Download Lecture Notes 18.5: Lorentz Transformation of EM Fields, the EM

UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 18.5
Prof. Steven Errede
LECTURE NOTES 18.5


The Lorentz Transformation of E and B Fields:


We have seen that one observer’s E -field is another’s B -field (or a mixture of the two),
as viewed from different inertial reference frames (IRF’s).
What are the mathematical rules / physical laws of {special} relativity that govern the


transformations of E  B in going from one IRF(S) to another IRF(S') ???
In the immediately preceding lecture notes, the reader may have noticed some tacit / implicit
assumptions were made, which we now make explicit:
1) Electric charge q (like c, the speed of light) is a Lorentz invariant scalar quantity.
No matter how fast/slow an electrically-charged particle is moving, the strength of its
electric charge is always the same, viewed from any/all IRF’s: e  1.602 1019 Coulombs.
{n.b. electric charge is also a conserved quantity, valid in any/all IRF’s.}
Since the speed of light c is a Lorentz invariant quantity, then since c  1
 o o then so is
c  1  o o ; the scalar EM properties of the vacuum -  o and o are separately Lorentz
invariant scalar quantities, i.e.
2
 o  8.85 1012 Farads/meter
o  4 107 Henrys/meter
same in any/all IRF’s




2) The Lorentz transformation rules for E and B are the same, no matter how the E and B fields
are produced - e.g. from sources: q (charges) and/or currents I, or from fields:
 

e.g.   E   B t , etc.
The Relativistic Parallel-Plate Capacitor:
The simplest possible electric field: Consider a large  -plate capacitor at rest in IRF(S0).
It has surface charge density  0 (Coul/m2) on the top/bottom plates respectively and has plate
dimensions  0 and w0 {in IRF(S0)!} separated by a small distance d   0 , w0 .
In IRF(S0):
Electric field as seen in IRF(S0):



E0  E0 yˆ  0 yˆ E0  0
o
o

No B -field is present in IRF(S0):

B0  0  no currents present!

n.b. E0 is non-zero only in the
gap region between  -plates
{i.e. neglect fringe field}
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
1
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 18.5
Prof. Steven Errede
Now consider examining this same capacitor setup from a different IRF(S), one which is
moving to the right at speed v0 (as viewed from the rest frame IRF(S0) of the  -plate capacitor)

i.e. v  v0 xˆ is the velocity of IRF(S) relative to IRF(S0).
Viewed from the moving frame IRF(S), the  -plate capacitor is moving to the left
(i.e. along the  x̂ -axis) with speed –v0.
The plates along the direction of motion have Lorentz-contracted by a factor of  0  1
1   v0 c  ,
2
i.e. the length of plates as seen by an observer in IRF(S) is:    0  0  1   v0 c   0 .
2
{n.b. the plate separation d and plate width w are unchanged in IRF(S), since both d and w are
 to direction of motion!!}
Since:  
And:
0 
Qtot
Q
 tot
Area   w
but: Qtot = Lorentz invariant quantity
Qtot
Qtot

A0  0  w0
but: w = w0 and d = d0 since both  to direction of motion.
Thus: Qtot    w   0 0 w0
But:
  0 
     0 0     0   0  
 0 
 
   0 
 0  0  0 0


Thus:    0 0 but since:  0  1     0
The surface charge density on the plates of capacitor in IRF(S)
is higher than in IRF(S0) by factor of
 0  1 1   v0 c 
To an observer in IRF(S), the plates of the  -plate capacitor are moving in the v0 xˆ direction.

Thus the electric field E in IRF(S) is:




E   yˆ   0 0 yˆ   0 E0 where:    0 0
o
o




E    0 E0 , E0  0 yˆ
o



The superscript  is to explicitly remind us that E    0 E0 is for E -fields  to the direction

of motion. Here, v  v0 xˆ between IRF’s.
2
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
2
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 18.5
Prof. Steven Errede
Now consider what happens when we rotate the {isolated}  -plate capacitor by 90o
 
in IRF(S0), then E0  0 xˆ in rest frame IRF(S0), but in the moving frame IRF(S):
0
The electric field in IRF(S) is:
 

E  xˆ  E 
o
but:

Qtot
Q
Q
 tot  tot   0
Area w  0 w0





E   xˆ  0 xˆ  E0



E   E0
o
o
   0 


 w  w0 

E -field in IRF(S0)
The plate separation distance d is Lorentz-contracted in IRF(S): d  d 0  0 but has no effect on


E {in IRF(S)}, because E does not depend on d! Why??? Because here, the  -plates were first
charged up (e.g. from a battery) and then disconnected from the battery!
 Potential difference: V  IRF  S    V0  IRF  S0   
n.b. IRF choice has intimate
connection to gauge invariance!!!
V0
V  V0
V 
  V    V0 


