Is time universal? NO (and its proof)

[Raphael]

Something to think about:

Allow that the clocks are spaced on the train in such a way so that clock A to mark the front of the train, and clock J marks the end of the train. Likewise, allow the embankment clock A to mark the beginning of the embankment at the start station, and clock J to mark the end of the embankment before the end station.

Both frames agree that if they were in a common frame, that the length of the train is equal to length of the embankment.

If the trip begins when the train's engine leaves the start station and the trip ends when the last car enters the end station, then both frames agree the total trip takes 27 microseconds. In addition, both frames agree that when half the total trip time has elapsed, the middle of the train is half the distance between the stations.

Scenario one:
Now, if the train came to an emergancy stop when the middle of the train was at the middle of the embankment at t=13.5, how many clock receipts will we find for TB-EJ meeting?

Scenario two: (editted)
There is a bank of lights on board the train labelled EA through EJ; when clock TB receives a receipt from an embankment clock, the corresponding light is turned on. When the the middle of the train is at the middle of the embankment at t=13.5 how many lights are on?

 

[Neddy Bate]

</em>Careful Neddy, you're making a common mistake.

What is the tick rate of clock AT in the embankment frame? Step me through how you calculate that rate from the receipts.

Key point:
In the embankment frame, the time on clock BT is not a reliable indicator of the time on clock AT.

Pete,
I see what you mean about clock TB having nothing to do with the rate of clock TA in the embankment frame. The idea that the 'other' frame is never properly synchronized is still new to me. I just realized it a few days ago when I was thinking about the light-signal-method for synchronizing clocks.

Other than that, I think I just interpreted the frames opposite from what you intended.

From Pete's table:​

<table border=1 cellpadding=4 align=center><tr><td></td><td>EA</td><td>EB</td><td>EC</td><td>ED</td><td>EE</td><td>EF</td><td>EG</td><td>EH</td><td>EI</td><td>EJ</td></tr><tr><td>TA</td><td>20,400</td><td>22,401</td><td>24,402</td><td>26,403</td><td>28,404</td><td>30,405</td><td>32,406</td><td>34,407</td><td>36,408</td><td>38,409</td></tr></table>
So, the above represents train-clock TA ticking 400, 401, 402... (time dilated) in the embankment frame, not the train frame. I think that is what I had backwards.

From Nedd's table:​

<table border=1 cellpadding=4 align=center><tr><td></td><td>E,A</td><td>E,B</td><td>E,C</td><td>E,D</td><td>E,E</td><td>E,F</td><td>E,G</td><td>E,H</td><td>E,I</td><td>E,J</td></tr><tr><td>A,T</td><td>20,400</td><td>21,402</td><td>22,404</td><td>23,406</td><td>24,408</td><td>25,410</td><td>26,412</td><td>27,414</td><td>28,416</td><td>29,418</tr></table>
The above was supposed to represent train-clock TA ticking 400, 402, 404... in the train frame, not the embankment frame.

Interesting that we don't agree on this even though we are looking at the same table.

It seems ambiguous, in a way. I don't know why I didn't realize that I just had it backwards.

 

[Pete]

I see! Yes, that's an easy mistake to make - I should have formatted the table better. In fact, I'll go back and modify it now to use colours, I think.

 

[Pete]

Here is the updated table of the receipts generated as each pair of clocks pass. The format is EmbankmentTime, TrainTime.

