Why does a change in direction of motion change the time shown on clocks? Consider the following scenario.

In inertial reference frame F0 their are four clocks, A, B, C, and D. All four clocks are in the x,y plane of F0. Clock A is located at ( -L*sqrt(2)/2, 0). Clock D is located at (L*sqrt(2)/2, 0). Clocks B and C are located at (0,L*(1-sqrt(2)/2)). The clocks are all synchronized. At time t =0, all four clocks accelerate at an identical rate to speed V as measured in F0. Clocks A and D accelerate in the positive y direction, perpendicular to the x-axis. Clock B accelerates at a minus 45 degree angle relative to the x-axis in the positive y direction, and Clock C accelerates at a plus 45 degree angle relative to the x-axis in the positive y direction.
At time t = T as measured in F0, after clocks A and B each travel a distance L, they meet at (-L*sqrt(2)/2, L). Clocks A and B show the identical time when they meet at that point. At time t = T as measured in F0, after clocks C and D each travel a distance L, they meet at (L*sqrt(2)/2, L). Clocks C and D show the identical time when they meet at that point.
As measured in frame F0, and in frame F1 that is moving with speed V relative to frame F0 in the y direction, both clock A and clock D always show the identical times. Per Einstein, since clocks B and C have a relative velocity with respect to each other, each of their clocks run at a different rate. So when clocks A and B meet, clock B measures that clock C has run at a slower rate than clock B and therefore doesn't show the same time as clocks A and B.
The questions I have is what happens if Clock B changes its direction of motion when A and B meet so that it travels in the same direction as clock A and Clock C changes its direction of motion when it meets clock D to travel in the same direction as clock D? All the clocks now are in sync, all showing the same time. How did the difference in times shown on clocks B and C disappear? If only clock B changes its direction when it meets clock A while clock C continues moving in a straight line, then there is always a difference in times shown on clocks B and C. Can someone explain the "physics" in these two situations? What are the time differences if neither clock B and C change their direction of travel versus the time difference if only one of the clocks changes its direction of travel versus if both clocks B and C change their direction of travel.
The Lorentz transform shows that relativistic events can be observed at everyday velocities when using relativistic distances. So let's say the speed of the clocks as measured in F0 is 3 meters per second. When moving at 3 meters per second, why does a change in direction suddenly cause a distant clock to change its time. Why does the amount of time that changes depend on the distance L and not simply the acceleration that occurred?

Thanks,
David Seppala
Bastrop TX

On Thursday, August 19, 2021 at 8:39:18 AM UTC-7, sep...@yahoo.com wrote:
> Why does a change in direction of motion change the time shown on clocks?

It doesn't.

> In inertial reference frame F0 their are four clocks, A, B, C, and D. All four clocks are in the x,y plane of F0. Clock A is located at ( -L*sqrt(2)/2, 0). Clock D is located at (L*sqrt(2)/2, 0). Clocks B and C are located at (0,L*(1-sqrt(2)/2)). The clocks are all synchronized. At time t =0, all four clocks accelerate at an identical rate to speed V as measured in F0. Clocks A and D accelerate in the positive y direction, perpendicular to the x-axis. Clock B accelerates at a minus 45 degree angle relative to the x-axis in the positive y direction, and Clock C accelerates at a plus 45 degree angle relative to the x-axis in the positive y direction.

> At time t = T as measured in F0, after clocks A and B each travel a distance L, they meet at (-L*sqrt(2)/2, L). Clocks A and B show the identical time when they meet at that point. At time t = T as measured in F0, after clocks C and D each travel a distance L, they meet at (L*sqrt(2)/2, L). Clocks C and D show the identical time when they meet at that point.

> As measured in frame F0, and in frame F1 that is moving with speed V relative to frame F0 in the y direction, both clock A and clock D always show the identical times. Per Einstein, since clocks B and C have a relative velocity with respect to each other, each of their clocks run at a different rate.

