# LOG#094. Group theory(XIV).

Group theory and the issue of mass: Majorana fermions in 2D spacetime

We have studied in the previous posts that a mass term is “forbidden” in the bivector/sixtor approach and the Dirac-like equation due to the gauge invariance. In fact,

$-i\overline{\Gamma}^\mu\partial_\mu$

as an operator has an “issue of mass” over the pure Dirac equation $i\Gamma^\mu\Psi=0$ of fermionic fields. This pure Dirac equation provides

$\overline{\Gamma_\mu}\Gamma_\nu\partial_\nu\partial_\mu\Psi=\partial_0^2\Psi+\nabla\times (\nabla\times \Psi)=\partial_0^2-\Delta\Psi+\nabla (\nabla\cdot \Psi)=0$

Therefore, $\Psi$ satisfies the wave equation

$\square^2\Psi=\partial^\mu\partial_\mu\Psi=0$

where $\square^2=\Delta-\partial_0^2$ if there are no charges or currents! If we introduce a mass by hand for the $\Psi$ field, we obtain

$(i\Gamma^\mu\partial_\mu-m )\Psi=0$

and we observe that it would spoil the relativistic invariance of the field equation!!!!!!! That is a crucial observation!!!!

A more general ansatz involves the (anti)linear operator V:

$i\Gamma^\mu\partial_\mu\Psi-mV\Psi=0$

A plane wave solution is something like an exponential function $\sim e^{\pm ipx}$ and it obeys:

$p^2=p_\mu p^\mu=-m^2$

If we square the Dirac-like equation in the next way

$i\overline{\Psi}^\nu\partial_\nu (i\Gamma^\mu\partial_\mu\Psi)=-\square \Psi=-m^2\Psi=i\overline{\Psi}^\nu\partial_\nu (mV\Psi)$

and a bivector transformation

$i\overline{\Gamma}^\mu\partial_\mu (V\Psi)-m\Psi=0$

$V(i\overline{\Gamma}^\mu\partial_\mu (v\Psi))-mV\Psi=0$

$Vi\overline{\Gamma}^\mu\partial_\mu (V\Psi)=mV\Psi=i\Gamma^\mu \partial_\mu \Psi$

from linearity we get

$Vi\overline{\Gamma}^\mu=i\Gamma^\mu$

$V^2=I_3$

$V\tilde{S}_aV=-S_a$

$V\tilde{S}_aV^{-1}=-\tilde{S}_a$

if $a=1,2,3$. But this is impossible! Why? The Lie structure constants are “stable” (or invariant) under similarity transformations. You can not change the structure constants with similarity transformations.

In fact, if V is an antilinear operator

$V=\tilde{V}\kappa=iV$ where $\kappa$ is a complex conjugation of the multiplication by the imaginary unit. Then, we would obtain

$\tilde{V}\tilde{V}^*=-I_3$

and

$\tilde{V}\tilde{S}_a^*\tilde{V}^*=\tilde{S}_a$

or equivalently

$\tilde{V}\tilde{S}_a^*\tilde{V}^{-1}=-\tilde{S}_a$

And again, this is impossible since we would obtain then

$\det (V\tilde{V}^*)=\det (V)\det (\tilde{V}^*)=\det \tilde{V}\det \tilde{V}^*>0$

and this contradicts the fact that $\det (-I_3)=-1$!!!

Remark: In 2d, with Pauli matrices defined by

$\sigma=(\sigma_1,\sigma_2,\sigma_3)$

and $\epsilon\tilde{\sigma}^*\epsilon^{-1}=-\tilde{\sigma}$

where

$\epsilon=\begin{pmatrix}0 & 1\\ -1 & 0\end{pmatrix}$

and we have

$\epsilon^2=(\eta\epsilon)(\eta\epsilon)^*=-I_2$

with

$\det (\epsilon)=\det (-I_2)$ such that the so-called Majorana equation(s) (as a generalization of the Weyl equation(s) and the Lorentz invariance in 4d) provides a 2-component field equation describing a massive particle DOES exist:

$i\sigma^\mu\partial_\mu\Psi-m\eta\epsilon\Psi^*=0$

In fact, the Majorana fermions can exist only in certain spacetime dimensions beyond the 1+1=2d example above. In 2D (or 2d) spacetime you write

$\boxed{i\sigma^\mu\partial_\mu\Psi-m\eta\epsilon\Psi^*=0}$

and it is called the Majorana equation. It describes a massive fermion that is its own antiparticle, an option that can not be possible for electrons or quarks (or their heavy “cousins” and their antiparticles). The only Standard Model particle that could be a Majorana particle are the neutrinos. There, in the above equations,

$\sigma^\mu=(I_2,\vec{\sigma})$ and $\eta$ is a “pure phase” often referred as the “Majorana phase”.

Gauge formalism and mass terms for field equations

Introduce some gauge potential like

$A=A^R+iA^I=\begin{pmatrix}A_1^R+iA_1^I\\ A_2^R+iA_2^I\\ A_3^R+iA_3^I\end{pmatrix}$

It is related to the massive bivector/sixtor field with the aid of the next equation

$\Psi=i\overline{\Gamma}^\nu\partial_\nu A=(i\partial_0+\nabla\times)(A^R+iA^I)=-\dot{A}^O+\nabla\times A^R+i(\dot{A}^R+\nabla\times A^I)$

It satisfies a massive wave equation

$\square^2A+m^2A=0$

This would mean that

$i\Gamma^\mu\partial_\mu (i\overline{\Gamma}^\nu \partial_\nu A)=(-\partial_0^2-\nabla\times\nabla\times)A=(-\partial_0^2+\Delta-\nabla (\nabla\cdot))A=- m^2A$

and then $\nabla (\nabla\cdot A)=0$. However, it would NOT BE a Lorentz invariant anymore!

