Relativity Acceleration Question

[sep...@yahoo.com]

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

[Dono.]

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?

[Dono.]

On Saturday, September 18, 2021 at 3:29:56 PM UTC-7, Dono. wrote:

> 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}

[Al Coe]

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

[Dono.]

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

[sep...@yahoo.com]

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

[Dono.]

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.

[Al Coe]

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

[Maciej Wozniak]

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

[Maciej Wozniak]

[sep...@yahoo.com]

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

[Dono.]

On Saturday, September 18, 2021 at 2:13:34 PM UTC-7, sep...@yahoo.com wrote:

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?

[Al Coe]

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

[sep...@yahoo.com]

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?

[Maciej Wozniak]

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

[Al Coe]

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

[sep...@yahoo.com]

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

[Al Coe]

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

[sep...@yahoo.com]

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

[sep...@yahoo.com]

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

[Al Coe]

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

[sep...@yahoo.com]

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

[Al Coe]

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

[sep...@yahoo.com]

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

[Al Coe]

On Sunday, September 19, 2021 at 2:32:33 PM UTC-7, sep...@yahoo.com wrote:
> Answer the question...

That question has been answered 6 times. For the 7th time, we can obviously set up two identical scenarios with two different rockets in terms of two different systems of inertial coordinates. This has nothing to do with your alleged contradiction, which necessarily involves a single scenario in terms of two different systems.

Remember, you began this thread by asking a brain-dead question that did not involve any contradiction, and I suggested that you please stop with the brain-dead trivialities and get on with your latest alleged contradiction. So you described your alleged contradiction, and I instantly debunked it, and now you have regressed back to your brain-dead trivialities that have nothing to do with your alleged contradiction. Your brain-dead questions have all been answered (explicitly), and your alleged contradiction has been thoroughly debunked. What part of the debunking of your alleged contradiction do you not understand?

Special Relativity: 820 ... Barnpole Dave: 0

[sep...@yahoo.com]

Okay, if you think you already answered that question then you agree that the identical rocket in F1 would take 10**8 seconds to go from zero velocity relative to F1 to V=c*sqrt(3)/2 as measured by observers in F1. So if the rocket from F0 has zero velocity with respect to F1 (it is just starting the second leg of its journey: returning to zero velocity relative to F0) and at the same point in time and at the same x coordinate the identical rocket of F1 starts accelerating toward F0, which rocket has a change in velocity of V=c*sqrt(3)/2 first?
David Seppala
Bastrop TX

[Al Coe]

On Sunday, September 19, 2021 at 3:11:36 PM UTC-7, sep...@yahoo.com wrote:
> [Consider a] rocket [initially at rest] in F1 would take 10**8 seconds to go from

> zero velocity relative to F1 to V=c*sqrt(3)/2 as measured by observers in F1.

Again, you stupidly refuse to specify the acceleration profile, so your scenario is under-defined, because we don't know the distance travelled.

> If the rocket from F0 has zero velocity with respect to F1 (it is just starting the

> second leg of its journey: returning to zero velocity relative to F0) and at the
> same point in time and at the same x coordinate the identical rocket of F1 starts
> accelerating toward F0, which rocket has a change in velocity of V=c*sqrt(3)/2 first?

Again, if two rockets have the same initial speed (0) in terms of F1 and are subjected to the same acceleration profile in terms of F1, then by definition they follow the same trajectory, so it is totally brain-dead to be treating them as two rockets... and, again, this trivial fact does not support your alleged contradiction at all.

Remember, your (thoroughly debunked) alleged contradiction was that the acceleration intervals in terms of F0 should be equal to the intervals in terms of F1, because the accelerations are physically symmetrical between F0 and F1, and that is indeed true for constant proper acceleration, in which case the intervals are indeed the same for both systems (as shown to you explicitly). On the other hand, if you insist on applying constant coordinate accelerations (or any other profile), those are not physically symmetrical in terms of F0 and F1, and hence the intervals are correspondingly different (as shown to you explicitly).

What part of this do you not understand? Maybe it would help for you to actually state your alleged contradiction again, to refresh your memory.

[sep...@yahoo.com]

I repeatedly said in F1, the acceleration profile is that the rocket accelerated at a constant rate as measured in F1 at a rate of 3*sqrt(3)/2 meters per second squared as measured in F1. The rocket took 10**8 seconds to accelerate from zero velocity relative to F1 to a velocity of c*sqrt(3)/2 relative to F1. What don't you understand?
David Seppala
Bastrop TX

[Al Coe]

On Sunday, September 19, 2021 at 4:10:24 PM UTC-7, sep...@yahoo.com wrote:
> I repeatedly said in F1, the acceleration profile is that the rocket accelerated at a
> constant rate as measured in F1 at a rate of 3*sqrt(3)/2 meters per second squared
> as measured in F1.

But you also said the rocket accelerates at that rate in terms of F0. You can't have it both ways. On the first leg, going from rest in F0 to rest in F1, are you accelerating at constant rate in terms of F0 or in terms of F1? And on the second leg, accelerating from rest in F1 to rest in F0, are you accelerating at constant rate in terms of F0 or in terms of F1?

> The rocket took 10**8 seconds to accelerate from zero velocity relative to F1 to a velocity
> of c*sqrt(3)/2 relative to F1. What don't you understand?

Nothing. What's missing is the specification of the acceleration profile, or at least a specification of the distance traveled during the acceleration. You need to specify the acceleration for each leg. Constant acceleration in terms of F0 is not constant coordinate acceleration in terms of F1, so you need to choose, and whichever choice you make, it is no longer symmetrical between F0 and F1. Do you understand this?

[sep...@yahoo.com]

Al,

When I said:
>> I repeatedly said in F1, the acceleration profile is that the rocket accelerated at a
> > constant rate as measured in F1 at a rate of 3*sqrt(3)/2 meters per second squared
> > as measured in F1.

You replied:

> But you also said the rocket accelerates at that rate in terms of F0. You can't have it both ways.

