There is actually a third thing in this post, but I'm not going to list that in the title. Also, the two things in the title are separate and unrelated. Follow the jump to see it all.

The electric field is an example of a vector, or more specifically, a polar vector. This means that it transforms properly under a reflection. By contrast, the magnetic field is a pseudovector/axial vector, which means that it transforms under a reflection and then additionally flips direction. Axial vectors are typically formed as the cross products of polar vectors, while the cross product of a polar vector and an axial vector is a polar vector. (I'm going to guess that the cross product of two axial vectors is another axial vector, but I'm not sure about that at all.) In the example of the electromagnetic field, the magnetic field is typically formed as a cross product between a current density and the radial vector. Furthermore, any axial vector in 3 dimensions can be reexpressed as a 2-index antisymmetric tensor, and vice versa: using Einstein summation, $v_{ij} = \epsilon_{ijk} v_k$ and $v_i = \frac{1}{2} \epsilon_{ijk} v_{jk}$. So what would the magnetic field look like if it was constructed as a tensor rather than a vector?

The components of the electromagnetic field tensor don't make a whole lot of sense if the magnetic field is a vector, but they make more sense if the magnetic field is a tensor. In the latter case, $F_{ij} = B_{ij}$, and by antisymmetry, $F_{(0, 0)} = 0$ so the electric field is stuffed into the remainder of the tensor $F_{(i, 0)} = E_i$ imposing antisymmetry. So that's nice. But what would AP Physics-style equations look like?

Let us start by noting the definitions of the magnetic field \[ B_{ij} = \epsilon_{ijk} B_k \\ B_i = \frac{1}{2} \epsilon_{ijk} B_{jk} . \] One quick calculation is to find the tensor equivalent of $\vec{B}^2 = B_i B_i$. Noting that $B_i = \frac{1}{2} \epsilon_{ijk} B_{jk} = \frac{1}{2} \epsilon_{ilm} B_{lm}$ and noting that $\epsilon_{ijk} \epsilon_{ilm} = \delta_{jl} \delta_{km} - \delta_{jm} \delta_{kl}$, then $(\delta_{jl} \delta_{km} - \delta_{jm} \delta_{kl})(B_{jk} B_{lm}) = B_{lk} B_{lk} - B_{mk} B_{km}$. As $B_{ij} = -B_{ji}$ then this is just $2B_{ij} B_{ij}$, so \[ B_i B_i = \frac{1}{2} B_{jk} B_{jk} . \] This is useful for calculating quantities like the magnetic energy density which is proportional to $\vec{B}^2$. This also makes sense given that the electromagnetic Lagrangian density $\mathcal{L} = \frac{1}{8\pi} \left(\vec{E}^2 - \vec{B}^2 \right)$ in vector form becomes $\mathcal{L} = \frac{1}{8\pi} \left(E_i E_i - \frac{1}{2} B_{jk} B_{jk} \right)$ in terms of the magnetic field tensor, which generalizes nicely to the special relativity expression $\mathcal{L} = -\frac{1}{8\pi} \left(\frac{1}{2} F_{\mu \nu} F^{\mu \nu} \right)$.

The Biot-Savart law for the vector magnetic field for a point charge is \[ \vec{B} = \frac{q\vec{v} \times \vec{x}}{cr^3} \] where $r \equiv \sqrt{x_i x_i}$. This can be rewritten in index notation as \[ B_i = \frac{q\epsilon_{ijk} v_j x_k}{cr^3} . \] Using now that $B_{ij} = \epsilon_{ijk} B_k$ and $B_k \propto \epsilon_{klm} v_l x_m$, then as $\epsilon_{ijk} \epsilon_{klm}$ = $\delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}$, then $(\delta_{il} \delta_{jm} - \delta_{im} \delta_{jl})(v_l x_m) = v_i x_j - v_j x_i$, so \[ B_{ij} = \frac{q(v_i x_j - v_j x_i)}{cr^3} \] is the Biot-Savart law for the magnetic field tensor.

