Let there be an inertial reference frame F0 and a rocket at rest in F0. The rocket can accelerate in the x direction or the negative x direction, and the acceleration rate is constant as produced by the rocket. Let the rocket accelerate in the positive x-direction to a velocity V relative to F0 and say it takes t1 seconds to reach velocity V as measured in F0. Then when the rocket has velocity V it accelerates in the opposite direction at the same rate as before returning to zero velocity with respect to F0 in t2 seconds. Do the observers in F0 measure that t1 equals t2?
Thanks,
David Seppala
Bastrop TX

On Saturday, September 18, 2021 at 2:13:34 PM UTC-7, sep...@yahoo.com wrote:
> Do the observers in F0 measure that t1 equals t2?
> David Seppala
> Bastrop TX

Isn't that obvious to you, crank?

On Saturday, September 18, 2021 at 2:55:48 PM UTC-7, Dono. wrote:
> On Saturday, September 18, 2021 at 2:13:34 PM UTC-7, sep...@yahoo.com wrote:
> > Do the observers in F0 measure that t1 equals t2?
> > David Seppala
> > Bastrop TX
>
>
> Isn't that obvious to you, crank?
t1=t2=\frac{v * \gamma(v)}{c}
Hint: The arithmetic requires that you are familiar with hyperbolic motion

On Saturday, September 18, 2021 at 3:29:56 PM UTC-7, Dono. wrote:
> On Saturday, September 18, 2021 at 2:55:48 PM UTC-7, Dono. wrote:
> > On Saturday, September 18, 2021 at 2:13:34 PM UTC-7, sep...@yahoo.com wrote:
> > > Do the observers in F0 measure that t1 equals t2?
> > > David Seppala
> > > Bastrop TX
> >
> >
> > Isn't that obvious to you, crank?
> t1=t2=\frac{v * \gamma(v)}{c}
> Hint: The arithmetic requires that you are familiar with hyperbolic motion

Typo correction: t1=t2=\frac{v * \gamma(v)}{a}

On Saturday, September 18, 2021 at 2:13:34 PM UTC-7, sep...@yahoo.com wrote:
> Do the observers in F0 measure that t1 equals t2?

Obviously. You can't possibly not have known the answer to that question. You must surely have some non-obvious question waiting in the wings, so why not just go ahead and type your latest alleged contradiction, so we can debunk it and update the score?

Score so far....
Special Relativity: 813 .... Barnpole Dave: 0

On Saturday, September 18, 2021 at 3:34:10 PM UTC-7, Al Coe wrote:
> On Saturday, September 18, 2021 at 2:13:34 PM UTC-7, sep...@yahoo.com wrote:
> > Do the observers in F0 measure that t1 equals t2?
> Obviously. You can't possibly not have known the answer to that question. You must surely have some non-obvious question waiting in the wings, so why not just go ahead and type your latest alleged contradiction, so we can debunk it and update the score?
>
> Score so far....
> Special Relativity: 813 .... Barnpole Dave: 0

I think that I cut him short this time. He's choking on the answer (hopefully, for good)

On Saturday, September 18, 2021 at 5:34:10 PM UTC-5, Al Coe wrote:
> On Saturday, September 18, 2021 at 2:13:34 PM UTC-7, sep...@yahoo.com wrote:
> > Do the observers in F0 measure that t1 equals t2?
> Obviously. You can't possibly not have known the answer to that question. You must surely have some non-obvious question waiting in the wings, so why not just go ahead and type your latest alleged contradiction, so we can debunk it and update the score?
>
> Score so far....
> Special Relativity: 813 .... Barnpole Dave: 0

I of course thought the two times were equal, but then I had trouble with the following scenario.
There are two inertial reference frames, F0 and F1. They have a relative velocity of V = c*sqrt(3)/2 along the x-axis.
If a rocket initially at rest accelerates from F0 to F1 at a constant rate and then decelerates back to F0 at the same rate then the time of the first leg, t1 equals the time of the return leg, t2.

Now if F1 measures the time it takes this rocket to go from rest in F1 to zero velocity with respect to F0, then observers in F1 must measure the same elapsed time as frame F0 observers do for the rocket to travel back and forth between the two frames, since whether the rocket is moving in the positive x direction or the negative x direction the times are the same, same rocket and same change in velocity. How do the Lorentz transforms show the identical elapsed times for the two frames for the journey of this rocket?

David Seppala
Bastrop TX

On Saturday, September 18, 2021 at 4:01:14 PM UTC-7, sep...@yahoo.com wrote:
> How do the Lorentz transforms show the identical elapsed times for the two frames for the journey of this rocket?
>
I told you, imbecile, you need to learn the equations of hyperbolic motion.

On Saturday, September 18, 2021 at 4:01:14 PM UTC-7, sep...@yahoo.com wrote:
> There are two inertial reference frames, F0 and F1. They have a relative velocity of
> V = c*sqrt(3)/2 along the x-axis. If a rocket initially at rest accelerates from F0 to F1
> at a constant rate and then decelerates back to F0 at the same rate then the time of
> the first leg, t1 equals the time of the return leg, t2.

Right, we covered that before, noting that you defined t1 and t2 in terms of inertial coordinates F0 (not the proper times).

> Now if F1 measures the time it takes this rocket to go from rest in F1 to zero velocity
> with respect to F0, then observers in F1 must measure the same elapsed time as frame
> F0 observers do for the rocket to travel back and forth between the two frames... How
> do the Lorentz transforms show the identical elapsed times for the two frames for the
> journey of this rocket?