Explain what happens in the following:
A company builds 10 identical rockets and ships them to 10 different inertial reference frames. In some frames the rocket is aligned in the x direction, in some the y direction, in some the z direction, in some the -x direction, and some the x-y direction, and so forth, and each of these frames measures the rate of acceleration, and the time it takes the rocket as measured in their reference frame to change velocity from zero to V=c*sqrt(3)/2 relative to their frame. Explain why all frames can't measure that time to be 10**8 seconds as measured in their own frame. Please explain how each of the different inertial reference frames measures the acceleration to be different in their inertial reference frame when all the rockets are identical? Which frame will measure the shortest time for the rocket in their frame to accelerate from 0 to V=c*sqrt(3)/2? Which frame will measure in their frame the longest time?
David Seppala
Bastrop TX

[sep...@yahoo.com]

Al,
If you think if frame F0 observers measure the acceleration to be a constant 3*sqrt(3)/2 meters per second squared and you say F1 can't measure an acceleration of an identical rocket accelerating in his frame to be that same acceleration, then tell me how the rocket must be modified so that F1 measures the acceleration to be 3*sqrt(3)/2 meters per second squared. Does the rocket have to have more thrust or less thrust, does it have to have varying acceleration as it goes from zero to V=c*sqrt(3)/2? Is it possible for anyone to build a rocket such that as measured in F1 it accelerates at a constant rate of 3*sqrt(3)/2 meters per second squared as measured in F1 or do you think frame F0 is the only frame that can have an acceleration like that?
David Seppala
Bastrop TX

[Al Coe]

On Sunday, September 19, 2021 at 4:57:26 PM UTC-7, sep...@yahoo.com wrote:
> If you think if frame F0 observers measure the acceleration to be a constant 3*sqrt(3)/2
> meters per second squared and you say F1 can't measure an acceleration of an

> identical rocket accelerating in his frame to be that same acceleration...

No, you nitwit. You are not listening. An object can accelerate at a given constant rate in terms of any system of inertial coordinates you like, but if an object is accelerating at a constant rate in terms of F0, it is not accelerating at a constant rate in terms of F1, and vice versa. Do you understand this?

> tell me how the rocket must be modified so that F1 measures the acceleration to
> be 3*sqrt(3)/2 meters per second squared.

Again, let's take this e v e n m o r e s l o w l y. A rocket can have a constant coordinate acceleration in terms of F0, and a rocket can have a constant coordinate acceleration in terms of F1, but if the rocket has constant coordinate acceleration in terms of F0 it does not have constant coordinate acceleration in terms of F1, and vice versa. Do you understand this?

[Odd Bodkin]

Constant acceleration is not determined by engine construction, David.
Whether the acceleration is constant for any particular rocket is an
accident of reference frame.

--
Odd Bodkin — Maker of fine toys, tools, tables

[sep...@yahoo.com]

No, I do not understand that. If a rocket has a constant acceleration rate in F0, and an identical rocket is in F1 don't the F1 observers measure that the acceleration rate of that identical rocket is constant?
David Seppala
Bastrop TX

[Al Coe]

On Sunday, September 19, 2021 at 7:18:02 PM UTC-7, sep...@yahoo.com wrote:
> > A rocket can have a constant coordinate acceleration in terms of F0, and a rocket
> > can have a constant coordinate acceleration in terms of F1, but if the rocket has
> > constant coordinate acceleration in terms of F0 it does not have constant coordinate
> > acceleration in terms of F1, and vice versa. Do you understand this?
>
> No, I do not understand that. If a rocket has a constant acceleration rate in F0, and an
> identical rocket is in F1 don't the F1 observers measure that the acceleration rate of that
> identical rocket is constant?

You can have a rocket with constant acceleration in terms of F0, and another rocket with constant acceleration in terms of F1, but the first rocket does not have constant acceleration in terms of F1, and the second does not have constant acceleration in terms of F0. Now do you understand?

Try this: You can have a rocket at rest in terms of F0, and you can have another rocket at rest in terms of F1, but the first rocket is not at rest in terms of F1, and the second is not at rest in terms of F0. You see? A rocket can't be at rest in terms of both F0 and F1. Likewise a rocket can't have constant acceleration in terms of both F0 and F1. Proper acceleration is an invariant, but coordinate acceleration is dependent on the coordinate system, just as is velocity.

Understand?

[Maciej Wozniak]

On Monday, 20 September 2021 at 05:02:41 UTC+2, Al Coe wrote:

> You can have a rocket with constant acceleration in terms of F0, and another rocket with constant acceleration in terms of F1, but the first rocket does not have constant acceleration in terms of F1, and the second does not have constant acceleration in terms of F0. Now do you understand?

In the meantieme in the real world - the clocks of GPS keep indicating

[sep...@yahoo.com]

Al,
Let's say there no rocket was launched in F0, and therefore no info from F0 as to how the rocket accelerates. Now if an identical rocket was in F1 and it accelerated from rest in F1 to a velocity V=c*sqrt(3)/2 relative to F1, what do you think observers in F1 would measure as the elapsed time of that journey? Do you really believe that question can't be answered unless they have knowledge of what times and distances were measured in some other inertial reference frame?
Thanks,
David Seppala

[Odd Bodkin]

A rocket1 initially at rest in F0 will have an acceleration profile in F0
that is identical to the acceleration profile in F1 of an identical rocket2
initially at rest in F1.

This does not mean of course that the acceleration profile of rocket1 will
be the same in F0 and in F1. It also does not mean that the acceleration
profile of rocket2 will be the same in F1 and in F0.

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

[Maciej Wozniak]

On Monday, 20 September 2021 at 14:51:13 UTC+2, bodk...@gmail.com wrote:


> A rocket1 initially at rest in F0 will have an acceleration profile in F0
> that is identical to the acceleration profile in F1 of an identical rocket2
> initially at rest in F1.

