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What is Time Dilation? – Explanation and Example

 

One of the most important concepts that we have learned so far is that – time is relative. The rate at which time passes depends on your velocity,  the faster you move the more slowly time gets passes. But how exactly time can be slower or faster?

The above discussion brings light to the most important concept of physics called Time Dilation. You can understand time dilation as – “A moving clock rounds more slowly than a clock at rest.”


It means that, if someone moving in a spacecraft observes that time interval between two events $T_{0}$, then the observer at rest will get to know that, the same interval has a longer duration of time T.

This can bring a thought that, if the clock in spacecraft runs slowly, why don’t we notice? Well, the answer to this question can be explained as As we are going slowly, that’s why we can’t observe any noticeable change. Even if we go at very high speed, we wouldn’t be able to notice it, because our own bio-clocks would also be running just as slowly.

This can be explained by a mathematical expression as:

T = $\frac{T_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$


Here, v is the velocity of the spacecraft and c is the velocity of light.

The time interval $T_{0}$ observed on the spacecraft is called the proper time and the time interval recorded by the stationary observer is called the dilated time.


The term $\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$ is called the Lorentz Factor, it is the factor by which time changes for an object while that object is moving. As the Lorentz Factor is always remains smaller than 1 for the moving spacecraft, T is always greater than $T_{0}$. And that’s why the clock on the spacecraft rounds at a slower rate than the stationary clock on the ground.


For example: In a spaceship moving at 95% of the speed of light, 10 years passes. How much time will pass on earth?


Using the above relation we can solve this problem as,


T=$\frac{10}{\sqrt{1-\frac{(0.95c)^{2}}{c^{2}}}}$
 
T=$\frac{10}{\sqrt{1-0.95^{2}}}$
 
T=$\frac{10}{0.312}$
 
T=32 years


10 years spent on the spacecraft is equal to 32 years spent by a stationary observer on earth. This is how the relativity of time works.

 

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