Current couplings

From the Ampère’s law

$\partial_t E=\nabla\times B-j$

and where we have absorbed the multiplicative constant into the definition of the current $j$, we observe that $i\Gamma^\mu\partial_\mu\Psi$ can NOT be interpreted as the Dirac form of the Maxwell equations since $j=(j_1,j_2,j_3)$ have 3 spatial components of a charge current 4d density $J=j^\mu=(j^0,\mathbf{j})=(j^0,j^1,j^2,j^3)$ so that

$\partial_t\Psi=-i\nabla\times \Psi-\mathbf{j}$ and

$\nabla\cdot (\partial_t \Psi)=\nabla\cdot (-i\nabla\times \Psi-\mathbf{j})$

or

$\nabla\cdot \dot{\Psi}=-i\nabla\cdot (\nabla\times \Psi)-\mbox{div}\mathbf{j}=\dot{\rho}$

If the continuity equation $\dot{\rho}+\mbox{div}\mathbf{j}=0$ holds. In the absence of magnetic charges, this last equation is equivalent to $\mbox{div} (\dot{E})=\dot{\rho}$ or $\nabla\cdot E=\rho$.

Remark: Even though the bivector/sixtor field couples to the spatial part of the 4D electromagnetic current, the charge distribution is encoded in the divergence of the field $\Psi$ itself and it is NOT and independent dynamical variable as the current density (in 4D spacetime) is linked to the initial conditions for the charge distribution and it fixes the actual charge density (also known as divergence of $\Psi$ at any time; $\Psi$ is a bispinor/bivector and it is NOT a true spinor/vector).

Dirac spinors under Lorentz transformations

A Lorentz transformation is a map

$X'=\Lambda X$

A Dirac spinor field is certain “function” transforming under certain representation of the Lorentz group. In particular,

$\Psi'(x')=Q_D(\Lambda)\Psi (x)$ for every Lorentz transformation belonging to the group $SO(1,3)^+$. Moreover,

$x'_\mu=\Lambda^\mu_{\;\;\; \nu}$

and Dirac spinor fields obey the so-called Dirac equation (no interactions are assumed in this point, only “free” fields):

$i\gamma^\mu\partial_\mu\Psi-m\Psi=0$

This Dirac equation is Lorentz invariant, and it means that it also hold in the “primed”/transformed coordinates since we have

$i\gamma^\mu (\partial_{\mu '}\Psi' (x'))=i\gamma^\mu\partial_{\mu '}(Q_D\Psi (x))=mQ_D\Psi$

and

$i\gamma^\mu\Lambda^{\;\;\; \nu}_\mu\partial_\nu Q_D\Psi=mQ_D\Psi$

Using that $\Lambda^\alpha_{\;\;\; \nu}\Lambda^{\nu}_{\;\;\; \mu}=\delta^\alpha_{\;\;\; \mu}$

we get the transformation law

$\boxed{Q_D^{-1}\gamma^\alpha Q_D=\Lambda^\alpha_{\;\;\; \nu}\gamma^\nu}$

Covariant Dirac equations are invariant under Lorentz transformations IFF the transformation of the spinor components is performed with suitable chosen operators $Q_D$. In fact,

$Q^{-1}\Gamma^\alpha Q=\Lambda^\alpha_{\;\;\; \nu}\Gamma^\nu$

$Q^T\Gamma^\alpha Q=\Lambda^\alpha_{\;\;\; \nu}\Gamma\nu$

$Q^*\Gamma^\alpha Q=\Lambda^\alpha_{\;\;\; \nu}\Gamma\nu$

DOES NOT hold for $\Psi$ bispinors/bivectors. For bivector fields, you obtain

$i\Gamma^\mu\partial_\mu\Psi=-i\mathbf{j}$

and

$i\Gamma^\mu_{ab}\partial_{\mu '}\Psi '(x')=-iJ'_a (x')$

This last equation implies that

$i\Gamma_{ab}^\mu\Lambda_{\mu}^{\;\;\; \nu}\partial_\nu Q_{bc}\Psi_c(x)=-i\Lambda^a_{\;\;\; \nu}j^\nu (x)=-i\Lambda^a_{0}\mbox{div} E(x)-i\Lambda^a_ cJ^c(x)$

$j^\mu=(\mbox{div}E,j^1,j^2,j^3)=(j^0,j^1,j^2,j^3)$

with

$\mbox{div} E=\delta^\nu_c\delta_\nu\Psi_c$ since $\mbox{div} B=\nabla\cdot B=0$ because there are no magnetic monopoles.

