2
$\begingroup$

This question is based on Example 15.12 in John R. Taylor's Classical Mechanics. It concerns the trajectory of a particle experiencing electromagnetic forces as observed in two different inertial reference frames.

In the noted example, Taylor compares the electromagnetic fields of an infinite line charge in two reference frames related by a Lorentz boost along the axis of the line charge (i.e. the z-axis in cylindrical coordinates). In one frame, the line charge is at rest; in the other, it moves uniformly along its length thereby producing a current. In the first frame, there is only a radial electric field, but in the second, there is also a magnetic field circulating around the line charge in the $\phi$-direction.

The example does not analyze this system further, but I noticed something unusual about the behavior of a point charge in these fields. Here's the issue:

A point charge starting at rest in the frame with both an electric and magnetic field will experience acceleration mainly in the radial direction, but it will also experience acceleration along the axis of the line charge (in the z-direction) due to its radial velocity coupling to the magnetic field. Additionally, its radial acceleration will degrade over time due to its z-velocity also coupling to the magnetic field.

Contrast this with the same particle in the other frame. It will have an initial velocity in the negative z-direction, but because there is no magnetic field, it will only experience an acceleration in the radial direction. In this frame, the component of velocity along the z-direction should remain constant and the radial acceleration should not degrade.

These differences should not be possible because the trajectory of the particle needs to be consistent between frames. If it hits a bullseye in one frame, it won't in the other if the trajectories are inconsistent. For example, consider a target at rest with respect to the particle that begins at rest in the frame with both an electric and magnetic field. The target will be missed in that frame but will be hit in the frame where it and the particle move with a constant z-velocity. Therefore, there must be an aspect of this particle's motion that is not being accounted for and which would rectify this discrepancy between frames.

Question: How can this analysis be corrected to guarantee consistent trajectories between frames of reference?

$\endgroup$
8
  • $\begingroup$ Are you aware that 3-acceleration has a complicated and unintuitive Lorentz transformation? I suspect that that resolves the discrepancy. $\endgroup$
    – Ghoster
    Feb 19, 2023 at 7:29
  • $\begingroup$ @Ghoster Totally, but this problem doesn't require transforming either the acceleration or the force. You can calculate the fields in both frames and confirm that they are related through the appropriate transformation (which is the purpose of the book's example). From those fields, you can get the equations of motion. Then, consider a target directly above the particle that starts at rest in the E&B frame. The particle will be deflected and miss it. In the other frame, the target moves at a constant speed with the particle and remains always above it. So in that frame it gets hit. $\endgroup$
    – Geoffrey
    Feb 19, 2023 at 19:05
  • $\begingroup$ @Geoffrey If what you say were true, no one would accept SR. Something is wrong with your analysis. Positions, times, fields, forces, accelerations, etc. all transform consistently to ensure that it doesn’t matter which frame is used. It does not hit the target in one frame and miss the target in another. $\endgroup$
    – Ghoster
    Feb 19, 2023 at 20:54
  • $\begingroup$ @Geoffrey I looked into solving the equations of motion analytically and didn’t make much progress. I suggest that you numerically integrate the trajectory in the two frames and then show that they are Lorentz transforms of each other. $\endgroup$
    – Ghoster
    Feb 19, 2023 at 20:55
  • $\begingroup$ @Ghoster Yeah, I know that it doesn't make sense. That's why I'm asking the question: the analysis seems right, but it can't be. I suspect that the issue arises from radiative braking caused by the energy loss due to the acceleration of the charge. I suspect that it loses more energy in the "moving" frame than the one where it starts at rest and that accounts for its effective deflection away from the target. Since I haven't included the time-like component's equation in this analysis, it seems like a likely culprit. $\endgroup$
    – Geoffrey
    Feb 19, 2023 at 21:06

1 Answer 1

2
$\begingroup$

Electromagnetism is a relativistic theory. If you're going to talk about EM and Lorentz transforms, remember to use relativistic kinematics, too.

In the frame where the line charge has no current, the radial acceleration of the point charge is reduced with time, because the faster a particle is, the harder it is to make it go even faster. It's not $\mathbf F=q(\mathbf E+\mathbf v\times\mathbf B)$ that fails. It is $\mathbf F=m\mathbf a!$ (Thought experiment/mnemonic device: if $1\;\mathrm{kg}$ is moving at $0.9999c$ and you apply a steady $1\;\mathrm N$ of force in the direction of motion, is it possible that the mass accelerates at $1\;\mathrm{m}\,\mathrm{s}^{-2}$ steadily? No!) In fact, $\mathbf F=\mathbf{\dot p}=\frac{d(\gamma m\mathbf v)}{dt}$ doesn't even have to be parallel to $\mathbf a=\mathbf{\dot v}.$ The radial force applied to the particle actually does cause it to slow down in the $z$ direction. (Mnemonic: the light speed limit suggests that, at some point, if you push something, its velocity in other directions must reduce in order to "make room" for more velocity in the direction of the force.)

Note that $F^\mu=ma^\mu$ is true for the four-force $F^\mu$ and four-acceleration $a^\mu.$ These are different from the "Newtonian" 3-force and 3-acceleration $\mathbf F,\mathbf a$ (and $\mathbf F,\mathbf a$ aren't even identical to any components of $F^\mu,a^\mu$), and $\mathbf F,\mathbf a$ are the quantities you are worried about.

So, qualitatively, both frames predict that the radial acceleration of the particle will drop with time and that the $z$-velocity will change. (NB: the frames agree on what "radial" and "$z$-direction" mean, but not on what "acceleration" and "velocity" mean. Be careful making comparisons!)

$\endgroup$
1
  • $\begingroup$ Thanks for the answer. I went back and worked out the details by hand when I had some time, and the appropriate dependence just fall right out of the algebra. When I was learning SR people made such a big deal about the gamma factor not being related to "relativistic mass" that I guess I sort of blocked out the part where the conserved momentum should be $m\gamma u^i$, meaning that acceleration in one direction can affect the transverse velocities through $\gamma$. The intuition that the particle slows down to conserve momentum because the "mass increases" is something I never got. $\endgroup$
    – Geoffrey
    Mar 2, 2023 at 21:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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