Div_and_Curl_of_B.pdf

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W3007 – Electricity and Magnetism 1
Divergence and Curl of �B
The Biot-Savart law for a stationary volume current density �J is
�B(�r) =
µ0
4π
∫ �J(�r ′)× �ˆ
�2
dτ ′ . (1)
I am using �� to denote the relative vector �r − �r ′, which Griffiths denote by ‘script-r’ (I could not
find how to write ‘script-r’ with LaTeX!). The integral is performed through the volume where
the currents are, but as usual we can extend the integral to all space for free, since where �J is zero
the contribution to the integral vanishes anyway.
Before starting computing derivatives, let’s be totally explicit about the notation. �B is a
function of the ‘observation position’ �r = (x, y, z). �J is a function of the ‘source position’ �r ′ =
(x′, y′, z′), over which we are integrating. �� = �r − �r ′ is the vector connecting the two. By �∇ we
denote the gradient operator w.r.t. to �r. Likewise, �∇′ is the gradient operator w.r.t. �r ′.
Now, we are interested in the divergence and the curl of �B. Before computing them, it is useful
to massage the Biot-Savart law to recast �B into a nicer form. We notice that
�ˆ
�2
= −�∇1
�
, (2)
so that
�B(�r) = −µ0
4π
∫
�J × �∇1
�
dτ ′ . (3)
We now use the product rule
�∇× (f �g) = f �∇× �g − �g × �∇f , (4)
with �g = �J and f = 1/�, to get
�B(�r) =
µ0
4π
∫ [
�∇×
(
�J
1
�
)
− 1
�
�∇× �J
]
dτ ′ . (5)
The second piece vanishes, because �J does not depend on �r (only on �r ′ !), and so its curl w.r.t. �r
is zero. In the first piece we can take the curl out of the integral, because it is a curl w.r.t to
variables (the ‘unprimed’ coordinates) over which we are not integrating. In conclusion we have
�B(�r) =
µ0
4π
�∇×
∫ �J
�
dτ ′ . (6)
�B is the curl of something else:
�B = �∇× �A , �A = µ0
4π
∫ �J
�
dτ ′ . (7)
2 Divergence and Curl of �B
Now we immediately see that the divergence of �B vanishes,
�∇ · �B = 0 , (8)
because the divergence of a curl is always zero. As to the curl of �B, the computation is somewhat
complicated. We use the product rule
�∇× (�∇× �A) = �∇(�∇ · �A)−∇2 �A , (9)
where by the Laplacian of a vector function �A, we simply mean the vector whose components are
the Laplacian of the corresponding components of �A: (∇2 �A)x = ∇2(Ax), etc. We thus have
�∇× �B = �∇(�∇ · �A)−∇2 �A . (10)
The first piece is
�∇(�∇ · �A) = µ0
4π
�∇
∫
�∇ ·
�J
�
dτ ′ (11)
Notice that inside the integral we have the divercence w.r.t. �r, whereas we are integrating in �r ′
(this is in fact why we could bring the divergence inside the integral). �J does not depend on �r, so
that we have
�∇ ·
�J
�
= �J · �∇1
�
(12)
= − �J · �∇′1
�
(13)
= −�∇′ · ( �J 1
�
)
+
1
�
�∇′ · �J , (14)
Where in the first step we used the product rule
�∇ · (f �g) = f �∇ · �g + �g · �∇f ; (15)
in the second step we used that since 1/� is a function of the difference �r−�r ′, derivatives w.r.t. �r are
the same as derivatives w.r.t. �r ′; in the third step we used product rule eq. (15) once again. Now,
the second piece in eq. (14) vanishes for stationary currents: the continuity equation (=local charge
conservation) relates the time-derivative of the charge density to the divergence of the current
density. For magnetostatics there is no time-dependence, and as a consequence the divergence of
�J vanishes.
In summary, eq. (11) becomes
�∇(�∇ · �A) = −µ0
4π
�∇
∫
�∇′ · ( �J 1
�
)
dτ ′ (16)
= −µ0
4π
�∇
∮
S
�J
�
· d�a ′ (17)
= 0 . (18)
W3007 – Electricity and Magnetism 3
In the first step we used Gauss’s theorem to rewrite the volume integral as a surface integral. In
the second step we realized that, since the volume integral was extended to all space, the boundary
surface S is at infinity, and there the current density vanishes (if we are dealing with a localized
system of currents).
In eq. (10) we are thus left with the second piece:
�∇× �B = −∇2 �A (19)
= −µ0
4π
∇2
∫ �J
�
dτ ′ (20)
= −µ0
4π
∫
�J ∇21
�
dτ ′ (21)
= −µ0
4π
∫
�J(�r ′)(−4π)δ3(�r − �r ′) dτ ′ (22)
= µ0J(�r) . (23)
The Laplacian was w.r.t. �r, and could therefore be brought inside the integral for free. Also, it
only acted on 1/�, because �J was a function of �r ′ alone. Finally we used the fundamental property
of the delta-function—that when integrated with another function, it yields the latter computed
where the argument of the delta-function vanishes.
In conclusion, we have
�∇ · �B = 0 , �∇× �B = µ0J . (24)