What is the velocity of observer O' in the Lorentz Transformation problem?

AI Thread Summary
The discussion centers on a problem involving the Lorentz Transformation and the concept of simultaneity between two observers, O and O'. Observer O records two flashes occurring at different times and positions, while observer O' perceives them as simultaneous. The key challenge is finding the relative velocity u of observer O' using the Lorentz equations, particularly noting that for simultaneity in O', the time difference Δt' must equal zero. Participants express confusion about the implications of simultaneity and the constraints on u, with some clarifying that the velocity must remain less than the speed of light. The conversation highlights the learning curve associated with advanced concepts in special relativity.
Arman777
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Homework Statement


According to observer ##O##, a blue flash occurs at ##x_b =10.4m## when ##t_b =0.124 μs##, and a red flash occurs at ##x_r =23.6m## when ##t_r =0.138 μs##. According to observer ##O'##, who is in motion relative to ##O## at velocity ##u##, the two flashes appear to be simultaneous. Find the velocity ##u##.

Homework Equations

The Attempt at a Solution


I didnt understand the concept of simultaneity I think.
The difference in position will be , ##Δx=13.2m## and the difference in time ##Δt=0.014μs##

Now I tried to use the lorentz transformation.
##Δx'=γ(Δx-uΔt)##
##Δt'=γ(Δt-uΔx/c^2)##

To find ##u## I guess ll need some relationship between ##Δx'## and ##Δt'##. I am kind of stuck.

I just solved a problem where if I had given the ##u## and ##Δx##, ##Δt##. I can find the ##Δx'## and ##Δt'## by just using the lorentz transformation but here something simple bugs me, that I can't see right now.
 
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Hint: That two events are simultaneous in a given frame means that the time difference between them is zero in that frame.
 
Orodruin said:
Hint: That two events are simultaneous in a given frame means that the time difference between them is zero in that frame.
So ##Δt'=0## ? I tried that but it gives me ##u=Δtc^2/Δx## which is larger then ##c## ?

Wait its not I guess...Okay thanks lol my stupid mistakes...
 
Please show your actual numerical results.
 
My mistake sorry for that. Its not larger then ##c##. I thought that its at first, so I started to think some other stuff to find the solution..
 
Arman777 said:
So ##Δt'=0## ? I tried that but it gives me ##u=Δtc^2/Δx## which is larger then ##c## ?

Wait its not I guess...Okay thanks lol my stupid mistakes...

You can check from this equation the constraint on the events in one frame to make it possible for them to be simultaneous in another frame:

##|u| < c \ \Leftrightarrow \ |\Delta x| > c |\Delta t|##

Which is the definition of spacelike-separated events.
 
PeroK said:
You can check from this equation the constraint on the events in one frame to make it possible for them to be simultaneous in another frame:

##|u| < c \ \Leftrightarrow \ |\Delta x| > c |\Delta t|##

Which is the definition of spacelike-separated events.
We didnt learn those stuff actually. And Idk that's because its advenced or something else. We are learning just basics. Once I asked a question about the doppler affect and Orodruin replied but I couldn't understand a thing.

We didnt learn anything about these stuff. But there's a open course about special relativity but again its kind of advanced so I didnt take it.
 
Arman777 said:
We didnt learn those stuff actually. And Idk that's because its advenced or something else. We are learning just basics. Once I asked a question about the doppler affect and Orodruin replied but I couldn't understand a thing.

We didnt learn anything about these stuff. But there's a open course about special relativity but again its kind of advanced so I didnt take it.

Well, you've learned it now!
 
PeroK said:
Well, you've learned it now!
Well i guess :)
 
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