Solve Relativity Question: Find Distance Between Events A & B

  • Thread starter Thread starter fzksfun
  • Start date Start date
  • Tags Tags
    Relativity
AI Thread Summary
In the discussion, a physics problem involves finding the distance between two events in different reference frames using Lorentz transformations. The events occur at the same spatial point in frame S, with event A happening 1.90 seconds before event B, while in frame S', event A occurs 2.45 seconds before event B. The user initially struggles with the calculations but is advised to set the spatial coordinate x to zero since the events are at the same point in frame S. The discussion also mentions the invariant quantity Δs², which relates time and space in special relativity, but emphasizes that focusing on the first approach is sufficient. The user expresses gratitude for the clarification regarding the spatial coordinate.
fzksfun
Messages
3
Reaction score
0

Homework Statement


In reference frame S two events occur at the same point; event A occurs 1.90 seconds before
event B. In another frame, S′, event A occurs 2.45 seconds before event B. How far apart
are events A and B in frame S′?


Homework Equations


x'=gamma (x-vt)
t' = [t - vx/c^2] gamma


The Attempt at a Solution


I used the aforementioned mentioned Lorentz transformations. I know both T and t'. I'm trying to find x'. So I solved for x in both equations and tried to calculate x' but could not do so. Am I approaching the question correctly?
 
Physics news on Phys.org
You can do that, but you don't want to solve for x. In the frame S, to which the coordinate x belongs, the events are at the same point, so you can take x to be zero. You know then x, t, t' and you want to find x'. You have two equations: one that will give you v, and the other can be used to then find x'.

I don't know if you have learned this already, but it may also be useful that in (special relativistically) equivalent frames such as S and S', the quantity
\Delta s^2 = - c^2 \Delta t^2 + d^2
where d^2 = \Delta x^2 + \Delta y^2 + \Delta z^2 is the "ordinary" spatial distance given by Pythagoras' law, is a constant. So you could, for example, calculate \Delta s^2 in S first and then find \Delta x in S' from that.

If you have no idea what I just said in the second paragraph, please forget it and stick to the first one, that works as well :smile:
 
Thank you so much! I can't believe I didn't realize that x = 0!
 
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