Basic relativity problem - Lorentz Transformations

AI Thread Summary
The discussion revolves around a homework problem involving Lorentz transformations to determine the speed required for an observer to perceive two events occurring at the same spatial location. The events are defined with specific coordinates and a time interval, leading to confusion about the correct application of the Lorentz equations. Participants clarify that the speed (V) can be calculated as the distance between the events divided by the time interval, resulting in a value of 2E8 m/s. The conversation highlights that the gamma factor (g) is not necessary for this particular problem, as the observer's speed can be directly derived from the events' parameters. Overall, the thread emphasizes understanding when to apply relativistic concepts and when simpler calculations suffice.
Akhilleus
Messages
9
Reaction score
0

Homework Statement



Event A occurs at xA = 500m. Event B occurs 5 microseconds later at xB = 1500m. With what speed must an observer move in the positive x direction so that the events occur at the same point in space in the observer's frame?

Homework Equations



Lorentz transformation equations:
x' = g(x - Vt)
t' = g(t - (Vx)/c2)
g = 1/sqrt(1 - V2/c2)

The Attempt at a Solution



I understand conceptually that, as the observer approaches c, the distance between the two events will contract and the time will "slow down", but I'm unsure how to find these values. Is the V value in the above equations simply the distance divided by time when the events are observed at rest? If so, solving for x' leads to a value of 0. Any help would be greatly appreciated.
 
Physics news on Phys.org
Hi Akhilleus! :smile:

(have a square-root: √ :wink:)
Akhilleus said:
Is the V value in the above equations simply the distance divided by time when the events are observed at rest? If so, solving for x' leads to a value of 0.

v is the speed of one observer relative to the other (the other way round, it's minus v of course).

Explain how you got a value of 0. :confused:

(btw, this problem really has nothing to do with relativity! :biggrin:)
 
Using the distance between the two events divided by the time (5 microseconds) as V:

V = 1000m/(5 E-6)s = 2E8 m/s
g = 1/√(1 - (2E8)2/c2) = 1.8
x' = 1.8(1000m - (2E8)(5E-6)) = 0


So V is the speed of the moving observer as seen by a hypothetical second person at rest. Here, V is just distance between the two events divided by time. Would this be the same V required by the moving observer?
 
Akhilleus said:
So V is the speed of the moving observer as seen by a hypothetical second person at rest. Here, V is just distance between the two events divided by time. Would this be the same V required by the moving observer?

Yes, that's fine. :smile:

What's worrying you about that? :confused:

The question asks you to find a v such that x' = 0 …

that's what you've done! :smile:

(but why are you bothering with gamma? don't you see that you would have got the same result even if you knew no relativity?)
 
Ah, I see. This was a LOT easier than I was making it... haha.

I was caught up on the idea that the distance traveled would contract if he was traveling at 0.6c, and thought that if he didn't slow down, event B would happen behind him. I'm very intrigued by the effects of relativity but I'm still not sure when they apply and when they don't.
 
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