Chapter 1: Electromagnetic Field Tensor
Electric and magnetic fields are not separate entities—they are components of a single electromagnetic field tensor Fμν. What appears as a pure electric field in one frame becomes a mix of electric and magnetic fields in another. This unification is one of the greatest triumphs of special relativity.
Historical Motivation
Before Einstein, physicists knew that moving charges create magnetic fields and changing magnetic fields induce electric fields. But there was a puzzle: what happens when you change reference frames?
The Thought Experiment
Consider a charge at rest in frame S. It creates only an electric field. Now observe from frame S' moving at velocity v:
- In S: Pure electric field \(\vec{E}\), no magnetic field
- In S': The charge appears to be moving → creates a current → magnetic field appears!
The same physical situation looks different in different frames. This suggests E and B must be related components of a single entity.
Constructing the Field Tensor
We need to combine the 3 components of \(\vec{E}\) and 3 components of \(\vec{B}\)into a 4D object. The natural choice is an antisymmetric rank-2 tensor.
From the Four-Potential
The electromagnetic four-potential is:
\( A^\mu = (\phi/c, \vec{A}) = (\phi/c, A_x, A_y, A_z) \)
where \(\phi\) is the scalar potential and \(\vec{A}\) is the vector potential. The field tensor is defined as:
\( F^{\mu\nu} = \partial^\mu A^\nu - \partial^\nu A^\mu \)
The curl generalized to 4D spacetime
This automatically satisfies the antisymmetry property \(F^{\mu\nu} = -F^{\nu\mu}\).
Explicit Components
Using \(\vec{E} = -\nabla\phi - \partial\vec{A}/\partial t\) and \(\vec{B} = \nabla \times \vec{A}\):
Electric Components
\( F^{0i} = \partial^0 A^i - \partial^i A^0 = -E_i/c \)
Time-space components encode \(\vec{E}\)
Magnetic Components
\( F^{ij} = \partial^i A^j - \partial^j A^i = -\epsilon^{ijk}B_k \)
Space-space components encode \(\vec{B}\)
The Field Tensor Fμν
In contravariant form, the electromagnetic field 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 covariant form (with both indices down) is:
\( F_{\mu\nu} = \eta_{\mu\alpha}\eta_{\nu\beta}F^{\alpha\beta} = \begin{pmatrix} 0 & E_x/c & E_y/c & E_z/c \\ -E_x/c & 0 & -B_z & B_y \\ -E_y/c & B_z & 0 & -B_x \\ -E_z/c & -B_y & B_x & 0 \end{pmatrix} \)
Key Properties
- • Antisymmetric: \(F^{\mu\nu} = -F^{\nu\mu}\) (diagonal is zero)
- • 6 independent components: 3 from \(\vec{E}\) + 3 from \(\vec{B}\)
- • Transforms as a tensor: \(F'^{\mu\nu} = \Lambda^\mu_\alpha \Lambda^\nu_\beta F^{\alpha\beta}\)
The Dual Tensor
The dual (Hodge dual) of the field tensor is defined using the Levi-Civita symbol:
\( \tilde{F}^{\mu\nu} = \frac{1}{2}\epsilon^{\mu\nu\alpha\beta}F_{\alpha\beta} \)
Explicitly, the dual tensor swaps the roles of E and B:
\( \tilde{F}^{\mu\nu} = \begin{pmatrix} 0 & -B_x & -B_y & -B_z \\ B_x & 0 & E_z/c & -E_y/c \\ B_y & -E_z/c & 0 & E_x/c \\ B_z & E_y/c & -E_x/c & 0 \end{pmatrix} \)
Duality Transformation
The dual tensor is obtained by the replacement: \(\vec{E}/c \to \vec{B}\) and \(\vec{B} \to -\vec{E}/c\)
This is related to electromagnetic duality in vacuum Maxwell equations.
Lorentz Transformation of Fields
For a boost along the x-axis with velocity v, the fields transform as:
Parallel Components (unchanged)
\( E'_x = E_x \)
\( B'_x = B_x \)
Components parallel to motion don't change
Perpendicular Components (mix)
\( E'_y = \gamma(E_y - vB_z) \)
\( E'_z = \gamma(E_z + vB_y) \)
\( B'_y = \gamma(B_y + vE_z/c^2) \)
\( B'_z = \gamma(B_z - vE_y/c^2) \)
General Transformation Formulas
For arbitrary velocity \(\vec{v}\):
\( \vec{E}' = \gamma(\vec{E} + \vec{v} \times \vec{B}) - \frac{\gamma^2}{\gamma+1}\vec{v}(\vec{v}\cdot\vec{E})/c^2 \)
\( \vec{B}' = \gamma(\vec{B} - \vec{v} \times \vec{E}/c^2) - \frac{\gamma^2}{\gamma+1}\vec{v}(\vec{v}\cdot\vec{B})/c^2 \)
Physical Example: Moving Charge
Rest frame of charge: Pure electric field \(\vec{E} = \frac{q}{4\pi\epsilon_0 r^2}\hat{r}\), no magnetic field.