Since:  E 
but: d  d 0  0 
 
   E0 
d
d0
d0  0
d0
d  
d0 


V  IRF  S     0 V0  IRF  S0  
The  -plate capacitor is deliberately not connected to an external battery (which would keep
V = constant, but then we would have    0 in the  case and    0 in the  case.
Currents would then flow (transitorially) in both situations…
Note that we also want to hang on to/utilize the Lorentz-invariant nature of Qtot, which is
another reason why the battery is disconnected…
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
3
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 18.5
Prof. Steven Errede
But wait !!! This isn’t the entire story for the EM fields in IRF(S) !!!
In the first example of the “horizontal”  -plate capacitor, which was moving with relative

velocity v  v0 xˆ as seen by an observer in IRF(S), shown in the figure below:
The moving surface-charged plates of the  -plate capacitor as
viewed by an observer in IRF(S) constitute surface currents:

K    v0 xˆ (Amps/m) with:    0 0 (Coul/m2).
The two surface currents create a magnetic field in IRF(S)
between the plates of the  -plate capacitor !!!
Side / edge-on view:
B-fields produced (use right-hand rule):
Ampere’s Circuital Law:

C
 
w
Bd   o I encl  o  K  dw
0
 2 Bw  o K  w


K
 v
K
 v
B  y  0    o  zˆ   o 0 zˆ  y  0 
B  y  d    o  zˆ   o 0 zˆ
2
2
2
2


o K 
o v0
o K 
o v0
B  y  d   
zˆ  
zˆ
B  y  0   
zˆ  
zˆ  y  0 
2
2
2
2



Add B and B to get the total B -field:

 v
 v
Btot
 y  d    o 0 zˆ  o 0 zˆ  0
2
2

 v
 v
Btot  0  y  d    o 0 zˆ  o 0 zˆ   o v0 zˆ
In IRF(S):
2
2

 v
 v
  o 0 zˆ  o 0 zˆ  0
Btot  0  y 
2
2
4
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 18.5
Prof. Steven Errede

Thus, in IRF(S) {which moves with velocity v  v0 xˆ relative to IRF(S0)}
we have for the horizontal  -plate capacitor:
 

0
yˆ   0 E0 yˆ where:    0 0 and:  0 
E exists only in region 0  y  d : E  yˆ   0
o
o


B exists only in region 0  y  d : B   o v0 zˆ   0 o 0 v0 zˆ   1c  0  0 E0 zˆ
1
1   02
 0  v0 c
In IRF(S):


The fact that B exists / is non-zero only where E exists / is non-zero is not an accident / not a “mere”
 




coincidence! Note also that {here}: B  1c   E where:    0 x̂ and: E  Eyˆ   0 E0 yˆ   0 0 yˆ .
o
The space-time properties associated with rest frame IRF(S0) are rotated (Lorentz-transformed)
in going to IRF(S).

E0 in IRF(S0) only exists between plates in rest frame IRF(S0) {neglecting fringe field}
Gets space-time rotated (Lorentz-transformed) in going to moving frame IRF(S)


→ E and B in IRF(S) exist only between the capacitor plates in IRF(S).



E and B between plates in IRF(S) comes from / is associated with E0 between plates in IRF(S0)
Point is: EM field energy density uEM (x,y,z,t) must be non-zero in a given IRF in order to have
EM fields present at space-time point (x,y,z,t)!


In IRF(S0): u0  r0 , t0   12  o E02  21 o  02  E0 only
EM field energy densities u0 and u



≠ 0 only in region between plates of
u  r , t   12  o E 2  r , t   2 1o B 2  r , t 
||-plate capacitor.
In IRF(S):
 12  o 02 E02  12  o 02  02 E02  12  o 02 1   02  E02
n.b. If EM energy density u0 = 0 in one IRF(S0) → u = 0 in another IRF(S).
Poyntings Vector:
 
In IRF(S0): S0  r0 , t0  
 
In IRF(S): S  r , t  
1
 
 
E0  r0 , t0   B0  r0 , t0   0
1
 
 
E r,t  B r,t   
o
o
1
o
0
0
2
  0 o  0 v0  yˆ  zˆ    02 0 v0 xˆ   02 o E02 v0 xˆ

o
o
 xˆ
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
5
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 18.5
Prof. Steven Errede
Momentary Aside: Hidden Momentum Associated with the Relativistic Parallel Plate Capacitor.
Between plates of
 -plate capacitor
in IRF(S):
 

E  yˆ   0 0 yˆ where:    0 0 and:  0 
0
0

B   o v0 zˆ   0 o 0 v0 zˆ
1
1   02
 0  v0 c
xˆ  yˆ  zˆ
xˆ


2
 1  
 02 02 v0
2  0 v0
E  B   0
xˆ yˆ  zˆ  xˆ
Between plates of  -plate capacitor (only!): S 
 yˆ  zˆ   
0
0
0
zˆ  xˆ  yˆ
Poynting’s vector points in the direction of motion of the  -plate capacitor, as seen by an observer
in IRF(S), and only exists/is non-zero in the region between plates of the  -plate capacitor.


