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相対性理論 電気力学の部

相対性理論
相対性理論

電気力学の部

 運動学の部と同じように解るところまで数式を勉強します。

$$
\frac{1}{c^2}\frac{\partial E_x}{\partial t}=\frac{\partial B_z}{\partial y}-\frac{\partial B_y}{\partial z},
$$

$$
\frac{\partial B_x}{\partial t}=-\left(\frac{\partial E_z}{\partial y}-\frac{\partial E_y}{\partial z}\right),
$$

$$
\frac{1}{c^2}\frac{\partial E_y}{\partial t}=\frac{\partial B_x}{\partial z}-\frac{\partial B_z}{\partial x},
$$

$$
\frac{\partial B_y}{\partial t}=-\left(\frac{\partial E_x}{\partial z}-\frac{\partial E_z}{\partial x}\right),
$$

$$
\frac{1}{c^2}\frac{\partial E_z}{\partial t}=\frac{\partial B_y}{\partial x}-\frac{\partial B_x}{\partial y},
$$

$$
\frac{\partial B_z}{\partial t}=-\left(\frac{\partial E_y}{\partial x}-\frac{\partial E_x}{\partial y}\right),
$$

$$\left.\begin{array}{c}
\frac{1}{c^2}\frac{\partial E_x}{\partial \tau}=\frac{\partial}{\partial\eta}\left\{\beta\left(E_z-\frac{v}{c^2}E_y\right)\right\}-\frac{\partial}{\partial\zeta}\left\{\beta\left(B_y+\frac{v}{c^2}E_z\right)\right\},
\\
\frac{1}{c^2}\frac{\partial}{\partial \tau}\{\beta(E_y-vB_z)\}=\frac{\partial B_x}{\partial\zeta}-\frac{\partial}{\partial\xi}\left\{\beta\left(B_z-\frac{v}{c^2}E_y\right)\right\},
\\
\frac{1}{c^2}\frac{\partial}{\partial \tau}\{\beta(E_z+vB_y)\}=\frac{\partial}{\partial\xi}\left\{\beta\left(B_y+\frac{v}{c^2}E_z\right)\right\}-\frac{\partial}{\partial\eta}B_x,
\end{array}\right\}
$$

$$
\left.\begin{array}{c}
\frac{\partial B_x}{\partial\tau}=-\frac{\partial}{\partial\eta}\{\beta(E_z+vB_y)\}+\frac{\partial}{\partial\zeta}\{\beta(E_y-vB_z)\},
\\
\frac{\partial}{\partial\tau}\left\{\beta\left(B_y+\frac{v}{c^2}E_z\right)\right\}=-\frac{\partial}{\partial\zeta}E_x+\frac{\partial}{\partial\xi}\{\beta(E_z+vB_y)\},
\\
\frac{\partial}{\partial\tau}\left\{\beta\left(B_z-\frac{v}{c^2}E_y\right)\right\}=-\frac{\partial}{\partial\xi}\{\beta(E_y-vB_z)\}+\frac{\partial}{\partial\eta}E_x.
\end{array}\right\}
$$

$$\beta=\frac{1}{\sqrt{1-(v/c)^2}}$$

$$
\frac{1}{c^2}\frac{\partial E’_{\xi}}{\partial\tau}=\frac{\partial B’_{\zeta}}{\partial\eta}-\frac{\partial B’_{\eta}}{\partial\zeta},
$$

$$
\frac{\partial B’_{\xi}}{\partial\tau}=-\left(\frac{\partial E’_{\zeta}}{\partial\eta}-\frac{\partial E’_{\eta}}{\partial\zeta}\right),
$$

$$
\frac{1}{c^2}\frac{\partial E’_{\eta}}{\partial\tau}=\frac{\partial B’_{\xi}}{\partial\zeta}-\frac{\partial B’_{\zeta}}{\partial\xi},
$$

$$
\frac{\partial B’_{\eta}}{\partial\tau}=-\left(\frac{\partial E’_{\xi}}{\partial\zeta}-\frac{\partial E’_{\zeta}}{\partial\xi}\right),
$$

$$
\frac{1}{c^2}\frac{\partial E’_{\zeta}}{\partial\tau}=\frac{\partial B’_{\eta}}{\partial\xi}-\frac{\partial B’_{\xi}}{\partial\eta},
$$

