Seamus Mcknight

2022-09-24

Magnetic force and frames of reference

I'm having a hard time trying to understand the next situation:

Suppose that I have a magnet that creates a quite uniform magnetic filed $\overrightarrow{B}$. In the vicinity of this magnet there is a particle with charge $q$ that is moving with velocity $\overrightarrow{v}$ in some direction.

This is, in the frame of reference of the magnet, the particle is moving with velocity $\overrightarrow{v}$ and therefore the particle experiences a force given by: $\overrightarrow{F}=q\phantom{\rule{thinmathspace}{0ex}}(\overrightarrow{v}\times \overrightarrow{B})$. This would cause the path of the particle to curve around the magnetic field.

On the other hand, an observer situated in the particle would see the magnet moving with velocity ${\overrightarrow{v}}^{\prime}=-\overrightarrow{v}$. How would this observer account for the movement of the magnet caused by the force ${F}^{\prime}$?

I believe that the force ${\overrightarrow{F}}^{\prime}$ is given by the relativistic transformation:

${\overrightarrow{F}}^{\prime}=-\gamma \overrightarrow{F}$

Knowing that the force is perpendicular to the velocity (this is if, for instance, the particle is moving along the x axis, then the force is along the y axis).

I am in the early stages of understanding relativistic mechanics so please forgive me is the question looks silly. Thanks in advance.

I'm having a hard time trying to understand the next situation:

Suppose that I have a magnet that creates a quite uniform magnetic filed $\overrightarrow{B}$. In the vicinity of this magnet there is a particle with charge $q$ that is moving with velocity $\overrightarrow{v}$ in some direction.

This is, in the frame of reference of the magnet, the particle is moving with velocity $\overrightarrow{v}$ and therefore the particle experiences a force given by: $\overrightarrow{F}=q\phantom{\rule{thinmathspace}{0ex}}(\overrightarrow{v}\times \overrightarrow{B})$. This would cause the path of the particle to curve around the magnetic field.

On the other hand, an observer situated in the particle would see the magnet moving with velocity ${\overrightarrow{v}}^{\prime}=-\overrightarrow{v}$. How would this observer account for the movement of the magnet caused by the force ${F}^{\prime}$?

I believe that the force ${\overrightarrow{F}}^{\prime}$ is given by the relativistic transformation:

${\overrightarrow{F}}^{\prime}=-\gamma \overrightarrow{F}$

Knowing that the force is perpendicular to the velocity (this is if, for instance, the particle is moving along the x axis, then the force is along the y axis).

I am in the early stages of understanding relativistic mechanics so please forgive me is the question looks silly. Thanks in advance.

Ryan Underwood

Beginner2022-09-25Added 4 answers

There's an electric component to the Lorentz force:

$\overrightarrow{F}=q(\overrightarrow{E}+\overrightarrow{v}\times \overrightarrow{B})$

In the frame of the particle where its velocity is zero, the force acting on the stationary particle would be interpreted as being electric, since the magnetic field $\overrightarrow{{B}^{\prime}}$ only acts on moving charge, and is given by:

$\overrightarrow{{F}^{\prime}}={q}^{\prime}\overrightarrow{{E}^{\prime}}$

In your example, we can take $\overrightarrow{v}$ along $x$ and $\overrightarrow{B}$ along z giving a $\overrightarrow{{F}_{y}^{\prime}}$ along $-{y}^{\prime}$ so that the forces transform as :

$\overrightarrow{{F}_{y}^{\prime}}=\gamma {\overrightarrow{F}}_{y},\phantom{\rule{1em}{0ex}}{q}^{\prime}{\overrightarrow{E}}_{y}^{\prime}=\gamma q\overrightarrow{v}\times {\overrightarrow{B}}_{z}$

And this is consistent with how the fields should transform, remembering that charge is a relativistic invariant where $q={q}^{\prime}$

$\overrightarrow{F}=q(\overrightarrow{E}+\overrightarrow{v}\times \overrightarrow{B})$

In the frame of the particle where its velocity is zero, the force acting on the stationary particle would be interpreted as being electric, since the magnetic field $\overrightarrow{{B}^{\prime}}$ only acts on moving charge, and is given by:

$\overrightarrow{{F}^{\prime}}={q}^{\prime}\overrightarrow{{E}^{\prime}}$

In your example, we can take $\overrightarrow{v}$ along $x$ and $\overrightarrow{B}$ along z giving a $\overrightarrow{{F}_{y}^{\prime}}$ along $-{y}^{\prime}$ so that the forces transform as :

$\overrightarrow{{F}_{y}^{\prime}}=\gamma {\overrightarrow{F}}_{y},\phantom{\rule{1em}{0ex}}{q}^{\prime}{\overrightarrow{E}}_{y}^{\prime}=\gamma q\overrightarrow{v}\times {\overrightarrow{B}}_{z}$

And this is consistent with how the fields should transform, remembering that charge is a relativistic invariant where $q={q}^{\prime}$

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