distorted by the retardation. We don’t need to say anything about the dimensions or charge density of the object in its own rest frame.

The key point is, we are supposed to do the integrals for the retarded potentials using the apparent volume.

The charge density of the object does not change as it moves past us, but its apparent volume does. So when we integrate for the retarded potential, the result will pick up the factor from the apparent volume.

This is still true in the limit where the size of the object goes to zero and we have a point charge. Lienard-Wiechert Potentials (1)

For a point charge in arbitrary motion, the potential is just the electrostatic result times the apparent volume factor

q 1 q c φ(r ,t) = = λ = r − r ′ πε λ ˆ πε 4 0 1− λ ⋅v c 4 0 λc − λ ⋅v

The vector potential integral similarly becomes

µ ρ(r ′ ,t )v µ v A (r ,t) = 0 ∫ r dv ′ = 0 ∫ ρ(r ′ ,t )dv ′ 4π λ 4π λ r µ µ = 0 v q = 0 qv 4π λ 1− λˆ ⋅v c 4πλ 1− λˆ ⋅v c

These are the Lienard-Wiechert potentials, valid for a point in arbitrary motion.

We can simplify φ if we define a new vector u = cλˆ − v , so

q c φ(r ,t) = πε λ ⋅ 4 0 u ε ε It’s also useful to note that by inserting an 0 0 factor we can write the vector potential in terms of the scalar potential:

q 1 v A (r ,t) = µ ε v = φ(r ,t) 0 0 πε λ − λˆ ⋅ 2 4 0 1 v c c Lienard-Wiechert Potentials (2)

We can go from the retarded potentials to the Lienard- Wiechert potentials a different way, by just carefully in- tegrating over a moving delta-function charge density.

1 qδ3 (x ′ − v t ) x − x ′ φ(x ,t) = r d 3 x ′ t = t − πε ∫ λ r 4 0 c

Recall the rule for delta-functions of functions:

1 1 1 ∫ δ(ax)dx = ∫ δ(ax) d(ax) = ∫ δ(z)dz = a a a   δ[ ( )] = δ + df + = 1 ∫ f x dx ∫  f root x ... dx  dx root  df dx

If we rotate so the velocity is in the z direction, the x and y delta functions just give unity, and for z we need

d   ∆x 2 + ∆y2 + ∆z2    z ′ − v t −   dz ′   c   v 2∆z(−1) = 1+ c 2 ∆x 2 + ∆y2 + ∆z2 v ∆z = 1− c ∆x 2 + ∆y2 + ∆z2 Lienard-Wiechert Potentials (3)

We note that

∆z = vˆ ⋅λˆ ∆x2 + ∆y2 + ∆z 2 and find

d   ∆x 2 + ∆y2 + ∆z2   v v ⋅λˆ  z ′ − v t −   =1− vˆ ⋅ λˆ = 1− dz ′   c   c c

Using this, the retarded potential integral becomes

q 1 φ(x ,t) = πε λ − λˆ ⋅ 4 0 1 v c which is the same result we got the other way. Retarded Potentials and Lienard-Wiechert Potentials

The Retarded Potentials for an arbitrary time-dependent continuous charge-current distribution are

1 ρ(r ′, t ) µ J (r ′, t ) φ(r ,t) = r dv ′ A (r ) = 0 r dv ′ πε ∫ λ π ∫ λ 4 0 4 λ ′ = − λ = r - r tr t c

This is the same form as the static result, except evaluat- ed at the retarded time.

The Lienard-Wiechert Potentials for a point charge in arbitrary motion are

1 q 1 µ qv 1 φ(r ,t) = A( r ,t) = 0 πε λ − λˆ ⋅ π λ − λˆ ⋅ 4 0 1 v c 4 1 v c

The electric potential is just the static result times the apparent volume factor, evaluated at the retarded time.

