Download The fields of a current wire

The fields of a current wire
A charge q has a velocity v in the laboratory frame S. A
q v
constant current I flows in an infinitely long wire, parallel to
v at a distance r from q. There is no net charge density on
the wire in S.
I
a) Find the force F on q in S. Then use the appropriate
rules for the transformation of forces between different inertial
frames to find the force F′ in the frame S ′ where q is at rest. What can be inferred on the EM fields
in S ′ from the expression of F′ ?
b) From the knowledge of the charge and current density in S use the Lorentz transformations to
obtain the charge and current density in S ′ and the related EM fields. Do the same for the scalar
and vector potential. Check out the results using the direct transformation rules for the EM field.
c) Solving points a) and b) shows that in S ′ there is a net charge density on the wire. Recover
this result using the Lorentz transformations of velocity and length to obtain in S ′ the density of
both electrons (flowing in S with velocity ve ) and ions (at rest in S). (This last point corresponds
to the approach presented in E. M. Purcell, Electricity and Magnetism, Berkeley course on Physics,
Volume 2, par.5.9.)
1
Solution
a) In S, the wire generates a magnetic field with azimuthal symmetry Bφ = Bφ (r) = µ0 I/(2πr). The
Lorentz force on the charge is F = qv × B, i.e. Fr = −qBφ (r)v = µ0 qIv/(2πr).
The S ′ frame moves with velocity v with respect to S and
q v
S
the velocity of the charge v′ = 0 in S ′ . This has two consequences: first, the transformation law for the force1 on q in S ′
F = qv × B
λ=0
is particularly simple, Fr′ = γFr (where γ = (1 − β 2 )−1/2 and
β = v/c with v = |v|). Second, since q is at rest in S ′ , the
I
force Fr′ must be due to an electric field Er′ = Fr′ /q. We may
thus write the transformation formula Er′ = −γvBφ or, in a
q v′ = 0
generalized vector form,
S′
E′⊥ = γ(v × B) ,
(2)
λ′ < 0
F′ = γF = qE′
where the suffix refers to the direction perpendicular to v. In
I ′ = γI
the limit |v| ≪ c (for which F = F′ ) we get E′⊥ ≃ v × B which
is correct up to first order in β = v/c and may be called the Galilei transformation of the field.
The electric field2
Er′ = −γvBφ (r′ ) = −µ0 γvI/(2πr′ ) = −γvI/(2πǫ0 c2 r′ )
(3)
is generated by a linear charge density λ′ = −vγI/c2 along the wire, as it can be easily verified
by applying Gauss’s law. Thus the wire is negatively charged in S ′ . It might then seem that there
is a violation of the conservation of charge in S ′ . Actually, this a consequence of the somewhat
“pathological” nature of currents which are not closed in a loop, as for the infinite wire. In fact,
strictly speaking neither the infinite current wire is a steady system nor charge conservation is
apparent for such system as the flowing charge accumulates at infinity. If in S we introduce “return”
currents to close the loop, e.g. if we take the wire as the inner conductor of a coaxial cable, or at
some distance we introduce a second wire currying the current −I, then we find that the return
currents would appear as an opposite charge densities in S ′ , making charge conservation apparent.
Since the force is purely magnetic in S and purely electric in S ′ , at this point we cannot say much
about the magnetic field in S ′ .
b) We know that J = (J, cρ) is a four-vector. The transverse section A of the wire is invariant for a
Lorentz boost along the wire direction, thus also λ = Aρ and I = AJ transform as ρ and j. Therefore
1
The general transformation law, defined by dp/dt = F and dp′ /dt′ = F′ , is
Fk′ = Fk −
V F · v⊥ /c2
,
1 − V vk /c2
F′⊥ =
F⊥
,
γ(1 − V vk /c2 )
(1)
where “k” and “⊥” refer to the components parallel and perpendicular to the boost velocity V, respectively. In our
case, Fk = 0, v⊥ = 0, and V = vk so that γ(1 − V vk /c2 ) = 1/γ.
2
In general the complete transformation reads E′⊥ (r′ , t) = γv × B (r[r′ , t′ ], t[r′ , t′ ]) where r = r[r′ , t′ ] and t = t[r′ , t′ ]
according to the Lorentz transformation of the coordinates. Since in cylindrical coordinates Bφ depends on r only
and for the coordinates in the plane transverse to the boost velocity r′⊥ = r⊥ , in the present case the transformation
is trivial, r′ = r.
2
in S ′ the wire has a linear charge density
λ′ = γ(λ − vI/c2 ) = −γvI/c2 ,
(4)
which generates, from Gauss’s law, the radial electric field Er′ = λ′ /2πrǫ0 . We thus recovered the
result of point a). We also obtain the in S ′ the current intensity is
I ′ = γ(I − vλ) = γI ,
(5)
from which we obtain that the magnetic field Bφ′ = γBφ′ .