EM field linear momentum density in IRF(S): EM   o o S   02 o 02 v0 xˆ .
- Points in the direction of motion of the  -plate capacitor, as seen by an observer in IRF(S),
- Only exists/is non-zero in the region between plates of the  -plate capacitor.


- n.b. In rest frame IRF(S0): B0 = 0  Poynting’s vector S0  0 and thus 0EM  0 !



EM field linear momentum in IRF(S): pEM EM  Volume(IRF(S )) EM VS
Volume (in IRF(S)): VS  wd 
0
0
w0 d 0 , where:    0  0 , w  w0 and: d  d 0 {here}.
Volume (in IRF(S0): V0   0 w0 d 0 . Thus: VS  V0  0 .
 wd

 pEM   02 o 02 v0VS xˆ   02 o 02 v0  0 0 0 xˆ   0 0 02 v0   0 w0 d 0  xˆ   0 0 02 v0V0 xˆ



0
V0
1  


phidden  2 m  E where m = magnetic dipole moment
c
 
 

where: m   IA zˆ with: I   K d   Kw where: K   v0 . Note that: m  B .
The “hidden momentum” in IRF(S) is:


 
The cross-sectional area is: A  d   0  d 0 since:    0  0 and: d  d 0 , w  w0 .
 0 
Side view (in IRF(S)):
 0 0


m   IA zˆ   Kwd    v0  w0 0 d 0 zˆ
0


   0  0 v0 w0
0
0
d 0 zˆ    0 v0   0 w0 d 0  zˆ


E   0 0 yˆ
o
6
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 18.5
Prof. Steven Errede
 In IRF(S):
 xˆ
 0 
 0 02v0
1
1

ˆ
ˆ
phidden   2  0 v0   0 w0 d 0    0
z

y


w0 d 0  xˆ   0 o 02 v0V0 xˆ using 2   o o


0 
2

c
c o
o
c
V0
 
1  


Thus in IRF(S): phidden  2 m  B   0 o 02 v0V0 xˆ But: pEM   o E  B VS   0 0 02 v0V0 xˆ
c


Thus, we (again) see that: phidden   pEM for the relativistic  -plate capacitor !!!






1  

E0  B0  0  pEM  0 .
Note that in IRF(S0): B0  0  S0 
o
1  



But note that: phidden  2 m0  B0  0 , thus phidden   pEM is also valid in IRF(S0).
c


 The numerical value of hidden momentum is reference frame dependent, i.e. it is not a

Lorentz-invariant quantity, just as relativistic momentum p {in general} is reference frame
dependent/is not a Lorentz-invariant quantity.


Now let us return to task of determining the Lorentz transformation rules for E and B :
For the case of the relativistic horizontal  -plate capacitor, let us consider a third IRF(S′) that

travels to the right (i.e. in the  x̂ -direction) with velocity v   vxˆ relative to IRF(S).
 

yˆ and: B   o vzˆ
In IRF(S′), the EM fields are: E  
o

From use of Einstein’s velocity addition rule: v is the velocity of IRF(S′) relative to IRF(S0):
v  v0
1
with:   
and:      0
2
2
1  vv0 c

1 v c




We need to express E  and B {defined in IRF(S′)} in terms of E and B {defined in IRF(S)}.
v 
 
  0
E 
yˆ 
yˆ where:   
o


 
E 
yˆ
 0 o
o
1
1   v c 
2
 
but: E  yˆ {in IRF(S)} 
o

B   o vzˆ   o 0 vzˆ but:  0    0

and:  0 

1
and:  0 
2
0
1   v0 c 
        
E     E    yˆ
 0 
  0  o


  
B     o vzˆ but: B   0 o 0 v0 zˆ   o v0 zˆ {in IRF(S)}
 0 
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
7
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 18.5
Prof. Steven Errede
1   v0 c 
  
1  vv0 c 2
1
 vv 

   1  20  where:  
Now (after some algebra):   
2
2
2
c 

 0 
1 v c
1   v c 
1 v c
2
But: v 
 v  v0 
1  vv
0
 In IRF(S′):
c2 
  
1
and:     1  vv0 c 2  with:  
2
 0 
1 v c
        

E     E    yˆ   1  vv0 c 2  yˆ
o
 0 
  0  o
 vv

  
2
0


ˆ
B     o v z   1  vv0 c o 
 1  vv0 c 2
 0 




 zˆ  o  v  v0  zˆ




Compare these to the E and B fields in IRF(S):
 

E  yˆ   0 0 yˆ where:    0 0 and:  0 
In IRF(S):
0
0

B   o v0 zˆ   0 o 0 v0 zˆ
1
1   02
 0  v0 c

Using: o  1 c 2 o we can rewrite E  in IRF(S′) as:

 

