It is different relative to each frame.We know that the distance AB is the same in each frame as it is measured within that frame as a proper length.That the Spacetime interval AB is the same in each frame and that having the same spatial component, the time components must also be equal and be zero – for if it is zero in one it must be in the other.We know that in the embankment frame AM = BM both as times, distances and Spacetime intervals. Originally Posted by space at the centre Yes I can understand what you are saying and that works fine, except that it gives preference to the embankment frame.The problem here for me is giving that preferment to the embankment frame. No. IOW, you take two points, A’ and B’ which are at rest with respect to M’ ( in other words affixed to the train) and if the observer at midpoint M’ were to see flashes of light arriving from these two point at the same time, then M’ concludes that the light left A’ and B’ simultaneouly.What it does not mean is that because M determines that the events at A and B are simultaneous by this test that M’ must conclude the same for the same events. If Front-back is the time axis, then Left-Right is the space axis.

Conversely, the proper distance between A’ and B’ can is measured in the train frame. Now someone else plotting the same events from a different frame will draw out a different triangle, but the hypothenuse will be the same. Yes I can understand what you are saying and that works fine, except that it gives preference to the embankment frame.The problem here for me is giving that preferment to the embankment frame.Remember what Einstein said: “People travelling in this train will with advantage use the train as a rigid reference-body (co-ordinate system); they regard all events in reference to the train. As at this point, we have only established that this is true in the embankment frame.

In addition, the laser coming from the right hits one balloon (say red) on the train when it is about four railway ties from the embankment observer, splattering red paint on the embankment there. I can look at it differently:A and B are two events in Spacetime. AB is at rest in the embankment frame, so the only frame in which we measure its proper length is the embankment frame.

And in turn, the lasers would hit the embankment observer at different times when he was next to different cars, so one would get two separate cars splattered each with a different colored paint. Therefore A and B are simultaneous in the Embankment Frame.In the train frame it is the embankment, and hence point M, that is moving; while the point M’, is at rest, at location S, midway between A and B; so in that train frame the lights meet at point M’ on the Train, at event L. The other laser then hits the green balloon when it is just about even with the right red dot, splattering it with green paint. The fact that we chose one frame as the starting point has no bearing on this. This view is arrived at by logical reasoning, so please shew me where that falls down, and that is not done by pointing out that someone says differently, it is done by simple maths and logic.I refer you to posts 17 and 18 where SpeedFreek agreed with my reasoning.I came to Einstein’s simple guide with an open mind, knowing nothing and that is how I read it. It downs when you make erroneous assumptions about the scenario described.Very simply, its goes like this. in the embankment frame we note that lightning strikes point A on the embankment when it is next to point A’ on the train.

Exactly according to you does the train observer see? Are you claiming that he sees the flashes simultaneously? In which case, you are violating the first postulate. In the embankment frame, the train, and hence the point M’, is moving; while the Point M is at rest, at location S, midway between A and B; so in that embankment frame the lights meet at point M on the Embankment, at event L. If you claim that the train observer sees both lasers simultaneously, then according to him his balloons pop simultaneously and leave a mixed paint splatter at a single point of the embankment. Also the definition of simultaneity can be given relative to the train in exactly the same way as with respect to the embankment.”He then goes on to give a complete description of simultaneity as seen by the observer on the Embankment; and finishes by saying and vice versa; if that is not referring to how it is seen from the train, then what does it refer to.

Therefore A and B are simultaneous in the Train Frame.Yet in each case for the moving frame, M or M’ in that moving frame cannot be simultaneous. That the Spacetime interval AB is the same in each frame and that having the same spatial component, the time components must also be equal and be zero – for if it is zero in one it must be in the other. Thus the space-time interval is the same, even if the separations in time and in space differ. This leaves two different points on the track splattered with different colored paint and one car of the train splattered with two colors as a permanent record.

This also gives us the same time points for A and B in each case.Note that I am seeing Relativity of Simultaneity; that Simultaneity is relative to where it is seen from – in the local (stationary) frame we see it, in the moving frame we don’t. As events they can have no movement, each being a point in space at a point in time.Light from those events meet at event L, a point in space S, midway between A and B, at a particular time t.M and M’ are colocated at location S at the same time point as events A and B.In the embankment frame, the train, and hence the point M’, is moving; while the Point M is at rest, at location S, midway between A and B; so in that embankment frame the lights meet at point M on the Embankment, at event L. We know that in the embankment frame AM = BM both as times, distances and Spacetime intervals. Remember what Einstein said: “People travelling in this train will with advantage use the train as a rigid reference-body (co-ordinate system); they regard all events in reference to the train. Now we go to the train frame.

