Relativistic 1+1 transformation

[A23]

Let [math]\left(\begin{array}{c}ct\\x\end{array}\right)[/math] coordinate of a flat Minkowski spacetime and

[math]\left(\begin{array}{c}ct'\\x'\end{array}\right)[/math]

 

a curved spacetime.

 

We consider the tangent space [math]\left(\begin{array}{c}\Delta ct\\\Delta x\end{array}\right)[/math] with the coordinate transformation to the flat spacetime [math]\left(\begin{array}{c}\frac{\partial t'}{\partial ct}\Delta ct+c\frac{\partial t'}{\partial x}\Delta x \\ \frac{\partial x'}{\partial ct}\Delta ct+\frac{\partial x'}{\partial x}\Delta x\end{array}\right)[/math]

 

since the tangent space has to be locally flat minkowskian, the coordinate transformation between the local coordinates has to be a locally Lorentz transform and a rotation.

 

Considering a 1+1 spacetime only local Lorentz transform, we get the possible transformation with the integrability of :

 

[math]t'(x,t)=\int\frac{dt}{\sqrt{1-\frac{v(x,t)^2}{c^2}}}=\int\frac{-v(x,t)/c^2 dx}{\sqrt{1-\frac{v(x,t)^2}{c^2}}}[/math]

and

 

[math]x'(x,t)=\int\frac{dx}{\sqrt{1-\frac{v(x,t)^2}{c^2}}}=\int\frac{-v(x,t) dt}{\sqrt{1-\frac{v(x,t)^2}{c^2}}}[/math]

 

Hence not all possible coordinate transformation are possible because they wouldn't be relativistic.

For example linear transformation like [math]x'=x-2ct[/math] are not possible.

 

But what would be possible solution to this system except v constant ?

[A23]

Errata-question :

 

a)should the tangeantial space be defined with a matrix with covariant derivative ? (in order to be a tensor)

 

GR field equ. are invariant under all coordinate transformation, among whom all relativistic ones. Does this influences the set of solution ?