But the Biot-Savart law is not the fundamental definition of the magnetic field. In fact, it only gives the magnetic field for differential nonrelativistic charges and does not account at all for dynamic magnetic fields arising from the dynamic electric field (also known as the displacement current). Really, the Ampère and Gauss laws define the magnetic field. Let us first look at the Gauss law. In vector form, this reads \[ \nabla \cdot \vec{B} = \partial_i B_i = 0 . \] Noting the relation between the vector and the tensor and substituting that yields \[ \epsilon_{ijk} \partial_i B_{jk} = 0 . \] One interesting thing to note is that the Gauss law for the magnetic field vector deals with the divergence. With the magnetic field tensor, the Gauss law involves the Levi-Civita tensor, so the divergence of the vector now looks like the

What about the Ampère law now? In vector form, this reads \[ \nabla \times \vec{B} = \frac{1}{c} \left(4\pi \vec{J} + \frac{\partial \vec{E}}{\partial t} \right) \] and in component form this reads \[ \epsilon_{ijk} \partial_j B_k = \frac{1}{c} \left(4\pi J_i + \frac{\partial E_i}{\partial t} \right) . \] Using the fact that $B_k = \frac{1}{2} \epsilon_{klm} B_{lm}$ and that $\epsilon_{ijk} \epsilon_{klm}$ = $\delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}$, then $\epsilon_{ijk} \partial_j (\epsilon_{klm} B_{lm}) = (\delta_{il} \delta_{jm} - \delta_{im} \delta_{jl})(\partial_j B_{lm}) = \partial_j B_{ij} - \partial_j B_{ji} = 2\partial_j B_{ij}$ as $B_{ji} = -B_{ij}$, yielding \[ \partial_j B_{ij} = \frac{1}{c} \left(4\pi J_i - \frac{\partial E_i}{\partial t} \right) \] which generalizes nicely to the relativistic source-containing Maxwell equations $\partial_{\nu} F^{\mu \nu} = \frac{4\pi}{c} J^{\mu}$. Note also that as the Gauss law went from involving the divergence of a vector to the curl of a tensor, the Ampère law went from involving the curl of a vector to the divergence of a tensor; there is a nice symmetry present there.

Finally, what does the Faraday law look like? In vector form, it says \[ \nabla \times \vec{E} = -\frac{1}{c} \frac{\partial \vec{B}}{\partial t} \] and in component form it says \[ \epsilon_{ijk} \partial_j E_k = -\frac{1}{c} \frac{\partial B_i}{\partial t} . \] Substituting $B_i = \frac{1}{2} \epsilon_{ilm} B_{lm}$, and contracting both sides (kind of like taking the inner product of a quantum mechanical equation involving only kets with a bra) by $\epsilon_{ilm}$ yields that $\epsilon_{ilm} \epsilon_{ilm} = 6$ and $\epsilon_{ilm} \epsilon_{ijk} = \delta_{jl} \delta_{km} - \delta_{jm} \delta_{kl}$, so as $(\delta_{jl} \delta_{km} - \delta_{jm} \delta_{kl})(\partial_j E_k) = \partial_l E_m - \partial_m E_l$, while $\frac{6}{2} = 3$, so \[ \partial_i E_j - \partial_j E_i = -\frac{3}{c} \frac{\partial B_{ij}}{\partial t} . \] Maybe I messed up somewhere, but that factor of 3 is really weird; does it have to do with the fact that the magnetic field tensor has 2 indices each of which can now take on one of 3 values? Otherwise, though, this looks like a fairly straightforward rewrite of the Faraday law.

