Why do the electrons want the rest frame of the wire to appear neutral?

  • Thread starter Thread starter omega_minus
  • Start date Start date
  • Tags Tags
    Relativity
omega_minus
Messages
71
Reaction score
11
I have seen quite a few times the argument about a wire carrying current and a charged particle outside the wire. When the particle is at rest with the wire it feels no force, since the magnetic force on the wire requires relative motion of the particle and the field. It also feels no electric force since the wire is electrically neutral. When the particle moves along the wire it then is accelerated toward (or away) from the wire. In the lab frame of reference its due to a magnetic force but in the particle/current frame its because now the charge densities of the conductor have changed. Here is my question: From the lab frame the electrons carrying current have *somehow* lowered their charge density to a lower amount in their frame so as to appear to the lab frame (once Lorenz contracted) that the wire is neutral. I have read that this is because any excess static electric field would be neutralized by the electrons in the wire. However these explanations never bother to say why the electrons have "decided" to make the wire neutral in another frame of reference and not their own. It seems like they always "want" the rest frame of the wire to appear neutral. Why don't the electrons arrange themselves to make the wire neutral in their own frame? What makes the frame they now travel in so special that they would allow a static field to exist there but not try to neutralize it?
 
Physics news on Phys.org
omega_minus said:
It seems like they always "want" the rest frame of the wire to appear neutral.
It isn't the electrons that want the wire to be neutral, it is the experimenter. He does so by not applying a high voltage to both ends of the wire.

If the experimenter wanted to charge the wire in his frame so that it was neutral in the drift frame of the electrons then he could do so simply by applying a high voltage to both ends of the wire (in addition to the small voltage difference across the wire).
 
I started reading a National Geographic article related to the Big Bang. It starts these statements: Gazing up at the stars at night, it’s easy to imagine that space goes on forever. But cosmologists know that the universe actually has limits. First, their best models indicate that space and time had a beginning, a subatomic point called a singularity. This point of intense heat and density rapidly ballooned outward. My first reaction was that this is a layman's approximation to...
Thread 'Dirac's integral for the energy-momentum of the gravitational field'
See Dirac's brief treatment of the energy-momentum pseudo-tensor in the attached picture. Dirac is presumably integrating eq. (31.2) over the 4D "hypercylinder" defined by ##T_1 \le x^0 \le T_2## and ##\mathbf{|x|} \le R##, where ##R## is sufficiently large to include all the matter-energy fields in the system. Then \begin{align} 0 &= \int_V \left[ ({t_\mu}^\nu + T_\mu^\nu)\sqrt{-g}\, \right]_{,\nu} d^4 x = \int_{\partial V} ({t_\mu}^\nu + T_\mu^\nu)\sqrt{-g} \, dS_\nu \nonumber\\ &= \left(...
In Philippe G. Ciarlet's book 'An introduction to differential geometry', He gives the integrability conditions of the differential equations like this: $$ \partial_{i} F_{lj}=L^p_{ij} F_{lp},\,\,\,F_{ij}(x_0)=F^0_{ij}. $$ The integrability conditions for the existence of a global solution ##F_{lj}## is: $$ R^i_{jkl}\equiv\partial_k L^i_{jl}-\partial_l L^i_{jk}+L^h_{jl} L^i_{hk}-L^h_{jk} L^i_{hl}=0 $$ Then from the equation: $$\nabla_b e_a= \Gamma^c_{ab} e_c$$ Using cartesian basis ## e_I...
Back
Top