Classical Field Theory

Textbook: Florian Scheck, Classical Field Theory: On Electrodynamics, Non-Abelian Gauge Theories and Gravitation

Classical Field Theory(Scheck, Springer, 2012)

Ch1 Maxwell’s Equations

1. Maxwell’s equations in integration form:
   1. Faraday’s law of induction(1831): Magnetic flux $\Phi (t):={\iint }_{S}B(t,x)\cdot \hat{n}d\mathcal{A}$. Then ${\oint }_{\partial S}{E}^{\prime }(t,x^{\prime })\cdot ds=-f_F\frac{d}{dt}\Phi (t)$.
      1. Remark: Suppose the conducting loop moves relative to the inertial frame where $B$ is defined. Seen from a frame comoving with the loop(which is the frame we used to define $E^{\prime }$), there is $\frac{d}{dt}=\frac{\partial }{\partial t}+v\cdot \nabla$. Then $RHS=-f_F{\iint }_S\frac{d}{dt}B(t,x)\cdot \hat{n}d\mathcal{A}=-f_F{\iint }_S\left(\frac{\partial B}{\partial t}+\nabla \times (B\times v)+(\nabla \cdot B)v\right)\cdot \hat{n}d\mathcal{A}$. Applying Stokes’ theorem and rearranging the equation, we obtain ${\oint }_{\partial S}(E^{\prime }-f_F(v\times B))ds=-f_F{\iint }_S\frac{\partial B}{\partial t}\cdot \hat{n}d\mathcal{A}$. Now the intergrands on both sides refer to the same system of reference(lab frame). Then the electric field in lab frame is $E=E^{\prime }-f_F(v\times B)$.
         1. “Note, however, that the decoupling of magnetic and electric phenomena is only an apparent one because currents are due to charges which are in motion. As soon as the electric and magnetic quantities become time dependent, all phenomena intermix. ”
      2. Lorentz force $F(t,x)=q(E(t,x)+f_{F}v\times B(t,x))$ is defined w.r.t lab frame.
   2. Gauss’s law: ${\iint }_{\partial V}D(t,x^{\prime })\cdot \hat{n}d\mathcal{A}=f_G{\iiint }_V{\rho }_{free}(t,x)d\mathcal{V}=f_G{Q}_{free}$.
      1. Corollary: Experiment shows that there are no free magnetic charges. If there were, the inverse square law (and hence Gauss’s law) could be applied as well. Then $\nabla \cdot B(t,x)=\rho$. Since $\rho =0$ in reality, we know $\nabla \cdot B(t,x)=0$.
   3. The law of Biot and Savart(1822): $H(t,x)=\frac{f_{BS}}{4\pi }\iiint j(t,x^{\prime })\times \frac{x-x^{\prime }}{|x-x^{\prime }{|}^3}d\mathcal{V}$.
   4. The continuity equation: “The electric charge is conserved by all fundamental interactions.” $-\frac{d}{dt}{\iiint }_V\rho (t,x)d\mathcal{V}={\iint }_{\partial V}j(t,x^{\prime })\cdot \hat{n}d\mathcal{S}$. Then $\frac{\partial \rho (t,x)}{\partial t}+\nabla \cdot j(t,x)=0$.
2. Maxwell’s equations in local form:
   1. Induction law and Gauss’ law: $\nabla \times E(t,x)=-f_F\frac{\partial }{\partial t}B(t,x)$. $\nabla \cdot D(t,x)=f_G{\rho }_{free}(t,x)$.
