Maxwell’s Formulation – Differential

Forms on Euclidean Space

Wang JingSchool of Physical and Mathematical Sciences

Nanyang Technological Universityjwang14@e.ntu.edu.sg

Abstract

One of the greatest advances in theoretical physics of the nineteenthcentury was Maxwell’s formulation of the the equations of electromag-netism. This article uses differential forms to solve a problem relatedto Maxwell’s formulation. The notion of differential form encompassessuch ideas as elements of surface area and volume elements, the workexerted by a force, the flow of a fluid, and the curvature of a surface,space or hyperspace. An important operation on differential forms isexterior differentiation, which generalizes the operators div, grad, curlof vector calculus. the study of differential forms, which was initiatedby E.Cartan in the years around 1900, is often termed the exteriordifferential calculus.However, Maxwells equations have many very im-portant implications in the life of a modern person, so much so thatpeople 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 𝑅𝑛 is an expression

𝛼 =∑𝐼

𝑓𝐼𝑑𝑥𝐼

Here I stands for a multi-index (𝑖1, 𝑖2, · · · , 𝑖𝑘)of degree k,that is a ”vector”consisting of k integer entries ranging between 1 and 𝑛, The 𝑓𝐼 are smoothfunctions on 𝑅𝑛 called the coefficients of 𝛼, and 𝑑𝑥𝐼 is an abbreviation for

𝑑𝑥𝑖1𝑑𝑥𝑖2 · · · 𝑑𝑥𝑖𝑘

(The notion 𝑑𝑥𝑖1⋀𝑑𝑥𝑖2

⋀· · ·

⋀𝑑𝑥𝑖𝑘 is also often used to distinguish this

kind of product from another kind, called the tensor product) For instance

1

the expressions:𝛼 = sin𝑥1 + 𝑒𝑥4𝑑𝑥1𝑑𝑥2 + 𝑥2𝑥

25𝑑𝑥2𝑑𝑥3 + 6𝑑𝑥2𝑑𝑥4 + cot𝑥2𝑑𝑥5𝑑𝑥3

𝛽 = 𝑥1𝑥3𝑥5𝑑𝑥1𝑑𝑥6𝑑𝑥3𝑑𝑥2represent a 2-form on 𝑅5, resp. a 4-form on 𝑅6. The form 𝛼 consists of fourterms, corresponding to the multi-indices (1,5),(2,3),(2,4),and (5,3), whereas𝛽 consists of one term, corresponding to the multi-index(1,6,3,2). Note,however, that 𝛼 could equally well be regarded as a 2-form on 𝑅6 that doesnot involve the variable 𝑥6. To avoid such ambiguities it is good practice tostate explicitly the domain of definition when writing a differential form. A0-form on 𝑅𝑛 us simply a smooth function (no dx’s)

1.2 Exterior derivative

If 𝑓 is a 0-form, that is a smooth function, we define 𝑑𝑓 to be the 1-form

𝑑𝑓 =

𝑛∑𝑖=1

𝜕𝑓

𝜕𝑥𝑖𝑑𝑥𝑖

Then we have the product or leibniz rule:

𝑑(𝑓𝑔) = 𝑓𝑑𝑔 + 𝑔𝑑𝑓

If𝛼 =∑

𝐼 𝑓𝐼𝑑𝑥𝐼 is a k-form, each of the coefficient 𝑓𝐼 is a smooth functionand we define 𝑑𝛼 to be the k+1 form

𝑑𝛼 =∑𝐼

𝑑𝑓𝐼𝑑𝑥𝐼

The operation 𝑑 is called exterior differentiation. An operator of this sort iscalled a first-order partial differential operator, because it involves the firstpartial derivatives of the coefficients of a form.

Proposition 1.1. (i)𝑑(𝑎𝛼+𝑏𝛽) = 𝑎𝑑𝛼+𝑏𝑑𝛽𝑓𝑜𝑟𝑎𝑙𝑙𝑘−𝑓𝑜𝑟𝑚𝑠𝛼𝑎𝑛𝑑𝛽𝑎𝑛𝑑𝑎𝑙𝑙𝑠𝑐𝑎𝑙𝑎𝑟𝑠𝑎𝑎𝑛𝑑𝑏.(ii)𝑑(𝛼𝛽) = (𝑑𝛼)𝛽 + (−1)𝑘𝛼𝑑𝛽𝑓𝑜𝑟𝑎𝑙𝑙𝑘 − 𝑓𝑜𝑟𝑚𝛼𝑎𝑛𝑑𝑙 − 𝑓𝑜𝑟𝑚𝑠𝛽.

Proposition 1.2. 𝑑(𝑑𝛼) = 0 for any form 𝛼, In short,

𝑑2 = 0

1.3 Closed and exact forms and Hodge star operator

A form 𝛼 is 𝑐𝑙𝑜𝑠𝑒𝑑 if 𝑑𝛼 = 0. It is exact if 𝛼 = 𝑑𝛽 for some form 𝛽(of degreeone or less).

