We all are very well familiar with time and how important time is. But, time dilation is something that only very few people like scientists, physicists, etc know about. Besides, we are going to discuss time dilation, time dilation formula, its derivation and solved example in this topic. Moreover, after going through this topic you will be able to understand time dilation easily.

**Time Dilation**

It refers to a special relative state that time can pass at diverse rates in diverse reference frames. Also, it depends upon the velocity of one reference frame comparative to another.

In simple words, time dilation is a measure of the elapsed time that we measure using two clocks. Also, there are two reference frames referred to as the proper time (one-position time) and observer time (two-position time). Moreover, both of them are interrelated and we can find the time dilation of one if we know the velocity and speed of others.

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**Time Dilation Formula**

Suppose in a reference frame, the time between the events is called proper time or one-position time and is labeled \(\Delta t_{0}\). Moreover, in another reference frame, the observer will see the two events occur in a different position. Furthermore, in the observer’s reference frame the time between two events is known as observer time or two-position time and is labeled as \(\Delta t\). Besides, the observer time will always be larger than the proper time. We refer to this effect of time as time dilation. Most noteworthy, we measure both the \(\Delta t_{0}\) and \(\Delta t\) in seconds.

### Formula

Observer time = \(\frac{proper time}{\sqrt{1 -\left ( \frac{velocity}{speed of light} \right )^{2}}}\)

\(\Delta t\) = \(\frac{\Delta t_{0}}{\sqrt{1 -\left ( \frac{v}{c} \right )^{2}}}\)

**Derivation of the Formula**

\(\Delta t\) = refers to the two-position time or observer time in seconds

\(\Delta t_{0}\) = refers to the one-position time or proper time in seconds

v = refers to the velocity in meter per second (m/s)

c = refers to the speed of light that is 3.0 × \(10^{8}\) m/s

**Solved Example on Time Dilation Formula**

**Example 1**

Suppose that Kalpana boards a spaceship and flies past the earth at 0.800 times the speed of light. Moreover, her twin sister Alpana stays on the earth. In addition, the instant Kalpana’s ship passes the earth both the sisters start timers. Furthermore, Kalpana watches her timer and stop it when the timer passes 60.0 seconds. Now, calculate the time that would have passes in Alpana’s timer?

**Solution:**

In this, the two events that need to be concerned is the start and stop of Kalpana’s timer. Also, the two events happen in the same position in Kalpana’s reference time. Moreover, it means that Kalpana’s time aboard the spaceship is the proper time and the time difference aboard the ship is \(\Delta t_{0}\). Furthermore, the starting and stopping of Kalpana’s timer happen in two positions in Alpana’s reference frame, so in this way, the time difference on the Earth is the observer time \(\Delta t\). So, we can find the amount of time passes in Alpana’s reference frame using this formula:

\(\Delta t\) = \(\frac{\Delta t_{0}}{\sqrt{1 -\left ( \frac{v}{c} \right )^{2}}}\)

\(\Delta t\) = \(\frac{60.0 s}{\sqrt{1 -\left ( \frac{0.800 c}{c} \right )^{2}}}\)

\(\Delta t\) = \(\frac{60.0 s}{\sqrt{1 -\left ( 0.800c \right )^{2}}}\)

\(\Delta t\) = \(\frac{60.0 s}{\sqrt{1 – 0.640}}\)

\(\Delta t\) = \(\frac{60.0 s}{\sqrt 0.360}\)

\(\Delta t\) = \(\frac{60.0 s}{0.600}\)

\(\Delta t\) = 100 s

So, when Kalpana is in a reference frame moving at 0.800 c relative to Alpana’s reference frame, and Kalpana observes that 60.0 s had passed then her sister Alpana will observe that 100 s have passed in her timer.

Typo Error>

Speed of Light, C = 299,792,458 m/s in vacuum

So U s/b C = 3 x 10^8 m/s

Not that C = 3 x 108 m/s

to imply C = 324 m/s

A bullet is faster than 324m/s

I have realy intrested to to this topic

m=f/a correct this

M=f/g

Interesting studies

It is already correct f= ma by second newton formula…