In the meantime in the real world, however, the clocks of

[Odd Bodkin]

I think you are missing something from basic physics here, David, not just
relativity.
You have this idea that if a rocket burns for a certain period of time,
then all outcomes in terms of measurable quantities should come out the
same regardless of reference frame. This isn’t even true in high school
physics, and so this preconception should probably dropped from your head.

As an illustration of this, consider a car that starts from rest,
accelerating to speed v in time t, at which point the engine is turned off
and the car put in neutral. The car has mass m. What is the increase in the
kinetic energy of this car due to the burning of the fuel in the car
engine? (You should get an answer of mv^2/2.) Now look at the *same* car
from a different reference frame in which its initial speed is v.
Classically, it will then accelerate to 2v in time t. What is the increase
in the kinetic energy of the car as seen in this frame? You should get an
answer of 3mv^2/2. Notice that not only was it an identical car in both
cases, but that it was in fact the very same acceleration, the burning of
the same fuel, the same single trajectory. How do you account for the
difference in the increase in the kinetic energy?

This is intended to pry loose from your head the notion that if the engines
are the same, the burns are the same, then all the measurable physics
parameters should come out the same. Once you get rid of that notion, then
a lot of your assumptions about what should be an outcome in one of your
gedankens might get healthily removed.

[sep...@yahoo.com]

Let's say the rocket was designed by an engineer and not a mathematician. The engineer and fellow observers in an inertial reference frame design and test the rocket so that as it accelerates from zero to V=c*sqrt(3)/2 it always accelerates at a constant rate of 3*sqrt(3)/2 meters per second squared as measured in that inertial reference frame, and when it decelerates it always decelerates at a constant rate of -3*sqrt(3)/2 meters per second squared as measured in that inertial reference frame. Explain the scenario when the rockets are built like that.
David Seppala
Bastrop TX

[Odd Bodkin]

You can only design it, even as an engineer, to have that acceleration
profile in one reference frame. You can design it to have that acceleration
profile in the initial rest frame of the rocket (pre-firing) or in the
final rest frame of the rocket (post-firing), but not both.

The question is the same as asking the engineer to design a car so that it
will have an increase in kinetic energy of mv^2/2 in time t. You can do
that for a particular reference frame, but you cannot do it so that the
increase in kinetic energy will be the same in any inertial reference
frame. As I demonstrated to you above. This is basic physics and has
nothing to do with relativity. It’s just an unconscious bias you have to
rid yourself of.

[sep...@yahoo.com]

You wrote

> You can only design it, even as an engineer, to have that acceleration
> profile in one reference frame. You can design it to have that acceleration
> profile in the initial rest frame of the rocket (pre-firing) or in the
> final rest frame of the rocket (post-firing), but not both.
>

In my scenario the initial rest frame of the rocket (pre-firing) is F0 and the final rest frame of the rocket (post-firing) is also F0.
David Seppala
Bastrop TX

[Al Coe]

On Monday, September 20, 2021 at 5:39:14 AM UTC-7, sep...@yahoo.com wrote:
> Let's say no rocket was launched in F0, and therefore no info from F0 as

> to how the rocket accelerates.

That has no effect on the acceleration rate in terms of any specified system of coordinates.

> If an identical rocket ...

Identical to what?

> ... accelerated from rest in F1 to a velocity V=c*sqrt(3)/2 relative to F1, what

> do you think observers in F1 would measure as the elapsed time of that journey?

The question is underspecified because you have not specified the rate of acceleration. You previously specified a case in which the elapsed F1 coordinate time was T (for some specific value of T), in which case the elapsed F1 coordinate time would be T (duh), and of course this is still insufficient to specify the distance traveled, which is needed for the translation to any other system of coordinates, and if you add the specification that it accelerates at a constant rate in terms of F1 then it is fully specified, but in that case it did not accelerate at a constant rate in terms of F0 or any other system (remembering that your brain-dead stipulation about F0 above has no significance), so there is no contradiction.

You seem to have slid into complete incoherence, making nothing but brain-dead assertions. To pull yourself out of the ditch, I suggest you try to state what contradiction you are alleging. Or have you conceded that there is no contradiction here?

> Do you really believe that question can't be answered unless they have knowledge
> of what times and distances were measured in some other inertial reference frame?

You are profoundly confused. You can specify any trajectory you like in terms of any coordinate system you like, and obviously this does not entail any contradiction.

Remember, your brain-dead claim of a contradiction was based on your belief that after you specified a trajectory in terms of F0, the Lorentz transformation gives a description of that trajectory in terms of F1 that is logically contradictory. You argued that, by symmetry, the coordinate time intervals in terms of F1 should be the same as the coordinate time intervals in terms of F0, and you further argued that the Lorentz transformation says they are different... hence a logical flaw in special relativity. I explained why you were mistaken, by giving the explicit results, both for constant proper acceleration and for constant coordinate acceleration. At that point you regressed into this latest batch of completely brain-dead questions. Have you conceded that there is no contradiction here?

> Let's say ... observers in an inertial reference frame design and test the rocket so

> that as it accelerates from zero to V=c*sqrt(3)/2 it always accelerates at a constant
> rate of 3*sqrt(3)/2 meters per second squared as measured in that inertial reference
> frame, and when it decelerates it always decelerates at a constant rate of -3*sqrt(3)/2
> meters per second squared as measured in that inertial reference frame. Explain the
> scenario when the rockets are built like that.

That precise question was answered explicitly yesterday. I gave you the detailed quantitative result. And I went further and gave you the description of that same scenario in terms of another system of inertial coordinates that is moving at speed V relative to the first, and explained why there is no logical contradiction. Remember? The answer has not changed. What part didn't you understand?

[Maciej Wozniak]

On Monday, 20 September 2021 at 15:55:29 UTC+2, bodk...@gmail.com wrote:

> You can only design it, even as an engineer, to have that acceleration
> profile in one reference frame. You can design it to have that acceleration
> profile in the initial rest frame of the rocket (pre-firing) or in the
> final rest frame of the rocket (post-firing), but not both.