So what do the components of the magnetic field tensor mean? Note that $B_{xy} = B_z$, $B_{yz} = B_x$, and $B_{zx} = B_y$. Let us compare it to the angular momentum tensor. In vector form, the component $L_i$ was interpreted as rotation about the axis $\vec{e}_i$. Now, the component $L_{ij}$ can be interpreted as rotation in the $ij$-plane. What about the magnetic field? The Biot-Savart law suggests that the magnetic field component $B_{ij}$ arises from motion of a charged particle in the $ij$-plane. What about the Lorentz force? In vector form, it is \[ \vec{F} = q\left(\vec{E} + \frac{1}{c} \vec{v} \times \vec{B} \right) \] and in component form it is \[ F_i = q\left(E_i + \frac{1}{c} \epsilon_{ijk} v_j B_k \right) . \] It was said before though that $B_{ij} \equiv \epsilon_{ijk} B_k$, so \[ F_i = q\left(E_i + \frac{1}{c} v_j B_{ij} \right) \] means that a charged particle moving in $\vec{e}_j$ feeling a magnetic field $B_{ij}$ experiences a force in $\vec{e}_i$.

Switching gears now to angular momentum, the general equation for rigid body motion in component form for vector angular momentum and velocity is \[ L_a = I_{ab} \omega_b . \] Noting that $L_a = \frac{1}{2} \epsilon_{acd} L_{cd}$ and $\omega_b = \frac{1}{2} \epsilon_{bef} \omega_{ef}$, then \[ \epsilon_{acd} L_{cd} = \epsilon_{bef} I_{ab} \omega_{ef} . \] This looks a little complicated, and it also shows that the matrix representing the angular momentum tensor is not simply the matrix multiplication of the matrices representing the moment of inertia and angular velocity tensors, respectively. Furthermore, the rotational kinetic energy for vector angular velocity is \[ K = \frac{1}{2} I_{ab} \omega_a \omega_b \] so using $\omega_a = \frac{1}{2} \epsilon_{acd} \omega_{cd}$ and $\omega_b = \frac{1}{2} \epsilon_{bef} \omega_{ef}$, then \[ K = \frac{1}{2} \epsilon_{acd} \epsilon_{bef} I_{ab} \omega_{cd} \omega_{ef} . \] The product of the Levi-Civita tensors is complicated: $\epsilon_{acd} \epsilon_{bef} = \delta_{ab} (\delta_{ce} \delta_{df} - \delta_{cf} \delta_{de}) - \delta_{ae} (\delta_{bc} \delta_{df} - \delta_{cd} \delta_{bf})$ $+ \delta_{af} (\delta_{bc} \delta_{de} - \delta_{bd} \delta_{ce})$. Applying this to $I_{ab} \omega_{cd} \omega_{ef}$ yields $I_{aa} (\omega_{ed} \omega_{ed} - \omega_{fd} \omega_{df}) - I_{eb} (\omega_{bd} \omega_{ed} - \omega_{cc} \omega_{eb})$ $+ I_{fb} (\omega_{bd} \omega_{df} - \omega_{cb} \omega_{cf})$. Maybe it is a good thing after all that in 3 dimensions angular momentum and velocity can be represented as vectors rather than tensors.

The last thing I wanted to mention was another thing I had been thinking about when reading a little bit about classical field theories for my 8.06 paper. The simplest example of a classical field theory is taking $N$ coupled longitudinal harmonic oscillators for very large $N$ and approximating the system as a continuous elastic string, such that the equation of motion is no longer that of a harmonic oscillator for each individual position but is now a wave equation for the whole string. What I was wondering was whether it would be possible to reproduce nontrivial dispersion as seen for finite $N$ with periodic boundaries as well as band structures when there are two atoms in the basis. I don't think it is possible to do either of those in the continuum approximation without making the density periodically dependent on position, in which case solutions are probably no longer analytical. In either case, the issue is that the continuum approximation is typically only taken for low energies, where only the linear portion of only the acoustic band can be seen.

