On Saturday, August 7, 2021 at 3:45:15 PM UTC-7, sep...@yahoo.com wrote:
> Al I should have been clearer, I just assumed that since they are identical
> twins you would have assumed they run identically as I wanted to imply.

I already pointed out to you (multiple times now) the obvious fact that if the runners run identically then, by definition, they run identically, meaning they have the same elapsed proper times and the same elapsed relevant coordinate times.

> Let's say each twin runs a constant V for the 100 meters and finishes with
> an elapsed time on his clock of 10 seconds every time he runs, no matter
> what inertial frame he is in or what direction he runs in that inertial reference
> frame. If a twin does that in F0, and the other twin does that in F1, and then
> they decide to race 100 meters in inertial frame F3, the race will be a tie.

Again, let's take this slowly: If the runners run identically (in terms of the inertial coordinates in which the track is at rest) then, by definition, they run identically, meaning they have the same total elapsed proper times and the same total elapsed relevant coordinate times. For example, this was true in your original scenario... remember?

> Why can't they use the elapsed time shown on their clocks to determine who's going to
> win a 200 meter race if the race is run 100 meters in one inertial reference frame, and
> the other 100 meters in some other inertial reference frame?

Again, since you are stipulating that the runners are running identically in terms of the inertial coordinate systems in which the segments of track are at rest, they have the same total elapsed proper times and the same total elapsed relevant coordinate times. Could you estimate how many more times I'm going to have to tell you this before you pay attention?

> The elapsed time on their clocks are identical...

Right.

> and they each ran the same distance in each frame...

Right, and they each went through the same total relevant coordinate time, and hence tied (just like the runners on curved tracks who have staggered starting or ending lines due to the unequal shapes of their paths).

> but that is not sufficient info per [special relativity] to determine the winner.

The winner? What is wrong with you? You have explicitly stipulated conditions for which the outcome is a tie, i.e., they each run the specified track distance in the same track time.

Special Relativity: 794 ..... Barnpole Dave: 0

On Saturday, August 7, 2021 at 6:35:25 PM UTC-5, Al Coe wrote:
> On Saturday, August 7, 2021 at 3:45:15 PM UTC-7, sep...@yahoo.com wrote:
> > Al I should have been clearer, I just assumed that since they are identical
> > twins you would have assumed they run identically as I wanted to imply.
> I already pointed out to you (multiple times now) the obvious fact that if the runners run identically then, by definition, they run identically, meaning they have the same elapsed proper times and the same elapsed relevant coordinate times.
> > Let's say each twin runs a constant V for the 100 meters and finishes with
> > an elapsed time on his clock of 10 seconds every time he runs, no matter
> > what inertial frame he is in or what direction he runs in that inertial reference
> > frame. If a twin does that in F0, and the other twin does that in F1, and then
> > they decide to race 100 meters in inertial frame F3, the race will be a tie.
> Again, let's take this slowly: If the runners run identically (in terms of the inertial coordinates in which the track is at rest) then, by definition, they run identically, meaning they have the same total elapsed proper times and the same total elapsed relevant coordinate times. For example, this was true in your original scenario... remember?
> > Why can't they use the elapsed time shown on their clocks to determine who's going to
> > win a 200 meter race if the race is run 100 meters in one inertial reference frame, and
> > the other 100 meters in some other inertial reference frame?
> Again, since you are stipulating that the runners are running identically in terms of the inertial coordinate systems in which the segments of track are at rest, they have the same total elapsed proper times and the same total elapsed relevant coordinate times. Could you estimate how many more times I'm going to have to tell you this before you pay attention?
>
> > The elapsed time on their clocks are identical...
>
> Right.
>
> > and they each ran the same distance in each frame...
>
> Right, and they each went through the same total relevant coordinate time, and hence tied (just like the runners on curved tracks who have staggered starting or ending lines due to the unequal shapes of their paths).
>
> > but that is not sufficient info per [special relativity] to determine the winner.
>
> The winner? What is wrong with you? You have explicitly stipulated conditions for which the outcome is a tie, i.e., they each run the specified track distance in the same track time.
>
> Special Relativity: 794 ..... Barnpole Dave: 0
Al,
If each twin runs 100 meters in 10 seconds in frame F0, and each twin runs 100 meters in 10 seconds in frame F1, they do not end up in a tie unless they run the same exact path. If you read the original posting they one of the twins crosses the finish line first. Read the original post.
David Seppala
Bastrop TX

On Saturday, August 7, 2021 at 8:30:14 PM UTC-7, sep...@yahoo.com wrote:
> If each twin runs 100 meters in 10 seconds in frame F0, and each twin runs 100 meters
> in 10 seconds in frame F1, they do not end up in a tie...

Yes they do. That is a tie by definition, as has been pointed out to you repeatedly. Remember, runners on a track may have staggered starting or ending positions because the shapes of their paths are different, but what counts is how much track time it takes to traverse the track distance. That is the measure of performance for runners. You have specified that these are equal for the two runners, so they have tied. Do you understand this?

> If you read the original posting they one of the twins crosses the finish line first.

You are failing to correctly account for the amount of track time each runner takes to traverse the track distance. Each one takes a total of 20/sqrt(1-V^2/c^2) seconds of track time to traverse the 200 meters. Do you understand this?

> Read the original post.

Read each of my replies, in which I've explained this to you half a dozen times (so far). After doing that, if anything is still unclear to you, just ask.

Special Relativity: 795 ..... Barnpole Dave: 0

On Saturday, August 7, 2021 at 11:48:08 PM UTC-5, Al Coe wrote:
> On Saturday, August 7, 2021 at 8:30:14 PM UTC-7, sep...@yahoo.com wrote:
> > If each twin runs 100 meters in 10 seconds in frame F0, and each twin runs 100 meters
> > in 10 seconds in frame F1, they do not end up in a tie...
>
> Yes they do. That is a tie by definition, as has been pointed out to you repeatedly. Remember, runners on a track may have staggered starting or ending positions because the shapes of their paths are different, but what counts is how much track time it takes to traverse the track distance. That is the measure of performance for runners. You have specified that these are equal for the two runners, so they have tied. Do you understand this?
> > If you read the original posting they one of the twins crosses the finish line first.
> You are failing to correctly account for the amount of track time each runner takes to traverse the track distance. Each one takes a total of 20/sqrt(1-V^2/c^2) seconds of track time to traverse the 200 meters. Do you understand this?
>
> > Read the original post.
>
> Read each of my replies, in which I've explained this to you half a dozen times (so far). After doing that, if anything is still unclear to you, just ask.
>
> Special Relativity: 795 ..... Barnpole Dave: 0
Al,
You said:
"Remember, runners on a track may have staggered starting or ending positions because the shapes of their paths are different, but what counts is how much track time it takes to traverse the track distance. That is the measure of performance for runners. You have specified that these are equal for the two runners, so they have tied. Do you understand this?