This is Introduction to Relativity 101. Use units with c=1, and with a constant proper acceleration of a=1 sec^-1. In terms of F0 the rocket begins at rest at (say) t=0, x=1, and it reaches speed v=sqrt(3)/2 at event t=sqrt(3), x=2, so it is at rest in F1, and then it switches to constant proper acceleration -a and comes to rest in F0 again at event t=2sqrt(3), x=3. Inserting these coordinates into the Lorentz transformation, their coordinates in terms of F1 are t'=-sqrt(3), x'=2, and t'=0, x'=1, and t'=sqrt(3), x'=0. Hence each leg takes coordinate time sqrt(3) in terms of F1 as well.

Special Relativity: 815 .... Dave "Barnpole" Seppala: 0

On Sunday, 19 September 2021 at 00:34:10 UTC+2, Al Coe wrote:
> On Saturday, September 18, 2021 at 2:13:34 PM UTC-7, sep...@yahoo.com wrote:
> > Do the observers in F0 measure that t1 equals t2?
> Obviously. You can't possibly not have known the answer to that question. You must surely have some non-obvious question waiting in the wings, so why not just go ahead and type your latest alleged contradiction, so we can debunk it and update the score?
>
> Score so far....
> Special Relativity: 813 .... Barnpole Dave: 0

In the meantime in the real world, however, the clocks of
GPS keep indicating t'=t, just like all serious clocks always
did.

On Sunday, 19 September 2021 at 02:34:38 UTC+2, Al Coe wrote:
> On Saturday, September 18, 2021 at 4:01:14 PM UTC-7, sep...@yahoo.com wrote:
> > There are two inertial reference frames, F0 and F1. They have a relative velocity of
> > V = c*sqrt(3)/2 along the x-axis. If a rocket initially at rest accelerates from F0 to F1
> > at a constant rate and then decelerates back to F0 at the same rate then the time of
> > the first leg, t1 equals the time of the return leg, t2.
> Right, we covered that before, noting that you defined t1 and t2 in terms of inertial coordinates F0 (not the proper times).
> > Now if F1 measures the time it takes this rocket to go from rest in F1 to zero velocity
> > with respect to F0, then observers in F1 must measure the same elapsed time as frame
> > F0 observers do for the rocket to travel back and forth between the two frames... How
> > do the Lorentz transforms show the identical elapsed times for the two frames for the
> > journey of this rocket?
> This is Introduction to Relativity 101. Use units with c=1, and with a constant proper acceleration of a=1 sec^-1. In terms of F0 the rocket begins at rest at (say) t=0, x=1, and it reaches speed v=sqrt(3)/2 at event t=sqrt(3), x=2, so it is at rest in F1, and then it switches to constant proper acceleration -a and comes to rest in F0 again at event t=2sqrt(3), x=3. Inserting these coordinates into the Lorentz transformation, their coordinates in terms of F1 are t'=-sqrt(3), x'=2, and t'=0, x'=1, and t'=sqrt(3), x'=0. Hence each leg takes coordinate time sqrt(3) in terms of F1 as well.
>
> Special Relativity: 815 .... Dave "Barnpole" Seppala: 0

In the meantime in the real world, however, the clocks of GPS
keep indicating t'=t, just like all serious clocks always did.

On Saturday, September 18, 2021 at 7:34:38 PM UTC-5, Al Coe wrote:
> On Saturday, September 18, 2021 at 4:01:14 PM UTC-7, sep...@yahoo.com wrote:
> > There are two inertial reference frames, F0 and F1. They have a relative velocity of
> > V = c*sqrt(3)/2 along the x-axis. If a rocket initially at rest accelerates from F0 to F1
> > at a constant rate and then decelerates back to F0 at the same rate then the time of
> > the first leg, t1 equals the time of the return leg, t2.
> Right, we covered that before, noting that you defined t1 and t2 in terms of inertial coordinates F0 (not the proper times).
> > Now if F1 measures the time it takes this rocket to go from rest in F1 to zero velocity
> > with respect to F0, then observers in F1 must measure the same elapsed time as frame
> > F0 observers do for the rocket to travel back and forth between the two frames... How
> > do the Lorentz transforms show the identical elapsed times for the two frames for the
> > journey of this rocket?
> This is Introduction to Relativity 101. Use units with c=1, and with a constant proper acceleration of a=1 sec^-1. In terms of F0 the rocket begins at rest at (say) t=0, x=1, and it reaches speed v=sqrt(3)/2 at event t=sqrt(3), x=2, so it is at rest in F1, and then it switches to constant proper acceleration -a and comes to rest in F0 again at event t=2sqrt(3), x=3. Inserting these coordinates into the Lorentz transformation, their coordinates in terms of F1 are t'=-sqrt(3), x'=2, and t'=0, x'=1, and t'=sqrt(3), x'=0. Hence each leg takes coordinate time sqrt(3) in terms of F1 as well.
>
> Special Relativity: 815 .... Dave "Barnpole" Seppala: 0
I did not follow your comment. Can you tell me what is wrong with this calculation.
In F0, let the acceleration rate be 3/2 * sqrt(3) meters per second squared.. Let the rocket accelerate along the x-axis to a speed of V=sqrt(3)/2*c relative to F0. If we approximate c as c = 3*10**8 meters per second, F0 observers measure that it takes 10**8 seconds to reach that relative velocity. The distance traveled during that time is 1/2*a*t**2 or 3/4*sqrt(3)*10**16 meters. Once the velocity V is reached the rocket decelerates back to zero velocity with respect to F0. Since the time accelerating and the time decelerating is the same, the time to go from F0 to F1 and back to F0 is 2*10**8 seconds.
Now if F1 has a relative velocity of V=sqrt(3)/2 * c relative to F0, it must measure exactly the same round trip time for the journey of this rocket. I cannot use the Lorentz transform of events in F0 to result in the identical round trip times as measured in both F0 and F1. I was using t' = 2 * (t- V*L/c**2). Show me how to make the two times identical as measured by the two inertial reference frames so that F1 also measures 2*10**8 seconds for the round trip journey.