The electric field is an example of a vector, or more specifically, a polar vector. This means that it transforms properly under a reflection. By contrast, the magnetic field is a pseudovector/axial vector, which means that it transforms under a reflection and then additionally flips direction. Axial vectors are typically formed as the cross products of polar vectors, while the cross product of a polar vector and an axial vector is a polar vector. (I'm going to guess that the cross product of two axial vectors is another axial vector, but I'm not sure about that at all.) In the example of the electromagnetic field, the magnetic field is typically formed as a cross product between a current density and the radial vector. Furthermore, any axial vector in 3 dimensions can be reexpressed as a 2-index antisymmetric tensor, and vice versa: using Einstein summation, $v_{ij} = \epsilon_{ijk} v_k$ and $v_i = \frac{1}{2} \epsilon_{ijk} v_{jk}$. So what would the magnetic field look like if it was constructed as a tensor rather than a vector?

The components of the electromagnetic field tensor don't make a whole lot of sense if the magnetic field is a vector, but they make more sense if the magnetic field is a tensor. In the latter case, $F_{ij} = B_{ij}$, and by antisymmetry, $F_{(0, 0)} = 0$ so the electric field is stuffed into the remainder of the tensor $F_{(i, 0)} = E_i$ imposing antisymmetry. So that's nice. But what would AP Physics-style equations look like?

Let us start by noting the definitions of the magnetic field \[ B_{ij} = \epsilon_{ijk} B_k \\ B_i = \frac{1}{2} \epsilon_{ijk} B_{jk} . \] One quick calculation is to find the tensor equivalent of $\vec{B}^2 = B_i B_i$. Noting that $B_i = \frac{1}{2} \epsilon_{ijk} B_{jk} = \frac{1}{2} \epsilon_{ilm} B_{lm}$ and noting that $\epsilon_{ijk} \epsilon_{ilm} = \delta_{jl} \delta_{km} - \delta_{jm} \delta_{kl}$, then $(\delta_{jl} \delta_{km} - \delta_{jm} \delta_{kl})(B_{jk} B_{lm}) = B_{lk} B_{lk} - B_{mk} B_{km}$. As $B_{ij} = -B_{ji}$ then this is just $2B_{ij} B_{ij}$, so \[ B_i B_i = \frac{1}{2} B_{jk} B_{jk} . \] This is useful for calculating quantities like the magnetic energy density which is proportional to $\vec{B}^2$. This also makes sense given that the electromagnetic Lagrangian density $\mathcal{L} = \frac{1}{8\pi} \left(\vec{E}^2 - \vec{B}^2 \right)$ in vector form becomes $\mathcal{L} = \frac{1}{8\pi} \left(E_i E_i - \frac{1}{2} B_{jk} B_{jk} \right)$ in terms of the magnetic field tensor, which generalizes nicely to the special relativity expression $\mathcal{L} = -\frac{1}{8\pi} \left(\frac{1}{2} F_{\mu \nu} F^{\mu \nu} \right)$.

The Biot-Savart law for the vector magnetic field for a point charge is \[ \vec{B} = \frac{q\vec{v} \times \vec{x}}{cr^3} \] where $r \equiv \sqrt{x_i x_i}$. This can be rewritten in index notation as \[ B_i = \frac{q\epsilon_{ijk} v_j x_k}{cr^3} . \] Using now that $B_{ij} = \epsilon_{ijk} B_k$ and $B_k \propto \epsilon_{klm} v_l x_m$, then as $\epsilon_{ijk} \epsilon_{klm}$ = $\delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}$, then $(\delta_{il} \delta_{jm} - \delta_{im} \delta_{jl})(v_l x_m) = v_i x_j - v_j x_i$, so \[ B_{ij} = \frac{q(v_i x_j - v_j x_i)}{cr^3} \] is the Biot-Savart law for the magnetic field tensor.