These can also be written using u = cλˆ − v and the simi- larity as

q c v φ(r ,t) = A (r ,t) = φ(r ,t) πε λ ⋅ 2 4 0 u c Field of a Moving Point Charge (1)

To get the electric field, we evaluate E = −∇φ − ∂A ∂t

The gradient of the Lienard-Wiechert φ with respect to ′ ( ) the observation point x [not the source point x tr ] is

 qc 1  qc −1 ∇φ = ∇  = ∇(λc − λ ⋅ v )  πε λ − λ ⋅  πε 2 4 0 c v 4 0 (λc − λ ⋅v ) λ λ = x − x ′( t ) t = t − r r c

The gradient at the end can be expanded

∇(λ − λ ⋅ ) = ∇λ −(λ ⋅ ∇) − ( ⋅∇)( − ′ ( )) c v c v v x x tr

In tensor notation this is

∇(λ − λ ⋅ ) = ∂ λ − λ ∂ − ∂ ( − ′ ) c v c i j i vj vj i x j x j Field of a Moving Point Charge (2)

In tensor notation, the chain rule is

∂ ∂t ∂ ∂ = = r i ∂ ∂ ∂ xi xi tr

The first term in the gradient is then

∂t ∂ ∂t c∂ λ = c r [c(t − t )] = −c2 r i ∂ ∂ r ∂ xi tr xi

The next term involves the retarded acceleration a

∂t ∂ ∂t ∂t λ ∂ v = λ r v = λ r a = λ ⋅a r j i j j ∂ ∂ j j ∂ j ∂ xi tr xi xi

The next term is easy ∂ = δ = = v j i x j v j ij vi v

The last term is

∂t ∂ ∂t ∂t v ∂ x ′ = v r x ′ = v r v = v2 r j i j j ∂ ∂ j j ∂ j ∂ xi tr xi xi Field of a Moving Point Charge (3)

The full gradient term is then

∇(λ − λ ⋅ ) = ∇λ −(λ ⋅ ∇) − ( ⋅∇)( − ′ ( )) c v c v v x x tr ∂t ∂t ∂t = −c 2 r − λ ⋅ a r − v + v2 r ∂ ∂ ∂ xi xi xi ∂t = (v 2 − c2 − λ ⋅ a ) r − v ∂ xi ∂ tr Now all we need is ∂ xi

 2  ∑ (x − x ′ ) ∂ ∂  λ ∂  j j j  t =  t −  = t − ∂ r ∂   ∂   xi xi c xi  c 

−1 1 ∂ 2 = ∑ (x − x ′ ) λ j ∂ j j c 2 xi −1 1 ∂ = ∑ 2(x − x ′ ) (x − x ′ ) λ j j j ∂ j j c 2 xi Field of a Moving Point Charge (4) ∂ ∂ The first derivative x j xi turns into a Kronecker delta, and we use the chain rule on the second

∂t −1 1  ∂  r = ∑ 2(x − x ′ ) δ − x ′  ∂ λ j j j  ij ∂ j  xi c 2 xi −1 1  ∂t ∂x ′  = ∑ 2(x − x ′ ) δ − r j  λ j j j  ij ∂ ∂  c 2 xi tr

λ δ λ The first parentheses is j the ij turns it into i and in ∂ ′ ∂ = the second x j tr vj so

∂t −1 ∂t  r = λ − λ v r ∂ λ  i j j ∂  xi c  xi  ∂ ∂ We have tr xi on both sides, so we solve for it:

∂t ∂t λ = λ v r − λc r i j j ∂ ∂ xi xi ∂t λ λ −λ r = i = = ∂ λ − λ λ ⋅ − λ λ − λ ⋅ xi j v j c v c c v Field of a Moving Point Charge (5)

Put it all together for the gradient of φ

qc −1 ∇φ = ∇(λc − λ ⋅v ) πε 2 4 0 (λc − λ ⋅v )

qc −1  ∂t  = (v2 − c 2 − λ ⋅a ) r − v πε 2  ∂  4 0 (λc − λ ⋅v )  xi 

qc −1  −λ  = ( v2 − c 2 − λ ⋅a ) − v  πε 2 λ − λ ⋅ 4 0 (λc − λ ⋅v )  c v 

Factor out the denominator:

qc −1 ∇φ = [−λ (v2 − c 2 − λ ⋅a ) − v (λc − λ ⋅v )] πε 3 4 0 (λc − λ ⋅v ) qc 1 = [(λc − λ ⋅v )v − (c2 − v2 + λ ⋅a )λ ] πε 3 4 0 (λc − λ ⋅v ) Field of a Moving Point Charge (6)

To get the electric field, we also need ∂A ∂t

∂A ∂t ∂ ∂t ∂  v   v ∂ ∂ v  ∂t = r A = r φ =  φ + φ  r = ∂ ∂ ∂ ∂ ∂  2   2 ∂ ∂ 2  ∂ t t tr t tr c c tr tr c t  v ∂φ a  ∂t =  + φ  r  2 ∂ 2  ∂ c tr c t ∂φ ∂ The tr term can be written

