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0
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Lorentz transformation from the Minkowski metric
Here's a quick way to see that you can't do it, choose $\Lambda = g$. Then, $\Lambda^T=\Lambda=g$ and you can check that:
$$ \Lambda g \Lambda^T = g g g = g $$
But $g$ is definitely not a Lorentz tran …
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Could someone explain this leap in the Feynman lectures that 'derives' the fourth Lorentz tr...
Here's one way that I think is rather simple to derive the fourth equation for $t'$.
Begin by considering how observers in a frame may synchronize their clocks. One of the simplest ways is to imagine …