# Thread: Rods and contacts, Poles and barns.

1. ## Rods and contacts, Poles and barns.

I am starting this thread to continue the ongoing discussion that has taken over my thread to test my understanding of Special Relativity. (http://www.bautforum.com/showthread....g-it-right-now)

Now the question that occurs to me is that when the rod is approaching the gap between the contacts, not moving along the abscissa but along the ordinate, then its length (along the x dimension) will determine whether it meets both contacts.

From the rods frame the distance between the contacts is contracted, so it will overlap that separation and the rod will meet both contacts.

However from the contacts frame the it is the rod that is contracted and it cannot meet both contacts.

And this is where I have the problem: the ends of the rod meeting (touching) the contacts are two distinct events. As such they exist in Spacetime.

While I understand that the Spacetime coordinates of those two events will be unique to each frame, I cannot see how an event can exist in one frame and be absent from another.

Surely an event is an event and only its coordinates can change - not its very existence.

2. Good idea to have the new thread - I was thinking the same thing ! First off, here's my reply to Bob's previous post.
Originally Posted by Bob Angstrom
I get the point that you and RobA are making but I am unclear about semantic distinction if there is one between a simultaneity that can be directly observed or observed to be casually connected and a simultaneity that is calculated. A calculated simultaneity is what I would consider to be a simultaneity of relativity as opposed to an observed simultaneity which I refer to as "simultaneous."
Hmmm, I'm afraid I'm now a little unclear about your distinctions, so I'll lay out my definitions! Let's take, for example, Einstein's embankment/train/lightening strikes. We have 2 observers - one in the middle of the embankment, one on the train. The observer on the embankment clearly views the strikes as simultaneous, since he's midway between the strikes and gets the light from both at the same time - so that's "observed simultaneity" or "simultaneous", yes? For simplicity, we can take the "observed" as a given, and just say that for that observer/frame, the strikes are simultaneous. There are no other definitions of simultaneous.

The observer on the train says the strikes are not simultaneous (in his frame). Nevertheless, using relativity he can calculate how the embankment observer views the strikes, and find that they're simultaneous for that one. Could that be said to be "calculated simultaneity"? Simple answer: No. It's clear that observers in any frame can calculate what's true for the embankment observer, so ascribing any sort of "simultaneity" for that just leads to confusion. For every frame, we are only concerned with what is true in that frame. So, for the train observer, the strikes are not simultaneous.

How about the observer that's standing still on the embankment, but 3/4 of the way along instead of in the middle? We would say that the strikes are every bit as simultaneous as for the observer in the middle of the embankment. No difference.

Originally Posted by Bob Angstrom
And if two inertial frames have observations in common they can each calculate what may be simultaneous in the other frame or simultaneous in both so a calculated simultaneity is not unique to a single reference frame.
Bold mine.
With the definitions of simultaneity I use above, then there is no room for "simultaneous in both". If two events are simultaneous in one frame, then every observer in every other frame will find that they are not simultaneous in their frame. Also, it's impossible for two events to be both simultaneous and "causally connected". If two events are simultaneous in one frame, then clearly they happen at two separate locations "at the same time". This precludes any form of signal or connection to travel from one to the other.

3. Originally Posted by Bob Angstrom
I agree with your conclusions but there are no signal times to be considered. The signals to the doors are instant as part of the “props.” If one wants to consider signal time, the observer on top of the barn would have to anticipate the event by the considering the distances involved. This is superfluous to the point Feynman is trying to illustrate so Feynman makes the doors able to close and open instantly for the sake of discussion and is done with it. The prior setup of the experiment and any communication between doors A and B is over thinking the problem.
Actually, the prior setup and the communication is the core of the problem just look at how much time Einstein spent talking about the light's travel time in the train/embankment.

To a certain extent, I agree with you about sending the signal at the time, which is why in post #61 I promptly switched off that tack by saying "ie. you send a signal to A and B. Regardless, in the barn frame, the doors at A and B operate simultaneously,".

And that's the main point - that the barn doors are both closed simultaneously in the barn's frame. So, let's take this another way. Say you (the barn owner) know that every day at 3pm sharp, a pole shoots through your barn at relativistic speed. So, one morning you program your barn doors to close and reopen simultaneously at 3pm, so momentarily trapping that pole in your barn.

So, is that straightforward, with no overthinking about prior setup nor communication? Unfortunately, no. You see, it leaves the question hanging - How do both doors get their clocks synchronized, so that both read 3pm "at the same time". Maybe the previous day they were synchronized by sending a timing signal from one to the other. However it was done, if both the barn's clocks read 3pm simultaneously in the barn's frame, then they do not read the same time simultaneously in the pole's frame - and that allows the pole time to get through.

WARNING: You really may not want to read the following ....

Since we're getting to the Olympics, it reminds me of two athletes walking along:
One says to the other "Are you a pole-vaulter?"
To which the other replies "Nein, I'm German, but how do you know my name?"

4. Originally Posted by Grimble
And this is where I have the problem: the ends of the rod meeting (touching) the contacts are two distinct events. As such they exist in Spacetime.

While I understand that the Spacetime coordinates of those two events will be unique to each frame, I cannot see how an event can exist in one frame and be absent from another.

Surely an event is an event and only its coordinates can change - not its very existence.
Excellent question. The first part of the answer is in pzkpfw's post #48 in the old thread. An event is what happens at one point in spacetime, like the two cars hitting a lamppost. Simultaneity is to do with the relationship between two events; it is not "an event" itself.

So, with the rods and the targets, there are actually four events:
1) The front of the rod hits the first target "A"
2) The front of the rod hits the second target "B"
3) The rear of the rod hits target "A"
4) The rear of the rod hits target "B"

Each of these 4 events occur in all frames; None of them go missing. Furthermore, all frames agree that event (1) comes first, and event (4) comes last.

