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Let's say a charged particle is at rest in a uniform magnetic field. Now if we change our frame of reference the particle appears moving. Now according to classical physics the particle should move in a circle in this frame ( this can't be true because we know that the particle must move in a straight line in this frame.) Thus we conclude that the electromagnetic field changes. Can you use relativity to find the field in the moving frame so that physics becomes self consistent.
Note: If a source is strictly necessary to find the change in the field consider the source as an infinitely long sheet carrying a uniform areal current along it's area. It should be easy to see using ampere's circutal law that the field is uniform and can be calculated easily.

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When you change to a different reference frame, both E and B change. You've started in an E=0 special case, but in another frame E won't be zero.

$E^2-c^2B^2$ is the conserved quantity (There are several sign and unit conventions, but this is the one I'm used to). As you move to other frames, you'll get a non-zero $E$ and an increasing $B$. Together, those will be consistent with what the observer sees the charged particle do.

Let's say the particle starts at the origin, and the $B$ field is along $z$. Now pick a frame where it's moving along $x$. You'd expect the $B$ field to deflect it along $y$. But in that frame, you'll observe an $E$ field that provides exactly enough force to cancel that, giving a net force of zero and no deflection.

The components of $B$ and $E$ don’t transform separately in SR. Like $x$ and $t$ in the usual Lorentz transform, they mix together. The mathematical formulation of this starts by combining $E$ and $B$ into an antisymmetric tensor called $F$. This Physics SE question and answers take it from there.

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  • $\begingroup$ How do we get the direction of the fields in the moving frame? $\endgroup$ – Karthik Jan 24 at 12:38
  • $\begingroup$ Added a bit more on the transform. It’s a big topic.. $\endgroup$ – Bob Jacobsen Jan 24 at 16:54
  • $\begingroup$ Yeah the Faraday tensor. So we apply Lorentz transformation matrices to the Faraday tensor matrix to use it in other frames? $\endgroup$ – Karthik Jan 26 at 3:18
  • $\begingroup$ Yes. Two of them, transforming both axes. $\endgroup$ – Bob Jacobsen Jan 26 at 3:57

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