Does the magnetic field for only two particle exist? I have read the article about the relationship between electric fields and magnetic field, which involves special relativity. But i wouder does the effect of relativity always take place?
When a single electron is flying by another single electron, and when their distance minimizes, the displacement between two electrons is perpendicular to the direction of the speed, so the distance between two electrons should not be affected by relativity, which means the force between them can be simply derived by Coulomb's law. 
Is the reasoning above correct? If not, could you please point these problems out?
I am only a sophomore and my mother tongue is not English, so i would be extremely thankful if you can use simple math and simple language. :D 
 A: I think your reasoning is partly right. One thing to keep in mind is that the concept of $\textit{force}$ is quantity that is dependent on which observer is measuring. To calculate the force on some charged test particle you must first calculate the potential of the electromagnetic field that the charge is traveling in. 
Say that we have two electrons $e_1$ and $e_2$ and we want the forces on these as seen from an observer that is stationary w.r.t $e_1$. Then $e_2$ moves in the potential created by $e_1$, and in our from this potential in constant. Hence when $e_2$ pass by $e_1$ with some velocity $v$ the force on $e_2$ will be the one calculated with Coulomb's law and directed towards $e_1$. However in our frame $e_2$ moves and in the instant that is passes by $e_1$ the electric field generated by $e_2$ will not be able to reach $e_1$ instantaneously but it takes a time $\approx \Delta x/c$ if the separation is $\Delta x$. This effect makes the force on $e_1$ be the one as if $e_2$ still were at the position it were a time $\approx \Delta x/c$ ago(the $\approx$ sign is not important it is just that to be totally correct we should not use $\Delta x$). Since this position is slightly longer away that $\Delta x$ the force on $e_1$ will actually be smaller, and not directed in the direction of where the particle actually is at that instant in time, but directed towards where $e_2$ were a time $\Delta x/c$ ago.
(I have made an approximation on the time by setting it to $\Delta x/c$ which holds if $e_2$ is moving relativistically but not by so much. But the it is not important for the reasoning.)
The analysis can be carried out from the point of view of $e_2$ as well. A remark is that we do not get any magnetic force on $e_1$ by the fact that $e_2$ is moving since $e_1$ does not move in its frame of reference.
A: 
When a single electron is flying by another single electron, and when their distance minimizes, the displacement between two electrons is perpendicular to the direction of the speed, so the distance between two electrons should not be affected by relativity, which means the force between them can be simply derived by comloub's law. 

Classical physics says that the electric field around a moving electron is not the same spherically symmetric electric field that exist around a non-moving electron, and because of that the Coulomb's law does not apply. And in addition to that there is a magnetic field too.
Relativity says that moving electric field is length-contracted, and that explains the forces felt by a charge next to a moving charge. 
Here you can see a depiction of a length-contracted electric field. (About half way through the page)
http://physics.weber.edu/schroeder/mrr/MRRtalk.html
Oh yes, in the picture the arrows pointing up and down are not unchanged, they are extra long. Which means that electric field is extra strong in the direction perpendicular to the motion of the electric field.
