A Practical Introduction to Differential Forms. Alexia E. Schulz. William C. Schulz

Transcription

1 A Practical Introduction to Differential Forms Alexia E. Schulz and William C. Schulz December 2, 204 Transgalactic Publishing Company Flagstaff, Vienna, Cosmopolis

2 ii c 202 by Alexia E. Schulz and William C. Schulz

3 Contents iii

4 iv CONTENTS

5 Chapter Introduction and Basic Applications

6 2 CHAPTER. INTRODUCTION AND BASIC APPLICATIONS. Maxwell s equations in Space-Time In this section we will use the material in the preceding section to work out Maxwell s equations in Space-Time. We will do this twice. First we will work out the the theory in free space where we assume ǫ = µ = and then we will work it out in general without the constraings on ǫ and µ. As is the preceding section this is highly specialized material and needed by a small minority of readers. For Special Relativity, the coordinate system cdt, dx, dy, dz forms an othonormal coordinate system, with that order. Thinking of it as dx 0, dx, dx 0, dx for convenience, we have and cdt, cdt =, dx, dx =, dy, dy =, dz, dz =, dx i, dx j = 0 The matrix for this coordinate system is g ij = dx i, dx j = for i j We can perform the operation using the following permutations, all of which are even permutations have sgnσ = Using these and the formula where dx i... dx ir = s sgnσdx j... dx jr σ = 2 r r r 2 n i i 2 i r j j 2 j n r and s is the number of negative basis elements that is dx, dy, dz among dx i,...,dx ir, we have the following formulas = cdt dx dy dz cdt = dx dy dz dx dx dx = cdt dx = cdt dy dz dy = cdt dz dx dz = cdt dx dy cdt dx = dy dz cdt dy = dz dx cdt dz = dx dy cdt dy dz = dx cdt dz dx = dy cdt dx dy = dz dy dz = cdt dx dz dx = cdt dy dy dz = cdt dx

7 .. MAXWELL S EQUATIONS IN SPACE-TIME cdt dx dy dz = For example, for the 2nd entry we have the permutation 0 2 σ = 0 2 with s = 0 and sgnσ =, whereas for the fourth entry we have 0 2 σ = 2 0 with s = and sgnσ =, this being the the fourth permatation in the above list with the second and third entries swapped. For the seventh entry we have 0 2 σ = 0 2 so s = and sgnσ =, this being the second permutation in the list. The entries in the second column can be derived from those in the first, but beware since the permutation is reversed. The fourth row second column has 0 2 σ = 0 2 and here s = 2 and σ is the third entry in the list of permatations so sgnσ =. It is far easier to derive the second column using ω = r4 r ω. Now we will start with our four dimensional treatment of Electromagnetics. We recall Maxwell s Equations div D = ρ div B = 0 curle = B c t curlh = D c t c j D = ǫ E B = µ H For the initial development we will set ǫ = µ = so that D = E and B = H though we will retain the letters for comparison with future work. Also, E and H will have high indices and D and B will have low indices. This is not important for us; I m just maintaining conventions. The first thing I want to do is to express divb = 0 and curl E = B c t in a single equation involving differential forms. This can be done in several ways, but I want to stay as consistent with classical tensor analysis as possible. So we will use the coefficients from F µν = 0 E E 2 E E 0 B B 2 E 2 B 0 B E B 2 B 0