The discussion centres around how compatible the frames of each of the observers are. Since each observer regards the other as being length contracted, the rod considers the gap between the targets to be really small (much smaller than the rod), whereas the observer with the targets views the rod to be really small. Can they both be right? Relativity says they can.

Interestingly, although it's bit more of a stretch to visualise, the two frames actually disagree of the ordering of events (2) and (3). The observer with the rod reckons (2) happens before (3) (so the rod spans the gap), whereas the observer with the targets reckons (3) comes before (2) - so the rod sits entirely within the gap.

Is this a problem? Again, not so - and in fact we've already implicitly covered it although you may not have noticed ! Remember the Train and Embankment? The strikes were simultaneous for the embankment, but the train moving Left to Right reckoned the one in front (to the Right) occurred first. Well, what about if there was another train heading in the opposite direction from Right to Left? By symmetry, an observer on that train would reckon the strike on the Left would have occurred first!

So, in relativity, things are not just "simultaneous or not", but are not even absolutely ordered - unless light can make the trip from one event to the other.

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Originally Posted by RobA
So, is that straightforward, with no overthinking about prior setup nor communication? Unfortunately, no. You see, it leaves the question hanging - How do both doors get their clocks synchronized, so that both read 3pm "at the same time". Maybe the previous day they were synchronized by sending a timing signal from one to the other. However it was done, if both the barn's clocks read 3pm simultaneously in the barn's frame, then they do not read the same time simultaneously in the pole's frame - and that allows the pole time to get through.
How is that not overthinking the problem? There are no clocks on the doors in Feynman's experiment and the signals from the switch to the doors are “instant” as one of the props.

PS That was bad! “Are you a pole vaulter?” ...and the athlete from Warsaw says, “Yes, but my name is Jerzy.”

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Originally Posted by RobA
How about the observer that's standing still on the embankment, but 3/4 of the way along instead of in the middle? We would say that the strikes are every bit as simultaneous as for the observer in the middle of the embankment. No difference.
I agree with your previous statements but I find a big difference in the “simultaneous” observations of those on the embankment. One observer sees the lightning strikes as simultaneous but the other does not. The observer ¾ of the way along the length of the train does not see the lightning strikes as simultaneous but he can calculate the strikes as simultaneous just as can the moving observer on the train. I would call this a calculated or “relative” simultaneity for both. I understand you would call this “simultaneous” only for the stationary observers outside the train but not the observer on the train because the “at rest” reference frame is the one under consideration and your definition forbids the term “simultaneous” for calculations outside the observer's reference frame. That makes sense but my personal preference is to draw the line at what is observed to be simultaneous rather than calculated to be simultaneous. This seems less ambiguous than drawing the line at the reference frame because we can only agree that two events happened “at the same time” if we observe them to happen at the same time or if we can observe a causal connection between the two. I have Machian distrust of anything that can't be observed
Originally Posted by RobA
With the definitions of simultaneity I use above, then there is no room for "simultaneous in both". If two events are simultaneous in one frame, then every observer in every other frame will find that they are not simultaneous in their frame. Also, it's impossible for two events to be both simultaneous and "causally connected". If two events are simultaneous in one frame, then clearly they happen at two separate locations "at the same time". This precludes any form of signal or connection to travel from one to the other.
This is where you lose me. You said the lightning strikes for the two observers on the side of the tracks are simultaneous. They are observed to be simultaneous by only one observer. ( I just call this simultaneous) They can be calculated to be simultaneous reference frame wide so, in that sense, they are “simultaneous” for observers on embankment. (I call this relatively simultaneous). But the two events are casually connected by the length of the train. So the events are relatively simultaneous in the reference frame, observed simultaneous by the middle observer on the embankment, and causally connected. All in the same frame.

7. Originally Posted by Bob Angstrom
How is that not overthinking the problem? There are no clocks on the doors in Feynman's experiment and the signals from the switch to the doors are “instant” as one of the props.
OK, just this once I'm going to allow instant communications - but I'm still going to insist on clocks on the doors (they're just too useful to read the time) - marked as ever A and B. Let's also make the barn and pole longer than in the example, to give us nice figures to play with. Say the barn is 1 light minute long, the pole is 2 light minutes long and travelling at 0.99c.

Now, let's say at 2:59:30, the front of the pole passes A, and we'll also say the rod's front clock reads this as well. At 3pm, you hit the switch and the doors close, all instantly Since we're talking instantly, we can assume that the clocks at the doors both read 3pm when their associated door closes (that's actually pretty much all I wanted with those clocks!). Now, Relativity says that since a velocity of 0.99c gives gamma=7, the barn reckons the pole is 120-light-seconds / 7 = just 17 light seconds long. So, A will see the rear of the pole pass at around 2:59:47 - so clearly the front of the pole could not have exited past B yet.

So, let's switch now to considering the pole's frame. We can also assume that the poles clocks are synchronized, so what time does the rear clock read as it's passing A? Well, 2 light minutes = 120 light seconds @ 0.99c = 121 seconds. So the rear clock will read 3:01:31 (and note with some surprise the 2:59:47 reading on the barn's clock !). Clearly the front of the pole has gone way through and out the barn by this time, which bears out that, in the rod's frame, the rod is longer. How about when does it see the door at B closing? Well, it will regard the barn as 60 light seconds / 7 = 8.5 light seconds long, so it reckons the door at B would shut t 2:59:38.5 and reopen in time for it to pass as 2:59:38.6 (I think those are right, but very late at night!).

Which is "really" longer? Each in their own frame. If you accept the speed of light being constant, then you must also accept length contraction, and that any and all experiments and measurements are entirely consistent and frame neutral.

8. Originally Posted by Bob Angstrom
One observer sees the lightning strikes as simultaneous but the other does not. The observer ¾ of the way along the length of the train does not see the lightning strikes as simultaneous but he can calculate the strikes as simultaneous just as can the moving observer on the train. I would call this a calculated or “relative” simultaneity for both
Then we need to be even more clear about what we mean by simultaneous. It's not just a matter of observing the lights at the same time.

