2. Ch. 1, part 2. Lecture 2: Electrostatic energy, Green s theorem

Transcription

1 2. Ch., part 2 Lecture 2: Electrostatic energy, Green s theorem

2 Lecture 2 (Ch part 2) Energy, Green s theorem Electrostatic energy The work required to bring the charge q i from infinity to the point x i in the presence of existing chargesq ;:::;q i is Xi q j W i =q i jx i x j j j= The total work requiredto assemble N charges is then: W N = NX i=2 Xi q j q i jx i x j j Each pairwise interaction appears once in this double sum. f we sum overiandj independently, but exclude i= j, each pair is counted twice. Thus we can also write j= W N = 2 X i6=j q i q j jx i x j j (2.) Going to a continuum distribution, we have no way to exclude the interaction of an infinitesimal charge element with itself. o we leave it in andwrite: W = 2 ½(x)½(x 0 ) jx x 0 j d 3 xd 3 x 0 (2.2). For a smooth three dimensional charge distribution, the interaction of a charge element withthe nearby chargesis 2 ½(x)d3 x ½(x 0 ) jx x 0 j d3 x 0 ¼ 2=3 ½ 2 (x)d 3 x and vanishes for! 0. o the self-interaction does not contribute and the formula (2.2) is correct, unless ½(x) contains a point charge, or even a line charge. At the macroscopic level the charge distributions always have a finite extent, really, so there is no problem. At the fundamental level, there is a problem if the elementary particles (leptons, quarks) are truly point charges. Relativistic QED (quantum electrodynamics) deals with this difficulty by postulating a simple negative term that cancels the infinite self-energy inw. Few people believe that this renormalization procedure is the final word The field energy One reason why we like eq. (2.2) is that it can be transformed to W = jr j 2 d 3 x= jej 2 d 3 x (2.3)

3 Lecture 2 (Ch part 2) Energy, Green s theorem 3 which shows that the work done in assembling the charges is stored as field energy. To get thisexpressionwe rewrite eq. (2.2) as W = 2 ½(x) (x)d 3 x; substitute ½(x)= (=4¼)r 2 (x); and eliminate r 2 (x) by usingr ( r )= r 2 + jr j 2. What we actually get is W = jr j 2 d 3 da o it seems that eq. (2.3) is obtained only if the integral is over all space and there are no chargesat infinity, orinspecial casessuchas grounded conductors ( = 0) on the boundary. n practice, however, one can always find an such that the boundary termvanishes or is unimportant. Eq. (2.3) is a fundamental result. t is very useful to think that there really is an energy density w= jej2 that resides inthe fieldandis greaterwhere the fieldis stronger. Note that the fieldenergy isalwayspositive. Aconsequence of thisisthat An assembly of charges cannot be in equilibriumunder electrostatic forces alone. We already noted that for point charges (2.) is not the same as (2.2) or (2.3). For two unlike charges, (2.) isnegative, while (2.3) ispositive (and infinite). Jackson discusses this point ingreater detail Boundary values A number of important relations can be obtained by fiddling around with the field energy expressionand related integrals. They are very useful to discuss the properties of the E field in a confined geometry. Typically, the boundaries are made of conductors, each held at a known potential, or carrying a known total charge. The problemis to find the charge density ¾ on the surface of eachconductor andthe field E in the space betweenthem Green s first identity (Ár 2 Ã+rÁ rã)d 3 x= This is obtained simply by integrating the identity r (ÁrÃ)=Ár 2 Ã+rÁ rã da (2.4) and using the divergence theorem. t does not seem too interesting in itself, but it has many uses. Animportant corollary is the following.