8 4 CHAPTER. INTRODUCTION AND BASIC APPLICATIONS which is a representation of a standard Electromagnetic Tensor. We can then form the differential form summation convention in force F = 2 F µν dx µ dx ν = E dx 0 dx E 2 dx 0 dx 2 E dx 0 dx B dx 2 dx B 2 dx dx B dx dx 2 Now we compute df df = E x 2 E2 x B x 0 dx 0 dx 2 dx E x E x B 2 x 0 dx 0 dx dx E2 x E x 2 B x 0 dx 0 dx dx 2 B x B 2 x 2 B x dx dx 2 dx = curl E B dx 0 dx 2 dx curl E c t 2 B 2 dx 0 dx dx c t curl E B dx 0 dx dx 2 divb c t dx dx 2 dx = 0 in view of the two Maxwell Equations. Now we know from the converse of the Poincaré lemma that since df = 0 there must be a -form A for which da = F. This is the four dimensional vector potential. For historical reasons we write this A as A = φdx 0 A dx A 2 dx 2 A dx For comparison with other treatments we also define the three vector A = A, A 2, A. We now take the exterior derivative of A and compare it to F. da = A x 0 φ x dx 0 dx A2 x 0 φ x 2 dx 0 dx 2 A x 0 φ x dx 0 dx A x 2 A2 x dx 2 dx A x A x dx dx A2 x A x 2 dx dx 2 A = x 0 A gradφ dx 0 dx x 0 gradφ 2 dx 0 dx 2 2 A x 0 gradφ dx 0 dx curl A dx 2 dx curl A 2 dx dx curl A dx dx 2 Comparing this with F = E dx 0 dx E 2 dx 0 dx 2 E dx 0 dx B dx 2 dx B 2 dx dx B dx dx 2

9 .. MAXWELL S EQUATIONS IN SPACE-TIME 5 we see E i B i A = gradφ i = curl A i x 0 i or more succinctly E = gradφ c B = curl A These last equations are not properly part of a four-dimensional theory; they have been included to show that the terms involving A i actually come from the three dimensional vector potential and the φ term in A is the old scalar potential. This is comforting. And to get it to work out right requires considerable fussiness in the definitions, though we made it look easy. We have now taken care of two of Maxwell s equations and defined the potential A, so it is time to take on the other two equations and connect them to the source terms ρ and j = j, j 2, j. Then we have to come up with the potential equations but it turns out this is quite easy. First we define the source form J = cρdx 0 j dx j 2 dx 2 j dx Next we compute the dual form F. We get F = E dx 0 dx E 2 dx 0 dx 2 E dx 0 dx A t B dx 2 dx B 2 dx dx B dx dx 2 = E dx 2 dx E 2 dx dx E dx dx 2 B dx 0 dx B 2 dx 0 dx 2 B dx 0 dx However, for comparison with later work it turns out to be useful to use the equations D = E and H = B to rewrite F slightly as F = D dx 2 dx D 2 dx dx D dx dx 2 Next we need the exterior derivative of F: d F = D D x 0 H x 2 H2 x H dx 0 dx H 2 dx 0 dx 2 H dx 0 dx dx 0 dx 2 dx dx 0 dx dx 2 D x 0 H2 x H x 2 = D x 0 curlh dx 0 dx 2 dx D 2 x 0 H x H x dx 0 dx dx x D2 x 2 D x dx dx 2 dx D x 0 2 curl H 2 dx 0 dx dx

10 6 CHAPTER. INTRODUCTION AND BASIC APPLICATIONS D x 0 curl H dx 0 dx dx 2 div D dx dx 2 dx = ρ dx dx 2 dx c j dx 0 dx 2 dx c j2 dx 0 dx dx c j dx 0 dx dx 2 = c J This is the expression of the second pair of Maxwell s equations using differential forms. We can rewrite this slightly as δf = d F = ρ dx 0 c j dx c j2 dx 2 c j dx = c J Now we continue on to the potential equations. The first thing to deal with is the condition of Lorenz. This is very convenient. We form δa = d φdx 0 A dx A 2 dx 2 A dx = d φdx dx 2 dx A dx 0 dx 2 dx A 2 dx 0 dx dx A dx 0 dx dx φ = x 0 A x A2 x 2 A x dx 0 dx dx 2 dx φ = x 0 A x A2 x 2 A x φ = c t div A Thus the condition of Lorenz φ c t div A = 0 is expressed simply by δa = 0. We discuss later how A may be modified so that it satisfies the condition of Lorenz. Now we can derive the potential equations for φ and A. We have to compute A. A = δd dδa = δda 0 using the condition of Lorenz. Next we recall da = F so A = δda = δf = c J and that is the potential equation. However, to get actual use out of it and to see how it relates to ancient notations, we compute the d Alembertian on See the historical note on the Condition of Lorenz in an Appendix to this section.