Imagine the embankment has clocks at both ends - just where the strikes happen. All observers standing on the embankment accept that both the embankment's clocks are accurate and consistent - ie. at every instant, the clocks "agree". They judge that the strikes are simultaneous because the clocks on the embankment both record the same time (say 3pm) for when the strikes hit.

Likewise, there are clocks on the train - coincidentally they happen to be exactly where the strikes hit as well. All observers on the train accept that both the train's clocks are accurate and consistent - ie. at every instant, the clocks "agree". They judge that the strikes are not simultaneous because the clocks on the train record different times for when the strikes hit (eg the front one says that strike hit at 2:59:50, the rear one at 3:00:10).

All observers, no matter where they are along the embankment or the train, agree with their sets of clocks, and what is simultaneous by their set of clocks. Again, the situation is symmetrical and consistent. Light hitting at the same time is just for ease of visualisation for the observer midway between the strikes - it's not the definition of simultaneity.

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Originally Posted by RobA
OK, just this once I'm going to allow instant communications
Allowing instant communications simplifies the problem enormously and we can extend the concept of instant communication to any SR problem. We know there is an apparent delay between signal and receiver and the delay can be explained as 'waiting for the signal to arrive' or we can think of the signal as being instant but we can never observe it as instant because any interval of separation includes an interval of time delay at the constant ratio of c.
Originally Posted by RobA
Which is "really" longer? Each in their own frame. If you accept the speed of light being constant, then you must also accept length contraction, and that any and all experiments and measurements are entirely consistent and frame neutral.
It is easy to understand that a ratio can be constant for all observers but it goes against the Galilean principle to consider a speed to be the same for all observers despite their individual motions. SR quickly becomes complicated when we try to consider c as both a ratio and a speed. If time is a variable then space must also be a variable if the ratio of c is to be maintained. In this paper below Anthony Osborne offers an economical derivation the apparent length contraction in reference frames other than ones own by using c as a constant ratio rather than a speed and no consideration of signal speed is necessary. http://www.poams.org/wp-content/files/osbpope_sr.pdf There is no mathematical difference between SR and Osborne's but the explanation behind the math is different and there is a difference in understanding what makes events simultaneous.

10. Originally Posted by Bob Angstrom
Allowing instant communications simplifies the problem enormously and we can extend the concept of instant communication to any SR problem.
Except that allowing instant communication can also lead you astray very quickly. The only reason I allowed it in this situation is because I didn't really Having the observer on top of the barn and sending a signal "instantly" at 3pm is actually indistinguishable from having the clocks pre-programmed to close the doors at 3pm (sneaky, wasn't I ). The implications of pre-programmed clocks that I referred to earlier also apply to Instant communication. For example, take a look at my previous post #8, where the strikes are simultaneous at 3pm for the embankment observer, but for the train observer (travelling left to right) occur at 3:29:50 in front/right and 3:00:10 at the rear/left. Let's say Bob is in the middle of the embankment, and at 3pm Alice is just passing him on the train. Let's also introduce a new observer "Charlie" standing on the left end of the embankment. Now at 3pm, Alice and Bob both send an instant message to Charlie - who must therefore receive the message at 3pm - but whose 3pm?? Or does he receive Alice's first, at the train's 3pm 10 seconds before the lightening strike, then Bob's 10 seconds later just as the lightening strikes??

So occasionally I can allow so-called "instant" communication within a single frame, since SR allows a limited form of simultaneity within a frame - but instant communication between frames is a no-no.

Originally Posted by Bob Angstrom
It is easy to understand that a ratio can be constant for all observers but it goes against the Galilean principle to consider a speed to be the same for all observers despite their individual motions. SR quickly becomes complicated when we try to consider c as both a ratio and a speed. If time is a variable then space must also be a variable if the ratio of c is to be maintained. In this paper below Anthony Osborne offers an economical derivation the apparent length contraction in reference frames other than ones own by using c as a constant ratio rather than a speed and no consideration of signal speed is necessary. http://www.poams.org/wp-content/files/osbpope_sr.pdf There is no mathematical difference between SR and Osborne's but the explanation behind the math is different and there is a difference in understanding what makes events simultaneous.
Thanks for the link - interesting read. Since it seems to come out with all the same equations, then fair enough - it just becomes a matter of choice which you're more comfortable with. For example, Einstein initially ridiculed Minkowski Diagrams, until he saw they were actually really useful. For me, though, the model feels a little explaining-after-the-fact. For example, tau is "as observed by O" (but "observed" is built on light-travel-time!), and having the axes "r" and "tau" smacks of mixing frames. Still, I certainly have no problem with c as a ratio - indeed, the fact that Minkowski Diagrams work at all is because Space and Time are related/exchanged by the ratio of c. However, personally I'm also quite happy with c as a speed, and regarding photons as travelling at that rate in a vacuum.

Speaking of "what makes events simultaneous" in the quote, are you happy now with my definition of simultaneous in #8, or do you still hold to your statement in #6 that "The observer ¾ of the way along the length of the train does not see the lightning strikes as simultaneous "? I guess that also affects what you think of my first paragraph, and warnings about instant messaging!

11. Originally Posted by RobA
Thanks for the link - interesting read. Since it seems to come out with all the same equations, then fair enough - it just becomes a matter of choice which you're more comfortable with.
Except that choosing units of distance and time so that 1 meter is equivalent to 1/3 * 10^(-8) seconds to all observers as just some random constant of proportion makes no sense, not without something to justify it, such as the speed of something that is measured the same in all frames, for instance. Otherwise, who's time, what frame, between which observers and which events? There is no logical reason given for why this should be or how this ratio should be applied as a constant rather than a speed. (The speed itself is a constant of course, yes, and gives us a ratio of distance to time that light will travel, but that's all.) The scenario for deriving time dilation from this is also nonsense, mostly because there is no actual scenario given for how to derive the time on the moving clock X to begin with, just some roundabout "rectilinear" diagram and equations tailored to give the correct result without any explanation given. That's as far as I read, no need to read further.

