Chapter 25, Problem 29QAP

To determine

(a)

A graph of ? versus v/c and determine the maximum value of v/c for which ? is less than 1% larger than unity (1.0)

Answer to Problem 29QAP

And, maximum value of v/c for which ? is less than 1% larger than unity (1.0) is 0.14

Explanation of Solution

Given:

v/c	?	v/c	?
0	1	0.7	1.40028
0.1	1.005038	0.8	1.666667
0.2	1.020621	0.9	2.294157
0.3	1.048285	0.99	7.088812
0.4	1.091089	0.999	22.36627
0.5	1.154701	0.9999	70.71245
0.6	1.25

Graph:

From the given values, we can draw graph.

Formula used:

Relativistic gamma:

  $\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

After rearranging

  $\left(\frac{v}{c}\right)=\sqrt{1-\frac{1}{{\gamma }^2}}$

Calculation:

Maximum value of v/c for which ? is less than 1% larger than unity (1.0)

So, 0.00999 < 0.01 (1 % of unity)

At, $\gamma =1.00999$

  $\left(\frac{v}{c}\right)=\sqrt{1-\frac{1}{{\gamma }^2}}=\sqrt{1-\frac{1}{{(1.00999)}^2}}=0.14$

Conclusion:

Maximum value of v/c for which ? is less than 1% larger than unity (1.0) is 0.14

To determine

(b)

What numerical value does the slope of your graph take at v/c = 0.1, v/c = 0.5, v/c = 0.9, v/c = 0.999?

Answer to Problem 29QAP

At v/c = 0.1, slope = 0.101

At v/c = 0.5, slope = 0.768

At v/c = 0.9, slope = 10.867

At v/c = 0.999, slope = 11177.54

Explanation of Solution

Formula used:

Relativistic gamma:

  $\gamma =\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$

  $\because \left[\begin{array}{l}y(x)=\frac{1}{\sqrt{1-x^2}}\\ \text{diffrentiating}\text{w}\text{.r}\text{.t}x\text{gives}\text{slope}\text{of}\text{the}\text{equation}\\ \frac{dy}{dx}=\frac{x}{{\left(1-x^2\right)}^{\frac{3}{2}}}\end{array}\right]$

Similarly,

  $slope=\frac{d\gamma }{d\left(\frac{v}{c}\right)}=\frac{\left(\frac{v}{c}\right)}{{\left(1-\frac{v^2}{c^2}\right)}^{\frac{3}{2}}}$

Calculation:

  $slope\left(at\frac{v}{c}=0.1\right)=\frac{\left(\frac{v}{c}\right)}{{\left(1-\frac{v^2}{c^2}\right)}^{\frac{3}{2}}}=\frac{\left(0.1\right)}{{\left(1-{0.1}^2\right)}^{\frac{3}{2}}}=0.1015$

  $slope\left(at\frac{v}{c}=0.5\right)=\frac{\left(\frac{v}{c}\right)}{{\left(1-\frac{v^2}{c^2}\right)}^{\frac{3}{2}}}=\frac{\left(0.5\right)}{{\left(1-{\left(0.5\right)}^2\right)}^{\frac{3}{2}}}=0.7698$

  $slope\left(at\frac{v}{c}=0.9\right)=\frac{\left(\frac{v}{c}\right)}{{\left(1-\frac{v^2}{c^2}\right)}^{\frac{3}{2}}}=\frac{\left(0.9\right)}{{\left(1-{\left(0.9\right)}^2\right)}^{\frac{3}{2}}}=10.867$

  $slope\left(at\frac{v}{c}=0.999\right)=\frac{\left(\frac{v}{c}\right)}{{\left(1-\frac{v^2}{c^2}\right)}^{\frac{3}{2}}}=\frac{\left(0.999\right)}{{\left(1-{\left(0.999\right)}^2\right)}^{\frac{3}{2}}}=11177.54$

Conclusion:

Thus, at v/c = 0.1, slope = 0.101

At v/c = 0.5, slope = 0.768

At v/c = 0.9, slope = 10.867

At v/c = 0.999, slope = 11177.54