91影视

Consider the axes so that E points lie in the z-direction, and B lies in the yz plane at an angle $鈭�$.

As E lies in the z-direction, write the coordinate of E in S frame.

$\mathrm{E}(0,0,\mathrm{E})$

Similarly, asBlies in theyzdirection, write the coordinates atBinSframe.

$\mathrm{B}(0,\mathrm{B}\mathrm{肠辞蝉蠒},\mathrm{B}\mathrm{肠辞蝉蠒})$

Write the coordinates of E in $\overline{S}$ frame.

localid="1654602630736" $\overline{\mathrm{E}}=(0,-\mathrm{纬惫叠}\mathrm{蝉颈苍蠒},\mathrm{纬}(\mathrm{E}+\mathrm{vB}\mathrm{肠辞蝉蠒}\left)\right)$

Write the coordinates of B in $\overline{S}$frame.

$\overline{\mathrm{B}}=\left(\mathrm{o},\mathrm{y}\left(\mathrm{Bcos}鈭�+\frac{\mathrm{v}}{{\mathrm{c}}^2}\mathrm{E}\right),\mathrm{y}\mathrm{Bcos}鈭�\right)$

Using equation 12.109, write the complete set of transformation rules for an electric field.

$$\begin{gathered} \overline{\mathrm{E}}={\mathrm{E}}_{\mathrm{x}} \\ \overline{\mathrm{E}}=\mathrm{y}\left({\mathrm{E}}_{\mathrm{y}}-{\mathrm{vB}}_{\mathrm{c}}\right) \\ \overline{\mathrm{E}}=\mathrm{y}\left({\mathrm{E}}_{\mathrm{y}}+{\mathrm{vB}}_{\mathrm{c}}\right) \end{gathered}$$

Similarly, write the complete set of transformation rules for a magnetic field.

$$\begin{gathered} \overline{{\mathrm{B}}_{\mathrm{x}}}={\mathrm{B}}_{\mathrm{x}} \\ \overline{{\mathrm{B}}_{\mathrm{y}}}=\mathrm{y}\left({\mathrm{B}}_{\mathrm{y}}+\frac{\mathrm{v}}{{\mathrm{c}}^2}{\mathrm{E}}_{\mathrm{z}}\right) \\ \overline{{\mathrm{B}}_{\mathrm{z}}}=\mathrm{y}\left({\mathrm{B}}_{\mathrm{z}}-\frac{\mathrm{v}}{{\mathrm{c}}^2}{\mathrm{E}}_{\mathrm{y}}\right) \end{gathered}$$

Equate $\overline{\frac{{\mathrm{E}}_{\mathrm{y}}}{\overline{{\mathrm{B}}_{\mathrm{y}}}}}-\overline{\frac{{\mathrm{E}}_x}{\overline{{\mathrm{B}}_z}}}$:

Substitute localid="1654600759816" ${\mathrm{B}}_{\mathrm{z}}=\mathrm{纬B蝉颈苍蠒},{\mathrm{E}}_{\mathrm{z}}=\mathrm{纬}(\mathrm{E}+\mathrm{vB肠辞蝉蠒}),{\mathrm{B}}_{\mathrm{y}}=\mathrm{y}\left(\mathrm{Bcos}鈭�+\frac{\mathrm{v}}{{\mathrm{c}}^2}\mathrm{E}\right)$and ${\mathrm{E}}_{\mathrm{y}}=-\mathrm{纬惫叠蝉颈苍蠒}$ in the above obtained expression.

$$\begin{gathered} \frac{\left({\mathrm{E}}_{\mathrm{y}}-{\mathrm{vB}}_{\mathrm{x}}\right)}{\left({\mathrm{B}}_{\mathrm{y}}+{\displaystyle \frac{\mathrm{v}}{{\mathrm{c}}^2}}{\mathrm{E}}_{\mathrm{z}}\right)}=\frac{\left({\mathrm{E}}_{\mathrm{z}}-{\mathrm{vB}}_{\mathrm{y}}\right)}{\left({\mathrm{B}}_{\mathrm{z}}+{\displaystyle \frac{\mathrm{v}}{{\mathrm{c}}^2}}{\mathrm{E}}_y\right)} \\ \frac{\left(-\mathrm{yvB}\sin 鈭�\right)}{\mathrm{y}\left(\mathrm{B}\cos 鈭�+{\displaystyle \frac{\mathrm{v}}{{\mathrm{c}}^2}}\mathrm{E}\right)}=\frac{\mathrm{y}\left(\mathrm{E}+\mathrm{vB}\cos 鈭�\right)}{\mathrm{yB}\sin 鈭�} \\ -{\mathrm{vB}}^2{\sin }^2鈭�=\left(\mathrm{B}\cos 鈭�+\frac{\mathrm{v}}{{\mathrm{c}}^2}\mathrm{E}\right)\left(\mathrm{E}+\mathrm{vB}\cos 鈭�\right) \\ 0=\mathrm{EB}\cos 鈭�+{\mathrm{vB}}^2{\cos }^2鈭�+\frac{\mathrm{v}}{{\mathrm{c}}^2}{\mathrm{E}}^2+\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}\mathrm{EB}\cos 鈭� \end{gathered}$$

