
#1
Sep2306, 06:38 AM

P: 144

In this 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 occured 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? 



#2
Sep2306, 08:46 AM

P: 144





#3
Sep2306, 08:53 AM

P: 1,545

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. 



#4
Sep2306, 09:01 AM

P: 29

Simultaneous events in SR...?
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. 



#5
Sep2306, 09:09 AM

P: 1,545





#6
Sep2306, 09:09 AM

P: 144

Or is it the same? 



#7
Sep2306, 09:11 AM

P: 144

And seen from A it will be simultaneous to A, and not simultaneous to B (seen from A). Have I understood this correct now? 



#8
Sep2306, 09:13 AM

P: 1,545

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. 



#9
Sep2506, 12:51 AM

P: 1,504

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




#10
Sep2506, 08:05 AM

P: 1,545

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 



#11
Sep2506, 08:37 AM

Sci Advisor
P: 8,470





#12
Sep2506, 08:45 AM

P: 1,504





#13
Oct606, 11:04 PM

P: 70

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... http://www.physicsforums.com/showpos...4&postcount=80 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. 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 


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