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?

1 Answer 1


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.

  • 1
    $\begingroup$ oh, i see now. Thank you @AccidentalTaylorExpansion $\endgroup$
    – megamence
    Commented Jan 25, 2021 at 21:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.