Novas and special relativity-check my work?

  • Thread starter Thread starter bcjochim07
  • Start date Start date
  • Tags Tags
    Work
bcjochim07
Messages
366
Reaction score
0
Novas and special relativity--check my work?

Homework Statement


Suppose Earth astronomers see two novas occur simultaneously, one in constellation Orion and the other in Lyra. Both nova are the same distance from Earth (2.5e3 cy) and are in exactly opposite directions from Earth. Observers are on board an aircraft flying at 1000 km/h on a line from Orion toward Lyra.

a) for the observers on the aircraft, how much time separates the nova?


Homework Equations


v=1000 km/hr = 277.78 m/s =9.26e7*c
Beta = 9.26e7
Gamma= approximately 1

The Attempt at a Solution



a) I will say that the astronomers see the novas both occur at t1(Orion)=t2(Lyra)=0 and that the origin of the Earth's frame is on the Earth.
Applying the Lorentz transform:
t1'=1*(0-(9.26e7*c)(-2.5e3cy)/c^2) = 0.002315 y
t2'=1*(0-(9.26e7*c)(2.5e3cy)/c^2) = -0.002315 y

Therefore, the interval measured by the observers on the aircraft is 0.00463 y = 40.56 hr.
The observers on the aircraft see the nova on Lyra occur first.

Is this correct?
 
Physics news on Phys.org


bcjochim07 said:
Beta = 9.26e7

You mean 9.26e(-7) right?...Other than that, everything looks correct!:approve:
 


Hi I was wondering how you went from 0.00463y to 40.56 hr
Thanks
 


Uhh, conversion? .00463 year * 365 day/year * 24 hour/day?
 


Of course :-p I was thinking to complicated about it lol Thx
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top