Originally posted by ...
yes, i know what u mean
but what i mean is if ur travelling at the speed of light, what will happen to you just b4 u reach there..
cos apparently if u travel at high high speed, things will demented...
and apparently u can see clocks stationary and stuff...
As you speed by Earth at some arbritary fraction of light speed, you will observe an Earthbound clock to run slow and contract in length in the direction of motion. The mass of objects appears the same regardless of your velocity.
Originally posted by Xayma
Then why cant we reacht he speed of light, if mass doesnt not increase to the point of infinity, then any continuous application of force will eventually bring the object to above the speed of light.
Recall that 4-acceleration is the proper time derivative of 4-velocity A<sup>u</sup> = dU<sup>u</sup>/dt<sub>0</sub> where 4-velocity is the derivative of 4-position with respect to proper time, and 4-position is a 4-dimensional vector in space-time of the form x<sup>u</sup> = (ct, x, y, z) where c is the speed of light, t is the time coordinate and x, y, z are the three spacial coordinates. Therefore U<sup>u</sup> = d/dt<sub>0</sub> (ct, x, y, z) = (c dt/dt<sub>0</sub>, dx/dt<sub>0</sub>, dy/dt<sub>0</sub>, dz/dt<sub>0</sub>). It's clear as day then that U<sup>u</sup> = gamma (c, v<sup>i</sup>) where v<sup>i</sup> = (dx/dt, dy/dt, dz/dt) is the ordinary 3-velocity. In the rest frame of the particle it is obvious that U<sup>0</sup> = (c, 0, 0, 0). Working through the algebra it can be shown that the magnitude of A<sup>0</sup> is equal to the proper acceleration alpha felt by the observer on the ship. Furthermore, the proper acceleration can be related to the coordinate acceleration a by
alpha = gamma<sup>3</sup> a
(too tedious to prove here). Clearly for constant alpha, a must diminish as gamma diverges. As you can see this nothing to do with relativistic mass.