- #106
plover
Homework Helper
- 191
- 2
zoobyshoe said:Somehow, I missed any statement of the results. What actual times do you get for the flashes from the train frame? Please show how you moved from - or + 346,410.15 to a value in seconds from t=0 for each flash. As I mourned to Doc Al, this number, 346,410.15 carries no meaning for me, + or -.
As I said (see previous post for derivation):
So at rest the length of the train is: 2 * 1.15 * L. Now why should this be so?
The train as measured in the embankment frame has length 2L. However, the train is in motion in the embankment frame, and thus appears contracted in that frame. So, since the value we started out with measures the contracted length of the (moving) train in the embankment frame, it makes sense that in the train frame, where the train is at rest, the length - no longer being contracted - is measured to be greater.
The train as measured in the embankment frame has length 2L. However, the train is in motion in the embankment frame, and thus appears contracted in that frame. So, since the value we started out with measures the contracted length of the (moving) train in the embankment frame, it makes sense that in the train frame, where the train is at rest, the length - no longer being contracted - is measured to be greater.
So if we're setting L = 300,000km then the rest length of half the train is: x' = (1.15)(300,000km) = 345,000km.
That's one significance of the number.
So far no one's calculated a value for the time of the flashes in the train frame. It would be (for the flash at B):
[tex]
\beta\ =\ .5c/c\ =\ .5
[/tex]
[tex]
\gamma\ =\ \frac{1}{\sqrt{1 - (0.5)^2}}\ =\ \sqrt{4/3}\ =\ 1.15
[/tex]
[tex]
t'\ =\ \frac{-\gamma\beta L}{c}
\ =\ \frac{-1.15 \cdot 0.5 \cdot 300,000\ \textrm{km}}{300,000\ \textrm{km}/\textrm{s}}
\ =\ -0.575\ \textrm{s}
[/tex]
So the event "flash at B" has coordinates\beta\ =\ .5c/c\ =\ .5
[/tex]
[tex]
\gamma\ =\ \frac{1}{\sqrt{1 - (0.5)^2}}\ =\ \sqrt{4/3}\ =\ 1.15
[/tex]
[tex]
t'\ =\ \frac{-\gamma\beta L}{c}
\ =\ \frac{-1.15 \cdot 0.5 \cdot 300,000\ \textrm{km}}{300,000\ \textrm{km}/\textrm{s}}
\ =\ -0.575\ \textrm{s}
[/tex]
(x, t) = (L km, 0 s) = (300000 km, 0 s)
in the embankment frame, and coordinates(x', t') = (345000 km, -0.575 s)
in the train frame.