91影视

Using equation 12.69, write the expression for the Minkowski force.

${\mathbf{K}}_{\mathbf{渭}\mathbf{}}{\mathbf{K}}^{\mathbf{渭}}\mathbf{=}\mathbf{鈥�}\left(k^0\right)\mathbf{}\mathbf{+}\mathbf{}\mathbf{K}\mathbf{路}\mathbf{K}$ 鈥︹�� (1)

Here,kis the ordinary force which is given by,

$$\begin{gathered} \mathbf{K}\mathbf{路}\mathbf{K}=\frac{1}{\sqrt{1-{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}}\mathrm{F}路\frac{1}{\sqrt{1-{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}}\mathrm{F} \\ \mathbf{K}\mathbf{路}\mathbf{K}\mathbf{}=\frac{{\mathrm{F}}^2}{1-\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}} \end{gathered}$$

Write the zeroth component ofKusing equation 12.70.

$$\begin{gathered} {\mathrm{K}}^0=\frac{{\mathrm{dp}}^0}{\mathrm{dt}} \\ {\mathrm{K}}^0=\frac{1}{\mathrm{c}}\frac{\mathrm{dE}}{\mathrm{dT}} \end{gathered}$$ 鈥︹�� (2)

Here,Eis the energy which is given by,

$$\begin{gathered} \mathrm{E}=\frac{{\mathrm{mc}}^2}{\mathrm{纬}} \\ \mathrm{E}=\frac{{\mathrm{mc}}^2}{\sqrt{1-{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}} \end{gathered}$$

Substitute $\frac{{\mathrm{mc}}^2}{\sqrt{1-\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}$for$E$in equation (2).

鈥︹�� (3)

$$\begin{gathered} {\mathrm{K}}^0=\frac{1}{\sqrt{1-\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}\frac{\mathrm{d}}{\mathrm{dt}}\left(\frac{{\mathrm{mc}}^2}{\sqrt{1-\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}\right) \\ {\mathrm{K}}^0=\frac{{\mathrm{mc}}^2}{\sqrt{1-\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}\left[-\frac{1}{2}\frac{\left(-1{\displaystyle \frac{1}{{\mathrm{c}}^2}}\right)}{{\left(1-\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}\right)}^{{\displaystyle \frac{3}{2}}}}2\mathrm{u}路\mathrm{a}\right] \\ {\mathrm{K}}^0=\frac{\mathrm{m}}{\mathrm{c}}\frac{\mathrm{u}路\mathrm{a}}{{\left(1-\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}\right)}^2} \end{gathered}$$

It is known that:

$\mathrm{F}-\frac{\mathrm{m}}{\sqrt{1-{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}}\left[\mathrm{a}+\frac{\mathrm{u}\left(\mathrm{u}路\mathrm{a}\right)}{{\mathrm{c}}^2-{\mathrm{u}}^2}\right]$

Multiply byuon both sides in the above expression.

Substitute$\mathrm{u}F\cos 胃$, for$\mathrm{u}路\mathrm{F}$in equation (3).

$K^0=\frac{\mathrm{m}}{c}\frac{\left(\mathrm{u}F\cos 胃\right)}{c\sqrt{1鈥�{\displaystyle \frac{{\mathrm{u}}^2}{c^2}}}}$

Substitute $\frac{\mathrm{m}}{c}\frac{\left(\mathrm{u}F\cos 胃\right)}{c\sqrt{1鈥�{\displaystyle \frac{{\mathrm{u}}^2}{c^2}}}}$for ${\mathrm{K}}^0$and $\frac{{\mathrm{F}}^2}{1鈥�\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}$for $\mathbf{K}\mathbf{路}\mathbf{K}$in equation (1).

localid="1654669882880" $$\begin{gathered} {\mathrm{K}}_{\mathrm{渭}}{\mathrm{K}}^{\mathrm{渭}}=-{\left(\frac{\mathrm{m}}{\mathrm{c}}\frac{\left(\mathrm{uF}\mathrm{肠辞蝉胃}\right)}{\mathrm{c}\sqrt{1鈥�{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}}\right)}^2+\left(\frac{{\mathrm{F}}^2}{1鈥�\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}\right) \\ {\mathrm{K}}_{\mathrm{渭}}{\mathrm{K}}^{\mathrm{渭}}=-\frac{{\mathrm{u}}^2{\mathrm{F}}^2{\cos }^2\mathrm{胃}}{{\mathrm{c}}^2\left(1鈥�\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}\right)}+\frac{{\mathrm{F}}^2}{1鈥�\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}} \\ {\mathrm{K}}_{\mathrm{渭}}{\mathrm{K}}^{\mathrm{渭}}=\frac{{\mathrm{F}}^2}{1鈥�\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}\left[1鈥�\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}{\cos }^2\mathrm{胃}\right] \\ {\mathrm{K}}_{\mathrm{渭}}{\mathrm{K}}^{\mathrm{渭}}={\mathrm{F}}^2\left[\frac{1鈥�\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}{\cos }^2\mathrm{胃}}{1鈥�\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}\right] \end{gathered}$$

${\mathrm{K}}_{\mathrm{渭}}{\mathrm{K}}^{\mathrm{渭}}={\mathrm{F}}^2\left[\frac{1鈥�\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}{\cos }^2\mathrm{胃}}{1鈥�\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}\right]$

Therefore, it is proved that .