# Galilean transformation and differentiation

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

• Good idea. Just did! – Christina Daniel Mar 17 at 4:58

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 $$\frac{\partial t}{\partial x'}=0$$