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Originally Posted by grav
Except that choosing units of distance and time so that 1 meter is equivalent to 1/3 * 10^(-8) seconds to all observers as just some random constant of proportion makes no sense, not without something to justify it, such as the speed of something that is measured the same in all frames, for instance.
We can measure departure and arrival times of a signal and that is a ratio and a ratio that is the same for all observers. But we can't really call it a speed unless we can observe something speeding between source and sink. And, beyond that, there is no need to call it a speed when all we need is a ratio to serve as a dimensional constant to define our units of length and space.

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Originally Posted by RobA
Having the observer on top of the barn and sending a signal "instantly" at 3pm is actually indistinguishable from having the clocks pre-programmed to close the doors at 3pm (sneaky, wasn't I ).
That was sneaky but it works.
Originally Posted by RobA
For example, tau is "as observed by O" (but "observed" is built on light-travel-time!), and having the axes "r" and "tau" smacks of mixing frames.
Osborne's model does have an after-the-fact feel but the 'fact' is from the perspective of light where emission and absorption are simultaneous events. Put a clock on a photon and it will arrive at the receiver at the same instant it left the sender by its proper time. Our observations add an element of distance with an obligatory amount of time proportional to c. This has nothing to do with light-travel-time which is neither observed nor observable. The model adheres to what we observe and not what we imagine to be observing, that is events between source and sink, and his c is a ratio but not a speed.
Originally Posted by RobA
Speaking of "what makes events simultaneous" in the quote, are you happy now with my definition of simultaneous in #8, or do you still hold to your statement in #6 that "The observer ¾ of the way along the length of the train does not see the lightning strikes as simultaneous "?
I don't understand why you say the strikes are simultaneous for the observer ¾ along the length of the train but they are not simultaneous for the observer on the train. An observer ¾th of the way down the tracks should see the nearest strike first and judge the strikes not to be simultaneous just as the moving observer in the middle of the train judges the strikes not to be simultaneous.

If I understand the problem in #8 correctly there is an instant when all four clocks line up perpendicular, like the four corners of a rectangle, and all four clocks read exactly 3:00:00 which is when both lightning strikes 'light up' all four clocks at exactly 3:00:00. All observers should read the same time for the lightning strikes.
The observers should agree on the location of the strikes and the time of the strikes when the clocks are 'lit up' at exactly 3:00:00 but they should not all agree that the clocks are synchronous with their local time. They may calculate the clocks to be synchronous but they should not all observe them to be so. If the observer on the train sees the strikes at different times, they should also observe the strikes at different locations along the tracks from where the observers on the embankment see them but that is not the case.
Originally Posted by RobA
The implications of pre-programmed clocks that I referred to earlier also apply to InsLet's also introduce a new observer "Charlie" standing on the left end of the embankment. Now at 3pm, Alice and Bob both send an instant message to Charlie - who must therefore receive the message at 3pm - but whose 3pm?? Or does he receive Alice's first, at the train's 3pm 10 seconds before the lightening strike, then Bob's 10 seconds later just as the lightening strikes??
I explained above that Alice should see both strikes at 3:00:00 as measured by the clocks "lit up" at opposite ends of the train but she should not observe them as simultaneous.
If Bob (stationary) and Alice (moving) send a message to Charlie across equal distances at exactly 3:00, Charlie should receive both messages at exactly 3:00 which is also when lightning strikes at both ends of the train. Charlie should see both senders hit the SEND button at exactly 3:00 by their clocks but their clocks are running ahead of the “correct” time relative to his. So from Charlie's point of view, they must have sent the messages before 3:00 by his clock. He should receive both messages and observe the nearby lightning strike simultaneously at 3:00:00 by his clock. And the signals came from Bob and Alice before either one observed the first lightning strike.

14. Originally Posted by Bob Angstrom
We can measure departure and arrival times of a signal and that is a ratio and a ratio that is the same for all observers. But we can't really call it a speed unless we can observe something speeding between source and sink. And, beyond that, there is no need to call it a speed when all we need is a ratio to serve as a dimensional constant to define our units of length and space.
Departure and arrival times of a signal is just d/v, the distance travelled divided by the speed of the signal. It has everything to do with speed. Speed = distance / time

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Originally Posted by Bob Angstrom
I explained above that Alice should see both strikes at 3:00:00 as measured by the clocks "lit up" at opposite ends of the train but she should not observe them as simultaneous.
On second thought, Alice should see the lightning strikes as simultaneous if she is exactly midway between the two on the moving train but she should still see them after she sent the message at 3:00:00 by her local clock. It is my understanding that the train's time (Alice) and the embankment time(Bob) all read 3:00:00 at the instant of both strikes.
Last edited by Bob Angstrom; 2012-Jun-23 at 01:33 AM. Reason: added, by her local clock

16. Originally Posted by Bob Angstrom
I don't understand why you say the strikes are simultaneous for the observer ¾ along the length of the train but they are not simultaneous for the observer on the train. An observer ¾th of the way down the tracks should see the nearest strike first and judge the strikes not to be simultaneous just as the moving observer in the middle of the train judges the strikes not to be simultaneous.
First off, "¾ along the length of the train" implies to me an observer on the train, but since we're talking about an observer on the embankment, I'm switching to "¾ along the length of the embankment".

The definition of "Simultaneous" is that the events happened "at the same time". That does not mean that the observer receives the light at the same time - that'll only happen for the observer midway between the events. No, "at the same time" means that the clocks of the reference frame read the same time for both events. If the embankment is lined with clocks, and these clocks are all synchronized, then the strikes were simultaneous if when they struck, all the clocks were showing the same time at that instant.