But the Biot-Savart law is not the fundamental definition of the magnetic field. In fact, it only gives the magnetic field for differential nonrelativistic charges and does not account at all for dynamic magnetic fields arising from the dynamic electric field (also known as the displacement current). Really, the Ampère and Gauss laws define the magnetic field. Let us first look at the Gauss law. In vector form, this reads \[ \nabla \cdot \vec{B} = \partial_i B_i = 0 . \] Noting the relation between the vector and the tensor and substituting that yields \[ \epsilon_{ijk} \partial_i B_{jk} = 0 . \] One interesting thing to note is that the Gauss law for the magnetic field vector deals with the divergence. With the magnetic field tensor, the Gauss law involves the Levi-Civita tensor, so the divergence of the vector now looks like the

*curl*of a*tensor*which is now a**scalar**(0). Another nice thing is that this easily generalizes to relativity: the magnetic field Gauss law as well as the Faraday law, which are the source-free Maxwell equations, are written as $\epsilon_{\alpha \beta \gamma \delta} \partial^{\beta} F^{\gamma \delta} = 0$.What about the Ampère law now? In vector form, this reads \[ \nabla \times \vec{B} = \frac{1}{c} \left(4\pi \vec{J} + \frac{\partial \vec{E}}{\partial t} \right) \] and in component form this reads \[ \epsilon_{ijk} \partial_j B_k = \frac{1}{c} \left(4\pi J_i + \frac{\partial E_i}{\partial t} \right) . \] Using the fact that $B_k = \frac{1}{2} \epsilon_{klm} B_{lm}$ and that $\epsilon_{ijk} \epsilon_{klm}$ = $\delta_{il} \delta_{jm} - \delta_{im} \delta_{jl}$, then $\epsilon_{ijk} \partial_j (\epsilon_{klm} B_{lm}) = (\delta_{il} \delta_{jm} - \delta_{im} \delta_{jl})(\partial_j B_{lm}) = \partial_j B_{ij} - \partial_j B_{ji} = 2\partial_j B_{ij}$ as $B_{ji} = -B_{ij}$, yielding \[ \partial_j B_{ij} = \frac{1}{c} \left(4\pi J_i - \frac{\partial E_i}{\partial t} \right) \] which generalizes nicely to the relativistic source-containing Maxwell equations $\partial_{\nu} F^{\mu \nu} = \frac{4\pi}{c} J^{\mu}$. Note also that as the Gauss law went from involving the divergence of a vector to the curl of a tensor, the Ampère law went from involving the curl of a vector to the divergence of a tensor; there is a nice symmetry present there.

Finally, what does the Faraday law look like? In vector form, it says \[ \nabla \times \vec{E} = -\frac{1}{c} \frac{\partial \vec{B}}{\partial t} \] and in component form it says \[ \epsilon_{ijk} \partial_j E_k = -\frac{1}{c} \frac{\partial B_i}{\partial t} . \] Substituting $B_i = \frac{1}{2} \epsilon_{ilm} B_{lm}$, and contracting both sides (kind of like taking the inner product of a quantum mechanical equation involving only kets with a bra) by $\epsilon_{ilm}$ yields that $\epsilon_{ilm} \epsilon_{ilm} = 6$ and $\epsilon_{ilm} \epsilon_{ijk} = \delta_{jl} \delta_{km} - \delta_{jm} \delta_{kl}$, so as $(\delta_{jl} \delta_{km} - \delta_{jm} \delta_{kl})(\partial_j E_k) = \partial_l E_m - \partial_m E_l$, while $\frac{6}{2} = 3$, so \[ \partial_i E_j - \partial_j E_i = -\frac{3}{c} \frac{\partial B_{ij}}{\partial t} . \] Maybe I messed up somewhere, but that factor of 3 is really weird; does it have to do with the fact that the magnetic field tensor has 2 indices each of which can now take on one of 3 values? Otherwise, though, this looks like a fairly straightforward rewrite of the Faraday law.