It isn't even a difficult task for a professional; but how would
a woodworker know.

[Odd Bodkin]

Which you can do. The construction or operation of the engine would be
different on the two legs, because in one case, you are trying to build the
rocket to have constant acceleration as viewed in the frame in which the
rocket was at rest pre-firing (F0), and in the other case you are trying to
build the rocket to have constant acceleration as viewed in the frame in
which the rocket was at rest post-firing (F0). I don’t think there’s any
particularly simple way to characterize the difference in the rocket design
for the two cases.

[Dono.]

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

According to your own post above there is only ONE frame (F0) in your exercise, so stop babbling about "F1". You obviously do not want to 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?

[sep...@yahoo.com]

Al,
If the F0 rocket has velocity V=c*sqrt(3)/2 with respect to F0 and zero velocity relative to F1, and decelerates at a constant rate of -3*sqrt(3)/2 meters per second squared as measured in F0, and if an identical F1 rocket has zero relative velocity relative to F1 and a velocity of |V|=c*sqrt(3)/2 with respect to F0, and that identical F1 rocket accelerates at a constant rate of 3*sqrt(3)/2 meters per second squared toward F0 as measured in F1, and they travel side by side, which rocket's speed changes by c*sqrt(3)/2 first?
Please clarify that for me.
David Seppala
Bastrop TX

[Al Coe]

On Monday, September 20, 2021 at 9:09:52 AM UTC-7, sep...@yahoo.com wrote:
> If a rocket initially has velocity V=c*sqrt(3)/2 with respect to F0 and zero velocity relative to F1,

> and decelerates at a constant rate of -3*sqrt(3)/2 meters per second squared as measured

> in F0, and if another rocket has zero relative velocity relative to F1 and a velocity of
> |V|=c*sqrt(3)/2 with respect to F0, and that rocket accelerates at a constant rate of

> 3*sqrt(3)/2 meters per second squared toward F0 as measured in F1, and they travel
> side by side, which rocket's speed changes by c*sqrt(3)/2 first?

So, you have two rockets initially at rest in terms of F1, and one of them accelerates at the constant rate A in terms of F0 and the other accelerates at constant rate A in terms of F1. You want to know which rocket will be the first (in terms of F1) to reach the speed V relative to F1.

For the rocket that accelerates from rest in F1 at constant rate A in terms of F1, the F1 coordinate time to reach the F1 speed V is simply T = V/A.

For the rocket that accelerates from rest in F1 at constant rate A in terms of F0, it begins at F0 speed V and takes T = V/A time in terms of F0 to reach F0 speed 0, and it travels an F0 spatial distance of (1/2)AT^2. So, your question is, how much F1 coordinate time does that take? Without loss of generality we can set the beginning of the acceleration at the common origin, and in terms of F0 coordinates x,t the end of the acceleration is at x=(1/2)AT^2 and t=T. In terms of F1 coordinates x',t' the end of that acceleration is at t' = T[(1-vV/2)/sqrt(1-v^2)], and since v=V, this is t' = T[(1-v^2/2)/sqrt(1-v^2)], and since you specified V=sqrt(3)/2, this gives the F1 time t' = (5/4)T, so the rocket that accelerated at constant A in terms of F1 completes its acceleration first in terms of F1. Of course, in terms of F0 the rocket that accelerated at constant A in terms of F0 completes its acceleration first. Remember, when comparing which of two separate events occurs "first", the answer depends on the system of coordinates, due to the relativity of simultaneity. Do you understand this?

Special Relativity: 822 ... Barnpole Dave: 0

[Maciej Wozniak]

On Monday, 20 September 2021 at 20:03:21 UTC+2, Al Coe wrote:

> So, you have two rockets initially at rest in terms of F1, and one of them accelerates at the constant rate A in terms of F0 and the other accelerates at constant rate A in terms of F1. You want to know which rocket will be the first (in terms of F1) to reach the speed V relative to F1.
>
> For the rocket that accelerates from rest in F1 at constant rate A in terms of F1, the F1 coordinate time to reach the F1 speed V is simply T = V/A.

[sep...@yahoo.com]

The two rockets are identical, so they accelerate in an identical fashion. Explain why two identical rockets that accelerate in the identical fashion that start accelerating at the same time and have the same initial velocity don't reach V at the same time. You omitted that from your explanation, instead you talked about coordinate differences of the two frames. The two identical rockets must both reach V simultaneously.
David Seppala
Bastrop TX

[Al Coe]

On Monday, September 20, 2021 at 1:06:40 PM UTC-7, sep...@yahoo.com wrote:
> The two rockets ... accelerate in an identical fashion.

No they do not. You yourself specified that one rocket accelerates at a constant rate in terms of F0 and the other accelerates at a constant rate in terms of F1. Those are two different acceleration profiles, whether you describe them both in terms of F0 or both in terms of F1. Now, the description of the constant F0 acceleration in terms of F0 is congruent (in reverse time) to the description of the constant F1 acceleration in terms of F1, but this does not signify (as you stupidly imagine) that they are the same trajectories. Understand?

> Explain why two identical rockets...

The construction of the rockets is irrelevant, nitwit. For each rocket, the thrust must be modulated to produce the desired acceleration (as fuel burn reduces mass, etc.), and the throttles must be modulated differently for the two rockets (in terms of any given system of coordinates), in one to produce constant F0 acceleration, and in the other to produce constant F1 acceleration. Again, the two rockets do not follow the same trajectory, so to map "the same time" between them is coordinate-dependent, and hence the comparison of the throttle positions "at the same time" is coordinate dependent. Again, congruence between descriptions of one object in terms of one frame and another object in terms of another frame does not imply that they are the same.

> that accelerate in the identical fashion...

