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Given $x=x’-vt$ and $t=t’$, why is $\frac{\partial t}{\partial x’}=0$ instead of $1/v$? $t$ seems to depend on $x’$ because if $t$ changes, $x’$ changes. Also, in this problem, $dx=dx’$ as well, but I do not currently see why that is the case, either.

Source: edx.org / MIT 8.04.1.x / Week 3 / Problem Set 3 / Question 3 / Part A. link

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  • $\begingroup$ Good idea. Just did! $\endgroup$ – Christina Daniel Mar 17 at 4:58
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The equation $x=x'-vt$ describes a coordinate transformation, not something physically moving at some velocity $v$.

With Galilean transformations lengths are invariant, therefore $\text dx=\text dx'$. Also time does not depend on the spatial coordinate system used, so $\frac{\partial t}{\partial x'}=0$

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