Need for SR to explain magnetic forces?

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Here's my problem. We know the magnetic force is just F=q(v x B). If we have a stationary charged particle in a magnetic field, it will not feel a force. If we change to a moving frame, the particle now has a velocity, but the idea that it feels a force by changing frames is ridiculous so Einstein invents special relativity. My question is, is that really necessary? The magnetic field now also appears to be moving, so the difference in speeds between it and the particle is still zero, hence no need yet to move away from Galilean relativity. Is my thinking correct or am I missing something?
 

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WannabeNewton
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Is my thinking correct or am I missing something?
It is certainly incorrect. Firstly a magnetic field does not move in the sense that it has a speed. A magnetic field can propagate through electromagnetic waves but that is quite a different concept. What one can have is the source of the magnetic field moving with some speed.

Let's say we have a permanent magnet and a charged particle that are both stationary in a frame ##O##. We then boost to a frame ##O'## that is moving with respect to ##O## at some speed. Then yes in this frame the magnet and the charged particle are still at rest with respect to one another. However the ##v## that appears in the Lorentz force law ##\vec{F} = q(\vec{v} \times \vec{B})## is not the velocity of the charged particle relative to the source of the magnetic field. It is the velocity of the particle relative to the chosen reference frame.

Hence ##\vec{F}\neq 0## in ##O'## whereas ##\vec{F} = 0## in ##O## if one naively follows Galilean relativity since the particle has some speed with respect to ##O'##.
 
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PeterDonis
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the vv that appears in the Lorentz force law F⃗ =q(v⃗ ×B⃗ )\vec{F} = q(\vec{v} \times \vec{B}) is not the velocity of the charged particle relative to the source of the magnetic field. It is the velocity of the particle relative to the chosen reference frame.
Not only that, the ##\vec{B}## that appears in the law is also relative to the chosen reference frame; it changes if you change frames. Also, if the field in frame ##O## is a pure magnetic field, then in frame ##O'## there will be an electric field as well, so the force law has to include a term for that. All this ensures that the actual force felt by the object is invariant under changes of frame--all that changes is how we describe the fields that cause the force.
 
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pervect
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Here's my problem. We know the magnetic force is just F=q(v x B). If we have a stationary charged particle in a magnetic field, it will not feel a force.
If you have a stationary charged particle, and E=0, it will not feel a force, regardless of the value of B. But you forgot to specify the part where E=0, I will assume that this was implied and that you didn't realize you needed to specify this explicitly.

If we change to a moving frame, the particle now has a velocity, but the idea that it feels a force by changing frames is ridiculous so Einstein invents special relativity. My question is, is that really necessary? The magnetic field now also appears to be moving, so the difference in speeds between it and the particle is still zero, hence no need yet to move away from Galilean relativity. Is my thinking correct or am I missing something?
If you happen to know how the E and B fields transform (either by the tensor equations, or the non-tensor version - see for example http://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity) you can compute what happens in the moving frame without solving Maxwell's equations again.

When you start with the E and B fields in your "rest frame" with ##\vec{E}=0## and ##\vec{B}= \vec{B_0}## (using the wiki notation), you'll find that the transformation law that takes E in the rest frame to E' in the primed (moving) frame, i.e. ##E'_{\perp} = \gamma \left( E_{\perp} + v \times \vec{B_0} \right)## gives you a non-zero value for the E field in the moving frame. This should cancel the vxB force, because if the force is zero in one frame it should be zero in all. But I haven't calculated the details.

Now we see the need to specify E - and we note that the value of E is not the same in the moving and stationary frame in general and in this problem in particular.
 

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