David Seppala
Bastrop TX

On Saturday, September 18, 2021 at 2:13:34 PM UTC-7, sep...@yahoo.com wrote:
> Let there be an inertial reference frame F0 and a rocket at rest in F0. The rocket can accelerate in the x direction or the negative x direction, and the acceleration rate is constant as produced by the rocket. Let the rocket accelerate in the positive x-direction to a velocity V relative to F0 and say it takes t1 seconds to reach velocity V as measured in F0. Then when the rocket has velocity V it accelerates in the opposite direction at the same rate as before returning to zero velocity with respect to F0 in t2 seconds. Do the observers in F0 measure that t1 equals t2?
> Thanks,
> David Seppala
> Bastrop TX

Seppalotto

There is only PNE framr (F0) in your exercise above, so stop babbling about "F1". You obviously do not want t learn about hyperbolic motion, you are here , once again, just to troll. This is a simple exercise, you have been given the tools to solve it, you have been given the answer : t1=t2=(v/a)*gamma(v), so why don't you crawl back in the shithole you came from?

On Sunday, September 19, 2021 at 7:44:11 AM UTC-7, sep...@yahoo.com wrote:
> > Use units with c=1, and with a constant proper acceleration of a=1 sec^-1.
> > In terms of F0 the rocket begins at rest at (say) t=0, x=1, and it reaches speed
> > v=sqrt(3)/2 at event t=sqrt(3), x=2, so it is at rest in F1, and then it switches
> > to constant proper acceleration -a and comes to rest in F0 again at event
> > t=2sqrt(3), x=3. Inserting these coordinates into the Lorentz transformation,
> > their coordinates in terms of F1 are t'=-sqrt(3), x'=2, and t'=0, x'=1, and
> > t'=sqrt(3), x'=0. Hence each leg takes coordinate time sqrt(3) in terms of F1
> > as well.
> >
> I did not follow your comment. Can you tell me what is wrong with this calculation.
> In [terms of] F0, let the acceleration rate be 3/2 * sqrt(3) meters per second squared.

No, you said it was a rocket maintaining "constant acceleration", and of course you doltishly neglected to state whether that was constant proper acceleration (as measured by an accelerometer on the rocket) or constant coordinate acceleration in terms of some specified system of coordinates, but your nitwit attempt to construct a contradiction explicitly asserted symmetry, i.e., the acceleration from rest in F0 to rest in F1 in terms of F0 is symmetrical to the acceleration from rest in F1 to rest in F0 in terms of F1. This applies only if you stipulate constant *proper* acceleration, not constant coordinate acceleration in terms of one or the other coordinate systems. With this in mind, the analysis is as given above.

> Now if F1 has a relative velocity of V=sqrt(3)/2 * c relative to F0, it must measure
> exactly the same round trip time for the journey of this rocket.

That would be true if the rocket was undergoing constant proper acceleration, because then it would be symmetrical, and the correct analysis is as given above. With constant coordinate acceleration in terms of F0, the proper acceleration and the coordinate acceleration in terms of F1 are not constant and the situation is not symmetrical. In that case letting T be the coordinate time between consecutive events in terms of F0, the corresponding time in terms of F1 is 2T - A sqrt(3) T^2/2, which would equal T only if A = 2/sqrt(3). Thus, in general, if you have constant coordinate acceleration in terms of F0, the leg coordinate times in terms of F1 are equal to each other (obviously), but not generally equal to the leg coordinate times in terms of F0, because the situation is not symmetrical. Do you understand this?

