I'm reading a solved problem which states that we have a bidimensional metric space whose metric is
$$
ds^2 = dv^2 – v^2 du^2
$$
and we want to find a coordinate transformation such that we get the Minkowski's bidimensional metric space
$$
ds^2 = dx^2 – dt^2\ .
$$
The solution involves equating both metrics,
$$
dv^2 – v^2 du^2 = dx^2 – dt^2\ .
$$
Then, for $x = x(u, v)$ and $t = t(u, v)$ we have
$$
dx = \dfrac{\partial x}{\partial u} du + \dfrac{\partial x}{\partial v} dv \quad\text{and}\quad dt = \dfrac{\partial t}{\partial u} du + \dfrac{\partial t}{\partial v} dv\ .
$$
Substituting these expressions in the equality of the metrics, we arrive at a system of partial differential equations.
The solution keeps going stating that we can solve the system with the method of variable separation,
$$
x(u, v) = U(u) V(v) \quad\text{and}\quad t(u, v) = T(u)S(v)\ ,
$$
leading to the system
$$
\left\{ \begin{array}{l} U'^2 V^2 – T'^2 S^2 = -v^2 \\ V'^2 U^2 – S'^2 T^2 = 1 \\ UU' VV' = TT'SS' \end{array} \right.
$$
I understand the following step, which is realizing from the third equation that
$$
\dfrac{UU'}{TT'} = \dfrac{SS'}{VV'} = \text{constant}\ ,
$$
but suddenly the proposed solution is like "from this we instantly get"
$$
x(u,v) = v \cosh u \quad\text{y}\quad t(u,v) = v \sinh u\ .
$$
I wonder if someone could give a more detailed step by step solution for this system of differential equations, or any hint about how to proceed. Any help on this would be appreciated.
Best Answer
I'm presenting my overall thought process as I went through the derivation. There might well be a simpler way to go about it. We start with $$ ds^2 = dv^2 - v^2 du^2 = v^2 [ \frac{dv^2}{v^2} - du^2 ] = v^2 d ( \ln v - u ) d ( \ln v + u ) $$ Define $$ u+\ln v = U , \qquad u - \ln v = V ~~\implies~~ u = \frac{1}{2}(U+V) , \qquad \ln v = \frac{1}{2} ( U - V ) $$ The metric is $$ ds^2 = - e^U e^{-V} dU dV = d(e^U) d(e^{-V}) $$ Next, we define $$ e^U = x + t , \qquad e^{-V} = x - t . $$ The metric is now $$ ds^2 = (dx+dt)(dx-dt) = dx^2 - dt^2. $$ This is the final form of the metric you want. We can now eliminate $U$ and $V$ and directly determine the relationship between $(t,x)$ and $(u,v)$ $$ v^2 = e^{U} e^{-V} = (x+t)(x-t) = x^2-t^2 \implies v = \sqrt{x^2-t^2} $$ and $$ u = \frac{1}{2} ( U + V ) = \frac{1}{2} \ln(x+t) - \frac{1}{2} \ln (x-t) = \frac{1}{2} \ln \frac{x+t}{x-t} . $$ Inversely, $$ x = v \cosh u , \qquad t = v \sinh u . $$