<table border=1 cellpadding=4 align=center><tr><td></td><td>EA</td><td>EB</td><td>EC</td><td>ED</td><td>EE</td><td>EF</td><td>EG</td><td>EH</td><td>EI</td><td>EJ</td></tr><tr><td>TA</td><td>20,400</td><td>22,401</td><td>24,402</td><td>26,403</td><td>28,404</td><td>30,405</td><td>32,406</td><td>34,407</td><td>36,408</td><td>38,409</td></tr><tr><td>TB</td><td>21,402</td><td>23,403</td><td>25,404</td><td>27,405</td><td>29,406</td><td>31,407</td><td>33,408</td><td>35,409</td><td>37,410</td><td>39,411</td></tr><tr><td>TC</td><td>22,404</td><td>24,405</td><td>26,406</td><td>28,407</td><td>30,408</td><td>32,409</td><td>34,410</td><td>36,411</td><td>38,412</td><td>40,413</td></tr><tr><td>TD</td><td>23,406</td><td>25,407</td><td>27,408</td><td>29,409</td><td>31,410</td><td>33,411</td><td>35,412</td><td>37,413</td><td>39,414</td><td>41,415</td></tr><tr><td>TE</td><td>24,408</td><td>26,409</td><td>28,410</td><td>30,411</td><td>32,412</td><td>34,413</td><td>36,414</td><td>38,415</td><td>40,416</td><td>42,417</td></tr><tr><td>TF</td><td>25,410</td><td>27,411</td><td>29,412</td><td>31,413</td><td>33,414</td><td>35,415</td><td>37,416</td><td>39,417</td><td>41,418</td><td>43,419</td></tr><tr><td>TG</td><td>26,412</td><td>28,413</td><td>30,414</td><td>32,415</td><td>34,416</td><td>36,417</td><td>38,418</td><td>40,419</td><td>42,420</td><td>44,421</td></tr><tr><td>TH</td><td>27,414</td><td>29,415</td><td>31,416</td><td>33,417</td><td>35,418</td><td>37,419</td><td>39,420</td><td>41,421</td><td>43,422</td><td>45,423</td></tr><tr><td>TI</td><td>28,416</td><td>30,417</td><td>32,418</td><td>34,419</td><td>36,420</td><td>38,421</td><td>40,422</td><td>42,423</td><td>44,424</td><td>46,425</td></tr><tr><td>TJ</td><td>29,418</td><td>31,419</td><td>33,420</td><td>35,421</td><td>37,422</td><td>39,423</td><td>41,424</td><td>43,425</td><td>45,426</td><td>47,427</td></tr></table>

 

[MacM]

Look at the data - from the data, you can determine the distance between moving clocks without making any assumptions about time dilation.

Here's an example:
On the embankment, we find the distance between two train clocks by noting where they are at some instant. For example, at t=34&mu;s, TA is at EH, TJ is half way between EC and ED, and the rest are spaced evenly in between.
The space between train clocks is clearly half the space between embankment clocks.

Likewise, time dilation can also be directly observed in the data without any assumptions about length contraction.

And yes, in both frames the moving clocks tick slower and are closer together, and this is as predicted by special relativity.

Sorry Pete but you are in denial. You can make such statements ONLY because you have assumed SRT valid.

It is not valid for reason I have elucidated. I'll repeat. You are taking the tick of one second in each frame and declaring them equal when it is known by emperical data that the clock that accelerated is ticking slower.

Applying those altered tick rates as being equal via d = v*t you get two different distances. It is that simple. If there were no emperical data to the contrary you would be on equal footing to argue that view.

But data actually exists and you are ignoring it to make your claim. Your chart only reflects your predetermination that length contraction is a physical reality. But accepting that you must ignore the actual known reality that the clocks are not ticking at the same rate.

 

[Pete]

That's right, I'm in denial.

 

[Neddy Bate]

Here is the updated table of the receipts generated as each pair of clocks pass. The format is EmbankmentTime, TrainTime.

Pete,
I don't think the colors make it unambiguous, because I could have done the same thing to my table and it still would have been backwards from yours. My chart also had the EA=20 and TA=400. It was not the identites that were reversed, it was the interpretation of the reference frame. (Refer to this post. I edited the post, but left the table for posterity.

I think what needs to be stated is something like this: "Horizontal rows represent the embankment frame, and vertical columns represent the train frame." Or, more to the point, "All clocks readings are taken from the 'opposite' frame, not their own.

It is strange that something like that makes any difference, but it does seem to explain the difference between your table and mine. What a bizarre stipulation to have to make. No wonder I was confused.

 

[Pete]

From Pete's table:​

<table border=1 cellpadding=4 align=center><tr><td></td><td>EA</td><td>EB</td><td>EC</td><td>ED</td><td>EE</td><td>EF</td><td>EG</td><td>EH</td><td>EI</td><td>EJ</td></tr><tr><td>TA</td><td>20,400</td><td>22,401</td><td>24,402</td><td>26,403</td><td>28,404</td><td>30,405</td><td>32,406</td><td>34,407</td><td>36,408</td><td>38,409</td></tr></table>
So, the above represents train-clock TA ticking 400, 401, 402... (time dilated) in the embankment frame, not the train frame. I think that is what I had backwards.