As always, you have misunderstood the relativity of simultaneity. In terms of both F0 and F1, B and C have the same speed and tick at the same rate. They each run slow in terms of the inertial coordinates in which the other is at rest, but (when moving at speed V) neither of them is at rest in either F0 or F1. Also, as always, you were wrong to say "per Einstein", you should really say "per the diseased brain of David Seppala".

> So when clocks A and B meet, clock B measures that clock C has run at a slower rate than clock B and therefore doesn't show the same time as clocks A and B.

As always, you've again witlessly used the word "when" without grasping it's ambiguity in the context of the relativity of simultaneity. See above.

> The questions I have...

As always, your "questions" (which are really assertions of contradiction, passive-aggressively posed as "questions") are based on false premises, due to your misunderstanding of the relativity of simultaneity. See above.

> What happens if Clock B changes its direction of motion when A and B meet so that it travels in the same direction as clock A and Clock C changes its direction of motion when it meets clock D to travel in the same direction as clock D?

All the clocks read the same values at those events, and instantly changing directions does not change the values of any of the clocks.

>All the clocks now are in sync, all showing the same time. How did the difference in times shown on clocks B and C disappear?

They were never different. As always, you misunderstood the relativity of simultaneity. See above. Also, see each of your previous threads.

Special Relativity: 807 ..... Barnpole Dave: 0

On Thursday, August 19, 2021 at 11:54:04 AM UTC-5, Al Coe wrote:
> On Thursday, August 19, 2021 at 8:39:18 AM UTC-7, sep...@yahoo.com wrote:
> > Why does a change in direction of motion change the time shown on clocks?
> It doesn't.
> > In inertial reference frame F0 their are four clocks, A, B, C, and D. All four clocks are in the x,y plane of F0. Clock A is located at ( -L*sqrt(2)/2, 0). Clock D is located at (L*sqrt(2)/2, 0). Clocks B and C are located at (0,L*(1-sqrt(2)/2)). The clocks are all synchronized. At time t =0, all four clocks accelerate at an identical rate to speed V as measured in F0. Clocks A and D accelerate in the positive y direction, perpendicular to the x-axis. Clock B accelerates at a minus 45 degree angle relative to the x-axis in the positive y direction, and Clock C accelerates at a plus 45 degree angle relative to the x-axis in the positive y direction.
>
> > At time t = T as measured in F0, after clocks A and B each travel a distance L, they meet at (-L*sqrt(2)/2, L). Clocks A and B show the identical time when they meet at that point. At time t = T as measured in F0, after clocks C and D each travel a distance L, they meet at (L*sqrt(2)/2, L). Clocks C and D show the identical time when they meet at that point.
>
> > As measured in frame F0, and in frame F1 that is moving with speed V relative to frame F0 in the y direction, both clock A and clock D always show the identical times. Per Einstein, since clocks B and C have a relative velocity with respect to each other, each of their clocks run at a different rate.
> As always, you have misunderstood the relativity of simultaneity. In terms of both F0 and F1, B and C have the same speed and tick at the same rate. They each run slow in terms of the inertial coordinates in which the other is at rest, but (when moving at speed V) neither of them is at rest in either F0 or F1. Also, as always, you were wrong to say "per Einstein", you should really say "per the diseased brain of David Seppala".
> > So when clocks A and B meet, clock B measures that clock C has run at a slower rate than clock B and therefore doesn't show the same time as clocks A and B.
> As always, you've again witlessly used the word "when" without grasping it's ambiguity in the context of the relativity of simultaneity. See above.
>
> > The questions I have...
>
> As always, your "questions" (which are really assertions of contradiction, passive-aggressively posed as "questions") are based on false premises, due to your misunderstanding of the relativity of simultaneity. See above.
>
> > What happens if Clock B changes its direction of motion when A and B meet so that it travels in the same direction as clock A and Clock C changes its direction of motion when it meets clock D to travel in the same direction as clock D?
>
> All the clocks read the same values at those events, and instantly changing directions does not change the values of any of the clocks.
> >All the clocks now are in sync, all showing the same time. How did the difference in times shown on clocks B and C disappear?
> They were never different. As always, you misunderstood the relativity of simultaneity. See above. Also, see each of your previous threads.
>
> Special Relativity: 807 ..... Barnpole Dave: 0
You replied:
As always, you have misunderstood the relativity of simultaneity. In terms of both F0 and F1, B and C have the same speed and tick at the same rate. They each run slow in terms of the inertial coordinates in which the other is at rest, but (when moving at speed V) neither of them is at rest in either F0 or F1.