∂φ ∂  qc 1  =   ∂ ∂  πε  tr tr 4 0 λc − λ ⋅v qc −1 ∂ = (λc − λ ⋅v ) πε 2 ∂ 4 0 (λc − λ ⋅v ) tr

∂ ∂ The tr term at the end of this is ∂ ∂ (λc − λ ⋅v ) = (λc − λ v ) ∂ ∂ j j t r t r ∂λ ∂v ∂λ = c − λ j − v j ∂ j ∂ j ∂ t r t r t r Field of a Moving Point Charge (7)

For the first term in the above, we use

∂λ  ∂t  λ = c(t − t ) ⇒ = c −1 r ∂  ∂  t r t r

For the second term, we use ∂ v = a ∂ j j t r

For the third term, we use ∂λ ∂ j = (x − x ′ ) = −v ∂ ∂ j j j t r t r ∂ ∂ So the tr term is

∂ ∂λ ∂v ∂λ (λc − λ ⋅v ) = c − λ j − v j ∂ ∂ j ∂ j ∂ t r t r t r t r  ∂t  = c2  −1 − λ a − v (−v )  ∂  j j j j t r ∂t = v 2 − c2 − λ ⋅a + c2 ∂ t r Field of a Moving Point Charge (8)

Thus we have

∂φ qc −1 ∂ = (λc − λ ⋅v ) ∂ πε 2 ∂ tr 4 0 (λc − λ ⋅v ) tr

qc −1  ∂t  =  v2 − c2 − λ ⋅a + c2  πε 2  ∂  4 0 (λc − λ ⋅v ) tr

Plug it all back in for the time derivative of A

∂A  v  ∂φ  a   ∂t  =  + [φ]  r  ∂ 2  ∂  2 ∂ t  c  tr  c   t      v qc −1  ∂t     v2 − c2 − λ ⋅a + c 2    2  πε 2  ∂    c 4 0 (λc − λ ⋅v ) tr   ∂t  =      r   ∂t   a  q c    +  2  πε   c  4 0 λc − λ ⋅v   Field of a Moving Point Charge (9)

Factor out the obvious term and note that we have both ∂ ∂ ∂ ∂ t tr and tr t in one term so we separate it out

 v  [(c2 − v2 + λ ⋅a )] ∂A qc 1  c2   ∂t  =    r  ∂t 4πε λ − λ ⋅ 2 a  ∂t  0 ( c v )  + [λc − λ ⋅v ]   c2  qc −1 +v πε 2 4 0 (λc − λ ⋅v )

Now we evaluate the derivative of the retarded time with respect to un-retarded time:

∂t ∂  λ 1 ∂ 1 1 ∂λ r =  t −  = 1− ( λ λ ) = 1− 2λ i ∂t ∂t  c  c ∂t i i c 2λ i ∂t 1 ∂t 1 ∂t = 1− λ (−v ) r =1+ λ ⋅v r λc i i ∂t λc ∂t

We have the derivative on both sides, so we solve for it:

∂t 1 ∂t r = 1+ λ ⋅v r ∂t λc ∂t ∂ λ tr = 1 = c ∂t λ ⋅v λc − λ ⋅v 1− λc Field of a Moving Point Charge (10)

Plug this into ∂A ∂t:

 v  [(c2 − v2 + λ ⋅a )] ∂A qc 1  c2   λc  =     ∂t 4πε λ − λ ⋅ 2 a  λc − λ ⋅v  0 ( c v )  + [λc − λ ⋅v ]   c2  qc −1 +v πε 2 4 0 (λc − λ ⋅v )

Now we can collect some terms

 λv  [c2 − v2 + λ ⋅a ] ∂A qc 1  c  =   ∂t 4πε 3  λa  0 (λc − λ ⋅v )  + − v [λc − λ ⋅v ]   c   Field of a Moving Point Charge (11)

∂A The electric field is E = −∇φ − and we already found ∂t qc 1 ∇φ = [(λc − λ ⋅v )v − (c 2 − v 2 + λ ⋅a )λ ] πε 3 4 0 (λc − λ ⋅v )

So we have

 (λc − λ ⋅v )v     − 2 − 2 + λ ⋅ λ   (c v a )  − = qc 1  λ  E v 2 2 4πε λ − λ ⋅ 3  + (c − v + λ ⋅a )  0 ( c v )  c    λa    + − v (λc − λ ⋅v )   c   Field of a Moving Point Charge (12)