Well, it's ticking in both frames, of course. But yes, when we compare its readings with several embankment frame clocks we're timing its ticks in the embankment frame (because we're assuming that the clocks we're comparing it against are synchronized).

I think what needs to be stated is something like this: "Horizontal rows represent the embankment frame, and vertical columns represent the train frame." Or, more to the point, "All clocks readings are taken from the 'opposite' frame, not their own.

I don't think it's necessary to spell it out... although it is a valuable insight to reach!

The clock readings occur in both frames. The difference in interpretation (the choice of frame) depends only on which clocks you assume to be synchronized. If you look at a rows to get the timing of a single train clock, you're implicitly assuming that the embankment clocks are synchronized, so the timing you calculate will the timing in the embankment frame.

 

[Raphael]

I don't know why I didn't realize that I just had it backwards.

Neddy,

All embankment times are given from the embankment frame, and all train times are given from the train frame.

Make two tables. Label table one "TRAIN" and table two "EMBANKMENT".

Across the top of both give all the columns a heading of "MY CLOCKS" and label the individual columns A through J. Down the side of each chart label all the row "THEIR CLOCK", and the individual rows A through J.

Take the "TRAIN" chart. Now fill in this chart with the train data from Pete's chart. (For another example, on the TRAIN chart where MY CLOCK D intersects with THEIR CLOCK G place 412 from the intersection of TD,EG on Pete's chart.)

Then take the EMBANKMENT chart and fill in the embankment data from Pete's Chart. (MY CLOCK E, THEIR CLOCK B is EE,TB 29).

This will allow you to compare the data from each frame without the confusion of a dual chart.

 

[D H]

Pete, Nice little thought experiment that demonstrates length contraction, time dilation, and reciprocity.

As viewed from the embankment frame, the moving train has undergone length contraction (train clocks are spaced two per embankment clock) and time dilation ( TA's clock advanced 9 seconds while moving from EA to EJ, during which embankment time advanced 18 seconds).

As viewed from the train frame, the moving embankment has undergone length contraction (embankment clocks are spaced two per train clock) and time dilation (EA's clock advanced 9 seconds while moving from TA to TJ, during which train time advanced 18 seconds).

Cool!


Three questions for MacM:

What is the MacM version of this table?

Which frame is moving?

Would the MacM table be different if the embankment and train frames were both moving at 1/4c (relative to the "rest" frame), but toward each other?

 

[Billy T]

Sorry Pete but you are in denial. You can make such statements ONLY because you have assumed SRT valid.
It is not valid for reason I have elucidated. I'll repeat. You are taking the tick of one second in each frame and declaring them equal when it is known by emperical data that the clock that accelerated is ticking slower. ...

Yes Pete's table is based on the predictions of SRT. Yes the real world might be different (as you think) from SRT discription of it. Yes only experimental results will decide if or if not SRT is the best currently available discription (as most well educated physicist believe it to be).

You some time ago spoke of "velocity history" and currently try to find a difference between the two frames by the text I made bold above. You (if I understand your "SRT's reciprocity is physical nonsense, impossible etc") need to find some way to identify the only frame to exhibit time dilation / length contraction.

(You do still contend, don't you?, that both clocks can not be ticking slower than the other, despite Pete's table showing exactly how this happens. You do still contend that "SIMULTANEITY" has nothing to do with measuring tick rates, don't you? You do still contend that "tick rate" is "tick rate" without the need to speak of "start counting ticks event" and "stop counting ticks event" don't you? You do still contend that it is physically impossible for the meter stick in both frames to be shorter than the one in the other, don't you?) If you have changed your non SRT (naive, intuitively reasonable, view) / position on any of this please tell us.

I have tried to get you to undersand that the "velocity history" that could have been different between the two frames only more than 1000 years ago has nothing to due with the current comparision between the two frames (by two entirely different proofs) but you refute proofs with "that is crap" or just ignore them.

Proof one: Make both the frames have accelerations exactly equal but opposite. I.e in your preferred frame, what ever it is, have two different frames both accelerate exactly the same way except for opposite directions 1000 years ago for one year and then both become inertial frames. I.e. Pete forgot to tell you that the embankment frame and the trian frame both had the same (except for sign) accelerations 1000 years ago.

You {MacM} say: "it is known by emperical data that the clock that accelerated is ticking slower." and I now ask you:

(A) "Would it be that only the embankment clocks are ticking slower in this "equal and opposite accelerations" case?" OR
(B) "Would it be that only the train clocks are ticking slower in this "equal and opposite accelerations" case?"