Clocks B and C have a relative velocity with respect to each other so you say "They run slow in terms of the inertial coordinates in which the other is at rest", so when Clock B reads t = T' in his inertial reference frame, what time does Clock C show as measured in the inertial reference frame of Clock B? Does it equal T', or is the clock C reading less than T', or is the clock C reading greater than T' as measured in the inertial reference frame of Clock B?

David Seppala
Bastrop TX

On Thursday, August 19, 2021 at 2:02:05 PM UTC-7, sep...@yahoo.com wrote:
> Clocks B and C have a relative velocity with respect to each other so you say "They run
> slow in terms of the inertial coordinates in which the other is at rest" so when Clock B
> reads t = T' in his inertial reference frame, what time does Clock C show as measured in
> the inertial reference frame of Clock B?

You have not defined (or even previously mentioned) the parameter T', so I can only take it as an arbitrary reading on clock B assuming it was set to 0 when coinciding with C and abruptly being given the speed V at 45 deg, meaning the relative speed between B and C is w = Vsqrt(2)/(1+V^2/2), and hence in terms of the inertial coordinates in which B is at rest, the clock C reads T' sqrt(1-w^2) when clock B reads T'.

Needless to say, the meeting of D and C is not simultaneous with the meeting of of A and B in terms of this system of coordinates, so if T' is the reading on B at its meeting with A, then T' sqrt(1-w^2) is *not* the reading on C at its meeting with D. Duh. In that case C reads T' at its meeting with D, and in terms of C's inertial rest coordinates B is reading T' sqrt(1-w^2) at that time.

Special Relativity: 809 ...... Barnpole Dave: 0

On Thursday, August 19, 2021 at 4:31:28 PM UTC-5, Al Coe wrote:
> On Thursday, August 19, 2021 at 2:02:05 PM UTC-7, sep...@yahoo.com wrote:
> > Clocks B and C have a relative velocity with respect to each other so you say "They run
> > slow in terms of the inertial coordinates in which the other is at rest" so when Clock B
> > reads t = T' in his inertial reference frame, what time does Clock C show as measured in
> > the inertial reference frame of Clock B?
> You have not defined (or even previously mentioned) the parameter T', so I can only take it as an arbitrary reading on clock B assuming it was set to 0 when coinciding with C and abruptly being given the speed V at 45 deg, meaning the relative speed between B and C is w = Vsqrt(2)/(1+V^2/2), and hence in terms of the inertial coordinates in which B is at rest, the clock C reads T' sqrt(1-w^2) when clock B reads T'.
>
> Needless to say, the meeting of D and C is not simultaneous with the meeting of of A and B in terms of this system of coordinates, so if T' is the reading on B at its meeting with A, then T' sqrt(1-w^2) is *not* the reading on C at its meeting with D. Duh. In that case C reads T' at its meeting with D, and in terms of C's inertial rest coordinates B is reading T' sqrt(1-w^2) at that time.
>
> Special Relativity: 809 ...... Barnpole Dave: 0

You said, "Needless to say, the meeting of D and C is not simultaneous with the meeting of of A and B in terms of this system of coordinates"
When you say "this system of coordinates", which system are you referring to? They are simultaneous in terms of the rest frame of clock A and clock D. If clock B starts accelerating 1 second before clock C, and clock C is always running slower than clock B, and Clock B changes its direction when clock A and clock B meet so that it has zero velocity with respect to clock A and clock D, what points in time and space did clock C suddenly gain time as measured by clock B travelers?

