Is there any difference between a molecule having $\vec\mu=0$ and being Non-Polar?
[Physics] Difference between Non-Polar and Dipole moment $\vec\mu$=0
dipole-momentmoleculesphysical-chemistry
Related Solutions
A polarizability tensor relates the vector components of the applied field to the vector components of the induced polarization by a scaling factor (nine relations in 3-space). For a field $\mathbf F$, polarization $\mathbf p$ and polarizability tensor $\alpha$, one can represent this is through the matrix equation:
$$ \quad \begin{bmatrix} p_x\\ p_y\\ p_z \end{bmatrix} \ = \ \begin{bmatrix} \alpha_{xx} & \alpha_{xy} & \alpha_{xz}\\ \alpha_{yx} & \alpha_{yy} & \alpha_{yz}\\ \alpha_{zx} & \alpha_{zy} & \alpha_{zz}\\ \end{bmatrix} \ \begin{bmatrix} F_x\\ F_y\\ F_z \end{bmatrix} $$
Tensors have some special (transformation) properties not shared by all matrices, but let's not worry about that now. To see which the tensor elements connect specific field and polarizability elements, just do the matrix multiplication:
$$p_x = \alpha_{xx}F_x + \alpha_{xy}F_y + \alpha_{xz}F_z\\ p_y = \alpha_{yx}F_x + \alpha_{yy}F_y + \alpha_{yz}F_z\\ p_z = \alpha_{zx}F_x + \alpha_{zy}F_y + \alpha_{zz}F_z$$
You may notice that the first index of $\alpha_{jk}$ always corresponds to the polarization direction, and the second index always corresponds to the field direction. For example, the component $\alpha_{xy}$ 'dots' into the $y$ direction of the applied field and results in a polarization in the $x$ direction, while $\alpha_{yx}$ 'dots' into the $x$ field direction and corresponds to a polarization in the $y$ direction. Imagining this 'dotting' with the closest component helps remember which index relates to the field/polarization.
Incidentally, the note above makes a good case for index notation:
$$p_j = \alpha_{jk} F_k.$$
A few more examples may be a helpful reference for relating index notation, matrix notation, and the physical system...
For an isotropic object/medium, the polarization always points in the same direction as the electric field scaled by a constant. This is given by a diagonal matrix:
$$\alpha = c\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{bmatrix}$$
or in index notation: $p_k = c F_k$.
An object whose polarization is in the same direction as the applied field but is more polarizable in a given direction (say $z$) is given by:
$$p = \begin{bmatrix}c_1 & 0 & 0\\ 0 & c_1 & 0\\ 0 & 0 & c_2\end{bmatrix}$$
And any polarizability tensor with off diagonal elements result in polarization induced in a different direction than the applied field, as stated above.
At sufficiently high temperature, a diatomic molecule may have a thermodynamic degree of freedom that corresponds to bond length oscillation. That would change the '5/2' to '6/2'.
Best Answer
$\vec\mu$ is just the electric dipole moment. However, a molecule can be polar with $\vec\mu=0$, as polarity has to do with charge separation, so a particle with any form of multipole moment is polar.
Molecules like methane, carbon dioxide, and perchlorate have $\vec\mu=0$, but have some level of charge separation, making them polar (these have quadrupole moments, not sure about higher order moments).
Actually, all molecules are polar by this definition, just that many aren't polar enough for this to matter. Generally, when we call a molecule "polar", we are talking about only $\vec\mu$.