So you are saying if the two runners start at the same starting point in F0 and they each run 100 meters in F0 and then they each run the last hundred meters in F1 and they finish at the same point in F1 but one finishes before the other, the race is a tie. Now you say that if they run the exact same distance in F0, and they run the exact same distance in F1 and their speeds in each frame are identical to each other, and they start at the same point in F0 simultaneously and they finish at the same point in F1 simultaneously the race is not a tie. Please explain.
David Seppala
Bastrop TX

On Sunday, August 8, 2021 at 6:25:17 AM UTC-7, sep...@yahoo.com wrote:
> So you are saying if the two runners start at the same starting point in F0...

No, your scenario does not have the runners starting at the same "point" (event), you specified that they begin at spatially separate events that are simultaneous in terms of F0 but not in terms of F1, and the runner that begins the race first in terms of F1 is also the runner that completes the race first in terms of F1, and the runners end at the same spatial location in terms of F1 but not in the same spatial location in terms of F0. You could change your scenario so that the runners begin at the same event, but they would then have to run at different angles (not straight in the y direction) to reach the separate spatial locations at which they step onto the conveyor belt, but then they would undergo different accelerations, and as always the relevant measure of performance is total track time to complete the total track distance.

> and they each run 100 meters in F0 and then they each run the last hundred meters in F1...

Every object is "in" every system of coordinates all the time, so all your statements are strictly gibberish. Also, note that the runners are never even at rest in either of the systems you say they are "in", so your verbiage is wrong even if we make allowance for sloppy usage. What you need to specify is that there are two segments of track, one segment at rest in the inertial coordinate system F0, and the other segment on some kind of conveyor belt so it is at rest in the inertial coordinate system F1.

> and they finish at the same point in F1...

They both finish at x'=0, but of course they finish at different values of x, just as they started at the same value of t but at different values of t'.

> but one finishes before the other, the race is a tie.

Right, in the scenario you specified, each runner completes the total track distance in the same total amount of track time. Remember, you already agreed that they could run on different days, so one runner could conduct their entire run before the other even starts to run. They run different paths, and just like runners on different lanes on a curved track, the beginning and/or ending events for different lanes may be staggered both spatially and temporally. The relevant measure is the amount of track time required to traverse the same track distance.

> Now you say...

"Now"? I'm not a saying anything that I haven't been telling you for many messages.

> that if they run the exact same distance in F0, and they run the exact same distance in F1
> and their speeds in each frame are identical to each other, and they start at the same point
> in F0 simultaneously...

So *now* you are proposing yet another scenario, this time with the runners starting the race at the same "point" (same spatial position and presumably at the same time, although your brain couldn't manage to express that), and then they follow exactly the same path, proceeding directly in the y direction, side by side, and stepping onto the track at the same event, and then each run the same track distance in the same track time, it will again be a tie by definition. Of course, since they step onto the conveyor belt at the same event, if they run in opposite directions they will be spatially separate when they end.

> and they finish at the same point in F1 simultaneously the race is not a tie.

Nope, in order for them to finish at the same time and place in this scenario (per above) they must run in the same direction on the conveyor belt, and finish at the same event. Understand?

Special Relativity: 796 ..... Barnpole Dave: 0

On Sunday, August 8, 2021 at 10:36:14 AM UTC-5, Al Coe wrote:
> On Sunday, August 8, 2021 at 6:25:17 AM UTC-7, sep...@yahoo.com wrote:
> > So you are saying if the two runners start at the same starting point in F0...
>
> No, your scenario does not have the runners starting at the same "point" (event), you specified that they begin at spatially separate events that are simultaneous in terms of F0 but not in terms of F1, and the runner that begins the race first in terms of F1 is also the runner that completes the race first in terms of F1, and the runners end at the same spatial location in terms of F1 but not in the same spatial location in terms of F0. You could change your scenario so that the runners begin at the same event, but they would then have to run at different angles (not straight in the y direction) to reach the separate spatial locations at which they step onto the conveyor belt, but then they would undergo different accelerations, and as always the relevant measure of performance is total track time to complete the total track distance.
>
> > and they each run 100 meters in F0 and then they each run the last hundred meters in F1...
>
> Every object is "in" every system of coordinates all the time, so all your statements are strictly gibberish. Also, note that the runners are never even at rest in either of the systems you say they are "in", so your verbiage is wrong even if we make allowance for sloppy usage. What you need to specify is that there are two segments of track, one segment at rest in the inertial coordinate system F0, and the other segment on some kind of conveyor belt so it is at rest in the inertial coordinate system F1.
>
> > and they finish at the same point in F1...
>
> They both finish at x'=0, but of course they finish at different values of x, just as they started at the same value of t but at different values of t'.
> > but one finishes before the other, the race is a tie.
> Right, in the scenario you specified, each runner completes the total track distance in the same total amount of track time. Remember, you already agreed that they could run on different days, so one runner could conduct their entire run before the other even starts to run. They run different paths, and just like runners on different lanes on a curved track, the beginning and/or ending events for different lanes may be staggered both spatially and temporally. The relevant measure is the amount of track time required to traverse the same track distance.
>
> > Now you say...
>
> "Now"? I'm not a saying anything that I haven't been telling you for many messages.
> > that if they run the exact same distance in F0, and they run the exact same distance in F1
> > and their speeds in each frame are identical to each other, and they start at the same point
> > in F0 simultaneously...
>
> So *now* you are proposing yet another scenario, this time with the runners starting the race at the same "point" (same spatial position and presumably at the same time, although your brain couldn't manage to express that), and then they follow exactly the same path, proceeding directly in the y direction, side by side, and stepping onto the track at the same event, and then each run the same track distance in the same track time, it will again be a tie by definition. Of course, since they step onto the conveyor belt at the same event, if they run in opposite directions they will be spatially separate when they end.
> > and they finish at the same point in F1 simultaneously the race is not a tie.
> Nope, in order for them to finish at the same time and place in this scenario (per above) they must run in the same direction on the conveyor belt, and finish at the same event. Understand?
>
> Special Relativity: 796 ..... Barnpole Dave: 0
I guess you have trouble reading English.
In the last post, when I said, "they start at the same point in F0 simultaneously"
You posted, " (same spatial position and presumably at the same time, although your brain couldn't manage to express that)"