We can factor the big bracket

  λa    (λc − λ ⋅v ) v + − v  −qc 1  c  E =   4πε λ − λ ⋅ 3   λv   0 ( c v )  −(c2 − v2 + λ ⋅a ) λ +    c  

We can simpify this if we define

u = λˆ c − v λu = λ c − λv λ ⋅u = λc − λ ⋅v

Then we have

−qc 1  λa λu  E = (λ ⋅u ) −(c 2 − v 2 + λ ⋅a ) πε 3   4 0 (λ ⋅u ) c c

Factor out λ c

−q λ E = [(λ ⋅u )a − (c2 − v2 + λ ⋅a )u ] πε 3 4 0 (λ ⋅u )

Change the sign and re-arrange a bit

q λ E = [(c2 − v 2 )u + (λ ⋅a )u −(λ ⋅u )a ] πε 3 4 0 (λ ⋅u ) Field of a Moving Point Charge (13)

The last two terms are a double-cross product, and we get

q λ E = [(c2 − v 2 )u + λ × (u × a )] πε 3 4 0 (λ ⋅u )

This, finally, is the electric field from a point charge with arbitrary velocity and acceleration.

If the acceleration is zero, the cross term is zero. If the velocity is also zero, u = cλˆ , and we get back to Coulomb, as we should.

But we have new terms that depend on the velocity and acceleration.

Note that the acceleration term has an extra power of distance λ, so it dominates at large distances if the accel- eration is non-zero.

This is the radiation term! Field of a Moving Point Charge (14)

For the magnetic field, recall that for Lienard-Wiechert,

v A (r ,t) = φ(r ,t) c2

 v  1 B = ∇ × A = ∇ × φ = ε [φ∂ v + v ∂ φ]  c2  c2 ijk i j j i

We can do the first term by the chain rule

∂t ∂ ∂t ∂ v = r v = r a i j ∂ ∂ j ∂ j xi tr xi and we worked out before that

∂ −λ −λ tr = = ∂ xi λc − λ ⋅v λ ⋅u

The second term is just the divergence of φ:

qc 1 ∇φ = [(λc − λ ⋅v )v − (c 2 − v 2 + λ ⋅a )λ ] πε 3 4 0 (λc − λ ⋅v ) qc 1 = [(λ ⋅u )v − (c2 − v2 + λ ⋅a )λ ] πε 3 4 0 (λ ⋅u ) Field of a Moving Point Charge (15)

If we plug in φ in the form using u , we have

  q c   −λ      i a   πε   j   4 0 λ ⋅u λ ⋅u  1 B = ε   2 ijk   (λ ⋅ )   c  qc 1 u vi   +v      j  πε 3  2 2    4 0 (λ ⋅ ) −( − + λ ⋅ )λ    u  c v a i   

Factor out the obvious term

 − λ ⋅ λ  ( u ) i a j = 1 qc 1 ε   B ijk c2 4πε λ ⋅ 3  + λ ⋅ − 2 − 2 + λ ⋅ λ  0 ( u )  v j [( u )vi (c v a ) i ]

ε λ The ijkv j vi term gives zero, and factor out i :

1 q λ B = i ε [−(λ ⋅u )a −(λ ⋅a )v − (c2 − v2 )v ] πε 3 ijk j j j c 4 0 (λ ⋅u ) Field of a Moving Point Charge (16)

Revert to vector notation, and change the order a bit

1 q λ B = ×[−(c2 − v2 )v − (λ ⋅a )v − (λ ⋅u )a ] πε 3 c 4 0 (λ ⋅u )

This looks a lot like the electric field we got before:

q λ E = [(c2 − v 2 )u + (λ ⋅a )u −(λ ⋅u )a ] πε 3 4 0 (λ ⋅u )

We have a factor of 1 c, we cross with vector λ rather than multiplying by scalar λ, and we have −v instead of u in the first two terms in the brackets.

But we are crossing with λ again anyway, and u = λˆ c − v , so the difference between them is parallel to λ . Thus it won’t contribute to the cross product. So we can just re- place −v with u without changing the cross product!

λˆ We therefore have B = × E c

If the velocity and acceleration is zero, then E is the Cou- lomb field, and is radial. The cross product will then be zero as well, as it should be for a charge at rest.