One of the greatest advances in theoretical physics of the nineteenth century was Maxwell’s formulation of the the equations of electromag- netism. This article uses differential forms to solve a problem related to Maxwell’s formulation. The notion of differential form encompasses such ideas as elements of surface area and volume elements, the work exerted by a force, the flow of a fluid, and the curvature of a surface, space or hyperspace. An important operation on differential forms is exterior differentiation, which generalizes the operators div, grad, curl of vector calculus. the study of differential forms, which was initiated by E.Cartan in the years around 1900, is often termed the exterior differential calculus.However, Maxwells equations have many very im- portant implications in the life of a modern person, so much so that people use devices that function off the principles in Maxwells equa- tions every day without even knowing it Published by: Wang Jing School of Physical and Mathematical Sciences Nanyang Technological University jwang14@e.ntu.edu.sg

1. Maxwell’s Formulation – Differential
Forms on Euclidean Space
Wang Jing
School of Physical and Mathematical Sciences
Nanyang Technological University
jwang14@e.ntu.edu.sg
Abstract
One of the greatest advances in theoretical physics of the nineteenth
century was Maxwell’s formulation of the the equations of electromag-netism.
This article uses differential forms to solve a problem related
to Maxwell’s formulation. The notion of differential form encompasses
such ideas as elements of surface area and volume elements, the work
exerted by a force, the flow of a fluid, and the curvature of a surface,
space or hyperspace. An important operation on differential forms is
exterior differentiation, which generalizes the operators div, grad, curl
of vector calculus. the study of differential forms, which was initiated
by E.Cartan in the years around 1900, is often termed the exterior
differential calculus.However, Maxwells equations have many very im-portant
implications in the life of a modern person, so much so that
people use devices that function off the principles in Maxwells equa-tions
every day without even knowing it.
1 Introductions to differential forms
1.1 Elementary properties
A differential form of degree k or a k-form on 푅푛 is an expression
훼 =
Σ︁
퐼
푓퐼푑푥퐼
Here I stands for a multi-index (푖1, 푖2, · · · , 푖푘)of degree k,that is a ”vector”
consisting of k integer entries ranging between 1 and 푛, The 푓퐼 are smooth
functions on 푅푛 called the coefficients of 훼, and 푑푥퐼 is an abbreviation for
푑푥푖1푑푥푖2 · · · 푑푥푖푘
(The notion 푑푥푖1
⋀︀
푑푥푖2
⋀︀
· · ·
⋀︀
푑푥푖푘 is also often used to distinguish this
kind of product from another kind, called the tensor product) For instance
1

2. the expressions:
훼 = sin 푥1 + 푒푥4푑푥1푑푥2 + 푥2푥25
푑푥2푑푥3 + 6푑푥2푑푥4 + cot 푥2푑푥5푑푥3
훽 = 푥1푥3푥5푑푥1푑푥6푑푥3푑푥2
represent a 2-form on 푅5, resp. a 4-form on 푅6. The form 훼 consists of four
terms, corresponding to the multi-indices (1,5),(2,3),(2,4),and (5,3), whereas
훽 consists of one term, corresponding to the multi-index(1,6,3,2). Note,
however, that 훼 could equally well be regarded as a 2-form on 푅6 that does
not involve the variable 푥6. To avoid such ambiguities it is good practice to
state explicitly the domain of definition when writing a differential form. A
0-form on 푅푛 us simply a smooth function (no dx’s)
1.2 Exterior derivative
If 푓 is a 0-form, that is a smooth function, we define 푑푓 to be the 1-form
푑푓 =
Σ︁푛
푖=1
휕푓
휕푥푖
푑푥푖
Then we have the product or leibniz rule:
푑(푓푔) = 푓푑푔 + 푔푑푓
If훼 =
Σ︀
퐼 푓퐼푑푥퐼 is a k-form, each of the coefficient 푓퐼 is a smooth function
and we define 푑훼 to be the k+1 form
푑훼 =
Σ︁
퐼
푑푓퐼푑푥퐼
The operation 푑 is called exterior differentiation. An operator of this sort is
called a first-order partial differential operator, because it involves the first
partial derivatives of the coefficients of a form.