Lab frame (charge moving at v):
- Electric field is enhanced perpendicular to motion by factor \(\gamma\)
- Magnetic field appears: \(\vec{B} = \vec{v} \times \vec{E}/c^2\)
- This is exactly the Biot-Savart field of a moving charge!
Electromagnetic Invariants
Two scalar quantities are invariant under Lorentz transformations:
First Invariant (Scalar)
\( \mathcal{F} = F_{\mu\nu}F^{\mu\nu} = 2\left(B^2 - \frac{E^2}{c^2}\right) \)
The sign tells us whether the field is "magnetically dominated" (\(\mathcal{F} > 0\)) or "electrically dominated" (\(\mathcal{F} < 0\)).
Second Invariant (Pseudoscalar)
\( \mathcal{G} = F_{\mu\nu}\tilde{F}^{\mu\nu} = -\frac{4}{c}\vec{E}\cdot\vec{B} \)
When \(\vec{E} \perp \vec{B}\), this is zero in all frames. If non-zero, E and B can never be perpendicular.
Physical Implications
- • If \(E > cB\) in one frame, it's true in ALL frames (electrically dominated)
- • If \(E < cB\) in one frame, it's true in ALL frames (magnetically dominated)
- • If \(\vec{E} \cdot \vec{B} = 0\) in one frame, it's true in ALL frames
- • For electromagnetic waves: \(E = cB\) and \(\vec{E} \perp \vec{B}\) (both invariants = 0)
Special Cases
Plane Wave
\(\mathcal{F} = 0\), \(\mathcal{G} = 0\)
E and B equal, perpendicular
Static Charge
\(\mathcal{F} < 0\), \(\mathcal{G} = 0\)
Pure E field
Static Magnet
\(\mathcal{F} > 0\), \(\mathcal{G} = 0\)
Pure B field
Electromagnetic Stress-Energy Tensor
The energy and momentum of the electromagnetic field are encoded in the stress-energy tensor:
\( T^{\mu\nu} = \frac{1}{\mu_0}\left(F^{\mu\alpha}F^\nu_\alpha - \frac{1}{4}\eta^{\mu\nu}F_{\alpha\beta}F^{\alpha\beta}\right) \)
Energy Density
\( T^{00} = u = \frac{1}{2}\left(\epsilon_0 E^2 + \frac{B^2}{\mu_0}\right) \)
Energy per unit volume
Poynting Vector
\( T^{0i}/c = S^i/c^2 = \frac{1}{\mu_0 c}(\vec{E} \times \vec{B})^i \)
Energy flux / momentum density
Maxwell Stress Tensor
\( T^{ij} = \epsilon_0\left(E^iE^j - \frac{1}{2}\delta^{ij}E^2\right) + \frac{1}{\mu_0}\left(B^iB^j - \frac{1}{2}\delta^{ij}B^2\right) \)
Momentum flux = force per unit area (electromagnetic pressure and tension)
Lorentz Force in Covariant Form
The Lorentz force on a charged particle takes a beautiful covariant form using the field tensor:
\( \frac{dp^\mu}{d\tau} = qF^{\mu\nu}u_\nu \)
where \(u^\mu = \gamma(c, \vec{v})\) is the four-velocity
Expanding the Components
Time Component (μ = 0)
\( \frac{d(\gamma mc)}{d\tau} = \frac{q}{c}\vec{E}\cdot\vec{v}\gamma \)
Rate of change of energy = power delivered
Space Components (μ = 1,2,3)
\( \frac{d(\gamma m\vec{v})}{d\tau} = q\gamma(\vec{E} + \vec{v}\times\vec{B}) \)
The familiar Lorentz force!
Converting from proper time \(\tau\) to coordinate time t (using \(d\tau = dt/\gamma\)), we recover the standard form:
\( \vec{F} = q(\vec{E} + \vec{v}\times\vec{B}) \)
Key Takeaways
Unification
E and B are not separate—they're components of a single tensor \(F^{\mu\nu}\). The split depends on the observer's frame.
Transformation
Under Lorentz boosts, E and B mix. A pure E field in one frame has both E and B in another.
Invariants
\(B^2 - E^2/c^2\) and \(\vec{E}\cdot\vec{B}\) are the same in all frames. They classify field configurations.
Covariance
All electromagnetic equations can be written in manifestly covariant form using \(F^{\mu\nu}\).
Learning Path: Electromagnetic Field Tensor
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