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Let us consider a Lorentz transformation of four vectors from frame S to S' where S' is moving with relative velocity $\textbf{v}$ with respect to S. The boost is given by $$t'=\gamma(t-vx), \quad x'=\gamma(x-vt), \quad y'=y, \quad z'=z.$$ The inverse transformation is given by $$t=\gamma(t'+vx'), \quad x=\gamma(x'+vt'), \quad y=y', \quad z=z'.$$ Now here comes the crucial part. Notice that $$\frac{\partial x}{\partial x'} = \gamma, \quad \frac{\partial x'}{\partial x} = \gamma. \tag{1}$$ I have thought about this for a while, but more thoughts always lead me to the same conclusion that this is true.

However, then we have a problem: $$\frac{\partial}{\partial x'} = \frac{\partial x}{\partial x'} \frac{\partial}{\partial x} = \gamma \frac{\partial}{\partial x} \tag{2}.$$ This seems fine. Continue: $$\frac{\partial}{\partial x} = \frac{\partial x'}{\partial x} \frac{\partial}{\partial x'} = \gamma \frac{\partial}{\partial x'} \tag{3}.$$ This also seems fine. Continue: $$ \frac{\partial}{\partial x'}=\gamma \frac{\partial}{\partial x} = \gamma \frac{\partial x'}{\partial x} \frac{\partial}{\partial x'} = \gamma^2 \frac{\partial}{\partial x'}$$

where we have reached a contradiction since $\gamma^2 \neq 1$.

Where's have I messed up in (1), (2), (3)?

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  • $\begingroup$ how did you get (1)? $\endgroup$
    – user65081
    Nov 30, 2019 at 0:52
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    $\begingroup$ You neglected the time derivatives. $\endgroup$
    – G. Smith
    Nov 30, 2019 at 0:52
  • $\begingroup$ Can you explain how you got (2)? $\endgroup$
    – WillO
    Nov 30, 2019 at 0:53
  • $\begingroup$ @WillO, no, I mean you obtain (2) from (1) $\endgroup$
    – user65081
    Nov 30, 2019 at 0:57
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    $\begingroup$ (1) is correct. (2) and (3) are incorrect. $\endgroup$
    – G. Smith
    Nov 30, 2019 at 0:57

1 Answer 1

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You forgot that $x'$ is not only function of $x$, but also of $t$: $x' = x'(x, t)$. Similarly, $x=x(x', t')$. Hence: $$ \frac{\partial}{\partial x'} = \frac{\partial x}{\partial x'} \frac{\partial}{\partial x} + \frac{\partial t}{\partial x'} \frac{\partial}{\partial t}$$ and $$ \frac{\partial}{\partial x} = \frac{\partial x'}{\partial x} \frac{\partial}{\partial x'} + \frac{\partial t'}{\partial x} \frac{\partial}{\partial t'}$$

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