Synchronized clocks in a moving frame

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Homework Help Overview

The discussion revolves around the concept of synchronized clocks in a moving frame, specifically addressing the implications of time dilation and the Lorentz transformation in the context of special relativity. Participants are examining the relationships between time intervals measured in different reference frames, particularly focusing on a scenario involving a moving train.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to understand the application of the time dilation equation Δt' = Δt/γ in a moving frame, questioning the implications of measuring time intervals for events occurring at different positions. Some participants raise concerns about the applicability of certain equations in this context and suggest that additional terms may be necessary for accurate calculations.

Discussion Status

Participants are actively engaging with the original poster's confusion, providing clarifications about the meaning of Δt' and discussing the limitations of specific equations in the scenario presented. There is an exploration of different interpretations regarding the synchronization of clocks and the effects of motion on time measurement.

Contextual Notes

There appears to be some confusion regarding the definitions and implications of time intervals in different frames, as well as the assumptions made about the motion of the clocks. The discussion reflects a need for clearer understanding of the Lorentz transformations and their application to the problem at hand.

davon806
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Homework Statement


It is not a homework question.I am trying to understand the concept in the attached diagram but I stuck.Hope someone can help

Homework Equations

The Attempt at a Solution


Please see the lower part of the note,stating that Δt' = Δt/γ.However,imagine you are in the moving frame S'(the stationary train).Yes just a stationary train.If you shine a beam towards the head and back of the train,they should arrive at the same time(in your frame).Hence Δt' = 0,but this also implies Δt = 0.What's wrong?

Thanks!
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That equation does not apply to time differences that do not happen at one position (in your reference frame). If you calculate time differences for the two ends of the train, you get an additional term in the transformation.
 
Thx for your reply...but i don't quite understand :d, could you explain a bit more?
 
You try to use a formula for a system where it does not apply.
It's like trying to use g=9.81 m/s^2 as gravitational acceleration on the Moon: it's simply not the right thing.
There is a more general formula that works both on Earth and Moon: F=GMm/r^2. In the same way, there are the more general Lorentz transformations that can be used here.
 
Sorry, I have just seen it. So, in my case, what does Δt' mean exactly?
 
Difference in time between two events (as defined in the text) in the frame that got a prime.
 
I find this derivation somewhat confusing.
It says that that clock A starts before clock B and then takes tB - tA as a positive quantity.
Also, it states that in the observer's frame the clocks in the moving frame are moving
with speed v. Then the diagram shows the moving frame moving to the left with
speed v.
The hand on both clocks are always in the same (relative) position which doesn't
help with the explanation.
 

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