What this means for each observer is that they effectively subtract x/c from their time "now" when they get the light, to find the time that each strike hit ("x" being the distance to the strike). For example, an embankment 10 light seconds long. The clocks strike at 3pm. The observer at the centre gets the light from the strikes at 3:00:05. He knows he's 5 light seconds from each end, so subtracts 5 seconds from the time he gets the light. Another observer 80% along gets the light from the nearer strike at 3:00:02, and from the further at 3:00:08. He knows he's 2 light seconds from the nearer, and 8 light seconds from the further, so subtracts these to find again that both strikes happened at 3pm. All observers along the embankment can do this.

However, suppose that observer 80% along got the light from another pair of strikes at 4:00:10? Then he would say that the nearer one happened at 4:00:08 and the further at 4:00:02 - therefore, those would not have been simultaneous.

Again, Simultaneous is about when the clocks in the reference frame read the same time - not about when an observer happens to get light from the events at the same time.
However, in the special case where an observer in a reference frame is midway between the events, then that observer will indeed also get the light from both at the same time - but that's the special case.

Originally Posted by Bob Angstrom
If I understand the problem in #8 correctly there is an instant when all four clocks line up perpendicular, like the four corners of a rectangle, and all four clocks read exactly 3:00:00 which is when both lightning strikes 'light up' all four clocks at exactly 3:00:00.
Then you don't understand the relativity of simultaneity. You may not realise it (and may even disagree!), but you have just described a universal time (aka a "preferred frame"). Suppose the embankment is 10 light seconds long in its rest frame, and we have 2 pairs of strikes - one at 4:00:00, and the other at 4:00:10, so all observers agree that the light from the first also meets the second. If all frames agree with these times, then they must also all agree that the embankment is 10 light seconds long. In other words, if true, then there is no such thing as length contraction, time dilation, etc.

Originally Posted by Bob Angstrom
They may calculate the clocks to be synchronous but they should not all observe them to be so.
Sorry, I don't understand this. Maybe it's just the word "calculated" - it depends what you reckon they're calculating. Let's take the observer 80% along again; For him, they happened at 3pm - no calculation (I don't count subtracting the light seconds as "calculated").

Observers strung along the embankment all have clocks, they're all synchronized so they all agree. That is their reality.
Observers strung along the train also all have clocks, they're all synchronized so they all agree. That is their reality.

And that's it - reality. What they observe (allowing for light travel time) is what is real. That's their frame of reference, and it's entirely consistent view of the universe.
But it turns out that, even when the middle observer's clocks agree, the rest of the embankment clocks and the train clocks don't - hence the different timings for the lightening strikes.

Originally Posted by Bob Angstrom
All observers should read the same time for the lightning strikes.
But they don't. That would imply events are simultaneous across frames.

Actually, since you're upto tackling poams, have a re-read of sections 1 and 2 of On The Electrodynamics of Moving Bodies again.

17. Bold mine
Originally Posted by Bob Angstrom
On second thought, Alice should see the lightning strikes as simultaneous if she is exactly midway between the two on the moving train but she should still see them after she sent the message at 3:00:00 by her local clock. It is my understanding that the train's time (Alice) and the embankment time(Bob) all read 3:00:00 at the instant of both strikes.
No, Alice is opposite Bob at 3pm. Considering the embankment's frame, that's when the lightening struck. It then takes time for the light to reach Bob, during which the train has moved forward. So, Alice gets the light from the front before the light from the back.

Alice's and Bob's clocks both read 3pm as they pass, and the lightening strikes are both at 3pm in the embankment frame. However, the strikes occur at different times in the train frame (eg the front strike hits at 2:59:50, the rear one at 3:00:10).

18. Originally Posted by Bob Angstrom
... I don't understand why you say the strikes are simultaneous for the observer ¾ along the length of the train but they are not simultaneous for the observer on the train. An observer ¾th of the way down the tracks should see the nearest strike first and judge the strikes not to be simultaneous just as the moving observer in the middle of the train judges the strikes not to be simultaneous. ...
The thing to remember about that "moving" observer is that in their frame, they are not moving.

Further, they are (as you note) in the middle - of where they consider the flashes to have occured.

So, if the flashes were simultaneous, in their frame, they should see the flashes at the same time.

(I think half the problem with the train-embankment experiment, is it's too hard to give up the idea that the embankment is "really" still and the train is "really" moving. That skews the thinking.

I suggest: convert the train to a rocket and the embankment to another rocket. Both observers (Alice and Bob) can see they are in relative motion, but they are in deep space with no external reference so can't determine which is "really" moving and which is "really" still. They both calculate physics and observe reality as though they are still. Alice thinks she's the one who's still and Bob is moving. Bob thinks he is still and Alice is moving.

Both consider the flashes to have occured at the ends of their own rocket. Both consider themselves as at rest. Both consider themselves as at the centre of the two flashes. In this case, simultaneous flashes should be seen by an observer at the same time. Since only one could actually see the flashes at the same time, they can't both consider the flashes to have been simultaneous.)

It's not just about seeing flashes at different times. An observer 1/3 of the distance between the two sources will see one flash twice as soon as the other, if they were simultaneous (in their frame). They would still consider the source events to have been simultaneous.

To write-off the experience of a "moving" observer not seeing the flashes at the same time as simply being a result of them no longer being in the middle of the sources, ignores that in their own frame they are not moving.

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Originally Posted by pzkpfw
I suggest: convert the train to a rocket and the embankment to another rocket. Both observers (Alice and Bob) can see they are in relative motion, but they are in deep space with no external reference so can't determine which is "really" moving and which is "really" still. They both calculate physics and observe reality as though they are still. Alice thinks she's the one who's still and Bob is moving. Bob thinks he is still and Alice is moving.

