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Given $x=x’-vt$ and $t=t’$, why is $\frac{\partial t}{\partial x’}=0$ instead of $1/v$? Maybe the answer has something to do with the fact that $dx’=dx$ in this Galilean transformation. Is $dx’=dx$ always the case for Galilean transformations?

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    $\begingroup$ In Galilean transformation x,y,z,t are independent in every frame $(x,y,z,t)$ I think. $\endgroup$ – Nosrati Mar 17 at 3:48
  • $\begingroup$ However, if $t$ changes, $x’$ changes. So how are $x’$ and $t$ independent variables? They seem dependent to me. $\endgroup$ – Christina Daniel Mar 17 at 3:51
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Partial derivatives are only defined when you specify a convention regarding what's held constant, or that convention is obvious in context. For example, $\frac{\partial t}{\partial x^\prime}=0$ is derived from $t=t^\prime$ and assumes you're holding $t^\prime$ constant, and we can express this by writing $\left(\frac{\partial t}{\partial x^\prime}\right)_{t^\prime}=0$. By contrast, from $t=\frac{x^\prime-x}{v}$ we get $\left(\frac{\partial t}{\partial x^\prime}\right)_x=\frac{1}{v}$. (Of course, we can't define $\frac{\partial t}{\partial x^\prime}$ with a convention that holds either $t$ or $x^\prime$ constant.)

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In short, you’re mixing up inputs and outputs of the coordinate transformations and hence confusing which variables are independent and which ones are dependent. The symbols $x$, $t$, $x'$ and $t'$ in your equations stand for different things depending on the context, so it might be helpful to give these different entities different names.

We have the forward map $\phi:(x,t)\mapsto(x+vt,t)$. Let $\phi_1$ and $\phi_2$ stand for the two components of $\phi$, i.e., $\phi_1:(x,t)\mapsto x+vt$ and $\phi_2:(x,t)\mapsto t$. We also have the backward map $\psi = \phi^{-1}:(x',t')\mapsto(x'-vt',t')$ with component functions $\psi_1$ and $\psi_2$. In any particular reference frame, the two coordinates are independent. If you’re talking about the forward map $(x',t')=\phi(x,t)$, then $x$ and $t$ are the independent variables while $x'$ and $t'$ are dependent, and vice-versa for the backward map $(x,t)=\psi(x',t')$. You have to commit to one or the other: one of the frames is designated as the reference frame and the variables that represent its coordinates are independent, while the variables that represent coordinates in the other frame are dependent on them.

In the comment to your question, you write that “if $t$ changes, $x'$ changes.” In this context, $t$ is an independent variable, so you’re implicitly talking about the forward map, so $x'$ means $\phi_1(x,t)$. On the other hand, when you differentiate with respect to $x'$, you’re saying that $x'$ is an independent variable, which means that you’re instead talking about the backward map. In that context, $t'$ is also an independent variable, so from $t=t'$ we have $${\partial t\over\partial x'}={\partial t'\over\partial x'}=0.$$ Using the function names that we’ve introduced, in this context the dependent variable $x$ stands for $\psi_1(x',t')$ and the dependent variable $t$ stands for $\psi_2(x',t')$. We of course have $\partial\psi_2/\partial x'=0$, but what of the equation $x=x'-vt$. It now reads $$\psi_1(x',t') = x'-v\psi_2(x',t').$$ Solving for $\psi_2$ and differentiating produces $${\partial\psi_2\over\partial x'} = \frac1v\left(1-{\partial\psi_1\over\partial x'}\right), v\ne0,$$ but the right-hand side of this also vanishes since $\partial\psi_1/\partial x'=1$.

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