Relativistic electrodynamics

In summary, the conversation discusses the calculation of magnetic and electric fields in a wire with a current using Ampere's law and Gauss' law in two different frames of reference, O and O'. The tricky part is determining the charge density and current density in O' relative to O, which involves considering the drift velocity and length contraction. There is uncertainty about how to approach this calculation and whether or not the velocity of the electrons carrying the current needs to be taken into account.
  • #1
bcoats
9
0
If someone could help me with this one that would be great. My prof has not been in his office in quite some time...

-A wire of cross-sectional area A carries a current I, and has zero net electric charge in frame O.
(a) Find the magnetic field a distance r from the axis of the wire
(b) Find the charge density ρ' and current density j' in frame O' moving with velocity v parallel to axis of wire relative to O.
(c) Using Ampere's law and Gauss' law with the charge and current densities found in (b), calculate the electric and magnetic fields in O' at a point a distance r from the axis of the wire.
(d) Verify that these fields agree with the ones obtained by transforming from O.

OK, so here's where I'm at:
(a) I expect B to be the same as in non-relativistic e.d., μ0*I/2πr.

(b) This is the tricky part. I don't know how to look at ρ. If simply looking at the ρ+ (charge density for positive charges), then ρ' should be greater than ρ0 by a factor of gamma due to length contraction. But if we are looking at ρ- (negative charges), then it's more complicated since it increases by more than a factor of gamma if v is moving in the same direction as the electrons. But if we just use the invariant, and ρ is the net charge density, then ρ0=0, and we have j^2=(c^2)(ρ'^2)+(j'^2). I'm assuming that j'=I'/A (where A is the cross-sectional area). So then we have an equation relating j to ρ', but how to solve for either?

(c) Of course, I can't figure this out until I figure out (b), but I think that ε0*E'*A=q', meaning that E'=q'/(ε0*2*pi*r*l)=ρ'/(ε0*2*pi*r). So once I figure out ρ' I should have that part. Then I have B'=μ0*I'/2πr. I' should be less than I by a factor of gamma due to time dilation, I think. So I believe B'=μ0*(I/γ)/2πr.

Anyways, if anyone gets a chance to help me, thanks so much. It's due tomorrow, but posts after that will still help me understand.

Ben
 
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  • #2
Is v meant to be an arbitrary velocity, or the drift velocity of the current charge carriers? If it is arbitrary, the drift velocity in O should still come into play. You will need th relativistic velocity addition formula to find the velocity of the drifting charges in O'
 
  • #3
v is the arbitrary linear velocity of frame O' wrt frame O. I thought about using relativistic velocity addition to determine the velocity of the electrons in the O' frame...but how can you figure that out? I suppose that the velocity of the electrons carrying the current should be some function of the current itself (among other things), but we've never talked about anything like that in class...although that's not exactly unusual.
 
  • #4
bcoats said:
v is the arbitrary linear velocity of frame O' wrt frame O. I thought about using relativistic velocity addition to determine the velocity of the electrons in the O' frame...but how can you figure that out? I suppose that the velocity of the electrons carrying the current should be some function of the current itself (among other things), but we've never talked about anything like that in class...although that's not exactly unusual.
I think you need the drift velocity to do the problem. There has to be a different effective density of the opposite charges in the wire, so different length contractions come into play. The current is related to the charge density and the drift velocity.
 

1. What is the concept of Relativistic Electrodynamics?

Relativistic Electrodynamics is a branch of physics that studies the interaction between electric and magnetic fields and how they are affected by the principles of special relativity. It explains how electric and magnetic fields are interconnected and how they can be transformed into each other in different frames of reference.

2. How does Relativistic Electrodynamics differ from Classical Electrodynamics?

Relativistic Electrodynamics takes into account the effects of motion and time dilation on electric and magnetic fields, while Classical Electrodynamics only considers stationary frames of reference. It also introduces the concept of four-vectors and tensors to describe electromagnetic phenomena in four-dimensional spacetime.

3. What is the relationship between Relativistic Electrodynamics and Einstein's theory of Special Relativity?

Relativistic Electrodynamics is built upon the principles of Special Relativity, which states that the laws of physics should remain the same for all observers in uniform motion. It extends these principles to explain the behavior of electromagnetic fields and their transformation between different frames of reference.

4. Can Relativistic Electrodynamics explain the constancy of the speed of light?

Yes, Relativistic Electrodynamics is based on the assumption that the speed of light is constant for all observers, regardless of their relative motion. This is a fundamental principle of Special Relativity and is essential in understanding the behavior of electromagnetic fields in different frames of reference.

5. How is Relativistic Electrodynamics applied in modern technology?

Relativistic Electrodynamics has many practical applications, including in the development of particle accelerators, GPS systems, and medical imaging technologies. It also plays a crucial role in the study of high-energy physics and the behavior of particles at near-light speeds.

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