Both consider the flashes to have occured at the ends of their own rocket. Both consider themselves as at rest. Both consider themselves as at the centre of the two flashes. In this case, simultaneous flashes should be seen by an observer at the same time. Since only one could actually see the flashes at the same time, they can't both consider the flashes to have been simultaneous.)
A simpler model of essentially the same problem works for me.
Originally Posted by pzkpfw
To write-off the experience of a "moving" observer not seeing the flashes at the same time as simply being a result of them no longer being in the middle of the sources, ignores that in their own frame they are not moving.
I agree with your observations that Alice (the moving observer) should see the lightning strikes from both front and rear at the same time. I realized that my original statement was wrong and corrected it in post 15.

20. Originally Posted by Bob Angstrom
A simpler model of essentially the same problem works for me.
No. If it worked for you you would not be making incorrect statements.

Originally Posted by Bob Angstrom
I agree with your observations that Alice (the moving observer) should see the lightning strikes from both front and rear at the same time. I realized that my original statement was wrong and corrected it in post 15.
No, that's not what I said (my references to Alice and Bob were in the context of my own explanation (the standard case)), not a reference to your own hypothetical.

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There can be (and is in this problem) an instant where it is 3pm in both Bob's frame and in Alice's frame. The clocks may be running much faster in one reference frame than the other but there is an instant when they all read the same in both reference frames. Why would that not be possible?

As I said before,
If I understand the problem in #8 correctly there is an instant when all four clocks line up perpendicular, like the four corners of a rectangle, and all four clocks read exactly 3:00:00 which is when both lightning strikes 'light up' all four clocks at exactly 3:00:00.

This is not a "preferred” frame because it applies to a single instant in time when all clocks in both reference frames all read the same. At another point in time they will be different because clocks in one frame run faster than clocks in the other.
The same scenario could also apply to Alice and Bob in rockets passing in deep space and that would be a less confusing example. If the situation above is not the way you view the setup, then how do you see it?
Originally Posted by RobA
Again, Simultaneous is about when the clocks in the reference frame read the same time -not about when an observer happens to get light from the events at the same time.
When I think of “simultaneous” events I generally think of events that are observed to be simultaneous as opposed to events that are simultaneous by the clock. I consider the latter to be “relatively” simultaneous. If you prefer to simply call it simultaneous, that is fine with me but you also say “simultaneous” is “not about when an observer happens to get light from the events at the same time.”
So what do you call two events that are observed at the same time?
I tend to concentrate my attention on subjective observations so “events” to me are things like observed lightning flashes and I tend to forget there is also an objective view such as the lightning strikes themselves and that may lead to some confusion.
This is all just a matter of semantics so far but I do see a material difference between “observed” simultaneity and “relative” simultaneity when it comes to quantum events where observation of an event changes the event itself – even a lightning strike.
Originally Posted by RobA
Sorry, I don't understand this. Maybe it's just the word "calculated" - it depends what you reckon they're calculating. Let's take the observer 80% along again; For him, they happened at 3pm - no calculation (I don't count subtracting the light seconds as "calculated"
.
I do consider subtracting light seconds as “calculated.” An observer 80% along the track does not observe the flashes to be simultaneous but he can subtract seconds and conclude that the lightning strikes were themselves were simultaneous. He didn't observe them to be simultaneous but he "calculated" them to be so.
Originally Posted by RobA
But it turns out that, even when the middle observer's clocks agree, the rest of the embankment clocks and the train clocks don't - hence the different timings for the lightening strikes.
You have mentioned this discrepancy in the clocks a few times but I don't understand where it comes from. When the middle observers clocks agree, the rest of the clocks in both reference frames should agree and they should also be able to agree on the timing of the lightning strikes as “relatively” simultaneous.[/QUOTE]
Originally Posted by RobA
But they don't. That would imply events are simultaneous across frames.
I don't understand why an event common to both reference frames can't be simultaneous in both.
Originally Posted by RobA
No, Alice is opposite Bob at 3pm. Considering the embankment's frame, that's when the lightening struck. It then takes time for the light to reach Bob, during which the train has moved forward. So, Alice gets the light from the front before the light from the back.
No, it hasn't worked that way since Michaelson and Morley. Alice should observe the light from the front and the light from the back at the same time. See "pzkpfw's" explanation in #18.

22. Originally Posted by Bob Angstrom
There can be (and is in this problem) an instant where it is 3pm in both Bob's frame and in Alice's frame. The clocks may be running much faster in one reference frame than the other but there is an instant when they all read the same in both reference frames. Why would that not be possible?
It's not possible because that's the way the maths works out.

Think of it this way - clocks are synchronized by sending light, and the clocks then add x/c to arrive at the frame's time. This is what I did in the previous post : "Another observer 80% along gets the light from the nearer strike at 3:00:02, and from the further at 3:00:08. He knows he's 2 light seconds from the nearer, and 8 light seconds from the further, so subtracts these to find again that both strikes happened at 3pm.". Now consider the person passing on the train at the same time - you have to bring in Length Contraction. That means his "x" measurements won't agree with observers on the embankment, so he'll be adding different values for x/c - his times CANNOT agree.

Length contraction, Time dilation and Relativity of Simultaneity all work together to form a consistent whole.

Originally Posted by Bob Angstrom
The same scenario could also apply to Alice and Bob in rockets passing in deep space and that would be a less confusing example. If the situation above is not the way you view the setup, then how do you see it?
Yes, I agree with pzkpfw - it's great to get away from the "stationary bias". It is exactly how I view the setup, and every statement I make about the embankment frame is entirely symmetrical with the train frame (and vice-versa)

