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From what I've read on magnetism as a result of incorporating special relativity into electrostatics, I have got the impression that:

a) If you have two parallel and (for simplicity) infinite streams of moving particles with equal velocity, in the particles' frame they observe each other to be stationary. The Coulomb force on any one particle is given by $qE$ where E is resultant field.

b) Viewing the same situation from a laboratory reference frame, we observe a length contraction for both streams. Then the particles appear closer to each other, so the density appears to us to be more than what the particle observes. So we predict a larger Coulomb force of repulsion than the actual force that the particle feels, and introduce the magnetic field as a 'correction'. For a section of aforementioned infinite streams of length x and separation r: $$ F_{net} = (F_{coul})_{frame} = (F_{coul})_{obs} + F_b \implies F_b = (F_{coul})_{frame} - (F_{coul})_{obs} $$ $$ F_b = \frac{1}{2\pi\epsilon_0r}\lambda_{frame}^2x - \frac{1}{2\pi\epsilon_0r}\lambda_{obs}^2x $$ Now by apparent length contraction of the particle stream in the lab frame: $\lambda_{frame} = \lambda_{obs}\sqrt{1 - v^2/c^2} $ $$ \implies F_b = -\frac{1}{2\pi\epsilon_0r}\lambda_{obs}^2x(1-\frac{c^2 - v^2}{c^2}) = -\frac{\lambda_{obs}^2v^2x}{2\pi\epsilon_0rc^2} = -\frac{i^2x}{2\pi\epsilon_0rc^2} =-\mu_0\frac{i^2x}{2\pi r} $$ I.e. the force is attractive in nature and this agrees with $ c^2 = \frac{1}{\mu_0\epsilon_0} $ and what was predicted by Biot-Savart's law.

The question arises what happens if there is no stream but just, say, two parallel particles moving with the same velocity? There is no question of any charge density observed or otherwise. But in the particle's frame of reference there is just the Coulomb force, and in our frame of reference there is also the Coulomb force, which should not change as there is no length contraction observed. What am I getting wrong here? Surely one moving charged particle generates a magnetic field (which I was told is true in any frame where the charge moves), and the field would act on the parallel particle? How do we explain this with special relativity?

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  • $\begingroup$ P.S. am a high school student, not familiar with vector calculus $\endgroup$
    – Sid
    Commented Jul 26, 2023 at 11:19
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    $\begingroup$ Try reading about Lienard-Wiechert potentials for point charges: we can derive the expression for the electric and magnetic field from this, and thus know about the magnetic field of this point charge. This formulation is consistent with special relativity. In terms of relativistic formulas, you can read about how the fields transform under coordinate transformation, and use those formulae to get your answer. $\endgroup$
    – V Govind
    Commented Jul 26, 2023 at 11:52
  • $\begingroup$ You get everything wrong, like many people. There is no magnetic force. We explain away such things. It's particularly easy in this case. $\endgroup$
    – stuffu
    Commented Jul 26, 2023 at 21:58
  • $\begingroup$ Magnetism is a result of incorporating special relativity into electrostatics. So there is no magnetism. Just special relativity and electrostatics. $\endgroup$
    – stuffu
    Commented Jul 27, 2023 at 1:22
  • $\begingroup$ @stuffu yes that's exactly what I thought. But I'm unable to explain any magnetism in the two parallel particles case by considering relativistic effects in electrostatics. That's what the question is. $\endgroup$
    – Sid
    Commented Jul 27, 2023 at 2:15

2 Answers 2

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So imagine two point, positive charges at rest, in a frame S: the interaction is purely Coulombic, and the force on one another has magnitude $\frac{q}{4\pi\epsilon_0r^2}$, $r$ is the distance between them. Now consider the events from a frame, which I denote as $\bar S$, moving along the x-axis at velocity v: the charges are moving according to this frame. Now, the transformation for fields in special relativity is as follows: suppose one has a frame $S$ with electric and magnetic fields denoted by $(E_x, E_y, E_z)$ and $(B_x, B_y, B_z)$ and another frame $\bar S$ moving at speed v, along the x-axis, relative to S, where the fields are given by $(\bar{E_x},\bar{E_y},\bar{E_z})$ and $(\bar{B_x},\bar{B_y},\bar{B_z})$. Then,

$\bar{E_x} = E_x, \bar{B_x}=B_x$

$\bar{E_y} = \gamma(E_y-vB_z), \bar{B_z}=\gamma(B_z-\frac{v}{c^2}E_y)$

$\bar{E_z} = \gamma(E_z+vB_y), \bar{B_y}=\gamma(B_y+\frac{v}{c^2}E_z)$

This derivation can be found in textbooks like Griffiths' Introduction to Electrodynamics or Feynman's Lectures. In our case, from $S$ there is only an electric field, which will have three components, and a magnetic field which is zero. However, when looking from $\bar S$, there are electric as well as magnetic fields due to the transformation equation I have stated. So yeah, the Coulombic field does change, and we also have a magnetic field here!