They do not, nitwit. You yourself specified that one accelerates at a constant rate in terms of F0 and the other accelerates at a constant rate in terms of F1. Those are two different acceleration profiles. A trajectory that has constant coordinate acceleration in terms of one system does not have constant coordinate acceleration in terms of the other. The throttles must be modulated differently (in terms of any given common system of coordinates) for the two rockets, in one to produce constant F0 acceleration, and in the other to produce constant F1 acceleration. Do you understand this?

> You omitted that from your explanation...

No, nothing was omitted from the explanation. You stipulated that the rockets are subject to different acceleration profiles.

> instead you talked about coordinate differences of the two frames.
> The two identical rockets must both reach V simultaneously.

Simultaneously in terms of what system of coordinates? As explicitly shown to you (and you ignored), those events are not simultaneous in either F0 or F1. In terms of the respective coordinate systems, each one occurs ahead of the other, with the elapsed coordinate times differing by the factor 5/4. Now do you finally understand?

> The two identical rockets must both reach V simultaneously.

Again, simultaneously in terms of what coordinate system? There is of course a coordinate system in terms of which those events occur simultaneously, but it is neither F0 nor F1. Do you understand this?

[Odd Bodkin]

No, that cannot be true at the same time the following is true: “So, you

have two rockets initially at rest in terms of F1, and one of them
accelerates at the constant rate A in terms of F0 and the other accelerates

at constant rate A in terms of F1. “ They cannot be identical and have that
behavior. There is no point in presenting stipulations that are
contradictory.

> Explain why two identical rockets that accelerate in the identical
> fashion that start accelerating at the same time and have the same
> initial velocity don't reach V at the same time. You omitted that from
> your explanation, instead you talked about coordinate differences of the
> two frames. The two identical rockets must both reach V simultaneously.
> David Seppala
> Bastrop TX
>

[sep...@yahoo.com]

You are wrong. If a rocket is shipped to F0 and the observers in F0 measure the acceleration rate as 3*sqrt(3)/2 meters per second squared as it accelerates from 0 to V=c*sqrt(3)/2, and and an identical rocket is shipped to F1 and the observers in F1 measure the acceleration rate as 3*sqrt(3)/2 meters per second squared as it accelerates from 0 to V=c*sqrt(3)/2, and an identical rocket is shipped to F2 and the observers in F2 measure the acceleration rate as 3*sqrt(3)/2 meters per second squared as it accelerates from 0 to V=c*sqrt(3)/2, and ... an identical rocket is shipped to FN and the observers in FN measure the acceleration rate as 3*sqrt(3)/2 meters per second squared as it accelerates from 0 to V=c*sqrt(3)/2, if all these rockets are placed side by side, with the same velocity, they all accelerate at the same rate, and they all reach the same change in velocity simultaneously. Every inertial reference frame observes that if the rockets all started their acceleration simultaneously then each one always has the same velocity (zero) relative to each of the other rockets.
David Seppala
Bastrop TX

[Al Coe]

On Monday, September 20, 2021 at 2:17:24 PM UTC-7, sep...@yahoo.com wrote:

On Monday, September 20, 2021 at 2:17:24 PM UTC-7, sep...@yahoo.com wrote:
> > The two identical rockets must both reach V simultaneously.
> Again, simultaneously in terms of what coordinate system?

You forgot to answer the crucial question. I ask again: simultaneously in terms of what coordinate system?
**Prediction: You will never answer that question.

> If a rocket is shipped to F0 and the observers in F0 measure the acceleration rate
> as 3*sqrt(3)/2 meters per second squared as it accelerates from 0 to V=c*sqrt(3)/2,

Translation: In terms of F0, a rocket accelerates at the constant rate A from speed 0 to V.

> and an identical rocket is shipped to F1 and the observers in F1 measure the
> acceleration rate as 3*sqrt(3)/2 meters per second squared as it accelerates

> from 0 to V=c*sqrt(3)/2....

Translation: In terms of F1, another rocket accelerates at constant rate A from speed 0 to V. Do you really mean this? You're now specifying that each rocket begins from rest in a different frame, so their initial trajectories don't even match. You previously said the first rocket accelerated from V to 0 in terms of F0 while the second accelerated from 0 to -V in terms of F1. Get your story straight.

> ... and if these rockets are placed side by side, with the same velocity, they all

> accelerate at the same rate, and they all reach the same change in velocity
> simultaneously.

Simultaneously in terms of what coordinate system? [**Prediction: You will never answer that question.]

Look, the case you are trying to consider is with two rockets initially at rest in terms of F1, and one of them accelerates at the constant rate A in terms of F0 and the other accelerates at constant rate A in terms of F1. You want to know which rocket will be the first (in terms of F1 or F0, you refuse to say) to reach the speed -V relative to F1 (or 0 relative to F0).

Answer: For the rocket that accelerates from rest in F1 at constant rate A in terms of F1, the F1 coordinate time to reach the F1 speed V is T = V/A.

For the rocket that accelerates from rest in F1 at constant rate A in terms of F0, it begins at F0 speed V and takes T = V/A time in terms of F0 to reach F0 speed 0, and it travels an F0 spatial distance of (1/2)AT^2. Your question is, how much F1 coordinate time does that take? Without loss of generality we can set the beginning of the acceleration at the common origin, and in terms of F0 coordinates x,t the end of the acceleration is at x=(1/2)AT^2 and t=T. In terms of F1 coordinates x',t' the end of that acceleration is at t' = T[(1-vV/2)/sqrt(1-v^2)], and since v=V, this is t' = T[(1-v^2/2)/sqrt(1-v^2)], and since you specified V=sqrt(3)/2, this gives the F1 time t' = (5/4)T, so the rocket that accelerated at constant A in terms of F1 completes its acceleration first in terms of F1.

Of course, in terms of F0 the rocket that accelerated at constant A in terms of F0 completes its acceleration first. Do you dispute any of this? If so, which part?

