Uniqueness of relations connecting EM field values in two inertial frames In Einstein's 1905 paper, he has derived the relation connecting the EM field in two inertial frames.
It can be seen in this link http://hermes.ffn.ub.es/luisnavarro/nuevo_maletin/Einstein_1905_relativity.pdf
and also in the third answer of this question. Einstein has shown that the relations satisfy all the conditions. Are these solutions unique ? Can it be proven? How?
 A: These are the inputs that Einstein used:

*

*The definition of Lorentz transformations.


*Maxwell's equations with no charges or currents.


*The assumption that Maxwell's equations should have the same form after a Lorentz transformation.


*The assumption that the transformation of the fields is linear in the fields.
Einstein didn't state all of these inputs quite so explicitly, but he did use all of them. (He used the 4th one implicitly, near the end of his analysis.) The question asks whether these inputs uniquely determine how the fields are affected by Lorentz transformations.
 Notation 
To make the analysis less monotonous, I'll write things a little differently than Einstein did: instead of using different letters to denote the different coordinates, I'll use an index. A Lorentz transformation is a coordinate transformation, so we'll be dealing with two coordinate systems (one original, and one transformed). Let $x^0,x^1,x^2,x^3$ denote the four cooridnates in one system, and let $\bar x^0,\bar x^1,\bar x^2,\bar x^3$ denote the four cooridnates in the other system. I'll use these abbreviations for partial derivatives:
$$
\newcommand{\pl}{\partial}
 \pl_a\equiv \frac{\pl}{\pl x^a}
\hspace{2cm}
 \bar\pl_b\equiv \frac{\pl}{\pl \bar x^b},
$$
and I'll use this abbreviation for the derivatives of the new coordinates $\bar x$ with respect to the original coordinates $x$:
$$
 \Lambda_a^b\equiv \pl_a\bar x^b.
$$
According to Einstein's definition (see the footnote on page 8, which used a different notation), a Lorentz transformation is a coordinate transformation for which
$$
 \sum_{a,b}\eta^{ab}\Lambda_a^c\Lambda_b^d = \eta^{cd}
\tag{1}
$$
with
$$
 \eta^{ab}=
 \begin{cases}
  1 & \text{ if }a=b=0, \\
  -1 & \text{ if }a=b\neq 0, \\
  0 & \text{ if }a\neq b.
 \end{cases}
\tag{2}
$$
(I'm using units where the speed of light is equal to $1$.) The fact that (1) really is equivalent to Einstein's definition is demonstrated in Qmechanic's answer to another question.
Einstein used different letters for the different components of the electric and magnetic fields. I'll use indices instead. The study of Lorentz transforms is easier if we arrange the components of the electric and magnetic fields into a single two-index quantity called the Faraday tensor, denoted $F_{ab}$. By definition, $F_{ab}=-F_{ba}$, and:

*

*The three components of the electric field are the components of $F_{ab}$ with one index equal to $0$.


*The three components of the magnetic field are the components of $F_{ab}$ with no index equal to $0$.
The explicit relationship between $F_{ab}$ and the more familiar way of writing things is shown in my answer to another question, where you can also find a little more detail about how the field transforms under a Lorentz boost. Using this notation, Maxwell's equations may be written
\begin{gather}
 \pl_a F_{bc}+\pl_b F_{ca}+\pl_c F_{ab}=0
\tag{3}
\\
 \sum_{a,b}\eta^{ab}\pl_a F_{bc}=0.
\tag{4}
\end{gather}
If we were including a charge/current density $J_c$, it would replace the $0$ on the right-hand side of equation (4).
 Analysis 
By definition, a coordinate transformation $x\to\bar x(x)$ must be smooth and smoothly invertible. The basic rules of calculus tell us how partial derivatives transformation under arbitrary coordinate transformations: for any coordinate transformation whatsoever,
$$
 \pl_a = \sum_b\Lambda_a^b\,\bar \pl_b
\tag{5}
$$
using the notation defined above. The identity (5) is just plain calculus. It doesn't know anything about special relativity. To deduce how $F_{ab}$ transforms, multiply equation (3) by $\Lambda_{A}^a\Lambda_{B}^b\Lambda_{C}^c$ and sum over $a,b,c$ to get
$$
 \bar\pl_A \bar F_{BC}+\bar\pl_B \bar F_{CA}+\bar \pl_C \bar F_{AB}=0
\tag{6}
$$
with
$$
 \bar F_{AB}\equiv \sum_{a,b}\Lambda_A^a\Lambda_B^b F_{ab}.
\tag{7}
$$
When the coordinate transformation is specialized to a Lorentz transformation (1), equations (4), (5), and (7) imply
$$
 \sum_{A,B}\eta^{AB}\bar \pl_A \bar F_{BC}=0.
\tag{8}
$$
According to the assumption listed above as input number 3, the fact that equations (3) and (4) have the same form as (6) and (8) means that (7) is at least one valid way to transform the field under the coordinate transformation. Equation (7) is the same transformation rule that Einstein derived, despite the different-looking notation.
The question asks whether (7) is the only way to satisfy the inputs 1,2,3,4 that were listed above. We can prove this using input number 4, which we haven't used yet. Equations (3) and (4) are linear in the field, so if $F$ and $F'$ are any two solutions, then $F+F'$ is also a solution. If we only used inputs 1,2,3, we could include an extra term $F'_{ab}$ on the right-hand side of (7), where $F'$ is any other solution of Maxwell's equations, because the preceding analysis does not exclude this possibility. And since equations (3) and (4) are linear, this is the only unexcluded possibility. However, the assumption listed above as input number 4 does exclude this possibility, so we conclude that the rule (7) is uniquely determined by inputs 1,2,3,4.
 How arbitrary coordinate transformations affect the field 
In a more modern perspective, $F_{ab}$ are the components of a two-form, and (7) is the correct transformation rule under arbitrary coordinate transformations, not just Lorentz transformations. Arbitrary coordinate transformations preserve the form of equation (3), but most of them do not preserve the form of equation (4).