$$
\frac{\partial B’_{\zeta}}{\partial\tau}=-\left(\frac{\partial E’_{\eta}}{\partial\xi}-\frac{\partial E’_{\xi}}{\partial\eta}\right).
$$

$$
E’_{\xi}=\psi(v)E_x,
\space
B’_{\xi}=\psi(v)B_x,
$$

$$
E’_{\eta}=\psi(v)\beta(E_y-vB_z),
\space
B’_{\eta}=\psi(v)\beta\{B_y+(v/c^2)E_z\},
$$

$$
E’_{\zeta}=\psi(v)\beta(E_z+vB_y),
\space
B’_{\zeta}=\psi(v)\beta\{B_z-(v/c^2)E_y\}.
$$

$$\psi(v)\cdot\psi(-v)=1.$$

$$\psi(v)=\psi(-v).$$

$$\psi(v)=1$$

$$E’_{\xi}=E_x,\space B’_{\xi}=B_x,$$

$$
E’_{\eta}=\beta(E_y-vB_z),\space B’_{\eta}=\beta\left(B_y+\frac{v}{c^2}E_z\right),
$$

$$
E’_{\zeta}=\beta(E_z-vB_y),\space B’_{\zeta}=\beta\left(B_z+\frac{v}{c^2}E_y\right).
$$

$$
E_x=E_{x0}\space sin\space\varPhi,\space B_x=B_{x0}\space sin\space\varPhi,
$$

$$
E_y=E_{y0}\space sin\space\varPhi,\space B_y=B_{y0}\space sin\space\varPhi,
$$

$$
E_z=E_{z0}\space sin\space\varPhi,\space B_z=B_{z0}\space sin\space\varPhi,
$$

$$
\varPhi=\omega\{t-(lx+my+nz)/c\}.
$$

$$
E’_{\xi}=E_{x0}\space sin\space\varPhi’,\space B’_{\xi}=B_{x0}\space sin\space\varPhi’,
$$

$$
E’_{\eta}=\beta(E_{y0}-vB_{z0})\space sin\space\varPhi’,\space B’_{\eta}=\beta\left(B_{y0}+\frac{v}{c^2}E_{z0}\right)\space sin\space\varPhi’,
$$

$$
E’_{\zeta}=\beta(E_{z0}-vB_{y0})\space sin\space\varPhi’,\space B’_{\zeta}=\beta\left(B_{z0}+\frac{v}{c^2}E_{y0}\right)\space sin\space\varPhi’,
$$

$$\varPhi’=\omega’\{\tau-(l’\xi+m’\eta+n’\zeta)/c\}.$$

$$\omega’=\omega\beta\left(1-l\frac{v}{c}\right),$$

$$l’=\left.\left(l-\frac{v}{c}\right)\middle/\left(1-l\frac{v}{c}\right)\right.,$$

$$m’=\left. m\middle/\beta\left(1-l\frac{v}{c}\right)\right.,$$

$$n’=\left. n\middle/\beta\left(1-l\frac{v}{c}\right)\right. .$$

$$\nu’=\left.\nu\left(1-\frac{v}{c}cos\space\varphi\right)\middle/\sqrt{1-\left(\frac{v}{c}\right)^2}\right.$$

$$\nu’=\nu\sqrt{\left.\left(1-\frac{v}{c}\right)\middle/\left(1+\frac{v}{c}\right)\right.}.$$

$$cos\space\varphi’=\left.\left(cos\space\varphi-\frac{v}{c}\right)\middle/\left(1-\frac{v}{c}cos\space\varphi\right)\right..$$

$$cos\space\varphi’=-\frac{v}{c}.$$

$$
(A’)^2=A^2\left.\left(1-\frac{v}{c}cos\space\varphi\right)^2\middle/\left\{1-\left(\frac{v}{c}\right)^2\right\}\right..
$$

$$\varphi=0$$

$$
(A’)^2=A^2\left.\left(1-\frac{v}{c}\right)\middle/\left(1+\frac{v}{c}\right)\right..
$$

 この式のところで「観測者が速さcで光源に向かって走ればこの人から見たとき光源は無限に強い明るさに輝いて見えることになる」とあったのですが直ぐに理解できなかったので補足しておきます。