So what do the components of the magnetic field tensor mean? Note that $B_{xy} = B_z$, $B_{yz} = B_x$, and $B_{zx} = B_y$. Let us compare it to the angular momentum tensor. In vector form, the component $L_i$ was interpreted as rotation about the axis $\vec{e}_i$. Now, the component $L_{ij}$ can be interpreted as rotation in the $ij$-plane. What about the magnetic field? The Biot-Savart law suggests that the magnetic field component $B_{ij}$ arises from motion of a charged particle in the $ij$-plane. What about the Lorentz force? In vector form, it is \[ \vec{F} = q\left(\vec{E} + \frac{1}{c} \vec{v} \times \vec{B} \right) \] and in component form it is \[ F_i = q\left(E_i + \frac{1}{c} \epsilon_{ijk} v_j B_k \right) . \] It was said before though that $B_{ij} \equiv \epsilon_{ijk} B_k$, so \[ F_i = q\left(E_i + \frac{1}{c} v_j B_{ij} \right) \] means that a charged particle moving in $\vec{e}_j$ feeling a magnetic field $B_{ij}$ experiences a force in $\vec{e}_i$.

Switching gears now to angular momentum, the general equation for rigid body motion in component form for vector angular momentum and velocity is \[ L_a = I_{ab} \omega_b . \] Noting that $L_a = \frac{1}{2} \epsilon_{acd} L_{cd}$ and $\omega_b = \frac{1}{2} \epsilon_{bef} \omega_{ef}$, then \[ \epsilon_{acd} L_{cd} = \epsilon_{bef} I_{ab} \omega_{ef} . \] This looks a little complicated, and it also shows that the matrix representing the angular momentum tensor is not simply the matrix multiplication of the matrices representing the moment of inertia and angular velocity tensors, respectively. Furthermore, the rotational kinetic energy for vector angular velocity is \[ K = \frac{1}{2} I_{ab} \omega_a \omega_b \] so using $\omega_a = \frac{1}{2} \epsilon_{acd} \omega_{cd}$ and $\omega_b = \frac{1}{2} \epsilon_{bef} \omega_{ef}$, then \[ K = \frac{1}{2} \epsilon_{acd} \epsilon_{bef} I_{ab} \omega_{cd} \omega_{ef} . \] The product of the Levi-Civita tensors is complicated: $\epsilon_{acd} \epsilon_{bef} = \delta_{ab} (\delta_{ce} \delta_{df} - \delta_{cf} \delta_{de}) - \delta_{ae} (\delta_{bc} \delta_{df} - \delta_{cd} \delta_{bf})$ $+ \delta_{af} (\delta_{bc} \delta_{de} - \delta_{bd} \delta_{ce})$. Applying this to $I_{ab} \omega_{cd} \omega_{ef}$ yields $I_{aa} (\omega_{ed} \omega_{ed} - \omega_{fd} \omega_{df}) - I_{eb} (\omega_{bd} \omega_{ed} - \omega_{cc} \omega_{eb})$ $+ I_{fb} (\omega_{bd} \omega_{df} - \omega_{cb} \omega_{cf})$. Maybe it is a good thing after all that in 3 dimensions angular momentum and velocity can be represented as vectors rather than tensors.

The last thing I wanted to mention was another thing I had been thinking about when reading a little bit about classical field theories for my 8.06 paper. The simplest example of a classical field theory is taking $N$ coupled longitudinal harmonic oscillators for very large $N$ and approximating the system as a continuous elastic string, such that the equation of motion is no longer that of a harmonic oscillator for each individual position but is now a wave equation for the whole string. What I was wondering was whether it would be possible to reproduce nontrivial dispersion as seen for finite $N$ with periodic boundaries as well as band structures when there are two atoms in the basis. I don't think it is possible to do either of those in the continuum approximation without making the density periodically dependent on position, in which case solutions are probably no longer analytical. In either case, the issue is that the continuum approximation is typically only taken for low energies, where only the linear portion of only the acoustic band can be seen.