Chapter 12: Q37P (page 549)

In classical mechanics, Newton’s law can be written in the more familiar form $F = ma$ . The relativistic equation, $F = \frac{dp}{dt}$ , cannot be so simply expressed. Show, rather, that
$F = \frac{m}{\sqrt{1 - u^{2} / c^{2}}} [a + \frac{u (u ‐ a)}{c^{2} ‐ u^{2}}]$
where $a = \frac{du}{dt}$ is the ordinary acceleration.

Short Answer

Expert verified

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

Step by step solution

Expression for the realistic momentum:

Write the expression for the realistic momentum is given by,

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

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

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

$F = \frac{d}{dt} [\frac{mu}{\sqrt{1 - \frac{u^{2}}{c^{2}}}}] F = m [\frac{1}{\sqrt{1 - \frac{u^{2}}{c^{2}}}} \frac{du}{dt} + (- \frac{1}{2}) \frac{- \frac{1}{c^{2}} (2 u) \frac{du}{dt}}{(1 - \frac{u^{2}}{c^{2}})}]$

On further solving,

$F = [\begin{matrix} \end{matrix} \frac{1}{\sqrt{1 - \frac{u^{2}}{c^{2}}}} a + \frac{\frac{1}{c^{2}} u (u - a)}{(1 - \frac{u^{2}}{c^{2}})}] F = [\begin{matrix} \end{matrix} \frac{1}{\sqrt{1 - \frac{u^{2}}{c^{2}}}} a + \frac{\frac{1}{c^{2}} u (u ‐ a)}{{(\frac{c^{2} - u^{2}}{c^{2}})}^{\frac{3}{2}}}] F = \frac{m}{\sqrt{1 - \frac{u^{2}}{c^{2}}}} [a + \frac{u (u - a)}{(\frac{c^{2} - u^{2}}{c^{2}})}]$

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

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Short Answer

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Expression for the realistic momentum:

Show that :

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In classical mechanics, Newton’s law can be written in the more familiar form F=ma. The relativistic equation, F=dpdt, cannot be so simply expressed. Show, rather, thatF=m1-u2/c2[a+uu‐ac2‐u2]where a=dudt is the ordinary acceleration.

Short Answer

Step by step solution

Expression for the realistic momentum:

Show that :

One App. One Place for Learning.

Most popular questions from this chapter

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Study anywhere. Anytime. Across all devices.