E  
yˆ  o vv0 yˆ    yˆ    v   o v0  yˆ


o
o 

 Bz
in IRF(S )
 Ey
in IRF(S )
Very Useful Table(s):
xˆ  yˆ  zˆ yˆ  xˆ   zˆ
yˆ  zˆ  xˆ zˆ  yˆ   xˆ
zˆ  xˆ  yˆ xˆ  zˆ   yˆ

 

where: v  vxˆ and: B  Bz zˆ  v  B  vBz  xˆ  zˆ   vBz yˆ

 yˆ


E   E y yˆ    E y   vBz  yˆ    E y  vBz  yˆ
Or simply: E y    E y  vBz 

Likewise, again using: o  1 c 2 o we can rewrite B in IRF(S′) as:

v  
B  o  v  v0  zˆ  o v0 zˆ  o vzˆ     o v0 zˆ    2   zˆ

c  o 

 B  Bz zˆ
in IRF(S )

 Ey
in IRF(S )

v


v
v


B  Bz zˆ   Bz zˆ   2 E y zˆ    Bz  2 E y  zˆ Or simply: Bz    Bz  2 E y 
c
c
c




Thus, we now know how the Ey and Bz fields transform.
8
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 18.5
Prof. Steven Errede
Next, in order to obtain the Lorentz transformation rules for Ez and By, we align the capacitor
plates parallel to x-y plane instead of x-z plane as shown in the figure below:
In IRF(S) the {now} rotated fields are:


E rot  zˆ  Ezrot zˆ
o

B rot   o v0 yˆ  Byrot yˆ
Use the right-hand rule to get correct sign !!!


The corresponding E  and B fields in IRF(S') are (repeating the above methodology):
Ez    Ez  vBy 
and:
v


By    By  2 Ez 
c


As we have already seen by orienting the plates of capacitor parallel to y-z plane:

Ex  Ex ← n.b. there was no accompanying B -field in this case!
 Thus we are not able to deduce the Lorentz transformation for Bx (  to direction of motion)
from the  -plate capacitor problem…
So alternatively, let us consider the long solenoid problem, with solenoid {and current flow}
oriented as shown in the figure below:
In IRF(S):
In IRF(S):

E 0

B  Bx xˆ  o nIxˆ

We want to view this from IRF(S'), which is moving with velocity v  vxˆ relative to IRF(S).
In IRF(S): n  N L = # turns/unit length, N = total # of turns, L = length of solenoid in IRF(S).
Viewed by an observer in IRF(S'), the solenoid length contracts: L  L  in IRF(S')
 In IRF(S'): n 
N
N
1
1


  n = # turns/unit length in IRF(S'), where:  
2
2
L
L
1 
1 v c
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
9
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 18.5
Prof. Steven Errede
However, time also dilates in IRF(S') relative to IRF(S) – affects currents:
I

dQ  dt  dQ  dt 
dQ
1
 dt  1
 
   I in IRF(S'). But:   
in IRF(S)  I  
 I  I
dt

dt   dt   dt  dt  
 dt   


1 
B  Bx xˆ  o nI xˆ  o  n  I  xˆ  o nIxˆ  Bx xˆ  B  Bx  Bx
 


 
Longitudinal / parallel-to-boost direction B-field does not change !!!


Thus, we now have a complete set of Lorentz transformation rules for E and B , for a Lorentz

transformation from IRF(S) to IRF(S'), where IRF(S') is moving with velocity v   vxˆ relative
to IRF(S):
Ex  Ex
Bx  Bx
  1 1  2
E y    E y  vBz 
v


By    By  2 Ez 
c



Ez    Ez  vBy 
v


Bz    Bz  2 E y 
c


v
c
Just stare at/ponder these relations for a while – take your (proper) {space-}time…
Do you possibly see a wee bit of Maxwell’s equations afoot here ??? ;)
Two limiting cases warrant special attention:


v
v
1.) If B  0 in lab IRF(S), then in IRF(S') we have: B   2  Ez yˆ  E y zˆ   2  Ez yˆ  E y zˆ 
c
c


1 

But: v   vxˆ  B   2 v  E  in IRF(S') !!!
c


2.) If E  0 in lab IRF(S), then in IRF(S') we have: E    v  Bz yˆ  By zˆ   v  Bz yˆ  By zˆ 
  

But: v   vxˆ  E   v  B in IRF(S') ← i.e. the magnetic part of Lorentz force law !!!