   2. Law of Biot and Savart: $H(t,x)=-\frac{f_{BS}}{4\pi }\iiint j(t,x^{\prime })\times {\nabla }_x\left(\frac{1}{|x-x^{\prime }|}\right)d\mathcal{V}=\frac{f_{BS}}{4\pi }{\nabla }_x\times \iiint \frac{j(t,x^{\prime })}{|x-x^{\prime }|}d\mathcal{V}$. Applying the identity $\nabla \times (\nabla \times A)=\nabla (\nabla \cdot A)-{\nabla }^{2}A$, one obtains $\nabla \times H(t,x)=\frac{f_{BS}}{4\pi }{\nabla }_x\times \left({\nabla }_x\times \iiint \frac{j(t,x^{\prime })}{|x-x^{\prime }|}d\mathcal{V}\right)=\underset{T_1}{\underbrace{\frac{f_{BS}}{4\pi }{\nabla }_x\left(\iiint {\nabla }_x\cdot \left(\frac{j(t,x^{\prime })}{|x-x^{\prime }|}\right)d\mathcal{V}\right)}}\underset{T_2}{\underbrace{-\frac{f_{BS}}{4\pi }\iiint j(t,x^{\prime }){\nabla }_x^2\left(\frac{1}{|x-x^{\prime }|}\right)d\mathcal{V}}}$. Recall that ${\nabla }_x^2\left(\frac{1}{|x-x^{\prime }|}\right)=-4\pi \delta (x-x^{\prime })$, where $\delta$=Dirac function. Then $T_2=f_{BS}j(t,x)$. Using $\nabla \cdot (\varphi A)=\varphi (\nabla \cdot A)+A\cdot \nabla \varphi$ , we have \nabla_{x} \cdot\left( j(t,x') \cdot \frac{1}{|x-x'|} \right) = j(t,x') \cdot \nabla_{x}\left( \frac{1}{|x-x'|} \right)+\left( \frac{1}{|x-x'|} \right) \nabla_{x} \cdot j(t,x')= j(t,x') \cdot \nabla_{x}\left( \frac{1}{|x-x'|} \right)$$$$\nabla_{\color{blue} x'} \cdot\left( j(t,x') \cdot \frac{1}{|x-x'|} \right) = j(t,x') \cdot \nabla_{\color{blue} x'}\left( \frac{1}{|x-x'|} \right)+\left( \frac{1}{|x-x'|} \right) \nabla_{\color{blue} x'} \cdot j(t,x')Note: $\nabla \cdot j=0$ is because of that $j(t,x^{\prime })$ is irrelevant to $x$, not $\nabla \cdot j=-\frac{\partial \rho }{\partial t}=0$. Since ${\nabla }_x\left(\frac{1}{|x-x^{\prime }|}\right)=-{\nabla }_{x^{\prime }}\left(\frac{1}{|x-x^{\prime }|}\right)$, T_{1}=\frac{f_{BS}}{4\pi} \nabla_{x}\left( \iiint -\nabla_{\color{blue}x'} \cdot \left( \frac{ j(t,x')}{|x-x'|} \right) + \left( \frac{1}{|x-x'|} \right) \nabla_{\color{blue} x'} \cdot j(t,x') d \mathcal{V} \right)$$$$= -\frac{f_{BS}}{4\pi} \nabla_{x} \oint_{|x'|\to \infty} \left( \frac{ j(t,x')}{|x-x'|} \right) d\mathcal{A} - \frac{f_{BS}}{4\pi} \nabla_{x}\left( \iiint \frac{1}{|x-x'|} \frac{ \partial \rho }{ \partial t } d\mathcal{V} \right)$$$$=- \frac{f_{BS}}{4\pi} \frac{ \partial }{ \partial t } \nabla_{x}\left( \iiint \frac{\rho(t,x')}{|x-x'|} d\mathcal{V} \right) = \frac{f_{BS}}{4\pi} \frac{ \partial }{ \partial t } \cdot \frac{4\pi}{f_{G}}D(t,x) = \frac{f_{BS}}{f_{G}} \frac{ \partial }{ \partial t } D(t,x)Eventually $\nabla \times H(t,x)=\frac{f_{BS}}{f_G}\frac{\partial }{\partial t}D(t,x)+f_{BS}j(t,x)$.
      1. Comparing to Coulomb’s law: $E(x)=\frac{f}{4\pi }\int \rho (x^{\prime })\frac{x-x^{\prime }}{|x-x^{\prime }{|}^3}d\mathcal{V}=-{\nabla }_x\varphi$, where $\varphi =\frac{f}{4\pi }\int \rho (x^{\prime })\cdot \frac{1}{|x-x^{\prime }|}d\mathcal{V}$. Note ${\nabla }^2\frac{1}{|x-x^{\prime }|}=-4\pi \delta (x-x^{\prime })$. Then ${\nabla }_x\cdot E=f\rho$.
