A Is the current density a vector?

##J = \nabla * I##
not sure what you mean by polar vector. You can express it in polar coordinates, if you like.

 

https://en.wikipedia.org/wiki/Current_density

 

Since ##\vec j=\sigma \vec E## (Ohm's Law), and ##\sigma## is a scalar, then ##\vec j## is polar since ##\vec E## is polar.

 

It's usually handled like a vector, but in some contexts the current density is expressed as a 3-form:

https://en.wikipedia.org/wiki/Differential_form

 

A 3-form in spacetime... related via the Hodge-dual in spacetime.
An odd (or twisted) 2-form in space...related via the Hodge-dual in space.
(https://books.google.com/books?id=4OTUBwAAQBAJ&pg=PA147&lpg=PA147&dq=current+twisted+2-form)

 

Of course, it depends on which quantity is flowing. In hydrodynamics (the only application of Newtonian continuum theory which comes to my mind) the most common flow vector is the particle-number flow. You start by defining the particle-number density, ##n(t,\vec{x})## which gives the number of particles per volume at time ##t## around position ##\vec{x}##. This is a scalar under rotations and space reflections. Then you define the flow-velocity field ##\vec{v}(t,\vec{x})## which gives the momentary velocity of the particles momentarily around position ##\vec{x}##. That's a polar vector. Now the current density is the number of particles per unit time going through an oriented surface element ##\mathrm{d} \vec{F}## (a vector with the magnitude of the area of the surface element directed perpendicular to the surface element; which of the two directions you choose is arbitrary and defines in which sense you count particles as entering and leaving a volume element; usually you orient the surface element vector in the direction pointing out of the volume of interest). This number of particles is
$$\mathrm{d} N = \mathrm{d} t \mathrm{d} \vec{F} \cdot n \vec{v}.$$
because all particles in a little volume ##\mathrm{d} V = \mathrm{d} t \vec{v} \cdot \mathrm{d} \vec{F}## will go through the surface element within a little time ##\mathrm{d} t##, and you get the number of particles by multiplying this volume by the particle-number density.

Now the surface-element vector is obviously a polar vector, because ##\mathrm{d} N## must be a scalar and ##\vec{v}## is a polar vector and ##n## a scalar. Thus
$$\vec{j}=n \vec{v}$$
must be a polar vector too.

 

You cannot use electrodynamics as examples, because that's a relativistic field theory, and Galilei symmetry doesn't apply.

Of course, velocity is a vector under rotations. The behavior of the quantities under Galilei boosts is obvious:
##t \rightarrow t'=t, \quad \vec{x} \rightarrow \vec{x}'=\vec{x}-\vec{w} t##
This implies
$$n'(t',\vec{x}')=n(t,\vec{x})=n(t',\vec{x}+\vec{w} t), \quad \vec{v}'(t',\vec{x}')=\vec{v}(t,\vec{x})-\vec{w} = \vec{v}(t',\vec{x}'+\vec{w}t)-\vec{w}.$$

 

It depends on what kind of tensors you are discussing. In non-relativistic physics only tensor-transformation under rotations is relevant, and thus you discuss tensors as objects invariant under rotations, i.e., the tensor components transforming under SO(3) transformations.

Further, sometimes (in quantum theory) you are interested in the behavior under space reflections (parity) ##\vec{x} \rightarrow -\vec{x}##, and then you have to distinguish between tensors under O(3) and pseudotensors under O(3). E.g., for tensors of rank 1, i.e., vectors \vec{V} the corresponding components all transform under SO(3) (rotations) via the fundamental representation, i.e., ##V^j \rightarrow {R^j}_k V^k## (if you use a Cartesian basis) with ##({R^j}_k)=\hat{R} \in \mathrm{SO}(3)##. Then you have to consider also parity transformations, which are either realized by ##\vec{V} \rightarrow -\vec{V}## (polar vector) or ##\vec{V} \rightarrow \vec{V}## (axial vector).