Griffiths Example 12.13: The Electric Field of a Point Charge in Uniform Motion
A point charge q is at rest in IRF(S0). An observer is in IRF(S), which moves to the right

(i.e. in the  x̂ direction) at speed v0 relative to IRF(S0). What is the E -field of the electric
charge q, as viewed from the moving frame IRF(S)?
10
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 18.5
Prof. Steven Errede
In the rest frame IRF(S0) of the point charge q, the {static} electric field of the point charge q is:


r0
1 q
2
2
2
2
E0 
rˆ0 where: r0  x0  y0  z0 and: rˆ0 
r0
4 o r02
x0
q

E0x 
and: r0  x0 xˆ  y0 yˆ  z0 zˆ
3
4 o x 2  y 2  z 2 2

E0 y 
E0z 
q
0
0

y0
4 o x 2  y 2  z 2 3 2
 0 0 0
q
z0
4 o x 2  y 2  z 2 3 2
 0 0 0
 0  1 1   u0 c 




But: E   0 E0 and: E  E0
0
2
Then in IRF(S), which is moving to right (i.e. in the  x̂ direction) at speed v0 relative to IRF(S):
Ex  E0x 
E y   0 E0 y 
Ez   0 E0z 
q
x0
4 o x 2  y 2  z 2 3 2
 0 0 0
q
 0 y0
4 o x 2  y 2  z 2
 0 0 0
q
3
 0 z0
2
n.b. These relations are expressed
in terms of the IRF(S0) coordinates
(x0, y0, z0) of the field point P.
4 o x 2  y 2  z 2 3 2
 0 0 0

However we want/need the IRF(S) E expressed in terms of the IRF(S) coordinates (x, y, z) of
the field point P.  Use the inverse Lorentz transformation on the coordinates:
 In IRF(S) at the IRF(S) present time t:
Observation/field point in IRF(S) at present time t:
Inverse Lorentz Transformation:

x0   0  x  v0t    0 Rx
R  Rx xˆ  Ry yˆ  Rz zˆ
y0  y  R y
 0  1 1   v0 c 
2
z0  z  Rz
Then: x02  y02  z02  r02   02 Rx2  Ry2  Rz2 =  02 R 2 cos 2   R 2 sin 2 
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
11
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 18.5
Prof. Steven Errede
From the above figure: Rx  R cos   Rx2  R 2 cos 2  .
Then since: R 2  Rx2   Ry2  Rz2  
R
2
y
 Rz2   R 2 sin 2 
Since: R 2  R 2  cos 2   sin 2   .
 In the moving frame IRF(S):
Ex 
Ey 
Ez 
 0 Rx
q
Picked up Lorentz factor
4 o  2 R 2 cos 2   R 2 sin 2  3 2
0

 0 Ry
q
4 o  2 R 2 cos 2   R 2 sin 2  3 2
0

Picked up Lorentz factor
 0 Rz
q
Or:

E
q

0
0
4 o  2 cos 2   sin 2  3 2
0

q
4 o

R  Rx xˆ  Ry yˆ  Rz zˆ




 R 
R
Rˆ
R
ˆ
ˆ
 3  where: 3  2 since: R  RR or: R 
R
R
R
R 
0


 03 2  cos 2   1 2 sin 2  
0


3
2
Rˆ
R2
with:
  v0 2 
 1 

 02   c  
1

q
Rˆ
q

Thus: E 
3
2
4 o 
4 o
  v0  2  2  2 R
2
1  sin   1     sin  




 c 


Thus:
from
Lorentz transformation of field
  v0  2 
 1    
 c 

E
0
3


E
from
Lorentz transformation of coordinate
4 o  2 R 2 cos 2   R 2 sin 2  2
0



 0R
q
E
with:
4 o  2 R 2 cos 2   R 2 sin 2  3 2

0
q
  v0  2 
1    
 c 
3
2
2
4 o 

 v0 
2
1    sin  
 c

  v0  2 
 1    
 c 
2


 v0 
2
2
2
 1  sin   sin     sin  
c


2
 Oliver Heaviside’s 1888 expression for the retarded
 
Rˆ
electric field Eret  r , t  expressed in terms of the IRF(S)

R2
present time t !!! See/compare with Physics 436 Lect. Notes
12, p. 21-24 / Griffiths Ch. 10, Example 10.4, p. 439-440.
In the moving frame IRF(S), the unit vector R̂ points along the line from the present position of
the charged particle at time t !!!
 
 E  R because the  0 factor is present in the numerator of each of the xˆ , yˆ , zˆ components !!!
12
3
Rˆ
R2
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 18.5
Prof. Steven Errede
Griffiths Example 12.14: Magnetic Field of a Point Charge in Uniform Motion

A point electric charge q moves with constant velocity v  vzˆ in the lab frame IRF(S).

Find the magnetic field B in IRF(S) associated with this moving point electric charge.

Note that in the rest frame of the electric charge {IRF(S0)} that B0  0 everywhere.