You seem to have trouble following the scenario. So I will adjust things to try to make things clearer.
Let's say in F0, the two identical twin runners start at x=0, and start running simultaneously. One twin runs at a positive 45 degree angle relative to the x axis toward the line at y=L and the other twin runs at a negative 45 degree angle relative to the x-axis toward the line at y=L. They both started running at the same point in space and time from x=0. The distance they each run is 100 meters as measured in F0. They both run at the identical speed as measured in F0. When they reach y=L simultaneously as measured in F0, they go into frame F1. At the two points in F1 that they land on, there are two lines, each at an angle to the x' axis, one a positive angle and the other a negative angle, each angle being the identical angle as measured in F1 (but in different directions) with each line being 100 meters in length angled such that the two lines meet at y' = L2'.
If we use elapsed time run in each frame segment of the race, both twins running at identical speeds, each twin ran each race leg in the same amount of time, and they both ran the equal amount of distance in each reference frame. But per Einstein's concepts they did not end up simultaneously where the two lines in F1 meet.
Please explain where one twin got the lead.
David Seppala
Bastrop TX

On Sunday, August 8, 2021 at 9:05:34 AM UTC-7, sep...@yahoo.com wrote:
> You seem to have trouble following the scenario.

No, you stipulated that the runners begin by running directly in the y direction in terms of F0, and then they step onto the conveyor belt at x'=-100 and x'=+100 respectively, which means they started at separate locations in terms of x as well. Do you understand this?

> So I will adjust things to try to make things clearer.

Translation: Having completely misunderstood your own original scenario, you are now (as always) going to switch to yet another scenario.

> Let's say in F0, the two identical twin runners start at x=0, and start running simultaneously. One twin runs at a positive 45 degree angle relative to the x axis toward the line at y=L and the other twin runs at a negative 45 degree angle relative to the x-axis toward the line at y=L. They both started running at the same point in space and time from x=0.

You see, you have just adopted my suggestion for how you would need to revise your scenario in order to make the runners begin at the same event, which means they can't start by running directly in the y direction, they need to run at an angle. Now you parrot this back to me as if you are "clarifying" your scenario, oblivious to the fact that I already explained why this revised scenario is perfectly consistent with special relativity -- just as is every other scenario. And you will doubtless overlook the fact that I also pointed out in this revised scenario the runners will undergo different accelerations when they step onto the conveyor belt.

> The distance they each run is 100 meters as measured in F0. They both run at the
> identical speed as measured in F0.

Again, you are just parroting the corrected scenario that I explained to you in my previous message.

> When they reach y=L simultaneously as measured in F0, they go into frame F1.

Again, what you mean is they start running on a conveyor belt that is at rest in terms of F1. The runners themselves are never at rest in terms of either F0 or F1. And, again, you are just parroting the corrected scenario that I explained to you in my previous message.

> At the two points in F1 that they land on, there are two lines, each at an angle
> to the x' axis, one a positive angle and the other a negative angle, each angle
> being the identical angle as measured in F1 (but in different directions) with
> each line being 100 meters in length angled such that the two lines meet at y' = L2'.

Yawn.

> If we use elapsed time run in each frame segment of the race, both twins running
> at identical [constant] speeds [in terms of the inertial coordinates in which the track
> is at rest], each twin ran each race leg in the same amount of [track] time, and they
> both ran the equal amount of distance in each [track] reference frame.

You overlooked (as I predicted you would) that in this scenario the runners undergo different accelerations when they step onto the conveyor belt, because one is angling in the negative x direction and the other is angling in the positive x direction, and the conveyor belt is moving in the positive x direction. Hence the lanes are not spatio-temporally or dynamically symmetrical. But, yes, aside from the important dynamical impulse difference in accelerations at the transition, each runner ran the same total track distance in the same track time.

> But per [special relativity] they did not end up simultaneously where the two lines in F1 meet.
> Please explain where one twin got the lead.

Again, "got the lead" is ambiguous due to the relativity of simultaneity. This was explained to you previously, remember? Once again: In terms of F1, the runners reach the conveyor belt at different times (t'), and they accordingly reach the end point at different times. In terms of F0, the runners reach the conveyor belt at the same time (t), but the end point is moving toward one runner and away from the other, so the closing speed for the former is greater, and hence he intersects with that ending x' locus at an earlier value of t. Now do you understand?

