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umby
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In Newtonian physics, is the current density, usually called j, a vector?
Namely, is it a polar vector in the sense of a tensor of rank 1?
Namely, is it a polar vector in the sense of a tensor of rank 1?
umby said:In Newtonian physics, is the current density, usually called j, a vector?
Namely, is it a polar vector in the sense of a tensor of rank 1?
Andy Resnick said: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
vanhees71 said: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
$$n'(t',\vec{x}')=n(t,\vec{x})=n(t',\vec{x}+\vec{w} t), \quad \vec{v}'(t',\vec{x}')=\vec{v}(t,\vec{x})-\vec{w} = \vec{v}(t',\vec{x}'+\vec{w}t)-\vec{w}.$$
In non-relativistic physics, is the velocity of a point mass a tensor of rank 1? Is the kinetic energy of a point mass a tensor of rank 0?vanhees71 said: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).
I probably have some conceptual problem.vanhees71 said: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.
Then the question is spontaneous: can you please give me some examples of quantities which are not tensors in Newtonian mechanics?vanhees71 said: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.
vanhees71 said: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.
Current density is a physical quantity that describes the flow of electric current through a given area. It is a vector quantity, meaning it has both magnitude and direction.
Current density is calculated by dividing the magnitude of the current by the cross-sectional area through which it is flowing. The direction of the current density vector is always perpendicular to the cross-sectional area.
Current density is represented as a vector because it has both magnitude and direction. This is important for understanding how the current is flowing through a given area and can be used in calculations to determine the strength and direction of the electric field.
No, current density and electric current are not the same. Electric current is the total amount of charge flowing through a given area per unit time, while current density describes the distribution of this current through a specific area.
Current density and resistivity are inversely related. This means that as the resistivity of a material increases, the current density will decrease. This relationship is described by Ohm's Law: J = σE, where J is current density, σ is resistivity, and E is electric field strength.