Special relativity: 816 .... Dave "Barnpole" Seppala: 0

On Sunday, September 19, 2021 at 12:51:18 PM UTC-5, Al Coe wrote:
> On Sunday, September 19, 2021 at 7:44:11 AM UTC-7, sep...@yahoo.com wrote:
> > > Use units with c=1, and with a constant proper acceleration of a=1 sec^-1.
> > > In terms of F0 the rocket begins at rest at (say) t=0, x=1, and it reaches speed
> > > v=sqrt(3)/2 at event t=sqrt(3), x=2, so it is at rest in F1, and then it switches
> > > to constant proper acceleration -a and comes to rest in F0 again at event
> > > t=2sqrt(3), x=3. Inserting these coordinates into the Lorentz transformation,
> > > their coordinates in terms of F1 are t'=-sqrt(3), x'=2, and t'=0, x'=1, and
> > > t'=sqrt(3), x'=0. Hence each leg takes coordinate time sqrt(3) in terms of F1
> > > as well.
> > >
> > I did not follow your comment. Can you tell me what is wrong with this calculation.
> > In [terms of] F0, let the acceleration rate be 3/2 * sqrt(3) meters per second squared.
>
> No, you said it was a rocket maintaining "constant acceleration", and of course you doltishly neglected to state whether that was constant proper acceleration (as measured by an accelerometer on the rocket) or constant coordinate acceleration in terms of some specified system of coordinates, but your nitwit attempt to construct a contradiction explicitly asserted symmetry, i.e., the acceleration from rest in F0 to rest in F1 in terms of F0 is symmetrical to the acceleration from rest in F1 to rest in F0 in terms of F1.. This applies only if you stipulate constant *proper* acceleration, not constant coordinate acceleration in terms of one or the other coordinate systems. With this in mind, the analysis is as given above.
> > Now if F1 has a relative velocity of V=sqrt(3)/2 * c relative to F0, it must measure
> > exactly the same round trip time for the journey of this rocket.
> That would be true if the rocket was undergoing constant proper acceleration, because then it would be symmetrical, and the correct analysis is as given above. With constant coordinate acceleration in terms of F0, the proper acceleration and the coordinate acceleration in terms of F1 are not constant and the situation is not symmetrical. In that case letting T be the coordinate time between consecutive events in terms of F0, the corresponding time in terms of F1 is 2T - A sqrt(3) T^2/2, which would equal T only if A = 2/sqrt(3). Thus, in general, if you have constant coordinate acceleration in terms of F0, the leg coordinate times in terms of F1 are equal to each other (obviously), but not generally equal to the leg coordinate times in terms of F0, because the situation is not symmetrical. Do you understand this?
>
> Special relativity: 816 .... Dave "Barnpole" Seppala: 0
Initially, I was told (and thought) that in F0 if the rocket accelerated from F0 to a velocity V with respect to F0 and then decelerated back to zero velocity relative to F0 that the time of the first half of the journey would equal the time of the second half of the journey. Let's say the first leg to accelerated from F0 to a velocity V = c*sqrt(3)/2 takes 10**8 seconds as measured in F0. So now please detail what time F1 measures if an identical rocket has zero velocity with respect F1 and accelerates from F1 to a velocity V = -c*sqrt(3)/2 with respect to F1 (or in other words has zero velocity with respect to F0).
How long do observers in F1 say it takes that identical rocket to accelerate so that its velocity changes by c*sqrt(3)/2?
Thanks,
David Seppala
Bastrop TX

On Sunday, 19 September 2021 at 19:51:18 UTC+2, Al Coe wrote:

> No, you said it was a rocket maintaining "constant acceleration", and of course you doltishly neglected to state whether that was constant proper acceleration (as measured by an accelerometer on the rocket) or constant coordinate acceleration in terms of some specified system of coordinates, but your nitwit attempt to construct a contradiction explicitly asserted symmetry, i.e., the acceleration from rest in F0 to rest in F1 in terms of F0 is symmetrical to the acceleration from rest in F1 to rest in F0 in terms of F1.. This applies only if you stipulate constant *proper* acceleration, not constant coordinate acceleration in terms of one or the other coordinate systems. With this in mind, the analysis is as given above.
> > Now if F1 has a relative velocity of V=sqrt(3)/2 * c relative to F0, it must measure
> > exactly the same round trip time for the journey of this rocket.
> That would be true if the rocket was undergoing constant proper acceleration,

In the meantime in the real world - the clocks of GPS keep
indicating t'=t, just like all serious clocks always did.

On Sunday, September 19, 2021 at 11:12:04 AM UTC-7, sep...@yahoo.com wrote:
> Initially, I was told (and thought) that in F0 if the rocket accelerated from F0 to a
> velocity V with respect to F0 and then decelerated back to zero velocity relative to
> F0 that the time of the first half of the journey would equal the time of the second
> half of the journey.

Yes, that is obvious. The mystery is why you are hung up on that trivial fact, and why you think it is relevant to your fallacious reasoning.

> Let's say the first leg to accelerated from F0 to a velocity V = c*sqrt(3)/2 takes
> 10**8 seconds as measured in F0.

You haven't specified the acceleration profile, so that is underspecified. The distance traveled during that leg is not specified. I've given you the explicit answers for (1) constant proper acceleration, and (2) constant F0 coordinate acceleration. If you want the answer for some other acceleration profile, you need to specify which one.

> So now please detail what time F1 measures if an identical rocket has zero velocity
> with respect F1 and accelerates from F1 to a velocity V = -c*sqrt(3)/2 with respect to
> F1 (or in other words has zero velocity with respect to F0).

Again, you have not specified the scenario, because you refuse to consistently state the acceleration profile. Are you talking about constant proper acceleration? Or constant F0 coordinate acceleration? Or constant F1 coordinate acceleration? Or some other acceleration profile? Of course, you could avoid needing to specify the acceleration by just specifying the distances and times. But you refuse to do that too. I've given you the detailed answers for the two most obvious cases (proper and coordinate). If you want the answer for some other case, you need to tell me what case you have in mind.

> How long do observers in F1 say it takes that identical rocket to accelerate
> so that its velocity changes by c*sqrt(3)/2?

There is no limit, in principle, to the rate of change of velocity. We can have impulse forces that can accelerate an object from one state of motion to another almost instantaneously, and these can take place over arbitrarily short spatial distances. No matter how fiercely your diseased brain is urging you to refuse to specify the acceleration, there simply is no way for you to avoid specifying the acceleration (or at least the times and distances). And, once you specify it, the answer is trivial.

