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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?

Sid
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2 Answers2

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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!

V Govind
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  • 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. – Sid Jul 27 '23 at 02:14
  • 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? – Sid Jul 27 '23 at 13:15
  • 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 $\vec{F'}-\vec{F'}_{coulomb} = -q\vec{E'}v^2/c^2$?? – Sid Jul 27 '23 at 14:59
  • 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 – Sid Jul 27 '23 at 15:01
  • Yes, your first statement is correct, provided that the field and force under consideration is perpendicular to the velocity of the frame. – V Govind Jul 27 '23 at 16:36
  • And for the second comments, I am not aware if there is a particular reason why the electric field transforms. But yeah, in the transformation of x component of electric field for a point charge, Lorentz transform of space-time coordinates do play the role. – V Govind Jul 27 '23 at 16:38
  • But E' is $\gamma kq/r^2$ so wouldn't the magnetic field due to a single moving particle be $\gamma \mu_0qv/ 4\pi r^2$? Afaik the expression doesn't have the Lorentz factor – Sid Jul 28 '23 at 04:53
  • There is a factor of $\gamma$, but I must warn you that these formulae hold true only when you are considering a direction that is perpendicular to the uniform velocity of the charge. Otherwise the field terms will have a factor $\frac{1-\frac{v^2}{c^2}}{(1-\frac{v^2sin^2\theta}{c^2})^{3/2}}$, where $\theta$ is the angle between velocity and the vector pointing from the charge to the point where you need the field – V Govind Jul 28 '23 at 14:16
  • Yes, I understand the formulae are only for this specific case of $E \bot v$. So Biot Savart's law is only valid for nonrelativistic drift speeds? – Sid Jul 28 '23 at 14:20
  • Let us continue this discussion in chat. – V Govind Jul 28 '23 at 14:24
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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.

stuffu
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  • This does make things clearer, however then wouldn't the observed 'magnetic' force be $F_{coulomb}(1-\gamma)$? – Sid Jul 27 '23 at 13:19
  • 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? – Sid Jul 27 '23 at 15:26
  • @Sid Yes to second comment. – stuffu Jul 27 '23 at 21:51
  • @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. – stuffu Jul 27 '23 at 21:53
  • 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. – stuffu Jul 28 '23 at 01:07
  • What I meant was we observe an extra force on a moving particle on top of the electrostatic force experienced by a charge stationary in our frame. We call that magnetism. Have I got something wrong? – Sid Jul 28 '23 at 04:50
  • @Sid I guess it's right. Not sure. I just know relativity. Maybe it's so that classical physics does not know about length contraction, so it gets the force between moving charged rods wrong? But manages to get point charges right by good luck. :) – stuffu Jul 28 '23 at 07:21