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Chapter 5: Problem 23

(a) What is \(\gamma\) if \(v=0.250 c ?\) (b) If \(v=0.500 c ?\)

Short Answer

Expert verified
(a) When \(v = 0.250c\), the Lorentz factor \(\gamma\) is approximately equal to \(1.032\). (b) When \(v = 0.500c\), the Lorentz factor \(\gamma\) is approximately equal to \(1.155\).

Step by step solution

01

(a) Calculating \(\gamma\) when \(v = 0.250c\)

To calculate the Lorentz factor at \(v = 0.250c\), we need to substitute the given value of \(v\) into the formula for \(\gamma\): \[\gamma = \frac{1}{\sqrt{1-\frac{(0.250c)^2}{c^2}}}\] Now, we can simplify the expression.
02

Simplifying the expression for \(v = 0.250c\)

Simplify the expression by squaring the value of \(v\) and canceling the \(c^2\) terms: \[\gamma = \frac{1}{\sqrt{1-\frac{0.0625c^2}{c^2}}}\] \[\gamma = \frac{1}{\sqrt{1-0.0625}}\] Now we can find the value of \(\gamma\).
03

Finding the value of \(\gamma\) for \(v = 0.250c\)

Subtract and find the square root: \[\gamma = \frac{1}{\sqrt{0.9375}}\] \[\gamma \approx 1.032\] So, the Lorentz factor \(\gamma\) when \(v = 0.250c\) is approximately equal to \(1.032\).
04

(b) Calculating \(\gamma\) when \(v = 0.500c\)

To calculate the Lorentz factor at \(v = 0.500c\), we need to substitute the given value of \(v\) into the formula for \(\gamma\): \[\gamma = \frac{1}{\sqrt{1-\frac{(0.500c)^2}{c^2}}}\] Now, we can simplify the expression.
05

Simplifying the expression for \(v = 0.500c\)

Simplify by squaring the value of \(v\) and canceling the \(c^2\) terms: \[\gamma = \frac{1}{\sqrt{1-\frac{0.25c^2}{c^2}}}\] \[\gamma = \frac{1}{\sqrt{1-0.25}}\] Now we can find the value of \(\gamma\).
06

Finding the value of \(\gamma\) for \(v = 0.500c\)

Subtract and take the square root: \[\gamma = \frac{1}{\sqrt{0.75}}\] \[\gamma \approx 1.155\] So, the Lorentz factor \(\gamma\) when \(v = 0.500c\) is approximately equal to \(1.155\).

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