Edit: Okay, I am so sorry, I understood your question now, and I'm removing the previous edit. As someone else has remarked here, there is a transformation law for the perpendicular force acting on the particle, which goes as $\bar F = \frac{F}{\gamma}.$ Here, what's important is that from one of the frames, the particle must be at rest, and F is the force on the particle as measured from this frame and $\bar F$ in the other frame. Now for your case, imagine two charges moving at some velocity $\bar u$ along the x-axis relative to a frame $\bar S$, one above the other. $\bar S$ itself is moving at velocity v along the x-axis relative to another frame $S$. What's special about S is that the charges are at rest from this frame(This translates to $\bar u = -v$, but I'll leave you to worry about how that happens!). So, the force between the charges is purely electrostatic, and I will call the direction of the force along the y-axis, and $F=\frac{q^2}{4\pi\epsilon_0y^2}$, say, on the upper charge. Now, from our rule, $\bar F = \frac{F}{\gamma}$, we find that in the frame $\bar S$, the force is reduced. However, we know that by applying relativity to a single charge, the electric field from $\bar S$ should be $\gamma E$, where $E$ is the field from $S$. so we expect the force to also increase by $\gamma$, which does not happen. This hints at the presence of another frame-dependent force acting opposite to the electric field in this frame, and it's of course what we call a 'magnetic force!' Hope this helped!

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  • $\begingroup$ I was looking for a reason for why any magnetic field arises at all if one incorporates special relativity in to the electrostatics of two charged particles. $\endgroup$
    – Sid
    Commented Jul 27, 2023 at 2:14
  • $\begingroup$ Thanks for the edit. I understand the case for a wire- the change in charge density is evident. What I failed to understand is the case of parallel particles where there is no charge density to speak of. Or can we model it as a sphere rather than a point of charge? $\endgroup$
    – Sid
    Commented Jul 27, 2023 at 13:15
  • $\begingroup$ Ohh thanks so much that makes a lot of sense! Just to make sure I have the right understanding, say I have two frames S (particles') and S' (my frame): $\vec{E'} = \gamma\vec{E} ; \vec{F'} = \vec{F}/\gamma$ Then I see an additional 'magnetic force' of $\vec{F'}-\vec{F'}_{coulomb} = -q\vec{E'}v^2/c^2$?? $\endgroup$
    – Sid
    Commented Jul 27, 2023 at 14:59
  • $\begingroup$ Also, can the perceived rotation of spacetime axes between relatively moving frames be considered as the (intuitive) reason for change in electric field? I was not familiar with this at all, hence asking $\endgroup$
    – Sid
    Commented Jul 27, 2023 at 15:01
  • $\begingroup$ Yes, your first statement is correct, provided that the field and force under consideration is perpendicular to the velocity of the frame. $\endgroup$
    – V Govind
    Commented Jul 27, 2023 at 16:36
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http://www.sciencebits.com/Transformation-Forces-Relativity

The last equation is the only relevant thing in this case.

F′y=Fy/γ

Clock hand of a moving clock experiences a reduced force in the frame where the clock is moving. Because of time dilation.

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  • $\begingroup$ This does make things clearer, however then wouldn't the observed 'magnetic' force be $F_{coulomb}(1-\gamma)$? $\endgroup$
    – Sid
    Commented Jul 27, 2023 at 13:19
  • $\begingroup$ And, what I understand is- the same event of $dp_y$ appears to take more time for the moving observer so observed force is less? $\endgroup$
    – Sid
    Commented Jul 27, 2023 at 15:26
  • $\begingroup$ @Sid Yes to second comment. $\endgroup$
    – stuffu
    Commented Jul 27, 2023 at 21:51
  • $\begingroup$ @Sid Observed magnetic force? What is that? "Magnetism" agrees with general force transformation. Like magnetic force cancels out Coulomb force at speed c, and so on. $\endgroup$
    – stuffu
    Commented Jul 27, 2023 at 21:53
  • $\begingroup$ If we go with the idea that magnetism is replaced by the general force transformation of SR, then what is "an observed magnetic field"? If we do not go with the idea that magnetism is replaced by the general force transformation of SR, then there seems to be a special case when the almost general force transformation law does not apply. At least In this simple case it's better to go with the SR. $\endgroup$
    – stuffu
    Commented Jul 28, 2023 at 1:07

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