Proposition 1.3. Every exact form is closed.

2

Proof: If 𝛼 = 𝑑(𝑑𝛽) then 𝑑𝛼 = 𝑑(𝑑𝛽) = 0 by last section’s secondproposition. The binomial coefficient 𝐶𝑘

𝑛 is the number of ways of selectingk(unordered) objects from a collection of 𝑛 objects. Equivalently, 𝐶𝑘

𝑛 is thenumber of ways of partitioning a pile of n objects into a pile of 𝑘 objectsand a pile of 𝑛− 𝑘 objects. Thus we see that

𝐶𝑘𝑛 = 𝐶𝑘

𝑛−𝑘

This means that in a certain sense there are as many 𝑘 − 𝑓𝑜𝑟𝑚𝑠. this is anatural way to turn 𝑘 − 𝑓𝑜𝑟𝑚𝑠 into 𝑛− 𝑘 − 𝑓𝑜𝑟𝑚𝑠. This is the Hodge staroperator. Hodge star of 𝛼 denoted by *𝛼 (or sometimes 𝛼*) and is definedas follows. If 𝛼 =

∑𝐼 𝑓𝐼𝑑𝑥𝐼 than

*𝛼 =∑𝐼

𝑓𝐼 * 𝑑𝑥𝐼

with*𝑑𝑥𝐼 = 𝜀𝐼𝑑𝑥𝐼𝑐

Here, for any increasing multi-index 𝐼, 𝐼𝑐 denote the complementary increas-ing multi-index, which consists of all numbers between 1 and n that do notoccur in 𝐼. The factor 𝜖𝐼 us a sign,

𝑣𝑎𝑟𝑒𝑝𝑠𝑖𝑙𝑜𝑛𝐼 =

{1 if 𝑑𝑥𝐼𝑑𝑥𝐼𝑐 = 𝑑𝑥1𝑑𝑥2 · · · 𝑑𝑥𝑛−1 if 𝑑𝑥𝐼𝑑𝑥𝐼𝑐 = −𝑑𝑥1𝑑𝑥2 · · · 𝑑𝑥𝑛

(1)

In other words, *𝑑𝑥𝐼 is the product of all the 𝑑𝑥′𝑗𝑠 that do not occur in𝑑𝑥𝐼 , times a factor ±1 which is chosen in such a way that 𝑑𝑥𝐼(*𝑑𝑥𝐼) is thevolume form:

𝑑𝑥𝐼(*𝑑𝑥𝐼) = 𝑑𝑥1𝑑𝑥2 · · · 𝑑𝑥𝑛Example. On 𝑅2 we have 𝑑𝑥 = 𝑑𝑦 and 𝑑𝑦 = −𝑑𝑥. On 𝑅3 we have

∙ *𝑑𝑥 = 𝑑𝑦𝑑𝑧, *(𝑑𝑥𝑑𝑦) = 𝑑𝑧,

∙ *𝑑𝑦 = −𝑑𝑥𝑑𝑧 = 𝑑𝑧𝑑𝑥, *(𝑑𝑥𝑑𝑧) = −𝑑𝑦,

∙ *𝑑𝑧 = 𝑑𝑥𝑑𝑦, *(𝑑𝑦𝑑𝑧) = 𝑑𝑥.

This is the reason that 2-forms on 𝑅3 are sometimes written as 𝑓𝑑𝑥𝑑𝑦 +𝑔𝑑𝑧𝑑𝑥+ℎ𝑑𝑦𝑑𝑧, 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 𝑅4 we have

∙ *𝑑𝑥1 = 𝑑𝑥2𝑑𝑥3𝑑𝑥4 * 𝑑𝑥3 = 𝑑𝑥1𝑑𝑥2𝑑𝑥4

∙ *𝑑𝑥2 = −𝑑𝑥1𝑑𝑥3𝑑𝑥4 * 𝑑𝑥4 = −𝑑𝑥1𝑑𝑥2𝑑𝑥3 and

∙ *(𝑑𝑥1𝑑𝑥2) = 𝑑𝑥3𝑑𝑥4 * (𝑑𝑥2𝑑𝑥3) = 𝑑𝑥1𝑑𝑥4

∙ *(𝑑𝑥1𝑑𝑥3) = −𝑑𝑥2𝑑𝑥4 * (𝑑𝑥2𝑑𝑥4) = −𝑑𝑥1𝑑𝑥3

∙ *(𝑑𝑥1𝑑𝑥4) = 𝑑𝑥2𝑑𝑥3 * (𝑑𝑥3𝑑𝑥4) = 𝑑𝑥1𝑑𝑥2

3

1.4 div,grad and curl

A vector field on an open subset U if 𝑅𝑛 is a smooth map 𝐹 :→ 𝑅𝑛. Wecan write 𝐹 in components as