If $\tilde{\Lambda}$ is the inverse of the 3d matric $\Lambda^a_{\;\;\; b}$, then we have

$\tilde{\Lambda}^d_a\Lambda^a_b=\delta^d_b$

In this case, we obtain that

$i (\tilde{\Lambda}^d_a\Gamma^\mu_{ab}\Lambda^\nu_\mu Q_{bc}+\tilde{\Lambda}^d_a\Lambda^a_0\delta^\nu_c)\partial_\nu\Psi_c (x)=-i\tilde{\Lambda}^d_a\Lambda^a_cJ^c=-ij^d$

so

$\Gamma^\nu_{dc}=\tilde{\Lambda}^d_a\Gamma^\mu_{ab}\Lambda_{\mu}^\nu Q_{bc}+\tilde{\Lambda}^d_a\Lambda^a_0\delta^\nu_c$

That is, for rotations we obtain that

$\Lambda^a_{\;\;\; b}=Q_{ab}$ $\tilde{\Lambda}^{-1}=Q^{-1}$ $\Lambda^a_{\;\;\; 0}\;\;\forall a=1,2,3$

and so

$\boxed{\Gamma^\nu=\Lambda_{\mu}^{\;\;\; \nu}Q^{-1}\Gamma^\mu Q}$

This means that for the case of pure rotations both bivector/bispinors and current densities transform as vectors under the group SO(3)!!!!

Conclusions of this blog post:

1st. A mass term analogue to the Marjorana or Dirac spinor equation does NOT arise in 4d electromagnetism due to the interplay of relativistic invariance and gauge symmetries. That is, bivector/bispinor fields in D=4 can NOT contain mass terms for group theoretical reasons: Lorentz invariance plus gauge symmetry.

2nd. The Dirac-like equation $i\Gamma^\mu \partial_\mu \Psi=0$ can NOT be interpreted as a Dirac equation in D=4 due to relativistic symmetry, but you can write that equation at formal level. However, you must be careful with the meaning of this equation!

3rd. In D=2 and other higher dimensions, Majorana “mass” terms arise and you can write a “Majorana mass” term without spoiling relativistic or gauge symmetries. Majorana fermions are particles that are their own antiparticles! Then, only neutrinos can be Majorana fermions in the SM (charged fermions can not be Majorana particles for technical reasons).

4th. The sixtor/bivector/bispinor formalism with $F=E+iB$ has certain uses. For instance, it is used in the so-called Ungar’s formalism of special relativity, it helps to remember the electromagnetic main invariants and the most simple transformations between electric and magnetic fields, even with the most general non-parallel Lorentz transformation.

# LOG#033. Electromagnetism in SR.

The Maxwell’s equations and the electromagnetism phenomena are one of the highest achievements and discoveries of the human kind. Thanks to it, we had radio waves, microwaves, electricity, the telephone, the telegraph, TV, electronics, computers, cell-phones, and internet. Electromagnetic waves are everywhere and everytime (as far as we know, with the permission of the dark matter and dark energy problems of Cosmology). Would you survive without electricity today?

The language used in the formulation of Maxwell equations has changed a lot since Maxwell treatise on Electromagnetis, in which he used the quaternions. You can see the evolution of the Mawell equations “portrait” with the above picture. Today, from the mid 20th centure, we can write Maxwell equations into a two single equations. However, it is less know that Maxwell equations can be written as a single equation $\nabla F=J$ using geometric algebra in Clifford spaces, with $\nabla =\nabla \cdot +\nabla\wedge$, or the so-called Kähler-Dirac-Clifford formalism in an analogue way.

Before entering into the details of electromagnetic fields, let me give some easy notions of tensor calculus. If $x^2=\mbox{invariant}$, how does $x^\mu$ transform under Lorentz transformations? Let me start with the tensor components in this way:

$x^\mu e_\mu=x^{\mu'}e_{\mu'}=\Lambda^{\mu'}_{\;\; \nu}x^\mu e_{\mu'}=\Lambda^{\mu'}_{\;\; \mu}x^\mu e_{\mu'}$

Then:

$e_\mu=\Lambda^{\mu'}_{\;\; \mu} e_{\mu'}\rightarrow e_{\mu'}=\left(\Lambda^{-1}\right)_{\;\; \mu'}^{\mu}e_\mu=\left[\left(\Lambda^{-1}\right)^T\right]^{\;\; \mu}_{\nu}e_\mu$

Note, we have used with caution:

1st. Einstein’s convention: sum over repeated subindices and superindices is understood, unless it is stated some exception.

2nd. Free indices can be labelled to the taste of the user segment.

3rd. Careful matrix type manipulations.

We define a contravariant vector (or tensor (1,0) ) as some object transforming in the next way:

$\boxed{a^{\mu'}=\Lambda^{\mu'}_{\;\; \nu}a^\nu}\leftrightarrow\boxed{a^{\mu'}=\left(\dfrac{\partial x^{\mu'}}{\partial x^\nu}\right)a^\nu}$

where $\left(\dfrac{\partial x^{\mu'}}{\partial x^\nu}\right)$ denotes the Jabobian matrix of the transformation.
In similar way, we can define a covariant vector ( or tensor (0,1) ) with the aid of the following equations

$\boxed{a_{\mu'}=\left[\left(\Lambda^{-1}\right)^{T}\right]_{\mu'}^{\:\;\; \nu}a_\nu}\leftrightarrow\boxed{a_{\mu'}=\left(\dfrac{\partial x^{\nu}}{\partial x^{\mu'}}\right)a_\nu}$

Note: $\left(\dfrac{\partial x^{\nu}}{\partial x^{\mu'}}\right)=\left(\dfrac{\partial x^{\mu'}}{\partial x^{\nu}}\right)^{-1}$

Contravariant tensors of second order ( tensors type (2,0)) are defined with the next equations:

