Galilean transformation of the wave equation, derivatives [closed]

Question

So I'm trying to show that when the wave function
$ (-\frac{1}{c^2}\frac{d^2}{dt^2} + \frac{d^2}{dx^2})\phi(t,x) = 0 $
undergoes the Galilean transformation
$ t' = t $
$ x' = x-Vt $
the resulting differential equation is
$ [-\frac{1}{c^2}\frac{d^2}{dt^2} + \frac{d^2}{dx^2} -\frac{V^2}{c^2}\frac{d}{dx'} + \frac{2V}{c^2}\frac{d}{dt'}\frac{d}{dx'}]\phi(t',x')=0 $

I've started by saying that $\frac{d}{dx} = \frac{dx'}{dx} \frac{d}{dx'} + \frac{dt'}{dx}\frac{d}{dt'}$ and I know that I'm supposed to get $\frac{d}{dx} = \frac{d}{dx'}$ but when I do it, $\frac{dt'}{dx} = 0$ must be true for that second term to cancel out. I could understand if I'm taking the derivative using $t'=t$, but if I rearrange the transformation $ x' = x-Vt $ to be $t=\frac{x-x'}{V}=t'$, then $\frac{dt'}{dx} = \frac{1}{V}$. What am I doing wrong here?

Answer 1

When you perform a coordinate transformation, you write the new variables as functions of the old ones:

$$t'(x,t) = t$$ $$x'(x,t) = x - Vt$$

Differentiation then proceeds as usual: $$\frac{\partial}{\partial x} t'(x,t) = \lim_{h\rightarrow 0} \frac{t'(x+h,t)-t'(x,t)}{h} = \lim_{h\rightarrow 0} \frac{t - t}{h} = 0$$ $$ \frac{\partial}{\partial t} t'(x,t) = \lim_{h\rightarrow 0} \frac{t'(x,t+h)-t'(x,t)}{h} = \lim_{h\rightarrow 0} \frac{t+h-t}{h} = 1$$ Similarly, $$\frac{\partial }{\partial x} x'(x,t) = 1$$ $$\frac{\partial}{\partial t} x'(x,t) = - V$$

I chose to write out the difference quotients explicitly to make it obvious that $t'$ and $x'$ are to be considered functions with two slots - one for the old position and one for the old time.

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If you rearrange that second equation to yield

$$t(x,x') = \frac{x-x'}{V}=t'(x,x')$$

Then you are writing the new time as a function of the old position and the new position. This is a different function from the one I wrote above, and is not what we are looking for when performing coordinate changes.

Fundamentally, this misunderstanding can arise when you think only about the quantity with respect to which you are differentiating and forget to also specify which quantities are being held constant. We should really write

$$\left(\frac{\partial t'}{\partial x}\right)_t = \lim_{h\rightarrow 0} \frac{t'(x+h,t)-t'(x,t)}{h} = \lim_{h\rightarrow 0} \frac{t-t}{h}=0$$ which means the partial derivative of $t'$ with respect to $x$, holding $t$ constant. Contrast this with

$$\left(\frac{\partial t'}{\partial x}\right)_{x'} = \lim_{h\rightarrow 0} \frac{t'(x+h,t+\frac{h}{V})-t'(x,t)}{h} = \lim_{h\rightarrow 0}\frac{t+\frac{h}{V}-t}{h} = \frac{1}{V}$$

which means the partial derivative of $t'$ with respect to $x$ holding $x'$ constant.

Note that the $t\rightarrow t+ \frac{h}{V}$ shift comes from the fact that if $x'(x,t)=x-Vt$ is being held constant, then when $x\rightarrow x+h$ we must also have that $t \rightarrow t + \frac{h}{V}$ to compensate.