Special Relativity: 817 .... Dave "Barnpole" Seppala: 0

On Sunday, September 19, 2021 at 1:31:55 PM UTC-5, Al Coe wrote:
> On Sunday, September 19, 2021 at 11:12:04 AM UTC-7, sep...@yahoo.com wrote:
> > Initially, I was told (and thought) that in F0 if the rocket accelerated from F0 to a
> > velocity V with respect to F0 and then decelerated back to zero velocity relative to
> > F0 that the time of the first half of the journey would equal the time of the second
> > half of the journey.
> Yes, that is obvious. The mystery is why you are hung up on that trivial fact, and why you think it is relevant to your fallacious reasoning.
> > Let's say the first leg to accelerated from F0 to a velocity V = c*sqrt(3)/2 takes
> > 10**8 seconds as measured in F0.
> You haven't specified the acceleration profile, so that is underspecified.. The distance traveled during that leg is not specified. I've given you the explicit answers for (1) constant proper acceleration, and (2) constant F0 coordinate acceleration. If you want the answer for some other acceleration profile, you need to specify which one.
> > So now please detail what time F1 measures if an identical rocket has zero velocity
> > with respect F1 and accelerates from F1 to a velocity V = -c*sqrt(3)/2 with respect to
> > F1 (or in other words has zero velocity with respect to F0).
> Again, you have not specified the scenario, because you refuse to consistently state the acceleration profile. Are you talking about constant proper acceleration? Or constant F0 coordinate acceleration? Or constant F1 coordinate acceleration? Or some other acceleration profile? Of course, you could avoid needing to specify the acceleration by just specifying the distances and times. But you refuse to do that too. I've given you the detailed answers for the two most obvious cases (proper and coordinate). If you want the answer for some other case, you need to tell me what case you have in mind.
> > How long do observers in F1 say it takes that identical rocket to accelerate
> > so that its velocity changes by c*sqrt(3)/2?
> There is no limit, in principle, to the rate of change of velocity. We can have impulse forces that can accelerate an object from one state of motion to another almost instantaneously, and these can take place over arbitrarily short spatial distances. No matter how fiercely your diseased brain is urging you to refuse to specify the acceleration, there simply is no way for you to avoid specifying the acceleration (or at least the times and distances). And, once you specify it, the answer is trivial.
>
> Special Relativity: 817 .... Dave "Barnpole" Seppala: 0
Al,
Just respond to this situation. In F0 there is a rocket that accelerates from F0 to V=c*sqrt(3)/2 with respect to F0 in 10**8 seconds as measured by observers in F0. In F1 there is an identical rocket. That rocket accelerates from F1 to V=c*sqrt(3)/2 with respect to F1.
Why do you need more information to determine how much time as measured by observers in F1 it takes for that identical rocket to change velocity with respect to F1 by V = c*sqrt(3)/2. What makes you think that the time it takes a rocket at rest in an inertial reference frame to accelerate to V as measured in the rocket's original inertial rest frame depends on which inertial reference frame in the universe it starts its acceleration in?

David Seppala
Bastrop TX

On Sunday, September 19, 2021 at 11:47:44 AM UTC-7, sep...@yahoo.com wrote:
> Just respond to this situation.

I have responded with the complete analyses of every scenario. Remember? Just scroll up to refresh your memory from an hour ago. (You seem to be falling into complete cognitive disarray.)

> In [terms of] F0 there is a rocket that accelerates from [rest] to V=c*sqrt(3)/2 with respect
> to F0 in 10**8 seconds...

Yet again, that is insufficient specification. You need to specify the acceleration profile. There are infinitely many different ways that an object can accelerate from one state of motion to another in a given time, e.g., with constant proper acceleration or constant coordinate acceleration or an impulse acceleration or any of infinitely many other possibilities, and each of them leads to different results.

> In F1 there is an identical rocket. That rocket accelerates from F1 to V=c*sqrt(3)/2 with respect to F1.

Yet again, you need to specify the acceleration profile. Is it constant proper acceleration? Constant coordinate acceleration? (I've given you the explicit answers for both of those alternatives.) Or is it some other acceleration profile? If so, what?

> Why do you need more information to determine how much time as measured
> by observers in F1 it takes for that identical rocket to change velocity with respect
> to F1 by V = c*sqrt(3)/2.

Because it depends on the acceleration profile, obviously. I've given you the explicit answers for two alternatives, one with constant proper acceleration and the other with constant coordinate acceleration. You saw that the answers are different, and you saw why they are different. Remember?

> What makes you think that the time it takes a rocket at rest in an inertial reference
> frame to accelerate to V as measured in the rocket's original inertial rest frame
> depends on which inertial reference frame in the universe it starts its acceleration in?

You've gone completely insane. You have actually specified (in some of your statements) the coordinate time to accelerate, so that it specified, but this does not specify the distance traveled, because that depends on the acceleration profile (e.g., you could have virtually zero acceleration for most of the time, and then at the end of the time interval an impulse acceleration up to V, and the distance traveled would be arbitrarily small). Both the distance and the time are relevant to the description of events in terms of the two different systems of coordinates. This has been explicitly shown to you for the two simplest alternatives (constant proper and constant coordinate acceleration). Do you understand this?