𝐹 (𝑥) =

⎛⎜⎜⎜⎝𝐹1(𝑥)𝐹2(𝑥)

...𝐹𝑛(𝑥)

⎞⎟⎟⎟⎠𝛼 = 𝐹 · 𝑑𝑥,the

𝑑 * 𝛼 = 𝑑(𝐹 · *𝑑𝑥) = 𝑑𝑖𝑣𝐹𝑑𝑥1𝑑𝑥2 . . . 𝐷𝑥𝑛

Al alternative way of writing this identity is obtained by applying * to bothsides, which gives

𝑑𝑖𝑣𝐹 = *𝑑 * 𝛼The correspondence between vector fields and 1-forms behaves in an

interesting way with respect to exterior differentiation and the Hodge staroperator. For each function 𝑓 the 1-form 𝑑𝑓 =

∑𝑛𝑖=1

𝜕𝑓𝜕𝑥𝑖

𝑑𝑥𝑖 is associated tothe vector field

𝐹 (𝑥) =𝑛∑

𝑖=1

𝜕𝑓

𝜕𝑥𝑖𝑒𝑖

⎛⎜⎜⎜⎝𝑓𝑟𝑎𝑐𝜕𝑓𝜕𝑥1𝑓𝑟𝑎𝑐𝜕𝑓𝜕𝑥2

...𝑓𝑟𝑎𝑐𝜕𝑓𝜕𝑥𝑛

⎞⎟⎟⎟⎠In three dimensions *𝑑𝛼 is a 1-form and so is associated to a vector field,

namely

𝑐𝑢𝑟𝑙F = (𝜕𝐹3

𝜕𝑥2− 𝜕𝐹2

𝜕𝑥3)𝑒1 − (

𝜕𝐹3

𝜕𝑥1− 𝜕𝐹1

𝜕𝑥3𝑒2 + (

𝜕𝐹2

𝜕𝑥1− 𝜕𝐹1

𝜕𝑥2)𝑒3,

the 𝑐𝑢𝑟𝑙 of F. Thus, for n=3, if 𝛼 = F · 𝑑𝑥, then

𝑐𝑢𝑟𝑙F · 𝑑𝑥 = *𝑑𝛼.

2 Maxwell’s Equations

2.1 Maxwell’s Equation

The differential forms of Maxwells equations as found by Heaviside, whilecompletely valid, are now considered somewhat archaic, and have been re-placed by the more useful (equivalent) integral forms. Each law is namedaccording to the person(s) who originally discovered the connections repre-sented by the equation. Here are the four equations:

𝐺𝑎𝑢𝑠𝑠′𝑠𝑙𝑎𝑤𝑓𝑜𝑟𝑒𝑙𝑒𝑐𝑡𝑟𝑖𝑐𝑖𝑡𝑦 :

∮𝑐𝑙𝑜𝑠𝑒𝑑𝑠𝑢𝑟𝑓𝑎𝑐𝑒

−→𝐸 · 𝑑

−→𝐴 =

𝑄𝑒𝑛𝑐

𝜖0

4

𝐺𝑎𝑢𝑠𝑠′𝑠𝑙𝑎𝑤𝑓𝑜𝑟𝑚𝑎𝑔𝑛𝑒𝑡𝑖𝑠𝑚 :

∮𝑐𝑙𝑜𝑠𝑒𝑑𝑠𝑢𝑟𝑓𝑎𝑐𝑒

−→𝐵 · 𝑑

−→𝐴 = 0

𝐹𝑎𝑟𝑎𝑑𝑎𝑦′𝑠𝑙𝑎𝑤 :

∮ −→𝐸 · 𝑑−→𝑠 = −𝑑∅𝐵

𝑑𝑡

𝐴𝑚𝑝𝑒𝑟𝑒−𝑀𝑎𝑥𝑤𝑒𝑙𝑙𝑙𝑎𝑤 :

∮ −→𝐵 · 𝑑−→𝑠 = 𝜇0𝜖0

𝑑∅𝐸𝑑𝑡

+ 𝜇0𝑖𝑒𝑛𝑐

Note:∮is used to specify a closed loop integral, also known as a line integral.

It simply means that in the calculations, we must go all the way around theloop; we cant stop part way through or the equations wont be valid.

2.2 Gauss’s law for electricity

Gauss law for electricity, more commonly simply refered to as Gausslaw,

states that the closed surface integral of−→𝐸 · 𝑑

−→𝐴 is equal to the charge

enclosed by the surface divided by the electric permittivity of the materialthe charge is in. Generally, the electric permittivity, denoted 𝜖, is taken tobe the electric permittivity of free (empty) space, and is written 𝜖0. (𝜖0 ≈8.85 · 1012𝐹/𝑚).

