By Edoardo Sernesi

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**Example text**

Then, one can show that these light signals will not reach simultaneously the observer at O which is moving with velocity v. Here, the moving O seems to lie in OY another inertial frame. The light signal takes t1 time to reach from Y to O as (c−v) OX and the time t2 taken by the light signal from X to O is (c+v) . [c is the velocity of light in all direction. 5 The Relativistic Concept of Space and Time 19 One can note that t1 = OX OY = = t2 , as c + v = c − v , O X = OY (c − v) (c + v) Therefore, time is not absolute.

Ds is the invariant spacetime interval between two events which are infinitely close to each other. Note 2: Suppose a point moves from (x, y, z, t) to any neighbouring point, then we know the spacetime interval, ds 2 = c2 dt 2 − dx 2 − dy 2 − dz 2 or, ds dt 2 = c2 − dx dt 2 + dy dt 2 + dz dt 2 = c2 − v 2 If c2 > v 2 , then ds 2 > 0, then the distance or interval is positive. It is time-like interval. If c2 < v 2 , then ds 2 < 0, then the distance or interval is negative. It is space-like interval.

This means side face of the square is seen inclined at an angle θ with the front side where tan θ = vl c l = vc . 9c. Calculate its lifetime as measured by an observer in laboratory. What will be the distance traversed by it before decaying? Also find the distance traversed without taking relativistic effect. 44 4 Mathematical Properties of Lorentz Transformations Hint: We use the result of time dilation, t1 = t 2 1− v2 . 9 × 3 × 108 . 58 µ s. 2 The proper lifetime of π + mesons is αμ s. If a beam of these mesons of velocity f c produced ( f < 1), calculate the distance, the beam can travel before the flux of the mesons beam is reduced to e−β times the initial flux, a being a constant.