# Transformation of the operators $\mathbf\nabla$ and $\partial/\partial t$ under Galilean transformation

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 $$\label{eq:d_operator_K} d=\mathbf\nabla\cdot d\mathbf r +\partial_tdt$$ 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 Vis zero. $$\label{eq:diferentials} \begin{cases} d\mathbf r = d\mathbf r' + \mathbf V dt \\ dt=dt' \end{cases}$$ 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 $$\label{eq:d_operator_KK} d=\mathbf\nabla'\cdot d\mathbf r +\partial_{t'}dt'$$ 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\boxed{ \begin{aligned} \mathbf\nabla' &=\mathbf\nabla \\ \partial_{t'}=&\mathbf\nabla + \mathbf V \partial_t \end{aligned} } 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? • Your second equation in the box is clearly wrong as a scalar quantity is set equal to a vector. – Jon Feb 12 '18 at 12:47 ## 1 Answer 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.

• Oh, yes. You are right, I didn't applied properly the distributive property. Do you know any references about this topic? – Gabriel Sandoval Feb 12 '18 at 14:23
• I think a classical one is the Goldstein's book books.google.it/… – Jon Feb 12 '18 at 14:32