11 .. MAXWELL S EQUATIONS IN SPACE-TIME 7 functions: f = δd dδf f = δdf 0 = d x 0 dx0 f x dx f x 2 dx2 f x dx f = d x 0 dx dx 2 dx f x dx0 dx 2 dx f x 2 dx0 dx dx f x dx0 dx dx 2 2 f = x f x 2 2 f x 22 2 f x 2 dx 0 dx dx 2 dx 2 f = x f x 2 2 f x 22 2 f x 2 The classical d Alembertian is so we see that p = c 2 t 2 2 x 2 f = p f 2 x 22 2 x 2 Next we use the formula proved in an appendix to this section that A = φdx 0 A dx A 2 dx 2 A dx Now comparing this with = φdx 0 A dx A 2 dx 2 A dx we see A = c J = c cρdx0 j dx j 2 dx 2 j dx φ = ρ p φ = ρ A i = c ji p A i = c ji The equations on the right are the classical potential equations. Next we discuss the matter of how to arrange for the condition of Lorenz to hold. Recall that that df = 0 from which, as a consequence of the converse of the Poincaré lemma, there is -form A with da = F. Suppose now we take any differentiable function G and add its differential to A. Then with A = A dg we have da = da dg = da ddg = da 0 = F So the question is, which G should be added to A so that the condition of Lorenz holds. Recall the condition of Lorenz is δa = 0, We need δa = 0 δa dg = 0 δdg = δa δdg dδg = δa since δg is 0 G = δa

12 8 CHAPTER. INTRODUCTION AND BASIC APPLICATIONS This is the same type of equation which we found before, and solving it one has G so that δa dg = 0 as required. GGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGGG Next we are interested in the codifferential δ. We have, as usual, with wedges omitted for ease of reading, Then A = φcdt A dx A 2 dy A dz δa = d A = d φdxdydz A cdtdydz A 2 cdtdxdy A cdtdzdx = φ c t A x A 2 y A cdtdxdydz z = φ c t A x A 2 y A z Cognoscenti will immediately recognize that the familiar condition of Lorenz, div A φ c t = 0, is here expressed by δa = 0. Next we will compute the d Alembertian f of a function. We have δf = 0 so f = δd dδf = δdf 0 f f f f = d cdt dx dy c t x y z dz f f f f = d dxdydz cdtdydz cdtdzdx c t x y z cdtdxdy 2 f = c 2 t 2 2 f x 2 2 f y 2 2 f z 2 cdtdxdydz = c 2 2 f t 2 2 f x 2 2 f y 2 2 f z 2 The minus signs come from swapping differentials around and sometimes from the action of. This is the negative of the classical d Albertian An obnoxious calculation shows that if p f = c 2 2 f t 2 2 f x 2 2 f y 2 2 f z 2 A = A 0 cdt A dx A 2 dy A dz then A = A 0 cdt A dx A 2 dy A dz

13 .. MAXWELL S EQUATIONS IN SPACE-TIME 9 We know of no intelligent way to prove this but in view of its importance we will do the calculation in an appendix to this chapter, which should be skimmed from a distance. We now want to apply this equipment to the problem of potential equations in four dimensional Space-Time. We first consider the case when ǫ = µ = where everything is simple. When these are not, there are certain complications which we will discuss later. For ease of reference we repeat here Maxwell s equations. along with div D = ρ curl E = c t D = ǫe div B = 0 curl H = D c t c j B = µh curl A = B B dφ = E c Let us set as is usual F = da. We then have A = φcdt A dx A 2 dy A dz A F = da = φ A 2 cdtdx φ A cdtdy φ cdtdz c t x c t y c t z A y A 2 A dydz z z A A2 dzdx x x A dxdy y = E cdtdx E 2 cdtdy E cdtdz B dydz B 2 dzdx B dxdy Next we take the of both sides. We remember that here we have ǫ = and µ = so that E = D and B = H. Then F = F A t = E dydz E 2 dzdx E dxdy B cdtdx B 2 cdtdy B cdtdz = D dydz D 2 dzdx D dxdy H cdtdx H 2 cdtdy H cdtdz D H d F = c t y H 2 D 2 H cdtdydz z c t z H x D H2 c t x H D cdtdxdy y x D 2 y D z = ρ dxdydz c j cdtdydz c j 2 cdtdzdx c j cdtdxdy cdtdzdx dxdydz δf = d F = ρ cdt c j dx c j 2 dy c j dz Now the coup; recall that we may take δa = 0 the condition of Lorenz so that we have A = δd dδa = δda 0 = δf = ρ cdt c j dx c j 2 dy c j dz