Proposition 1.1. (i)푑(푎훼+푏훽) = 푎푑훼+푏푑훽푓표푟푎푙푙푘−푓표푟푚푠훼푎푛푑훽푎푛푑푎푙푙푠푐푎푙푎푟푠푎푎푛푑푏.
(ii)푑(훼훽) = (푑훼)훽 + (−1)푘훼푑훽푓표푟푎푙푙푘 − 푓표푟푚훼푎푛푑푙 − 푓표푟푚푠훽.
Proposition 1.2. 푑(푑훼) = 0 for any form 훼, In short,
푑2 = 0
1.3 Closed and exact forms and Hodge star operator
A form 훼 is 푐푙표푠푒푑 if 푑훼 = 0. It is exact if 훼 = 푑훽 for some form 훽(of degree
one or less).
Proposition 1.3. Every exact form is closed.
2

3. Proof: If 훼 = 푑(푑훽) then 푑훼 = 푑(푑훽) = 0 by last section’s second
proposition. The binomial coefficient 퐶푘
푛 is the number of ways of selecting
k(unordered) objects from a collection of 푛 objects. Equivalently, 퐶푘
푛 is the
number of ways of partitioning a pile of n objects into a pile of 푘 objects
and a pile of 푛 − 푘 objects. Thus we see that
퐶푘
푛 = 퐶푘
푛−푘
This means that in a certain sense there are as many 푘 − 푓표푟푚푠. this is a
natural way to turn 푘 − 푓표푟푚푠 into 푛 − 푘 − 푓표푟푚푠. This is the Hodge star
operator. Hodge star Σ︀
of 훼 denoted by *훼 (or sometimes 훼*) and is defined
as follows. If 훼 =
퐼 푓퐼푑푥퐼 than
*훼 =
Σ︁
퐼
푓퐼 * 푑푥퐼
with
*푑푥퐼 = 휀퐼푑푥퐼푐
Here, for any increasing multi-index 퐼, 퐼푐 denote the complementary increas-ing
multi-index, which consists of all numbers between 1 and n that do not
occur in 퐼. The factor 휖퐼 us a sign,
푣푎푟푒푝푠푖푙표푛퐼 =
{︃
1 if 푑푥퐼푑푥퐼푐 = 푑푥1푑푥2 · · · 푑푥푛
−1 if 푑푥퐼푑푥퐼푐 = −푑푥1푑푥2 · · · 푑푥푛
(1)
In other words, *푑푥퐼 is the product of all the 푑푥′
푗푠 that do not occur in
푑푥퐼 , times a factor ±1 which is chosen in such a way that 푑푥퐼 (*푑푥퐼 ) is the
volume form:
푑푥퐼 (*푑푥퐼 ) = 푑푥1푑푥2 · · · 푑푥푛
Example. On 푅2 we have 푑푥 = 푑푦 and 푑푦 = −푑푥. On 푅3 we have
∙ *푑푥 = 푑푦푑푧, *(푑푥푑푦) = 푑푧,
∙ *푑푦 = −푑푥푑푧 = 푑푧푑푥, *(푑푥푑푧) = −푑푦,
∙ *푑푧 = 푑푥푑푦, *(푑푦푑푧) = 푑푥.
This is the reason that 2-forms on 푅3 are sometimes written as 푓푑푥푑푦 +
푔푑푧푑푥+ℎ푑푦푑푧, in contravention of our rule to write the variables in increas-ing
order. In higher dimensions it is better to stich to the rule.
On 푅4 we have
∙ *푑푥1 = 푑푥2푑푥3푑푥4 * 푑푥3 = 푑푥1푑푥2푑푥4
∙ *푑푥2 = −푑푥1푑푥3푑푥4 * 푑푥4 = −푑푥1푑푥2푑푥3 and
∙ *(푑푥1푑푥2) = 푑푥3푑푥4 * (푑푥2푑푥3) = 푑푥1푑푥4
∙ *(푑푥1푑푥3) = −푑푥2푑푥4 * (푑푥2푑푥4) = −푑푥1푑푥3
∙ *(푑푥1푑푥4) = 푑푥2푑푥3 * (푑푥3푑푥4) = 푑푥1푑푥2
3

4. 1.4 div,grad and curl
A vector field on an open subset ^U
if 푅푛 is a smooth map 퐹 :→ 푅푛. We
can write 퐹 in components as
퐹(푥) =
⎛
⎜⎜⎜⎝
퐹1(푥)
퐹2(푥)
...