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 surfaceschosen are spheres and cylinders, because mathematically, symmetry makesapplying Gauss law much easier, but theoretically, any closed surface can bechosen and it will give the exact same results.

Imagine a point charge of +Q floating in space. Centered around thischarge, construct a spherical Gaussian surface of radius R. Since the chargeis centered in the sphere, the E field points radially outward and has thesame magnitude at all points on the sphere. Remember that 𝐸 = 1

4𝜋𝜖0𝑄𝑟2.

Since in this example, 𝑟 = 𝑅, this equation becomes 𝐸 = 14𝜋𝜖0

𝑄𝑅2

From the definition of electric flux, ∅𝐸 =∮

𝑐𝑙𝑜𝑠𝑒𝑑𝑠𝑢𝑟𝑓𝑎𝑐𝑒

−→𝐸 ·𝑑

−→𝐴 , so applying

Gauss’law is a way of finding the electric flux through a surface due to

a charge 𝑄 · 𝑑−→𝐴 is a unit vector normal to the surface at all points, and

represents a tiny portion of the surface area of the Gaussian surface. The

closed surface integral of 𝑑−→𝐴 is the surface area, A.

Again from the definition of electric flux,

∅𝐸 =

∮𝑐𝑙𝑜𝑠𝑒𝑑𝑠𝑢𝑟𝑓𝑎𝑐𝑒

−→𝐸 · 𝑑

−→𝐴

𝐸 =1

4𝜋𝜖0

𝑄

𝑅2

5

∅𝐸 =

∮𝑐𝑙𝑜𝑠𝑒𝑑𝑠𝑢𝑟𝑓𝑎𝑐𝑒

(1

4𝜋𝜖0

𝑄

𝑅2) · 𝑑

−→𝐴

Since−→𝐸 is pointing radially outward everywhere, it is always parallel to

𝑑−→𝐴 , and

−→𝐸 · 𝑑

−→𝐴 becomes (

−→𝐸 )𝑑

−→𝐴 . Since

−→𝐸 is constant at all points on the

sphere, it can be moved outside the integral:

∅𝐸 = (1

4𝜋𝜖0

𝑄

𝑅2)

∮𝑐𝑙𝑜𝑠𝑒𝑑𝑠𝑢𝑟𝑓𝑎𝑐𝑒

𝑑−→𝐴

∅𝐸 = (1

4𝜋𝜖0

𝑄

𝑅2)𝐴

where A is the surface area of the sphere. However, the surface area of asphere is simply 4𝜋𝑅2, so this becomes

∅𝐸 = (1

4𝜋𝜖0

𝑄

𝑅2)(4𝜋𝑅2)

∅𝐸 =𝑄

𝜖0

But this, of course, is simply Gauss’law! ∅𝐸 is independent of the radius

of the sphere, which may seem strange, since−→𝐸 clearly decreases at a rate

∝ 1∖𝑅∖𝑅2; however, since−→𝐸 points away from the charge, no matter how

large the radius of the sphere is, the electric field will still penetrate it atsome point, and the flux will have to be the same. Mathematically, it works

because ∅𝐸 is−→𝐸 multiplied by the surface area of the Gaussian surface;−→

𝐸 ∝ 1∖𝑅∖𝑅2, and 𝐴 ∝ 𝑅2, so their product, ∅𝐸 must be independent of 𝑅.Imagine that, instead placing a charge of +𝑄 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 fieldis entering the surface! But the electric field will have to leave the Gaussiansurface on the other side, creating a positive flux. Since all the field lines thatenter the surface must leave again - they don’t just stop - the net electricflux will be zero, as predicted by Gauss’law.

Using arguments of symmetry, it is also possible to prove Gauss’law forGaussian surfaces of other shapes, such as cylinders. It can also be usedinreverse; by dividing both sides of the equation by 𝐴 after integrating, theelectric field caused by various charge configurations can be found for allpoints in space. An example of this is finding the electric field at all pointsin space caused by an infinitely large plane of charge density 𝜌. Its doneusing a cylindrical Gaussian surface rather than a spherical one, and whilethe idea of an infinitely large plane is ridiculous, the results hold true as longas the distance from the plane at which the electric field is being calculatedis significantly smaller than the size of the plane, and not near the edge.

6

2.3 Gauss’s law for magnetism

Gauss’law for magnetism is remarkably similar to Gauss law for electricityin form, but means something rather different. Imagine that a magnetwas placed in space, and that a spherical Gaussian surface was constructedaround it. Remember from the section on magnetism that magnetic fieldsflow,by convention, from the North pole of a magnet to the South pole.

