How Does Relativity Affect the Perception of Time Between Supernovae?

  • Thread starter Thread starter doublemint
  • Start date Start date
  • Tags Tags
    Intro Relativity
AI Thread Summary
The discussion explores how relativity affects the perception of time between two supernovae observed by aliens. When Alpha goes supernova, followed by Beta ten years later, the aliens perceive the explosions as occurring 120 light years apart. Initial calculations using Lorentz Transformations suggested a time difference of -24 years, indicating a misunderstanding of the reference frame. The user later realizes that applying the spacetime interval provides the correct approach to understanding the time perception. This highlights the complexities of time perception in relativistic contexts.
doublemint
Messages
138
Reaction score
0
1. The star Alpha goes supernova. Then ten years later and 100 light years away, star beta goes supernova. Aliens passing by observes that the explosions occur 120 light years away. According to them, what is the time between the explosions?


2. So I think I am supposed to used the Lorentz Transformations.
x'=\gamma(x-vt) and x=\gamma(x'+vt')
Then I solved for v in both equations since we do not know the speed at which the aliens are traveling at. I equate them together to get t'
\gammat'=[\gammat(x-\gammax')]/(\gammax-x')
I solved for it and got -24 years..
So does it mean that it took 24 years inbetween explosions according to the aliens reference frame?

Thank You

DoubleMint
 
Physics news on Phys.org
Nvm I've got it! I had to use the spacetime interval..obviously XD
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top