퐹푛(푥)
⎞
⎟⎟⎟⎠
훼 = 퐹 · 푑푥,the
푑 * 훼 = 푑(퐹 · *푑푥) = 푑푖푣퐹푑푥1푑푥2 . . .퐷푥푛
Al alternative way of writing this identity is obtained by applying * to both
sides, which gives
푑푖푣퐹 = *푑 * 훼
The correspondence between vector fields and 1-forms behaves in an
interesting way with respect to exterior differentiation Σ︀and the Hodge star
operator. For each function 푓 the 1-form 푑푓 =
푛
푖=1
휕푓
휕푥푖
푑푥푖 is associated to
the vector field
퐹(푥) =
Σ︁푛
푖=1
휕푓
휕푥푖
푒푖
⎛
⎜⎜⎜⎝
푓푟푎푐휕푓휕푥1
푓푟푎푐휕푓휕푥2
...
푓푟푎푐휕푓휕푥푛
⎞
⎟⎟⎟⎠
In three dimensions *푑훼 is a 1-form and so is associated to a vector field,
namely
푐푢푟푙F = (
휕퐹3
휕푥2
−
휕퐹2
휕푥3
)푒1 − (
휕퐹3
휕푥1
−
휕퐹1
휕푥3
푒2 + (
휕퐹2
휕푥1
−
휕퐹1
휕푥2
)푒3,
the 푐푢푟푙 of F. Thus, for n=3, if 훼 = F · 푑푥, then
푐푢푟푙F · 푑푥 = *푑훼.
2 Maxwell’s Equations
2.1 Maxwell’s Equation
The differential forms of Maxwells equations as found by Heaviside, while
completely valid, are now considered somewhat archaic, and have been re-placed
by the more useful (equivalent) integral forms. Each law is named
according to the person(s) who originally discovered the connections repre-sented
by the equation. Here are the four equations:
퐺푎푢푠푠′푠푙푎푤푓표푟푒푙푒푐푡푟푖푐푖푡푦 :
∮︁
푐푙표푠푒푑푠푢푟푓푎푐푒
−→퐸
−→퐴
=
· 푑
푄푒푛푐
휖0
4

5. 퐺푎푢푠푠′푠푙푎푤푓표푟푚푎푔푛푒푡푖푠푚 :
∮︁
푐푙표푠푒푑푠푢푟푓푎푐푒
−→퐵
−→퐴
= 0
· 푑
퐹푎푟푎푑푎푦′푠푙푎푤 :
∮︁
−→퐸
· 푑−→푠
= −
푑∅퐵
푑푡
퐴푚푝푒푟푒 −푀푎푥푤푒푙푙푙푎푤 :
∮︁
−→퐵
· 푑−→푠
= 휇0휖0
푑∅퐸
푑푡
+ 휇0푖푒푛푐
∮︀
is used to specify a closed loop integral, also known as a line integral.
Note:
It simply means that in the calculations, we must go all the way around the
loop; we cant stop part way through or the equations wont be valid.
퐴
2.2 Gauss’s 퐸
law for electricity
Gauss law for electricity, more commonly simply refered to as Gausslaw,
−−states that the closed surface integral of
· 푑
is equal to →→the charge
enclosed by the surface divided by the electric permittivity of the material
the charge is in. Generally, the electric permittivity, denoted 휖, is taken to
be the electric permittivity of free (empty) space, and is written 휖0. (휖0 ≈
8.85 · 1012퐹/푚).
We are free to choose our surfaceits an imaginary construct for the pur-poses
of doing the math, not a real entity. The most common surfaces
chosen are spheres and cylinders, because mathematically, symmetry makes
applying Gauss law much easier, but theoretically, any closed surface can be
chosen and it will give the exact same results.
