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Addendum 1 (Illustration added at the beginning):

What is the total electric field exerted by Line Charges 1 & 3 at Line Charge 2?

Suppose I had 3 identical parallel line charges parallel to the x-axis of infinite length in the same geometric plane.

The first and third line charges are equidistant from the second.

Now suppose I have a series of observers that have different velocities which vary only along the x-axis (i.e. differences in the x-component of velocity only).

For our first observer, the "center of charge" is stationary, and for the rest of the observers the "center of charge" is moving at some velocity directed along the x-axis. The first observer would observe no component of the electric field directed along the x-axis.

However, the others would observe such electric fields directed along the x-axis.

How do we ensure there is no acceleration of the second line charge despite the electric field exerted on it in the x-direction? Is its momentum affected due to the varying "electric potential" therefore "mass" in some compensating way, ensuring no acceleration, as in (mv)' - m'v = mv' = 0?

Addendum 2 (Added complementary information):

One should consider how a neutral wire with an electric current can be represented by the superposition of two line charges with equal and opposite linear charge densities while having different velocities. Notice that in the bottom half of the illustration below labeled "Test charge frame:" (in red) it shows an electric field with a component parallel to the wire that exists because the positive charges contribute to it and yet the negative charges do not negate that contribution. From http://physics.weber.edu/schroeder/mrr/mrrtalk.html Point charge moving toward a wire, in both frames

I've included information about this because some may doubt that electric field components in the x-direction are seen by observers having different component velocities in the x-direction than those of the line charges.

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    $\begingroup$ I don't understand exactly the situation you're trying to explain (if you could, try to describe it better or include some figures), but keep in mind that you need Lorentz Transformations and Special Relativity if you want to change the reference frame when electromagnetic fields are present. $\endgroup$ – ErickShock Apr 14 '19 at 18:40
  • $\begingroup$ What is the distance of wire 1 and 3? Is the same as 1 and 2? Are they in one plane? What is the speed of the first and other observers? Why did you state "others would observe such electric fields directed along the x-axis"? $\endgroup$ – DanielTuzes Apr 14 '19 at 20:10
  • $\begingroup$ @DanielTuzes, Let's have the lines exist in the XY plane for example. Let the y positions of these lines be -d, 0, and +d respectively, where d is a function of time as specfied in the inertial frame observer who sees the least possible system momentum where line 2 is stationary. $\endgroup$ – Kevin Marinas Apr 14 '19 at 20:21
  • $\begingroup$ @DanielTuzes, The observers I specified I believe were sufficent to illustrate problem I encountered as to the electric fields that would or would not be produced at the x-axis as a function of the observer. The length contraction differences between these frames are confined to the x-direction for members of the set of observers as there is no transverse length contraction. The first observer sees no x-component of the velocity of the line charges. Each of the other observers will see the three lines moving in the x-direction at some common rate. $\endgroup$ – Kevin Marinas Apr 14 '19 at 20:31
  • $\begingroup$ @ErickShock, Per your suggestion, I added an illustration figure with some additional information about the problem. Hopefully, I have stated the problem clearly enough that proceeding to apply Lorentz Transformations and Special Relativity will make sense. Also, I am seeking an intuition for the result that I should expect from the application of these. $\endgroup$ – Kevin Marinas Apr 14 '19 at 22:50
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As I said in my comment, you need special relativity to transform the components of the $\mathbf E$ and $\mathbf B$ fields. They transform according to $$ \begin{align} \mathbf E'_{||} &= \mathbf E_{||} \\ \mathbf B'_{||} &= \mathbf B_{||} \\ \mathbf E'_{\perp} &= \gamma (\mathbf E_\perp + \mathbf v \times \mathbf B_\perp) \\ \mathbf B'_{\perp} &= \gamma \left ( \mathbf B_\perp -\frac{1}{c^2} \mathbf v \times\mathbf E_\perp \right ) \end{align} $$ where the primed quantities denote the (total) fields at a frame moving with velocity $\mathbf v$ relative to your original frame ($\mathbf v$ is the velocity of any observer), $\gamma$ is the Lorentz factor and the subscript $||$ denotes the component parallel to $\mathbf v$ and $\perp$ the perpendicular component. As you can see, the parallel components remain the same, so no acceleration in the $x$-direction will be measured by any observer. The cool stuff happens in the second figure, where a magnetic field is created whose force exactly counterbalances the electric force in order to don't accelerate the charge.

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  • $\begingroup$ In my examples, the line charges are oriented in the x-direction and their velocities have components in the x-direction and the y-direction. If I alter the x component of the velocity of the line charges, wouldn't that affect the x component of the electric field they produce since they also have a y component to their velocity? If instead I simply changed the x-velocity of the observer by the opposite amount, why wouldn't I get essentially that same result with the change in relative velocity being the same? Is the picture I included in "Addendum 2" wrong in some way? $\endgroup$ – Kevin Marinas Apr 15 '19 at 5:09
  • $\begingroup$ @Kevin Marinas First, no. The parallel components of the field doesn't change when you alter the $x$ component of their velocity. And yes, the second image is not right. SR demands that both "lines" contract, increasing their charge density. But since they contract in the same way, the net effect keeps the wire neutral. $\endgroup$ – ErickShock Apr 16 '19 at 3:37
  • $\begingroup$ The second image involves a boost in the y direction not the x direction. So its charge density should not change. The test charge in that picture is moving perpendicular to the wire and experiences a Lorentz Force parallel to that wire. Whether that Lorentz force is magnetic or electric depends on the type of observer, lab frame or test charge frame respectively. In the test charge frame, it seems that the diagonal movement of the + charge is what gives rise to the x component of the electric field. $\endgroup$ – Kevin Marinas Apr 17 '19 at 4:28
  • $\begingroup$ I have no problems understanding that, if I were an observer seeing a charge at rest, then no matter how I boosted my frame, the component of the electric field in the direction of my boost should not change. However, after that boost, I am in a frame where the charge is moving. If I boost again, but in a non-colinear direction, should the electric field really be unchanged in the direction of that second boost? I would think that the electric field would be essentially unchanged in the direction of the "combined boost" while changing perpendicular to that. $\endgroup$ – Kevin Marinas Apr 17 '19 at 5:11

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