   3. Local equations in all systems of units: \nabla \cdot B (t,x)=0;\qquad \nabla \times E(t,x) + f_{F} \frac{ \partial }{ \partial t }B(t,x)=0;$$$$\nabla \cdot D(t,x)=f_{G} \rho(t,x);\qquad \nabla \times H(t,x)-\frac{f_{BS}}{f_{G}} \frac{ \partial }{ \partial t } D(t,x) = f_{BS}j(t,x).Note: These equations are supplemented by Lorentz force and the relation between $D,E$, $B,H$. “The quantities of the first group concern and describe the radiation field, those of the second group concern matter whose building blocks are electrons, ions and atomic nuclei. This distinction is physically meaningful: Matter, a priori, is described by a kind of dynamics other than the Maxwell fields. ""The Lorentz force, with its typical dependence on velocity, finally gives an important hint about the spacetime symmetries of Maxwell’s equations. ”
      1. By substituting them into the continuity equation, we find that $f_F=\frac{f_{BS}}{f_G}$ holds.
      2. Specifically, in SI unit, $f_F=f_G=f_{BS}=1$; in CGS unit $f_F=\frac{1}{c},f_G=4\pi ,f_{BS}=\frac{4\pi }{c}$.
      3. Wave equation: Suppose there isn’t any external source. Take curl of (b): $0=-{\nabla }^{2}E(t,x)+{\mu }_0{f}_F\frac{\partial }{\partial t}\nabla \times H(t,x)=\left({\nabla }^2-f_F^2{\epsilon }_0{\mu }_0\frac{{\partial }^2}{\partial t^2}\right)E(t,x)$. Substitute the general solution of the partial differential equation $E(t,x)=\mathcal{E}{e}^{-i\omega t}{e}^{ik\cdot x}$ to obtain $f_F^2{\mu }_0{\epsilon }_0=\frac{k^2}{{\omega }^2}=\frac{1}{c^2}$.
3. Scalar potentials and vector potentials:
   1. Construction of a vector field from its source and curl: Given smooth functions $f(t,x)=\nabla \cdot A(t,x),g(t,x)=\nabla \times A(t,x)$ and suppose the vector field $A$ is smooth. We first verify that $A$ can be decomposed to $A_1+A_2$ s.t. $\nabla \cdot A_1(t,x)=f(t,x),\nabla \times A_1(t,x)=0;$ $\nabla \cdot A_2(t,x)=0,\nabla \times A_2(t,x)=g(x,t).$Additionally, we can choose $A_1,A_2$ to have some properties, e.g. simply connected and some boundary conditions. (in order to ensure the existence of $\Phi ,C$ below) Since $A_1$ has a vanishing curl, there exists $\Phi$ s.t $A_1=-\nabla \Phi$, or ${\nabla }^2\Phi =-f(t,x)$. A solution of $\Phi$ is that $\Phi (t,x)=\frac{1}{4\pi }\iiint d^3{x}^{\prime }\frac{f(t,x^{\prime })}{|x-x^{\prime }|}$. Similarly $A_2=\nabla \times C$. Here $C$ can be chosen to be source-free(if not, we may add $B$ s.t $C^{\prime }=C+B$ is source-free. Suppose $\nabla \times B=0$, then $B=-\nabla h$ and $A_2^{\prime }=\nabla \times C^{\prime }=\nabla \times C=A_2$. That ${\nabla }^{2}h=\nabla \cdot C$ is always soluble yields the existence of $B$. ) Therefore $g(x,t)=\nabla \times (\nabla \times C)=-{\nabla }^{2}C$. A solution of $C$ is that $C(t,x)=\frac{1}{4\pi }\iiint d^3{x}^{\prime }\frac{g(t,x^{\prime })}{|x-x^{\prime }|}$. Finally we have $A(t,x)=-{\nabla }_x\left(\frac{1}{4\pi }\iiint d^3{x}^{\prime }\frac{f(t,x^{\prime })}{|x-x^{\prime }|}\right)+{\nabla }_x\times \left(\frac{1}{4\pi }\iiint d^3{x}^{\prime }\frac{g(t,x^{\prime })}{|x-x^{\prime }|}\right).$Furthermore, all vector field $A_1+A_2$ constructs the set $\{A(t,x)+\nabla \chi (t,x):\chi (t,x)\text{ smooth solution of }{\nabla }^2\chi (t,x)=0\}$.
   2. Scalar potentials and vector potentials: $B$ can be represented as $B=\nabla \times A$. Then $\nabla \times \left(E+\frac{\partial }{\partial t}A\right)=0$, yielding that $E=-\frac{\partial }{\partial t}A-\nabla \Phi$. Here $\Phi$ is called the scalar potential and $A$ the vector potential.