$\boxed{b^{\mu'\nu'}=\Lambda^{\mu'}_{\;\; \lambda}\Lambda^{\nu'}_{\;\; \sigma}b^{\lambda\sigma}=\Lambda^{\mu'}_{\;\; \lambda}b^{\lambda\sigma}\Lambda^{T \;\; \nu'}_{\sigma}\leftrightarrow b^{\mu'\nu'}=\dfrac{\partial x^{\mu'}}{\partial x^\lambda}\dfrac{\partial x^{\nu'}}{\partial x^\sigma}b^{\lambda\sigma}}$

Covariant tensors of second order ( tensors type (0,2)) are defined similarly:

$\boxed{c_{\mu'\nu'}=\left(\left(\Lambda\right)^{-1}\right)^{T \;\;\lambda}_{\mu'}\left(\left(\Lambda\right)^{-1T}\right)^{\;\; \sigma}_{\nu'}c_{\lambda\sigma}=\left(\Lambda^{-1T}\right)^{\;\; \lambda}_{\mu'}c_{\lambda\sigma}\Lambda^{-1 \;\; \nu'}_{\sigma}\leftrightarrow c_{\mu'\nu'}=\dfrac{\partial x^{\lambda}}{\partial x^{\mu'}}\dfrac{\partial x^{\sigma}}{\partial x^{\nu'}}c_{\lambda\sigma}}$

Mixed tensors of second order (tensors type (1,1)) can be also made:

$\boxed{d^{\mu'}_{\;\; \nu'}=\Lambda^{\mu'}_{\;\; \lambda}\left(\left(\Lambda\right)^{-1T}\right)^{\;\;\;\; \sigma}_{\nu'}d^{\lambda}_{\;\;\sigma}=\Lambda^{\mu'}_{\;\; \lambda}d^{\lambda}_{\;\;\sigma}\left(\left(\Lambda\right)^{-1}\right)^{\sigma}_{\;\;\; \nu'}\leftrightarrow d^{\mu'}_{\;\; \nu'}=\dfrac{\partial x^{\mu'}}{\partial x^{\lambda}}\dfrac{\partial x^{\sigma}}{\partial x^{\nu'}}d^{\lambda}_{\;\; \sigma}}$

We can summarize these transformations rules in matrix notation making the transcript from the index notation easily:

1st. Contravariant vectors change of coordinates rule: $X'=\Lambda X$

2nd. Covariant vectors change of coordinates rule: $X'=\Lambda^{-1T} X$

3rd. (2,0)-tensors change of coordinates rule: $B'=\Lambda B \Lambda^T$

4rd. (0,2)-tensors change of coordinates rule: $C'=\Lambda^{-1T}C\Lambda^{-1}$

5th. (1,1)-tensors change of coordinates rule: $D'=\Lambda D \Lambda^{-1}$

Indeed, without taking care with subindices and superindices, and the issue of the inverse and transpose for transformation matrices, a general tensor type (r,s) is defined as follows:

$\boxed{T^{\mu'_1\mu'_2\ldots \mu'_r}_{\nu'_1\nu'_2\ldots \nu'_s}=L^{\nu_s}_{\nu'_s}\cdots L^{\nu_1}_{\nu'_1}L^{\mu'_r}_{\mu_r}\cdots L^{\mu'_1}_{\mu_1}T^{\mu_1\mu_2\ldots\mu_r}_{\nu_1\nu_2\ldots \nu_s}}$

We return to electromagnetism! The easiest examples of electromagnetic wave motion are plane waves:

$x=x_0\exp (iKX)=x_0\exp (ix^\mu p_\mu)$

where $\phi=XK=KX=X\cdot K=x^\mu p_\mu=\mathbf{k}\cdot\mathbf{r}-\omega t$

Indeed, the cuadrivector K can be “guessed” from the phase invariant ($\phi=\phi'$ since the phase is a dot product):

$K=\square \phi$

where $\square$ is the four dimensional nabla vector defined by

$\square=\left(\dfrac{\partial}{c\partial t},\dfrac{\partial}{\partial x},\dfrac{\partial}{\partial y},\dfrac{\partial}{\partial z}\right)$

and so

$K^\mu=(K^0,K^1,K^2,K^3)=(\omega/c,k_x,k_y,k_z)$

Now, let me discuss different notions of velocity when we are considering electromagnetic fields, beyond the usual notions of particle velocity and observer relative motion, we have the following notions of velocity in relativistic electromagnetism:

1st. The light speed c. It is the ultimate limit in vacuum and SR to the propagation of electromagnetic signals. Therefore, it is sometimes called energy transfer velocity in vacuum or vacuum speed of light.

2nd. Phase velocity $v_{ph}$. It is defined as the velocity of the modulated signal in a plane wave, if $\omega =\omega (k)=\sqrt{c^2\mathbf{k}-K^2}$, we have

$v_{ph}=\dfrac{\omega (\mathbf{k})}{k}$ where k is the modulus of $\mathbf{k}$. It measures how much fast the phase changes with the wavelength vector.

From the definition of cuadrivector wave length, we deduce:

$K^2=\omega^2\left(\dfrac{1}{v_p}-\dfrac{1}{c^2}\right)$

Then, we can rewrite the distinguish three cases according to the sign of the invariant $K^2$:

a) $K^2>0$. The separation is spacelike and we get $v_p.

b)$K^2=0$. The separation is lightlike or isotropic. We obtain $v_p=c$.

c)$K^2<0$. The separation is timelike. We deduce that $v_p>c$. This situation is not contradictory with special relativity since phase oscillations can not transport information.

