--- title: The Electromagnetic Field Tensor module: Covariant Electromagnetism moduleNumber: 4 lessonNumber: 2 order: 402 summary: > The antisymmetric derivative of the four-potential is the field-strength tensor F, gauge invariant by construction, with the electric and magnetic fields as its components. Its dual exchanges E and B, and its two contractions form the Lorentz invariants that classify a field as electric, magnetic, or radiative in every frame. topics: [Covariant Electromagnetism] draft: false sources: - book: Schutz ref: "Ch. 4 — the electromagnetic field tensor and its components" - book: Carroll ref: "Lecture Notes on General Relativity §1 — Special Relativity and Flat Spacetime (electromagnetism, field strength), arXiv:gr-qc/9712019" --- The [four-potential](/relativity/covariant-electrodynamics/four-current-and-the-four-potential) carries a gauge redundancy, so it is not itself observable. The observable fields come from a particular antisymmetric combination of its derivatives, one that is invariant under $A^\mu \to A^\mu + \partial^\mu \chi$. That combination is a rank-two tensor, and packing the six numbers of $\vec E$ and $\vec B$ into its components is the step that makes electromagnetism manifestly Lorentz covariant.[^carroll2] The conventions of the previous lesson carry over: signature $(-,+,+,+)$, $\partial_\mu = (\tfrac{1}{c}\partial_t, \nabla)$, and $A^\mu = (\phi/c, \vec A)$. ## Building the field strength The gauge shift $A^\mu \to A^\mu + \partial^\mu \chi$ adds $\partial^\mu \partial^\nu \chi$ to $\partial^\mu A^\nu$. Because partial derivatives commute, $\partial^\mu \partial^\nu \chi = \partial^\nu \partial^\mu \chi$, the antisymmetric part is unchanged. Define the **electromagnetic field tensor**, or field strength, as $$ F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu. $$ Under a gauge transformation, $$ F^{\mu\nu} \to \partial^\mu(A^\nu + \partial^\nu \chi) - \partial^\nu(A^\mu + \partial^\mu \chi) = F^{\mu\nu} + (\partial^\mu \partial^\nu - \partial^\nu \partial^\mu)\chi = F^{\mu\nu}. $$ The field strength is gauge invariant, which is the property that qualifies it to hold physical fields. It is antisymmetric, $F^{\mu\nu} = -F^{\nu\mu}$, so its diagonal vanishes and it has six independent components — exactly the count of $\vec E$ and $\vec B$ together. > **Definition (Field-strength tensor).** $F^{\mu\nu} = \partial^\mu A^\nu - > \partial^\nu A^\mu$. It is antisymmetric and gauge invariant, with six > independent components. Its lower-index form is $F_{\mu\nu} = \partial_\mu > A_\nu - \partial_\nu A_\mu$, obtained by lowering both indices with $\eta$. ## Reading off E and B Evaluate the components. For the time–space entries, with $A^0 = \phi/c$ and $\partial^0 = -\tfrac{1}{c}\partial_t$, $$ F^{0i} = \partial^0 A^i - \partial^i A^0 = -\frac{1}{c}\frac{\partial A_i}{\partial t} - \partial_i\!\left(\frac{\phi}{c}\right) = \frac{1}{c}\left(- \frac{\partial A_i}{\partial t} - \partial_i \phi\right) = \frac{E_i}{c}. $$ The time–space block of $F^{\mu\nu}$ is the electric field, scaled by $1/c$. For the space–space entries, with spatial indices raised trivially, $$ F^{ij} = \partial^i A^j - \partial^j A^i = \partial_i A_j - \partial_j A_i = \epsilon^{ijk} B_k, $$ using $\vec B = \nabla \times \vec A$. The space–space block is the magnetic field, arranged antisymmetrically. Assembled, the tensor is $$ F^{\mu\nu} = \begin{pmatrix} 0 & E_x/c & E_y/c & E_z/c \\ -E_x/c & 0 & B_z & -B_y \\ -E_y/c & -B_z & 0 & B_x \\ -E_z/c & B_y & -B_x & 0 \end{pmatrix}. $$ The electric field occupies the first row and column; the magnetic field fills the spatial $3 \times 3$ block. The single object $F^{\mu\nu}$ holds both fields; "electric" and "magnetic" are the names of its time–space and space–space parts in one chosen frame. $$ % caption: The antisymmetric field-strength tensor with the electric field in % its time–space border (scaled by one over c) and the magnetic field in the % spatial block; the vanishing diagonal and the sign flip across it encode the % antisymmetry. \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0] \definecolor{acc}{HTML}{4A6FA5} \draw[black] (0,0) rectangle (4.8,4.8); % grid \foreach \i in {1,2,3} \draw[black] (1.2*\i,0) -- (1.2*\i,4.8); \foreach \i in {1,2,3} \draw[black] (0,1.2*\i) -- (4.8,1.2*\i); % shade time-space border (E) : first row and first column \fill[acc!12] (1.2,3.6) rectangle (4.8,4.8); \fill[acc!12] (0,0) rectangle (1.2,3.6); % diagonal zeros \foreach \i in {0,1,2,3} \node[black] at (0.6+1.2*\i,4.2-1.2*\i) {0}; % E labels top row \node[black!70] at (1.8,4.2) {Ex/c}; \node[black!70] at (3.0,4.2) {Ey/c}; \node[black!70] at (4.2,4.2) {Ez/c}; % B labels in spatial block \node[black!70] at (3.0,3.0) {Bz}; \node[black!70] at (4.2,3.0) {By}; \node[black!70] at (4.2,1.8) {Bx}; % axis labels \node[black, anchor=south] at (0.6,4.9) {t}; \node[black, anchor=south] at (1.8,4.9) {x}; \node[black, anchor=south] at (3.0,4.9) {y}; \node[black, anchor=south] at (4.2,4.9) {z}; \node[acc, anchor=west] at (5.1,4.2) {E in the border}; \node[black, anchor=west] at (5.1,2.4) {B in the block}; \end{tikzpicture} $$ Lowering both indices flips the sign of the time–space entries only, since each factor of $\eta_{00} = -1$ appears once there and twice (or zero times) elsewhere: $$ F_{\mu\nu} = \begin{pmatrix} 0 & -E_x/c & -E_y/c & -E_z/c \\ E_x/c & 0 & B_z & -B_y \\ E_y/c & -B_z & 0 & B_x \\ E_z/c & B_y & -B_x & 0 \end{pmatrix}. $$ ## Fields mix under a boost Because $\vec E$ and $\vec B$ are components of a single tensor, a change of frame mixes them. Under a Lorentz transformation the field strength transforms as a rank-two tensor, $$ F'^{\mu\nu} = \Lambda^\mu{}_\alpha \Lambda^\nu{}_\beta F^{\alpha\beta}. $$ A boost $\Lambda$ rotates the time axis into a space axis, so it rotates entries of the electric block into the magnetic block and back. A field that is purely electric in one frame acquires a magnetic part in another. The next lesson works out the transformation explicitly; the point here is structural. There is no frame-independent split of the field into "electric" and "magnetic" pieces, any more than there is a frame-independent split of a four-vector into time and space. There is only $F^{\mu\nu}$, and observers slice it differently. $$ % caption: A boost rotates the time direction into a spatial direction, so it % carries entries of the electric time–space border into the magnetic spatial % block: a purely electric field in one frame has a magnetic part in another. \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0] \definecolor{acc}{HTML}{4A6FA5} % left tensor: pure E \draw[black] (0,0) rectangle (2.4,2.4); \fill[black!8] (0.6,1.2) rectangle (2.4,1.8); \fill[black!8] (0,0) rectangle (0.6,1.8); \node at (1.2,2.1) {E}; \node[black] at (1.5,0.9) {0}; \node[black!70, anchor=north] at (1.2,-0.15) {one frame}; % arrow \draw[->, very thick] (2.9,1.2) -- (4.5,1.2) node[above, midway, black!70] {boost}; % right tensor: E and B \draw[black] (5.0,0) rectangle (7.4,2.4); \fill[black!8] (5.6,1.2) rectangle (7.4,1.8); \fill[black!8] (5.0,0) rectangle (5.6,1.8); \fill[acc!14] (5.6,0) rectangle (7.4,1.2); \node at (6.2,2.1) {E}; \node[acc] at (6.5,0.6) {B}; \node[black!70, anchor=north] at (6.2,-0.15) {other frame}; \end{tikzpicture} $$ ## The dual tensor A second antisymmetric tensor is built from $F$ using the totally antisymmetric Levi-Civita symbol $\epsilon^{\mu\nu\alpha\beta}$ (with $\epsilon^{0123} = +1$). The **dual field tensor** is $$ \tilde F^{\mu\nu} = \frac{1}{2}\epsilon^{\mu\nu\alpha\beta} F_{\alpha\beta}. $$ Contracting with $\epsilon$ exchanges the roles of the two blocks. The dual is obtained from $F$ by the substitution $$ \frac{\vec E}{c} \to \vec B, \qquad \vec B \to -\frac{\vec E}{c}, $$ so that its components are $$ \tilde F^{\mu\nu} = \begin{pmatrix} 0 & B_x & B_y & B_z \\ -B_x & 0 & -E_z/c & E_y/c \\ -B_y & E_z/c & 0 & -E_x/c \\ -B_z & -E_y/c & E_x/c & 0 \end{pmatrix}. $$ The dual is the tool for writing the two source-free Maxwell equations compactly, as the [covariant Maxwell lesson](/relativity/covariant-electrodynamics/covariant-maxwell-and-the-stress-energy-tensor) shows: $\partial_\mu \tilde F^{\mu\nu} = 0$ is Faraday's law and $\nabla \cdot \vec B = 0$ together. The exchange $\vec E \leftrightarrow c\vec B$ (with a sign) is the electric–magnetic duality of the source-free equations. $$ % caption: The dual tensor is F with its blocks swapped, sending E over c to B % and B to minus E over c; this exchange is the electric–magnetic duality of the % source-free Maxwell equations. \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0] \definecolor{acc}{HTML}{4A6FA5} % F \draw[black] (0,0) rectangle (2.4,2.4); \fill[acc!14] (0,1.2) rectangle (2.4,2.4); \fill[black] (0,0) rectangle (2.4,1.2); \node[acc] at (1.2,1.8) {E block}; \node[black!70] at (1.2,0.6) {B block}; \node[black!70, anchor=north] at (1.2,-0.2) {F}; % arrows swapping \draw[->, black, thick] (2.7,1.7) to[out=20,in=160] (4.9,1.7); \draw[->, black, thick] (4.9,0.7) to[out=200,in=340] (2.7,0.7); \node[black!70, anchor=south] at (3.8,1.85) {swap}; % dual \draw[black] (5.2,0) rectangle (7.6,2.4); \fill[black] (5.2,1.2) rectangle (7.6,2.4); \fill[acc!14] (5.2,0) rectangle (7.6,1.2); \node[black!70] at (6.4,1.8) {B block}; \node[acc] at (6.4,0.6) {E block}; \node[black!70, anchor=north] at (6.4,-0.2) {dual F}; \end{tikzpicture} $$ ## The two invariants From the six components of the field two Lorentz scalars can be built, and only two independent ones. They are the full contractions of $F$ with itself and with its dual. The first is $$ F_{\mu\nu} F^{\mu\nu} = 2\left(B^2 - \frac{E^2}{c^2}\right). $$ The time–space entries contribute $-2E^2/c^2$ (each carries a sign from lowering one time index) and the space–space entries contribute $+2B^2$. The second invariant comes from the dual: $$ F_{\mu\nu} \tilde F^{\mu\nu} = -\frac{4}{c}\,\vec E \cdot \vec B. $$ These two combinations, $$ E^2 - c^2 B^2 \quad\text{and}\quad \vec E \cdot \vec B, $$ have the same value in every inertial frame. They classify the field. - **$\vec E \cdot \vec B$ is invariant.