Special Relativity: 797 ..... Barnpole Dave: 0

On Sunday, August 8, 2021 at 12:29:20 PM UTC-5, Al Coe wrote:
> On Sunday, August 8, 2021 at 9:05:34 AM UTC-7, sep...@yahoo.com wrote:
> > You seem to have trouble following the scenario.
> No, you stipulated that the runners begin by running directly in the y direction in terms of F0, and then they step onto the conveyor belt at x'=-100 and x'=+100 respectively, which means they started at separate locations in terms of x as well. Do you understand this?
> > So I will adjust things to try to make things clearer.
> Translation: Having completely misunderstood your own original scenario, you are now (as always) going to switch to yet another scenario.
> > Let's say in F0, the two identical twin runners start at x=0, and start running simultaneously. One twin runs at a positive 45 degree angle relative to the x axis toward the line at y=L and the other twin runs at a negative 45 degree angle relative to the x-axis toward the line at y=L. They both started running at the same point in space and time from x=0.
> You see, you have just adopted my suggestion for how you would need to revise your scenario in order to make the runners begin at the same event, which means they can't start by running directly in the y direction, they need to run at an angle. Now you parrot this back to me as if you are "clarifying" your scenario, oblivious to the fact that I already explained why this revised scenario is perfectly consistent with special relativity -- just as is every other scenario. And you will doubtless overlook the fact that I also pointed out in this revised scenario the runners will undergo different accelerations when they step onto the conveyor belt.
> > The distance they each run is 100 meters as measured in F0. They both run at the
> > identical speed as measured in F0.
> Again, you are just parroting the corrected scenario that I explained to you in my previous message.
> > When they reach y=L simultaneously as measured in F0, they go into frame F1.
> Again, what you mean is they start running on a conveyor belt that is at rest in terms of F1. The runners themselves are never at rest in terms of either F0 or F1. And, again, you are just parroting the corrected scenario that I explained to you in my previous message.
> > At the two points in F1 that they land on, there are two lines, each at an angle
> > to the x' axis, one a positive angle and the other a negative angle, each angle
> > being the identical angle as measured in F1 (but in different directions) with
> > each line being 100 meters in length angled such that the two lines meet at y' = L2'.
> Yawn.
> > If we use elapsed time run in each frame segment of the race, both twins running
> > at identical [constant] speeds [in terms of the inertial coordinates in which the track
> > is at rest], each twin ran each race leg in the same amount of [track] time, and they
> > both ran the equal amount of distance in each [track] reference frame.
>
> You overlooked (as I predicted you would) that in this scenario the runners undergo different accelerations when they step onto the conveyor belt, because one is angling in the negative x direction and the other is angling in the positive x direction, and the conveyor belt is moving in the positive x direction. Hence the lanes are not spatio-temporally or dynamically symmetrical. But, yes, aside from the important dynamical impulse difference in accelerations at the transition, each runner ran the same total track distance in the same track time.
>
> > But per [special relativity] they did not end up simultaneously where the two lines in F1 meet.
> > Please explain where one twin got the lead.
> Again, "got the lead" is ambiguous due to the relativity of simultaneity. This was explained to you previously, remember? Once again: In terms of F1, the runners reach the conveyor belt at different times (t'), and they accordingly reach the end point at different times. In terms of F0, the runners reach the conveyor belt at the same time (t), but the end point is moving toward one runner and away from the other, so the closing speed for the former is greater, and hence he intersects with that ending x' locus at an earlier value of t. Now do you understand?
>
> Special Relativity: 797 ..... Barnpole Dave: 0

Al,
Please explain why the TWINS CANNOT use the elapsed times shown on each of their clocks to determine who won the race.
Each twin can use the elapsed time shown on his clock while in F0 to determine which twin transitioned into frame F1 first. Each twin can use the elapsed time shown on his clock to determine which twin ran the 100 meters in F1 faster. Why can't they use the total elapsed time shown from the beginning of the race to the end the race to determine if they meet at the end point of the race simultaneously?
We know that the two twins do not end up at the finish point of the race in F1 simultaneously. Each clock showed the same elapsed time for the leg run in frame F0 and each clock showed the same elapsed time for the leg run in frame F1. Each twin ran at exactly the same speed in each leg of the race.. Each twin ran the exact same distance in each of the two frames. Explain why running at the same speed, same elapsed times in each frame, same distance in each frame, the two twins didn't end up at the same point at the finish simultaneously. By the way your aside that their accelerations were different is a moot point. We can say that they changed their direction in frame F0 before they transitioned to frame F1, and in F1, they changed their direction slightly after they landed in frame F1.
David Seppala
Bastrop TX

On Sunday, August 8, 2021 at 11:42:44 AM UTC-7, sep...@yahoo.com wrote:
> Please explain why the TWINS CANNOT use the elapsed times shown on each of
> their clocks to determine who won the race.

What is wrong with you? It's been explained to you several times now that, given your stipulation of constant coordinate speeds, the runners *can* use their elapsed proper times (as would be shown on clocks that they carry) to determine who won, because in that case there is a fixed proportionality between proper times and coordinate times. For example, in the scenarios you've described, the runners have tied, and they show equal elapsed proper times on their wrist watches.

I've also explained why, if we more realistically allow the coordinate speeds to vary during the race, the simple proportionality no longer applies, even on a single segment in a single system of inertial coordinates, but this is clearly beyond you, since you can't even grasp the trivial case of constant speeds.

> Each twin can use the elapsed time shown on his clock while in F0 to determine
> which twin transitioned into frame F1 first.

No, the co-moving clocks just show the elapsed proper times between when the respective runner starting and when he reaches the conveyor belt. This says nothing about the synchronization between those clocks, i.e., which things "happened first". They could even be running on separate days. You are free to synchronize them using the simultaneity of F0, or you could synchronize them using the simultaneity of F1, or you could synchronize them using any other simultaneity you like.

> Each twin can use the elapsed time shown on his clock to determine which twin
> ran the 100 meters in F1 faster.

Given your simplified assumption that each runner runs at a (possibly different) constant speed, yes, the elapsed proper times enable us to determine which runner was faster. This was explained to you before, remember? And this (again) does not tell us anything about what "happened first". We can initialize the clocks differently, to match the simultaneity of F0, or of F1, or of any other system we like.

> Why can't they use the total elapsed time shown from the beginning of the race to
> the end the race to determine if they meet at the end point of the race simultaneously?

They can (given your simplifying assumption). This is precisely what I've been explaining to you. In the scenario you described, it is trivial to determine that the runners do not arrive at the end of their respective runs at the same event, and it is trivial to determine that the completions of their respective runs are not simultaneous in terms of either F0 or F1, and they do not end at the same place in terms of F0, and they do not reach the conveyor belt at the same time in terms of F1, and so on. We can precisely specify when and where everything occurs in terms of F0 and in terms of F1 and in terms of any other specified system of coordinates. Do you understand this?

> We know that the two twins do not end up at the finish point of the race in F1 simultaneously.

Right.

> Each clock showed the same elapsed time for the leg run in frame F0 and
> each clock showed the same elapsed time for the leg run in frame F1.

Right again (assuming equal constant speeds).

> Each twin ran at exactly the same speed in each leg of the race.

Moreover, you are stipulating they both ran at equal *constant* speed. That's important, because you started out just giving the elapsed time for each leg without specifying constant speed, which introduced a deep ambiguity as explained previously.

> Each twin ran the exact same distance in each of the two frames.

Right again, meaning that they ran the same coordinate distance in each respective leg, but of course when on the conveyor belt they aren't going the same distance in terms of F0, and on the first leg they aren't going the same distance in terms of F1. And they have different accelerations when stepping onto the conveyor belt.

> Explain why running at the same speed, same elapsed times in each frame,
> same distance in each frame, the two twins didn't end up at the same point
> at the finish simultaneously.

