I'm currently reading the book ( been doing so for a few years on and off ) Im at a point where the Einstien is begining to talk about the general results the special thoery has "evinced".

It says the kinetic energy of a point mass 1/2* m*v^2 is replaced by the expression

m*c^2/(sqrt(1-(v/c)^2)) .....eq(1)

Then it goes on to say if a series is developed for the kinetic energy come you come to

mc^2 + 1/2*m*v^2 + 3/8*m*v^4/c^2+...

My question is how is this series developed, Is it developed from eq(1)? It seems to me that the first term is a result of eq (1), v=0.

So are the sucsessive terms specific results of eq(1) as well, if so ,or if not, how are they obtained?

Originally posted by KazetNagorra It's a Taylor series expansion around v = 0.

See also: http://www.wolframalpha.com/input/?i=m*c^2%2F%28sqrt%281-%28v%2Fc%29^2%29%29

ok, so the other terms have little signifigance if v is relatively small when compared to "c".

Given the first 2 terms

mc^2 + 1/2*m*v^2

I suppose I incorrectly thought there was physical significance to the remaining terms, as the series would lead me to believe, given the KE from classical mechanics, and the first term synonomous with Einstein and relativity.

I find it funny that we found the second term first, certainly the first term was well deserved to be found second leading to a new piece of the infinite jigsaw puzzle.

Originally posted by joe shmo ok, so the other terms have little signifigance if v is relatively small when compared to "c".

Given the first 2 terms

mc^2 + 1/2*m*v^2

I suppose I incorrectly thought there was physical significance to the remaining terms, as the series would lead me to believe, given the KE from classical mechanics, and the first term synonomous with Einstein and ...[text shortened]... erm was well deserved to be found second leading to a new piece of the infinite jigsaw puzzle.

The other terms do have physical significance, but their contribution is negligible if v << c.

The first term wasn't known in classical physics because you don't notice it when mass is constant.