1  

 In IRF(S) the electric charge q is moving with velocity v  vzˆ and: B   2 v  E .
c

  
1  v

c


q
But the electric field of moving point charge q in IRF(S) is: E 
4 o 
(See pages 3-6 above/Griffiths Example 12.13)
1  v c


2
  sin  

 
2
2
3
2
Rˆ
R2

E  Rˆ
 sin 
qv 1  v



c
o
1 
 1 ˆ



 where:   cos 1 vˆ Rˆ , v  Rˆ  v sin  ˆ .
B   2 vE 
3 
2 
2
c
4 
 2R 
2
1  v c sin  




2
 
 
If the point electric charge q is heading directly towards you (i.e. along the  ẑ direction)
then ̂ points counter-clockwise:
(out of page)
  o
NOTE: In the non-relativistic limit ( v  c ): B  
 4

  v  Rˆ 
q
  2  
  R 
The Biot-Savart Law for
a moving point charge !!!
NOTE ALSO: H.C. Øersted discovered the link between electricity and magnetism in 1820.
It wasn’t until 1905, with Einstein’s special relativity paper that a handful of humans on this
planet finally understood the profound nature of this relationship – a time span of 85 years –
approximately a human lifetime passed!
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
13
UIUC Physics 436 EM Fields & Sources II
14
Fall Semester, 2015
Lect. Notes 18.5
Prof. Steven Errede
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 18.5
Prof. Steven Errede
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
15
UIUC Physics 436 EM Fields & Sources II
16
Fall Semester, 2015
Lect. Notes 18.5
Prof. Steven Errede
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 18.5
Prof. Steven Errede
The Electromagnetic Field Tensor F  v




The relations for the Lorentz transformation of E and B in the lab frame IRF(S) to E  and B

in IRF(S'), where IRF(S') is moving with velocity v   vxˆ relative to IRF(S) are:
 components
Ex  Ex
B components
Bx  Bx
E y    E y   cBz 
By    By   Ez c 
  1 1  2
Ez    Ez   cBy 
Bz    Bz   E y c 
  vc
  components
B  components


It is readily apparent that the six components of E and B field certainly do not transform like
the spatial / 3-D vector parts of e.g. two separate contravariant 4-vectors ( E  and B  ), because


the (orthogonal) components of E and B  to the direction of the Lorentz transformation are


 
mixed together {as seen in the case for B  0 in IRF(S) resulting in B   c12 v  E  in IRF(S')
  

and the case for E  0 in IRF(S) resulting in E   v  B in IRF(S')}.


© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
17
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 18.5
Prof. Steven Errede
Note also that the form of the Lorentz transformation for  vs.  components is “switched”
{here} for the EM fields vs. 4-vectors!!!
Recall that true relativistic 4-vectors transform by the rule: a    v a v ,

e.g. for a Lorentz boost from IRF(S) → IRF(S') along v  vxˆ .
Specifically, recall that x    v x v is explicitly:
ct     ct   x 
Parallel components:
x    x   ct 
Perpendicular
y  y
Components:
z  z
Ex  Ex
Bx  Bx
Ez    Ez   cBy 
Bz    Bz   E y c 
E y    E y   cBz 
By    By   Ez c 

For a Lorentz transformation x    v x v from IRF(S) → IRF(S') along v   vxˆ ,
the   v tensor has the form:
Row index
v
 



 0

 0


0
0
0
0
1
0
0

0
 n.b.   v is a symmetric matrix
0

1
Column index


Thus, there is no way that the six components of the E and B fields can be construed as being
the spatial / 3-D vector components of contravariant 4-vectors, e.g. E  and B  .
(What would be their temporal components/ scalar counterparts: E 0 and B 0 = ???)


It turns out that the 3-D spatial vectors E and B are the six unique components of a 4  4
rank-two EM field strength tensor, F  v !!!
A 4  4 rank two tensor t  Lorentz transforms via two  -factors (one for each index):
t  v     t   v In matrix form: t   t T  t  because  is symmetric matrix
Where t  is:
Column # 0
 t 00
 10
t

t   20
t

Column  t 30

Row
18
1
2
t 01 t 02
t11
t12
21
22
t
t
t 31 t 32
3
t 03 

t13 
t 23 

t 33 
Row #
 t -t

sx - t
1
 
 sy - t
2

3
 sz - t
0
t - sx
t - sy
sx - sx
sx - s y
s y - sx
sy - sy
sz - sx
sz - s y
t - sz 

sx - sz 

s y - sz 

sz - sz 
n.b. Each
element
of tensor
t has
temporalspatial
meaning!
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 18.5
Prof. Steven Errede
The 16 components of the 4  4 second rank tensor t  need not all be different (e.g.   v )
– the 16 components of t  may have symmetry / anti-symmetry properties:
Symmetric 4  4 rank two tensor:
Anti-symmetric 4  4 rank two tensor:

t   t 
t   t 
For the case of a symmetric 4  4 rank two tensor t   t  , of the 16 total components,
10 are unique, but 6 are repeats:
t01 = t10, t02 = t20, t03 = t30
t12 = t21, t13 = t31
t32 = t23

tsym
.

