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Time dilation

Two identical and accurate clocks

Consider two identical accurate clocks, one stationary and the other moving at close to light speed. Due to the effects of special relativity the moving clock measures a shorter time than the stationary clock.

The time of the moving clock is correct as far as an observer moving in the same frame of reference as the clock and is called the ‘proper time’. However the time recorded by a stationary observer in the same time interval is longer or ‘dilated'.

The dilated time t’ can be calculated if the proper time t and the speed v are known using the relationship:

\(t\textquotesingle = \frac{t}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\)

The Russian Cosmonaut Sergei Avdeyev spent 747 days (64 540 800 seconds) aboard the Space Station Mir which travels at 7600 ms-1. Calculate how much time Sergei 'extended' his life by.

First put in the known numerical values:

\(t\textquotesingle = \frac{64,540,800}{\sqrt {1- \frac{7600^{2}}{(3\times 10^{8})^{2}}}}\)

When calculating this equation it is best to use the brackets facility on a calculator or to work outwards.

Start off with the \(\frac{v^{2}}{c^{2}}\) term and work through stage by stage.

\(t\textquotesingle = \frac{64,540,800}{\sqrt{1-6.41\times 10^{-10}}}\)

Which then gives:

\(t\textquotesingle = \frac{64,540,800}{0.9999999996795}\)

Remember to take the square root.

\(t\textquotesingle = 64,540,800.0207s\)

In other words after over two years being the fastest human being, Sergei only experienced 21 milliseconds less than colleagues on Earth.

Note in these examples since there is such a small difference we need to use a much larger number of significant figures than is usual.

Question

In considering relativity we think of ourselves at rest. Is this strictly true and if not, why not?