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I am taking a class in fluid mechanics right now and my book has this statement with no explanation:

What is the time derivate seen by an observer moving with a velocity $\mathbf{v}$ of a scalar field $f(\mathbf{x},t)$? $$\frac{df}{dt} = \lim_{\delta t \rightarrow 0} \frac{f(\mathbf{x}+\mathbf{c}\delta t, t+\delta t)-f(\mathbf{x},t)}{\delta t} \\ =\left(\frac{\partial f}{\partial t} \right)_{\mathbf{x}}+\mathbf{c}\cdot \nabla\mathbf{u}$$

My question has two parts:

  1. Why should the derivative with respect to time change if an observer is moving? Isn't the scalar field already set up in space, why would the way I observe it change it?
  2. What would happen if I am moving with a variable velocity, $\mathbf{c}(\mathbf{x}, t)$, would there be no change in the functional form of the derivative?
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Here $\frac{df}{dt}$ is a total derivative. A total derivative is a derivative with respect to all of its variables. You probably already know this but what is often overlooked is that the total derivative is defined on a one-dimensional path. The function $f(x,y,z,t)$ is defined on $\mathbb R^4$, which is 4-dimensional, so you have to write $x,y,z$ as a function of $t$ before you can take the total derivative. So you get $$\frac{d}{dt}f(x(t),y(t),z(t),t)=\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}+\frac{\partial f}{\partial z}\frac{dz}{dt}+\frac{\partial f}{\partial t}$$ So what have we gained by writing this so explicitly? Now we can notice the author has written $\mathbf x$ as a function of $t$. He has defined some path through space which the observer travels along. So compare this to the total derivative for an observer at a fixed point $\mathbf x(t)=\mathbf x_0$: $$\left.\frac{df}{dt}\right|_{\mathbf x(t)=\mathbf x_0}=\frac{\partial f}{\partial t}(\mathbf x_0)$$ This is probably closer to what you had in mind.

To answer your second question the formula stays the same for a time dependent velocity of the observer. To see this align the velocity of the observer with one of the axes. For example the x-axis. Now the formula is just the ordinary chain rule $\frac{df}{dt}=\frac{df}{dx}\frac{dx}{dt}$. There's no acceleration in there.

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    $\begingroup$ oh, i see now. Thank you @AccidentalTaylorExpansion $\endgroup$
    – megamence
    Jan 25, 2021 at 21:37

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