## Special Relativity

Special relativity explains the relationship between space and time. It's based on two postulates: 1) The laws of physics are same in all inertial frames of reference and 2) The speed of light in vacuum is same for all observers. On applying these postulates, one realizes that certain physical quantities when measured from different inertial frames of reference do not tally with each other. We explore three such quantities.

Consider a Geosynchronous satellite orbiting the Earth. It's near-circular orbital path has a radius of 42300 km. We measure the passage of time, length of the same object and frequency of waves tranmitted by the satellite and received by the Earth. Use the slider to vary the orbital speed of the satellite as a % of speed of light.

Or type the % of speed of light here:

### Description

Time Dilation $t = \frac{t_0}{\sqrt{1 - \frac{v^2}{c^2}}}$

According to a stationary observer on Earth, time appears to pass slower aboard the satellite. If the time period between two events on the satellite is 't0'' aboard the satellite, then it is 't' (as calculated above) for the observer on Earth. Observe that t0 < t. This phenomenon is known as time dilation.

Length contraction $l = l_0 \sqrt{1 - \frac{v^2}{c^2}}$

If the length of an object on Earth is measured to be 'l0'' by a stationary observer on Earth, then an observer aboard the satellite measured the same object's length to be 'l' as calculated above. Observe that l0 > l. This phenomenon is known as length contraction.

Relativistic Doppler Effect $f = f_0 \left( \frac{c-v}{c+v} \right)$

If the satellite transmits waves of frequency 'f0'', then a stationary receiver on the Earth will measure the same waves to have a frequency of 'f' as calculated above. This phenomenon is known as relativistic doppler effect.

Note:
1. Speaking in a strict sense, a satellit orbiting the Earth won't exactly show these effects in the same manner. The orbiting satellite analogy has been provided to get a brief idea of speed. Instead, consider a spaceship going away from Earth at the same speed.
2. This simulation is compatible with SI units
3. Gamma (γ) is defined as: $\gamma = \frac{1}{1 - \frac{v^2}{c^2}}$ Hence, $$t = \gamma t_0$$ and $$l = l_0 / \gamma$$
4. Gravitational effects on physical quantities have been ignored. They come under the purview of General Relativity.

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