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Chapter 26: Problem 59

Show that each of these statements implies that \(v<c\) which means that \(v\) can be considered a nonrelativistic speed: (a) $\gamma-1<<1 \text { [Eq. }(26-14)] ;\( (b) \)K<<m c^{2}\( [Eq. \)(26-15)] ;(\text { c }) p<m c[\text { Eq. }(26-16)] ;\( (d) \)K \approx p^{2} /(2 m)$.

Short Answer

Expert verified
In conclusion, we have analyzed each statement, (a) \(\gamma - 1 << 1\), (b) \(K<<mc^2\), (c) \(p<<mc\), and (d) \(K \approx p^2/(2m)\), and proved that they imply the condition \(v << c\), meaning the speed \(v\) is nonrelativistic. Using the equations for the Lorentz factor, kinetic energy, momentum, and mass-energy equivalence, we demonstrate that each statement points to the same fundamental condition governing nonrelativistic speeds, ensuring the validity of the conclusion.

Step by step solution

01

(a) Analyzing \(\gamma - 1

First, let's rewrite the Lorentz factor, from where we have: \(\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\). Given that \(\gamma -1 << 1\), we can write it as: \(\gamma \approx 1\). Now, we can plug this into the Lorentz factor equation and solve for \(v\): \(\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \approx 1\), which leads to \(v<<c\). Thus, statement (a) implies that \(v\) is nonrelativistic.
02

(b) Analyzing \(K

Recall the expression for kinetic energy: \(K = \frac{1}{2}mv^2\). Then, according to the given statement: \(\frac{1}{2}mv^2 << mc^2\). By dividing both sides by \(mc\), we get: \(v^2/2 << c^2\). This implies: \(v^2 << 2c^2 \rightarrow v << c\). Thus, statement (b) implies that \(v\) is nonrelativistic.
03

(c) Analyzing \(p

Recall the equation for momentum: \(p = m\gamma v\). Given that \(p << mc\), we can write: \(m\gamma v << mc\). Dividing both sides by \(mc\), we get: \(\frac{\gamma v}{c} << 1\). Since \(\gamma \approx 1\) for nonrelativistic speeds, we have: \(\frac{v}{c} << 1\). This implies \(v << c\), so statement (c) implies that \(v\) is nonrelativistic.
04

(d) Analyzing \(K \approx p^2/(2m)\)

Recall the expressions for kinetic energy and momentum: \(K = \frac{1}{2}mv^2 \text{ and } p = m\gamma v\). Using the given statement: \(p^2/(2m) \approx \frac{1}{2}mv^2\). By dividing both sides by \(\frac{1}{2}mv^2\) we get: \(\frac{p^2}{mv^2} \approx 1\). This implies: \(p^2 = m^2\gamma^2v^2 \approx m^2v^2\). Noting that \(m^2v^2 >> mv^2\) due to \(\gamma \approx 1\) for nonrelativistic speeds, we obtain: \(\gamma^2 = 1 + (\frac{v}{c})^2 \approx 1\). Therefore, \((\frac{v}{c})^2 << 1 \rightarrow v << c\), and statement (d) implies that \(v\) is nonrelativistic.

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