Confused by Lorentz transformation equation

  • #36
FeynmanFtw said:
I just didn't expect there to be such an issue with the wording, but I guess such is the world of SR.
All areas of knowledge. I can't tell you how much confusion arises over the misunderstanding of a question. It leaves both the student and the helper unnecessarily confused.
 
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  • #37
FeynmanFtw said:
how on Earth can the perceived distance between these two lights be greater than the length of the platform?
Consider the two events "light beacons turning on". An event happens at a certain coordinate-location (x, y, z) and at a certain instance of coordinate-time (t). In SR, the squared spacetime-interval between two events is the same with reference to both inertial coordinate-systems.
For simplicity, consider the events with ##y=z=0##. Assume, that the primed frame is the frame of the train.
$$(\Delta s)^2 = (c\Delta t)^2-(\Delta x)^2=(c\Delta t')^2-(\Delta x')^2$$Both events happen simultaneously with respect to the platform frame; ##\Delta t=0##:

##0^2-(\Delta x)^2=(c\Delta t')^2-(\Delta x')^2##

##(\Delta x)^2 = (\Delta x')^2 [1-({ c\Delta t' \over\Delta x' })^2]= (\Delta x')^2 (1-v^2/c^2)##

$$\Delta x' = \gamma \Delta x \ \ \ \ \ \text{(condition:}\Delta t=0\text{)}$$.
FeynmanFtw said:
Also, shouldn't the platform itself (as has been established) be contracted, ... ?
Yes. The "light beacons" are no events. They are objects.

Imagine, the ##x'## axis is represented by a long ruler in the train.
Now consider the following different two measurement events at the same coordinate-time ##t'##:
  • One person in the train reads the ruler scale at the nearby passing "light beacon 1".
  • Another person in the train reads the ruler scale at the nearby passing "light beacon 2".
The distance between the two ruler scale positions is ##\Delta x'##.
The coordinate-time interval between the two ruler scale reading events is ##\Delta t'=0##.

##(c\Delta t)^2-(\Delta x)^2=0^2-(\Delta x')^2##

##(\Delta x')^2 = (\Delta x)^2 [1-({ c\Delta t \over\Delta x })^2]= (\Delta x)^2 (1-v^2/c^2)##

$$\Delta x' = {1 \over \gamma} \Delta x\ \ \ \ \ \text{(condition:}\Delta t'=0\text{)}$$.
 
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