Already explained in detail multiple times. Again, in terms of F1 the runners reach the conveyor belt at different times, so they obviously reach the end spot at different times because they have the same speeds. On the other hand, in terms of F0 they reach the conveyor belt at the same time, but the end point is moving spatially toward one and away from the other, and their individual speeds are given by the relativistic composition, so again they reach the end spot at different times. The two descriptions are perfectly consistent, as are the descriptions we can give in terms of any other system of coordinates.

Maybe it would help you to consider two runners going from starting event E1 to ending event E2, and ask if they must show the same elapsed times on their clocks. Obviously not. You understand this, right? This is just the standard twins scenario. The have the same elapsed coordinate time, but different elapsed proper times. So, if you understand that, you should be able to grasp conversely that for different paths we can have equal elapsed proper times but different elapsed coordinate times. Understand?

> Your aside that their accelerations were different is a moot point.
> We can say that they changed their direction...

Changing direction is acceleration. Sheesh.

Special Relativity: 798 ..... Barnpole Dave: 0

On Sunday, August 8, 2021 at 2:57:29 PM UTC-5, Al Coe wrote:
> On Sunday, August 8, 2021 at 11:42:44 AM UTC-7, sep...@yahoo.com wrote:
> > Please explain why the TWINS CANNOT use the elapsed times shown on each of
> > their clocks to determine who won the race.
> What is wrong with you? It's been explained to you several times now that, given your stipulation of constant coordinate speeds, the runners *can* use their elapsed proper times (as would be shown on clocks that they carry) to determine who won, because in that case there is a fixed proportionality between proper times and coordinate times. For example, in the scenarios you've described, the runners have tied, and they show equal elapsed proper times on their wrist watches.
>
> I've also explained why, if we more realistically allow the coordinate speeds to vary during the race, the simple proportionality no longer applies, even on a single segment in a single system of inertial coordinates, but this is clearly beyond you, since you can't even grasp the trivial case of constant speeds.
> > Each twin can use the elapsed time shown on his clock while in F0 to determine
> > which twin transitioned into frame F1 first.
> No, the co-moving clocks just show the elapsed proper times between when the respective runner starting and when he reaches the conveyor belt. This says nothing about the synchronization between those clocks, i.e., which things "happened first". They could even be running on separate days. You are free to synchronize them using the simultaneity of F0, or you could synchronize them using the simultaneity of F1, or you could synchronize them using any other simultaneity you like.
> > Each twin can use the elapsed time shown on his clock to determine which twin
> > ran the 100 meters in F1 faster.
> Given your simplified assumption that each runner runs at a (possibly different) constant speed, yes, the elapsed proper times enable us to determine which runner was faster. This was explained to you before, remember? And this (again) does not tell us anything about what "happened first". We can initialize the clocks differently, to match the simultaneity of F0, or of F1, or of any other system we like.
> > Why can't they use the total elapsed time shown from the beginning of the race to
> > the end the race to determine if they meet at the end point of the race simultaneously?
> They can (given your simplifying assumption). This is precisely what I've been explaining to you. In the scenario you described, it is trivial to determine that the runners do not arrive at the end of their respective runs at the same event, and it is trivial to determine that the completions of their respective runs are not simultaneous in terms of either F0 or F1, and they do not end at the same place in terms of F0, and they do not reach the conveyor belt at the same time in terms of F1, and so on. We can precisely specify when and where everything occurs in terms of F0 and in terms of F1 and in terms of any other specified system of coordinates. Do you understand this?
> > We know that the two twins do not end up at the finish point of the race in F1 simultaneously.
> Right.
> > Each clock showed the same elapsed time for the leg run in frame F0 and
> > each clock showed the same elapsed time for the leg run in frame F1.
> Right again (assuming equal constant speeds).
> > Each twin ran at exactly the same speed in each leg of the race.
> Moreover, you are stipulating they both ran at equal *constant* speed. That's important, because you started out just giving the elapsed time for each leg without specifying constant speed, which introduced a deep ambiguity as explained previously.
> > Each twin ran the exact same distance in each of the two frames.
> Right again, meaning that they ran the same coordinate distance in each respective leg, but of course when on the conveyor belt they aren't going the same distance in terms of F0, and on the first leg they aren't going the same distance in terms of F1. And they have different accelerations when stepping onto the conveyor belt.
> > Explain why running at the same speed, same elapsed times in each frame,
> > same distance in each frame, the two twins didn't end up at the same point
> > at the finish simultaneously.
> Already explained in detail multiple times. Again, in terms of F1 the runners reach the conveyor belt at different times, so they obviously reach the end spot at different times because they have the same speeds. On the other hand, in terms of F0 they reach the conveyor belt at the same time, but the end point is moving spatially toward one and away from the other, and their individual speeds are given by the relativistic composition, so again they reach the end spot at different times. The two descriptions are perfectly consistent, as are the descriptions we can give in terms of any other system of coordinates.
>
> Maybe it would help you to consider two runners going from starting event E1 to ending event E2, and ask if they must show the same elapsed times on their clocks. Obviously not. You understand this, right? This is just the standard twins scenario. The have the same elapsed coordinate time, but different elapsed proper times. So, if you understand that, you should be able to grasp conversely that for different paths we can have equal elapsed proper times but different elapsed coordinate times. Understand?
>
> > Your aside that their accelerations were different is a moot point.
> > We can say that they changed their direction...
>
> Changing direction is acceleration. Sheesh.
>
> Special Relativity: 798 ..... Barnpole Dave: 0
Al,
You say the twins can use the elapsed time shown on their clocks to determine who reaches the end point of the race first. In F0, both twins clocks read the same time t1 when they reach the transition to F1. Observers in F1 say, the two twins did not arrive in F1 simultaneously, so they did not reach the end point of the race simultaneously, since they each run the 100 meters in F1 at the same speed and elapsed time. Correct?
David Seppala
Bastrop TX