I won't even ask for the "emperical data" that supports your view that it is "impossible for both to be ticking slower than the other" but I would appreciate it if you said a few words as to why you selected "A" or "B" alternative above.

Proof two: perhaps not logically as strong as proof one because it requires that you conceed that embros held at liquid N2 temperatures do not grow, age, etc. but this does does seem to be a well established "emprirical fact"

Identical twin embros are both held at liquid N2 temp and one is transported far from the other with acceleration "-a" and then later with acceleration "2a" so it is headed back towards his twin brother in an inertial frame moving at high speed wrt his never acclerated, still frozen, brother but will not be side by side for 10 months. During the first month of these two different frames, foster mothers are found and the twin embros are implanted in their wombs 8.5 months before they will be delivered by C section exactly when they are passing each other. They are both in their own frames born side by side at t=0 two weeks early (compared to the normal 9 months gestation).

SRT predicts that when each is blowing out the 15 candles on his birthday cake that the other is blowing out only 10. You {MacM} say this is not possible. I assume that you think that the never accelerated twin is 15 simultaneously with the 10th birthday of the one that had that complex acceleration while a frozen embro, but not the converse "impossible fact" that the accelerated embro can also have his 15th birthday simulatneously (for hm) with is brother's 10th. Is this your view? I would like you to explain how the acceleration of the frozen embro produced the slower time advance for the accelerated embro only, if that is your view.

Like in Pete's table, this is "SRT reciprocity" possible according to SRT as they disagree that the other made the comparison "simultaneously" corrrectly. That is, being 10years old is simultanious with one who's 15th it is, but is not accpeted as simlultaneous by the other with his 15th birthday.

 

[Neddy Bate]

</em>All embankment times are given from the embankment frame, and all train times are given from the train frame.

Raphael, I assumed the same thing, but it turned out to be the opposite.

<table border=1 cellpadding=4 align=center><tr><td></td><td>EA</td><td>EB</td><td>EC</td><td>ED</td><td>EE</td><td>EF</td><td>EG</td><td>EH</td><td>EI</td><td>EJ</td></tr><tr><td>TA</td><td>20,400</td><td>22,401</td><td>24,402</td><td>26,403</td><td>28,404</td><td>30,405</td><td>32,406</td><td>34,407</td><td>36,408</td><td>38,409</td></tr><tr><td>TB</td><td>21,402</td><td>23,403</td><td>25,404</td><td>27,405</td><td>29,406</td><td>31,407</td><td>33,408</td><td>35,409</td><td>37,410</td><td>39,411</td></tr><tr><td>TC</td><td>22,404</td><td>24,405</td><td>26,406</td><td>28,407</td><td>30,408</td><td>32,409</td><td>34,410</td><td>36,411</td><td>38,412</td><td>40,413</td></tr><tr><td>TD</td><td>23,406</td><td>25,407</td><td>27,408</td><td>29,409</td><td>31,410</td><td>33,411</td><td>35,412</td><td>37,413</td><td>39,414</td><td>41,415</td></tr><tr><td>TE</td><td>24,408</td><td>26,409</td><td>28,410</td><td>30,411</td><td>32,412</td><td>34,413</td><td>36,414</td><td>38,415</td><td>40,416</td><td>42,417</td></tr><tr><td>TF</td><td>25,410</td><td>27,411</td><td>29,412</td><td>31,413</td><td>33,414</td><td>35,415</td><td>37,416</td><td>39,417</td><td>41,418</td><td>43,419</td></tr><tr><td>TG</td><td>26,412</td><td>28,413</td><td>30,414</td><td>32,415</td><td>34,416</td><td>36,417</td><td>38,418</td><td>40,419</td><td>42,420</td><td>44,421</td></tr><tr><td>TH</td><td>27,414</td><td>29,415</td><td>31,416</td><td>33,417</td><td>35,418</td><td>37,419</td><td>39,420</td><td>41,421</td><td>43,422</td><td>45,423</td></tr><tr><td>TI</td><td>28,416</td><td>30,417</td><td>32,418</td><td>34,419</td><td>36,420</td><td>38,421</td><td>40,422</td><td>42,423</td><td>44,424</td><td>46,425</td></tr><tr><td>TJ</td><td>29,418</td><td>31,419</td><td>33,420</td><td>35,421</td><td>37,422</td><td>39,423</td><td>41,424</td><td>43,425</td><td>45,426</td><td>47,427</td></tr></table>

Notice how the embankment times are stacked up in columns under the Ex headers. They are relative to the train frame, and are dilated (i.e. 20, 21, 22).

