Two black holes of mass `m=10M_s` are traveling at 80% the speed of light and collide head on and merge into a new black hole. If `M_s` is the mass of our sun, what is the mass of the black hole?

Expert Answers
Charlie Hooper eNotes educator| Certified Educator

We must apply conservation of relativistic 4-momentum, `P^(mu)=[E_(t o t)/c,p]` . Where` p=gamma*mv` is the relativistic spatial momentum.

`P_i^(mu)=P_f^(mu)`

`P_(i,1)^(mu)+P_(i,2)^(mu)=P_f^(mu)`

`[E_(m,t o t)/c,gamma*mv]+[E_(m,t o t)/c,gamma*m(-v)]=[E_( t ot)/c,0]`

Notice the first two black holes have the same momentum but are traveling in the opposite direction. The final black hole has no spatial momentum. Setting the spatial components equal to each other is redundant, so set the zeroth energy components equal to each other.

`E_(m,t o t)/c+E_(m,t o t)/c=E_(t o t)/c`

`2E_(m,t o t)=E_(t o t)`

We can use the equation:

`E_(m,t o t)=gamma*mc^2`

Where `gamma` is the lorentz factor, for the initial black holes it is:

`gamma=1/sqrt(1-v^2/c^2)=1/sqrt(1-(0.8c)^2/c^2)`

`gamma=5/3`

`gamma=1 ` when the object is at rest.

`2E_(m,t o t)=E_(t o t)`

`2gamma*mc^2=Mc^2`

Where M is the mass of the final black hole.

`2gamma*m=M`

Then the final mass of the black hole M is equal to:

`M=2gamma*m=2(5/3)(10M_s)~~33.3M_s`

This is not merely the sum of the initial two masses. A huge amount of kinetic energy was converted into the final mass and therefore relativity was essential in in the treatment of this problem.

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