 光源に向かっているのでvはマイナスの値になります。vがcに近づくほど分母は1+v/cなので0に近づきます。分子は1-v/cなので2に近づきます。従って限り無く2に近い値を限り無く0に近い値で割るので無限に大きい振幅になるということです。

$$w=(\vec E\cdot\vec D+\vec H\cdot\vec B)/2$$

$$(x-lct)^2+(y-mct)^2+(z-nct)^2=R^2$$

$$\left(\beta\xi-l\frac{v}{c}\beta\xi\right)^2+\left(\eta-m\frac{v}{c}\beta\xi\right)^2+\left(\zeta-n\frac{v}{c}\beta\xi\right)^2=R^2$$

$$
\frac{S’}{S}=\left.\sqrt{1-\left(\frac{v}{c}\right)^2}\middle/\left(1-\frac{v}{c}cos\space\varphi\right)\right.
$$

$$
\frac{E’}{E}=\frac{w’S’}{wS}=\frac{(A’)^2S’}{(A)^2S}=\frac{1-(v/c)cos\space\varphi}{\sqrt{1-(v/c)^2}}
$$

$$\varphi=0$$

$$\frac{E’}{E}=\sqrt{\frac{1-(v/c)}{1+(v/c)}}.$$

$$
A’=A\{1-(v/c)cos\space\varphi\}/\sqrt{1-(v/c)^2},
$$

$$
cos\space\varphi’=\{cos\space\varphi-(v/c)\}/\{1-(v/c)cos\space\varphi\},
$$

$$
\nu’=\nu\cdot\{1-(v/c)cos\space\varphi\}/\sqrt{1-(v/c)^2}
$$

$$
A^{\prime\prime}=A’,\space cos\space\varphi^{\prime\prime}=-cos\space\varphi’,\space\nu^{\prime\prime}=\nu’
$$

$$
A^{\prime\prime\prime}=A^{\prime\prime}\frac{1+(v/c)cos\space\varphi^{\prime\prime}}{\sqrt{1-(v/c)^2}}=A\frac{1-2(v/c)cos\space\varphi+(v/c)^2}{1-(v/c)^2},
$$

$$
cos\space\varphi^{\prime\prime\prime}=\frac{cos\space\varphi^{\prime\prime}+(v/c)}{1+(v/c)cos\space\varphi^{\prime\prime}}=-\frac{\{1+(v/c)^2\}cos\space\varphi-2(v/c)}{1-2(v/c)cos\space\varphi+(v/c)^2},
$$

$$
\nu^{\prime\prime\prime}=\nu^{\prime\prime}\frac{1+(v/c)cos\space\varphi^{\prime\prime}}{\sqrt{1-(v/c)^2}}=\nu\frac{1-2(v/c)cos\space\varphi+(v/c)^2}{\{1-(v/c)\}^2},
$$

$$
P=2w\frac{\{cos\space\varphi-(v/c)\}^2}{1-(v/c)^2}.
$$

$$P=2w\space cos^2\varphi$$

 ようやく「力」が出てきました。

$$
\frac{\partial D_x}{\partial t}+\rho u_x=\frac{\partial H_z}{\partial y}-\frac{\partial H_y}{\partial z},\space\frac{\partial B_x}{\partial t}=-\frac{\partial E_z}{\partial y}+\frac{\partial E_y}{\partial z},
$$

$$
\frac{\partial D_y}{\partial t}+\rho u_y=\frac{\partial H_x}{\partial z}-\frac{\partial H_z}{\partial x},\space\frac{\partial B_y}{\partial t}=-\frac{\partial E_x}{\partial z}+\frac{\partial E_z}{\partial x},
$$

$$
\frac{\partial D_z}{\partial t}+\rho u_z=\frac{\partial H_y}{\partial x}-\frac{\partial H_x}{\partial y},\space\frac{\partial B_z}{\partial t}=-\frac{\partial E_y}{\partial x}+\frac{\partial E_x}{\partial y}.
$$

$$
\rho=\frac{\partial D_x}{\partial x}+\frac{\partial D_y}{\partial y}+\frac{\partial D_z}{\partial z}
$$

$$
\frac{\partial D’_\xi}{\partial\tau}+\rho’ u_\xi=\frac{\partial H’_\zeta}{\partial \eta}-\frac{\partial H’_\eta}{\partial \zeta},\space\frac{\partial B’_\xi}{\partial\tau}=-\frac{\partial E’_\zeta}{\partial \eta}+\frac{\partial E’_\eta}{\partial \zeta},
$$

