Velocity frame of reference in lorentz force equation

[Atheer]

Hi, I am new here, but there is a question that really bugs me :-(. In Lorentz force why was the velocity defined as relative to the observer and not relative to thesystem generating the magnetic field. By defining it relativ eto the system relativity is not needed to explain how an observer moving with the charge explains the force on the charge.

 

[Dale]

It is relative to any inertial frame. Not just the one where "the system generating the magnetic field" is at rest. That is why relativity is needed.

Btw, there may not be such a frame when the system has moving parts, such as a generator or motor.

 

[Atheer]

I mean if we are in the reference of frame of the moving charge, and we use in the lorentz law qVXB the velocity difference between the wire and the charge instead which is V in this case instead of V=0 as is normally done.

 

[Dale]

Your sentence structure is a little confusing, so I am not certain exactly what you are trying to say. However, why should you be forced to use the reference frame where some charge is at rest or the reference frame where some wire is at rest? What if you have a system with two wires or two charges moving with respect to each other? Then your analysis breaks down.

No, the proposal doesn't make sense. Either the principle of relativity holds and you can use Maxwell's equations in any inertial frame, or the principle of relativity does not hold and you can only use Maxwell's equations in one particular reference frame. You cannot have that Maxwell's equations hold only in the frame of wires or only in the frame of charges because then you get scenarios with multiple wires or charges that cannot be analyzed.

 

[Dale]

The link [mentor's note: DaleSpam is referring to a link posted in a previous post which as been removed] reference to aisn't working on my mobile device. I will try again later.

In the meantime, please address my point. How can such a restriction possibly be valid given that some systems will have multiple wires or multiple charges?

 

[Atheer]

Absolutely, i am thinking about your point, which is very interesting, as i limited my thinking to a wire carrying a current with a moving charge beside. Looking from the two frames of reference (the lab reference and the moving charge frame of reference) I noticed that I can get same results defining v as the velocity difference beteen wire and charge without using relativity. I did not think about what will happen if two wires are there and looking again from the two frames of reference and used the way i defined the velocity. :-)
Thanks for your efforts. I read some of your answers on different posts and i thought that was great.

 

[PeterDonis]

if we are in the reference of frame of the moving charge, and we use in the lorentz law qVXB

The Lorentz force law also includes the electric field: $F = q \left( E + v \times B \right)$. In the frame of the moving charge, $v = 0$, but the field is transformed as well so $E$ is nonzero. Whenever you change frames, you have to transform everything, not just velocities.

 

[Atheer]

If I am in the lab and i move a dc current carrying wire (a long wire along its axis) near a charge at rest the charge will experience force right?

 

[Nugatory]

If I am in the lab and i move a dc current carrying wire (a long wire along its axis) near a charge at rest the charge will experience force right?

Yes. However, you could approach the exact same problem by saying that the wire is at rest and the charge is moving; that's just choosing to use a reference frame in which the speed of the wire is zero and speed of the particle is -v instead of using a reference frame in which the speed of the wire is +v and the speed of the particle is zero. For that matter, you could choose to use a reference frame in which the jet aircraft flying over the lab at 1000 km/hr is at rest; in this frame the speed of the particle is 1000 km/hr and the speed of the wire is 1000 km/hr+v.

In all three of these frames, the force on the particle will be given by $\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})$, where $\vec{v}$ is the velocity of the particle in that frame (more precisely, the velocity of the particle relative to an observer who is at rest in that frame).

 

[PeterDonis]

the force on the particle will be given by $\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})$, where $\vec{v}$ is the velocity of the particle in that frame

And also where $\vec{E}$ and $\vec{B}$ are the electric and magnetic fields in that frame; those fields transform when you change frames just as $\vec{v}$ does.

 

[Nugatory]

And also where $\vec{E}$ and $\vec{B}$ are the electric and magnetic fields in that frame; those fields transform when you change frames just as $\vec{v}$ does.

Yep - that's the answer to the question that I was thinking atheer would be asking next...