4 Lecture 2 (Ch part 2) Energy, Green s theorem 4 Uniqueness theorem The solution of Poisson s equation is unique if is given on the boundary (Dirichlet boundary condition). t is unique up to an additive constant = is given on the boundary (Neumann boundary condition). t is also unique if the boundary conditions are Dirichlet on part of the boundary, Neumann on the rest. This is proven by assuming that there are two distinct solutions, and 2, and showing thatu = 2 must vanish. Just use 2.4 withá= Ã=U (Ur 2 U+rU ru)d 3 x= da and note that r 2 U =0 in all and that eitheru =0 (Dirichlet) or@u==0 (Neumann) on. t follows that jruj vanishes andu is constant in. The constant must be zero in the Dirichlet case, or for mixed boundary conditions, but is arbitrary in the pure Neumann case. By a modification of this argument one can prove that, in charge-free space, the quantity R jrªj 2 d 3 x is minimal, subject to the appropriate boundary conditions, whenªcoincides with the potential, solution of Laplace s equation r 2 =0. ee Jackson s problem Green s theorem and integral equations Write downgreen s first identity forá;ã andã;á: (Ár 2 à rá rã)d 3 x = (Ãr 2 Á rã rá)d 3 x = da da andsubtract side by side. The result is Green s secondidentity, orgreen s theorem: (Ár 2 à Ãr 2 Á)d 3 x= This gives a very useful formula if we choose Ã= R = jx x 0 j à Ã@Á da (2.5) so that r 2 Ã= 4¼±(x x 0 ), and we identifyáwith the potential. fxis inside we get 4¼ (x 0 ) à R r2 d 3 R da

5 Lecture 2 (Ch part 2) Energy, Green s theorem 5 Recalling that r 2 = 4¼½ and interchanging the names ofxandx 0 : There is a lot of mathematicalphysics in this equation: ½(x 0 ) R d3 x 0 + 4¼ R 0 (x0 da 0 (2.6) 0 R ² f is all of space and there are no charges at infinity, it simply gives back Coulomb s law ½(x 0 ) R d3 x 0 ² f@ = isgivenonthe boundary (Neumann), we obtainanintegralequationfor on. This equation may be easier to solve than the original Poisson equation. Once (x) is known forxon, it can be found everywhere by integration (in principle, if not in practice). ² f is given on the boundary (Dirichlet), we can similarly get an integral equation for@ = on andthenfind (x) everywhere. The integralequation = ½(x 0 R d3 x 0 @ 2 # 4¼ R 0 (x0 ) da 0 0 R ² These integral equations show that it is not possible to assign both = arbitrarily onthe boundary. Physically, could be the surface of conductorswhich are held at fixed potentials. The charges in these conductors shift about and determine the surface charge density¾, which is given by4¼¾=@ =. Note thatnis directed into the conductor (contrary to the usualpractice). ² We can view(=4¼)@ = as a surface charge density and (=4¼) as a dipole layer that cause, respectively, a discontinuity in the field and in the potential, insucha way that jumps to zero in crossing.

6 Lecture 2 (Ch part 2) Energy, Green s theorem Green s function We can solve the boundary value problem anew for every½(x), but there is a better way. Consider for instance the Dirichlet case. We can construct the general solution if we can solve the particular case when is at zero potential and½(x) is a unit point charge,½(x)=±(x x 0 ). This special solutiong(x;x 0 ) is the Green function. t obeys the equation r 2 G(x;x 0 ) = 4¼±(x x 0 ) G(x;x 0 ) = 0 whenxis on Once G has been found (this is easier said than done), we can use Green s theorem with Ã=G andá= and proceed as in the derivation of eq.(2.6). n other words, simply write G in place of =R in (2.6). What we find is anexplicit expressionfor (x): n particular, if =0 on, we get G(x;x 0 )½(x 0 )d 3 x 0 + (x 0 (2.7) 4¼ 0da0 G(x;x 0 )½(x 0 )d 3 x 0 which is very much like Coulomb s law. n fact,= jx x 0 j is nothing but the Green function for all space. imilarly, one can define and use a Green function G N (x;x 0 ) for Neumann boundary conditions. Jackson discussesthiscase indetail. f we putg==r+f and we use r 2 G= 4¼±(x x 0 ) and r 2 (=R)= 4¼±(x x 0 ), we see thatf satisfies Laplace s equation r 2 F =0everywhere in. We can regard F as an electric potential due to (fictitious) charges located outside. The idea is to continue F andgacross asif they were not discontinuousat. n some casesthis approachleadsto a simple solutionof the boundary value problem (methodof images, Chapter2). There are many applications of Green s theorems. A useful one is the mean value theorem, Jackson s Problem.0 (assigned): n charge-free space, (x) is equal to the average of on any sphere centered at x. A consequence of this is that (x) cannot have maxima or minima in charge-free space. One could say that this is obvious, because in charge-free space satisfies Laplace s equation, r 2 =0, which in cartesian coordinates =0 so that it is impossible for all the second derivatives to have the same sign. However, the mean value theorem shows that it is also impossible to have a point where all the second and third derivatives vanish but all the fourth derivatives are positive, for instance.