source of magnetic effect

This article is a memo after study, the level is limited, if there is an error, please understand.
The parameters without ' are the parameters of the static reference frame X, and the parameters with ' are the parameters of the moving reference frame X'.

1. Lorentz transformation

令$$k=\sqrt{1-u^2/c^2}$$
$$x^{\prime }=\frac{x-ut}{k}$$
$$y^{\prime }=y$$
$$z^{\prime }=z$$
$$t^{\prime }=\frac{t-ux/c^2}{k}$$

2. Speed ​​change

$$v_x^{\prime }=\frac{d{x}^{\prime }}{d{t}^{\prime }}=\frac{dx-ud}{dt-\frac{u}{c^2}dx}=\frac{v_x-u}{1-\frac{u}{c^2}{v}_x}$$
$$v_y^{\prime }=\frac{d{y}^{\prime }}{d{t}^{\prime }}=k\frac{dy}{dt-\frac{u}{c^2}dx}=k\frac{v_y}{1-\frac{u}{c^2}{v}_x}$$
$$v_z^{\prime }=\frac{d{z}^{\prime }}{d{t}^{\prime }}=k\frac{dz}{dt-\frac{u}{c^2}dx}=k\frac{v_z}{1-\frac{u}{c^2}{v}_x}$$

Transformation of three momentums

由$m^{\prime }=\frac{m_0}{\sqrt{1-\frac{v^{\prime 2}}{c^2}}}$和$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$得$$m^{\prime }=m\frac{\sqrt{1-\frac{v^2}{c^2}}}{\sqrt{1-\frac{v^{\prime 2}}{c^2}}}$$
Now to $v^{\prime }$ and$v_x^{\prime }$用K系电影电影电影。
$$v^{\prime 2}=v_x^{\prime 2}+v_y^{\prime 2}+v_z^{\prime 2}=\frac{(v_x-u{)}^2+(k{v}_y{)}^2+(k{v}_z{)}^2}{(1-\frac{u}{c^2}{v}_x{)}^2}$$
化简得$$\sqrt{1-\frac{v^{\prime 2}}{c^2}}=\frac{k\sqrt{1-\frac{v^2}{c_2}}}{1-\frac{u}{c^2}{v}_x}$$
因此$$P_x^{\prime }=m^{\prime }{v}_x^{\prime }=\frac{m}{k}(1-\frac{u}{c^2}{v}_x)v_x^{\prime }=\frac{m}{k}(v_x-u)=\frac{P_x-\frac{u}{c^2}E}{k}$$
$$P_y^{\prime }=P_y$$
$$P_z^{\prime }=P_z$$

4. Transformation of force

$$f_x^{\prime }=\frac{d{P}_x^{\prime }}{d{t}^{\prime }}=(Need\; to\; put{K}^{\text{'}}variable\; becomesKvariable,\; after\; simplification)=f_x-\frac{u{f}_y{v}_y}{c^2(1-\frac{u}{c^2}{v}_x)}-\frac{u{f}_z{v}_z}{c^2(1-\frac{u}{c^2}{v}_x)}$$
$$f_y^{\prime }=(Need\; to\; put{K}^{\text{'}}variable\; becomesKvariable,\; after\; simplification)=\frac{k{f}_y}{1-\frac{u}{c^2}{v}_x}$$
$$f_z^{\prime }=(Need\; to\; put{K}^{\text{'}}variable\; becomesKvariable,\; after\; simplification)=\frac{k{f}_z}{1-\frac{u}{c^2}{v}_x}$$

5. The source of the magnetic effect

In the above figure, for the convenience of calculation, it is assumed that the origins of the two coordinate systems coincide at this time. The K' system moves in the positive direction of the x-axis relative to the K system at the speed u. $q_0$At the origin, relative to the K system, $q_1$在$r(r_x,r_y,r_z)$ place. Relative to the K system at the speed$v_1((v_1{)}_x,(v_1{)}_y,(v_1{)}_z)$ movement.. Now examine$q_1的受力情况$
$$f_x^{\prime }=f_x-\frac{u{f}_y{v}_y}{c^2(1-\frac{u}{c^2}{v}_x)}-\frac{u{f}_z{v}_z}{c^2(1-\frac{u}{c^2}{v}_x)}$$
whereuc $$\frac{u}{c^2}{f}_y{v}_y=\frac{u}{c^2}{E}_y{q}_1{v}_y=\frac{u}{c^2}\frac{q_0}{4pi{e}_0{r}_y^2}{q}_1{v}_y$$
Because $$\frac{1}{c^2}=m_0{e}_0$$
Let $$\frac{u}{c^2}{f}_y{v}_y=q_1{v}_y(\frac{m_{0}u{q}_0}{4\pi r_y^2})$$
同理，$$\frac{u}{c^2}{f}_z{v}_z=q_1{v}_z\left(\frac{m_{0}u{q}_0}{4\pi r_z^2}\right)$$
则$$f_x^{\prime }=f_x-q_1\frac{v_y}{1-\frac{u}{c^2}{v}_x}(\frac{q_0{m}_{0}u}{4\pi r_y^2})-q_1\frac{v_z}{1-\frac{u}{c^2}{v}_x}\left(\frac{q_0{m}_{0}u}{4\pi r_z^2}\right)=f_x-q_1{v}_y^{\prime }\left(\frac{q_0{m}_{0}u}{4\pi r_y^{2}k}\right)-q_1{v}_z^{\prime }\left(\frac{q_0{m}_{0}u}{4\pi r_z^{2}k}\right)$$

Same calculation，
$$f_y^{\prime }=\frac{f_{y}k}{1-\frac{u}{c^2}{v}_x}=\frac{f_y}{k}(1+\frac{u{v}_x^{\prime }}{c^2})=\frac{f_y}{k}+q_1{v}_x^{\prime }\frac{q_0{m}_{0}u}{4\pi r_y^{2}k}$$

$$f_z^{\prime }=\frac{f_{z}k}{1-\frac{u}{c^2}{v}_x}=\frac{f_z}{k}(1+\frac{u{v}_x^{\prime }}{c^2})=\frac{f_z}{k}+q_1{v}_x^{\prime }\frac{q_0{m}_{0}u}{4\pi r_z^{2}k}$$
Note that the following three formulas are used here to replace the K-system speed with the K'-system speed $v_x^{\prime },v_y^{\prime },v_z^{\prime }$。
$v_x=\frac{v_x^{\prime }+u}{1+\frac{u}{c^2}{v}_x^{\prime }}$
$v_y=k\frac{v_y^{\prime }}{1+\frac{u}{c^2}{v}_x^{\prime }}$
$v_z=k\frac{v_z^{\prime }}{1+\frac{u}{c^2}{v}_x^{\prime }}$
The above about $f_x^{\prime },f_y^{\prime },f_z^{\prime }$The meaning of the three formulas is that when a charged particle $q_0$speed $When\; u$ moves in the negative direction of the x-axis, the charged particle q 1 q_1moving at the speed v'$q_1$at $q_0$The force in the generated electric field is $f^{\prime }(f_x^{\prime },f_y^{\prime },f_z^{\prime })$ , each direction can be written as$f+q{v}^{\prime }B$form
where$$B_i=\frac{q_0{m}_{0}u}{4\pi r_i^{2}k}$$