Synchronized clocks in a moving frame

[davon806]

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!

 

[mfb]

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.

 

[davon806]

Thx for your reply...but i don't quite understand :d, could you explain a bit more?

 

[mfb]

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.

 

[davon806]

Sorry, I have just seen it. So, in my case, what does Δt' mean exactly?

 

[mfb]

Difference in time between two events (as defined in the text) in the frame that got a prime.

 

[J Hann]

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.