I Derivative operators in Galilean transformations

[TomServo]

TL;DR Summary

I'm confused about how the derivative with respect to time transforms under a Galilean transformation.

I'm studying how derivatives and partial derivatives transform under a Galilean transformation.

On this page:

http://www.physics.princeton.edu/~mcdonald/examples/wave_velocity.pdf

Equation (16) relies on $\frac{\partial t'}{\partial x}=0$ but $\frac{\partial x'}{\partial t}=-v$

But this seems like a contradiction to me. If you swap primed/unprimed you get $\frac{\partial t}{\partial x'}=0$ but $\frac{\partial x}{\partial t'}=v$, in which case you have $x=vt+x_0$ and $t=t'=\frac{x-x_0}{v}$. Thus $\frac{dt'}{dx}=\frac{\partial t'}{\partial x}=\frac{1}{v}$, in violation of Eq. (16).

So where have I gone wrong? Thanks.

 

[Orodruin]

$t$ is not given by $(x-x_0)/v$, that can be true only for a very particular world-line and that is not what you are considering, you are considering the transformation of coordinates. $\partial x/\partial t’ = v$ is a partial differential, not a total differential.

 

[vanhees71]

The Galilei transformation reads
$$t'=t, \quad \vec{x}'=\vec{x}-\vec{v} t.$$
You consider $t$ and $\vec{x}$ as independent variables when it comes to (non-relativistic) field equations. Thus you have
$$\frac{\partial t'}{\partial t}=1, \quad \vec{\nabla} t'=0, \quad \partial_t \vec{x}'=-\vec{v}, \quad \vec{\nabla} \otimes \vec{x}=\hat{1}.$$

 

[TomServo]

$t$ is not given by $(x-x_0)/v$, that can be true only for a very particular world-line and that is not what you are considering, you are considering the transformation of coordinates. $\partial x/\partial t’ = v$ is a partial differential, not a total differential.

Could you further explain what you mean here? I know what worldlines are, but it seems to me (just algebraically) that the $t=\frac{x-x’}{v}$ relation holds in general. After all, I’m just solving the transformation equation for t. I know this is wrong, but I’m trying to understand why the algebra leads me astray (or seems to).

 

[TomServo]

And where I wrote $x_0$ originally I meant $x’$.

 

[Orodruin]

Then you are treating x’ as a constant when differentiating with respect to x. That is incorrect. What is being kept constant when you take the partial with respect to x is t, not x’.