Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A Relativistic conduction current density?

  1. Jul 1, 2016 #1

    I am an electrical engineer rather than a physicist, however, I am trying to understand the physics of a twin wire transmission line in terms of the charge and current density. Lets say we have a lossless, infinite length, twin wire transmission line, a step current is induced into the line and the signal wavefront propagates at the speed of light from the start to the line to infinite. My physical understanding is that conventional current moves on average at drift velocity (a few mm/second) even though the signal propagates at the speed of light. This is due to the fact that the lines are electrically neutral and the electric field associated with the wavefront travelling at the speed of light is terminated by mobile charges moving slightly toward or away from the conductor surface, bearing in mind that the lattice is fixed positive charge. This effectively supplies the surface charges terminating the transverse electric field.

    Now, my thoughts were confirmed by the book 'The Power and Beauty of Electromagnetic Fields', by Frederic R. Morgenthaler. (Chapter 18 - TEM Transmission Lines - pg. 189-191)

    However, as my knowledge of relativity is rather limited, I am struggling to understand one of his equations;


    which he calls the relativistic current-density as c is the speed of light. He says this equation "arises from imposition of the Lorentz-gauge and conservation of charge", without any derivation.

    I do not know where this comes from and physically its distinction from my classical understanding of convection current density j=ρu where u is the drift velocity. It appears that his equation is a fictional current density relating to the relativistic signal propagating at the speed of light?

    Can anyone shed any light on this equation or provide a derivation?

    Thanks in advance, apologies for any ignorance on my behalf...


  2. jcsd
  3. Jul 1, 2016 #2


    User Avatar
    Science Advisor
    2016 Award

    This cannot be true! Your syntax checker should realize this immideately!

    A current density is a vector, and of course the Maxwell theory is a fully covariant relativistic field theory, although usually not written in a manifestly covariant way. This is for the reason that it's easier for us to formulate problems (particularly those of electrical engineering!) in the good old 1+3-dimensional formalism of 3D vector calculus.

    If you describe the currents as the flow of a charged fluid (which is a very good model for many purposes, among them the flow of electrons in a metal etc.) then it's described by a charge density ##\rho(t,\vec{x})## and a current density ##\vec{j}(t,\vec{x})##. Relativistically it's very convenient to define a four-vector density,
    $$j^{\mu}=(c \rho,\vec{j}).$$
    The reason is that this is a Minkowski-vector field, and it has well-defined simply transformation properties under Lorentz transformations:
    $$j^{\prime \mu}(x^{\prime \mu})={\Lambda^{\mu}}_{\nu} j^{\nu}(x) = {\Lambda^{\mu}}_{\nu} j^{\nu}(\Lambda^{-1} x'),$$
    where ##x=(x^0,\vec{x})=(c t,\vec{x})## is the time-position four vector.

    Now if ##\vec{v}(t,\vec{x})## is the usual velocity field of the fluid, the current density is given by
    $$\vec{j}=\rho \vec{v},$$
    but this is a bit ugly from the point of view of the Lorentz transformation properties. The problem is that ##c \rho## transforms as the time component of the four-vector ##j^{\mu}## while ##\vec{v}## has complicated transformation properties, but we know that ##j^{\mu}## is a four-vector. Thus we can characterize the charge density by the scalar quantity
    $$j_{\mu} j^{\mu}=(c^2 -\vec{v}^2) \rho^2 = \frac{c^2}{\gamma^2} \rho^2 = c^2 \rho_0^2.$$
    Thus if we define
    $$\rho=\gamma \rho_0$$
    we rather get
    $$(j^{\mu})=\rho_0 \gamma (c,\vec{v}) = c \rho_0 u^{\mu},$$
    $$u^{\mu}=\gamma (1,\vec{v})$$
    is the four-velocity of the particle. Obviously by construction ##\rho_0## is a scalar field and the flow-velocity four-vector ##u^{\mu}## is thus a four-vector since ##j^{\mu}## is a four-vector to begin with.

    The physical meaning of ##\rho_0## is also simple. If you boost to a reference frame where ##\vec{v}=0## at the considered moment at the consideres place, this means your fluid cell at this space-time point is momentarily at rest, and in this reference frame thus you have ##\rho=\rho_0##. Thus ##\rho_0## is the charge density of the fluid in the local restframe of the fluid, which is a scalar.
  4. Jul 2, 2016 #3


    Staff: Mentor

    I agree with @vanhees71 the syntax of this equation is wrong. Depending on the context j is either a three vector or a four vector. In either context ##\rho_0## is a scalar and c is a constant. And there is no context in which the product of a scalar and a constant is a vector.
  5. Jul 18, 2016 #4
    Thanks for the responses. Yes, this is one of the reasons for my question, the equation is particularly strange. Though, I did think that it was my understanding that was at fault because it was found in a major IEEE publication written by an MIT Professor!

    Thanks for your help.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Relativistic conduction current density?
  1. Relativistic Currents (Replies: 16)