That is whole point behind this exercise, to determine whehter or not this is the case. Or are you claiming that he sees them at different times, but from that concludes that the strikes occurred simultaneously? In which case you are violating the second postulate.To demonstrate what I mean, let’s rig our experiment thusly: Instead of lightning flashes, we will have lasers which are mechanically triggered when each end of the train reaches its respective red dot. with each observer are two paint balloons, one with red paint and one with green paint. One laser is aimed at the red balloons and the other at the green.

This view is arrived at by logical reasoning, so please shew me where that falls down, and that is not done by pointing out that someone says differently, it is done by simple maths and logic.I refer you to posts 17 and 18 where SpeedFreek agreed with my reasoning.I came to Einstein’s simple guide with an open mind, knowing nothing and that is how I read it. That makes no sense. All it is doing is saying that events that occur buy cheap essays online

at a particular point of the train or embankment have to occur at the same point according to both frames. this does not give preference to one frame over the other, because they both agree on these events. Now according to the embankment observer the laser arrive simulataneously popping both balloon, splattering red and green paint on both the observer and the second from the last car. We purport that these strikes occur simultaneously in this frame and while M who is midway between A and B is next to M’ who is riding the train. You cannot just assume that M’ is midway between A and B when the lightning strikes these points according the train frame. Therefore A and B are simultaneous in the Embankment Frame.In the train frame it is the embankment, and hence point M, that is moving; while the point M’, is at rest, at location S, midway between A and B; so in that train frame the lights meet at point M’ on the Train, at event L. As events they can have no movement, each being a point in space at a point in time.Light from those events meet at event L, a point in space S, midway between A and B, at a particular time t.[/b]M and M’ are colocated at location S at the same time point as events A and B.[/b] According to the embankment frame. whether or not this is true in the train frame is as yet to be determined.

Then every event which takes place along the line also takes place at a particular point of the train. What is meant here is that in order to determine simultaneity for the train you perform the same type of test as you would for the embankment. You are assuming the conclusion We know that the distance AB is the same in each frame as it is measured within that frame as a proper length.

The lasers are powerful enough to pop each balloon in an infinitesimal amount of time. I can look at it differently:A and B are two events in Spacetime. In this sense space-time is different for the two observers in different frames. But above I said, “Yes and no”, where does the “yes” come in. We also note that it strikes point B on the embankment when it is next to point A’ . Then every event which takes place along the line also takes place at a particular point of the train.

So I have to agree that there is nothing absolute about what is seen. No it doesn’t. If we plot out our two event by their space and time coordinates, they can be drawn as two legs of a right triangle, the hypothenuse of this triangle is known as the space-time interval.

Einstein’s ‘vice versa’.I am not saying I am right and you are wrong! I am asking for you to shew me what is wrong with the above reasoning. Also the definition of simultaneity can be given relative to the train in exactly the same way as with respect to the embankment.”He then goes on to give a complete description of simultaneity as seen by the observer on the Embankment; and finishes by saying and vice versa; if that is not referring to how it is seen from the train, then what does it refer to. Einstein’s ‘vice versa’.I am not saying I am right and you are wrong! I am asking for you to shew me what is wrong with the above reasoning.

And, as I pointed out in an earlier thread, Einstein then goes on in later sections to show that lengths lengths measured in one frame will not measure the same from another. Thus in this analogy, the time separation and space separation of our two events does depend on which way the direction our observer faces. It is the space-time interval that is the same across frames.I hope that this sheds some light(no pun intended) on the subject. So what happens if you stop the train and bring the observers back to together? Do separate paint splashes magically merge together or mixed ones separate apart?

That would be a good trick. The time axis and space axis “turn” with him as he changes directions. Again, this is a permanent physical record. No we do not.

This also gives us the same time points for A and B in each case.Note that I am seeing Relativity of Simultaneity; that Simultaneity is relative to where it is seen from – in the local (stationary) frame we see it, in the moving frame we don’t. The only way we can remain consistent between frames as to what happens where, relative to the track and train and also staying true to the postulates of Relativity is for the second animation to represents what happens in the train frame as seen from the train frame. Originally Posted by space at the centre Originally Posted by Janus Events according to the embankment frame (according to anyone located anywhere at rest with respect to the embankment):Same events according to the train( according to anyone on the train): No, the second one is according to how the observer on the embankment will conclude that the observer on the train will see it. Therefore A and B are simultaneous in the Train Frame.

But the spatial component is not the same between frames.