I'm want to know how are the transformations of the operators $\mathbf\nabla$ and $\partial/\partial and $\partial/\partial t$ when the transformation of the Galilean relativity is applied.

This is what I've tried:

Galilean Transformation \begin{cases} \mathbf {r}' = \mathbf r - \mathbf Vt \\ t'=t \end{cases}

The "total differential" operator can be written like this $$ d=\partial_xdx+\partial_ydy+\partial_zdz+\partial_tdt$$ or more compactly

\begin{equation} \label{eq:d_operator_K} d=\mathbf\nabla\cdot d\mathbf r +\partial_tdt \end{equation} Since $\mathbf r=\mathbf r'+\mathbf Vt$, then

\begin{align*} d\mathbf r & = d\mathbf r' + d(\mathbf V t) \\ & = d\mathbf r' + t\,d\mathbf V + V\,dt \end{align*} The velocity $\mathbf V$ is constant, so the differential $d\mathbf V$$ is zero.

\begin{equation} \label{eq:diferentials} \begin{cases} d\mathbf r = d\mathbf r' + \mathbf V dt \\ dt=dt' \end{cases} \end{equation} Substituting the differential above into the total differential operator, gives

\begin{align} \label{eq:d_operator} d &= \mathbf\nabla\cdot \left(d\mathbf r' + \mathbf V dt\right) +\partial_t dt \\ & = \mathbf\nabla\cdot d\mathbf r' + \left(\mathbf V dt +\partial_t\right)dt \end{align} The same $d$ operator, in terms of the moving coordinate system is written like this \begin{equation} \label{eq:d_operator_KK} d=\mathbf\nabla'\cdot d\mathbf r +\partial_{t'}dt' \end{equation}

Here is the doubt I think the operators $d$, no matter in which coordinete system are written, must to be equals. So I can match coefficients and get this $$\mathbf\nabla'\cdot d\mathbf r +\partial_{t'}dt=\mathbf\nabla\cdot d\mathbf r' + \left(\mathbf V dt +\partial_t\right)dt$$ \begin{equation} \boxed{ \begin{aligned} \mathbf\nabla' &=\mathbf\nabla \\ \partial_{t'}=&\mathbf\nabla + \mathbf V \partial_t \end{aligned} } \end{equation}

Can you tell me whether my argument is right or wrong?

And can you tell me some references where I can take a look about this topic?

  • $\begingroup$ Your second equation in the box is clearly wrong as a scalar quantity is set equal to a vector. $\endgroup$
    – Jon
    Feb 12, 2018 at 12:47

1 Answer 1


It should be $$ \nabla'\cdot d{\bf r'}+dt'\partial_{t'}=\nabla\cdot(d{\bf r}+{\bf V}dt)+dt\partial_t = \nabla\cdot d{\bf r}+(\partial_t+{\bf V}\cdot\nabla)dt $$ and you get the Euler time derivative as it should.

  • $\begingroup$ Oh, yes. You are right, I didn't applied properly the distributive property. Do you know any references about this topic? $\endgroup$ Feb 12, 2018 at 14:23
  • $\begingroup$ I think a classical one is the Goldstein's book books.google.it/… $\endgroup$
    – Jon
    Feb 12, 2018 at 14:32

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