Simultaneous events in SR ?

[P3X-018]

In http://members.tripod.com/conduit9SR/SR3.html" example of relativity with the space trains, if we are in train B (were will observe train A is moving), then why can't the events be observed simultaniuosly?
If we ARE in the middle of B, then the light SHOULD take the same time to reach from the front and from the back to the middle if the events occurred simultaneously, otherwise we wouldn't be in the middle.
If the light didn't take the same time from from the front and the back to reach the middle, then wouldn't we KNOW that WE are moving.

Because how can the events be observed simultaneously by A if A is moving relative to B?

 

[P3X-018]

That illustration is a strawman argument and is wrong. It's obvious B sees the light fronts meet at the midpoint A as well simultaneously. B just
doesn't see the light fronts meet at its midpoint simultaneously.

The correct analysis should be that the event is simultaneous in time
and place for both A and B.

And the fact that the light fronts don't meet at B's center means
the speed of light is not c relative to B. So, either the light fronts
are traveling outside of train B through absolute rest space and train
A is stationary in that space or the light fronts only bounced off
train A and its velocity became conditioned to be c relative to
its frame.

The event should be considered from A, and A only! Because A will say it didn't occur simultaneously for B. B CANNOT say it didn't occur simultaneously for I am self.. Correct?

 

[RandallB]

The event should be considered from A, and A only! Because A will say it didn't occur simultaneously for B. B CANNOT say it didn't occur simultaneously for I am self.. Correct?

NO

The example you site where both trains are “one light minute long” is incorrect so expect to be confused by it. The one and only reference frame that would see the two events described there as simultaneous; would be the frame of an observer traveling in the same direction as B at one half the speed of B as measured by A. This observer would also be seen by B as moving at that same speed but in the opposite direction.
Note the problem defines the relative that the speed between A & B as 0.5c half that speed for the observer as seen from both A&B would not be 0.25c but about 0.27c.

If you want to double check it on your I’d recommend using 0.8C for the speed between A&B; the speed of the extra observer that will see these events as simultaneous would be 0.5c as measured from A or B. The numbers are a little easier to work with that way.

 

[neophysique]

B train is moving relative to the light sources while A train is stationary,
that's what that illustration shows. Since B can tell it is moving relative
to the light sources by observing the A frame's results, it can deduce
the actual time the lights struck its front and rear by editing out its velocity
with respect to the light sources to conclude that the events were simultaneous as in A. Or better yet, just watch the A frame.

 

[RandallB]

B train is moving relative to the light sources while A train is stationary,
that's what that illustration shows. Since B can tell it is moving relative
to the light sources by observing the A frame's results, it can deduce
the actual time the lights struck its front and rear by editing out its velocity
with respect to the light sources to conclude that the events were simultaneous as in A. Or better yet, just watch the A frame.

Just watching from the A frame will not help. It is important to understand that with both trains “one light minute long” the events are not and cannot be simultaneous in A.

 

[P3X-018]

B train is moving relative to the light sources while A train is stationary,
that's what that illustration shows. Since B can tell it is moving relative
to the light sources by observing the A frame's results, it can deduce
the actual time the lights struck its front and rear by editing out its velocity
with respect to the light sources to conclude that the events were simultaneous as in A. Or better yet, just watch the A frame.

But then the speed of light seen from B (where B is considered in rest) can't be c in all directions, since it takes the light a little longer to reach the middle from the back then from the front. But I think http://en.wikipedia.org/wiki/Relativity_of_simultaneity" discribes this problem a little better.
Or is it the same?

 

[P3X-018]

Just watching from the A frame will not help. It is important to understand that with both trains “one light minute long” the events are not and cannot be simultaneous in A.

So seen from B it will be simultaneous to B, and not simultaneous to A (seen from B).

And seen from A it will be simultaneous to A, and not simultaneous to B (seen from A).

Have I understood this correct now?

 

[RandallB]

So seen from B it will be simultaneous to B, and not simultaneous to A (seen from B).

And seen from A it will be simultaneous to A, and not simultaneous to B (seen from A).

Have I understood this correct now?

NO
The events can be simultaneous to one and only one observer, the one I discribed in Post #3. It cannot be A or B.

Added Edit:
Note that the Web site you’re looking at tries to fix the problem with the page you pointed out in his following page on “how to measure a fish. He says the “train Car” in B is longer that the one in A, meaning it is longer than “one light minute long” as measured in its own frame! It makes his explanation a little hard to follow or understand since it doesn’t use space train cars of the same size.
As I suggested earlier build your own example using 0.8 & 0.5 c and fixed length trains.

 

[lightarrow]

The two events are simultaneously in A reference frame, and are NOT simultaneously in B ref. frame. That's all.

The event should be considered from A, and A only! Because A will say it didn't occur simultaneously for B. B CANNOT say it didn't occur simultaneously for I am self.. Correct?

Correct. In the same way, B sees that the two events occurred simultaneously in A and not simultaneously in his ref. frame. They agree completely about it!