14 0 CHAPTER. INTRODUCTION AND BASIC APPLICATIONS Using the fact that in rectangular coordinates we have A = φcdt A dx A 2 dy A dz we have, with the classical d Alembertian p A i = c 2 2 A i t 2 2 A i x 2 2 A i y 2 2 A i z 2 = A i we have the potential equations in four space p φ = ρ p A i = c j i which is the same result we got in the section on Maxwell s equations in three dimensional space. HHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH We now want to extend this result to the case where ǫ or µ are not one. This is surprisingly difficult. However, we suspect it can be done because the corresponding equations in three space have the same form for the potential φ and the vector potential components A i. Recall that the form of the equation is ǫµ 2 A i c 2 t 2 2 A i x 2 2 A i y 2 2 A i z 2 = µ c j i This suggests an electromagnetic wave moving at a speed of k = c ǫµ Now we must steer the boat, like Odysseus, between Charybdis and Skilla. Charybdis refers to the fact that for the d Alembertian to come out with k = ǫµ 2 c 2 in it, we are going to have to modify the operator by changing c to k. Skilla refers to the fact that Maxwell s equations have c not k in them. Thus replacing all the c s by k s won t work; we must steer more subtlely. Nevertheless the fact that the equations for φ and A i have the same form in -space suggests that it is possible to navigate successfully, which we will now do. It is worth noting first that the methodology we are about to introduce cannot be the final electromagnetic word, because there are materials for which the relationship between D and E is not given by a constant ratio, and there are two reasons for this. First, ǫ may vary over space, or over time, but even worse, the two vectors do not point the same way in some materials and circumstances so that the constant ǫ must be replaced by a matrix ǫ ij so that D i = ǫ ij E j. Naturally the same is true of B and H. So, to be explicit, we assume in what follows that we are working over a region where c, ǫ and µ are all constant 2 2 We of course know c is constant over the entire age of the entire universe because we have measured it for 50 years and across the solar system.

15 .. MAXWELL S EQUATIONS IN SPACE-TIME There is some possibility of confusion in what follows if we use dx 0. To avoid this, we will revert to using dt, dx, dy, dz. In the following calculations the operator uses the same equations as before but the constant c in those equations is replaced by the constant k. Thus changes to dx 0 dx 2 = dx dx cdtdy = dzdx k dtdy = dzdx Except for some trivial algebra the calculation goes as before, and since the ideas are the same we will just present the calculations in the most efficient order. Recalling that k = c/ ǫµ, the potential form A is A = φcdt A dx A 2 dx A dx First we go after the codifferential δa, A = d A = δa = d A = = c k φkdt A dx A 2 dx A dx = ǫµφkdt A dx A 2 dx A dx ǫµ φdxdydz A kdtdydz A 2 kdtdzdx A kdtdxdy ǫµ φ k t A x A2 y A kdtdxdydz z ǫµ φ c t A x A2 y A = 0 z by the condition of Lorenz, which we are assuming, as before. One of the the positive aspects of the four dimensional treatment is that the condition of Lorenz is so simply expressed: δa = 0 Next we have, and note we here use c not k so that we can use Maxwell s equations, F = da = c φ cdtdx A 2 x c t A t A y A2 z = E cdtdx E 2 cdtdy E cdtdz dydz A z A x B dydz B 2 dzdx B dxdy = ǫµ E kdtdx E 2 kdtdy E kdtdz B dydz B 2 dzdx B dxdy F = ǫµ E dydz E 2 dzdx E dxdy B kdtdx B 2 kdtdy B kdtdz φ cdtdy A y c t dzdx A 2 x A y φ cdtdy z dxdy