And the train times are layed out in rows across from the Tx headers. They are relative to the embankment frame, and are dilated (i.e. 400, 401, 402).

 

[Pete]

Notice how the embankment times are stacked up in columns under the Ex headers. They are relative to the train frame, and are dilated (i.e. 20, 21, 22).

Neddy, these times are still meaningful in the embankment frame.
In the embankment frame:
At t=20&mu;s, clock EA reads 20 (of course), and clock TA reads 400.
At t=21&mu;s, clock EA reads 21 (of course), and clock TB reads 402.


"But wait!" I hear you say. "If this is the embankment frame, then how come 2&mu;s passed on the train clocks while only 1&mu;s passed on the embankment clocks?"
The thing is that we've looked at the times on two different train clocks. But we don't know that the train clocks are synchronized in this frame, so this doesn't tell us how fast any train clock is ticking. Comparing 400 on TA with 402 on TB is not meaningful in the embankment frame.

 

[Neddy Bate]

</em>Scenario one:
Now, if the train came to an emergancy stop when the middle of the train was at the middle of the embankment at t=13.5, how many clock receipts will we find for TB-EJ meeting?

Let us assume that T has virtually zero mass. We can assume this because our thought experiment relies only on two relatively moving frames of reference, and mass has not yet been incorporated.

The "emergency stop" can be considered to be an acceleration of frame T so that its velocity goes from V>0 to V=0 (relative to the embankment) virtually instantaneously.

There is no reading t=13.5 on this table, but let us consider t=29 in the embankment frame. The train accelerates and joins frames with the embankment at this exact time. We see from the table that the following clocks are co-located (at the same point in space) just before the acceleration:

EA:TJ (29,418)
...(TI is half-way between these two embankment clocks)
EB:TH (29,415)
...(TG is half-way between these two embankment clocks)
EC:TF (29,412)
...(TE is half-way between these two embankment clocks)
ED:TD (29,409)
...(TC is half-way between these two embankment clocks)
EE:TB (29,406)

So, the length contraction of the train is evident from the embankment frame. But once the acceleration brings the two frames together, the length contraction vanishes. This brings up the following questions for those more knowledgeable in this theory:

By what mechanism do the train-clocks instantly re-arrange themselves to be twice as far apart as they are just before the acceleration? Can this re-arrangement be considered to be motion at infinite speed? Does this violate any known laws of physics?

Also, the clocks in frame T are not synchronized just before the acceleration. They read 418, 415, 412, 409, and 406. After the acceleration ,they will be ticking in synch with the embankment clocks. Do the train-clocks simply proceed to tick from these various readings, or do they also become synchronized to each other during the acceleration?

 

[Pete]

Hi Raphael,
I'm interested in exactly how this "emergency stop" happens...

Let's say that there is some "superbrake" that can stop some part of the train effectively instantly.

Does every wheel on the train have a superbrake? If not, where is the superbrake (front, back, middle)?
How are the superbrakes activated?
Does the driver (or some machine) send a signal? Where is the driver (front, back, middle)?
Are the superbrakes timed in some way to stop every part of the train simultaneously?


Actually, all these questions come down to the same thing:

Do all parts of the train stop simultaneously? Simultaneously in which frame?

 

[Neddy Bate]

...How are the superbrakes activated?
Does the driver (or some machine) send a signal? Where is the driver (front, back, middle)?
Are the superbrakes timed in some way to stop every part of the train simultaneously?

Actually, all these questions come down to the same thing:

Do all parts of the train stop simultaneously? Simultaneously in which frame?

Suppose the clocks along the embankment (which are synchronized in that frame) are alarm clocks. They are programmed to go off at t=29 and instantly 'squeeze' the train like a linear version of a disk brake. The 'super-brake' is applied to the whole train at one instant in the embankment frame.

Q: How does the super-brake stop the train when doing so would require the train to enlarge to twice its length in the process?

Of course the brake does not apply 'at one instant' in the train frame. I am not sure what that has to do with the above problem.