On further solving,

localid="1654602636380" $$\begin{gathered} \mathrm{EB}\cos 鈭�+{\mathrm{vB}}^2+\frac{\mathrm{v}}{{\mathrm{c}}^2}{\mathrm{E}}^2+\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}\mathrm{EB}\cos 鈭�=0 \\ \mathrm{EB}\cos 鈭�\left(1+\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}\right)+{\mathrm{vB}}^2+\frac{\mathrm{v}}{{\mathrm{c}}^2}{\mathrm{E}}^2=0 \\ -\mathrm{EB}\cos 鈭�\left(1+\frac{{\mathrm{v}}^2}{{\mathrm{E}}^2}\right)=\mathrm{v}\left({\mathrm{B}}^2+\frac{{\mathrm{E}}^2}{{\mathrm{c}}^2}\right) \\ \frac{\mathrm{v}}{1+\frac{{\mathrm{v}}^2}{{\mathrm{E}}^2}}=\frac{-\mathrm{EB}\cos 鈭�}{{\mathrm{B}}^2-\frac{{\mathrm{E}}^2}{{\mathrm{c}}^2}} \end{gathered}$$ 鈥︹�� (1)

Calculate the cross product of $\mathbf{E}\mathbf{脳}\mathbf{B}$.

localid="1654602416573" $$\begin{gathered} \mathbf{E}\mathbf{脳}\mathbf{B}=\left|\begin{array}{ccc}\stackrel{铃�}{\mathrm{x}}& \stackrel{铃�}{\mathrm{y}}& \stackrel{铃�}{\mathrm{z}}\\ \mathrm{o}& \mathrm{o}& \mathrm{E}\\ \mathrm{o}& \mathrm{Bcos}鈭�& \mathrm{Bsin}鈭�\end{array}\right| \\ \mathbf{E}\mathbf{脳}\mathbf{B}=\stackrel{铃�}{\mathrm{x}}\left(\mathrm{o}脳\mathrm{Bcos}鈭�-\mathrm{Esin}鈭�\right)-\stackrel{铃�}{\mathrm{y}}\left(\mathrm{Bsin}鈭�-\mathrm{E}脳\mathrm{o}\right)+\stackrel{铃�}{\mathrm{z}}\left(\mathrm{o}脳\mathrm{Bcos}鈭�+\mathrm{o}脳\mathrm{o}\right) \\ \mathbf{E}\mathbf{脳}\mathbf{B}=-\mathrm{EBcos}鈭�\stackrel{铃�}{\mathrm{x}} \end{gathered}$$

Substitute $\mathbf{E}\mathbf{脳}\mathbf{B}$for $-\mathrm{EBcos}鈭�\stackrel{铃�}{\mathrm{x}}$ in equation (1).

localid="1654602578114" $\frac{\mathrm{v}}{1+\frac{{\mathrm{v}}^2}{c^2}}=\frac{\mathbf{E}\mathbf{脳}\mathbf{B}}{{\mathrm{B}}^2-\frac{{\mathrm{E}}^2}{{\mathrm{c}}^2}}$

No, there can be no frame in which E is perpendicular to B because as the dot product of E and B is invariant and not equal to zero, it can鈥檛 be zero in $\overline{\mathrm{S}}$frame.

Therefore, in a differential inertial system $\overline{S}$moving to relative to S with velocity v is

$\frac{\mathrm{v}}{1+\frac{{\mathrm{v}}^2}{c^2}}=\frac{\mathbf{E}\mathbf{脳}\mathbf{B}}{{\mathrm{B}}^2-\frac{{\mathrm{E}}^2}{{\mathrm{c}}^2}}$

and no, there can be no frame in which the two are perpendicular.