Special Relativity: 817 ..... Barnpole Dave: 0

On Sunday, September 19, 2021 at 2:07:13 PM UTC-5, Al Coe wrote:
> On Sunday, September 19, 2021 at 11:47:44 AM UTC-7, sep...@yahoo.com wrote:
> > Just respond to this situation.
> I have responded with the complete analyses of every scenario. Remember? Just scroll up to refresh your memory from an hour ago. (You seem to be falling into complete cognitive disarray.)
>
> > In [terms of] F0 there is a rocket that accelerates from [rest] to V=c*sqrt(3)/2 with respect
> > to F0 in 10**8 seconds...
>
> Yet again, that is insufficient specification. You need to specify the acceleration profile. There are infinitely many different ways that an object can accelerate from one state of motion to another in a given time, e.g., with constant proper acceleration or constant coordinate acceleration or an impulse acceleration or any of infinitely many other possibilities, and each of them leads to different results.
> > In F1 there is an identical rocket. That rocket accelerates from F1 to V=c*sqrt(3)/2 with respect to F1.
> Yet again, you need to specify the acceleration profile. Is it constant proper acceleration? Constant coordinate acceleration? (I've given you the explicit answers for both of those alternatives.) Or is it some other acceleration profile? If so, what?
> > Why do you need more information to determine how much time as measured
> > by observers in F1 it takes for that identical rocket to change velocity with respect
> > to F1 by V = c*sqrt(3)/2.
> Because it depends on the acceleration profile, obviously. I've given you the explicit answers for two alternatives, one with constant proper acceleration and the other with constant coordinate acceleration. You saw that the answers are different, and you saw why they are different. Remember?
> > What makes you think that the time it takes a rocket at rest in an inertial reference
> > frame to accelerate to V as measured in the rocket's original inertial rest frame
> > depends on which inertial reference frame in the universe it starts its acceleration in?
> You've gone completely insane. You have actually specified (in some of your statements) the coordinate time to accelerate, so that it specified, but this does not specify the distance traveled, because that depends on the acceleration profile (e.g., you could have virtually zero acceleration for most of the time, and then at the end of the time interval an impulse acceleration up to V, and the distance traveled would be arbitrarily small). Both the distance and the time are relevant to the description of events in terms of the two different systems of coordinates. This has been explicitly shown to you for the two simplest alternatives (constant proper and constant coordinate acceleration). Do you understand this?
>
> Special Relativity: 817 ..... Barnpole Dave: 0
Al,
Let's say as measured in F0 the rocket accelerates at a constant rate of 3/2*sqrt(3) meters/second squared. F0 observers measure that it takes 10*8 seconds for the rocket to go from zero velocity relative to F0 to V= c*sqrt(3)/2 (using c=3*10*8 meters as the velocity of light). The distance traveled as measured in F0 during that acceleration is 3/4*sqrt(3)*10**16 meters.
Now tell me if that same identical rocket was in a different inertial reference frame, say F1, and it once again accelerated just as it did before, with nothing changed on the rocket, you seem to have the view that the time and distance would somehow be different as measured by F1 observers when nothing changes on the rocket. Is that correct or do you think observers in F1 would make the identical measurements as observers in F0 did when their identical rocket accelerated?
David Seppala
Bastrop TX

On Sunday, September 19, 2021 at 2:24:17 PM UTC-5, sep...@yahoo.com wrote:
> On Sunday, September 19, 2021 at 2:07:13 PM UTC-5, Al Coe wrote:
> > On Sunday, September 19, 2021 at 11:47:44 AM UTC-7, sep...@yahoo.com wrote:
> > > Just respond to this situation.
> > I have responded with the complete analyses of every scenario. Remember? Just scroll up to refresh your memory from an hour ago. (You seem to be falling into complete cognitive disarray.)
> >
> > > In [terms of] F0 there is a rocket that accelerates from [rest] to V=c*sqrt(3)/2 with respect
> > > to F0 in 10**8 seconds...
> >
> > Yet again, that is insufficient specification. You need to specify the acceleration profile. There are infinitely many different ways that an object can accelerate from one state of motion to another in a given time, e.g., with constant proper acceleration or constant coordinate acceleration or an impulse acceleration or any of infinitely many other possibilities, and each of them leads to different results.
> > > In F1 there is an identical rocket. That rocket accelerates from F1 to V=c*sqrt(3)/2 with respect to F1.
> > Yet again, you need to specify the acceleration profile. Is it constant proper acceleration? Constant coordinate acceleration? (I've given you the explicit answers for both of those alternatives.) Or is it some other acceleration profile? If so, what?
> > > Why do you need more information to determine how much time as measured
> > > by observers in F1 it takes for that identical rocket to change velocity with respect
> > > to F1 by V = c*sqrt(3)/2.
> > Because it depends on the acceleration profile, obviously. I've given you the explicit answers for two alternatives, one with constant proper acceleration and the other with constant coordinate acceleration. You saw that the answers are different, and you saw why they are different. Remember?
> > > What makes you think that the time it takes a rocket at rest in an inertial reference
> > > frame to accelerate to V as measured in the rocket's original inertial rest frame
> > > depends on which inertial reference frame in the universe it starts its acceleration in?
> > You've gone completely insane. You have actually specified (in some of your statements) the coordinate time to accelerate, so that it specified, but this does not specify the distance traveled, because that depends on the acceleration profile (e.g., you could have virtually zero acceleration for most of the time, and then at the end of the time interval an impulse acceleration up to V, and the distance traveled would be arbitrarily small). Both the distance and the time are relevant to the description of events in terms of the two different systems of coordinates. This has been explicitly shown to you for the two simplest alternatives (constant proper and constant coordinate acceleration). Do you understand this?
> >
> > Special Relativity: 817 ..... Barnpole Dave: 0
> Al,
> Let's say as measured in F0 the rocket accelerates at a constant rate of 3/2*sqrt(3) meters/second squared. F0 observers measure that it takes 10*8 seconds for the rocket to go from zero velocity relative to F0 to V= c*sqrt(3)/2 (using c=3*10*8 meters as the velocity of light). The distance traveled as measured in F0 during that acceleration is 3/4*sqrt(3)*10**16 meters.
> Now tell me if that same identical rocket was in a different inertial reference frame, say F1, and it once again accelerated just as it did before, with nothing changed on the rocket, you seem to have the view that the time and distance would somehow be different as measured by F1 observers when nothing changes on the rocket. Is that correct or do you think observers in F1 would make the identical measurements as observers in F0 did when their identical rocket accelerated?
> David Seppala
> Bastrop TX
Al,
Right now, I am not comparing the coordinates of F0 with those of F1, I'm only asking if there is an identical rocket in each inertial reference frame, would observers in each of these inertial reference frames measure that the rocket in their own reference frame each took 10**8 to accelerate from 0 to V.
David Seppala
Bastrop TX

On Sunday, September 19, 2021 at 12:24:17 PM UTC-7, sep...@yahoo.com wrote:
> Let's say as measured in F0 the rocket accelerates at a constant rate of 3/2*sqrt(3) meters/second squared.

So you're back to constant coordinate acceleration in terms of F0. The answer for this scenario was already given above.

> F0 observers measure that it takes 10*8 seconds for the rocket to go from zero velocity
> relative to F0 to V= c*sqrt(3)/2 (using c=3*10*8 meters as the velocity of light).

Again, the general scenario with constant coordinate acceleration was explained previously. In terms of F0 it starts from rest at (x,t)=(0,0) and undergoes constant coordinate acceleration A for a duration of coordinate time T, during which it travels a distance AT^2/2, so it is at (x,t) = (AT^2/2, T) with speed V=AT. It then decelerates at constant (negative) coordinate acceleration A for a coordinate time T, at the end of which it is at (x,t) = (AT^2, 2T) and 0 speed. Each leg takes a coordinate time of T. Now, apply the Lorentz transformation to the coordinates of these three events, to give their coordinates in terms of F1. You find that each leg takes a coordinate time of T(2 - sqrt(3)V/2).

> Now tell me if that same identical rocket was [at rest] in a different inertial reference frame,
> say F1, and it once again accelerated just as it did before, with nothing changed on the rocket,
> you seem to have the view that the time and distance would somehow be different as measured
> by F1 observers when nothing changes on the rocket. Is that correct...

No, that is obviously not correct. When you say "accelerated just as it did before" you are lying, right? It was subjected to constant coordinate acceleration in terms of F0 before, so do you mean it will be subjected to constant coordinate acceleration in terms of F0 again? No, that's obviously not what you mean. Your diseased brain is dimly imagining that it will undergo constant coordinate acceleration in terms of F1. But those are different, i.e., something undergoing constant coordinate acceleration in terms of F0 is not undergoing constant coordinate acceleration in terms of F1, and vice versa. Your entire nitwit attempt to construct a contradiction is based on arguing for symmetry, but that requires constant proper acceleration, not constant coordinate acceleration.

Again, the explicit answers for both scenarios have been provided to you, multiple times.

> Right now, I am not comparing the coordinates of F0 with those of F1...

Right, you are digressing into even more infantile idiocy. Inertial coordinate systems are equally suitable for the expression of physical laws, but the fact that we can set up two identical dynamical situations in terms of two different inertial coordinate systems does not at all imply that constant coordinate acceleration in one system is constant coordinate acceleration in the other, and that's what you would need in order to construct your nitwit attempt at contradiction. Your problem is not with special relativity, it is with basic logic and rationality.

Again, the explicit answer for your nitwit case of constant coordinate acceleration has been provided to you. (See above.) What part of the explanation do you not understand?

Special Relativity: 818 .... Barnpole Dave: 0

On Sunday, September 19, 2021 at 3:05:03 PM UTC-5, Al Coe wrote:
> On Sunday, September 19, 2021 at 12:24:17 PM UTC-7, sep...@yahoo.com wrote:
> > Let's say as measured in F0 the rocket accelerates at a constant rate of 3/2*sqrt(3) meters/second squared.
> So you're back to constant coordinate acceleration in terms of F0. The answer for this scenario was already given above.
> > F0 observers measure that it takes 10*8 seconds for the rocket to go from zero velocity
> > relative to F0 to V= c*sqrt(3)/2 (using c=3*10*8 meters as the velocity of light).
> Again, the general scenario with constant coordinate acceleration was explained previously. In terms of F0 it starts from rest at (x,t)=(0,0) and undergoes constant coordinate acceleration A for a duration of coordinate time T, during which it travels a distance AT^2/2, so it is at (x,t) = (AT^2/2, T) with speed V=AT. It then decelerates at constant (negative) coordinate acceleration A for a coordinate time T, at the end of which it is at (x,t) = (AT^2, 2T) and 0 speed. Each leg takes a coordinate time of T. Now, apply the Lorentz transformation to the coordinates of these three events, to give their coordinates in terms of F1. You find that each leg takes a coordinate time of T(2 - sqrt(3)V/2).
>
> > Now tell me if that same identical rocket was [at rest] in a different inertial reference frame,
> > say F1, and it once again accelerated just as it did before, with nothing changed on the rocket,
> > you seem to have the view that the time and distance would somehow be different as measured
> > by F1 observers when nothing changes on the rocket. Is that correct...
>
> No, that is obviously not correct. When you say "accelerated just as it did before" you are lying, right? It was subjected to constant coordinate acceleration in terms of F0 before, so do you mean it will be subjected to constant coordinate acceleration in terms of F0 again? No, that's obviously not what you mean. Your diseased brain is dimly imagining that it will undergo constant coordinate acceleration in terms of F1. But those are different, i.e., something undergoing constant coordinate acceleration in terms of F0 is not undergoing constant coordinate acceleration in terms of F1, and vice versa. Your entire nitwit attempt to construct a contradiction is based on arguing for symmetry, but that requires constant proper acceleration, not constant coordinate acceleration.
>
> Again, the explicit answers for both scenarios have been provided to you, multiple times.
>
> > Right now, I am not comparing the coordinates of F0 with those of F1...
>
> Right, you are digressing into even more infantile idiocy. Inertial coordinate systems are equally suitable for the expression of physical laws, but the fact that we can set up two identical dynamical situations in terms of two different inertial coordinate systems does not at all imply that constant coordinate acceleration in one system is constant coordinate acceleration in the other, and that's what you would need in order to construct your nitwit attempt at contradiction. Your problem is not with special relativity, it is with basic logic and rationality.
>
> Again, the explicit answer for your nitwit case of constant coordinate acceleration has been provided to you. (See above.) What part of the explanation do you not understand?
>
> Special Relativity: 818 .... Barnpole Dave: 0