16 2 CHAPTER. INTRODUCTION AND BASIC APPLICATIONS = = ǫµ D dydz D 2 dzdx D dxdy ǫ µ H cdtdx H 2 cdtdy H cdtdz ǫµ µ ǫ F where F = ǫ µ F = H cdtdx H 2 cdtdy H cdtdz D dydz D 2 dzdx D dxdy Now we find d F, SIGNS CHANGED TO HERE; START HERE!! d F = D H c t y H 2 cdtdydz D 2 H z c t z H cdtdzdx x D H2 c t x H D cdtdxdy y x D 2 y D dxdydz z = D curl H c t cdtdydz D 2 curl H c t 2 cdtdzdx D curl H c t cdtdxdy div Ddxdydx = ρ dxdydz c j cdtdydz j 2 cdtdzdx j cdtdxdy ǫµ = ρ dxdydz j kdtdydz j 2 kdtdzdx j kdtdxdy c Finally, we have A = δd dδa = δda 0 conditon of Lorenz: δ A = 0 = δf = d F µ = d ǫ F µ = ǫ d F µ ǫµ = ǫ ρ dxdydz j kdtdydz j 2 kdtdzdx j kdtdxdy c µ ǫµ = ρ kdt j dx j 2 dy j dz ǫ c µ = ǫµ ρ cdt j dx j 2 dy j dz ǫ ǫµ c = ρ ǫ cdt µ c j dx j 2 dy j dz

17 .. MAXWELL S EQUATIONS IN SPACE-TIME This decodes in the usual way, with p =, to p φ = c 2 2 φ t 2 2 φ x 2 2 φ y 2 2 φ z 2 = ρ ǫ p A i = 2 A i c 2 t 2 2 A i x 2 2 A i y 2 2 A i z 2 = µ c j i Appendix I Derivation of d Alembertian for Functions and Forms We are here going to derive a couple of formulas for the d Alembertian which we used in the main part of the Chapter. This calculation is of almost no interest and unless you are really interested you can skip it. The result can be generalized, which we discuss after the derivation. We will first calculate the d Alembertian on functions, which is fairly simple. We revert to using x 0, x, x 2, x instead of cdt, dx, dy, dz so that the symmetry of the situation is more obvious. Remembering that δ = d f is 0 since f Λ 4 and thus d f = 0, we have f = δd dδf = d df 0 f = d x 0 dx0 f x dx f x 2 dx2 f x dx f = d x 0 dx dx 2 dx f x dx0 dx 2 dx f x 2 dx0 dx dx f x dx0 dx dx 2 2 f = x f x 2 2 f x 22 2 f x 2 dx 0 dx dx 2 dx = 2 f x 02 2 f x 2 2 f x f x 2 Our next job is to compute the d Alembertian on the -form A = A 0 dx 0 A dx A 2 dx 2 A dx For the final step of this calculation we need, for clarity, to distinguish the d Alembertian on functions from that on -forms. Thus, temporarily, we will refer to the d Alembertian on functions f Λ 0 by 0 f. We need to compute dδ δda, which we do in stages. δa = d A = d A 0 dx 0 A dx A 2 dx 2 A dx = da 0 dx dx 2 dx A dx 0 dx 2 dx A 2 dx 0 dx dx A dx 0 dx dx 2 A0 = x 0 A x A 2 x 2 A dx 0 x dx dx 2 dx = A 0 x 0 A x A 2 x 2 A x dδa = 2 A 0 x 02 2 A x 0 x 2 A 2 x 0 x 2 2 A x 0 x dx 0