From the definition of magnetic flux, ∅𝐵 =∮

𝑐𝑙𝑜𝑠𝑒𝑑𝑠𝑢𝑟𝑓𝑎𝑐𝑒

−→𝐵 · 𝑑

−→𝐴 . Part of

the magnetic field will not pierce the Gaussian surface - this portion of thefield clearly will not contribute to the flux through the surface, so it can beignored. The rest of the magnetic field lines will leave through the surfacefrom the North pole of the magnet, but because the field flows from theNorth pole to the South pole, the same field lines will enter the surfaceagain somewhere on the surface to go to the South pole. Since the fluxgoing out is equal to the flux coming in, the net flux is zero, as indicated byGauss 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 entersthe 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 fluxwill be zero. Again, this matches what was predicted by Gauss law.

Pretend that a special magnet with only a North pole, and no Southpole, existed. This would be called a magnetic monopole. All the magneticfield lines would point away from this theoretical magnetic monopole, justlike the electric field lines point away from a positive charge 𝑄. If a Gaussiansurface was constructed around this monopole, there would obviously be apositive flux going through the surface, because the magnetic field is leaving,and it isnt coming back in! Gauss law for magnetism, however, very clearlysays 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 everbeen observed, and if one is ever found, it will mean that Gauss law formagnetism is incorrect.

2.4 Faraday’s law

According to the definition of magnetic flux, ∅𝐵, a magnetic field passingthrough an area 𝐴 will create magnetic flux. Imagine that a circular loop

of wire of radius R is placed in a magnetic field−→𝐵 , perpendicular to the

direction of the field. The flux through the loop is clearly the strength of

the magnetic field multiplied by the area of the loop:∅𝐵 =−→𝐵 (𝜋𝑅2). Now

imagine that the magnetic field began changing with time at a rate of 𝑑−→𝐵𝑑𝑡 .

The change in flux with time would be 𝑑∅𝐵𝑑𝑡 = (𝜋𝑅2)𝑑

−→𝐵𝑑𝑡 . The flux could also

7

be changed by altering the area of the loop, but since changing the area of theloop in real applications is not as practical as changing the magnetic field,and since the mathematics are largely similar, only the case of changingmagnetic fields will be examined. As was observed by Faraday, when ∅𝐵through the loop is changing, a voltage is induced in the loop in an attemptby the system to ”fight” the change. A current will then flow in the loop asdetermined by the Ohm’law, 𝑉 = 𝐼𝑅, where 𝑅 is the resistance of the loop.

Consider again the scenario above. Faraday’s law contains the integral

of−→𝐸 ·𝑑−→𝑠 . The 𝑑−→𝑠 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 resultsother 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 wentaround, and in the other direction, we would be decreasing in potential!From Faraday’law, we have ∮ −→

𝐸 · 𝑑−→𝑠 = −𝑑∅𝐵𝑑𝑡

2.5 Ampere-Maxwell law

Ampere observed that current flowing through a wire created a magneticfield around the wire, and formulated the equation∮ −→

𝐵 · 𝑑−→𝑠 = 𝜇0𝑖𝑒𝑛𝑐

𝑖𝑒𝑛𝑐, meaning current enclosed, is perhaps a deceptive notation. Currentcannot be enclosed; rather, what is meant is the current that passes through theinterior of the closed loop. 0is a constant called the magnetic permeabilityof free space; if there is a material present instead of simply space, 𝜇0 isreplaced with 𝜇 for the material.

Ampere’s law is used by simply selecting any closed loop, traversing itwith small elements 𝑑−→𝑠 , and solving the resulting equation. It is key tonote that any closed loop can be selected a flat disc, or perhaps a shapemore similar to a grocery bag and it will give the same results.

Ampere’s law predicted the magnetic field very accurately, but Maxwellnoticed that there was a piece missing. He noted that a capacitor is madeof two conducting plates separated by some distance 𝑑, and that while thecapacitor was charging, positive charge accumulated on one plate, and neg-ative charge accumulated on the other plate, but that no current passedbetween the plates. A capacitor is essentially a gap in a circuit, but becauseof its nature, the circuit is still complete. However, using Amperes law tofind the magnetic field at a point in space, it was possible to select one closedloop passing through the capacitor, so that no current passed through the

8

closed loop. This would indicate that there was no magnetic field at thatpoint. However, another closed loop could be selected for the same pointthat passed through one of the wires connected to the capacitor the lawleaves us free to choose our own closed loop and since current flows in thewire, the law would clearly indicate that there was a magnetic field at thatpoint! Clearly this could not be, so something had to be missing.

Maxwell named the missing term displacement current, even though it isnot really a current at all, but rather is the changing electric field within thecapacitor. Since charge is accumulating on the plates of the capacitor, thereis a changing electric field between the two plates. By introducing the term𝜇0𝜖0

𝑑𝑝ℎ𝑖𝑒𝑑𝑡 , Maxwell completed the equation, now called the Ampere-Maxwell

law: ∮ −→𝐵 · 𝑑−→𝑠 = 𝜇0𝜖0

𝑑∅𝐸𝑑𝑡

+ 𝜇0𝑖𝑒𝑛𝑐

When there is no changing electric field, 𝑓𝑟𝑎𝑐𝑑∅𝐸𝑑𝑡 = 0 and the law simplybecomes Amperes law.