 t 00 t 01 t 02
 01 11 12
t
t
t
  02 12 22
t
t
t
 03 13 23
t
t
t
t 03 

t13 
t 23 

t 33 
For the case of an anti-symmetric 4  4 rank two tensor t   t  , of the 16 total
components, only 6 are unique, 6 are repeats (but with a minus sign!) and the 4 diagonal
elements must all be  0 (i.e. t00 = t11 = t22 = t33  0):
t01 = t10, t02 = t20, t03 = t30
t12 = t21, t13 = t31
t32 = t23

tanti
sym.
 0
 01
t
  02
 t
 03
 t
t 01
t 02
0
t12
t12
0
t
t 23
13
t 03 

t13 
t 23 

0 


So it would seem that the six components of the spatial / 3-D E and B vectors can be
represented by an anti-symmetric 4  4 rank two contravariant tensor !!!
v
  v
Let’s investigate how the Lorentz transformation rule t     t   for a 4  4 rank two

anti-symmetric tensor tanti
works (6 unique non-zero components). Starting with t 01   0  t  1 :
sym.
Column #
0
 



v
   or     
 0

 0
1
2


0
0
3
0
0
1
0
Row #
0

0
0

1
0
1
2
3

tanti
sym.
 0
 01
t
  02
 t
 03
 t
t 01
t 02
0
t12
t12
0
t
t 23
13
t 03 

t13 
t 23 

0 
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
19
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
 00
Row #

We see that  0   0 , unless   0 or 1:
Column #
Row #
We also see that 1  0 , unless   0 or 1:
Column #
 There are only 4 non-zero terms in the sum:
Lect. Notes 18.5
Prof. Steven Errede
 01

10
11


t 01   0  t  1   0 0t 00 10   0 0t 0111   01t10 10   01t1111
But: t00 = 0 and t11 = 0, whereas: t01 = t10.
 t 01    0 0 11   0110  t 01             t 01    2   2  2  t 01   2 1   2  t 01
01
01
But:  2  1 1   2   t   t
v
  v
One (e.g. you!!!) can work through the remaining 5 of the 6 distinct cases for t     t  
(or for completeness’ sake, why not work out all 16 cases explicitly!!!) . . .
The complete set of six rules for the Lorentz transformation of a 4  4 rank two contravariant
v
  v
anti-symmetric tensor t     t   are:
t 01  t 01
↔
Ex  Ex
t 02    t 02   t12 
↔
E y    E y   cBz 
t 03    t 03   t 31 
↔
Ez    Ez   cBy 
t 23  t 23
↔
Bx  Bx
t 31    t 31   t 03 
↔
By    By   Ez c 
t 12    t12   t 02 
↔
Bz    Bz   E y c 
As can be seen, the above Lorentz transformation rules for this tensor are indeed precisely
those we derived on physical grounds for the EM fields!
Thus, we can now explicitly construct the so-called electromagnetic field tensor F  v
– a 4  4 rank two contravariant anti-symmetric tensor:
F v
20
 F 00  0
 10
F   F 01
  20
 F   F 02
 30
03
 F  F
F 01
F 02
F 11  0
F 12
F 21   F 12
F 22  0
F 31
F 32   F 23
  0
 
F 13   F 31    Ex

  Ey
F 23
 
F 33  0    Ez
F 03
Ex
Ey
0
cBz
cBz
0
cBy
cBx
Ez 

cBy 
cBx 

0 
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
UIUC Physics 436 EM Fields & Sources II
F v
 0

 Ex

 Ey

  Ez
Fall Semester, 2015
Ex
Ey
0
cBz
cBz
0
cBy
cBx
Ez 

cBy 
cBx 

0 
Lect. Notes 18.5
Prof. Steven Errede
E-field representation of
F  v (SI units: Volts/m).
Note that there exist alternate/equivalent forms of the field tensor F  v .
If we divide our F  v by c, we get Griffith’s definition of F  v :
i.e.
1 

FGriffiths
 FOurs
c
Ex c E y c Ez c 
 0


0
Bz
 Ex c
 By 


  E y c  Bz
0
Bx 


0 
By
 Bx
  Ez c
B-field representation of
F  v (SI units: Tesla)


Note that there also exists an equivalent, but different way of embedding E and B in an
anti-symmetric 4  4 rank two tensor:
Instead of comparing:
a) Ex  Ex ,
with: a') t 01  t 01 ,
E y    E y   cBz  ,
t 02    t 02   t12  ,
Ez    Ez   cBy 
t 03    t 03   t 31 
and also comparing:
b) cBx  cBx , cBy    cBy   z  , cBz    cBz   y 
with: b') t 23  t 23 ,
t 31    t 31   t 03  ,
t 12    t12   t 02 
We could instead compare {a) with b')} and {b) with a')}, and thereby obtain the so-called
anti-symmetric rank-two dual tensor G  v :
G v
 0

cBx

 cBy

 cBz
cBx
cBy
0
Ez
 Ez
0
Ey
Ex
cBz 

Ey 
 Ex 

0 
E-field representation of
G  v (SI units: Volts/m)
Note further that the G  v elements can be obtained directly from F  v by carrying out




a duality transformation (!!!): E  cB, cB   E


Additionally note that a duality transformation leaves the Lorentz transformation of E and B
unchanged !!! (i.e.  duality = 90o).
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
21
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 18.5
Prof. Steven Errede
Again, note that our representation of the dual tensor G  v is simply related to Griffiths by
dividing ours by c, i.e.:
n.b. For the sake of
compatibility with
Griffiths book, we will use
his definitions of the field
strength tensors. Please
be aware that these tensor
definitions do vary in
different books!
1 v
v
GGriffiths
 Gours
c
 0