18 4 CHAPTER. INTRODUCTION AND BASIC APPLICATIONS 2 A 0 x x 0 2 A x 2 2 A 2 x x 2 2 A x x dx 2 A 0 Now we do the other term x 2 x 0 2 A x 2 x 2 A 2 x A x 2 x dx 2 2 A 0 x x 0 2 A x x 2 A 2 x x 2 2 A x 2 dx da = da 0 dx 0 A dx A 2 dx 2 A dx A = x 0 A 0 x dx 0 dx A2 x 0 A 0 x 2 dx 0 dx 2 A x 0 A 0 x dx 0 dx A x 2 A 2 x dx 2 dx A x A x dx dx A2 x A x 2 dx dx 2 A da = x 0 A 0 x dx 2 dx A2 x 0 A 0 x 2 dx dx A x 0 A 0 x dx dx 2 A x 2 A 2 x dx 0 dx A x A x dx 0 dx 2 A2 x A x 2 dx 0 dx d da = 2 A x x 0 2 A 0 x 2 2 A 2 x 2 x 0 2 A 0 x A x x 0 2 A 0 x 2 dx dx 2 dx 2 A x 02 2 A 0 x 0 x 2 A 2 x 2 x 2 A x A x 2 2 A x x dx 0 dx 2 dx 2 A 2 x 02 2 A 0 x 0 x 2 2 A x x 2 2 A 2 x 2 2 A 2 x 2 2 A x x 2 dx 0 dx dx 2 A x 02 2 A 0 x 0 x 2 A x x 2 A x 2 2 A x 22 2 A 2 x 2 x dx 0 dx dx 2 δda = d da = 2 A x x 0 2 A 0 x 2 2 A 2 x 2 x 0 2 A 0 x A x x 0 2 A 0 x 2 dx 0 2 A x 02 2 A 0 x 0 x 2 A 2 x 2 x 2 A x A x 2 2 A x x dx 2 A 2 x 02 2 A 0 x 0 x 2 2 A x x 2 2 A 2 x 2 2 A 2 x 2 2 A x x 2 dx 2 2 A x 02 2 A 0 x 0 x 2 A x x 2 A x 2 2 A x 22 2 A 2 x 2 x dx A = dδa δda = 2 A 0 x A 0 x 2 2 A 0 x 22 2 A 0 x 2 dx 0 2 A x A x 2 2 A x 22 2 A x 2 dx 2 A 2 x A 2 x 2 2 A 2 x 22 2 A 2 x 2 dx 2 2 A x A x 2 2 A x 22 2 A x 2 dx = 0 A 0 dx 0 0 A dx 0 A 2 dx 2 0 A dx This is the formula that we used in the main text. It may appear completely miraculous. However, something like it is true in general; the difference is that there are lower order terms, so the each term then becomes

19 .. MAXWELL S EQUATIONS IN SPACE-TIME 5 A lower order terms dx. This is interesting but too difficult for us to investigate. Appendix II Tensor form of the Electromagnetic Equations Persons who have been through Electromagnetics using Maxwell s equations with a tensor approach might be interested in comparing the two treatments. Therefore I provide a very brief look at the tensor approach. We begin by specifying the metric tensor: g ij = We will use the following descriptions of the Electric and Magnetic fields: E = E, E 2, E B = B, B 2, B There is no significance to the position of the indices on E i or b j. Next we specify the electromagnetic tensors F µν and F µν which are connected in the usual way by F µν = g ρµ g σν F ρσ. Since g µν is summetric the order of its indices does not matter. The electromagnetic tensors are F µν = 0 E E 2 E E 0 B B 2 E 2 B 0 B E B 2 B 0 F µν = 0 E E 2 E E 0 B B 2 E 2 B 0 B E B 2 B 0 The values in F µν are selected so that they eventually mesh with Maxwell s equations. There is more than one way to do this but those give above are fairly standard. More notation: µ = x and ν = g µν µ µ. Also the action of µ on a tensor is often indicated by adding, µ to its lower indices, for example µ F ρσ = F ρσ,µ In tensorland there is a theorem that if F µν,ρ F ρµ,ν F νρ,µ = 0 then there is a Tensor A so that F µν = µ A ν ν A µ. This is nothing but the converse of the Poincare lemma applied to F = F µν dx µ dx ν. This is where the vector potential A = A, A 2, A enters the game. We define a four-vector A by A µ = φ, A, A 2, A Signs are chosen so the usual dimensional formulas come out right. Hopefully we will now get that F µν,ρ F ρµ,ν F νρ,µ = 0 is the tensor form of the two Maxwell equations divb = 0 and curl E = c Also the formula F µν = µ A ν ν A µ should be the tensor form of B = curl A B t.

