- #1
myoho.renge.kyo
- 37
- 0
A. Einstein states the following on page 49 in The Principle of Relativity:
"Between the quantities x, t, and T, which refer to the position of the clock, we have evidently, x = vt and
T = (t - v*x/c^2)/sqrt(1 - v^2/c^2).
Therefore,
T = t*sqrt(1 - v^2/c^2) = t - (1 - sqrt(1 - v^2/c^2))*t
whence it follows that the time marked by the clock (viewed in the stationary system) is slow by 1 - sqrt(1 - v^2/c^2) seconds per second, or - neglecting magnitudes of fourth and higher order - by
(1/2)*(v^2/c^2).
From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by (1/2)*(t*v^2/c^2) (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B."
it is important to note here that "the time marked by the clock (viewed in the stationary system) is slow by 1 - sqrt(1 - v^2/c^2) seconds per second, or - neglecting magnitudes of fourth and higher order - by
(1/2)*(v^2/c^2)." for example, if the observer moving with the clock at A measures the value of t using the clock at A (the observer at A and the clock at A are at rest relatively to each other), and the stationary observer at B measures the value of t using the stationary clock at B, the value of t obtained by the obsever at A equals the value of t obtained by the observer at B.
but if the stationary observer at B measures the value of t using the stationary clock at B, and then the same stationary observer at B measures the value of t using the moving clock at A, then the value of t obtained by the stationary observer at B using the stationary clock at B is not equal to the value of t obtained by the same oberver using the moving clock at A. the moving clock lags behind the stationary clock "by (1/2)*(t*v^2/c^2) (up to magnitudes of fourth and higher order)." but i don't know what magnitudes of fourth and higher orde is.
"Between the quantities x, t, and T, which refer to the position of the clock, we have evidently, x = vt and
T = (t - v*x/c^2)/sqrt(1 - v^2/c^2).
Therefore,
T = t*sqrt(1 - v^2/c^2) = t - (1 - sqrt(1 - v^2/c^2))*t
whence it follows that the time marked by the clock (viewed in the stationary system) is slow by 1 - sqrt(1 - v^2/c^2) seconds per second, or - neglecting magnitudes of fourth and higher order - by
(1/2)*(v^2/c^2).
From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by (1/2)*(t*v^2/c^2) (up to magnitudes of fourth and higher order), t being the time occupied in the journey from A to B."
it is important to note here that "the time marked by the clock (viewed in the stationary system) is slow by 1 - sqrt(1 - v^2/c^2) seconds per second, or - neglecting magnitudes of fourth and higher order - by
(1/2)*(v^2/c^2)." for example, if the observer moving with the clock at A measures the value of t using the clock at A (the observer at A and the clock at A are at rest relatively to each other), and the stationary observer at B measures the value of t using the stationary clock at B, the value of t obtained by the obsever at A equals the value of t obtained by the observer at B.
but if the stationary observer at B measures the value of t using the stationary clock at B, and then the same stationary observer at B measures the value of t using the moving clock at A, then the value of t obtained by the stationary observer at B using the stationary clock at B is not equal to the value of t obtained by the same oberver using the moving clock at A. the moving clock lags behind the stationary clock "by (1/2)*(t*v^2/c^2) (up to magnitudes of fourth and higher order)." but i don't know what magnitudes of fourth and higher orde is.
Last edited: