Chapter 15, Problem 36A

A ship whizzes by you at 0.60c. Someone aboard is making a 3-minute egg for breakfast.

a. What cooking time will you measure for the egg?

b. Why should you not be surprised when the egg turns out to be perfectly cooked, rather than overcooked?

To determine

The cooking time for egg.

Answer to Problem 36A

The cooking time for egg is $3.8\text{ min}$ .

Explanation of Solution

Given:

The speed of ship whizzes is $v=0.6c$ .

The time for making egg is $t_0=3\text{ min}$ .

Formula used:

The expression for relative time in a frame of reference is,

  $t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}$

Here, $t_0$ is the proper time, $v$ is the speed of the object relative to an observer, and $c$ is the speed of the light.

Calculation:

The cooking time for egg is,

  $\begin{array}{l}t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}\\ t=\frac{\left(\text{3 min}\right)}{\sqrt{1-\frac{{\left(0.6c\right)}^2}{c^2}}}\\ t=3.8\text{ min}\end{array}$

Conclusion:

Thus, the cooking time for egg is $3.8\text{ min}$ .

To determine

To explain: The reason for not being surprised when the egg turns out to be perfectly cooked, rather than overcooked.

Explanation of Solution

Introduction:

The expression for relative time in a frame of reference is,

  $t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}$

Here, $t_0$ is the proper time, $v$ is the speed of the object relative to an observer, and $c$ is the speed of the light.

The occurrence of Time dilation is negligible for day to speed of the vehicles, but it becomes a matter of concern when it speed of the object approaches light’s speed.

When the moving events are seen from others reference frame, there is a time dilation.

The cook and the egg both are in same reference frame, so the egg is perfectly cooked.

The cooking time of egg is not changed, when the egg is “perfectly cooked”, instead of over cooking.

Conclusion:

Thus, the cooking time is not changed, when the egg turns out to be perfectly cooked, rather than overcooked.