20 6 CHAPTER. INTRODUCTION AND BASIC APPLICATIONS and E = gradφ c A t. We now run a few random checks to see this is true. F 2, F,2 F 2, = B x B 2 x 2 B x = div B F 02, F 0,2 F 2,0 = E2 x E x 2 B x 0 = curle B c t verifying that the the condition F µν,ρ F ρµ,ν F νρ,µ = 0 is indeed identical to the two Maxwell equations. Next we look at the the potential A. E = F 0 = 0 A A 0 = A c t = gradφ A c t φ x B = F 2 = A 2 2 A = A2 x A x 2 = curla So we see F µν = µ A ν ν A µ is indeed equivalent to B = curl A and E = A gradφ c t. If you are interested in these matters it would be useful for you to work out a few more of the possible choices of indices and verify that all is as it should be. We have expressed two of the four Maxwell equations in tensor form. Next we need to express the other two. For this we need the current tensor j ν = cρ, j, j 2, j where the current is j = j, j 2, j and ρ is the charge density. The reason for the c in the charge density part of j ν is that current is moving charge with respect to t not x 0 which perturbs the formulas. If we were starting from scratch we could fix this but it is too late; the formula as given is standard. The tensor formulation of the second two Maxwell s equations is simply µ F µν = c jν as we will now show by calculating a small sample of the possible index choices. µ F µ0 = F F F x x2 x = E x E2 x 2 E x = div E = ρ = c j0

21 .. MAXWELL S EQUATIONS IN SPACE-TIME 7 µ F µ 0 2 F F F = x0 x2 x = E B c t x 2 B 2 x = c E t curlb = c j showing that the boxed equation above is indeed equivalent to the Maxwell s equations with ǫ = µ =, D = E, H = B Notice the corollary div E = ρ curl B = c E t c j ν j ν = c ν µ F µν = 0 because F µν is skew symmetric. This then gives ρ c t j x j x j x = 0 ρ c t div j = 0 which is the equation of continuity. The last thing we must treat is the potential equations which in the tensor formulation are quite easy. First note µ = x 0, x, x 2, x µ = x 0, x, x 2, x since µ = g µν ν. Next the condition of Lorenz is easily described as µ A µ = 0 Condition of Lorenz indeed this equation decodes as φ c t A x A x A x = 0 φ c t div A = 0 Now for the potential equations. First recall the equations F µν = µ A ν ν A µ from which it is easy to get F µν = µ A ν ν A µ using g µν. We then have c jν = µ F µν = µ µ A ν ν A µ = µ µ A ν ν µ A µ

22 8 CHAPTER. INTRODUCTION AND BASIC APPLICATIONS so, by the condition of Lorenz µ A µ = 0, we have the potential equation which decodes as µ µ A ν = c jν 2 φ c 2 t 2 2 φ x 2 2 φ x φ x 2 = ρ 2 A i c 2 t 2 2 A i x 2 2 A i x A i x 2 = c ji which would be written more economically as p φ = ρ, p A i = c ji Appendix III Historical Note on the Condition of Lorenz The condition of Lorenz was first used by the Danish physicist Ludvig Lorenz Lorenz had the misfortune to have almost the same name as the much more famous Dutch physicist Hendrik Lorentz , a friend of Einstein and early exponent of relativity, who also used the condition. This is the source of the confusion in the spelling of the name when citing the condition. It was suggested that the condition be referred to as the Lorenz-Lorentz condition, but this suggestion did not make it into general use, possibly due to considerations of euphony.

23 Bibliography [] Misner, Thorne and Wheeler, GRAVITATION, W.H. Freeman, San Francisco, 97 [2] Frankel, Theodore THE GEOMETRY OF PHYSICS, rd edition, Cambridge University Press, Cambridge,etc, 202 [] Gilkey, Peter B. INVARIANCE THEORY, THE HEAT EQUATION AND THE ATIZYA-SINGER INDEX THEOREM, 2nd Edition Boca Raton 995 [4] Nakahara, Mikio GEOMETRY, TOPOLOGY, AND PHYSICS, Institute of Physics Publishing, London, 200 [5] Lovelock, David and Rund, Hanno TENSORS, DIFFERENTIAL FORMS AND VARIATIONAL PRINCIPLES, Wiley Interscience [6] Roe, John, ELLIPTIC OPERATORS, TOPOLOGY AND ASYMP- TOTIC METHODS, 2nd Edition, Boca Raton, 998 [7] Rosenberg, S. THE LAPLACIAN ON A RIEMANNIAN MANI- FOLD, Cambridge, 997 [8] Schulz, William, THEORY AND APPLICATION OF GRASSMANN ALGEBRA, schulz/grassmann.pdf 9