1
$\begingroup$

By the idea in Section 13-6 of this chapter I came to up with an absurd question. Actually I understand completely the contents in the above text. The main objective of the above mentioned section is to explain that electric field and magnetic field are relative to one another and depend on the reference frame of the observer. As explained, in the reference frame S' in which the test charge is stationary and wire is moving past it, we see that due to length contraction (as due to the Special Relativity) the charge densities of positive charge and negative charge tranform thus resulting in a net charge density which is not zero which then results in Coulombic Forces between the wire and test charge. I am not diving in the details as all of it is explained very well in the above texts.
So, the question is very simple, if a test charge $q$ is kept near a wire without any current through it, we see no force acting in the scene as all the charges are distributed uniformly throughout the wire. But, when the negative charges start moving in the wire, which constitute current, the length of these should contract, due to the length contraction, thus resulting in a net negative charge density in the wire. Then why is it that we don't see any electrostatic forces between the test charge and the wire?

Note: In the given question, suggested as duplicate, it is already assumed that there is no force on a charge around a current carrying wire.

The current carrying wire doesn't apply any magnetic force on nearby charge q ( positive stationary charge) because it has 0 velocity in lab frame. We found that there is no force on q by wire.

But I am asking the reason for that. Also I have supplied arguments from my side that why (according to me) the charge should experience force. All I am asking is the correction to my wrong idea about the theory.

$\endgroup$
9
  • $\begingroup$ The longitudinal separation between free electrons in a length of current-carrying wire mustn't be treated as contracting like a moving rod. In the latter case we compare the length of the rod measured in the frame in which it is stationary with the separation between its ends when their positions in our frame are marked at the same time according to synchronised clocks in our frame. For the electrons we never consider their separation in the frame in which they are stationary when they are moving in our frame (that in which the wire and $q$ are stationary). $\endgroup$ Commented Jul 4, 2023 at 17:43
  • $\begingroup$ Suppose that we apply a field and set the free electrons in our wire moving at almost the same time. They will rapidly accelerate and reach their 'final' drift speed at almost the same time. If one electron is close to ion A when the field is applied, and another is close to ion B, each will have travelled the same distance from 'its' ion in the same time (as measured in our frame). There will be no contraction. [I'm using a very simple model of a wire as a line of positive ions and free electrons with the same mean spacing as the ions.] $\endgroup$ Commented Jul 4, 2023 at 18:00
  • $\begingroup$ @PhilipWood may I ask why don't we consider the separation of electrons in the frame in which they are stationary? What's the problem in that? $\endgroup$ Commented Jul 4, 2023 at 18:26
  • $\begingroup$ The point I'm trying to make is that in the contraction of the moving rod we DO consider the separation of the ends of the rod in the frame in which its is stationary, while it is moving in our frame, but for the electrons in the wire it is not relevant to consider the separation of electrons in the frame in which they are stationary, while they are moving in our frame. See my second comment above. $\endgroup$ Commented Jul 4, 2023 at 18:52
  • $\begingroup$ Maybe I understand what you are trying to explain but then I have started to question my understanding of the above mentioned text I read before. How come there we are applying length contraction? $\endgroup$ Commented Jul 4, 2023 at 19:25

1 Answer 1

0
$\begingroup$

I have same question while reading the Feynman Lectures recently. Here is my thought.

Let's say that there are two reference frame S and S'. The wire is at rest in S. The moving conduction electrons which constitute current are at rest in the S'. And S' has speed $u$ relative to S in the x-axis direction, which means that the moving conduction electrons have speed $u$ relative to the wire.

The key point is the conduction electrons are initially stationary in S and are suddenly accelerated when an electric field appears in the wire.

Let's focus on S first. Before the acceleration, assume two conduction electrons b and c are separated by $l$ in the wire. Their coordinates are ${x_b}$ and ${x_c }$ respectively, the y and z coordinates are omitted for convenience. At the time $t$, an electric field is suddenly established in the wire, and the two electrons begin to accelerate. So in frame S, the two events b and c start to accelerate are $(x_b,t)$, $(x_c ,t)$ respectively, and $x_b+l=x_c $.

Let's switch the perspective to frame S'. The x coordinate where b starts to accelerate is: $$x'_b = \frac{x_b-ut} {\sqrt{1-u^2/c^2}}$$ The x coordinate where c starts to accelerate is: $$x'_c = \frac{x_c-ut} {\sqrt{1-u^2/c^2}} = x'_b+\frac{l} {\sqrt{1-u^2/c^2}}$$ In addition, in frame S', the time when b and c start to accelerate is different.The time b starts to accelerate is: $$t'_b = \frac{t-ux_b/c^2} {\sqrt{1-u^2/c^2}}$$ The time c starts to accelerate is: $$t'_c = \frac{t-ux_c/c^2} {\sqrt{1-u^2/c^2}} = t'_b-\frac{ul} {\sqrt{1-u^2/c^2}}$$ If we assume that the acceleration process of conduction electrons is completed almost instantly. In S' the events that b and c complete acceleration are $(x'_b,t'_b)$, $(x'_c ,t'_c)$, and the electrons come to rest after the acceleration in S'. So the distance between b and c in S' is: $$x'_c-x'_b = \frac{l} {\sqrt{1-u^2/c^2}}$$ Thus, the distance between b and c in S is: $$\frac{l} {\sqrt{1-u^2/c^2}}\cdot\sqrt{1-u^2/c^2} = l$$ The negative charge density won't change in the wire. In summary, the length contraction occurs when switching reference frames, not when the electrons accelerate.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.