Special Relativity: 823 ... Barnpole Dave: 0

[sep...@yahoo.com]

On Monday, September 20, 2021 at 5:23:44 PM UTC-5, Al Coe wrote:
> On Monday, September 20, 2021 at 2:17:24 PM UTC-7, sep...@yahoo.com wrote:
> On Monday, September 20, 2021 at 2:17:24 PM UTC-7, sep...@yahoo.com wrote:
> > > The two identical rockets must both reach V simultaneously.
> > Again, simultaneously in terms of what coordinate system?
> You forgot to answer the crucial question. I ask again: simultaneously in terms of what coordinate system?

The rockets all have zero relative velocity relative to each other so all coordinate systems say they reach each and every velocity simultaneously.

> **Prediction: You will never answer that question.
> > If a rocket is shipped to F0 and the observers in F0 measure the acceleration rate
> > as 3*sqrt(3)/2 meters per second squared as it accelerates from 0 to V=c*sqrt(3)/2,
> Translation: In terms of F0, a rocket accelerates at the constant rate A from speed 0 to V.
> > and an identical rocket is shipped to F1 and the observers in F1 measure the
> > acceleration rate as 3*sqrt(3)/2 meters per second squared as it accelerates
> > from 0 to V=c*sqrt(3)/2....
>
> Translation: In terms of F1, another rocket accelerates at constant rate A from speed 0 to V. Do you really mean this? You're now >specifying that each rocket begins from rest in a different frame, so their initial trajectories don't even match.

I didn't say their trajectories match if they start their accelerations with different initial velocities. I said that each inertial reference frame measures the same acceleration every rocket that starts with zero velocity relative to that inertial reference frame and accelerates identically to every rocket in that same inertial reference frame.

> You previously said the first rocket accelerated from V to 0 in terms of F0 while the second accelerated from 0 to -V in terms of F1. Get your story straight.

Yes, the first rocket is going from frame F1 to F0 (V to 0 as measured by observers in frame F0) and the second rocket is going from F1 to F0 (0 to -V as measured by observers in frame F1) so the rockets are traveling side by side, with identical instantaneous velocities.

>
> > ... and if these rockets are placed side by side, with the same velocity, they all
> > accelerate at the same rate, and they all reach the same change in velocity
> > simultaneously.
> Simultaneously in terms of what coordinate system? [**Prediction: You will never answer that question.]
>
> Look, the case you are trying to consider is with two rockets initially at rest in terms of F1, and one of them accelerates at the constant rate A in terms of F0 and the other accelerates at constant rate A in terms of F1. You want to know which rocket will be the first (in terms of F1 or F0, you refuse to say) to reach the speed -V relative to F1 (or 0 relative to F0).
>
> Answer: For the rocket that accelerates from rest in F1 at constant rate A in terms of F1, the F1 coordinate time to reach the F1 speed V is T = V/A.
>
> For the rocket that accelerates from rest in F1 at constant rate A in terms of F0, it begins at F0 speed V and takes T = V/A time in terms of F0 to reach F0 speed 0, and it travels an F0 spatial distance of (1/2)AT^2. Your question is, how much F1 coordinate time does that take? Without loss of generality we can set the beginning of the acceleration at the common origin, and in terms of F0 coordinates x,t the end of the acceleration is at x=(1/2)AT^2 and t=T. In terms of F1 coordinates x',t' the end of that acceleration is at t' = T[(1-vV/2)/sqrt(1-v^2)], and since v=V, this is t' = T[(1-v^2/2)/sqrt(1-v^2)], and since you specified V=sqrt(3)/2, this gives the F1 time t' = (5/4)T, so the rocket that accelerated at constant A in terms of F1 completes its acceleration first in terms of F1.
>
> Of course, in terms of F0 the rocket that accelerated at constant A in terms of F0 completes its acceleration first. Do you dispute any of this? If so, which part?

When you say in terms of F0 the rocket that accelerated at a constant A in terms of F0 completes its acceleration first, that applies to both identical rockets accelerating side by side. They are identical rockets, accelerating identically, with the same identical starting velocity.
David Seppala
Bastrop TX

[Al Coe]

On Monday, September 20, 2021 at 3:48:48 PM UTC-7, sep...@yahoo.com wrote:
> > You forgot to answer the crucial question. I ask again: simultaneously in terms
> > of what coordinate system?
> The rockets all have zero relative velocity relative to each other so all coordinate
> systems say they reach each and every velocity simultaneously.

No, the rockets do not all follow the same trajectory and they do not have zero velocity relative to each other. They begin on the same trajectory, and then one undergoes constant coordinate acceleration in terms of F0 and the other undergoes constant coordinate acceleration in terms of F1. Those produce different trajectories, as has been shown to you in detail. Understand?

> I didn't say their trajectories match if they start their accelerations with different
> initial velocities. I said that each inertial reference frame measures the same
> acceleration every rocket that starts with zero velocity relative to that inertial
> reference frame and accelerates identically to every rocket in that same inertial
> reference frame.

That's the completely brain-dead triviality that we exposed previously. Why you continue to repeat such pointless and irrelevant trivialities, as if they have some bearing on the answer to your question, is a mystery.

> > You previously said the first rocket accelerated from V to 0 in terms of F0 while the second accelerated from 0 to -V in terms of F1. Get your story straight.
>
> Yes, the first rocket is going from frame F1 to F0 (V to 0 as measured by observers in frame F0) and the second rocket is going from F1 to F0 (0 to -V as measured by observers in frame F1) so the rockets are traveling side by side, with identical instantaneous velocities.

Nope, your misconception has been explained to you repeatedly. You just aren't paying attention. Again, the rockets are not subjected to the same acceleration profile. One undergoes constant acceleration in terms of F0, and the other in terms of F1. Coordinate acceleration is coordinate dependent. Remember, your original question was how to reconcile the facts with the Lorentz transformation, which entails relativity of simultaneity.