3 Differential forms and Maxwell’s equation

3.1 Relationship

We denoted 𝐸:

𝐸 = 𝐸𝑥𝑑𝑥+ 𝐸𝑦𝑑𝑦 + 𝐸𝑧𝑑𝑧

The electric field is a one-form because its duality product with a vector is sscalar. An example of a two-from is the density of electric current J whichis integrable over a surface,

J = 𝐽𝑥𝑦𝑑𝑥⋀

𝑑𝑦 + 𝐽𝑦𝑧𝑑𝑦⋀

𝑑𝑧 + 𝐽𝑧𝑥𝑑𝑧⋀

𝑑𝑥

The surface is defined by a two-dimensional manifold 𝑆 which can be ap-proximated by a chain of two-simplexes which are bi-vectors

ΔS𝑖𝑗 = (1/2)Δr𝑖⋀

Δr𝑗 .

he current I through the surface is expressed as the integral

I = lim∑

J | ΔS𝑖𝑗 = J | S

Finally, as an example of a zero-form which is not integrable over a spaceregion is the scalar potential scalar ∅. To summarize, various basic electro-magnetic quantities can be expressed in 3D Euclidean differential forms asfollows:

∙ Zero-forms:scalar potential ∅, magnetic scalar potential ∅𝑚;

9

∙ One-forms:electric field E, magnetic field H, vector potential A,magnetic vector potential A𝑚;

∙ Two-forms:electric flux density D, magnetic flux density B, electriccurrent density J, magnetic current density J𝑚;

∙ Three-forms:electric charge density 𝜚 ;

4 Solution of problem 2.18

4.1 Question 2.18

Maxwell’s formulation of the equations of electromagnetism:

𝑐𝑢𝑟𝑙𝐸 = −1

𝑐

𝜕𝐵

𝜕𝑡(1)

𝑐𝑢𝑟𝑙𝐻 =4𝜋

𝑐𝐽 +

1

𝑐

𝜕𝐷

𝜕𝑡(2)

𝑑𝑖𝑣𝐷 = 4𝜋𝜌(3)

𝑑𝑖𝑣𝐵 = 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, 𝜌 is the density of electic charge, B isthe magnetic induction and D is the dielectric displacement. E, H, B, Jand D are vector fields and 𝜌 is a fuction on 𝑅3 and all depend o time t.𝛼 = (𝐸1𝑑𝑥1 + 𝐸2𝑑𝑥2 + 𝐸3𝑑𝑥3)𝑑𝑥4 + 𝐵1𝑑𝑥2𝑑𝑥3 + 𝐵2𝑑𝑥3𝑑𝑥1 + 𝐵3𝑑𝑥1𝑑𝑥2(5)𝛽 = −(𝐻1𝑑𝑥1+𝐻2𝑑𝑥2+𝐻3𝑑𝑥3)𝑑𝑥4+𝐷1𝑑𝑥2𝑑𝑥3+𝐷2𝑑𝑥3𝑑𝑥1+𝐷3𝑑𝑥1𝑑𝑥2(6)𝛾 = 1

𝑐 (𝐽1𝑑𝑥2𝑑𝑥3 + 𝐽2𝑑𝑥3𝑑𝑥1 + 𝐽3𝑑𝑥1𝑑𝑥2)𝑑𝑥4 − 𝜌𝑑𝑥1𝑑𝑥2𝑑𝑥3(7)

