Question

In classical mechanics, Newton’s law can be written in the more familiar form $\mathbf{F}\mathbf{=}\mathbf{ma}$. The relativistic equation, $\mathbf{F}\mathbf{=}\frac{\mathbf{dp}}{\mathbf{dt}}$, cannot be so simply expressed. Show, rather, that

$\mathbf{F}\mathbf{=}\frac{\mathbf{m}}{\sqrt{\mathbf{1}\mathbf{-}{\mathbf{u}}^{\mathbf{2}}\mathbf{/}{\mathbf{c}}^{\mathbf{2}}}}\left[a+\frac{u\left(u‐a\right)}{c^2‐u^2}\right]$

where $\mathbf{a}\mathbf{=}\frac{\mathbf{du}}{\mathbf{dt}}$ is the ordinary acceleration.

Short answer

It is proved that $\mathrm{F}=\frac{\mathrm{m}}{\sqrt{1{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}}\left\{\left.\mathrm{a}+\frac{\mathrm{u}\left(\mathrm{u}‐\mathrm{a}\right)}{{\mathrm{c}}^2-{\mathrm{u}}^2}\right\}\right.$

Step-by-step solution

Expression for the realistic momentum:

Write the expression for the realistic momentum is given by,

$\mathbf{p}\mathbf{=}\frac{\mathbf{mu}}{\sqrt{\mathbf{1}\mathbf{+}{\displaystyle \frac{{\mathbf{u}}^{\mathbf{2}}}{{\mathbf{c}}^{\mathbf{2}}}}}}$ …… (1)

Here, m is the mass, u is the ordinary velocity, and c is the speed of light.

Show that :

From the given problem, it is given that:

$$\begin{gathered} \mathrm{F}=\frac{\mathrm{dp}}{\mathrm{dt}} \\ \mathrm{Also}, \\ \mathrm{a}=\frac{\mathrm{du}}{\mathrm{dt}} \end{gathered}$$ .......(2)

Substitute the value of equation (1) in equation (2).

$$\begin{gathered} \mathrm{F}=\frac{\mathrm{d}}{\mathrm{dt}}\left[\frac{\mathrm{mu}}{\sqrt{1-{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}}\right] \\ \mathrm{F}=\mathrm{m}\left[\frac{1}{\sqrt{1-\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}\frac{\mathrm{du}}{\mathrm{dt}}+\left(-\frac{1}{2}\right)\frac{-{\displaystyle \frac{1}{{\mathrm{c}}^2}}\left(2\mathrm{u}\right){\displaystyle \frac{\mathrm{du}}{\mathrm{dt}}}}{\left(1-{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}\right)}\right] \end{gathered}$$

On further solving,

$$\begin{gathered} \mathrm{F}=\left[\begin{array}{c}\\ \end{array}\frac{1}{\sqrt{1-{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}}\mathrm{a}+\frac{{\displaystyle \frac{1}{{\mathrm{c}}^2}}\mathrm{u}\left(\mathrm{u}-\mathrm{a}\right)}{\left(1-\frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}\right)}\right] \\ \mathrm{F}=\left[\begin{array}{c}\\ \end{array}\frac{1}{\sqrt{1-{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}}\mathrm{a}+\frac{{\displaystyle \frac{1}{{\mathrm{c}}^2}}\mathrm{u}\left(\mathrm{u}‐\mathrm{a}\right)}{{\left(\frac{{\mathrm{c}}^2-{\mathrm{u}}^2}{{\mathrm{c}}^2}\right)}^{{\displaystyle \frac{3}{2}}}}\right] \\ \mathrm{F}=\frac{\mathrm{m}}{\sqrt{1-{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}}\left[\mathrm{a}+\frac{\mathrm{u}\left(\mathrm{u}-\mathrm{a}\right)}{\left(\frac{{\mathrm{c}}^2-{\mathrm{u}}^2}{{\mathrm{c}}^2}\right)}\right] \end{gathered}$$

Therefore, it is proved that $\mathrm{F}=\frac{\mathrm{m}}{\sqrt{1-{\displaystyle \frac{{\mathrm{u}}^2}{{\mathrm{c}}^2}}}}\left[\mathrm{a}+\frac{\mathrm{u}\left(\mathrm{u}-\mathrm{a}\right)}{\left(\frac{{\mathrm{c}}^2-{\mathrm{u}}^2}{{\mathrm{c}}^2}\right)}\right]$