Question about forces due to magnetism

[DCN]

The force on a particle moving through a magnetic field is given by:
$$\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
F} = q\,\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} \times \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
B} $$
Is the velocity ##\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} ## relative to the source of magnetic field or some third party observer?

 

[Dale]

The velocity is relative to any inertial reference frame. It is not necessarily relative to the source, particularly if the source is moving non inertially.

 

[DCN]

If that's the case, then please explain this.

Let's say that there is a test charge ##q## whose position from the origin is given by the radial vector ##\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
r} ## and there is a source charge ##Q## at the origin moving to the right along the positive ##x##-axis with a constant velocity ##{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over v} _s}## with respect to ##q##.

Now, in the rest frame of ##q##, ##q## has a velocity of magnitude ##0##, so it will not feel any magnetic effects. The force it will feel will be solely from the electric field generated by ##Q##. The force in the rest frame of ##q## is then:
$$\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
F} = {{Qq} \over {4\pi {\varepsilon _0}}}{{\hat r} \over {{{\left\| {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
r} } \right\|}^2}}}$$
If we consider a difference reference frame in which we are moving to the right with a velocity of ##{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} _o}##, then ##q## will have a velocity of ## - {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} _o}## and ##Q## will have a velocity of ##{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} _s} - {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} _o}##. The magnetic field then produced by ##Q## is:
$$\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
B} = {{{\mu _0}Q} \over {4\pi }}{{\left( {{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} }_s} - {{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} }_o}} \right) \times \hat r} \over {{{\left\| {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
r} } \right\|}^2}}}$$
The force on ##q## due to the magnetic field will then be:
$$\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
F} = - q\,{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} }_o} \times \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
B} $$
$$\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
F} = - {{{\mu _0}Qq} \over {4\pi {{\left\| {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
r} } \right\|}^2}}}\,{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} }_o} \times \left( {\left( {{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} }_s} - {{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} }_o}} \right) \times \hat r} \right)$$
So the total force the particle will experience in the moving reference frame is:
$${{Qq} \over {4\pi {\varepsilon _0}}}{{\hat r} \over {{{\left\| {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
r} } \right\|}^2}}} - {{{\mu _0}Qq} \over {4\pi {{\left\| {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
r} } \right\|}^2}}}\,{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} }_o} \times \left( {\left( {{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} }_s} - {{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} }_o}} \right) \times \hat r} \right)$$
Which is clearly not equal to the force found in the stationary reference frame. Am I missing something here?

 

[Vanadium 50]

Yes. The fields have to transform as well when you change frames.

 

[DCN]

How does that work?

 

[Vanadium 50]

I don't understand the question. There is a Lorentz transformation for electromagnetic fields. You change frames, you need to use it to determine how the fields transform.

 

[jtbell]

How does that work?

https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity#The_E_and_B_fields

For another way of writing out the transformation:

https://en.wikipedia.org/wiki/Class...special_relativity#Field_tensor_and_4-current

This page is the first result of a Google search for "transform electric and magnetic fields".