# Does special relativity causes magnetism? [closed]

And if yes, according to amperes law when the magnetic force beetween two current-flowing conductors the force is in linear reverse connection beetween the distance and the force, but if magnetism is also electrical force, in the coulombs law there is a quadratic connection beetween the force and the distance, is this a paradox or not?

• Please, consider rewriting your question in a clearer way. Separate your ideas and your questions. Separate sentences. Right now, it is very unclear. I'm sorry but I don't understand what you are asking. – FGSUZ Apr 2 '19 at 20:14
• I also cannot understand the question. Please revise for clarity. – Dale Apr 2 '19 at 20:42
• I think what you are asking about is when two charges (electrons) are moving relative to each other at relativistic speeds. In this case, the two charges will see a distance between each other shorter then an observer at rest would see. This is length contraction. Because they feel the other charge closer, they will feel it to have a stronger EM field (effect). – Árpád Szendrei Apr 3 '19 at 17:03

Suppose you have a charge in a laboratory on a space ship moving at constant velocity. In the field of the lab, the charge is not moving, so there is only an electro static field with potential $$V=q/4\pi \epsilon_0 r$$ where $$r$$ is the distance from the charge. Of course the Electric field is the Gradient of $$V$$ where the derivatives are taken with respect to the coordinates of the Lab Frame.

It can be shown that $$(V/c, A_x, A_y,A_z)$$ transforms as a 4-vector between Inertial Frames. Here $$V$$ is the electro static potential and $$\vec{A}$$ is the Magnetic Vector Potential.

You have no Vector Potential if you have no moving charge, so all components are zero in the lab frame.

In the frame in which the spaceship is moving at constant speed, we have different 4-Vector Potential, after performing the Lorentz Transform: $$(V'/c=\gamma V/c, A_z'=-\gamma\beta V/c)$$.

Now $$\vec{E}= -\nabla'V$$ and $$\vec{B}=\nabla' \times \vec{A}$$ where the prime on Del implies taking the derivatives with respect to coordinates of the frame in which the space ship is moving.

Here the Magnetic Potential is non zero, so we have a magnetic field. What is a pure electric field in one frame is a mixed electric and magnetic field in another. That will allow you to calculate the fields for simple cases.

As to why the fields are adjusted, I think its a combination of three effects.

The field drops off by distance as $$(1/r^2)$$.

$$\frac{d}{dt}(\hat{r}/r^2)=\frac{-2}{r^3}\frac{dr}{dt}\hat{r}+\frac{1}{r^2}\frac{d\hat{r}}{dt}$$

Effect 1: the time variance of the inverse square property is proportional to the velocity.

Effect 2: $$\hat{r}$$, the unit vector pointing from charge to observation point also changes with time. Further the component perpendicular to the velocity remains unchanged over time. The component of vector $$\vec{v}$$ along unit vector $$\hat{r}$$ = $$\hat{r} \times (\vec{v} \times \hat{r})$$. Between the asymmetric variation of the electric field between axes parallel and perpendicular to the velocity, we expect field changes to deviate spatially according to velocity direction.

Effect 3: The most relativistic effect, Eletromagnetic fields travel at the speed of light, $$c$$. So the change in the field at an observation point is delayed. First the charge changes, then eventually, the field at the observation point changes.

Blend these 3 effects together, you get the Force on a charge being proportional to the charge , with one component proportional to the speed of the velocity of the charge with special consideration for force components as the fall along directiosn parallel and perpendicular to the velocity.

Hope that helps.