 Bx

  By

  Bz


0
 Ez c E y c 
0
Ez c
 Ex c 

0 
 E y c Ex c
Bx
By
Bz
B-field representation of
G  v (SI units: Tesla)
With Einstein’s 1905 publication of his paper on special relativity, physicists of that era soon


realized that the E and B -fields were indeed intimately related to each other, because of the
unique structure of space-time that is associated with the universe in which we live.
Prior to Øersted’s 1820 discovery that there was indeed a relation between electric vs.
magnetic phenomena, physicists thought that electricity and magnetism were separate / distinct
entities. Maxwell’s theory of electromagnetism “unified” these two phenomena as one, but it
was not until Einstein’s 1905 paper, that physicists truly understood why they were so-related!
It is indeed amazing that the electromagnetic field is a 4  4 rank two anti-symmetric
contravariant tensor, F  v (or equivalently, G  v ) !!! (six components of which are unique.)
The contravariant and covariant forms of the field strength tensor are:
Ex c E y c Ez c 
 Ex c  E y c  Ez c 
 0
 0




0
Bz
0
Bz
 Ex c
 By 
 Ex c
 By 



and: F v  
  E y c  Bz
0
0
Bx 
Bx 
 E y c  Bz




0 
0 
By
By
 Bx
 Bx
  Ez c
  Ez c
F v
The contravariant and covariant forms of the dual field strength tensor are:
G v
 0

 Bx

  By

  Bz

 0


0
 Ez c E y c 
 Bx
G v  
and:
  By
0
Ez c
 Ex c 


0 
 E y c Ex c
  Bz
Bx
By
Bz
 Bx
 By
 Bz 

0
 Ez c E y c 
0
Ez c
 Ex c 

0 
 E y c Ex c
An arbitrary relativistic contravariant 4-vector x  is related to its covariant version x by:
x  g  v x v and: x   g  v xv , where the “flat” space-time metric g  v  g  v {here} is defined:
g v  gv
22
 1

0

0

0
0
1
0
0
0
0
1
0
0

0
with: g g   

0

1
 is the 4-D space-time version of
the 3-D Kroenecker -function  ij ,
i.e.   0 if    and
  1 for   0,1, 2,3 .
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
UIUC Physics 436 EM Fields & Sources II
Fall Semester, 2015
Lect. Notes 18.5
Prof. Steven Errede
Likewise, the field strength tensors F  v and F v , G  v and G v are formally related to each other
by: F  g F  v g v or: F   g  F v g v and: G  g G  v g v or: G  g  G v g v .
The dual field strength tensor G  v is formally related to the field strength tensor F  v by:
G  12   F v where:      is the 4×4×4×4 = 256 element totally anti-symmetric
tensor of rank four, the 4-D generalization of the 2-D Levi-Cività symbol ( ij ) :    1 (1)
for even (odd) permutations of 0-1-2-3, respectively,    0 if any two indices are equal/same.
We have learned that the 4-vector product of any two (bona-fide) relativistic 4-vectors,
e.g. a  b  a b  = Lorentz invariant quantity (i.e. a  b  a b  = same value in all/any IRF’s)
Likewise, the tensor product A v B v  A v B  v of any two relativistic rank-2 tensors is also
a Lorentz invariant quantity.
Thus, the tensor products of EM field tensors F  v and G  v , namely: F  v F v , G  v G v ,
and: F  v G v  F v G  v are all Lorentz invariant quantities.
Specifically, it can be shown that (see Griffiths Problem 12.50, p. 537):
1


F  v F v  G  v G v  2  B 2  2 E 2    uM  uE  ← Lorentz invariant quantity !!!
c


And:
F  v G v  F v G  v  
4  
E  B ← Lorentz invariant quantity !!!
c


Obviously, F  v F v  F v F  v , G  v G v  G v G  v and F  vG v  F v G  v are the only Lorentz
invariants that can be formed / constructed using F  v and G  v !!!
Note that it is by no means obvious/clear how to form such Lorentz invariant EM field quantities


starting directly from the E and B fields themselves !!!

 
 n.b. Since EM energy density uEM  12  o E 2  2 1o B 2 , Poynting’s vector S  1o E  B ,


EM field linear momentum density EM   o o S and EM field angular momentum density

 
 EM  r EM cannot be formed from any linear combinations of F  v F v , G  v G v or F  v G v ,
 

then uEM , S , EM and  EM cannot be Lorentz invariant quantities!
© Professor Steven Errede, Department of Physics, University of Illinois at Urbana-Champaign, Illinois
2005-2015. All Rights Reserved.
23