** If $\vec E$ and $\vec B$ are perpendicular in one frame (as in a light wave), they are perpendicular in every frame. - **$E^2 - c^2 B^2$ is invariant.** Its sign labels the field as electrically dominated ($E > cB$), magnetically dominated ($E < cB$), or balanced ($E = cB$), the same label in all frames. Several statements follow at once. If a field is purely electric in some frame, then $\vec E \cdot \vec B = 0$ and $E^2 - c^2 B^2 > 0$ there, hence in every frame — so no boost can transform it into a purely magnetic field, though it can give it a magnetic part. If both invariants vanish, $\vec E \perp \vec B$ and $E = cB$ in every frame; this is the invariant signature of a radiation field, a light wave, which no observer can boost away. $$ % caption: The invariant E-squared minus c-squared B-squared classifies a field % in a frame-independent way: its sign is the same for every observer, so a field % dominated by one part cannot be boosted into a field dominated by the other. \begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0] \definecolor{acc}{HTML}{4A6FA5} % number line for the invariant \draw[->, black] (-4.2,0) -- (4.6,0) node[right, black!70] {invariant}; \draw[acc] (0,-0.12) -- (0,0.12); \node[acc, anchor=north] at (0,-0.22) {balanced}; \node[acc, anchor=north] at (0,-0.7) {$E = cB$ (radiation)}; % electric side \fill[black!8] (0.1,0.15) rectangle (4.1,0.6); \node[anchor=south] at (2.1,0.62) {electric dominated}; \node[black, anchor=north] at (2.1,-0.15) {no boost removes E}; % magnetic side \fill[black!8] (-4.1,0.15) rectangle (-0.1,0.6); \node[black!70, anchor=south] at (-2.1,0.62) {magnetic dominated}; \node[black, anchor=north] at (-2.1,-0.15) {no boost removes B}; \end{tikzpicture} $$ > **Worked example (Boosting away a field).** A field has $\vec E \cdot \vec B = > 0$ and $E < cB$ in the lab, so it is magnetically dominated. Ask whether some > frame sees a pure magnetic field. In that frame $E' = 0$, giving $\vec E' \cdot > \vec B' = 0$ (consistent) and $E'^2 - c^2 B'^2 = -c^2 B'^2 < 0$ (also > consistent, since $E^2 - c^2 B^2 < 0$ in the lab). Both invariants permit it, > and such a frame exists: boost with velocity $\vec v = (\vec E \times \vec > B)/B^2$, whose speed $v = E/(cB) \cdot c = E/B < c$ precisely because the field > is magnetically dominated. Had the field been electrically dominated, this > velocity would exceed $c$ and no such frame would exist. ## Summary - The field-strength tensor $F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu$ is antisymmetric and gauge invariant. Its six independent components are $\vec E$ (time–space block, scaled by $1/c$) and $\vec B$ (space–space block). - A boost transforms $F$ as a tensor and mixes its blocks; the split into electric and magnetic fields is frame dependent, while $F^{\mu\nu}$ is not. - The dual tensor $\tilde F^{\mu\nu} = \tfrac{1}{2}\epsilon^{\mu\nu\alpha\beta} F_{\alpha\beta}$ exchanges $\vec E/c \leftrightarrow \vec B$ (with a sign) and packages the source-free equations. - The two invariants $E^2 - c^2 B^2$ and $\vec E \cdot \vec B$ take the same value in every frame and classify the field; both vanish for a radiation field. [^carroll2]: Carroll, _Lecture Notes on General Relativity_, §1 (electromagnetism and the field strength), arXiv:gr-qc/9712019; Schutz, _A First Course in General Relativity_, Ch. 4. The component arrangement and the two invariants follow these treatments in the $(-,+,+,+)$ signature.