### Time Dilation

## Relative Clocks

Time runs slower within a gravitational field compared to a clock very distant to the gravitating object. The stronger the gravity the slow time will pass. The ultimate is a black hole where at the surface, the event horizon, time stops.

A clock held fixed within a gravitating ticks slower than a clock further from the gravitating object. Lets imagine the gravitating object is a planet. To calculate the time dilation effect of gravity let an object fall towards the planet. The object will increase in speed through space. Imagine an observer stationary at the surface of the planet as the object passes by. The velocity of the object as measured by the observer is the same as Newton's equations would predict. Applying the special relativistic time dilation to that velocity gives the time dilation effect due to gravity at that distance from the planet as expected by a distant observer.

For example the escape velocity of the Earth is 11km/s, or 11 000 m/s. This is the same velocity as a object that has fallen from a great distance to the surface of the Earth. The time dilation of an object moving at 11km/s is 0.999999999327 for every second passes by the distant clock. Or it loses 0.672 nanoseconds every second.

An observer of the surface of the Earth eperience 1 second. A falling object must also be experiencing the same distance through space time as the observer. But as the object falls is moves into regions of slower time. After 1 second, on the surface of the Earth, it increase its speed by 10 m/s. The clock is forced to run slower.

Using the time - velocity diagram. With dt = 1. (1-v^{2})^{1/2} = dτ The change in dτ with a change in v is dτ/dv = v (1-v^{2})^{-1/2} ~ v. So 11km/s is 3.66x10^{-5}c times dv is 10 m/s is 3.33*10^{-8}c is 1.11x10^{-12}. So time is running 1.11x10^{-12}sec slower after one seconds of free fall at 11 km/s. Or if falling from rest, v is 3.33*10^{-8}c and dv is 3.33*10^{-8}c so 3.33*10^{-8}c * 3.33*10^{-8}c is 1.1x10^{-15}.