Question about SR in Griffiths

In summary, the conversation discusses the topics of frames and transformations in the context of electromagnetism. The example of a parallel plate capacitor is used to demonstrate how the electric field and charge density change in different frames of reference. The concept of surface currents is also introduced. The confusion arises when considering two special cases, E=0 and B=0, in a particular frame S. It is clarified that S is just any inertial frame and not necessarily related to the initial example. The conversation ends with a request for further clarification and a note that Griffiths uses different physical situations to motivate the transformation equations.
  • #1
VortexLattice
146
0
Hi, I'm reading the section about "How the fields transform" in Griffiths' Intro to Electrodynamics book, but I'm a little confused and want to understand this fully.

He starts off with the example of a parallel plate capacitor where each plate has charge density [itex]\sigma _0[/itex]. So in the rest frame [itex]S_0[/itex] of the plates, the electric field in between the plates is [itex]E_0 = \sigma _0 /\epsilon_0[/itex]. Then he says, let's look at a frame (S) that's moving parallel to the plates (like, along them) at speed [itex]v_0[/itex]. Then, because the length of the plates is contracted in this frame but the charge is constant, the charge density increases in this frame and thus the E field increases to [itex]E = \gamma_0 E_0[/itex].

So this makes sense. Then he says he wants to look at magnetic fields. So he says, in frame S, there's a magnetic field due to the surface currents on each plate: [itex]K = \sigma v_0[/itex]. I assume these "surface currents" are from the fact that in frame S, it look like the charge on the plates is moving at the speed S is moving with respect to [itex]S_0[/itex]. If this is true, that makes sense too.

Then he introduces yet another frame, [itex]\overline{S}[/itex], that's moving at speed v relative to speed S. He derives the equations for the fields in this frame in terms of the fields in frame S. But this is what I'm confused about: He looks at two "special cases", when E = 0 and when B = 0, in frame S. But how can either equal zero? I guess B could be 0 if frame S isn't actually moving at all with respect to [itex]S_0[/itex], so it's basically the same as S, but I still can't see how E would be 0. I thought we basically chose to look at S because it necessarily has a nonzero magnetic field.

If anyone has a copy handy, it's on pages 525-532. Can anyone help me?

Thanks!
 
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  • #2
You are right, there cannot exist a frame where E=0 in this scenario since (E^2-B^2) is an Lorentz invariant and so if there exists some frame where E>B, then one cannot transform away E using a Lorentz transformation (and vice versa).

As I don't have a copy of the book, I can't really help much more than that, sorry.
 
  • #3
Griffiths uses two specific physical situations, a parallel plate capacitor and a solenoid, to motivate the transformation equations for general fields. After Griffiths writes down the general transformation equations, he then treats cases that don't necessarily have anything to do with the motivating cases.
 
  • #4
George Jones said:
Griffiths uses two specific physical situations, a parallel plate capacitor and a solenoid, to motivate the transformation equations for general fields. After Griffiths writes down the general transformation equations, he then treats cases that don't necessarily have anything to do with the motivating cases.

Ok, but he's still referring to some type of frame when he says "When B = 0 is frame S". So what is S here, just any type of frame now?
 
  • #5
VortexLattice said:
Ok, but he's still referring to some type of frame when he says "When B = 0 is frame S". So what is S here, just any type of frame now?


Yes, Griffiths means:

"Suppose there is an electromagnetic field and an inertial frame S such that in S, B = 0. Then, in another inertial frame S' that moves with velocity v with respect to S, ..."
 
  • #6
George Jones said:
Yes, Griffiths means:

"Suppose there is an electromagnetic field and an inertial frame S such that in S, B = 0. Then, in another inertial frame S' that moves with velocity v with respect to S, ..."

Ahhh, ok, thanks!
 

Related to Question about SR in Griffiths

1. What is SR in Griffiths?

SR stands for Special Relativity and it is a theory proposed by Albert Einstein in 1905. It describes the motion of objects in the absence of gravitational forces and is based on the principle of the constancy of the speed of light.

2. What are the main concepts of SR in Griffiths?

The main concepts of SR in Griffiths include the principle of relativity, time dilation, length contraction, and the relativity of simultaneity. These concepts help explain how time and space are relative and how they are affected by the speed of an observer.

3. How is SR in Griffiths different from classical mechanics?

Classical mechanics is based on the laws of motion proposed by Isaac Newton, which work well for everyday objects moving at slow speeds. However, SR in Griffiths takes into account the effects of high speeds and the constancy of the speed of light, making it a more accurate theory for objects moving at high speeds.

4. What are some real-life applications of SR in Griffiths?

SR in Griffiths has been used in various fields such as particle physics, astrophysics, and engineering. It is also used in the development of technologies such as GPS systems, particle accelerators, and nuclear power plants.

5. How does SR in Griffiths impact our understanding of the universe?

The principles of SR in Griffiths have led to a better understanding of the nature of time, space, and the universe. It has also helped in the development of other theories such as General Relativity, which describes the effects of gravity on the motion of objects.

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