 

[RandallB]

The two events are simultaneously in A reference frame, and are NOT simultaneously in B ref. frame. That's all.
Correct. In the same way, B sees that the two events occurred simultaneously in A and not simultaneously in his ref. frame. They agree completely about it!

NO
Not if you retain the length of both trains as “one light minute long” as defined in the linked page from the OP. The events described cannot be simultaneous in ether A or B frames, only to the observer frame I described in post #3

 

[JesseM]

But then the speed of light seen from B (where B is considered in rest) can't be c in all directions, since it takes the light a little longer to reach the middle from the back then from the front. But I think http://en.wikipedia.org/wiki/Relativity_of_simultaneity" discribes this problem a little better.
Or is it the same?

In wikipedia's example, the light beams are emitted from the center of the train, and example is trying to show that different frames disagree on whether the light hit the front of the train and the back of the train simultaneously. The important point to understand is that in relativity, different observers have their own set of clocks which they use to measure the times of different events, each observer using clocks which are at rest relative to themselves, and that each observer "synchronizes" different clocks within his own set using the assumption that light travels at the same speed in all directions. But this method of synchronization, sometimes called the "Einstein clock synchronization convention", actually ensures that each observer will consider the clocks of other observers to be out-of-sync. In the train example, an observer on board the train could actually use a flash set off at the center to synchronize two clocks at equal distances from the center--if he assumes both light beams from the center travel at the same speed, then the clocks should be "synchronized" if they read the same time at the moment the light reaches them, in his frame. But now consider the perspective of another observer in a different frame who sees the train moving forward. If she sees the flash set off at the midpoint of the two moving clocks on the train, and she assumes the light beams from the flash both travel at the same speed in her own frame, then since the clock at the back is moving towards the source of the flash and the clock and the back is moving away from it, she should conclude that the light will catch up to the clock at the back before it catches up to the clock at the front. Yet the observer inside the train had set the clocks to read the same time at the moment the light reached them, so she will conclude that his clocks are out-of-sync from the perspective of her own frame.

 

[lightarrow]

NO
Not if you retain the length of both trains as “one light minute long” as defined in the linked page from the OP. The events described cannot be simultaneous in ether A or B frames, only to the observer frame I described in post #3

Right. I missed the part where it's stated (implicitly) the two lenghts are the same at rest.

 

[pess5]

In http://members.tripod.com/conduit9SR/SR3.html" example of relativity with the space trains, if we are in train B (were will observe train A is moving), then why can't the events be observed simultaniuosly?

It really does seem strange, doesn't it? I don't think anyone has studied the relativity without challenging the loss of absolute simultaneity. The fact is, the orientation of the trains in spacetime is angularly rotated wrt one another due to the relative motion of the 2 trains. This "frame rotation" between frame perspectives produces the relativistic effects of contraction and dilation. The only way that speed c can remain invariant, is if there is disagreement of when and where events of afar occur. One thing that falls out of this all is that ... what is simultaneous in one frame is not simultaneous in another differing frame.

You should take a look at the Minkowski worldline diagram I recently posted in another thread, assuming you are up on Minkowski spacetime illustrations. I believe the illustration in the attached file of the following link will answer your question...

https://www.physicsforums.com/showpost.php?p=1104734&postcount=80"​

If we ARE in the middle of B, then the light SHOULD take the same time to reach from the front and from the back to the middle if the events occurred simultaneously, otherwise we wouldn't be in the middle.
If the light didn't take the same time from the front and the back to reach the middle, then wouldn't we KNOW that WE are moving.:

First, each train is inertial, and so the passengers of each train are equally obliged to believe they are at rest and the other train in motion. This is because one cannot FEEL their own inertia when inertial, since there is no accelerational force experienced.

Second, per train B passengers, the light in fact does travel identical length from each end of the train to the center pilot. The fact that one meteor strikes the fwd end of train 1st per train B doesns't matter. The light must travel from each impact point to the center of the train, and the ends of the train are equal distance from its center. Light from the 1st (fwd) collision arrives 1st from half the train's length, then light from the 2nd (aft) collision arrives last from half the train's length. So per A or B, the light paths from each end of one's own train are identical and the light travels at c all the way. In train A the events are simultaneous, but in train B they are asynchronous.

Because how can the events be observed simultaneously by A if A is moving relative to B?

By definition of the scenario, we start with the fact that the meteors strike the train A's ends AT ONCE per A. So this is a scenario "reqt". Then, it's a matter of determining how train B must see it. The Lorentz Transformations allow observers to disagree on when & where events occur, because they both agree on their disagreements :-)

I should point out something which many folks do not pick up on in regards to this scenario. If both (or either) those trains accelerated in such a way to bring the trains into a common frame at rest with each other, train B is larger than train A. Only at a specific relative speed does train B attain a length such that train A sees B as long as itself...

Also, train B doesn't experience train A as the same size as B. Hence, it is quite impossible for the meteors to strike both ends AT ONCE since both ends cannot possibly be aligned simultaneously. Train B must see one end get hit firts, then the other end later, if both trains are to be struck by a single meteor when the train ends do align.

pess