alex_ts
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- Homework Statement
- From the stationary position of the observed O, two nuclei of atoms A and B are launched simultaneously, which move (with respect to O) in the same direction with corresponding constant relativistic velocities ua=ac and ub=bc. c denotes the speed of light and for the constants a and b it is valid that 0<a<b<1. Observer O found that after time t, the two nuclei split simultaneously.
What is the time interval between the decays of the two nuclei with respect to nucleus A? What was the distance (relative to A) between nuclei A and B when A detected the fission of B?
- Relevant Equations
- uba=(ub-ua)/(1-uaub/c^2)
Dta=γa(tb-uba*xba/c^2) =>Dta=γa(t-uba*uba*t/c^2)=>Dta=γat(1-uba^2/c^2)
γ=1/(1-(u/c)^2)^1/2
The relative velocity of nucleus B with respect to nucleus A is given by the relativistic velocity addition formula:
uba=(ub-ua)/(1-uaub/c^2)
Dta=γa(tb-uba*xba/c^2) =>Dta=γa(t-uba*uba*t/c^2)=>Dta=γat(1-uba^2/c^2)
γa=1/(1-a^2)^(1/2)
and and we can replace uba with their respective expression.
And the distance at the moment of detection is: dab=uba*Dta and and we can replace uab and Dta with their respective expressions.
Is my solution correct?
uba=(ub-ua)/(1-uaub/c^2)
Dta=γa(tb-uba*xba/c^2) =>Dta=γa(t-uba*uba*t/c^2)=>Dta=γat(1-uba^2/c^2)
γa=1/(1-a^2)^(1/2)
and and we can replace uba with their respective expression.
And the distance at the moment of detection is: dab=uba*Dta and and we can replace uab and Dta with their respective expressions.
Is my solution correct?