   3. Gauge transformation: Note that for an arbitrary smooth function $\chi (t,x)$(gauge function), applying the gauge transformation $A(t,x)\mapsto A^{\prime }(t,x)=A(t,x)+\nabla \chi (t,x),\Phi (t,x)\mapsto {\Phi }^{\prime }(t,x)=\Phi (t,x)-\frac{\partial }{\partial t}\chi (t,x)$won’t change $E,B$.
      1. We can add restrictions to $\chi$ so as to get some specific forms of $A,\Phi$. e.g. $\nabla \cdot A(t,x)+\frac{\partial \Phi (t,x)}{\partial t}=0$(Lorentz gauge), $\nabla \cdot \mathbf{A}(t,x)=0$ (Coulomb gauge, here $\mathbf{A}(x)=(\Phi (x),A(x))$).
      2. Set $A$ to be the corresponding differential form of $\mathbf{A}$, i.e. $A=\phi dt+A_{\alpha }d{x}^{\alpha }$. Consider $F=dA$ to eliminate any DOF. Then $F=dA=\cdots =(-\nabla \phi -{\partial }_{t}A{)}_{\alpha }d{x}^{\alpha }\wedge dt+(\nabla \times A{)}_{\alpha }$. This 2-form can be expressed from $E,B$ using a matrix $F_{\mu \nu }=\left(\begin{array}{cccc}0& E_1& E_2& E_3\\ -E_1& 0& -B_3& B_2\\ -E_2& B_3& 0& -B_1\\ -E_3& -B_2& B_1& 0\end{array}\right)$.

Ch2 Symmetries and Covariance of the Maxwell Equations

1. The Maxwell equations in a fixed frame of reference:
   1. Rotations and discrete space time transformation:
      1. Rotations: $SO(3)\ni R:(t,x)\mapsto (t^{\prime }=t,x^{\prime }=Rx)$ of the frame of reference satisfies $R^{T}R=\mathrm{id},\det R=1$. Under this transformation we have $\mathbf{A}(t,x)=(\phi (t,x),A(t,x))\stackrel{R}{\mapsto }{\mathbf{A}}^{\prime }(t^{\prime },x^{\prime })=({\phi }^{\prime }(t^{\prime },x^{\prime }),A^{\prime }(t^{\prime },x^{\prime }))=(\phi (t,x),dRA(t,x))$ i.e. Let $R=\left(\begin{array}{cc}1& 0\\ 0& R\end{array}\right)$, $(R^{\ast }\mathbf{A})(v)=dR\mathbf{A}(R^{-1}v)$.
         1. Tentatively, this equation can be understood as that the values of $\phi ,A$ at point $p$ remain the same whatever the coordinate of $p$ is. Only their representations using basis $\left\{\frac{\partial }{\partial x_i}\right\}$(which is determined by the world coordinate $\{x_i\}$) change under $R$.
         2. Things are different when $R\in O(3)$. Vectors like $\ell =x\times p$(axial vector) and scalars like $x\cdot \ell$(pseudoscalar) go against the rule. To be more specific, ${\ell }^{\prime }(t^{\prime },x^{\prime })=(\det R)dR\ell (t,x)$, i.e. the value of $\ell$ is multiplied with $\det R$. This can be traced back to that they are defined on $T_{p}M$ and properties of $M$(e.g. orientation) may contribute to determine the values. They are intrinsic and thus can form fields(with orientation). However we define genuine scalar and vector fields to be dependent only on the point set structure of $M$, so as to rule out these “impure” fields.
      2. Space reflection: $(t,x{)}^T=(t^{\prime }=t,x^{\prime }=-x)$. Notice that the curl of a genuine(axial) vector field is an axial(genuine) vector field, $A^{\prime }(t^{\prime },x^{\prime })=(A(t,x){)}^T⟺{\nabla }^{\prime }\times A^{\prime }(t^{\prime },x^{\prime })=-(\nabla \times A(t,x){)}^T$. The Maxwell equations are invariant under space reflection iff $E,D,j$ are vector fields, $B,H$ are axial vector fields and $\rho$ is a scalar field, so as to be consistent with the laws based on experiments. Another view, notice that $F,v$ are both vectors in $F=q(E+v\times B)$(i.e. to “forget” about the experimental laws and take account of the Maxwell’s equations(including the supplements) only).
      3. Time reversal: $(t,x)\mapsto (t^{\prime }=-t,x^{\prime }=x)$. $\rho ,D,E$ are even under this transformation. $j,H,B$ are odd(e.g. $\nabla \cdot j=-\frac{\partial \rho }{\partial t}$ is the multiplication of an even and an odd). Maxwell’s equations are conserved.
      4. Charge conjugation: To reverse the signs of all charges, $C$. $\rho ,j,D,E,H,B$ are all odd under $C$. Maxwell’s equations are conserved.
   2. Maxwell’s equations and exterior forms:
      1. Definitions: ${\Lambda }^{p}M\ni \stackrel{p}{\omega }:T_{x}M\to \mathbb{R}$ is a p-form. The set of 1-form defined at $x$ is $T_x^{\ast }M$, the cotangent space. The exterior derivative $d:{\Lambda }^k(M)\to {\Lambda }^{k+1}(M)$ satisfies (1) that it behaves like the total differential when applied to 0-forms, (2) the Leibniz rule of wedge product. The Hodge star operator is defined as $(\star \omega )(e_{i_{k+1}},\cdots ,e_{i_n})={\epsilon }_{i_1\cdots i_n}\omega (e_{i_1},\cdots ,e_{i_k})$(here $\epsilon$=Levi-Civita symbol).
      2. Maxwell’s equations require ${\omega }_E,{\omega }_H$ to be 1-forms, ${\omega }_B,{\omega }_D,{\omega }_j$ 2-forms and ${\omega }_{\rho }$ a 3-form. It can now be displayed using differential forms: \mathrm{d} \, \overset{2}{\omega}_{B} = 0, \qquad \mathrm{d} \, \overset{1}{\omega}_{E}+\frac{ \partial }{ \partial t } \overset{2}{\omega}_{B}, \qquad \mathrm{d} \, \overset{2}{\omega}_{D}=\overset{3}{\omega}_{\rho}, \qquad \mathrm{d} \, \overset{1}{\omega}_{H}-\frac{ \partial }{ \partial t }\overset{2}{\omega}_{D} = \overset{2}{\omega}_{j} ;$$$$\overset{2}{\omega}_{D} = \varepsilon \star \overset{1}{\omega}_{E},\qquad \overset{1}{\omega}_{H} = \frac{1}{\mu} \star\overset{2}{\omega}_{B},\qquad F^{\flat}=q(\overset{1}{\omega}_{E}+\star (v^{\flat} \wedge \star \overset{2}{\omega}_{B})).
2. Lorentz covariance of Maxwell’s equations:
   1. Poincaré and Lorentz groups: Poincaré transformation $(\Lambda ,a):T_{x}M\to T_{x}M,x\mapsto x^{\prime }=\Lambda x+a$, where $M={\mathbb{R}}^4$. If it keeps the distance invariant, there must be ${\Lambda }^{T}g\Lambda =g$. All the $\Lambda$ satisfying this construct the Lorentz group.
      1. Calculation shows that $({\Lambda }_0^0{)}^2\ge 1$ and $(\det \Lambda {)}^2=1$. By pairing one of each situation we can divide Lorentz group into 4 (disjoint) branches, denoted by $\pm ,\uparrow /\downarrow$. $L_{+}^{\uparrow }$ is called the proper orthochronous Lorentz group which keeps the chirality and time direction. With the help of space reflection $\Pi =diag(1,-1,-1,-1)$ and time reversal $\mathsf{T}=(-1,1,1,1)$, the remaining three branches can be expressed as $L_{-}^{\uparrow }=\Pi L_{+}^{\uparrow },L_{-}^{\downarrow }=\mathsf{T}{L}_{+}^{\uparrow },L_{+}^{\downarrow }=\mathsf{T}\Pi L_{+}^{\uparrow }$respectively.
      2. Decomposition theorem: Every element of $L_{+}^{\uparrow }$ can be written (in a unique manner) as the product of a rotation $\mathcal{R}=\left(\begin{array}{cc}1& 0\\ 0& R\end{array}\right),R\in SO(3)$ and a Lorentz boost, i.e. $\Lambda =\mathcal{L}(v)\mathcal{R}$.
   2. Relativistic kinematics and dynamic: