Can anyone explain the time concept of relativity when two inertial reference frames continually measure the same elapsed time between events while also concluding that clocks in the two inertial reference frames are running at different rates?
For example, let there be three inertial reference frames, F0, F1 and F2. Let F1 have a velocity V= sqrt(3)/2 *c relative to F0 moving in the positive x direction and let F2 have a velocity V=sqrt(3)/2*c relative to F0 moving in the negative direction. Now at x=0 in F0, frames F1 and F2 have clocks that read t' = t'' = 0. At the point in space, both frame F1 and F2 observers place a clock that runs at twice the rate of a standard clock into frame F0 at x=0. Now each future event that occurs at x=0 in F0, observers in F1 and F2 measure that the same elapsed time has occurred as measured in their respective inertial reference frames. They also observe that both clocks that were placed in F0 at x=0 match the time shown on each clock that they pass in their reference frames, and the both of these clocks show the identical time. Yet even though these clocks show the same time and run at the same rate and match the clock reading of every clock they pass in frame F1 and F2, F1 and F2 observers say that clocks in the two frames do not run at the same rate. Please clarify that concept.
Thanks,
David Seppala
Bastrop TX

On Thursday, August 4, 2022 at 4:06:40 PM UTC-7, sep...@yahoo.com wrote:
> Can anyone explain the time concept of relativity when two inertial reference frames continually measure the same elapsed time between events while also concluding that clocks in the two inertial reference frames are running at different rates?
> For example, let there be three inertial reference frames, F0, F1 and F2. Let F1 have a velocity V= sqrt(3)/2 *c relative to F0 moving in the positive x direction and let F2 have a velocity V=sqrt(3)/2*c relative to F0 moving in the negative direction. Now at x=0 in F0, frames F1 and F2 have clocks that read t' = t'' = 0. At the point in space, both frame F1 and F2 observers place a clock that runs at twice the rate of a standard clock into frame F0 at x=0. Now each future event that occurs at x=0 in F0, observers in F1 and F2 measure that the same elapsed time has occurred as measured in their respective inertial reference frames. They also observe that both clocks that were placed in F0 at x=0 match the time shown on each clock that they pass in their reference frames, and the both of these clocks show the identical time. Yet even though these clocks show the same time and run at the same rate and match the clock reading of every clock they pass in frame F1 and F2, F1 and F2 observers say that clocks in the two frames do not run at the same rate. Please clarify that concept.
> Thanks,
> David Seppala
> Bastrop TX

The problem with relativity is the contradiction of a Twin Paradox.
Only slow time with the other twin faster exists.
Each doesn't see the other slower. One is slow one is fast by moving in space.
Why doesn't both see each other faster instead?
Where has the Twin Paradox been measured. Where is evidence for the claim?

On Thursday, August 4, 2022 at 4:06:40 PM UTC-7, sep...@yahoo.com wrote:
> Can anyone explain the time concept of relativity when two inertial reference frames
> continually measure the same elapsed time between events while also concluding
> that clocks in the two inertial reference frames are running at different rates?

Your question is garbled and based on false premises. For two relatively moving objects, the rate of elapsed proper time for each object runs slow in terms of the standard inertial coordinate system in which the other object is at rest. You can verify this using two rows of clocks, each row inertially synchronized in its own frame, moving past each other in opposite directions, and note that the elapsed time on each clock between passing two consecutive clocks in the other row is less than the difference between the times on those other clocks at those passings.

> Let there be three inertial reference frames, F0, F1 and F2. Let F1 have a velocity
> V= sqrt(3)/2 *c relative to F0 moving in the positive x direction and let F2 have a
> velocity V=sqrt(3)/2*c relative to F0 moving in the negative direction. Now at x=0
> in F0, frames F1 and F2 have clocks that read t' = t'' = 0. At the point in space, both
> frame F1 and F2 observers place a clock that runs at twice the rate of a standard
> clock into frame F0 at x=0.

To clarify, you are taking a clock that is constructed to run at twice the normal rate (in terms of its own rest frame), and you are placing it at rest in F0 at x=0. There is not point in placing multiple such clocks.

> Now each future event that occurs at x=0 in F0, observers in F1 and F2 measure
> that the same elapsed time has occurred as measured in their respective inertial
> reference frames.

To clarify, you are noting that dt/dt' = dt/dt" at constant x, which for the conditions you've specified is true. You could just begin with this banal observation, and dispense with all the rigamorole.

> They also observe that both clocks that were placed in F0 at x=0 match the time
> shown on each clock that they pass in their reference frames, and the both of these
> clocks show the identical time.

Right. And there was no reason to have two of them, since they are identically constructed and sitting side by side. Yawn.

> Yet even though these clocks show the same time and run at the same rate and
> match the clock reading of every clock they pass in frame F1 and F2, F1 and F2
> observers say that clocks in the two frames do not run at the same rate.

This was explained up above. For two rows of clocks, at rest and inertially synchronized in terms of F1 and F2, the elapsed time on each clock between passing two consecutive clocks in the other row is less than the difference in the times on those clocks as they pass. The same is true for rows of clocks in F0 and F1, and for rows at rest in F0 and F2. And of course this implies dt/dt' = dt/dt" at constant x, as explained above, i.e., the elapsed time on a clock at rest in F0 between passing two consecutive clocks at rest in F1 is the same as for two consecutive clocks in F2.

On 05-Aug-22 9:06 am, sepp623@yahoo.com wrote:
> Can anyone explain the time concept of relativity when two inertial reference frames continually measure the same elapsed time between events while also concluding that clocks in the two inertial reference frames are running at different rates?
> For example, let there be three inertial reference frames, F0, F1 and F2. Let F1 have a velocity V= sqrt(3)/2 *c relative to F0 moving in the positive x direction and let F2 have a velocity V=sqrt(3)/2*c relative to F0 moving in the negative direction. Now at x=0 in F0, frames F1 and F2 have clocks that read t' = t'' = 0. At the point in space, both frame F1 and F2 observers place a clock that runs at twice the rate of a standard clock into frame F0 at x=0. Now each future event that occurs at x=0 in F0, observers in F1 and F2 measure that the same elapsed time has occurred as measured in their respective inertial reference frames. They also observe that both clocks that were placed in F0 at x=0 match the time shown on each clock that they pass in their reference frames, and the both of these clocks show the identical time. Yet even though these clocks show the same time and run at the same rate and match the clock reading of every clock they pass in frame F1 and F2, F1 and F2 observers say that clocks in the two frames do not run at the same rate. Please clarify that concept.
> Thanks,
> David Seppala
> Bastrop TX
Yet another in a long line of your spurious attempts to show a
contradiction in special relativity.
Apply the Lorentz transform, and all will become clear.
Sylvia.

On Thursday, August 4, 2022 at 8:58:16 PM UTC-5, Sylvia Else wrote:
> On 05-Aug-22 9:06 am, sep...@yahoo.com wrote:
> > Can anyone explain the time concept of relativity when two inertial reference frames continually measure the same elapsed time between events while also concluding that clocks in the two inertial reference frames are running at different rates?
> > For example, let there be three inertial reference frames, F0, F1 and F2. Let F1 have a velocity V= sqrt(3)/2 *c relative to F0 moving in the positive x direction and let F2 have a velocity V=sqrt(3)/2*c relative to F0 moving in the negative direction. Now at x=0 in F0, frames F1 and F2 have clocks that read t' = t'' = 0. At the point in space, both frame F1 and F2 observers place a clock that runs at twice the rate of a standard clock into frame F0 at x=0. Now each future event that occurs at x=0 in F0, observers in F1 and F2 measure that the same elapsed time has occurred as measured in their respective inertial reference frames. They also observe that both clocks that were placed in F0 at x=0 match the time shown on each clock that they pass in their reference frames, and the both of these clocks show the identical time. Yet even though these clocks show the same time and run at the same rate and match the clock reading of every clock they pass in frame F1 and F2, F1 and F2 observers say that clocks in the two frames do not run at the same rate. Please clarify that concept.
> > Thanks,
> > David Seppala
> > Bastrop TX
> Yet another in a long line of your spurious attempts to show a
> contradiction in special relativity.
>
> Apply the Lorentz transform, and all will become clear.
>
> Sylvia.
When I apply the Lorentz transform that's when it becomes unclear. As stated in original post let F1 nd F2 have a velocity V=c*sqrt(3/2) relative to F0 but in opposite directions. Let there be two events in F0 at x=0, one that occurs 10 seconds later than the other as measured in F0. Observers in F1 and F2 each put a clock running at twice the standard rate at x=0 in F0 when event A occurs. When event B occurs at that same location in F0, but 10 seconds later the Lorentz transform gives the elapsed time between the two events as measured in frame F1 to be:
elapsed time in F1 = 2 * (10 seconds)
and the elapsed time in F2 between events A and B to be
2 * (10 seconds).
The two clocks running at twice the standard rate each show 20 seconds has elapsed. Observers in F1 measure that 20 seconds has elapsed between the two events and observers in F2 measure that 20 seconds has elapsed between the two events. Yet they don't agree that the F1 clocks are running at the same rate as the F2 clocks.
That is what is unclear clear to me. Please explain the logic.
Thanks,
David Seppala
Bastrop TX

On Thursday, August 4, 2022 at 6:58:16 PM UTC-7, Sylvia Else wrote:
> On 05-Aug-22 9:06 am, sep...@yahoo.com wrote:
> > Can anyone explain the time concept of relativity when two inertial reference frames continually measure the same elapsed time between events while also concluding that clocks in the two inertial reference frames are running at different rates?
> > For example, let there be three inertial reference frames, F0, F1 and F2. Let F1 have a velocity V= sqrt(3)/2 *c relative to F0 moving in the positive x direction and let F2 have a velocity V=sqrt(3)/2*c relative to F0 moving in the negative direction. Now at x=0 in F0, frames F1 and F2 have clocks that read t' = t'' = 0. At the point in space, both frame F1 and F2 observers place a clock that runs at twice the rate of a standard clock into frame F0 at x=0. Now each future event that occurs at x=0 in F0, observers in F1 and F2 measure that the same elapsed time has occurred as measured in their respective inertial reference frames. They also observe that both clocks that were placed in F0 at x=0 match the time shown on each clock that they pass in their reference frames, and the both of these clocks show the identical time. Yet even though these clocks show the same time and run at the same rate and match the clock reading of every clock they pass in frame F1 and F2, F1 and F2 observers say that clocks in the two frames do not run at the same rate. Please clarify that concept.
> > Thanks,
> > David Seppala
> > Bastrop TX
> Yet another in a long line of your spurious attempts to show a
> contradiction in special relativity.
>
> Apply the Lorentz transform, and all will become clear.
>
> Sylvia.

There is a contradiction sylvia... the dumb.

Mitchell Raemsch

On Friday, August 5, 2022 at 7:59:31 AM UTC-7, sep...@yahoo.com wrote:
> Observers in F1 measure that 20 seconds has elapsed between the two
> events and observers in F2 measure that 20 seconds has elapsed between
> the two events. Yet they don't agree that the F1 clocks are running at the
> same rate as the F2 clocks. That is what is unclear clear to me. Please explain the logic.

What part of the careful and complete explanation didn't you understand?

Perhaps you don't even understand enough to know what part of the explanation you don't understand, so let's start with the basics: Letting S1 and S2 denote the standard inertial coordinates in which F1 and F2 are at rest, do you understand that a standard clock at rest in S1 runs slow in terms of S2, and a standard clock at rest in S2 runs slow in terms of S1?

Symbolically, letting t1 and t2 denote the time coordinates of S1 and S2 respectively, we have dtau1/dt1 = dtau2/dt2 = 1 and dtau1/dt2 = dtau2/dt1 = sqrt(1-v^2). Do you understand this?

On Friday, August 5, 2022 at 11:33:04 AM UTC-7, Stan Fultoni wrote:
> > Observers in F1 measure that 20 seconds has elapsed between the two
> > events and observers in F2 measure that 20 seconds has elapsed between
> > the two events. Yet they don't agree that the F1 clocks are running at the
> > same rate as the F2 clocks. That is what is unclear clear to me. Please
> > explain the logic.

Making allowances for your disability, let me give you the explanation in simple words: Yes, a clock at constant x runs at the same rate in terms of S1 and S2, but a clock at constant x' does not, nor does a clock at constant x". Again, for any two objects with relative speed v, each clock runs slow by the factor sqrt(1-v^2/c^2) in terms of the standard inertial coordinates in which the other is at rest. This was explained to you explicitly in terms of three rows of clocks, each row being inertially synchronized in its own frame. There is nothing tricky or magical about this, once you grasp the skew of simultaneity. You can write down the exactly readings of every clock at each and every event. Understand?

On Friday, August 5, 2022 at 1:57:01 PM UTC-5, Stan Fultoni wrote:
> On Friday, August 5, 2022 at 11:33:04 AM UTC-7, Stan Fultoni wrote:
> > > Observers in F1 measure that 20 seconds has elapsed between the two
> > > events and observers in F2 measure that 20 seconds has elapsed between
> > > the two events. Yet they don't agree that the F1 clocks are running at the
> > > same rate as the F2 clocks. That is what is unclear clear to me. Please
> > > explain the logic.
> Making allowances for your disability, let me give you the explanation in simple words: Yes, a clock at constant x runs at the same rate in terms of S1 and S2, but a clock at constant x' does not, nor does a clock at constant x". Again, for any two objects with relative speed v, each clock runs slow by the factor sqrt(1-v^2/c^2) in terms of the standard inertial coordinates in which the other is at rest. This was explained to you explicitly in terms of three rows of clocks, each row being inertially synchronized in its own frame. There is nothing tricky or magical about this, once you grasp the skew of simultaneity. You can write down the exactly readings of every clock at each and every event. Understand?

If F1 observers place a clock (which I'll call Clock1) that runs at twice the standard rate into F0 at x=0, that clock will show the exact elapsed time for every event that occurs in frame F0 at x0 that standard clocks in F1 show as the elapsed time, since |V| = c*sqrt(3)/2. Likewise if F2 observers place a clock (which I'll call Clock2) that runs at twice the standard rate into F0 at x=0, that clock will show the exact elapsed time for every event that occurs in frame F0 at x0 that the standard clocks in F2 show as the elapsed time, since again |V|=c*sqrt(3)/2. So those two clocks now in F0 always show the same time, and always match the elapsed time between events that occur in F0 at x=0 as measured in each in either F1 or F2, but F1 and F2 observers do not agree that clocks in their respective frames run at the same rate even though observers in both frames agree that they measure they measure identical times for events that occur in F0 at x=0. So how can the two frames measure the identical elapsed times between these events if their clocks aren't running at the same rate?
If event A0 occurs in F0 at x=0 Clock1 reads 0 and Clock 2 reads 0.
If event A1 occurs in F0 at x=0 when Clock1 reads 10, Clock 2 also reads 10. Observers in F1 measure the time between events A0 and A1 is 10 seconds. Observers in F2 measure the time between the two events is also 10 seconds. So Clock 1 and Clock 2 both measure the identical elapsed time between events A0 and A1 as clocks in F1 and F2 measure.
If event A2 occurs in F0 at x=0 when Clock1 reads 30, Clock 2 also reads 30. Observers in F1 measure the time between events A0 and A1 is 30 seconds. Observers in F2 measure the time between the two events is also 30 seconds.
So why doesn't this data indicate that clocks in frames F1 and F2 aren't running at identical rates?
Thanks,
David Seppala
Bastrop TX

piątek, 5 sierpnia 2022 o 03:58:16 UTC+2 Sylvia Else napisał(a):
> On 05-Aug-22 9:06 am, sep...@yahoo.com wrote:
> > Can anyone explain the time concept of relativity when two inertial reference frames continually measure the same elapsed time between events while also concluding that clocks in the two inertial reference frames are running at different rates?
> > For example, let there be three inertial reference frames, F0, F1 and F2. Let F1 have a velocity V= sqrt(3)/2 *c relative to F0 moving in the positive x direction and let F2 have a velocity V=sqrt(3)/2*c relative to F0 moving in the negative direction. Now at x=0 in F0, frames F1 and F2 have clocks that read t' = t'' = 0. At the point in space, both frame F1 and F2 observers place a clock that runs at twice the rate of a standard clock into frame F0 at x=0. Now each future event that occurs at x=0 in F0, observers in F1 and F2 measure that the same elapsed time has occurred as measured in their respective inertial reference frames. They also observe that both clocks that were placed in F0 at x=0 match the time shown on each clock that they pass in their reference frames, and the both of these clocks show the identical time. Yet even though these clocks show the same time and run at the same rate and match the clock reading of every clock they pass in frame F1 and F2, F1 and F2 observers say that clocks in the two frames do not run at the same rate. Please clarify that concept.
> > Thanks,
> > David Seppala
> > Bastrop TX
> Yet another in a long line of your spurious attempts to show a
> contradiction in special relativity.
>
> Apply the Lorentz transform, and all will become clear.
>
> Sylvia.

The Lorentz transform is a consequence of convention only.

Simply: If you assume c = const in the common sense = SR,
then you get nothing new, because in geometry only distances are usable, not time.

r = ct' = ct' = t' = c'

this is evident in every equation of relativity pseudo-math.

try this:

ct = ?

c' = ?
t' = ?

ct' = ?

c't = ?

have you seen ever:
c't' = ?

no, because this duality doubles.. the stupidity,
but you know: stupidity = infinity in advance, so this combination is useless.. :)

On Friday, August 5, 2022 at 1:25:26 PM UTC-7, sep...@yahoo.com wrote:
> If a clock that runs at twice the standard rate (in terms of its rest frame) is at
> rest in F0 at x=0, that clock will show the same elapsed time for every event
> at x=0 as the standard clocks at rest in F1, and the standard clocks at rest in F2.

Correct. (I fixed several garbled and superfluous portions of your statement.)

> The elapsed coordinate time between events A0 and A1 are the same for F1 and F2.
> So why doesn't this imply that clocks at rest in frames F1 and F2 aren't running at
> identical rates?

Because those elapsed coordinate times are not elapsed proper times on individual clocks at rest in F1 or F2, they are differences between the readings of sequences of clocks (because the path x=0 is moving at speed v in terms of both F1 and F2).

Remember, each standard clock at rest in F1 runs at 1/7 the normal rate in terms of F2, and each standard clock at rest in F2 runs at 1/7 the normal rate in terms of F1. Also, each standard clock at rest in either F1 or F2 runs at 1/2 the normal rate in terms of F0, and each standard clock at rest in F0 runs at half the normal rate in terms of both F1 and F2. Which of these elementary facts do you still not understand?

On 06-Aug-22 12:59 am, sepp623@yahoo.com wrote:
> On Thursday, August 4, 2022 at 8:58:16 PM UTC-5, Sylvia Else wrote:
>> On 05-Aug-22 9:06 am, sep...@yahoo.com wrote:
>>> Can anyone explain the time concept of relativity when two inertial reference frames continually measure the same elapsed time between events while also concluding that clocks in the two inertial reference frames are running at different rates?
>>> For example, let there be three inertial reference frames, F0, F1 and F2. Let F1 have a velocity V= sqrt(3)/2 *c relative to F0 moving in the positive x direction and let F2 have a velocity V=sqrt(3)/2*c relative to F0 moving in the negative direction. Now at x=0 in F0, frames F1 and F2 have clocks that read t' = t'' = 0. At the point in space, both frame F1 and F2 observers place a clock that runs at twice the rate of a standard clock into frame F0 at x=0. Now each future event that occurs at x=0 in F0, observers in F1 and F2 measure that the same elapsed time has occurred as measured in their respective inertial reference frames. They also observe that both clocks that were placed in F0 at x=0 match the time shown on each clock that they pass in their reference frames, and the both of these clocks show the identical time. Yet even though these clocks show the same time and run at the same rate and match the clock reading of every clock they pass in frame F1 and F2, F1 and F2 observers say that clocks in the two frames do not run at the same rate. Please clarify that concept.
>>> Thanks,
>>> David Seppala
>>> Bastrop TX
>> Yet another in a long line of your spurious attempts to show a
>> contradiction in special relativity.
>>
>> Apply the Lorentz transform, and all will become clear.
>>
>> Sylvia.
> When I apply the Lorentz transform that's when it becomes unclear.
Then show your math.
Sylvia.

On Saturday, 6 August 2022 at 02:45:30 UTC+2, Sylvia Else wrote:
> On 06-Aug-22 12:59 am, sep...@yahoo.com wrote:
> > On Thursday, August 4, 2022 at 8:58:16 PM UTC-5, Sylvia Else wrote:
> >> On 05-Aug-22 9:06 am, sep...@yahoo.com wrote:
> >>> Can anyone explain the time concept of relativity when two inertial reference frames continually measure the same elapsed time between events while also concluding that clocks in the two inertial reference frames are running at different rates?
> >>> For example, let there be three inertial reference frames, F0, F1 and F2. Let F1 have a velocity V= sqrt(3)/2 *c relative to F0 moving in the positive x direction and let F2 have a velocity V=sqrt(3)/2*c relative to F0 moving in the negative direction. Now at x=0 in F0, frames F1 and F2 have clocks that read t' = t'' = 0. At the point in space, both frame F1 and F2 observers place a clock that runs at twice the rate of a standard clock into frame F0 at x=0. Now each future event that occurs at x=0 in F0, observers in F1 and F2 measure that the same elapsed time has occurred as measured in their respective inertial reference frames. They also observe that both clocks that were placed in F0 at x=0 match the time shown on each clock that they pass in their reference frames, and the both of these clocks show the identical time. Yet even though these clocks show the same time and run at the same rate and match the clock reading of every clock they pass in frame F1 and F2, F1 and F2 observers say that clocks in the two frames do not run at the same rate. Please clarify that concept.
> >>> Thanks,
> >>> David Seppala
> >>> Bastrop TX
> >> Yet another in a long line of your spurious attempts to show a
> >> contradiction in special relativity.
> >>
> >> Apply the Lorentz transform, and all will become clear.
> >>
> >> Sylvia.
> > When I apply the Lorentz transform that's when it becomes unclear.
> Then show your math.

Speaking of math, it's always good to remind that Your
insane bunch had to announce its oldest, very important
part false, as it didn't want to fit Your religious postulates.

On Friday, August 5, 2022 at 7:59:31 AM UTC-7, sep...@yahoo.com wrote:
> On Thursday, August 4, 2022 at 8:58:16 PM UTC-5, Sylvia Else wrote:
> > On 05-Aug-22 9:06 am, sep...@yahoo.com wrote:
> > > Can anyone explain the time concept of relativity when two inertial reference frames continually measure the same elapsed time between events while also concluding that clocks in the two inertial reference frames are running at different rates?
> > > For example, let there be three inertial reference frames, F0, F1 and F2. Let F1 have a velocity V= sqrt(3)/2 *c relative to F0 moving in the positive x direction and let F2 have a velocity V=sqrt(3)/2*c relative to F0 moving in the negative direction. Now at x=0 in F0, frames F1 and F2 have clocks that read t' = t'' = 0. At the point in space, both frame F1 and F2 observers place a clock that runs at twice the rate of a standard clock into frame F0 at x=0. Now each future event that occurs at x=0 in F0, observers in F1 and F2 measure that the same elapsed time has occurred as measured in their respective inertial reference frames. They also observe that both clocks that were placed in F0 at x=0 match the time shown on each clock that they pass in their reference frames, and the both of these clocks show the identical time. Yet even though these clocks show the same time and run at the same rate and match the clock reading of every clock they pass in frame F1 and F2, F1 and F2 observers say that clocks in the two frames do not run at the same rate. Please clarify that concept.
> > > Thanks,
> > > David Seppala
> > > Bastrop TX
> > Yet another in a long line of your spurious attempts to show a
> > contradiction in special relativity.
> >
> > Apply the Lorentz transform, and all will become clear.
> >
> > Sylvia.
> When I apply the Lorentz transform that's when it becomes unclear. As stated in original post let F1 nd F2 have a velocity V=c*sqrt(3/2) relative to F0 but in opposite directions. Let there be two events in F0 at x=0, one that occurs 10 seconds later than the other as measured in F0. Observers in F1 and F2 each put a clock running at twice the standard rate at x=0 in F0 when event A occurs. When event B occurs at that same location in F0, but 10 seconds later the Lorentz transform gives the elapsed time between the two events as measured in frame F1 to be:
> elapsed time in F1 = 2 * (10 seconds)
> and the elapsed time in F2 between events A and B to be
> 2 * (10 seconds).
> The two clocks running at twice the standard rate each show 20 seconds has elapsed. Observers in F1 measure that 20 seconds has elapsed between the two events and observers in F2 measure that 20 seconds has elapsed between the two events. Yet they don't agree that the F1 clocks are running at the same rate as the F2 clocks.
> That is what is unclear clear to me. Please explain the logic.
> Thanks,
> David Seppala
> Bastrop TX

So you're saying all the way along their travel,
instantaneously, at each point, both's clocks
are running.

Then, that it looks the two are approaching
also they are approaching. Then for example
the non-moving or slow-velocity is as likely to
meet, while, the fast-moving is as likely to pass.

So, it seems there is to be breaking down the
Lorentz Transform, into Lagrangians, all along
the path travel, some Lorentz travel path integral,
that results an accurate transform usually computed
instantaneously, accurate multi-body Lorentz transforms.

On Saturday, 6 August 2022 at 06:26:13 UTC+2, Ross A. Finlayson wrote:
> On Friday, August 5, 2022 at 7:59:31 AM UTC-7, sep...@yahoo.com wrote:
> > On Thursday, August 4, 2022 at 8:58:16 PM UTC-5, Sylvia Else wrote:
> > > On 05-Aug-22 9:06 am, sep...@yahoo.com wrote:
> > > > Can anyone explain the time concept of relativity when two inertial reference frames continually measure the same elapsed time between events while also concluding that clocks in the two inertial reference frames are running at different rates?
> > > > For example, let there be three inertial reference frames, F0, F1 and F2. Let F1 have a velocity V= sqrt(3)/2 *c relative to F0 moving in the positive x direction and let F2 have a velocity V=sqrt(3)/2*c relative to F0 moving in the negative direction. Now at x=0 in F0, frames F1 and F2 have clocks that read t' = t'' = 0. At the point in space, both frame F1 and F2 observers place a clock that runs at twice the rate of a standard clock into frame F0 at x=0. Now each future event that occurs at x=0 in F0, observers in F1 and F2 measure that the same elapsed time has occurred as measured in their respective inertial reference frames. They also observe that both clocks that were placed in F0 at x=0 match the time shown on each clock that they pass in their reference frames, and the both of these clocks show the identical time. Yet even though these clocks show the same time and run at the same rate and match the clock reading of every clock they pass in frame F1 and F2, F1 and F2 observers say that clocks in the two frames do not run at the same rate. Please clarify that concept.
> > > > Thanks,
> > > > David Seppala
> > > > Bastrop TX
> > > Yet another in a long line of your spurious attempts to show a
> > > contradiction in special relativity.
> > >
> > > Apply the Lorentz transform, and all will become clear.
> > >
> > > Sylvia.
> > When I apply the Lorentz transform that's when it becomes unclear. As stated in original post let F1 nd F2 have a velocity V=c*sqrt(3/2) relative to F0 but in opposite directions. Let there be two events in F0 at x=0, one that occurs 10 seconds later than the other as measured in F0. Observers in F1 and F2 each put a clock running at twice the standard rate at x=0 in F0 when event A occurs. When event B occurs at that same location in F0, but 10 seconds later the Lorentz transform gives the elapsed time between the two events as measured in frame F1 to be:
> > elapsed time in F1 = 2 * (10 seconds)
> > and the elapsed time in F2 between events A and B to be
> > 2 * (10 seconds).
> > The two clocks running at twice the standard rate each show 20 seconds has elapsed. Observers in F1 measure that 20 seconds has elapsed between the two events and observers in F2 measure that 20 seconds has elapsed between the two events. Yet they don't agree that the F1 clocks are running at the same rate as the F2 clocks.
> > That is what is unclear clear to me. Please explain the logic.
> > Thanks,
> > David Seppala
> > Bastrop TX
> So you're saying all the way along their travel,
> instantaneously, at each point, both's clocks
> are running.
>
> Then, that it looks the two are approaching
> also they are approaching. Then for example
> the non-moving or slow-velocity is as likely to
> meet, while, the fast-moving is as likely to pass.
>
> So, it seems there is to be breaking down the
> Lorentz Transform, into Lagrangians, all along
> the path travel, some Lorentz travel path integral,
> that results an accurate transform usually computed
> instantaneously, accurate multi-body Lorentz transforms.

Fortunately, we have GPS now, so we can be absolutely
sure that your insane mumble has nothing in common
with real clocks, real observers, real measurements or
real anything.