Imagine a point charge of +Q floating in space. Centered around this
charge, construct a spherical Gaussian surface of radius R. Since the charge
is centered in the sphere, the E field points radially outward and has the
same magnitude at all points on the sphere. Remember that 퐸 = 1
4휋휖0
푄
푟2 .
Since in this example, 푟 = 푅, this equation becomes 퐸 = 1
4휋휖0
푄
푅2
From the definition of electric flux, ∅퐸 =
∮︀
푐푙표푠푒푑푠푢푟푓푎푐푒
−→퐸
−→퐴
, so applying
·푑
퐴
Gauss’law is a way of finding the electric flux through a surface due to
−a charge 푄 · 푑
is a unit vector normal to the surface at all points, and
→퐴represents a tiny portion of the surface area of the Gaussian surface. The
−closed surface integral of 푑
is the surface area, A.
→Again from the definition of electric flux,
∅퐸 =
∮︁
푐푙표푠푒푑푠푢푟푓푎푐푒
−→퐸
−→퐴
· 푑
퐸 =
1
4휋휖0
푄
푅2
5

6. ∅퐸 =
∮︁
푐푙표푠푒푑푠푢푟푓푎푐푒
(
1
4휋휖0
−→퐴
푄
푅2 ) · 푑
Since
−→퐸
is pointing radially outward everywhere, it is always parallel to
−→퐴
, and
푑
−→퐸
−→퐴
becomes (
· 푑
−→퐸
)푑
−→퐴
. Since
−→퐸
is constant at all points on the
sphere, it can be moved outside the integral:
∅퐸 = (
1
4휋휖0
푄
푅2 )
∮︁
푐푙표푠푒푑푠푢푟푓푎푐푒
−→퐴
푑
∅퐸 = (
1
4휋휖0
푄
푅2 )퐴
where A is the surface area of the sphere. However, the surface area of a
sphere is simply 4휋푅2, so this becomes
∅퐸 = (
1
4휋휖0
푄
푅2 )(4휋푅2)
∅퐸 =
푄
휖0
퐸
But this, of course, is simply Gauss’law! ∅퐸 is independent of the radius
−of the sphere, which may seem strange, since
clearly decreases at a rate
→_ 1∖푅∖푅2; however, since
−→퐸
points away from the charge, no matter how
퐸
large the →radius of the sphere is, the electric field will still penetrate it at
some point, and the flux will have to be the same. Mathematically, it works
−because ∅퐸 is
multiplied by the surface area of the Gaussian surface;
−→퐸
_ 1∖푅∖푅2, and 퐴 _ 푅2, so their product, ∅퐸 must be independent of 푅.
Imagine that, instead placing a charge of +푄 inside the Gaussian surface,
we placed outside. Clearly the electric field still points away from the charge,
and at some point, the electric field will pass through the Gaussian surface.
On one side of the surface, this will give a negative flux - the electric field
is entering the surface! But the electric field will have to leave the Gaussian
surface on the other side, creating a positive flux. Since all the field lines that
enter the surface must leave again - they don’t just stop - the net electric
flux will be zero, as predicted by Gauss’law.
Using arguments of symmetry, it is also possible to prove Gauss’law for
Gaussian surfaces of other shapes, such as cylinders. It can also be usedin
reverse; by dividing both sides of the equation by 퐴 after integrating, the
electric field caused by various charge configurations can be found for all
points in space. An example of this is finding the electric field at all points
in space caused by an infinitely large plane of charge density 휌. Its done
using a cylindrical Gaussian surface rather than a spherical one, and while
the idea of an infinitely large plane is ridiculous, the results hold true as long
as the distance from the plane at which the electric field is being calculated
is significantly smaller than the size of the plane, and not near the edge.
6

7. 2.3 Gauss’s law for magnetism
Gauss’law for magnetism is remarkably similar to Gauss law for electricity
in form, but means something rather different. Imagine that a magnet
was placed in space, and that a spherical Gaussian surface was constructed
around it. Remember from the section on magnetism that magnetic fields
flow,by convention, from the North pole of a magnet ∮︀
to the South pole.