Originally Posted by Bob Angstrom
When I think of “simultaneous” events I generally think of events that are observed to be simultaneous as opposed to events that are simultaneous by the clock. I consider the latter to be “relatively” simultaneous. … So what do you call two events that are observed at the same time?
Then get out of that habit In Relativity, Simultaneous is ALWAYS by the clock. Events observed at the same time (ie. an observer receiving the light at the same time) I call utterly irrelevant (except where the observer is midway between the events, which highlights that they're simultaneous in that observer's frame).
Originally Posted by Einstein
we have settled what is to be understood by synchronous stationary clocks located at different places, and have evidently obtained a definition of “simultaneous,” or “synchronous,” and of “time.” The “time” of an event is that which is given simultaneously with the event by a stationary clock located at the place of the event, this clock being synchronous, and indeed synchronous for all time determinations, with a specified stationary clock.
Originally Posted by Bob Angstrom
I do consider subtracting light seconds as “calculated.” An observer 80% along the track does not observe the flashes to be simultaneous but he can subtract seconds and conclude that the lightning strikes were themselves were simultaneous. He didn't observe them to be simultaneous but he "calculated" them to be so.
OK, but "calculated" tends to be used for calculating between frames - ie. the calculations of applying the Lorentz transformations. For observers in a single frame, it's generally assumed that their clocks are synchronized within that frame - ie. each frame has a consistent time coordinate, and observers in that frame have their clocks set to that. The adding of x/c could be assumed to have been performed in the past, so observers need do no more than look at their clocks.
Originally Posted by Bob Angstrom
Originally Posted by RobA
No, Alice is opposite Bob at 3pm. Considering the embankment's frame, that's when the lightening struck. It then takes time for the light to reach Bob, during which the train has moved forward. So, Alice gets the light from the front before the light from the back.
No, it hasn't worked that way since Michaelson and Morley. Alice should observe the light from the front and the light from the back at the same time. See "pzkpfw's" explanation in #18.
Michelson and Morley proved that speed of light was constant - not that it was instant. This is exactly the way it works : (from here)
Originally Posted by Einstein
Just when the flashes 1 of lightning occur, this point M' naturally coincides with the point M, but it moves towards the right in the diagram with the velocity v of the train. If an observer sitting in the position M’ in the train did not possess this velocity, then he would remain permanently at M, and the light rays emitted by the flashes of lightning A and B would reach him simultaneously, i.e. they would meet just where he is situated. Now in reality (considered with reference to the railway embankment) he is hastening towards the beam of light coming from B, whilst he is riding on ahead of the beam of light coming from A. Hence the observer will see the beam of light emitted from B earlier than he will see that emitted from A. Observers who take the railway train as their reference-body must therefore come to the conclusion that the lightning flash B took place earlier than the lightning flash A.
Last edited by RobA; 2012-Jun-24 at 01:59 PM. Reason: added clarification to pzkpfw setup.

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Originally Posted by RobA
Michelson and Morley proved that speed of light was constant - not that it was instant.
People always measure only the average - two-way speed of light.

(c1 + c2) / 2 = c => c1 + c2 = 2c;
This is the equation of an ellipse (ellipsoid of revolution) with semi-axes: a = c

The beautiful illusion of constancy of c, which Einstein postulated.

24. Originally Posted by Hetman
People always measure only the average - two-way speed of light.

(c1 + c2) / 2 = c => c1 + c2 = 2c;
This is the equation of an ellipse (ellipsoid of revolution) with semi-axes: a = c

The beautiful illusion of constancy of c, which Einstein postulated.
No that is the equation of an average, the equation of an ellipse is (x/a)2 + (y/b)2 = 1.

You just came back from suspension, please don't start discussing ATM in Q&A, otherwise you will be out again soon.

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Originally Posted by RobA
OK, but "calculated" tends to be used for calculating between frames - ie. the calculations of applying the Lorentz transformations. For observers in a single frame, it's generally assumed that their clocks are synchronized within that frame - ie. each frame has a consistent time coordinate, and observers in that frame have their clocks set to that. The adding of x/c could be assumed to have been performed in the past, so observers need do no more than look at their clocks.
This is what I am asking. For the two observers Alice and Bob when they are side by side, midway between events and they look at their clocks what time do they see?
Originally Posted by RobA
It's not possible because that's the way the maths works out.

Think of it this way - clocks are synchronized by sending light, and the clocks then add x/c to arrive at the frame's time. This is what I did in the previous post : "Another observer 80% along gets the light from the nearer strike at 3:00:02, and from the further at 3:00:08. He knows he's 2 light seconds from the nearer, and 8 light seconds from the further, so subtracts these to find again that both strikes happened at 3pm.". Now consider the person passing on the train at the same time - you have to bring in Length Contraction. That means his "x" measurements won't agree with observers on the embankment, so he'll be adding different values for x/c - his times CANNOT agree.
If it is 3:00 pm in Alice's frame and 3:00 pm in Bob's frame at t=0 then it is 3:00 pm in both frames at that instant and that can serve as a starting point for later considerations. I don't understand why you say this is impossible. There is no math or conversions involved. It is simply two clock readings at the same instant. You must be reading something strange into my question because I am just asking what is the starting point by clock time in the model?
Originally Posted by RobA
Michelson and Morley proved that speed of light was constant - not that it was instant. This is exactly the way it works : (from here)
I would rephrase that to say, Michelson and Morley proved that c was constant and that it did not act like a typical speed in that you can not add or subtract a velocity to it. That makes me dubious about c as a speed but it makes sense with c as a dimensional constant and what you choose to call it makes no difference to the calculations.

26. Originally Posted by Bob Angstrom
This is what I am asking. For the two observers Alice and Bob when they are side by side, midway between events and they look at their clocks what time do they see?
For two observers in different frames, they can synchronize their clocks to read the same as they pass. That is because as they pass, they coincide in the same place. And they can then synchronize all of the other clocks in each of their own frames to read the same. Alice says that all of the clocks in her frame read the same. Bob says all of the clocks in his frame read the same. But when either observes the clocks in the other frame, they will see that the clocks in the other frame have a lesser reading in the direction of motion of the other frame. That is not an observation due to the time of light travel over a distance from a clock to a distant observer in the same frame, but the actual readings upon the clocks. This is relativity of simultaneity.