> > Look, the case you are trying to consider is with two rockets initially at rest in terms
> > of F1, and one of them accelerates at the constant rate A in terms of F0 and the other
> > accelerates at constant rate A in terms of F1. You want to know which rocket will be
> > the first (in terms of F1 or F0, you refuse to say) to reach the speed -V relative to F1
> > (or 0 relative to F0).
> >
> > Answer: For the rocket that accelerates from rest in F1 at constant rate A in terms of F1,
> > the F1 coordinate time to reach the F1 speed V is T = V/A.
> >
> > For the rocket that accelerates from rest in F1 at constant rate A in terms of F0, it
> > begins at F0 speed V and takes T = V/A time in terms of F0 to reach F0 speed 0, and
> > it travels an F0 spatial distance of (1/2)AT^2. Your question is, how much F1 coordinate
> > time does that take? Without loss of generality we can set the beginning of the acceleration
> > at the common origin, and in terms of F0 coordinates x,t the end of the acceleration is at
> > x=(1/2)AT^2 and t=T. In terms of F1 coordinates x',t' the end of that acceleration is at
> > t' = T[(1-vV/2)/sqrt(1-v^2)], and since v=V, this is t' = T[(1-v^2/2)/sqrt(1-v^2)], and since you
> > specified V=sqrt(3)/2, this gives the F1 time t' = (5/4)T, so the rocket that accelerated at
> > constant A in terms of F1 completes its acceleration first in terms of F1. Of course, in
> > terms of F0 the rocket that accelerated at constant A in terms of F0 completes its
> > acceleration first.
>

> When you say in terms of F0 the rocket that accelerated at a constant A in terms
> of F0 completes its acceleration first, that applies to both identical rockets accelerating
> side by side. They are identical rockets, accelerating identically, with the same identical
> starting velocity.

Nope, one rocket undergoes constant coordinate acceleration in terms of F0 and the other undergoes constant coordinate acceleration in terms of F1, resulting to diverging trajectories, as explained above. They do not accelerate side by side. You've been given the precise formulas for how they diverge, and why. Which of those results (quoted above) do you dispute?

To put the question another way: Do you claim that a single trajectory can have constant coordinate acceleration in terms of both F0 and F1?

[Maciej Wozniak]

On Monday, 20 September 2021 at 23:14:34 UTC+2, bodk...@gmail.com wrote:

> No, that cannot be true at the same time the following is true: “So, you
> have two rockets initially at rest in terms of F1, and one of them
> accelerates at the constant rate A in terms of F0 and the other accelerates
> at constant rate A in terms of F1. “ They cannot be identical and have that
> behavior. There is no point in presenting stipulations that are
> contradictory.

[sep...@yahoo.com]

Let me ask you the question in another way. Two identical rockets are built in some inertial reference frame. The designer built each rocket so that it would accelerate at a constant rate as measured in any inertial reference frame. One rocket is shipped to F0 and the other is shipped to F1. The designer asks observers in F0 to record the acceleration as a function of time and distance as measured in F0. The designer also asks observers in F1 to record the acceleration as a function of time and distance as measured in F1. That is done in F0 and F1 and the test results from both inertial reference frames are sent back to the designer. Are the two charts the designer received identical? If not, explain how they vary.
David Seppala
Bastrop TX

[Odd Bodkin]

No, I’m sorry, that’s not a reasonable physical stipulation. What you CAN
stipulate is that the designer built each rocket so that it would
accelerate at a constant rate in a SELECTED reference frame, say the
reference frame in which the rocket was at rest pre-firing. But it wouldn’t
be possible for the SAME designed rocket to accelerate at a constant rate
in any other frame beside that one.

> One rocket is shipped to F0 and the other is shipped to F1. The designer
> asks observers in F0 to record the acceleration as a function of time and
> distance as measured in F0. The designer also asks observers in F1 to
> record the acceleration as a function of time and distance as measured in
> F1. That is done in F0 and F1 and the test results from both inertial
> reference frames are sent back to the designer. Are the two charts the
> designer received identical?

Yes, they are, IF the observers in F1 are only measuring the acceleration
of the rocket fired from rest in F1 and IF the observers in F0 are only
measuring the acceleration of the rocket fired from rest in F0. Those two
profiles will be identical because they are both measuring a rocket fired
from rest in their frame of reference.

However, notice that the profile of the rocket fired from rest in F1 will
be tracked by the observers in F0 to have a different acceleration profile,
and vice versa.

> If not, explain how they vary.
> David Seppala
> Bastrop TX
>

[Maciej Wozniak]

On Tuesday, 21 September 2021 at 15:55:14 UTC+2, bodk...@gmail.com wrote:

> No, I’m sorry, that’s not a reasonable physical stipulation. What you CAN
> stipulate is that the designer built each rocket so that it would
> accelerate at a constant rate in a SELECTED reference frame, say the
> reference frame in which the rocket was at rest pre-firing. But it wouldn’t
> be possible for the SAME designed rocket to accelerate at a constant rate
> in any other frame beside that one.

What is impossible for an idiot woodworker doesn't have to be
sspecially difficult for a professional.

[sep...@yahoo.com]

You said:
> Yes, they are, IF the observers in F1 are only measuring the acceleration
> of the rocket fired from rest in F1 and IF the observers in F0 are only
> measuring the acceleration of the rocket fired from rest in F0. Those two
> profiles will be identical because they are both measuring a rocket fired
> from rest in their frame of reference.

So if the observers in F0 measure that the rocket accelerates at a constant acceleration rate as measured in F0, you say the identical rocket when fired in F1 would have the identical chart, but you also say that chart would not show a constant acceleration rate as measured in F1. Explain that logic.
David Seppala
Bastrop TX

[Dono.]

On Tuesday, September 21, 2021 at 7:17:58 AM UTC-7, sep...@yahoo.com wrote:

> So if the observers in F0 measure that the rocket accelerates at a constant acceleration rate as measured in F0, you say the identical rocket when fired in F1 would have the identical chart, but you also say that chart would not show a constant acceleration rate as measured in F1. Explain that logic.
> David Seppala
> Bastrop TX

Acceleration is frame dependent, imbecile

[Odd Bodkin]

No, that’s not what I said. Please reread what I said.
The rocket fired from rest in F0 and measured by observers in F0 will get a
profile, which matches the profile for the rocket fired from rest in F1 and
measured by the observers in F1.