3rd. Group velocity $v_g$. It is defined like the velocity that a “wave packet” or “pulse” has in its propagation. Therefore,

$v_g=\dfrac{dE}{dp}=\dfrac{d\omega}{dk}$

where we used the Planck relationships for photons $E=\hbar \omega$ and $p=\hbar k$, with $\hbar=\dfrac{h}{2\pi}$

4th. Particle velocity. It is defined in SR by the cuadrivector $U=\gamma (c,\mathbf{v})$

5th. Observer relative velocity, V. It is the velocity (constant) at which two inertial observes move.

There is a nice relationship between the group velocity, the phase velocity and the energy transfer, the lightspeed in vacuum. To see it, look at the invariant:

$K^2=\mathbf{k}^2-\omega^2/c^2$

Deriving this expression, we get $v_g=d\omega/dk=kc^2/\omega=c^2/v_{ph}$

so we have the very important equation

$\boxed{v_gv_{ph}=c^2}$

Other important concept in electromagnetism is “light intensity”. Light intensity can be thought like the “flux of light”, and you can imagine it both in the wave or particle (photon corpuscles) theory in a similar fashion. Mathematically speaking:

$\mbox{Light intensity=Flux of light}=\dfrac{\mbox{POWER}}{\mbox{Area}}\rightarrow I=\dfrac{\mathcal{P}}{A}=\dfrac{E/V}{tA/V}=\dfrac{uV}{tA}=uc$

so $I=uc$ where u is the energy density of the electromagnetic field and c is the light speed in vacuum. By Lorentz transformations, it can be showed that for electromagnetic waves, energy, wavelength, energy density and intensity change in the following way:

$E'=\sqrt{\dfrac{1+\beta}{1-\beta}}E$

$\lambda'=\sqrt{\dfrac{1-\beta}{1+\beta}}\lambda$

$u'= \dfrac{E'}{\lambda' NA}=\dfrac{1-\beta}{1+\beta}\dfrac{E}{N\lambda A}$

$I'=\dfrac{1-\beta}{1+\beta}I$

The relativistic momentum can be related to the wavelength cuadrivector using the Planck relation $P^\mu=\hbar K^\mu$. Under a Lorentz transformation, momenergy transforms $P'=\Lambda P$. Assign to the wave number vector $\mathbf{k}$ a direction in the S-frame:

$\vert \mathbf{k}\vert \left( \cos \theta, \sin \theta, 0 \right)=\dfrac{\omega}{c}\left(\cos\theta,\sin\theta,0\right)$

and then

$K^\mu=\dfrac{\omega}{c}\left(1,\cos\theta,\sin\theta,0\right)$

In matrix notation, the whole change is written as:

$\begin{pmatrix}\dfrac{\omega'}{c}\\ k'_x\\ k'_y\\k'_z\end{pmatrix}=\begin{pmatrix}\gamma & -\beta\gamma & 0 & 0\\ -\beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0& 1\end{pmatrix}\dfrac{\omega}{c}\begin{pmatrix}1\\ \cos\theta\\ \sin\theta\\ 0\end{pmatrix}$

so

$K'\begin{cases}\omega'=\gamma \omega(1-\beta\cos\theta)\\ \;\\ k'_x=\gamma\dfrac{\omega}{c}\\ \;\\ k'_y=\dfrac{\omega}{c}\sin\theta\\\;\\ k'_z=0\end{cases}$

Using the first two equations, we get:

$k'_x=\dfrac{\omega'}{c}\dfrac{\cos\theta-\beta}{1-\beta\cos\theta}$

Using the first and the third equation, we obtain:

$k'_y=\dfrac{\omega'}{c}\dfrac{\sin\theta}{\gamma\left(1-\beta\cos\theta\right)}$

Dividing the last two equations, we deduce:

$\dfrac{k'_x}{k'_y}=\dfrac{\sin\theta}{\gamma\left(\cos\theta-\beta\right)}=\dfrac{u'_y}{u'_x}=\tan\theta'$

This formula is the so-called stellar aberration formula, and we will dedicate it a post in the future.

If we write the first equation with the aid of frequency f (and $f_0$) instead of angular frequency,

$f=f_0\dfrac{1}{\gamma(1-\beta\cos\theta)}$

where we wrote the frequency of the source as $\omega'=2\pi f_0$ and the frequency of the receiver as $\omega=2\pi \nu$. This last formula is called the relativistic Doppler shift.

