The 1000km railcar between the two clocks is unnecessary. All we need is are two ends, moving at the same speed. But, a 1000km car would work too, if you can get the funding.The above will not work because the math disproves your claims.
Let $$S$$ be the frame of the moving 1000km (!!!) rail car with the clocks at both ends and $$S'$$ the frame of the rails.
To be clear, in my setup both the $$S$$ and the $$S'$$ clocks are the same distance apart in $$S$$.The $$S'$$ clocks are situated $$L=1000km$$ apart. we will label the clocks A and B:
This means that in $$S'$$, the distance between the $$S'$$ clocks is $$L\gamma$$.
As Neddy said, that's not a flaw, it's the expected result.$$t'_A=\gamma(t_A+vx_A/c^2)$$
$$t'_B=\gamma(t_B+vx_B/c^2)$$
The $$S$$ clocks are synchronized , so:
$$t_A=t_B=t_0$$
$$x_A=0$$
$$x_B=L$$
so:
$$t'_A=\gamma t_0$$
$$t'_B=\gamma t_0+\gamma vL/c^2$$
Therefore:
$$t'_B-t'_A=\gamma vL/c^2$$
No surprise here.
But:
$$x'_A=\gamma(x_A+vt_A)=\gamma vt_0$$
$$x'_B=\gamma(x_B+vt_B)=\gamma vt_0+\gamma L$$
So, your scenario has a flaw, the clocks on the rail car do not line up with the clocks on the ground, the clock at B in the railcar is measured by the platform as being way past to the right of the corresponding ground clock. No wonder the ground clock has accumulated the extra time, the railcar clock has passed it by quite a bit. The larger the relative speed, the larger the missalignment, both spatial and temporal
In $$S'$$, the clocks obviously don't line up at the same time - the events are not simultaneous. That's the whole point of the experiment.
The $$S'$$ clocks measure the two events to be non-simultaneous.
The $$S$$ clocks measure the events to be simultaneous.
The rail clock at B is measured by the $$S'$$ clocks as being way past to the right of the corresponding ground clock at the time that the A clocks line up.
The rail clock at A is measured by the $$S'$$ clocks as being still way off to the leftt of the corresponding ground clock at the time that the B clocks line up.
They are lined up at the same time in $$S$$.Tach said:They are not lined up in any frame.
They are lined up at different times in $$S'$$.