Problem 4.1. show that Maxwell’s equations are equivalent to

𝑑𝛼 = 0

𝑑𝛽 + 4𝜋𝛾 = 0

Answer: From (5), we know(1)𝑑𝛼 = 𝜕𝐸1

𝜕𝑥2𝑑𝑥2𝑑𝑥1 +

𝜕𝐸1𝜕𝑥3

𝑑𝑥3𝑑𝑥1 +𝜕𝐸2𝜕𝑥1

𝑑𝑥1𝑑𝑥2 +𝜕𝐸2𝜕𝑥3

𝑑𝑥3𝑑𝑥2 +𝜕𝐸3𝜕𝑥1

𝑑𝑥1𝑑𝑥3 +𝜕𝐸3𝜕𝑥2

𝑑𝑥2𝑑𝑥3)𝑑𝑥4 +𝜕𝐵1𝜕𝑥1

𝑑𝑥1𝑑𝑥2𝑑𝑥3𝜕𝐵1𝜕𝑡 𝑑𝑡𝑑𝑥2𝑑𝑥3 +

𝜕𝐵2𝜕𝑥2

𝑑𝑥2𝑑𝑥3𝑑𝑥1 +𝜕𝐵2𝜕𝑡 𝑑𝑡𝑑𝑥3𝑑𝑥1 +

𝜕𝐵3𝜕𝑥3

𝑑𝑥3𝑑𝑥1𝑑𝑥2 +𝜕𝐵3𝜕𝑡 𝑑𝑡𝑑𝑥1𝑑𝑥2

= (−𝜕𝐸1𝜕𝑥2

· 𝑐+ 𝜕𝐸2𝜕𝑥1

· 𝑐+ 𝜕𝐵3𝜕𝑡 )𝑑𝑥1𝑑𝑥2𝑑𝑡+ (−𝜕𝐸1

𝜕𝑥3· 𝑐+ 𝜕𝐸3

𝜕𝑥1· 𝑐+ 𝜕𝐵2

𝜕𝑡 )𝑑𝑥1𝑑𝑥3𝑑𝑡+

(−𝜕𝐸2𝜕𝑥3

· 𝑐+ 𝜕𝐸3𝜕𝑥2

· 𝑐+ 𝜕𝐵1𝜕𝑡 )𝑑𝑥2𝑑𝑥3𝑑𝑡+ (𝜕𝐵1

𝜕𝑥1+ 𝜕𝐵2

𝜕𝑥2+ 𝜕𝐵3

𝜕𝑥3)𝑑𝑥1𝑑𝑥2𝑑𝑥3

= 0⇐⇒−𝜕𝐸1

𝜕𝑥2· 𝑐+ 𝜕𝐸2

𝜕𝑥1· 𝑐+ 𝜕𝐵3

𝜕𝑡 = 0

−𝜕𝐸1𝜕𝑥3

· 𝑐+ 𝜕𝐸3𝜕𝑥1

· 𝑐+ 𝜕𝐵2𝜕𝑡 = 0

10

−𝜕𝐸2𝜕𝑥3

· 𝑐+ 𝜕𝐸3𝜕𝑥2

· 𝑐+ 𝜕𝐵1𝜕𝑡 = 0

𝜕𝐵1𝜕𝑥1

+ 𝜕𝐵2𝜕𝑥2

+ 𝜕𝐵3𝜕𝑥3

= 0⇐⇒𝑐𝑢𝑟𝑙𝐸 = −1

𝑐𝜕𝐵𝜕𝑡

𝑑𝑖𝑣𝐵 = 0(2)Using the same method as above, we get𝑑𝛽 = −(−𝜕𝐻1

𝜕𝑥2·𝑐+ 𝜕𝐻2

𝜕𝑥1·𝑐− 𝜕𝐷3

𝜕𝑡 )𝑑𝑥1𝑑𝑥2𝑑𝑡−(−𝜕𝐻1𝜕𝑥3

·𝑐+ 𝜕𝐻3𝜕𝑥1

·𝑐− 𝜕𝐷2𝜕𝑡 )𝑑𝑥1𝑑𝑥3𝑑𝑡−

(−𝜕𝐻2𝜕𝑥3

· 𝑐+ 𝜕𝐻3𝜕𝑥2

· 𝑐− 𝜕𝐷1𝜕𝑡 )𝑑𝑥2𝑑𝑥3𝑑𝑡+ (𝜕𝐷1

𝜕𝑥1+ 𝜕𝐷2

𝜕𝑥2+ 𝜕𝐷3

𝜕𝑥3)𝑑𝑥1𝑑𝑥2𝑑𝑥3

𝑑𝛽 + 4𝜋𝛾 =𝑑𝛽 + 4𝜋(1𝑐 (𝐽1𝑑𝑥2𝑑𝑥3 + 𝐽2𝑑𝑥3𝑑𝑥1 + 𝐽3𝑑𝑥1𝑑𝑥2)𝑑𝑥4 − 𝜌𝑑𝑥1𝑑𝑥2𝑑𝑥3) = 0⇐⇒−4𝜋𝜌+ 𝜕𝐷1