On Sunday, August 8, 2021 at 2:57:29 PM UTC-5, Al Coe wrote:
> On Sunday, August 8, 2021 at 11:42:44 AM UTC-7, sep...@yahoo.com wrote:
> > Please explain why the TWINS CANNOT use the elapsed times shown on each of
> > their clocks to determine who won the race.
> What is wrong with you? It's been explained to you several times now that, given your stipulation of constant coordinate speeds, the runners *can* use their elapsed proper times (as would be shown on clocks that they carry) to determine who won, because in that case there is a fixed proportionality between proper times and coordinate times. For example, in the scenarios you've described, the runners have tied, and they show equal elapsed proper times on their wrist watches.
>
> I've also explained why, if we more realistically allow the coordinate speeds to vary during the race, the simple proportionality no longer applies, even on a single segment in a single system of inertial coordinates, but this is clearly beyond you, since you can't even grasp the trivial case of constant speeds.
> > Each twin can use the elapsed time shown on his clock while in F0 to determine
> > which twin transitioned into frame F1 first.
> No, the co-moving clocks just show the elapsed proper times between when the respective runner starting and when he reaches the conveyor belt. This says nothing about the synchronization between those clocks, i.e., which things "happened first". They could even be running on separate days. You are free to synchronize them using the simultaneity of F0, or you could synchronize them using the simultaneity of F1, or you could synchronize them using any other simultaneity you like.
> > Each twin can use the elapsed time shown on his clock to determine which twin
> > ran the 100 meters in F1 faster.
> Given your simplified assumption that each runner runs at a (possibly different) constant speed, yes, the elapsed proper times enable us to determine which runner was faster. This was explained to you before, remember? And this (again) does not tell us anything about what "happened first". We can initialize the clocks differently, to match the simultaneity of F0, or of F1, or of any other system we like.
> > Why can't they use the total elapsed time shown from the beginning of the race to
> > the end the race to determine if they meet at the end point of the race simultaneously?
> They can (given your simplifying assumption). This is precisely what I've been explaining to you. In the scenario you described, it is trivial to determine that the runners do not arrive at the end of their respective runs at the same event, and it is trivial to determine that the completions of their respective runs are not simultaneous in terms of either F0 or F1, and they do not end at the same place in terms of F0, and they do not reach the conveyor belt at the same time in terms of F1, and so on. We can precisely specify when and where everything occurs in terms of F0 and in terms of F1 and in terms of any other specified system of coordinates. Do you understand this?
> > We know that the two twins do not end up at the finish point of the race in F1 simultaneously.
> Right.
> > Each clock showed the same elapsed time for the leg run in frame F0 and
> > each clock showed the same elapsed time for the leg run in frame F1.
> Right again (assuming equal constant speeds).
> > Each twin ran at exactly the same speed in each leg of the race.
> Moreover, you are stipulating they both ran at equal *constant* speed. That's important, because you started out just giving the elapsed time for each leg without specifying constant speed, which introduced a deep ambiguity as explained previously.
> > Each twin ran the exact same distance in each of the two frames.
> Right again, meaning that they ran the same coordinate distance in each respective leg, but of course when on the conveyor belt they aren't going the same distance in terms of F0, and on the first leg they aren't going the same distance in terms of F1. And they have different accelerations when stepping onto the conveyor belt.
> > Explain why running at the same speed, same elapsed times in each frame,
> > same distance in each frame, the two twins didn't end up at the same point
> > at the finish simultaneously.
> Already explained in detail multiple times. Again, in terms of F1 the runners reach the conveyor belt at different times, so they obviously reach the end spot at different times because they have the same speeds. On the other hand, in terms of F0 they reach the conveyor belt at the same time, but the end point is moving spatially toward one and away from the other, and their individual speeds are given by the relativistic composition, so again they reach the end spot at different times. The two descriptions are perfectly consistent, as are the descriptions we can give in terms of any other system of coordinates.
>
> Maybe it would help you to consider two runners going from starting event E1 to ending event E2, and ask if they must show the same elapsed times on their clocks. Obviously not. You understand this, right? This is just the standard twins scenario. The have the same elapsed coordinate time, but different elapsed proper times. So, if you understand that, you should be able to grasp conversely that for different paths we can have equal elapsed proper times but different elapsed coordinate times. Understand?
>
> > Your aside that their accelerations were different is a moot point.
> > We can say that they changed their direction...
>
> Changing direction is acceleration. Sheesh.
>
> Special Relativity: 798 ..... Barnpole Dave: 0
Al,
Both twins agree that they started simultaneously at the same point and time in F0. Both twins agree that when they finished the 100 meters each twin ran at the same speed in frame F0, both their clocks read the same time. Both twins agree that when they left frame F0, they both left F0 simultaneously. When they ended up in F1, do both twins agree that they landed simultaneously? If no, why not?
David Seppala
Bastrop TX

On Sunday, August 8, 2021 at 1:52:44 PM UTC-7, sep...@yahoo.com wrote:
> You say the twins can use the elapsed time shown on their clocks to determine who
> reaches the end point of the race first.

Sure, in fact, by dead reckoning, given the proper acceleration of every object as a function of proper time, it's possible to infer the time and place of every events in terms of any specified system of coordinates.

> In F0, both twins clocks read the same time t1 when they reach the transition to F1.

Not necessarily. They will read that way if you have initialized them so that they are synchronized in terms of F0, but of course they will not read that way if you initialize them so that they are synchronized in some other way (e.g., in terms of F1). You just appear to have no grasp at all of the relativity of simultaneity.

> Observers in F1 say, the two twins did not arrive in F1 simultaneously...

Stated correctly: The two runners do not arrive at the conveyor belt simultaneously in terms of the inertial coordinate system that you are calling F1.

> so they did not reach the end point of the race simultaneously, since they each run the
> 100 meters in F1 at the same speed and elapsed time. Correct?

Making allowances, yes, you have given a slightly muddled paraphrase of what I said, i.e., in terms of F1 they step onto the conveyor belt at different times, and they have equal speeds (opposite direction) to the end point that is equidistant, so they arrive at the end point at different times.

> Both twins agree that they started simultaneously at the same point and time in F0.

It would help you to stop talk about what people agree, just state the facts: You are stipulating that the runners begin at the same event (i.e., same time and place), which is a frame-invariant fact.

> Both twins agree that when they finished the 100 meters each twin ran at the
> same speed in frame F0, both their clocks read the same time.

State the facts: You are stipulating that both runners run at the same speed V for 100 meters in terms of F0. Each clock will show the same amount difference between what it showed at the start and what it shows at the 100 meter event. Whether they read the same at those events depends on whether they read the same at the start.

> Both twins agree that when they left frame F0, they both left F0 simultaneously.