 

[Pete]

Suppose the clocks along the embankment (which are synchronized in that frame) are alarm clocks. They are programmed to go off at t=29 and instantly 'squeeze' the train like a linear version of a disk brake. The 'super-brake' is applied to the whole train at one instant in the embankment frame.

Q: How does the super-brake stop the train when doing so would require the train to enlarge to twice its length in the process?

Of course the brake does not apply 'at one instant' in the train frame. I am not sure what that has to do with the above problem.

At which Embankment clock will clock TA stop at? At what time in the train frame will that happen?

At which Embankment clock will clock TB stop at? At what time in the train frame will that happen?

At which Embankment clock will clock TC stop at? At what time in the train frame will that happen?

...

In the train frame, if the part of the train at TA is stopped before the part of the train at TB, what is physically happening to the train? What about if the TB part is stopped before the TA part?

 

[Raphael]

Raphael, I assumed the same thing, but it turned out to be the opposite.

Trust me when I say they are-- or you can just make the charts I mentioned and think about how the data for a single frame's chart is collected.

Notice how the embankment times are stacked up in columns under the Ex headers. They are relative to the train frame, and are dilated (i.e. 20, 21, 22).

And the train times are layed out in rows across from the Tx headers. They are relative to the embankment frame, and are dilated (i.e. 400, 401, 402).

Yes, there is a correlation between the columns/row and the information. But it is an artifact of "mixing" frames and not related to the origin of the data.

Hi Raphael,
I'm interested in exactly how this "emergency stop" happens...

I am not concerned how the stop is accomplished, only that the middle of the train stops in a very short amount of distance/time. (Late Edit: relative to the middle of the embankment. So that the middle of the embankment would also stop in a very short amount of distance/time relative to the middle of the train.)

Actually, all these questions come down to the same thing:

Do all parts of the train stop simultaneously? Simultaneously in which frame?

I would like to assume a common thought experiment condition in which a person asleep in the frame in which the acceleration occurs would be unaware of a change in velocity when he awoke. Which implies that the clocks in his frame would need to be synchronized both before and after the acceleration.

 

[Neddy Bate]

</em>Neddy,
All embankment times are given from the embankment frame, and all train times are given from the train frame.

</em>Trust me when I say they are-- or you can just make the charts I mentioned and think about how the data for a single frame's chart is collected.

I agree. I never meant to imply that the printouts were only valid from one frame or the other. All the printed receipts show an embankment time and a train time, as taken from their own respective reference frames. Each receipt is printed at a moment that is simultaneous in both frames at that point in space.

Also, I have understood from the beginning that 20 is an embankment time, and that 400 is a train time. This was also the case with my table, which was similar to Pete's, but backwards in approach. I thought my table made more sense because of the well-known SR thought experiment which has a relatively moving light-clock ticking slower than a relatively stationary one.

</em>Yes, there is a correlation between the columns/row and the information. But it is an artifact of "mixing" frames and not related to the origin of the data.

Agreed. I have abandonded my table in favor of Pete's.

</em>Neddy, these times are still meaningful in the embankment frame.
In the embankment frame:
At t=20&mu;s, clock EA reads 20 (of course), and clock TA reads 400.
At t=21&mu;s, clock EA reads 21 (of course), and clock TB reads 402.

If I am finally understanding, the times are meaningful in both frames, but the rate is only meaningful in one frame.

</em>"But wait!" I hear you say. "If this is the embankment frame, then how come 2&mu;s passed on the train clocks while only 1&mu;s passed on the embankment clocks?"
The thing is that we've looked at the times on two different train clocks. But we don't know that the train clocks are synchronized in this frame, so this doesn't tell us how fast any train clock is ticking. Comparing 400 on TA with 402 on TB is not meaningful in the embankment frame.

I agree that TA and TB are not synchronized to each other in the embankment frame. I can also understand that therefore 400, 402, 404... is not really a rate of 2 train ticks per embankment tick.

What shall we call the rate of train ticks as perceived from the embankment? "Undefined"?

 

[Pete]

What shall we call the rate of train ticks as perceived from the embankment? "Undefined"?

All one need do to find the rate of train ticks in the embankment frame is compare one train clock across multiple embankment clocks... which (as your rightly pointed out) means looking across a row in my original table, or down a column in your suggested modification.

Both are valid, of course... it's just a different presentation.