This discussion is inspired in the works of Bogoslovsky
Review of Lorentz Transformations
We assume an homogeneous space-time. It means that the transformations
between two systems of coordinates (only two dimensions by simplicity) $S’$
and $S$, with relative velocity $v$ are linears:
$$\begin{pmatrix}
c t’ \\ x’ \end{pmatrix} = \begin{pmatrix}
\alpha (v) & \sigma (v)\\
\delta (v) & \lambda (v)
\end{pmatrix} \begin{pmatrix}
c t \\ x \end{pmatrix}$$
It can only depend on the relative velocity.
If we consider both observers start together at $x = x’ = 0$. We consider $S’$
is in rest and after a time $t$ for $S$ we have
$$\begin{pmatrix}
c t’ \\ 0
\end{pmatrix} = \begin{pmatrix}
\alpha & \sigma \\
\delta & \lambda
\end{pmatrix} \begin{pmatrix}
c t \\ v t
\end{pmatrix}$$
From the second component we have
$$\delta = – \lambda \beta$$
with $\beta = \frac{v}{c}$
Hence,
$$\begin{pmatrix}
c t’ \\ x’ \end{pmatrix} = \begin{pmatrix}
\alpha (v) & \sigma (v) \\
– \lambda (v) \beta & \lambda (v)
\end{pmatrix} \begin{pmatrix}
c t \\ x \end{pmatrix} \tag{1}$$
From it, we have for the spatial part $x’ = \lambda (v) (x – v t)$
The inverse transformation should be the same with the opposite direction in
the velocity ($- v$):
$$\begin{pmatrix}
c t \\ x
\end{pmatrix} = \begin{pmatrix}
\alpha (- v) & \sigma (- v)\\
\lambda (- v) \beta & \lambda (- v)
\end{pmatrix} \begin{pmatrix}
c t’ \\ x’ \end{pmatrix} \tag{2}$$
For the spatial part: $x = \lambda (- v) (x’ + v t’)$.
We put together both spatial parts:
$$x’ = \lambda (v) x – \lambda (v) \beta c t$$
$$x = \lambda (- v) x’ + \lambda (- v) \beta c t’$$
Inserting $x’$ in the equation for $x$ and arranging:
$$c t’ = \left( \frac{1}{\lambda (- v)} – \lambda (v) \right) \frac{1}{\beta}
x + \lambda (v) c t \tag{3}$$
From (1) we have for the temporal part:
$$c t’ = \alpha (v) c t + \sigma (v) x \tag{4}$$
equating (3) and (4):
$$\alpha (v) = \lambda (v)$$
$$\sigma (v) = \left( \frac{1}{\lambda (- v)} –
\lambda (v) \right) \frac{1}{\beta}$$
Hence,
$$\begin{pmatrix}c t’ \\ x’ \end{pmatrix} = \lambda (v) \begin{pmatrix} 1 & \left( \frac{1}{\lambda (- v) \lambda (v)} – 1 \right) \frac{1}{\beta}\\\beta & 1
\end{pmatrix} \begin{pmatrix}
c t \\ x \end{pmatrix} \tag{5}$$
Thus, the infinitesimal transformations are
$$c d t’ = \lambda (v) \left( c d t + \left( \frac{1}{\lambda (- v) \lambda
(v)} – 1 \right) \frac{1}{\beta} d x \right)$$
$$d x’ = \lambda (v) (d x – \beta c d t)$$
From this, we can get
$$\frac{d x’}{d t’} = \frac{\frac{d x}{d t} – \beta c}{1 + \left(
\frac{1}{\lambda (- v) \lambda (v)} – 1 \right) \frac{1}{\beta c} \frac{d x}{d
t}}$$
This is the velocity addition rule. We assume the speed of light $c$ as the
same in any system. Thus, considering $\frac{d x’}{d t’} = \frac{d x}{d t} =
c$ we have
$$c = \frac{c – \beta c}{1 + \left( \frac{1}{\lambda (- v) \lambda (v)} – 1
\right) \frac{1}{\beta c} c}$$
Therefore,
$$\frac{1}{\lambda (- v) \lambda (v)} = 1 – \beta^2$$
We define $\gamma = \frac{1}{\sqrt{1 – \beta^2}}$. Hence,
$$\lambda (- v) \lambda (v) = \gamma^2 \tag{6}$$
Using (6) in (5):
$$\begin{pmatrix}
c t’ \\ x’
\end{pmatrix} = \lambda (v) \begin{pmatrix}
1 & – \beta\\
– \beta & 1
\end{pmatrix} \begin{pmatrix}
c t \\ x \end{pmatrix} \tag{7}$$
We can define $\varphi (v) \gamma = \lambda (v)$, thus
$$\varphi (- v) \varphi (v) = 1 \tag{8}$$
and
$$\begin{pmatrix}
c t’ \\ x’
\end{pmatrix} = \varphi (v) \gamma \begin{pmatrix}
1 & – \beta \\
– \beta & 1
\end{pmatrix} \begin{pmatrix}
c t \\ x \end{pmatrix} \tag{9}$$
Thus, the transformations are:
$$c t’ = \varphi (v) \gamma (c t – \beta x) \tag{10}$$
$$x’ = \varphi (v) \gamma (x – \beta c t) \tag{11}$$
Under the assumption of isotropy $\varphi (v) = \varphi (- v)$. Using this in
(8), $\varphi (v) = 1$ and we get the Lorentz transformations:
$$\begin{pmatrix}
c t’ \\ x’ \end{pmatrix}= \gamma \begin{pmatrix}
1 & – \beta\\
– \beta & 1
\end{pmatrix} \begin{pmatrix}
c t \\ x \end{pmatrix} \tag{12}$$
This matricial equation contains the transformations
$$c t’ = \gamma (c t – \beta x) \tag{13}$$
$$x’ = \gamma (x – \beta c t) \tag{14}$$
Making a geometrical connection we can define $\tanh \theta = \beta =
\frac{v}{c}$:
$$\begin{pmatrix} c t’ \\ x’ \end{pmatrix} = \begin{pmatrix} \cosh \theta & – \sinh \theta\\ \sinh \theta & \cosh \theta \end{pmatrix} \begin{pmatrix}
c t \\ x \end{pmatrix} \tag{15}$$
If we take the line element $d s^2 = c^2 d t^2 – d x^2$, in the primed frame:
$$c^2 {d t’}^2 – {d x’}^2 = \gamma^2 (c d t – \beta d x)^2 – \gamma^2 (d x –
\beta c d t)^2$$
After some algebra:
$$c^2 {d t’}^2 – {d x’}^2 = c^2 d t^2 – d x^2 = d s^2$$
Lorentz transformations left the line element invariant.
The anisotropic assumption
Following Bogoslovsky, we introduce a conformal factor $e^{- b \theta}$ in (15):
$$\begin{pmatrix}
c t’ \\ x’ \end{pmatrix} = \begin{pmatrix}
e^{- b \theta} \cosh \theta & – e^{- b \theta} \sinh \theta\\
– e^{- b \theta} \sinh \theta & e^{- b \theta} \cosh \theta
\end{pmatrix} \begin{pmatrix}
c t \\ x \end{pmatrix} \tag{16}$$
Since $\theta = \tanh^{- 1} \beta = \frac{1}{2} \ln \left( \frac{1 + \beta}{1 – \beta} \right)$ we have $e^{- b \theta} = e^{- \frac{b}{2} \ln \left( \frac{1 + \beta}{1 –
\beta} \right)} = \left( \frac{1 – \beta}{1 + \beta} \right)^{\frac{b}{2}}$, therefore,
$$c t’ = \left( \frac{1 – \beta}{1 + \beta} \right)^{\frac{b}{2}} \gamma (c t –
\beta x)$$
$$x’ = \left( \frac{1 – \beta}{1 + \beta} \right)^{\frac{b}{2}} \gamma (x –
\beta c t)$$
The same result can be obtained if we don’t assume isotropy in (9). Using
$\tanh \theta = \beta = \frac{v}{c}$ we get
$$\begin{pmatrix}
c t’ \\ x’ \end{pmatrix} = \varphi (\theta) \begin{pmatrix}
\cosh \theta & – \sinh \theta\\ \sinh \theta & \cosh \theta
\end{pmatrix} \begin{pmatrix}
c t \\ x \end{pmatrix}$$
The product of two transformations is:
$$\varphi (\theta) \begin{pmatrix}
\cosh \theta & – \sinh \theta\\ \sinh \theta & \cosh \theta
\end{pmatrix} \varphi (\psi) \begin{pmatrix}
\cosh \psi & – \sinh \psi\\ \sinh \psi & \cosh \psi
\end{pmatrix} = \varphi (\theta) \varphi (\psi) \begin{pmatrix}
\cosh (\theta + \psi) & – \sinh (\theta + \psi)\\ \sinh (\theta + \psi) & \cosh (\theta + \psi)
\end{pmatrix}$$
To get a group the following should be satisfied:
$$\varphi (\theta) \varphi (\psi) = \varphi (\theta + \psi) \tag{17}$$
Dhasmana and Silagadze say that all the continuous solutions of (17) are of the form $\varphi (\theta) = e^{- b \theta}$, which is the Bogoslovsky term.
These transformations don’t left invariant the standard line element $d s^2 =
c^2 d t^2 – d x^2$:
$${c d t’}^2 {- d x’}^2 = \left( \frac{1 – \beta}{1 + \beta} \right)^b (c d t^2 – d x^2) = \left( \frac{1 – \beta}{1 + \beta} \right)^b d s^2$$
However they left invariant
$$d s^2 = \left( \frac{(c d t – d x)^2}{c d t^2 – d x^2} \right)^b (c d t^2 –
d x^2)$$
The new line element can be written as
$$d s^2 = (c d t – d x)^{2 b} (c^2 d t^2 – d x^2)^{1 – b}$$
invariant under
$$c t’ = \left( \frac{1 – \beta}{1 + \beta} \right)^{\frac{b}{2}} \gamma (c t – \beta x)$$
$$x’ = \left( \frac{1 – \beta}{1 + \beta} \right)^{\frac{b}{2}} \gamma (x – \beta c t)$$