From the definition of magnetic flux, ∅퐵 =
푐푙표푠푒푑푠푢푟푓푎푐푒
−→퐵
−→퐴
. Part of
· 푑
the magnetic field will not pierce the Gaussian surface - this portion of the
field clearly will not contribute to the flux through the surface, so it can be
ignored. The rest of the magnetic field lines will leave through the surface
from the North pole of the magnet, but because the field flows from the
North pole to the South pole, the same field lines will enter the surface
again somewhere on the surface to go to the South pole. Since the flux
going out is equal to the flux coming in, the net flux is zero, as indicated by
Gauss law for magnetism.
Suppose that instead the magnet was placed outside the Gaussian sur-face.
The same argument applies: any part of the magnetic field that enters
the surface will have to leave again through the surface, since it is closed.
The positive flux will equal the negative flux, theyll cancel, and the net flux
will be zero. Again, this matches what was predicted by Gauss law.
Pretend that a special magnet with only a North pole, and no South
pole, existed. This would be called a magnetic monopole. All the magnetic
field lines would point away from this theoretical magnetic monopole, just
like the electric field lines point away from a positive charge 푄. If a Gaussian
surface was constructed around this monopole, there would obviously be a
positive flux going through the surface, because the magnetic field is leaving,
and it isnt coming back in! Gauss law for magnetism, however, very clearly
says that the flux should be zero! This means that according to Gauss,
there can be no magnetic monopoles - all magnets must have two poles.
Although some people are looking for magnetic monopoles, none have ever
been observed, and if one is ever found, it will mean that Gauss law for
magnetism is incorrect.
퐵
2.4 Faraday’s →law
According to the definition of magnetic flux, ∅퐵, a magnetic field passing
through an area 퐴 will create magnetic flux. Imagine that a circular loop
−of wire of radius R is placed in a magnetic field
, perpendicular to the
퐵
direction of the field. The flux through the loop is clearly the strength of
−the magnetic field multiplied by the area of the loop:∅퐵 =
(휋푅2). Now
→−→퐵
푑푡 .
imagine that the magnetic field began changing with time at a rate of 푑
The change in flux with time would be 푑∅퐵
−→퐵
푑푡 . The flux could also
푑푡 = (휋푅2) 푑
7

8. be changed by altering the area of the loop, but since changing the area of the
loop in real applications is not as practical as changing the magnetic field,
and since the mathematics are largely similar, only the case of changing
magnetic fields will be examined. As was observed by Faraday, when ∅퐵
through the loop is changing, a voltage is induced in the loop in an attempt
by the system to ”fight” the change. A current will then flow in the loop as
determined by the Ohm’law, 푉 = 퐼푅, where 푅 is the resistance of the loop.
Consider again the scenario above. Faraday’s law contains the integral
of
−→퐸
·푑−→푠
. The 푑−→푠
represents an infinitely small portion of the loop of wire.
Recall that an electric field multiplied by a distance represents a voltage.
We can go around the loop in either direction and it won’t affect our results
other than a change in sign - but that change in sign is to be expected,
because in one direction, we would be increasing in potential as we went
around, and in the other direction, we would be decreasing in potential!
From Faraday’law, we have
∮︁
−→퐸
· 푑−→푠
= −
푑∅퐵
푑푡
2.5 Ampere-Maxwell law
Ampere observed that current flowing through a wire created a magnetic
field around the wire, and formulated the equation
∮︁
−→퐵
· 푑−→푠
= 휇0푖푒푛푐
푖푒푛푐, meaning current enclosed, is perhaps a deceptive notation. Currentcan
not be enclosed; rather, what is meant is the current that passes through the
interior of the closed loop. 0is a constant called the magnetic permeability
of free space; if there is a material present instead of simply space, 휇0 is
replaced with 휇 for the material.
Ampere’s law is used by simply selecting any closed loop, traversing it
with small elements 푑−→푠
, and solving the resulting equation. It is key to
note that any closed loop can be selected a flat disc, or perhaps a shape
more similar to a grocery bag and it will give the same results.
