Chapter 15, Problem 34A

Assume that your heart normally beats once every second, and that you are in a spaceship that moves past Earth at 0.6c.

a. What time do you measure between your own heartbeats?

b. Show that your heartbeats are measured by someone on Earth to be 1.25 s apart.

c. As you and the spaceship whiz past Earth, you make similar measurements on Earthlings who measure their own heart-beats to be 1 s apart. How much time do you measure between their heartbeats?

To determine

The time measured by a person’s own heartbeat.

Answer to Problem 34A

The time measured by a person’s own heartbeat is once every second.

Explanation of Solution

Given:

The speed of spaceship is $v=0.6c$ .

Calculation:

To propel the spaceships to relativistic speeds, the amounts of energy requiredare billions of times the energy that is used to put the space shuttles into orbit.

Given that the persons heart normally beats once every second. So, the time measured by a person’s own heartbeat is once every second since the body is at rest with respect to reference frame.

Conclusion:

Thus, the time measured by a person’s own heartbeat is once every second.

To determine

the time period of heartbeats by someone on Earth to be $2.5\text{ s}$ .

Answer to Problem 34A

The time period of heartbeats by someone on Earth is $2.5\text{ s}$ .

Explanation of Solution

Given info:

The speed of spaceship is $v=0.6c$ .

Formula used:

Show the expression for relative time in a frame of reference as follows:

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

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:

Refer to part (a).

The time measured by a person’s own heartbeat is once every second. That is, the time of heartbeats is $t_0=1.0\text{ s}$ .

Find the time period of heartbeats by someone on Earth as follows:

  $\begin{array}{l}t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}\\ t=\frac{\left(\text{1}\text{.0 s}\right)}{\sqrt{1-\frac{{\left(0.6c\right)}^2}{c^2}}}\\ t=1.25\text{ s}\end{array}$

Hence, the time period of heartbeats by someone on Earth is $t=1.25\text{ s}$ .

Conclusion:

Thus, the time period of heartbeats by someone on Earth is $1.25\text{ s}$ .

To determine

the time period of heartbeats by someone on Earth.

Answer to Problem 34A

The time period of heartbeats by someone on Earth is $2.5\text{ s}$ .

Explanation of Solution

Given info:

The time of heartbeats is $t_0=1.0\text{ s}$ .

The speed of spaceship is $v=0.6c$ .

Formula used:

Show the expression for relative time in a frame of reference as follows:

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

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:

Find the time period of heartbeats by someone on Earth as follows:

  $\begin{array}{l}t=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}\\ t=\frac{\left(\text{1}\text{.0 s}\right)}{\sqrt{1-\frac{{\left(0.6c\right)}^2}{c^2}}}\\ t=1.25\text{ s}\end{array}$

Hence, the time period of heartbeats by someone on Earth is $t=1.25\text{ s}$ .

Conclusion:

Thus, the time period of heartbeats by someone on Earth is $1.25\text{ s}$ .