I say NOTHING IN THE ROCKET CHANGED, therefore, no matter what inertial reference frame the rocket starts accelerating in, observers in that inertial reference frame will measure that it takes 10**8 seconds for the rocket to change from zero to V=c*sqrt(3)/2 with respect to that frame.
You say no. Explain what has to physically change in the rocket to make observers in each and every inertial frame that the rocket starts accelerating with zero velocity in measure the time to accelerate from zero to V=c*sqrt(3)/2 to be 10**8 seconds as measured in that inertial reference frame. What physical modifications to the rocket are necessary?
What physical modifications must be made to the rocket so that frame F1 measures the time for the rocket to accelerate from zero to V=c*sqrt(3)/2 is 10**8, with the acceleration being 3*sqrt(3)/2 meters/second squared as measured in F1. Why won't a rocket that is identical to the one used in F0 accelerate that way? Explain what needs to change in the rocket.
If the same rocket that was used in F0, was shipped to F1, and then the acceleration experiment was done in F1, what would observers in F1 measure as the time it takes the rocket to go from zero to V=c*sqrt(3)/2 relative to F1?

David Seppala
Bastrop TX

On Sunday, September 19, 2021 at 1:44:24 PM UTC-7, sep...@yahoo.com wrote:
> I say NOTHING IN THE ROCKET CHANGED...

For the 5th time, you can obviously set up two identical scenarios in terms of two different systems of inertial coordinates, but that obviously does nothing to support your alleged contradiction. Again, constant coordinate acceleration in terms of F0 is not constant coordinate acceleration in terms of F1, and vice versa, so your nitwit attempted contradiction fails.

> You say no.

No, you aren't paying attention. I say that the obvious and trivial fact that we can set up two identical scenarios in terms of two different coordinate systems has nothing to do with your alleged contradiction, which involves just one scenario in terms of two different coordinate systems. If you have one rocket undergoing constant coordinate acceleration in terms of F0, and another rocket undergoing constant coordinate acceleration in terms of F1, those rockets do not have constant coordinate accelerations in terms of the other system, so neither of them is symmetrical for the two systems.

> If the same rocket that was used in F0, was shipped to F1, and then the acceleration experiment was done in F1, what would observers in F1 measure as the time it takes the rocket to go from zero to V=c*sqrt(3)/2 relative to F1?

For the 6th time, you can obviously set up two identical scenarios in terms of two different systems of inertial coordinates, but that obviously does nothing to support your alleged contradiction. Constant coordinate acceleration in terms of F0 is not constant coordinate acceleration in terms of F1, and vice versa, so your nitwit attempted contradiction fails. Do you understand this?

Remember, you've been given the explicit answer to both scenarios -- constant proper acceleration, which gives symmetrical results (because it is symmetrical), and constant coordinate acceleration, which does not give symmetrical results (because it is not symmetrical). What part of this do you not understand?

Special Relativity: 819 .... Barnpole Dave: 0

On Sunday, September 19, 2021 at 4:11:42 PM UTC-5, Al Coe wrote:
> On Sunday, September 19, 2021 at 1:44:24 PM UTC-7, sep...@yahoo.com wrote:
> > I say NOTHING IN THE ROCKET CHANGED...
>
> For the 5th time, you can obviously set up two identical scenarios in terms of two different systems of inertial coordinates, but that obviously does nothing to support your alleged contradiction. Again, constant coordinate acceleration in terms of F0 is not constant coordinate acceleration in terms of F1, and vice versa, so your nitwit attempted contradiction fails.
>
> > You say no.
>
> No, you aren't paying attention. I say that the obvious and trivial fact that we can set up two identical scenarios in terms of two different coordinate systems has nothing to do with your alleged contradiction, which involves just one scenario in terms of two different coordinate systems. If you have one rocket undergoing constant coordinate acceleration in terms of F0, and another rocket undergoing constant coordinate acceleration in terms of F1, those rockets do not have constant coordinate accelerations in terms of the other system, so neither of them is symmetrical for the two systems.
> > If the same rocket that was used in F0, was shipped to F1, and then the acceleration experiment was done in F1, what would observers in F1 measure as the time it takes the rocket to go from zero to V=c*sqrt(3)/2 relative to F1?
> For the 6th time, you can obviously set up two identical scenarios in terms of two different systems of inertial coordinates, but that obviously does nothing to support your alleged contradiction. Constant coordinate acceleration in terms of F0 is not constant coordinate acceleration in terms of F1, and vice versa, so your nitwit attempted contradiction fails. Do you understand this?
>
> Remember, you've been given the explicit answer to both scenarios -- constant proper acceleration, which gives symmetrical results (because it is symmetrical), and constant coordinate acceleration, which does not give symmetrical results (because it is not symmetrical). What part of this do you not understand?
>
> Special Relativity: 819 .... Barnpole Dave: 0
Al,
Answer the question: If an identical rocket was shipped to F1, and it had zero velocity with respect to F1, and it accelerated to V=c*sqrt(3)/2 with respect to F1, would F1 observers measure that it took 10**8 seconds to reach V? If the answer is no, how long would it take that identical rocket to reach V relative to F1?
David Seppala
Bastrop TX