David Seppala
Bastrop TX

On Thursday, August 19, 2021 at 3:37:20 PM UTC-7, sep...@yahoo.com wrote:
> You said, "Needless to say, the meeting of D and C is not simultaneous with the meeting of of
> A and B in terms of this system of coordinates". When you say "this system of coordinates",
> which system are you referring to?

The system of coordinates that was specified in the preceding sentence, i.e.., "the inertial coordinates in which B is at rest".

> If Clock B changes its direction when clock A and clock B meet so that it
> has zero velocity with respect to clock A and clock D, what points in time and
> space did clock C suddenly gain time as measured by clock B travelers?

C does not "suddenly gain time as measured by B". The phrase "as measured by X" that you insist on using appears in the relatively literature only as sloppy (and somewhat misleading) shorthand for "in terms of the inertial coordinate system in which X is at rest". For example, if B abruptly accelerates so that it is at rest with A, it is then at rest in terms of an inertial coordinate system in which the meeting of A and B is simultaneous with the meeting of C and D. In the instant of acceleration, B's proper time doesn't advance, and it reads what you call T', and that event is simultaneous with the event at which C reads T' sqrt(1-w^2) in terms of the inertial coordinates in which B was at rest prior to the acceleration, and it is simultaneous with the event at which C reads T' in terms of the inertial coordinates in which B is at rest after the acceleration. This is the relativity of simultaneity.

Special Relativity: 810 ..... Barnpole Dave: 0

On Thursday, August 19, 2021 at 8:39:18 AM UTC-7, sep...@yahoo.com wrote:
> Why does a change in direction of motion change the time shown on clocks? Consider the following scenario.
>
> In inertial reference frame F0 their are four clocks, A, B, C, and D. All four clocks are in the x,y plane of F0. Clock A is located at ( -L*sqrt(2)/2, 0). Clock D is located at (L*sqrt(2)/2, 0). Clocks B and C are located at (0,L*(1-sqrt(2)/2)). The clocks are all synchronized. At time t =0, all four clocks accelerate at an identical rate to speed V as measured in F0. Clocks A and D accelerate in the positive y direction, perpendicular to the x-axis. Clock B accelerates at a minus 45 degree angle relative to the x-axis in the positive y direction, and Clock C accelerates at a plus 45 degree angle relative to the x-axis in the positive y direction.
> At time t = T as measured in F0, after clocks A and B each travel a distance L, they meet at (-L*sqrt(2)/2, L). Clocks A and B show the identical time when they meet at that point. At time t = T as measured in F0, after clocks C and D each travel a distance L, they meet at (L*sqrt(2)/2, L). Clocks C and D show the identical time when they meet at that point.
> As measured in frame F0, and in frame F1 that is moving with speed V relative to frame F0 in the y direction, both clock A and clock D always show the identical times. Per Einstein, since clocks B and C have a relative velocity with respect to each other, each of their clocks run at a different rate. So when clocks A and B meet, clock B measures that clock C has run at a slower rate than clock B and therefore doesn't show the same time as clocks A and B.
> The questions I have is what happens if Clock B changes its direction of motion when A and B meet so that it travels in the same direction as clock A and Clock C changes its direction of motion when it meets clock D to travel in the same direction as clock D? All the clocks now are in sync, all showing the same time. How did the difference in times shown on clocks B and C disappear? If only clock B changes its direction when it meets clock A while clock C continues moving in a straight line, then there is always a difference in times shown on clocks B and C. Can someone explain the "physics" in these two situations? What are the time differences if neither clock B and C change their direction of travel versus the time difference if only one of the clocks changes its direction of travel versus if both clocks B and C change their direction of travel.
> The Lorentz transform shows that relativistic events can be observed at everyday velocities when using relativistic distances. So let's say the speed of the clocks as measured in F0 is 3 meters per second. When moving at 3 meters per second, why does a change in direction suddenly cause a distant clock to change its time. Why does the amount of time that changes depend on the distance L and not simply the acceleration that occurred?
>
> Thanks,
> David Seppala
> Bastrop TX

Gamma is not influenced by direction...
only atomic speed matters there...

Mitchell Raemsch