Nugatory said:
You are misunderstanding the train experiment and where relativity of simultaneity comes in. [..]
Perhaps I should show you the train experiment.
I'm an (very) amateur, too in SR. But it's this simulation that leads me to discover Lorentz Length contraction, Lorentz Time dilation and relative simultaneity of events. (Tough Hendrik had devised it more then 120 years ago!)
Janus said:
Consider Einstein's Train example.
https://www.physicsforums.com/threads/length-contraction.817911/page-2#post-5135255
You have a train with an observer at the midpoint between the ends. you also have an observer standing along the tracks. Lightning strikes the end of the trains when, according to the track-side observer the train observer is passing him. Thus he sees the light from the strikes at the same time and determines that the strikes occurred simultaneously. Thus, according to the frame of the tracks, events look like this:
ANIMATION 1
Now consider how things look from the frame of the train. The lightning strikes the ends of the train, and each lightning strike has to happen when the respective end of the train is next to the same red dot as it was according to the track frame. The light from each flash must also arrive at the track-side observer at the same time just like in th efirst animation. The light from either flash must also hit the train observer when he is next to the same point of the tracks as it does according to the track frame. (In other words, any event that happens in any frame must happen in the other frame.)
Now here is where you have to take length contraction into account. In the track frame as shown above, the train is moving so it is length contracted. So it is the length contracted train that fits between the two red dots that mark where the lightning strikes occur. In the train frame, the train is its non-contracted proper length, and it is the tracks that are length contracted. Thus the distance between the red dots is shorter than the length of the train and the ends of the train do not align with these dots at the same time. Since the event of the lightning striking an end of the train when it is next to a red dot is common to both frames, this means the lighting strikes cannot occur at the same time in the train frame. And in order for the light from each strike to reach the track observer at the same time and to meet the requirements of all the other common events between frames, the light must expand outward at a constant speed from the points of the strikes. Thus from the frame of the train events look like this.
[PLAIN]http://home.earthlink.net/~jparvey/sitebuildercontent/sitebuilderpictures/trainsimul2.gif[B]ANIMATION 2[/B]
Supposed the velocity is 0.6c.
The length of the platform is 120c. Let's use a 8 * 3 * 5 number for rounds result.
Lets say the back of the platform is PB the front if PF
The back of the train: TB, front = TF.
See Animation 1: Platform rest frame. Platform will see
The length of the train is (of course) 120ls, remember the front and the back of the train fits the platform.
The length of the platform; 120 ls
The distance of the observer (R) from the back of the platform; halfway or 60ls.
(PB meets TB) and (PF meets TF) event happen at the same time. For the light reaches the observer (R) at the sametime.
Dividing the speed of light and distance, the light will reach the observer (R) at 60 seconds.
Supposed if (R) regardless of frame of reference receives the signal AT THE SAME TIME, R will shine a GREEN light. If not R will shine a Red light.
In this case, we see that the lights reach R AT THE SAME TIME. So R will show Green.
--- And remember. The speed of light is INVARIANT. So
PB and
PF will see that they are at the CENTER of the LIGHT CONE.
| --------------------------------------------------------
|
| Now let's take a look from the train frame, see animation 2
|
-->And remember. The speed of light is INVARIANT. So
TB and
TF will see that they are at the CENTER of the LIGHT CONE.
So, when TF reaches PF the light shines, but the
train, sorry the platform also moves at 0.6c back ward. So, the light will reaches R at 60/(1-0.6) = 150 seconds.
When TB reaches PB the light shines, so the light will reaches R at 60/(1+0.6) = 37.5 seconds.
No, this CAN'T be right. The light have to reach R AT THE SAME TIME. The proof? R lights Green lamp remember?
Let's study this picture.
Remember, all calculation below does NOT use Relativity, it uses ALGEBRA. So you shouldn't be afraid with SR.
Okay, let define
L, first
L is the distance from the observer (R) or half the platform size. Platform = 2L
Let's see the above diagram first.
L3 and L
R has already at L length from TF/PB traveling at 0.6c when the light shines at c.
So, what is L3 so that the light catches up R?
##L3 = V(L3+L); L3 = 0.6(L3+L); L3 = \frac{3}{2}L; \frac{L3}{L3+L} = V##
See below
What is L1 so that the light will meet R?
##(1+V)L1 = L; 1.6L1 = L; L1 = \frac{5}{8}L; L2 = \frac{3}{8}L; again \frac{L2}{L1} = V##
So at TF/PF, R has to travel along L4 distance before TB meets PB
##L4+L2 = L3; L4 = \frac{9}{8}L##
Let's add all length
##L_{Train} = L + L4 + L = \frac{25}{8}L; L_{Train} = \frac{25}{16}L_{Platform}##
Conclusion: The length of the train at rest must be ##\frac{25}{16}## length of the platform at rest.
No, that is wrong!
Conclusion: The length of the train at rest must be ##\frac{25}{16}## length of the platform at
traveling!
The length contraction works both ways.
Conclusion: The length of the train at rest is ##\sqrt{\frac{25}{16}} = \frac{5}{4} = 1.25## length of the platform at rest.
Where in Lorentz factor ##\gamma=\frac{1}{\sqrt{1-V^2}} = \frac{1}{\sqrt{0.64}} = 1.25##
How much time that TF takes when it meets PB until it meets PF?
##T_{Train} = L_{Platform, traveling}/V = \frac{120 / 1.25}{0.6} =160 seconds##
At platform frame, how much time that TF takes when it meets PB until it meets PF?
##T_{Platform} = L_{Platform, rest}/V = \frac{120}{0.6 } = 200 seconds##
So in this example you'll see that
1. The length is contracted for moving object
2. There's a relativity simultaneity of event.
- Where in platform frame TF/PF happens at the same time as TB/PB,
- In train frame TF/PF happens first, then TB/PB.
3. Time dilation. 160 secods for train frame = 200 seconds for platform frame.
That's all. I didn't want to mislead the OP. I hope the honorable mentors/advisors will immediately correct me if I'm wrong.