Now, we are going to introduce a very important object in electromagnetism: the electric charge and the electric current. We are going to make an analogy with the momenergy $\mathbb{P}=m\gamma\left(c,\mathbf{v}\right)$. The cuadrivector electric current is something very similar:

$\mathbb{J}=\rho_0\gamma\left(c,\mathbf{u}\right)=\rho\left(c,\mathbf{u}\right)=\left(\rho c,\mathbf{j}\right)$

where $\rho=\gamma \rho_0$ is the electric current density, and $\mathbf{u}$ is the charge velocity. Moreover, $\rho_0=nq$ and where $q$ is the electric charge and $n=N/V$ is the electric charge density number, i.e., the number of “elementary” charges in certain volume. Indeed, we can identify the components of such a cuadrivector:

$\mathbb{J}=\left(J^0,J^1,J^2,J^3\right)=\rho_0\left(c\gamma,\gamma\mathbf{v}\right)=\rho_0\gamma\left(c,\mathbf{u}\right)$. We can make some interesting observations. Suppose certain rest frame S where we have $\rho=\rho_++\rho_-=0$, i.e., a frame with equilibred charges $\rho_+=-\rho_-$, and suppose we move with the relative velocity of the electron (or negative charge) observer. Then $u=v(e)$ and $j_x=\rho_-v$, while the other components are $j_y=j_z=0$. Then, the charge density current transforms as follows:

$\begin{pmatrix}\rho'c\\ j'_x\\ j'_y\\j'_z\end{pmatrix}=\begin{pmatrix}\gamma & -\beta\gamma & 0 & 0\\ -\beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0& 1\end{pmatrix}\begin{pmatrix}0\\ \rho_- v\\ 0\\ 0\end{pmatrix}=\begin{pmatrix}-\gamma \beta \rho_- v\\ \gamma \rho_- v\\ 0\\ 0\end{pmatrix}$

and

$\rho'=-\gamma \beta^2\rho_-=\gamma\beta^2\rho_+$

$j'_x=\gamma\rho_- v=-\gamma \rho_+ v$

We conclude:

1st. Length contraction implies that the charge density increases by a gamma factor, i.e., $\rho_+\rightarrow \rho_+\gamma$.

2nd. The crystal lattice “hole” velocity $-v$ in the primed frame implies the existence in that frame of a current density $j'_x=-\gamma \rho_+ v$.

3rd. The existence of charges in motion when seen from an inertial frame (boosted from a rest reference S) implies that in a moving reference frame electric fields are not alone but with magnetic fields. From this perspective, magnetic fields are associated to the existence of moving charges. That is, electric fields and magnetic fields are intimately connected and they are caused by static and moving charges, as we do know from classical non-relativistic physics.

Remember now the general expression of the FORPOWER tetravector, or Power-Force tetravector, in SR:

$\mathcal{F}=\mathcal{F}^\mu e_\mu=\gamma\left(\dfrac{\mathbf{f}\cdot\mathbf{v}}{c},f_x,f_y,f_z\right)$

and using the metric, with the mainly plus convention, we get the covariant componets for the power-force tetravector:

$\mathcal{F}_\mu=\gamma\left(-\dfrac{\mathbf{f}\cdot\mathbf{v}}{c},f_x,f_y,f_z\right)$

We define the Lorentz force as the sum of the electric and magnetic forces

$\mathbf{f}_L=\mathbf{f}_e+\mathbf{f}_m=q\mathbf{E}+\mathbf{v}\times \mathbf{B}$

Noting that $(\mathbf{v}\times\mathbf{B})\cdot \mathbf{v}=0$, the Power-Force tetravector for the Lorentz electromagnetic force reads:

$\mathcal{F}_L=\mathcal{F}^\mu e_\mu=\gamma q\left(\dfrac{\mathbf{E}\cdot{\mathbf{v}}}{c},\mathbf{E}+\mathbf{v}\times\mathbf{B}\right)$

And now, we realize that we can understand the electromagnetic force in terms of a tensor (1,1), i.e., a matrix, if we write:

$\mathcal{F}=\dfrac{q}{c}\mathbb{F}\mathbb{U}$

so

$\begin{pmatrix}\mathcal{F}^0\\ \mathcal{F}^1\\ \mathcal{F}^2\\ \mathcal{F}^3\end{pmatrix}=\dfrac{q}{c}\begin{pmatrix}0 &E_x & E_y & E_z\\ E_x & 0 & cB_z& -cB_y\\ E_y & -cB_z & 0 & cB_x\\ E_z & cB_y& -cB_x& 0\end{pmatrix}\begin{pmatrix}\gamma c\\ \gamma v_x\\ \gamma v_y\\ \gamma v_z\end{pmatrix}$

Therefore, $\mathcal{F}^\mu=\dfrac{q}{c}F^\mu_{\;\; \nu}U^\nu\leftrightarrow \mathcal{F}=\dfrac{q}{c}\mathbb{F}\mathbb{U}$

where the components of the (1,1) tensor can be read:

$\mathbb{F}=\mathbf{F}^\mu _{\;\; \nu}=\begin{pmatrix}0& E_x& E_y& E_z\\ E_x & 0 & cB_z& -cB_y\\ E_y& -cB_z& 0 & cB_x\\ E_z & cB_y& -cB_x& 0\end{pmatrix}$

We can lower the indices with the metric $\eta=diag(-1,1,1,1)$ in order to have a more “natural” equation and to read the symmetry of the electromagnetic tensor $F_{\mu\nu}$ (note that we can not study symmetries with indices covariant and contravariant),

$\mathbf{F}_{\mu\nu}=\eta_{\mu \alpha}\mathbf{F}^{\alpha}_{\;\; \nu}$

with

$\mathbf{F}_{\mu\nu}=\begin{pmatrix}0& -E_x& -E_y& -E_z\\ E_x & 0 & cB_z& -cB_y\\ E_y& -cB_z& 0 & cB_x\\ E_z & cB_y& -cB_x& 0\end{pmatrix}$