Ampere’s law predicted the magnetic field very accurately, but Maxwell
noticed that there was a piece missing. He noted that a capacitor is made
of two conducting plates separated by some distance 푑, and that while the
capacitor was charging, positive charge accumulated on one plate, and neg-ative
charge accumulated on the other plate, but that no current passed
between the plates. A capacitor is essentially a gap in a circuit, but because
of its nature, the circuit is still complete. However, using Amperes law to
find the magnetic field at a point in space, it was possible to select one closed
loop passing through the capacitor, so that no current passed through the
8

9. closed loop. This would indicate that there was no magnetic field at that
point. However, another closed loop could be selected for the same point
that passed through one of the wires connected to the capacitor the law
leaves us free to choose our own closed loop and since current flows in the
wire, the law would clearly indicate that there was a magnetic field at that
point! Clearly this could not be, so something had to be missing.
Maxwell named the missing term displacement current, even though it is
not really a current at all, but rather is the changing electric field within the
capacitor. Since charge is accumulating on the plates of the capacitor, there
is a changing electric field between the two plates. By introducing the term
휇0휖0푑푝ℎ푖푒
푑푡 , Maxwell completed the equation, now called the Ampere-Maxwell
law: ∮︁
−→퐵
· 푑−→푠
= 휇0휖0
푑∅퐸
푑푡
+ 휇0푖푒푛푐
When there is no changing electric field, 푓푟푎푐푑∅퐸푑푡 = 0 and the law simply
becomes Amperes law.
3 Differential forms and Maxwell’s equation
3.1 Relationship
We denoted 퐸:
퐸 = 퐸푥푑푥 + 퐸푦푑푦 + 퐸푧푑푧
The electric field is a one-form because its duality product with a vector is s
scalar. An example of a two-from is the density of electric current J which
is integrable over a surface,
J = 퐽푥푦푑푥
⋀︁
푑푦 + 퐽푦푧푑푦
⋀︁
푑푧 + 퐽푧푥푑푧
⋀︁
푑푥
The surface is defined by a two-dimensional manifold 푆 which can be ap-proximated
by a chain of two-simplexes which are bi-vectors
ΔS푖푗 = (1/2)Δr푖
⋀︁
Δr푗 .
he current I through the surface is expressed as the integral
I = lim
Σ︁
J | ΔS푖푗 = J | S
Finally, as an example of a zero-form which is not integrable over a space
region is the scalar potential scalar ∅. To summarize, various basic electro-magnetic
quantities can be expressed in 3D Euclidean differential forms as
follows:
∙ Zero-forms:scalar potential ∅, magnetic scalar potential ∅푚;
9

10. ∙ One-forms:electric field E, magnetic field H, vector potential A,
magnetic vector potential A푚;
∙ Two-forms:electric flux density D, magnetic flux density B, electric
current density J, magnetic current density J푚;
∙ Three-forms:electric charge density 휚 ;
4 Solution of problem 2.18
4.1 Question 2.18
Maxwell’s formulation of the equations of electromagnetism:
푐푢푟푙퐸 = −
1
푐
휕퐵
휕푡
(1)
푐푢푟푙퐻 =
4휋
푐
퐽 +
1
푐
휕퐷
휕푡
(2)
푑푖푣퐷 = 4휋휌(3)
푑푖푣퐵 = 0(4)
Here c is the speed of light, E is the electirc field, H is the magnetic field,
J it the density of electic current, 휌 is the density of electic charge, B is
the magnetic induction and D is the dielectric displacement. E, H, B, J
and D are vector fields and 휌 is a fuction on 푅3 and all depend o time t.
