Math solution on Time Dilation

Time dilation is a concept from Einstein's theory of relativity that describes how time is experienced differently for observers in different states of motion. According to the theory of special relativity, time passes more slowly for an object moving at high speed compared to an object at rest.

The formula for time dilation is given by:

$$\mathrm{\Delta }{𝑡}^{\mathrm{\prime }}=\mathrm{\Delta }𝑡\sqrt{1-\frac{{𝑣}^2}{{𝑐}^2}}$$

where:

• $\mathrm{\Delta }{𝑡}^{\mathrm{\prime }}$ is the time interval measured by an observer moving with velocity $𝑣$.
• $\mathrm{\Delta }𝑡$ is the time interval measured by an observer at rest.
• $𝑣$ is the relative velocity between the observers.
• $𝑐$ is the speed of light in a vacuum ($𝑐\approx 3\times 10^8\text{m/s}$).

Example Problem

Let's solve an example problem step-by-step using this formula.

Problem

A spaceship travels at a speed of $2.4\times 10^8\text{m/s}$ relative to an observer on Earth. If the time interval measured by the observer on Earth is 2 hours, what is the time interval experienced by the astronaut on the spaceship?

Solution

1. Identify the given values:

   • Relative velocity, $𝑣=2.4\times 10^8\text{m/s}$
   • Time interval measured by the observer on Earth, $\mathrm{\Delta }𝑡=2\text{hours}$
   • Speed of light, $𝑐=3\times 10^8\text{m/s}$

2. Convert the time interval to seconds (if necessary):

   $$\mathrm{\Delta }𝑡=2\text{hours}=2\times 3600\text{s}=7200\text{s}$$

3. Substitute the values into the time dilation formula:

   $$\mathrm{\Delta }{𝑡}^{\mathrm{\prime }}=\mathrm{\Delta }𝑡\sqrt{1-\frac{{𝑣}^2}{{𝑐}^2}}$$
   $$\mathrm{\Delta }{𝑡}^{\mathrm{\prime }}=7200\text{s}\sqrt{1-\frac{(2.4\times 10^8\text{m/s}{)}^2}{(3\times 10^8\text{m/s}{)}^2}}$$

4. Calculate the term inside the square root:

   $$\frac{{𝑣}^2}{{𝑐}^2}=\frac{(2.4\times 10^8\text{m/s}{)}^2}{(3\times 10^8\text{m/s}{)}^2}=\frac{5.76\times 10^{16}}{9\times 10^{16}}=\frac{5.76}{9}=0.64$$

5. Continue with the calculation:

   $$\mathrm{\Delta }{𝑡}^{\mathrm{\prime }}=7200\text{s}\sqrt{1-0.64}=7200\text{s}\sqrt{0.36}=7200\text{s}\times 0.6$$
   $$\mathrm{\Delta }{𝑡}^{\mathrm{\prime }}=4320\text{s}$$

6. Convert the time interval back to hours (if necessary):

   $$\mathrm{\Delta }{𝑡}^{\mathrm{\prime }}=\frac{4320\text{s}}{3600\text{s/hour}}=1.2\text{hours}$$

So, the time interval experienced by the astronaut on the spaceship is 1.2 hours.

Conclusion

Using the time dilation formula, we found that for an observer moving at $2.4\times 10^8\text{m/s}$

m/s relative to Earth, a time interval of 2 hours on Earth would be experienced as only 1.2 hours on the spaceship. This illustrates how high velocities can significantly affect the passage of time.