Question

Show that

${\mathbf{k}}_{\mathbf{\mu }}{\mathbf{k}}^{\mathbf{\mu }}\mathbf{=}\frac{\mathbf{\theta }\mathbf{‐}\left({\displaystyle \frac{u^2}{c^2}}\right){\mathbf{cos}}^{\mathbf{2}}}{\mathbf{1}\mathbf{-}{\displaystyle \frac{{\mathbf{u}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}}$

Where$\mathbf{\theta }$is the angle between u and F.

Short answer

${\mathrm{K}}_{\mathrm{\mu }}{\mathrm{K}}^{°}={\mathrm{F}}^2\left[\begin{array}{c}1-\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}{\cos }^{2}0\\ 1-\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}\end{array}\right]$

It is proved that .

Step-by-step solution

Expression for the Minkowski force:

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

${\mathbf{K}}_{\mathbf{\mu }\mathbf{}}{\mathbf{K}}^{\mathbf{\mu }}\mathbf{=}\mathbf{‐}\left(k^0\right)\mathbf{}\mathbf{+}\mathbf{}\mathbf{K}\mathbf{·}\mathbf{K}$ …… (1)

Here,kis the ordinary force which is given by,

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

Determine the zeroth component of K:

Write the zeroth component ofKusing equation 12.70.

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

Here,Eis the energy which is given by,

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

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

…… (3)

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

Prove that :

It is known that:

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

Multiply byuon both sides in the above expression.

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

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

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

localid="1654669882880" $$\begin{gathered} {\mathrm{K}}_{\mathrm{\mu }}{\mathrm{K}}^{\mathrm{\mu }}=-{\left(\frac{\mathrm{m}}{\mathrm{c}}\frac{\left(\mathrm{uF}\mathrm{cos\theta }\right)}{\mathrm{c}\sqrt{1‐{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}}\right)}^2+\left(\frac{{\mathrm{F}}^2}{1‐\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}\right) \\ {\mathrm{K}}_{\mathrm{\mu }}{\mathrm{K}}^{\mathrm{\mu }}=-\frac{{\mathrm{u}}^2{\mathrm{F}}^2{\cos }^2\mathrm{\theta }}{{\mathrm{c}}^2\left(1‐\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}\right)}+\frac{{\mathrm{F}}^2}{1‐\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}} \\ {\mathrm{K}}_{\mathrm{\mu }}{\mathrm{K}}^{\mathrm{\mu }}=\frac{{\mathrm{F}}^2}{1‐\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}\left[1‐\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}{\cos }^2\mathrm{\theta }\right] \\ {\mathrm{K}}_{\mathrm{\mu }}{\mathrm{K}}^{\mathrm{\mu }}={\mathrm{F}}^2\left[\frac{1‐\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}{\cos }^2\mathrm{\theta }}{1‐\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}\right] \end{gathered}$$

${\mathrm{K}}_{\mathrm{\mu }}{\mathrm{K}}^{\mathrm{\mu }}={\mathrm{F}}^2\left[\frac{1‐\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}{\cos }^2\mathrm{\theta }}{1‐\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}\right]$

Therefore, it is proved that .