Download Electric and magnetic field transformations Picture: Consider inertial frames

Electric and magnetic field
transformations
Consider inertial frames F’ and F.
Suppose we know the electric and
magnetic fields, E’(x’,t’) and B’(x’,t’),
as observed in frame F’.
Picture:
Then what are the electric and
magnetic fields, E(x,t) and B(x,t), as
observed in frame F?
Suppose F’ moves with velocity v
relative to F. WLOG, take v to be in
the x direction; v = v i.
Fields are present, which are
measured, by the forces that they
produce, with respect to the two
different frames of reference, F and
F’.
Electric and magnetic field
transformations
To figure out E(x,t) and B(x,t):
(1) We already know the field tensor
F’μν for the inertial frame F’.
(2) We know how to transform the
field tensor,
Fμν = Λμρ Λνσ F’ρσ
where Λμρ is the Lorentz
transformation matrix.
(3) Finally deconstruct Fμν into E(x,t)
and B(x,t).
(1)
0
E’x/c
F’μν = −E’x/c
0
−E’y/c −B’z
−E’z/c
B’y
(2)
xμ = Λμρ x’ρ
γ
γv/c
Λμρ = γv/c γ
0
0
0
0
;
E’y/c E’z/c
B’z
−B’y
0
B’x
−B’x
0
Fμν = Λμρ Λνσ F’ρσ
0 0
0 0
1 0
0 1
(3)
{Ex, Ey, Ez } = { c F01 , c F02 , c F03}
{Bx, By, Bz } = { F23, F31, F12}
μν = 01
μν = 12
F01 = Λ0ρ Λ1σ F’ρσ = Λ00 Λ11 F’01 + Λ01Λ10 F’10
Ex = γ2 E’x + γ2 (v2/c2) (-E’x )
Ex = E’x
μν = 02
F02 = Λ0ρ Λ2σ F’ρσ = Λ00 Λ22 F’02 + Λ01 Λ22 F’12
Ey = γ E’y + c γ(v/c) B’z
F12 = Λ1ρ Λ2σ F’ρσ
= Λ10 Λ22 F’02 + Λ11 Λ22 F’12
Bz = γ (v/c) (Ey /c) + γ B’z
Bz = γ (B’z + (v/c2) Ey )
μν = 31
similarly
By = γ (B’y − (v/c2) Ez )
μν = 23
F23 = Λ2ρ Λ3σ F’ρσ
= Λ22 Λ33 F’23
Bx = B’x
Ey = γ (E’y + v B’z )
μν = 03
similarly
Ez = γ (E’z − v B’y )
Tables of coordinate transformations
and field transformations
Suppose reference frame F’ moves with
velocity v = v i with respect to frame F.
Picture:
Coordinate transformations (x μ ← x’ μ )
xμ = Λμρ x’ρ
Λμρ =
γ
γv/c
γv/c γ
0
0
0
0
0 0
0 0
1 0
0 1
t = γ (t’ + (v/c2) x’)
x = γ (x’ + v t’)
y = y’
z = z’
Field transformations (F μν← F’ μν)
Ex = E’x
Ey = γ (E’y + v B’z )
Ez = γ (E’z − v B’y )
Bx = B’x
By = γ (B’y − (v/c2) Ez )
Bz = γ (B’z + (v/c2) Ey )
Example.
Consider a line of charge that is at rest in
reference frame F’, lying along the x’ axis;
charge per unit length = λ.
Use the field transformations to calculate the
electric and magnetic fields in the frame F .
NOTA BENE: the line of charge moves in the x
direction relative to frame F.
Picture:
E’ = λ/(2πε0 r’) { (y’/r’) j + (z’/r’) k }
On the other hand B’ = 0 because there is no
current.
The transformed fields (relative velocity = v i):
Ex = E’x = 0
Ey = γ E’y
Ez = γ E’z
E = γλ/(2πε0 r’) { (y’/r’) j + (z’/r’) k }
E = γλ/(2πε0 r) { (y/r) j + (z/r) k } ;
this is the same as a line of charge with density γλ,
larger than λ because of Lorentz contraction.
Solution:
First, what are the fields E’ and B’?
By Gauss’s law, the electric field around a
charged line points radially away from the line,
with magnitude λ/(2πε0 r’) where r’ is the
perpendicular distance from the line.
Bx = B’x = 0
By = − γ (v/c2) Ez
Bz = γ (v/c2) Ey
B = γ(v/c2) λ/(2πε0 r’) { (−z’/r’) j + (y’/r’) k }
B = μ0 I /(2πr) { (−z/r) j + (y/r) k } ;
this is the same as a line of current with
I = γ λ v, by Ampere’s law.
Next lecture -calculate the electric and magnetic
fields of a single charge that moves
with constant velocity v.