Similarly

$\mathbf{F}^{\mu\nu}=\mathbf{F}^{\alpha}_{\;\; \beta}\eta^{\beta \nu}=\begin{pmatrix}0& E_x& E_y& E_z\\ -E_x & 0 & cB_z& -cB_y\\ -E_y& -cB_z& 0 & cB_x\\ -E_z & cB_y& -cB_x& 0\end{pmatrix}$

Please, note that $F_{\mu\nu}=-F_{\nu\mu}$. Focusing on the components of the electromagnetic tensor as a tensor type (1,1), we have seen that under Lorentz transformations its components change as $F'=LFL^{-1}$ under a boost with $\mathbf{v}=\left(v,0,0\right)$ in such a case. So, we write:

$\boxed{F'=\begin{pmatrix}\gamma & -\beta\gamma & 0 & 0\\ -\beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0& 1\end{pmatrix}\begin{pmatrix}0& E_x& E_y& E_z\\ E_x & 0 & cB_z& -cB_y\\ E_y& -cB_z& 0 & cB_x\\ E_z & cB_y& -cB_x& 0\end{pmatrix}\begin{pmatrix}\gamma & \beta\gamma & 0 & 0\\ \beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0& 1\end{pmatrix}}$

$\boxed{F'=\mathbf{F}^{\mu'}_{\;\; \nu'}=\begin{pmatrix}0& E_x& \gamma (E_y-vB_z)& \gamma (E_z+vB_y)\\ E_x & 0 & c\gamma(B_z-\frac{v}{c^2}E_y) & -c\gamma (B_y+\frac{v}{c^2}E_z)\\ \gamma (E_y-vB_z)& -c\gamma (B_z-\frac{v}{c^2}E_y) & 0 & cB_x\\ \gamma (E_z+vB_y) & c\gamma (B_y+\frac{v}{c^2}E_z)& -cB_x& 0\end{pmatrix}}$

From this equation we deduce that:

$\mbox{EM fields after a boost}\begin{cases}E_{x'}=E_x,\; \; E_{y'}=\gamma \left( E_y-vB_z\right),\;\; E_{z'}=\gamma \left(E_z+vB_y\right)\\ B_{x'}=B_x,\;\; B_{y'}=\gamma \left(B_y+\frac{v}{c^2}E_z\right),\;\;B_{z'}=\gamma \left(B_z-\frac{v}{c^2}E_y\right)\end{cases}$

Example: In the S-frame we have the fields $E=(0,0,0)$ and $B=(0,B_y,0)$. The Coulomb force is $f_C=qE=(0,0,0)$ and the Lorentz force is $f_L=(0,0,qvB_y)$. How are these fields seen from the S’-frame? It is easy using the above transformations. We obtain that

$E'=(0,0,\gamma vB_y)$, $B'=(0,\gamma B_y,0)$, $f'_C=qE'=(0,0,\gamma qvB_y)$, $f'_L=(0,0,0)$

Surprinsingly, or not, the S’-observer sees a boosted electric field (non null!), a boosted magnetic field,  a boosted non-null Coulomb force and a null Lorentz force!

We can generalize the above transformations to the case of a general velocity in 3d-space $\mathbf{v}=(v_x,v_y,v_z)$

$\mathbf{E}_{\parallel'}=\mathbf{E}_\parallel$ $\mathbf{B}_{\parallel'}=\mathbf{B}_{\parallel}$

$\mathbf{E}_{\perp'}=\gamma \left[\mathbf{E}_\perp+(\mathbf{v}\times \mathbf{B})_\perp\right]=\gamma \left[\mathbf{E}_\perp+(\mathbf{v}\times \mathbf{B})\right]$

$\mathbf{B}_{\perp'}=\gamma \left[\mathbf{B}_\perp-\dfrac{1}{c^2}(\mathbf{v}\times \mathbf{E})_\perp\right]=\gamma \left[\mathbf{B}_\perp-\dfrac{1}{c^2}(\mathbf{v}\times \mathbf{E})\right]$

The last equal in the last two equations is due to the orthogonality of the position vector to the velocity in 3d space due to the cross product. From these equations, we easily obtain:

$E_\parallel=\dfrac{(v\cdot E)v}{v^2}=\dfrac{\beta\cdot E}{\beta^2}$

$E_\perp=E-E_\parallel=E-\dfrac{(v\cdot E)v}{v^2}$

and similarly with the magnetic field. The final tranformations we obtain are:

$\boxed{E'=E_{\parallel'}+E_{\perp'}=\dfrac{(v\cdot E)v}{v^2}+\gamma \left[ E-\dfrac{(v\cdot E)v}{v^2}+v\times B\right]}$

$\boxed{B'=B_{\parallel'}+B_{\perp'}=\dfrac{(v\cdot B)v}{v^2}+\gamma \left[ B-\dfrac{(v\cdot B)v}{v^2}-v\times E\right]}$

Equivalently

$\boxed{E'=\gamma \left(E+v\times B\right)-\left(\gamma-1\right)\dfrac{\left(v\cdot E\right) v}{v^2}}$

$\boxed{B'=\gamma \left(B-v\times \dfrac{E}{c^2}\right)-\left(\gamma-1\right)\dfrac{\left(v\cdot B\right) v}{v^2}}$

In the limit where $c\rightarrow \infty$ or $\dfrac{v}{c}\rightarrow 0$, we get that

$E'=E+v\times B$ $B'=B-\dfrac{v\times E}{c^2}$

There are two invariants for electromagnetic fields:

$I_1=\mathbf{E}\cdot\mathbf{B}$ and $I_2=\mathbf{E}^2-c^2\mathbf{B}^2$

It can be checked that

$\mathbf{E}\cdot\mathbf{B}=\mathbf{E}'\cdot\mathbf{B}'=invariant$

and

$\mathbf{E}^2-c^2\mathbf{B}^2=\mathbf{E'}^2-c^2\mathbf{B'}^2=invariant$  under Lorentz transformations. It is obvious since, up to a multiplicative constant,

$I_1=\dfrac{1}{4} F^\star_{\mu\nu}F^{\mu\nu}=\dfrac{1}{8}\epsilon_{\mu\nu\sigma \tau}F^{\sigma \tau}F^{\mu\nu}=\dfrac{1}{2}tr \left(F^{\star T}F\right)$

$I_2=\dfrac{1}{2}F_{\mu\nu}F^{\mu\nu}=tr\left(F^TF\right)$

and where we have defined the dual electromagnetic field as

$\star F=F^\star_{\mu\nu}=\dfrac{1}{2}\epsilon_{\mu\nu \sigma \tau}F^{\sigma \tau}$

or if we write it in components ( duality sends $\mathbf{E}$ to $\mathbf{B}$ and $\mathbf{B}$ to $-\mathbf{E}$)

$\star F=F^\star_{\mu\nu}=\begin{pmatrix}0& -B_x& -B_y& -B_z\\ B_x & 0 & -cE_z& cE_y\\ B_y& cE_z& 0 & -cE_x\\ B_z & -cE_y& cE_x& 0\end{pmatrix}$

We can guess some consequences for the electromagnetic invariants:

1st. If $E\perp B$, then $E\cdot B=0$ and thus $E_\perp B$ in every frame! This fact is important, since it shows that plane waves are orthogonal in any frame in SR. It is also the case of electromagnetic radiation.

2nd. As $E\cdot B=\vert E\vert \vert B\vert \cos \varphi$ can be in the non-orthogonal case either positive or negative. If $E\cdot B$ is positive, then it will be positive in any frame and similarly in the negative case. Morevoer, a transformation into a frame with $E=0$ (null electric field) and/or $B=0$ (null magnetic field) is impossible. That is, if a Lorentz transformation of the electric field or the magnetic field turns it to zero, it means that the electric field and magnetic field are orthogonal.

3rd. If E=cB, i.e., if $E^2-c^2B^2=0$, then it is valid in every frame.

4th. If there is a electric field but there is no magnetic field B in S, a Lorentz transformation to a pure B’ in S’ is impossible and viceversa.

5th. If the electric field is such that $E>cB$ or $E, then they can be turned in a pure electric or magnetic frame only if the electric field and the magnetic field are orthogonal.

6th. There is a trick to remember the two invariants. It is due to Riemann. We can build the hexadimensional vector( six-vector, or sixtor) and complex valued entity

$\boxed{\mathbf{F}=\mathbf{E}+ic\mathbf{B}}$

The two invariants are easily obtained squaring F:

$F^2=\mathbf{E}^2-c^2\mathbf{B}^2+2ic\mathbf{E}\cdot\mathbf{B}=invariant$

We can introduce now a vector potencial tetravector:

$\mathbb{A}=A^\mu e_\mu=\left( A^0,A^1,A^2,A^3\right)=\left(\dfrac{V}{c},\mathbf{A}\right)=\left(\dfrac{V}{c},A_x,A_y,A_z\right)$

This tetravector is also called gauge field. We can write the Maxwell tensor in terms of this field:

$F_{\mu\nu}=\partial_\mu A_\nu-\partial _\nu A_\mu$

It can be easily probed that, up to a multiplicative constant in front of the electric current tetravector, the first set of Maxwell equations are:

$\boxed{\partial_\mu F^{\mu\nu}=j^\nu \leftrightarrow \square \cdot \mathbf{F}=\mathbb{J}}$

The second set of Maxwell equations (sometimes called Bianchi identities) can be written as follows:

$\boxed{\partial_\mu F^{\star \mu \nu}=\dfrac{1}{2}\epsilon^{\nu\mu\alpha\beta}\partial_\mu F_{\alpha\beta}=0}$

The Maxwell equations are invariant under the gauge transformations in spacetime:

$\boxed{A^{\mu'}=A^\mu+e\partial^\mu \Psi}$

where the potential tetravector and the function $\Psi$ are arbitrary functions of the spacetime.

Some elections of gauge are common in the solution of electromagnetic problems:

A) Lorentz gauge: $\square \cdot A=\partial_\mu A^\mu=0$

B) Coulomb gauge: $\nabla \cdot \mathbf{A}=0$

C) Temporal gauge: $A^0=V/c=0$

If we use the Lorentz gauge, and the Maxwell equations without sources, we deduce that the vector potential components satisfy the wave equation, i.e.,

$\boxed{\square^2 A^\mu=0 \leftrightarrow \square^2 \mathbb{A}=0}$

Finally, let me point out an important thing about Maxwell equations. Specifically, about its invariance group. It is known that Maxwell equations are invariant under Lorentz transformations, and it was the guide Einstein used to extend galilean relativity to the case of electromagnetic fields, enlarging the mechanical concepts. But, the larger group leaving invariant the Maxwell equation’s invariant is not the Lorentz group but the conformal group. But it is another story unrelated to this post.