𝜕𝑥1+ 𝜕𝐷2

𝜕𝑥2+ 𝜕𝐷3

𝜕𝑥3= 0

−𝜕𝐻2𝜕𝑥3

· 𝑐+ 𝜕𝐻3𝜕𝑥2

· 𝑐− 𝜕𝐷1𝜕𝑡 − 4𝜋𝐽1 = 0

−𝜕𝐻1𝜕𝑥3

· 𝑐+ 𝜕𝐻3𝜕𝑥1

· 𝑐− 𝜕𝐷2𝜕𝑡 − 4𝜋𝐽2 = 0

−𝜕𝐻1𝜕𝑥2

· 𝑐+ 𝜕𝐻2𝜕𝑥1

· 𝑐− 𝜕𝐷3𝜕𝑡 − 4𝜋𝐽3 = 0

⇐⇒𝑐𝑢𝑟𝑙𝐻 = 4𝜋

𝑐 𝐽 + 1𝑐𝜕𝐷𝜕𝑡

𝑑𝑖𝑣𝐷 = 4𝜋𝜌

Problem 4.2. Conclude that 𝛾 is closed and that 𝑑𝑖𝑣𝐽 + 𝜕𝜌/𝜕𝑡 = 0

Answer: Form equation (7)𝑑𝛾 = 1

𝑐 (𝜕𝐽1𝜕𝑥1

𝑑𝑥1𝑑𝑥2𝑑𝑥3+𝜕𝐽2𝜕𝑥2

𝑑𝑥1𝑑𝑥2𝑑𝑥3+𝜕𝐽3𝜕𝑥3

𝑑𝑥1𝑑𝑥2𝑑𝑥3)𝑑𝑥4+𝜕𝜌𝜕𝑡 𝑑𝑥1𝑑𝑥2𝑑𝑥3𝑑𝑡

When 𝛾 is closed, we know the former formula: =0that means:1𝑐 (

𝜕𝐽1𝜕𝑥1

+ 𝜕𝐽2𝜕𝑥2

+ 𝜕𝐽3𝜕𝑥3

) + 𝜕𝜌𝜕𝑡 = 0

⇐⇒𝑑𝑖𝑣𝐽 + 𝜕𝜌/𝜕𝑡 = 0

Problem 4.3. In vacuum one has 𝐸 = 𝐷 and 𝐻 = 𝐵. show that in vacuum𝛽 = *𝛼, Use the relative Hodge star of 𝛼 in EX.2.17

Answer: From equation (5)*𝛼 =𝐸1𝑑𝑥2𝑑𝑥3−𝐸2𝑑𝑥1𝑑𝑥3+𝐸3𝑑𝑥1𝑑𝑥2− (𝐵1𝑑𝑥1𝑑𝑥4+𝐵2𝑑𝑥2𝑑𝑥4+𝐵3𝑑𝑥3𝑑𝑥4)(9)𝑠𝑖𝑛𝑐𝑒𝐸 = 𝐷 𝑎𝑛𝑑 𝐻 = 𝐵*𝛼 =𝐷1𝑑𝑥2𝑑𝑥3 −𝐷2𝑑𝑥1𝑑𝑥3 +𝐷3𝑑𝑥1𝑑𝑥2 − (𝐻1𝑑𝑥1𝑑𝑥4 +𝐻2𝑑𝑥2𝑑𝑥4 +𝐻3𝑑𝑥3𝑑𝑥4)= 𝛽

Problem 4.4. Free space is a vacuum without charges or currents. Showthat the Maxwell’s equations in free space are equivalent to 𝑑𝛼 = 𝑑 * 𝛼 = 0

Answer: Since free space is a vacuum without charges or currents then𝐽 = 0 and 𝜌 = 0, E=D and H=B

11

From the first problem we know that 𝑑𝛼 = ⇐⇒ equation (1) and (4).and we only need to get that 𝑑 * 𝛼 = 0 ⇐⇒ 𝑑𝛽 + 4𝜋𝛾 = 0since 𝐽 = 0 and 𝜌 = 0, we know 𝑓 = 0that is 𝑑𝛽 = 0 ⇐⇒ 𝑑 * 𝛼 = 0 (from in vacuum 𝛽 = *𝛼)problem was done.

Problem 4.5. Let 𝑓, 𝑔 : 𝑅 −→ 𝑅 be any smooth functions and define

𝐸(𝑥) =0

𝑓(𝑥1 − 𝑥4)𝑔(𝑥1 − 𝑥4)

(2)

𝐵(𝑥) =0

−𝑔(𝑥1 − 𝑥4)𝑓(𝑥1 − 𝑥4)

(3)

Show that the corresponding 2-form 𝛼 satisfies the free Maxwell’s equations𝑑𝛼 = 𝑑*𝛼 = 0. Such serious are called electromagnetic waves. Explain why.In what direction do these waves travel.

Answer: Since 𝑓(𝑥1 − 𝑥4), we know 𝜕𝑓𝜕𝑥1

= − 𝜕𝑓𝜕𝑥4

(*)The same thing happens to 𝑔(𝑥1 − 𝑥4), that is

𝜕𝑔𝜕𝑥1

= − 𝜕𝑔𝜕𝑥4

(**)Bring E(x) and B(x) to 𝑑𝛼We get 𝑑𝛼 = (𝜕𝐸3

𝜕𝑥1− 𝜕𝐵2

𝜕𝑥4𝑑𝑥1𝑑𝑥3𝑑𝑥4 + (𝜕𝐸2

𝜕𝑥1+ 𝜕𝐵3

𝜕𝑥4𝑑𝑥1𝑑𝑥2𝑑𝑥4

From (*) and (**), we know 𝑑𝛼=0The same thing happens to 𝑑 * 𝛼 and 𝑑 * 𝛼 = 0The wave travels in the direction of x-axis, because for E and B the 1stdimension part are 0; At the same time, the 2nd and 3rd dimension ischanged only related to 𝑥1 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