Now let us obtain the same results through the transformation of the four-potential (cφ, A). In
S, obviously φ = 0 as there is no charge density, while the vector potential A satisfies
∇2 A = −µ0 J = −µ0 Iδ(r),
(6)
which is identical to Poisson’s equation for the electrostatic potential of a wire with a linear charge
density. We thus obtain that A is parallel to the wire and the only non-zero component is
µ0 I r Az = −
ln
,
2π
a
(7)
where a is an arbitrary constant. It is straightforward to verify that Bφ = −∂r Az .
We obtain for the scalar potential in S ′
r
r
γβI
λ′
φ = γ(φ − vAz ) = −γvAz = −
ln
ln
=−
,
2πǫ0 c2
a
2πǫ0
a
′
where we put λ′ = −βγI/c2 . The electric field is finally obtained as E′ = −∇φ′ , recovering the
above result. For the vector potential in S ′ , trivially A′z = γ(Az − vφ/c2 ) = γAz from which we get
Bφ′ = γBφ again.
The results may be checked by the explicit formulas for the transformation of the EM field,
γ2
β(β · E) ,
γ+1
γ2
B′ = γ (B − β × E/c) −
β(β · B) ,
γ+1
E′ = γ (E + β × Bc) −
(8)
(9)
which in our case give E′ = γβ × Bc and B′ = γB, in agreement with what was found above.
c) Let us first consider the linear charge densities of both ions (λi = Zeni A) and electrons (λe =
−ene A) in S. Since in S there is no net charge on the wire, λi = −λe .
Now let us find what are the charge densities for both species in S ′ , considering relativistic
kinematics only. On a wire segment of length ∆L there is an ion charge ∆Q = λi ∆L. In S ′ ,
the same segment has the same charge (since it is a Lorentz invariant) but the length undergoes
relativistic contraction, ∆L′ = ∆L/γ, so that we obtain a higher density λ′i = ∆Q/∆L′ = γλi . This
is a quite general result: a fluid moving at a velocity v has a higher density (by a factor γ) than the
density in its rest frame.
3
When considering the electrons, however, they are not at rest in S: the move along the wire with
velocity ve < 0 such that I = −ene ve A = λe ve = −λi ve . Thus, their density is already higher by a
factor γe = (1 − ve2 /c2 )−1/2 than the density λe0 in the frame where the electrons are at rest: we have
λe0 = λe /γe . In S ′ , the electrons drift with a velocity ve′ given by
ve′ =
ve − v
,
1 − ve v/c2
(10)
according to Lorentz transformations. Thus, the electron density in S ′ is
λ′e
=
γe′ λe0
γe′
= λe ,
γe
(11)
′
where γe′ = (1 − ve2 /c2 )−1/2 . This relativistic factor can be put in a convenient form by some algebra:
γe′
=
(ve − v)2 /c2
1−
(1 − ve v/c2 )2
−1/2
= (1 − ve v/c2 )(1 − 2ve v/c2 + ve2 v 2 /c2 − ve2 /c2 + 2ve v/c2 − v 2 /c2 )−1/2
−1/2
= (1 − ve v/c2 ) (1 − ve2 /c2 )(1 − v 2 /c2 )
= (1 − ve v/c2 )γe γ .
We thus obtain for the total charge density in S ′
γe′
′
′
′
λ = λi + λe = λi γ −
= λi γ 1 − 1 + ve v/c2 = λi γve v/c2 = −γvI/c2 ,
γe
(12)
(13)
as we previously found on the basis of Lorentz transformations for the forces, charge and current
densities and EM fields.
It may be interesting to notice that there is already a subtle issue of charge conservation with the
wire in S. In the material of which the wire is composed the charge densities of ions and electrons are
of course equal, so both charge and current densities are zero on an isolated wire without external
fields and disconnected from any voltage or current generator. Now suppose we drive a steady current
I by keeping electrons in motion along the wire with velocity ve . If the wire is still electrically neutral
as we assumed, then the charge densities of both ions and electrons are still the same. However,
while the charge density of ions at rest has not changed, for the charge density of electrons there
is the effect of “relativistic increase” by the factor γe ; but if the resulting density has not changed
(since it must still cancel the ion charge out), some electron charge must have disappeared or, more
correctly, moved out of the wire.3 We may find where the missing charge has gone only if we assume
a current loop instead of an “open” wire, adding specifications on boundary conditions and how the
circuit is closed at some distance.
3
Of course the effect is negligibly small for ordinary conduction in metals, for which typical electron velocities
ve ∼ 10−10 c. However, the issue if very important for relativistic hydrodynamics, i.e. for contexts where fluids do
move at velocities close to c.
4