$$
\frac{\partial D’_\eta}{\partial\tau}+\rho’ u_\eta=\frac{\partial H’_\xi}{\partial \zeta}-\frac{\partial H’_\zeta}{\partial \xi},\space\frac{\partial B’_\eta}{\partial\tau}=-\frac{\partial E’_\xi}{\partial \zeta}+\frac{\partial E’_\zeta}{\partial \xi},
$$

$$
\frac{\partial D’_\zeta}{\partial\tau}+\rho’ u_\zeta=\frac{\partial H’_\eta}{\partial \xi}-\frac{\partial H’_\xi}{\partial \eta},\space\frac{\partial B’_\zeta}{\partial\tau}=-\frac{\partial E’_\eta}{\partial \xi}+\frac{\partial E’_\xi}{\partial \eta}.
$$

$$u_\xi=\frac{u_x-v}{1-(u_xv/c^2)},\space u_\eta=\frac{u_y}{\beta\{1-(u_xv/c^2)\}},\space u_\zeta=\frac{u_z}{\beta\{1-(u_xv/c^2)\}},$$

$$\rho’=\frac{\partial D’_\xi}{\partial\xi}+\frac{\partial D’_\eta}{\partial\eta}+\frac{\partial D’_\zeta}{\partial\zeta}=\beta\left(1-\frac{u_xv}{c^2}\right)\rho.$$

$$
\mu\frac{d^2x}{dt^2}=\varepsilon E_x,
\space
\mu\frac{d^2y}{dt^2}=\varepsilon E_y,
\space
\mu\frac{d^2z}{dt^2}=\varepsilon E_z
$$

ここの説明に \(\mu\) は「その質量を表す」とあります。「力」に続き「質量」も登場してきました。

$$t=x=y=z=0$$

$$\tau=\xi=\eta=\zeta=0$$

$$\tau=\beta\{t-(v/c^2)x\},$$

$$\xi=\beta(x-vy),\space E’_\xi=E_x,$$

$$\eta=y,\space E’_\eta=\beta(E_y-vB_z),$$

$$\zeta=z,\space E’_\zeta=\beta(E_z+vB_y).$$

$$\left.\begin{array}{c}
\frac{d^2x}{dt^2}=\frac{\varepsilon}{\mu}\frac{1}{\beta^3}E_x,
\\
\frac{d^2y}{dt^2}=\frac{\varepsilon}{\mu}\frac{1}{\beta}(E_y-vB_z),
\\
\frac{d^2z}{dt^2}=\frac{\varepsilon}{\mu}\frac{1}{\beta}(E_z+vB_y).
\end{array}\right\}
$$

$$\mu\beta^3\frac{d^2x}{dt^2}=\varepsilon E_x =\varepsilon E’\xi,$$

$$\mu\beta^2\frac{d^2y}{dt^2}=\varepsilon\beta(E_y-vB_z) =\varepsilon E’\eta,$$

$$\mu\beta^2\frac{d^2z}{dt^2}=\varepsilon\beta(E_z+vB_y) =\varepsilon E’\zeta,$$

$$mass \times acceleration = power$$

$$longitudinal\space mass=\left.\mu\middle/\left\{\sqrt{1-\left(\frac{v}{c}\right)^2}\right\}^3\right.,$$

$$transverse\space mass=\left.\mu\middle/\left\{1-\left(\frac{v}{c}\right)^2\right\}\right.$$

$$W=\int\varepsilon E_xdx=\mu\int_0^v\beta^3vdv=\mu c^2\left\{\frac{1}{\sqrt{1-(v/c)^2}}-1\right\}.$$

$$A_m/A_e=v/c$$

$$P=\int E_xdx=\frac{\mu}{\varepsilon}c^2\left\{\frac{1}{\sqrt{1-(v/c)^2}}-1\right\}.$$

$$-\frac{d^2y}{dt^2}=\frac{v^2}{R}=\frac{\varepsilon}{\mu}vB\sqrt{1-(v/c)^2},$$

$$R=\mu v/\varepsilon B\sqrt{1-(v/c)^2}.$$

 光速が不変となった為に運動学の部では時間と空間が可変となり、電気力学の部では質量までも可変になってしまいました。当時最新の科学であったであろう電磁波と数百年来慣れ親しんだ力学を統一した理論で再構築しあらゆる科学に見直しを迫るというとてつもない偉業が書かれている一冊でした。