Two unstable particles move in the reference frame K along a straight line in the same direction with a velocity `v=0.990c`. The distance between them in this reference frame is equal to `l=120m`. At a certain moment both particles decay simultaneously in the reference frame fixed to them. What time interval between the moments of decay of the two particles will be observed in the frame K? Which particle decays later in the frame K?

Explanation

In `K_0`, the rest frame of the particles, the events corresponding to the decay of the particles are, lt brgt `A:(0,0,0,0)` and `(0,l_0,0,0)=B`
In the reference frame K, the corresponding coordinates are by Lorentz transformation
`A:(0,0,0,0), B: ((vl_0)/(c^2sqrt(1-beta^2)), (l_0)/(sqrt(1-beta^2)), 0,0)`
Now `l_0sqrt(1-beta^2)=l`
by Lorentz Fitzgerald contraction formula. Thus the time lag of the decay time of B is
`Deltat=(vl_0)/(c^2sqrt(1-beta^2))=(vl)/(c^2(1-beta^2))=(vl)/(c^2-v)`
B decays later (B is the forward particle in the direction of motion)