In a change of coordinates, what is the physical meaning of the new partial derivatives when one coordinate didn't change?

Question

In a change of coordinates where one coordinate, the time, remains the same, I don't understand why the new partial derivatives for that coordinate are different. Let me show you:
The variable $ u(x,t) $ satisfies a particular PDE.
$(x,t)$ was changed to $(X,T)$ where $$ X = f(x,t) \\ T = t $$ Applying the chain rule, the derivatives of u can be written as:
$$ \frac{\partial u}{\partial x} = \frac{\partial u}{\partial X} \frac{\partial X}{\partial x} + \frac{\partial u}{\partial T} \frac{\partial T}{\partial x} = \frac{\partial u}{\partial X} \frac{\partial f}{\partial x} \\ \frac{\partial u}{\partial t} = \frac{\partial u}{\partial X} \frac{\partial X}{\partial t} + \frac{\partial u}{\partial T} \frac{\partial T}{\partial t} = \frac{\partial u}{\partial X} \frac{\partial f}{\partial t} + \frac{\partial u}{\partial T} $$ So what what confuses me is the fact that, although $T = t$ in the new coordinate system, $\frac{\partial u}{\partial T} \neq \frac{\partial u}{\partial t}$. Can anyone explain the physical meaning behind that?

Answer 1

This is best explained by the notion of directional derivative. The derivative of $u \left(x , t\right)$ in the direction $\left(a , b\right)$ is the limit

\begin{equation}{\partial}_{\left(a , b\right)} u = \lim _{h \rightarrow 0} \frac{u \left(x+a h , t+b h\right)-u \left(x , t\right)}{h}\end{equation}

By definition,

\begin{equation}\frac{\partial u}{\partial t} = {\partial}_{\left(0 , 1\right)} u\end{equation}

But

\begin{equation}\frac{\partial u}{\partial T} = {\partial}_{\left({-{\left(\frac{\partial f}{\partial x}\right)}^{{-1}}} \frac{\partial f}{\partial t} , 1\right)} u\end{equation}

Indeed, suppose that one increases $ T$ by a small amount $\delta T$ while keeping $ X$ invariant, then $t$ increases by the amount $ \delta t = \delta T$, but

\begin{equation}0 = \delta X \approx \frac{\partial f}{\partial x} \delta x+\frac{\partial f}{\partial t} \delta t\end{equation}

it means that $x$ increases by an amount

\begin{equation}\delta x \approx-{\left(\frac{\partial f}{\partial x}\right)}^{{-1}} \frac{\partial f}{\partial t} \delta T\end{equation}

So $\frac{\partial }{\partial t}$ and $\frac{\partial }{\partial T}$ are directional derivatives in two different spacetime directions.