훼 = (퐸1푑푥1 + 퐸2푑푥2 + 퐸3푑푥3)푑푥4 + 퐵1푑푥2푑푥3 + 퐵2푑푥3푑푥1 + 퐵3푑푥1푑푥2(5)
훽 = −(퐻1푑푥1+퐻2푑푥2+퐻3푑푥3)푑푥4+퐷1푑푥2푑푥3+퐷2푑푥3푑푥1+퐷3푑푥1푑푥2(6)
훾 = 1
푐 (퐽1푑푥2푑푥3 + 퐽2푑푥3푑푥1 + 퐽3푑푥1푑푥2)푑푥4 − 휌푑푥1푑푥2푑푥3(7)
Problem 4.1. show that Maxwell’s equations are equivalent to
푑훼 = 0
푑훽 + 4휋훾 = 0
Answer: From (5), we know
(1)푑훼 = 휕퐸1
휕푥2
푑푥2푑푥1 + 휕퐸1
휕푥3
푑푥3푑푥1 + 휕퐸2
휕푥1
푑푥1푑푥2 + 휕퐸2
휕푥3
푑푥3푑푥2 + 휕퐸3
휕푥1
푑푥1푑푥3 +
휕퐸3
휕푥2
푑푥2푑푥3)푑푥4 + 휕퐵1
휕푥1
푑푥1푑푥2푑푥3
휕퐵1
휕푡 푑푡푑푥2푑푥3 + 휕퐵2
휕푥2
푑푥2푑푥3푑푥1 +
휕퐵2
휕푡 푑푡푑푥3푑푥1 + 휕퐵3
휕푥3
푑푥3푑푥1푑푥2 + 휕퐵3
휕푡 푑푡푑푥1푑푥2
= (−휕퐸1
휕푥2
· 푐+ 휕퐸2
휕푥1
· 푐+ 휕퐵3
휕푡 )푑푥1푑푥2푑푡+(−휕퐸1
휕푥3
· 푐+ 휕퐸3
휕푥1
· 푐+ 휕퐵2
휕푡 )푑푥1푑푥3푑푡+
(−휕퐸2
휕푥3
· 푐 + 휕퐸3
휕푥2
· 푐 + 휕퐵1
휕푡 )푑푥2푑푥3푑푡 + ( 휕퐵1
휕푥1
+ 휕퐵2
휕푥2
+ 휕퐵3
휕푥3
)푑푥1푑푥2푑푥3
= 0
⇐⇒
−휕퐸1
휕푥2
· 푐 + 휕퐸2
휕푥1
· 푐 + 휕퐵3
휕푡 = 0
−휕퐸1
휕푥3
· 푐 + 휕퐸3
휕푥1
· 푐 + 휕퐵2
휕푡 = 0
10

11. −휕퐸2
휕푥3
· 푐 + 휕퐸3
휕푥2
· 푐 + 휕퐵1
휕푡 = 0
휕퐵1
휕푥1
+ 휕퐵2
휕푥2
+ 휕퐵3
휕푥3
= 0
⇐⇒
푐푢푟푙퐸 = −1
푐
휕퐵
휕푡
푑푖푣퐵 = 0
(2)Using the same method as above, we get
푑훽 = −(−휕퐻1
휕푥2
·푐+휕퐻2
휕푥1
·푐−휕퐷3
휕푡 )푑푥1푑푥2푑푡−(−휕퐻1
휕푥3
·푐+휕퐻3
휕푥1
·푐−휕퐷2
휕푡 )푑푥1푑푥3푑푡−
(−휕퐻2
휕푥3
· 푐 + 휕퐻3
휕푥2
· 푐 − 휕퐷1
휕푡 )푑푥2푑푥3푑푡 + ( 휕퐷1
휕푥1
+ 휕퐷2
휕푥2
+ 휕퐷3
휕푥3
)푑푥1푑푥2푑푥3
푑훽 + 4휋훾 =
푑훽 + 4휋( 1
푐 (퐽1푑푥2푑푥3 + 퐽2푑푥3푑푥1 + 퐽3푑푥1푑푥2)푑푥4 − 휌푑푥1푑푥2푑푥3) = 0
⇐⇒
−4휋휌 + 휕퐷1
휕푥1
+ 휕퐷2
휕푥2
+ 휕퐷3
휕푥3
= 0
−휕퐻2
휕푥3
· 푐 + 휕퐻3
휕푥2
· 푐 − 휕퐷1
휕푡 − 4휋퐽1 = 0
−휕퐻1
휕푥3
· 푐 + 휕퐻3
휕푥1
· 푐 − 휕퐷2
휕푡 − 4휋퐽2 = 0
−휕퐻1
휕푥2
· 푐 + 휕퐻2
휕푥1
· 푐 − 휕퐷3
휕푡 − 4휋퐽3 = 0
⇐⇒
푐푢푟푙퐻 = 4휋
푐 퐽 + 1
푐
휕퐷
휕푡
푑푖푣퐷 = 4휋휌
Problem 4.2. Conclude that 훾 is closed and that 푑푖푣퐽 + 휕휌/휕푡 = 0
Answer: Form equation (7)
푑훾 = 1
푐 ( 휕퐽1
휕푥1
푑푥1푑푥2푑푥3+ 휕퐽2
휕푥2
푑푥1푑푥2푑푥3+ 휕퐽3
휕푥3
푑푥1푑푥2푑푥3)푑푥4+휕휌
휕푡 푑푥1푑푥2푑푥3푑푡
When 훾 is closed, we know the former formula: =0
that means:
1
( 휕퐽1
+ 휕퐽2
+ 휕퐽3
) + 휕휌
= 0
푐 휕푥1
휕푥2
휕푥3
휕푡 ⇐⇒
푑푖푣퐽 + 휕휌/휕푡 = 0
