Special Relativity - clocks on two trains

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  • Thread starter Mike_bb
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Mike_bb said:
I've never heard how to differ which inertial frame move relative to another.
You are struggling with the most basic things, that have nothing to do with Special Relativity. You should learn Newtonian physics first, as I have told you this in previous threads:
A.T. said:
Learn about kinematics in different reference frames
 
Mike_bb said:
See comment #44.
The calculations @Dale did in that comment were based on a different value of ##V## from the one you said you wanted to use. Can we assume that you can redo his calculations with the value of ##V## you wanted to use? The formulas are all there.

Also, as @Dale noted in that comment, nothing corresponding to your ##0.5## second ##\Delta T##, what you were calling the "effect of simultaneity", appears anywhere in his analysis. Do you understand why not?
 
PeterDonis said:
The calculations @Dale did in that comment were based on a different value of ##V## from the one you said you wanted to use. Can we assume that you can redo his calculations with the value of ##V## you wanted to use? The formulas are all there.

Also, as @Dale noted in that comment, nothing corresponding to your ##0.5## second ##\Delta T##, what you were calling the "effect of simultaneity", appears anywhere in his analysis. Do you understand why not?
Ok. If I need to, I will do it.
 
I wouldn’t recommend trying to redo the same analysis with different numbers. I would do the following:

Start with the left train’s frame and using units with ##c=1##. In that frame the left train has a worldline ##(t,x)=(t,0)## and the right train has a worldline ##(t,x)=(t,vt)##. Pick an event on the left worldline and find the event on the right worldline that happens at the same time in the left frame.

Then transform those events into the right frame and calculate the difference in time. All symbolically, don’t use any numbers.
 
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Handwritten derivation in special relativity with c = 1: events E1 and E2 given as (t,x) coordinates, transformed to primed frame coordinates with gamma factor, followed by algebra solving for gamma = 1/sqrt(1 - v^2).
 
Point Conception said:
Relativity of simualtanity can be shown with lightning strikes at both ends of a passing train car by a platform. In platform frame the lightning strikes are simultaneous . The observer in center of train car sees the time between strikes:View attachment 374901
Then with the spacial separation in trains frame you can calculate and show the invarience of the spacetime interval: So @Mike_bb is this about what you are showing and is your original question answered? Or Alternatively with left side train as rest frame and right side train at 2(v) SR
Equation for the Lorentz transformation of a spatial interval: Δx′ = γ(Δx − vΔt), with gamma on the right side multiplying the bracketed difference.
Equation: Δs² = c²(Δt)² − (Δx)², the spacetime interval formula of special relativity written with squared differences of time and space.
Equation of the spacetime interval in primed coordinates: Δs'^2 = (cΔt')^2 − Δx'^2, written as plain typeset mathematics on a white background.
Equation: Δt′ = γ v Δx / c², where γ multiplies the fraction v Δx over c squared.
 
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Point Conception said:
So @Mike_bb is this about what you are showing and is your original question answered?
Yes.
 
Mike_bb said:
Event1: reading the left clock Δt when distance between left and right clock is Δx
Event2: reading the right clock Δt′ when distance between left and right clock is Δx (Δx′=0)

The notation is confusing. ##\Delta t## is not a clock-reading. ##t## is a clock-reading and ##\Delta t## is the time that elapses between two events.