General procedure to show that something is a tensor

Question

Let $\phi({\bf x})$ be a scalar field.

Show that $\nabla \phi$ is a tensor of rank 1

I know that the definition of a first rank tensor is:

$$ A'_i=R_{ij}A_j \tag{1} $$

Hence something that is a 1st rank tensor has to satisfy (1)

Also $\nabla(\phi)_i=\partial_{x_{i}}\phi$ and transforms like $\nabla(\phi)'_i=\partial_{x_{i}}\phi\partial{x'_{i}}x$

Can I just say that here $\partial{x'_{i}}x=R_{ij}$ and hence complete the proof?

Answer 1

Remember that for a transformation of the form ${x'}^i = {x'}^i({\bf x})$ you have

\begin{eqnarray} (\nabla \phi({\bf x'}))_i &=& \frac{\partial\phi({\bf x'})}{\partial {x'}^i} = \frac{\partial x^j}{\partial {x'}^i}\frac{\partial \phi({\bf x})}{\partial x^j} \\ &=& \frac{\partial x^j}{\partial {x'}^i} (\nabla \phi({\bf x}))_j \end{eqnarray}

So

$$ (\nabla \phi')_i = R^j_i (\nabla \phi)_j $$

That is, $\nabla$ transforms as a rank 1 tensor