KAMP
Fisika => Mekanika Klasik => Topik dimulai oleh: trfrm pada Juli 20, 2018, 11:34:21 AM
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Excuse me ... . :)
In non-relativistic mechanics, the Galilean transformation will yield the Galilean velocity addition formula ... , whereas ... , in the special theory of relativity, the Lorentz’s transformation of space-time will yield the Einstein’s velocity addition formula ... . The Einstein’s velocity addition formula, which reduce to Galilean velocity addition formula if the magnitude of the velocities are much smaller than $c\approx299792458\,\textrm{m/s}$ (speed of light in vacuum) ... .
If the velocity of particle $A$ and $B$ are $\mathbf{v}_A$ and $\mathbf{v}_B$, respectively, relative to observer $O$ , and the velocity of observer $O'$ is $\mathbf{V}$ relative to observer $O$ , then
the velocity of $A$ relative to $O'$ is $\displaystyle\mathbf{v}'_A=\frac{\mathbf{v}_A+(\Gamma-1)(\mathbf{v}_A\cdot\mathbf{V})\mathbf{V}/|\mathbf{V}|^2-\Gamma\mathbf{V}}{\Gamma(1-\mathbf{v}_A\cdot\mathbf{V}/c^2)}$ , and
the velocity of $B$ relative to $O'$ is $\displaystyle\mathbf{v}'_B=\frac{\mathbf{v}_B+(\Gamma-1)(\mathbf{v}_B\cdot\mathbf{V})\mathbf{V}/|\mathbf{V}|^2-\Gamma\mathbf{V}}{\Gamma(1-\mathbf{v}_B\cdot\mathbf{V}/c^2)}$ ,
where $\Gamma:=(1-|\mathbf{V}|^2/c^2)^{-1/2}$ ... .
If we want such that $\mathbf{v}'_B=-\mathbf{v}'_A$ , we will get an equation
$\displaystyle\frac{|\mathbf{V}|^2(\mathbf{v}_B-\Gamma\mathbf{V})+(\Gamma-1)(\mathbf{v}_B\cdot\mathbf{V})\mathbf{V}}{c^2-\mathbf{v}_B\cdot\mathbf{V}}=\frac{|\mathbf{V}|^2(\Gamma\mathbf{V}-\mathbf{v}_A)+(1-\Gamma)(\mathbf{v}_A\cdot\mathbf{V})\mathbf{V}}{c^2-\mathbf{v}_A\cdot\mathbf{V}}$ ,
which has three components because the equation is 3-vector equation ... .
We are asked to find $\mathbf{V}$ alias $(V_x,V_y,V_z)$ from the vector equation (alias three scalar component equations) ... .
Does there exist other methods to find $(V_x,V_y,V_z)$ from three simultaneous non-linear scalar equation which containing $(V_x,V_y,V_z)$ ... ?
Does the solution for $\mathbf{V}$ of this vector equation must be numerical solution ... ? Does there exist an analytical solution of $\mathbf{V}$ ... ?
Thank you very much ... . :)
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Oh ... . I’m sorry very much ... . :( Really, this problem is not a homework problem ... . I just want know the brief clue of alternative solution to solve the equation ... . I just know that
$|\mathbf{V}|^2={V_x}^2+{V_y}^2+{V_z}^2$ ,
$\Gamma=[1-({V_x}^2+{V_y}^2+{V_z}^2)/c^2]^{-1/2}$ ,
$\mathbf{v}_A\cdot\mathbf{V}=v_{Ax}V_x+v_{Ay}V_y+v_{Az}V_z$ , and
$\mathbf{v}_B\cdot\mathbf{V}=v_{Bx}V_x+v_{By}V_y+v_{Bz}V_z$ ,
in Cartesian coordinate system ... .
Maybe ... , its analytic solution does not exist ... . If in special case (one-dimension), $\mathbf{v}_A=v_A\mathbf{i}$ , $\mathbf{v}_B=v_B\mathbf{i}$ , and $\mathbf{V}=V\mathbf{i}$ ... , then the vector equation becomes
$\displaystyle\frac{V^2(v_A-\Gamma{V})+(\Gamma-1)v_AV^2}{c^2-v_AV}=\frac{V^2(\Gamma{V}-v_B)+(1-\Gamma)v_BV^2}{c^2-v_BV}$ ,
alias $\displaystyle\frac{v_A-V}{c^2-v_AV}=\frac{V-v_B}{c^2-v_BV}$ ,
alias $c^2v_A-(v_Av_B+c^2)V+v_BV^2=-c^2v_B+(v_Av_B+c^2)V-v_AV^2$ ,
alias $(v_A+v_B)V^2-2(v_Av_B+c^2)V+c^2(v_A+v_B)=0$ ,
alias $\displaystyle\,V=\frac{v_Av_B+c^2\pm\sqrt{{v_A}^2{v_B}^2+c^2(c^2-{v_A}^2-{v_{B}}^2)}}{v_A+v_B}$ ... .
alias $\displaystyle\,V=\frac{v_Av_B+c^2\pm\sqrt{c^2-{v_A}^2}\sqrt{c^2-{v_B}^2}}{v_A+v_B}$ ... .
But ... , this is a special case in one-dimension, not 3-dimension ... .