No, that is not a fact. The runners reach the conveyor belt simultaneously in terms of F0, and not simultaneously in terms of F1. Every rational person agrees with this. [Please note, since your sentences are overflowing with fallacies and misconceptions, and we could not progress at all if I stopped to correct each one, and you would ignore the corrections anyway, I'm just going to disregard your idiotic talking about "leaving frames" and "entering frames". Suffice to say that you will never really understand special relativity until you stop saying things like that.]

> When they ended up in F1, do both twins agree that they landed simultaneously? If no, why not?

Again, every rational person agrees that the runners reach the conveyor belt simultaneously in terms of F0 and not simultaneously in terms of F1 (and not in terms of infinitely many other systems of inertial coordinates). You see, the coordinate axes are mutually skewed, so events that are at equal t are generally not at equal t'. This is called the relativity of simultaneity. Understand?

Special Relativity: 799 ..... Barnpole Dave: 0

On Sunday, August 8, 2021 at 5:01:13 PM UTC-5, Al Coe wrote:
> On Sunday, August 8, 2021 at 1:52:44 PM UTC-7, sep...@yahoo.com wrote:
> > You say the twins can use the elapsed time shown on their clocks to determine who
> > reaches the end point of the race first.
>
> Sure, in fact, by dead reckoning, given the proper acceleration of every object as a function of proper time, it's possible to infer the time and place of every events in terms of any specified system of coordinates.
> > In F0, both twins clocks read the same time t1 when they reach the transition to F1.
> Not necessarily. They will read that way if you have initialized them so that they are synchronized in terms of F0, but of course they will not read that way if you initialize them so that they are synchronized in some other way (e.g., in terms of F1). You just appear to have no grasp at all of the relativity of simultaneity.
>
> > Observers in F1 say, the two twins did not arrive in F1 simultaneously....
>
> Stated correctly: The two runners do not arrive at the conveyor belt simultaneously in terms of the inertial coordinate system that you are calling F1.
> > so they did not reach the end point of the race simultaneously, since they each run the
> > 100 meters in F1 at the same speed and elapsed time. Correct?
> Making allowances, yes, you have given a slightly muddled paraphrase of what I said, i.e., in terms of F1 they step onto the conveyor belt at different times, and they have equal speeds (opposite direction) to the end point that is equidistant, so they arrive at the end point at different times.
> > Both twins agree that they started simultaneously at the same point and time in F0.
> It would help you to stop talk about what people agree, just state the facts: You are stipulating that the runners begin at the same event (i.e., same time and place), which is a frame-invariant fact.
> > Both twins agree that when they finished the 100 meters each twin ran at the
> > same speed in frame F0, both their clocks read the same time.
> State the facts: You are stipulating that both runners run at the same speed V for 100 meters in terms of F0. Each clock will show the same amount difference between what it showed at the start and what it shows at the 100 meter event. Whether they read the same at those events depends on whether they read the same at the start.
> > Both twins agree that when they left frame F0, they both left F0 simultaneously.
> No, that is not a fact. The runners reach the conveyor belt simultaneously in terms of F0, and not simultaneously in terms of F1. Every rational person agrees with this. [Please note, since your sentences are overflowing with fallacies and misconceptions, and we could not progress at all if I stopped to correct each one, and you would ignore the corrections anyway, I'm just going to disregard your idiotic talking about "leaving frames" and "entering frames". Suffice to say that you will never really understand special relativity until you stop saying things like that.]
> > When they ended up in F1, do both twins agree that they landed simultaneously? If no, why not?
> Again, every rational person agrees that the runners reach the conveyor belt simultaneously in terms of F0 and not simultaneously in terms of F1 (and not in terms of infinitely many other systems of inertial coordinates). You see, the coordinate axes are mutually skewed, so events that are at equal t are generally not at equal t'. This is called the relativity of simultaneity. Understand?
>
> Special Relativity: 799 ..... Barnpole Dave: 0

Al,
You said, in reply to my post of >> In F0, both twins clocks read the same time t1 when they reach the transition to F1.<<
"Not necessarily. They will read that way if you have initialized them so that they are synchronized in terms of F0, but of course they will not read that way if you initialize them so that they are synchronized in some other way (e.g., in terms of F1). You just appear to have no grasp at all of the relativity of simultaneity."
I said in the current scenario, that both twins start at the same point in F0 at the same time. They both travel at the same speed, going toward, one at a positive 45 degrees angle with wrt to the x-axis, the other at 45 degree angle in the opposite direction. Both twins travel 100 meters in F0 at the same speed. When they arrive at the 100 meter marking, each twin shows an elapsed time of 10 seconds. Do the twins consider their clocks synchronized when they each reach 100 meters in F0?
David Seppala
Bastrop TX

On Sunday, August 8, 2021 at 3:17:45 PM UTC-7, sep...@yahoo.com wrote:
> > > In F0, both twins clocks read the same time t1 when they reach the transition to F1.
> > Not necessarily. They will read that way if you have initialized them so that they are
> > synchronized in terms of F0, but of course they will not read that way if you initialize
> > them so that they are synchronized in some other way (e.g., in terms of F1).
>
> I said in the current scenario, that both twins start at the same point in F0 at the same time.

Right, and if you set the clocks equal at that event, and if they subsequently move symmetrically in terms of F0, then they will be synchronized in terms of F0, but they will not be synchronized in terms of F1 (because they are not moved symmetrically in terms of F1). Do you understand this?

> When they arrive at the 100 meter marking, each twin shows an elapsed time of 10 seconds.
> Do the twins consider their clocks synchronized when they each reach 100 meters in F0?

Again, you need to stop talking about what various people "consider" and just state the facts. And you need to grasp the distinction between changes in values and the actual values. You have noted that each clock shows the same change in its value from the starting event to the conveyor belt, but this does not specify that the clocks have equal values. That's an independent stipulation. If you initialize the clocks to have equal values at the starting event, and then move them symmetrically in terms of F0, they will continue to be synchronized in terms of F0, but they will not be synchronized in terms of F1 or any other system of inertial coordinates. On the other hand, if after initializing them to be equal at the starting event you move them symmetrically in terms of F1, they will remain synchronized in terms of F1. You could also initialize them at the starting event in such a way that after the motions symmetrical in terms of F0 they are synchronized in terms of F1.

Again, the point is that the runners arrive at the conveyor belt simultaneously in terms of F0 and not simultaneously in terms of F1. Do you understand this now?

