**Time Dilation**

Time dilation is pretty much the basis for Einstein’s theory of special relativity. To explain what it is, it’s important to know what a *frame of reference *is.

A common analogy used is to imagine someone is on a train bouncing a basketball. When the train is moving, whoever is bouncing the ball will simply see the ball moving up and down. However, to someone who is *not * on the train, the ball will appear to move a bit like this:

This is an example of two different frames of reference, the one on the train and one off the train both observe different things. Now imagine that train is going really fast, like half of the speed of light, and replace the basketball with a device to measure time.

Since electromagnetic radiation (light) travels at the same speed in a vacuum, the ‘timer’ should measure light in some way. The timer, shown below, has an emitter that emits a photon or beam of light, and a receiver, that detects the light. For simplicity, I’m going to assume that L_{0} is very large compared to d_{0} so that the light travels directly up and down. The time taken for this process, which we call ∆t is equal to the distance the light travels, divided by the speed of light (c).

∆t_{0 }= 2L_{0}/c (equation 1)

So while the train is *not* moving, both observers can agree on what ∆t_{ }equals. When the train reaches a constant velocity, half the speed of light, the observer who didn’t get on the train will see train-person’s ‘timer’ do this:

Now the light has to travel much further to reach the receiver (rather than just straight up and down), and therefore the stationary observer sees the person on the having a slower time interval. Focusing on the frame on reference that the stationary observer is in, the time interval ∆t_{ }is the same as equation 1, but from the stationary observer’s point of view, the *time interval will be different*.

If you want to read the math part:

The distance, 2r, that the light travels will equal:

2r = 2√(L_{0}^{2 } + (0.5c(∆t_{0})/2)^{2} ) (equation 2)

D = 0.5c(∆t_{0}), because the train is moving at half the speed of light.

The new time measured by the stationary observer will be:

∆t’ = 2r/c (equation 3)

Then after substituting equation 2 into equation 3 and a bit of algebra (which I won’t do because there’s already too much maths here) eventually you get this:

t’ = ∆t_{0} / [√(1 - [(0.5c)^{2}/(c^{2})])]

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tagged as: relativity. science. time.

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