For instance, if Alice observes all of the clocks in her frame read 3:00, and Bob synchronizes all of the clocks in his own frame to read 3:00 at the moment he passes Alice at v = .6 c, Alice will not observe that all of the clocks in Bob's frame are set to 3:00 while Bob says they are. Instead, Alice will observe that only Bob's own clock is set to 3:00, while 20 light-seconds in the direction of motion of Bob's frame, the clocks are set to 2:59:45, at 80 proper light-seconds in the direction of motion, Alice says the clocks in Bob's frame are set to 2:59:00, 20 light_seconds opposite the direction of motion, they are set to 3:00:15, etc. In other words, the clocks in Bob's frame all read the same according to Bob, but are all read differently along the direction of motion according to Alice. It doesn't appear to make sense at first, I know, but that is relativity of simultaneity for you, and all of the mathematics and all of the readings for light travel and events work out perfectly without ever any paradoxes when time dilation, length contraction, and relativity of simultaneity are all accounted for.

If it is 3:00 pm in Alice's frame and 3:00 pm in Bob's frame at t=0 then it is 3:00 pm in both frames at that instant and that can serve as a starting point for later considerations. I don't understand why you say this is impossible. There is no math or conversions involved. It is simply two clock readings at the same instant. You must be reading something strange into my question because I am just asking what is the starting point by clock time in the model?
You cannot say that is the time for all of each of the frames due to relativity of simultaneity, but you can say that Alice and Bob themselves set both of their own local clocks to 3:00, yes. Each will then observe the clocks within the rest of the other frame to be set differently, however.

I would rephrase that to say, Michelson and Morley proved that c was constant and that it did not act like a typical speed in that you can not add or subtract a velocity to it. That makes me dubious about c as a speed but it makes sense with c as a dimensional constant and what you choose to call it makes no difference to the calculations.
Yes, c can be considered a dimensional constant of sorts, and is anyway since it a constant speed, a constant distance to time ratio for light travel, as well as a maximum speed and universal constant, but just not used in the way you have been trying to apply it. There is no observational distance applied in SR, which ironically really would then rely directly upon light speed to begin with rather than a universal constant if that was all there was to it, but that doesn't matter because it's not what SR uses, rather only the distance measured with rulers within a frame and actual times according to clocks that coincide with events, a clock that is stationary with the frame making the reading and at the same place as the event.

27. Originally Posted by Bob Angstrom
... When I think of “simultaneous” events I generally think of events that are observed to be simultaneous as opposed to events that are simultaneous by the clock. I consider the latter to be “relatively” simultaneous. If you prefer to simply call it simultaneous, that is fine with me but you also say “simultaneous” is “not about when an observer happens to get light from the events at the same time.”
So what do you call two events that are observed at the same time? ...
It's never going to help that you use you own defintions of things. Two events are simultaneous (in a frame) if they occur "at the same time". A way to determine "at the same time" is that if an observer (in that frame) were exactly in the middle of the two events they would see the events "at the same time". That does not mean an observer actually has to be in the middle. If they are not, they can calculate with distances and speed of light that if they were in the middle they would have seen the events at the same time.

That's it.

Originally Posted by Bob Angstrom
... I don't understand why an event common to both reference frames can't be simultaneous in both. No, it hasn't worked that way since Michaelson and Morley. Alice should observe the light from the front and the light from the back at the same time. See "pzkpfw's" explanation in #18.
No that's not what I wrote. You've very badly misinterpreted what I wrote (especially given all my other posts on this subject recently in threads in which you've been a participant).

The flashes might be simulatenous for my Alice, or they might be simultaneous for my Bob, but they can't be simultaneous for both.

Alice will only observe the light from the front and the light from the (equidistant) back at the same time if the flashes were simultaneous in her frame. And if that's the case, they won't be simultaneous for Bob. Bob in this case will see the flashes at different times, and this is quite different to the case of an observer (in the same frame as Alice) who happens to not be in the exact middle of two events and but can calculate that they were actually simultaneous. Because: in Bobs view of the World, he is at the centre of the two events (and stationary with respect to them) so in that case he should see the two flashes at the same time - if they were simultaneous in his frame. He doesn't, so they weren't.

(
Bob will only observe the light from the front and the light from the (equidistant) back at the same time if the flashes were simultaneous in his frame. And if that's the case, they won't be simultaneous for Alice. Alice in this case will see the flashes at different times, and this is quite different to the case of an observer (in the same frame as Bob) who happens to not be in the exact middle of two events and but can calculate that they were actually simultaneous. Because: in Alices view of the World, she is at the centre of the two events (and stationary with respect to them) so in that case she should see the two flashes at the same time - if they were simultaneous in her frame. She doesn't, so they weren't.
)
Last edited by pzkpfw; 2012-Jun-24 at 09:20 PM. Reason: Add detail

28. And now in mod colour: I'm tired of you pushing an ATM view in the Q&A section. Stick to questions, please, not assertions.

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Originally Posted by tusenfem
No that is the equation of an average, the equation of an ellipse is (x/a)2 + (y/b)2 = 1.
r_1 + r_2 = 2a; r_1 = c_1t; r_2 = c_2t; a = ct;

In the context of experiments designed to measure the speed of light, that is ellipsoid, and only the ellipsoid.

Poincare's ellipsoid, from which we get Einstein's sphere, after applying the Lorentz transformation.

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Originally Posted by pzkpfw
Bob will only observe the light from the front and the light from the (equidistant) back at the same time if the flashes were simultaneous in his frame. And if that's the case, they won't be simultaneous for Alice. Alice in this case will see the flashes at different times, and this is quite different to the case of an observer (in the same frame as Bob) who happens to not be in the exact middle of two events and but can calculate that they were actually simultaneous. Because: in Alices view of the World, she is at the centre of the two events (and stationary with respect to them) so in that case she should see the two flashes at the same time - if they were simultaneous in her frame. She doesn't, so they weren't.
)
I admit it has taken a while but things are beginning to gel in my mind in favor of yours and others contention that events can only be considered "simultaneous" within a single reference frame. My intention was never to reinvent the language but understand and I think progress has been made in that direction. Thanks to all for your efforts.

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