Special Relativity: 800 ..... Barnpole Dave: 0

On Sunday, August 8, 2021 at 5:33:31 PM UTC-5, Al Coe wrote:
> On Sunday, August 8, 2021 at 3:17:45 PM UTC-7, sep...@yahoo.com wrote:
> > > > In F0, both twins clocks read the same time t1 when they reach the transition to F1.
> > > Not necessarily. They will read that way if you have initialized them so that they are
> > > synchronized in terms of F0, but of course they will not read that way if you initialize
> > > them so that they are synchronized in some other way (e.g., in terms of F1).
> >
> > I said in the current scenario, that both twins start at the same point in F0 at the same time.
> Right, and if you set the clocks equal at that event, and if they subsequently move symmetrically in terms of F0, then they will be synchronized in terms of F0, but they will not be synchronized in terms of F1 (because they are not moved symmetrically in terms of F1). Do you understand this?
> > When they arrive at the 100 meter marking, each twin shows an elapsed time of 10 seconds.
> > Do the twins consider their clocks synchronized when they each reach 100 meters in F0?
> Again, you need to stop talking about what various people "consider" and just state the facts. And you need to grasp the distinction between changes in values and the actual values. You have noted that each clock shows the same change in its value from the starting event to the conveyor belt, but this does not specify that the clocks have equal values. That's an independent stipulation. If you initialize the clocks to have equal values at the starting event, and then move them symmetrically in terms of F0, they will continue to be synchronized in terms of F0, but they will not be synchronized in terms of F1 or any other system of inertial coordinates. On the other hand, if after initializing them to be equal at the starting event you move them symmetrically in terms of F1, they will remain synchronized in terms of F1. You could also initialize them at the starting event in such a way that after the motions symmetrical in terms of F0 they are synchronized in terms of F1.
>
> Again, the point is that the runners arrive at the conveyor belt simultaneously in terms of F0 and not simultaneously in terms of F1. Do you understand this now?
> > Special Relativity: 800 ..... Barnpole Dave: 0

If the twins start at x=0 in F0 and travel different directions along the x-axis, but travel at the same speed relative to F0, one twin says that time for the other twin is running slower than his own. Say twin 1 compares his time to the clock at rest at x=0. He determines that the clock at rest at x=0 is running say at half the rate of his own clock. Twin 2 compares his time to the clock at rest at x=0. He also determines that the clock at x=0 is running at half the rate of his own clock. Twin 1 says that when he passed x = -L his clock read T seconds. Twin 2 says that when he x=L his clock read T seconds. When twin 1 passes x = -10L he stops (zero speed relative to F0) and when twin 2 passes x=10L he stops (zero speed relative to F0). During the journey, per Einstein, each twin says the other twins clock was running slower. Even if one twin says he stops before the other twin, per Einstein, he still measures that his clock is running slower than the other twin's clock. Why don't the twins have to resync their clocks once they stop? If there were no clocks in F0 other than the clocks the two twins had, how must they adjust their times?
David Seppala
Bastrop TX

On Sunday, August 8, 2021 at 4:21:14 PM UTC-7, sep...@yahoo.com wrote:
> If the twins start at x=0 in F0 and travel different directions along the x-axis, but travel at
> the same speed relative to F0, one twin says that time for the other twin is running slower
> than his own.

Stated correctly: Each clock runs slow in terms of the inertial coordinates in which the other clock is at rest. Also, each clock runs equally slow (but not as slow) in terms of F0, and unequally slow in terms of F1.

> Say twin 1 compares his time to the clock at rest at x=0.

Stated correctly: The clock 1 runs slow in terms of F0. (Distant clocks to not compare to each other unambiguously.)

> He determines that the clock at rest at x=0 is running say at half the rate of his own clock.

Stated correctly: Suppose the clock is moving in terms of F0 such that the rate of the clock is half the rate of the time coordinate of F0.

> Twin 2 compares his time to the clock at rest at x=0. He also determines that the clock
> at x=0 is running at half the rate of his own clock.

Corresponding corrections as above.

> Twin 1 says that when he passed x = -L his clock read T seconds. Twin 2 says that when
> he x=L his clock read T seconds.

Yawn.

> When twin 1 passes x = -10L he stops (zero speed relative to F0) and when twin 2
> passes x=10L he stops (zero speed relative to F0).

Wider yawn.

> During the journey, per [special relativity], each twin says the other twins clock was
> running slower.

Just state the facts: Each clock runs slow in terms of the inertial coordinates in which the other is at rest. This is an objective fact.

> Even if one twin says he stops before the other twin, per [special relativity], he still
> measures that his clock is running slower than the other twin's clock.

That's garbled. Again, in terms of any system of inertial coordinates, a clock moving at speed v runs slow by the factor sqrt(1-v^2/c^2). This is an objective fact, and everyone agrees on this, and it explains to correct version of every one of your garbled statements. Understand?

> Why don't the twins have to resync their clocks once they stop?

Anyone can set their clocks any way they like. Do you understand this? Nothing forces anyone to initialize or synchronize separate clocks in any particular way. Understand?

> If there were no clocks in F0 other than the clocks the two twins had, how must they adjust their times?

You are typing complete gibberish now. There is no "must adjust" here. There are different systems of local inertial coordinates, with their well-defined operational meanings, for any state of motion, and you can set clocks to match any of these, or none of these, at your discretion. Nothing forces you to use any particular system of coordinates (including the inertial coordinates in which you are momentarily at rest at any given moment). In fact, we typically do not use coordinate systems in which we are at rest. Do you understand now why your statements are completely senseless?

It looks a lot like a same person with two different accounts is discussing with himself or herself.

On Monday, 9 August 2021 at 00:33:31 UTC+2, Al Coe wrote:

> Right, and if you set the clocks equal at that event, and if they subsequently move symmetrically in terms of F0, then they will be synchronized in terms of F0, but they will not be synchronized in terms of F1 (because they are not moved symmetrically in terms of F1). Do you understand this?

They will not be synchronized only in the terms of some
newspeak of some brainwashed maniacs. Do you understand
this?

Richard Hertz <hertz778@gmail.com> wrote:
> It looks a lot like a same person with two different accounts is
> discussing with himself or herself.
>

Why do nutjobs on Usenet suspect this with such regularity.
Sepalla is well known, uses his real name, lives in Texas, and has for
years and years.

--
Odd Bodkin -- maker of fine toys, tools, tables