I worked through the problem using Peter's notation.
I agree with Peter's 4-velocity.
A simple derivation. In Rindelr coordinates (t,x,y), we can write the motion of the sliding block as y=vt. We can say ##t = \gamma \tau##. This gives us the 4-velocity in Rindler coodinates as
U = (##dt/d\tau, dx/d\tau, dy/d\tau##) = (##\gamma, 0, \gamma \, v##).
Converting to Minkowskii coordinates (T,X,Y) we get in those coordinates Peter's result
U = ##( dT/d\tau, dX/d\tau, dY/d\tau)## = ##( \gamma \cosh \gamma g \tau, \gamma \sinh \gamma g \tau, \gamma v)##
Integrating the four velocity we find the position vs time of the block parameterized by proper time ##\tau##
##T(\tau) = (1/g) \sinh \gamma g \tau \quad X(\tau) = (1/g) [ \cosh \gamma g \tau -1 ] \quad Y(\tau) = \gamma v \tau ##
Peter's values for ##(\hat{e}_i)## in
https://www.physicsforums.com/threads/gravity-in-einsteins-train.835994/page-4#post-5254985 check out. Thus we can write
##T = (1/g) \sinh \gamma g \tau + \chi (e_1)^0 + \psi (e_2)^0##
##X = (1/g) [ \cosh \gamma g \tau -1 ] + \chi (e_1)^1 + \psi (e_2)^1##
##Y = \gamma v \tau + \chi (e_2)^1 + \psi (e_2)^2##
The notation here is slightly different than Peter's (it is the same notation as used in MTW). Here ##(e_i)^0## is the ##\partial_T## component of the basis vector ##e_i## in Peter's notation, where i=0,1,2. Similarly ##(e_i)^1## is the ##\partial_X## comonent and ##(e_i)^2## is the ##\partial_Y## component of said basis vector.
[edit for consistenh]
If this seems daunting, it's not. In vector notation we are just saying that the position at time ##\tau## is ##\vec{P_0 }+ \chi \vec{e_\chi} + \psi \vec{e_\psi}##, where ##\vec{P_0}## is the position of the block at time ##\tau##, ##\vec{e_\chi}## and ##\vec{e_\psi}## have at time ##\tau## the component values that Peter found. We have replaced the numeric indices (0,1,2) that Peter used with symbolic indices (##\tau, \chi, \psi)##. - We we multiply these purely spatial basis vector by the spatial coordinate value (##\chi, \psi##) to get the spatial displacement from the block position. Thus, by adding the initial block position ##\vec{P_0}## to the spatial displacement vector we calculate, we find the spatial position of the point specified by coordinates ##(\chi, \psi)## at time ##\tau##. This all happens on a surface of constant coordinate time ##\tau##. Of course, because of the relativity of simultaneity, a surface of constant coordinate time ##\tau## isn't a surface of constant cordinate time T, this is the reason our spatial vectors have components in the ##\partial_T## direction. This process specifies a momentarily co-moving spatial frame of reference, though it's a rotating co-moving frame and not an inertial co-moving frame.
Carrying this out yields the following equations which transform from ##\tau, \chi, \psi## coordinates to T,X,Y, coordinates:
##T = (1/g) \sinh \gamma g \tau + \chi \, \sinh \gamma g \tau + \psi \, \gamma v \cosh \gamma g \tau##
##X = (1/g) [ \cosh \gamma g \tau -1 ] + \chi \, \cosh \gamma g \tau + \psi \, \gamma v \sinh \gamma g \tau##
##Y = \gamma v \tau + \psi \, \gamma##
The remaining part requires a lot of computer algebra, plus a fair amount of manual simplification and collection of terms, but is concetually easy. We find the metric in the new coordinates by using the chain rule to find dT, dX, and dY in terms of ##d\tau##, ##d\chi##, ##d\psi##. Then we calculate the line element -dT^2 + dX^2 + dY^2
The result I get for the resulting metric is
##ds^2 = \alpha d\tau^2 + \beta (\psi d\chi- \chi d\psi) d\tau + d\chi^2 + d\psi^2##
where
\alpha =<br />
-{\frac { \left( \chi\,g+1 \right) ^{2}}{-{v}^{2}+1}}+{\frac {{v}^{2}<br />
}{-{v}^{2}+1}}+{\frac {{g}^{2}{\psi}^{2}{v}^{2}}{ \left( -{v}^{2}+1<br />
\right) ^{2}}}
\beta =\frac{ 2 g v}{1-v^2}
The physical interpretation of this is fairly straightforwards, it's just an expression for time dilation (the alpha coefficient) and rotation (the beta coefficient), with the spatial part of the metric being perfectly ordinary Euclidean space.
Fixed now.