 

Yes. The velocity is a vector. Under rotations the position-vector components ##\vec{x}=(x,y,z)^{\mathrm{T}}## transform with a matrix ##\hat{T} \in \mathrm{SO}(3)## and time doesn't change at all. Thus you have
$$t'=t, \quad \vec{x}'=\hat{T} \vec{x}.$$
Thus the velocity transforms like
$$\vec{v}'=\frac{\mathrm{d} \vec{x}'}{\mathrm{d} t'} = \frac{\mathrm{d}}{\mathrm{d} t} \hat{R} \vec{x}=\hat{R} \frac{\mathrm{d} \vec{x}}{\mathrm{d} t}=\hat{R} \vec{v},$$
i.e., the velocity transforms under the same rotation as the position vector.

Further, since ##\hat{T}^{-1}=\hat{T}^{\mathrm{T}}## (that's what's called an orthgonal matrix), you have
$$T'=\frac{m}{2} \vec{v}'^2=\frac{m}{2} (\hat{T} \vec{v})^2=\frac{m}{2} \vec{v}^2=T,$$
which means that the kinetic energy of the particle is a scalar (tensor of rank 0). Of course, I've used that (by definition!) the mass of the particle, ##m##, is a scalar too.

 

The quantities are not tensors with respect to the full Galilei group but only under rotations. Non-relativistic spacetime is not an affine space as is the case in special-relativistic spacetime (Minkowski space) but a fiber bundle. You have just copies of a 3D Euclidean affine spaces along the 1D oriented time axis. That's it. Thus the Newtonian quantities are not described as tensors under the full Galilei group but only under rotations in the Euclidean (configuration) spaces.

The velocity is a vector in this sense, and you can build valid quantities which transform in a specific way under Galilei boosts,
$$t'=t, \quad \vec{x}'=\vec{x}-\vec{w} t \; \Rightarrow \; \vec{v}'=\vec{v}-\vec{w},$$
etc. Starting from this you can derive the transformation laws for all other quantities (together with the assumption that the mass of a particle or the total mass of a composite system is a strict scalar under the full Galilei group).
E.g., momentum is a necessary building block for the Newtonian dynamics
$$\vec{p}=m \vec{v} \; \Rightarrow \; \vec{p}'=\vec{p}-m \vec{w}.$$
The 1st Law reads
$$\vec{F}=\dot{\vec{p}} \; \Rightarrow \; \vec{F}'=\vec{F}$$
under Galilei boosts, etc.

 

Lagrangians with gyroscopic forces generate Finsler structure on configuration manifold. But as a rule physical reasonable objects are tensors, this expresses the fact that the physical reasonable objects do not depend on corrdinate frames which we employ by our own wish

 

By an "affine space", do you essentially mean a vector space that has forgotten its origin?
If so, then (non-gravitational non-relativistic) Galilean spacetime is an affine space
like Euclidean space and Minkowski spacetime
in the Cayley-Klein classification of geometries (https://www.google.com/search?q=Cayley-Klein+classification+of+geometries ).

 

An affine space is a point manifold, where two (ordered) points define a vector. The usual Euclidean geometry is an affine space, where the vector space has the additional feature of having a scalar product (positive definite fundamental bilinear form). Also Minkowski spacetime is an affine space with the vector space having an indefinite fundamental bilinear form of signature (1,3) (or (3,1) if you prefer the east-coast convention). The symmetry groups of these spaces are semidirect products of translations an the linear mappings of the vector space that leave the fundamental form invariant. For the Euclidean space that the SO(3) (rotations) and the corresponding symmetry group of the affine Euclidean space is called ISO(3). For Minkowski space it's ISO(1,3) (more precisely ##\mathrm{ISO}(1,3)^{\uparrow}##). Here I use the point of view that only that part of the symmetry group of the space under consideration is that that is continuously connected with the identity operation. As the example of Minkowski space shows only this subgroup must be necessarily a symmetry group of the dynamics. In Nature only this group is realized as a symmetry. The weak interaction breaks P and T (and C) separately and also CP.

 

So, it seems to me that
(non-gravitational non-relativistic) Galilean spacetime is an also affine space [since two ordered points (t1,x1) and (t2,x2) define a vector]
with the additional feature of having a degenerate scalar product with signature (1,0) and a degenerate scalar product with signature (0,3).
There is a corresponding inhomogeneous Galilean group.

 

Which scalar product should that be?