- #1

- 38

- 7

Namely, is it a polar vector in the sense of a tensor of rank 1?

- A
- Thread starter umby
- Start date

- #1

- 38

- 7

Namely, is it a polar vector in the sense of a tensor of rank 1?

- #2

BiGyElLoWhAt

Gold Member

- 1,570

- 116

not sure what you mean by polar vector. You can express it in polar coordinates, if you like.

- #3

- 38

- 7

A "polar vector" is required to have components that "transform" in a certain way in passing from one coordinate system to another. Does the current density satisfy this rule?

- #4

BiGyElLoWhAt

Gold Member

- 1,570

- 116

- #5

- 5,779

- 1,078

- #6

- 7,548

- 2,158

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

Namely, is it a polar vector in the sense of a tensor of rank 1?

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

- #7

- 5,779

- 1,078

A 3-form in spacetime... related via the Hodge-dual in spacetime.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

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)

- #8

- 17,077

- 8,179

$$\mathrm{d} N = \mathrm{d} t \mathrm{d} \vec{F} \cdot n \vec{v}.$$

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.

- #9

- 38

- 7

I try to explain my doubt. First of all, let’s restrict ourselves to Newtonian physics, read Galilean transformation,

Consider two equally-oriented reference frames in relative motion, namely two observers in relative motion.

In the proposed example, if one observer sees a particle standing still, the other sees it moving. Then, the velocity is not a tensor of rank 1, because does not satisfies the above mentioned transformation law. That is why the velocity cannot appear in a law of nature.

The contrary is valid for the acceleration, which can be demonstrated to follow the transformation law.

The previous discussion can be easily extended to the current density (simply consider a charge fixed for one observer), which then does not appear to be a tensor.

However, if this reasoning is right, we still need to explain the relation proposed by robphy:

Actually, if we apply the Lorentz transformation to the previous problem, we get an interesting solution (imaging the relative motion along the x-axis). But this is why the four-current of the observer seeing the charge still, J =[0,0,0,ρc], transforms according to the Lorentz transformation as J' = [-vργ,0,0,ρcγ] for the second observer. in this case everything seems fine.

- #10

- 17,077

- 8,179

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

$$n'(t',\vec{x}')=n(t,\vec{x})=n(t',\vec{x}+\vec{w} t), \quad \vec{v}'(t',\vec{x}')=\vec{v}(t,\vec{x})-\vec{w} = \vec{v}(t',\vec{x}'+\vec{w}t)-\vec{w}.$$

- #11

- 38

- 7

To be invariant under rotation is enough to be a tensor?

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

$$n'(t',\vec{x}')=n(t,\vec{x})=n(t',\vec{x}+\vec{w} t), \quad \vec{v}'(t',\vec{x}')=\vec{v}(t,\vec{x})-\vec{w} = \vec{v}(t',\vec{x}'+\vec{w}t)-\vec{w}.$$

- #12

- 17,077

- 8,179

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).

- #13

- 38

- 7

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?

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).

- #14

- 17,077

- 8,179

$$t'=t, \quad \vec{x}'=\hat{T} \vec{x}.$$

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.

- #15

- 38

- 7

I probably have some conceptual problem.

$$t'=t, \quad \vec{x}'=\hat{T} \vec{x}.$$

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.

Let’s restrict our discussion to non-relativistic physics and Cartesian coordinate (I mean coincidence of contra- an co- variant components, etc…).

I know that a tensor is something that guarantees the covariance of a law of physics when you switch between admissible frames of reference. To achieve this, physical quantities involved in the physics laws must transform according the well-known rule of transformation of the tensor components under change of coordinate (read transformation matrix in this case of Cartesian systems).

Let’s consider two frames of reference with the same orientation but in relative motion along one axis (Galilean transformation). These frames are both inertial, and then the physics law must transform covariantly going from one frame to the other. The transformation matrix in this case is the identity. This means that the law of transformation of tensor components (whatever the tensor rank be) is that each component is invariant going from one frame to the other.

In one frame, a point mass can look static, i.e. its velocity components are all 0. In the other, this point mass looks moving along the axis of relative motion between the frames, i.e. its velocity has at least one component different from 0. Then, the velocity is not a tensor of rank 1.

Similarly, the kinetic energy of this point mass is 0 in one frame, and different from 0 in the other. Then the kinetic energy is not a tensor of rank 0.

Following this interpretation, the position vector is not a tensor, but the displacement vector between two points is a tensor of rank 1…. The acceleration is a tensor of rank 1 (it can be demonstrated that its components remain the same going from one frame to the other).

For this reason

F =m v

cannot be a law of physics, because contains a quantity which is not a tensor, namely the velocity, while

F = ma on the contrary is a correct law of physics (contains only tensor).

- #16

- 17,077

- 8,179

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.

- #17

- 38

- 7

Then the question is spontaneous: can you please give me some examples of quantities which are not tensors in Newtonian mechanics?

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.

- #18

wrobel

Science Advisor

- 745

- 456

- #19

- 5,779

- 1,078

By an "affine space", do you essentially mean a vector space that has forgotten its origin?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.

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 ).

- #20

- 17,077

- 8,179

- #21

- 5,779

- 1,078

(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.

- #22

- 17,077

- 8,179

Which scalar product should that be?

- Replies
- 29

- Views
- 609

- Replies
- 20

- Views
- 1K