- #1
snoopies622
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I thought I understood this matter, but the more I think about it the more confused I am becoming.
Suppose I have a very long wire which has a uniform negative charge per unit length ρ. I hold near it a particle with charge -q. Due to the electrical field around the wire there is a force that pushes the particle directly away from it, and if I release the particle it accelerates radially from the wire and contines on a particular radial trajectory.
Now suppose instead of simply letting go of the particle, I throw it in a direction parallel to the wire with initial velocity v. From the wire's frame of reference there is no magnetic field in the vicinity of the particle, so the component of the particle's acceleration that is perpendicular to the wire should be the same as in the case when it was released with initial velocity zero.
But is that what is seen?
I know that from the point of view of the particle, in addition to the electrical field around the wire there is a current running through it, and this creates a magnetic field which may also influence its radial acceleration (although I'm not even certain of that since in its own reference frame the particle's velocity parallel to the moving wire is always zero). But since the laws of electromagnetism are supposed to be the same and equally valid in every reference frame, can this magnetic field simply be ignored?
With special relativity there are the additional complications of an increased charge density in the wire from the moving particle's point of view, as well as a more massive moving particle (and smaller acceleration per unit force) from the wire's point of view. Maybe time dilation is a factor, too.
I simply don't know what influences are mutually consistent and which ones are mutually redundant. I was under the impression that special relativity was consistent with classical electromagnetism and conflicted with only classical mechanics, but then I have also read that magnetic fields can be ignored if one considers special relativity and the electric fields alone, which seems to imply the opposite.
Suppose I have a very long wire which has a uniform negative charge per unit length ρ. I hold near it a particle with charge -q. Due to the electrical field around the wire there is a force that pushes the particle directly away from it, and if I release the particle it accelerates radially from the wire and contines on a particular radial trajectory.
Now suppose instead of simply letting go of the particle, I throw it in a direction parallel to the wire with initial velocity v. From the wire's frame of reference there is no magnetic field in the vicinity of the particle, so the component of the particle's acceleration that is perpendicular to the wire should be the same as in the case when it was released with initial velocity zero.
But is that what is seen?
I know that from the point of view of the particle, in addition to the electrical field around the wire there is a current running through it, and this creates a magnetic field which may also influence its radial acceleration (although I'm not even certain of that since in its own reference frame the particle's velocity parallel to the moving wire is always zero). But since the laws of electromagnetism are supposed to be the same and equally valid in every reference frame, can this magnetic field simply be ignored?
With special relativity there are the additional complications of an increased charge density in the wire from the moving particle's point of view, as well as a more massive moving particle (and smaller acceleration per unit force) from the wire's point of view. Maybe time dilation is a factor, too.
I simply don't know what influences are mutually consistent and which ones are mutually redundant. I was under the impression that special relativity was consistent with classical electromagnetism and conflicted with only classical mechanics, but then I have also read that magnetic fields can be ignored if one considers special relativity and the electric fields alone, which seems to imply the opposite.