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A Is the current density a vector?

  1. Jun 27, 2016 #1
    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?
     
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  3. Jun 27, 2016 #2

    BiGyElLoWhAt

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    ##J = \nabla * I##
    not sure what you mean by polar vector. You can express it in polar coordinates, if you like.
     
  4. Jun 27, 2016 #3
    https://en.wikipedia.org/wiki/Pseudovector

    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?
     
  5. Jun 27, 2016 #4

    BiGyElLoWhAt

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  6. Jun 27, 2016 #5

    robphy

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    Since ##\vec j=\sigma \vec E## (Ohm's Law), and ##\sigma## is a scalar, then ##\vec j## is polar since ##\vec E## is polar.
     
  7. Jun 27, 2016 #6

    Andy Resnick

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    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
     
  8. Jun 27, 2016 #7

    robphy

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  9. Jun 28, 2016 #8

    vanhees71

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    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.
     
  10. Jun 28, 2016 #9
    Thank you very much for your answers.

    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.
    I think that in this case the law of transformation from one coordinate system to the other of a tensor of any rank is the identity, i.e. its components stay the same in the two reference systems.

    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: j = σ⋅E, where E is surely a tensor.

    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.
     
  11. Jun 28, 2016 #10

    vanhees71

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    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
    $$n'(t',\vec{x}')=n(t,\vec{x})=n(t',\vec{x}+\vec{w} t), \quad \vec{v}'(t',\vec{x}')=\vec{v}(t,\vec{x})-\vec{w} = \vec{v}(t',\vec{x}'+\vec{w}t)-\vec{w}.$$
     
  12. Jun 28, 2016 #11
    To be invariant under rotation is enough to be a tensor?
     
  13. Jun 29, 2016 #12

    vanhees71

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    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).
     
  14. Jun 29, 2016 #13
    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?
     
  15. Jun 29, 2016 #14

    vanhees71

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    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.
     
  16. Jun 29, 2016 #15
    I probably have some conceptual problem.

    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).
     
  17. Jun 29, 2016 #16

    vanhees71

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    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.
     
  18. Jun 29, 2016 #17
    Then the question is spontaneous: can you please give me some examples of quantities which are not tensors in Newtonian mechanics?
     
  19. Jun 29, 2016 #18
    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
     
  20. Jun 29, 2016 #19

    robphy

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    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 ).
     
  21. Jun 30, 2016 #20

    vanhees71

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    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.
     
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