In the previous article, I introduced the Minkowski metric as a generalization of the theorem of Pythagoras and stressed the importance of the distance between two points being invariant under a change of coordinates. In fact the entire special theory of relativity is nothing more complicated than the statement that the distance between two points in four-dimensional spacetime, and given by the Minkowski metric, is the same for all coordinate systems. A few readers have asked me to give a few more details of this point, and so that is what I will do in today's article.

Suppose that there are two observers, A and B, who are moving relative to each other at speed v. Maybe A is on the planet Earth and B is in a rocket ship on his way to Mars. According to A, the distance that B travels through spacetime is given by

ds2 = c2 dt2 - dx2 = c2 dt2 (1 - v2)

For convenience I am only working in one space dimension, but these equations are valid in the full three spatial dimensions as well. I have also switched the minus sign compared with yesterday's notes, because this arrangement is more common in modern physics as it gives a positive spacetime dimension for most systems. However I have left the speed-of-light in the equations to avoid confusion later.

According to B, his rocket ship is not moving at all in his coordinate system and so the distance he sees himself as travelling in his own frame of reference is

ds2 = c2 dt'2

However as I stressed yesterday, the two distances must be equal. The distance does not depend on the coordinate system used. Therefore it follows that

c2dt'2 = c2 dt2 - dx2 = c2 dt2 (1 - v2)

and that the time observed by A is related to the time observed by B by the formula

dt2 = dt'2 / (1 - v2/c2)

This is the most famous result of special relativity, known as time dilation.

A very similar calculation can also be done using the requirement that the four-momentum of an object must be the same in all coordinate systems. (Most people remember momentum from high school physics as the mass of the object multiplied by its velocity. This is essentially three-momentum, and the three-momentum together with the energy of the object forms a four-dimensional vector called the four-momentum). Using this requirement, it follows that

m2 c4 = E2 - p2c2

which for a stationary object reduces to the famous equation, E  = m c2.

That completes the review of special relativity, and now on to the general theory!