Problem 4.3. In vacuum one has 퐸 = 퐷 and 퐻 = 퐵. show that in vacuum
훽 = *훼, Use the relative Hodge star of 훼 in EX.2.17
Answer: From equation (5)
*훼 =
퐸1푑푥2푑푥3 −퐸2푑푥1푑푥3 +퐸3푑푥1푑푥2 −(퐵1푑푥1푑푥4 +퐵2푑푥2푑푥4 +퐵3푑푥3푑푥4)(9)
푠푖푛푐푒퐸 = 퐷 푎푛푑 퐻 = 퐵
*훼 =
퐷1푑푥2푑푥3 − 퐷2푑푥1푑푥3 + 퐷3푑푥1푑푥2 − (퐻1푑푥1푑푥4 + 퐻2푑푥2푑푥4 + 퐻3푑푥3푑푥4)
= 훽
Problem 4.4. Free space is a vacuum without charges or currents. Show
that the Maxwell’s equations in free space are equivalent to 푑훼 = 푑 * 훼 = 0
Answer: Since free space is a vacuum without charges or currents then
퐽 = 0 and 휌 = 0, E=D and H=B
11

12. From the first problem we know that 푑훼 = ⇐⇒ equation (1) and (4).
and we only need to get that 푑 * 훼 = 0 ⇐⇒ 푑훽 + 4휋훾 = 0
since 퐽 = 0 and 휌 = 0, we know 푓 = 0
that is 푑훽 = 0 ⇐⇒ 푑 * 훼 = 0 (from in vacuum 훽 = *훼)
problem was done.
Problem 4.5. Let 푓, 푔 : 푅 −→ 푅 be any smooth functions and define
퐸(푥) =
0
푓(푥1 − 푥4)
푔(푥1 − 푥4)
(2)
퐵(푥) =
0
−푔(푥1 − 푥4)
푓(푥1 − 푥4)
(3)
Show that the corresponding 2-form 훼 satisfies the free Maxwell’s equations
푑훼 = 푑 *훼 = 0. Such serious are called electromagnetic waves. Explain why.
In what direction do these waves travel.
Answer: Since 푓(푥1 − 푥4), we know 휕푓
휕푥1
= − 휕푓
휕푥4
(*)
The same thing happens to 푔(푥1 − 푥4), that is 휕푔
휕푥1
= − 휕푔
휕푥4
(**)
Bring E(x) and B(x) to 푑훼
We get 푑훼 = ( 휕퐸3
휕푥1
− 휕퐵2
휕푥4
푑푥1푑푥3푑푥4 + ( 휕퐸2
휕푥1
+ 휕퐵3
휕푥4
푑푥1푑푥2푑푥4
From (*) and (**), we know 푑훼=0
The same thing happens to 푑 * 훼 and 푑 * 훼 = 0
The wave travels in the direction of x-axis, because for E and B the 1st
dimension part are 0; At the same time, the 2nd and 3rd dimension is
changed only related to 푥1 and time t.
References
1. Differential Forms in Electromagnetics. Ismo V. Lindell
2. Introduction to differential forms Donu Arapura
3. Maxwells Equations Matt Hansen
12