However, the profile for the rocket fired from rest in F0 and measured by
observers in F0 will not match the profile for the rocket fired from rest
in F1 and measured by observers in F0. Same observers looking at two
different rockets.

I think one of the trade-offs you face when you present these hypothetical
cases is that you lose track of which observers are measuring what and then
you get yourself confused. The other problem you face is thinking that if
the mechanisms are the same, then the outcomes should be measured the same,
and the classical example of measured increase in kinetic energy (same car,
measured in two different reference frames) shows that thinking to be
faulty, but you have a really hard time letting go of it.

[Maciej Wozniak]

On Tuesday, 21 September 2021 at 16:47:48 UTC+2, bodk...@gmail.com wrote:

> No, that’s not what I said. Please reread what I said.
> The rocket fired from rest in F0 and measured by observers in F0 will get a
> profile, which matches the profile for the rocket fired from rest in F1 and
> measured by the observers in F1.

In the meantime in the real world, GPS clocks keep measuring

[sep...@yahoo.com]

Odd Bodkin -- maker of fine toys, tools, tables wrote:
>> ... IF the observers in F1 are only measuring the acceleration

>> of the rocket fired from rest in F1 and IF the observers in F0 are only
>> measuring the acceleration of the rocket fired from rest in F0. Those two
>> profiles will be identical because they are both measuring a rocket fired
>> from rest in their frame of reference.

So if observers in F0 measure that as the rocket accelerates from zero velocity relative to F0 to V=c*sqrt(3)/2 relative to F0, at a constant acceleration rate of 3*sqrt(3)/2 meters per second squared as measured in their inertial reference frame F0, and observers in F1 measure the acceleration of an identical rocket as measured in their reference frame going from zero velocity relative to F1 to V=c*sqrt(3)/2 relative to F1, you agree that as measured by observers in F1 the acceleration rate of the rocket would be a constant 3*sqrt(3)/2 meters per second squared as measured by observers in F1.. Those two profiles are identical, is that correct?
David Seppala
Bastrop TX

[Odd Bodkin]

That’s right, because they are both viewing and measuring a rocket fired
from rest in their respective frames.

Notice that the observers in F0 will NOT observe the rocket fired from rest
in F1 (fired from a moving star in F0 of course) to have a constant
acceleration of 3*sqrt(3)/2 meters per second squared.

--

[Al Coe]

On Tuesday, September 21, 2021 at 6:25:44 AM UTC-7, sep...@yahoo.com wrote:
> > Do you claim that a single trajectory can have constant coordinate acceleration
> > in terms of both F0 and F1?
>

> The designer built each rocket so that it would accelerate at a constant rate as

> measured in any inertial reference frame...

So your answer to my question is yes, you claim that a single trajectory can have constant coordinate acceleration in terms of both F0 and F1. Your claim is manifestly false. To prove this, just express the equation of motion x(t)=(1/2)At^2 in terms of the relatively moving system of coordinates x',t'. Thus, a trajectory that has constant coordinate acceleration in terms of F0 does not have constant coordinate acceleration in terms of F1, and vice versa.

Bear in mind that if we design a rocket with constant thrust, and if the rest mass of the rocket wasn't changing, then the rocket would undergo constant proper acceleration, not constant coordinate acceleration. In this case an accelerometer in the rocket would always measure the same acceleration, and indeed the elapsed coordinate time intervals would be the same for F0 and F1, as detailed previously. Of course, the mass of the rocket is not really constant, so we would need to modulate the thrust to maintain constant proper acceleration, but this would not give constant coordinate acceleration in terms of any system of inertial coordinates. We could also modulate the thrust so that the rocket maintains constant coordinate acceleration in terms of any particular system of inertial coordinates, such as those in which it was initially at rest (for example), but this would not be constant proper acceleration nor would it be constant coordinate acceleration in terms of any other system of inertial coordinates.

> One rocket is shipped to F0 and the other is shipped to F1. The designer asks observers
> in F0 to record the acceleration as a function of time and distance as measured in F0. The
> designer also asks observers in F1 to record the acceleration as a function of time and
> distance as measured in F1. That is done in F0 and F1 and the test results from both inertial
> reference frames are sent back to the designer. Are the two charts the designer received
> identical?

There aren't just two sets of results, there are four. Letting A and B denote the rockets that begin their acceleration from rest in F0 and F1 respectively, we have

(1) the description of A's trajectory in terms of F0
(2) the description of A's trajectory in terms of F1
(3) the description of B's trajectory in terms of F0
(4) the description of B's trajectory in terms of F1

Assuming the thrust of each ship is modulated so that it maintains constant coordinate acceleration in terms of the inertial coordinates in which it was initially at rest, results (1) and (4) will both show that each rocket maintained constant coordinate acceleration in terms of the inertial coordinates in which it was initially at rest (duh). However, (1) and (2) are different, and this is the relevant point, i.e., a rocket that has constant coordinate acceleration in terms of F0 does not have constant coordinate acceleration in terms of F1. Also, (2) and (4) are different, because a rocket that has constant coordinate acceleration in terms of F1 does not have constant coordinate acceleration in terms of F0.

Needless to say, your second rocket is superfluous, since the relevant fact is that a single trajectory with constant coordinate acceleration in terms of one system does not have constant coordinate acceleration in terms of another. That is already shown by the difference between (1) and (2) for rocket A. Introducing rocket B is pointless, other than to re-iterate the point by noting that (3) and (4) differ from each other as well. Now do you understand?

Special Relativity: 823 .... Barnpole Dave: 0

[sep...@yahoo.com]

Okay,
Now let's say the rocket is designed to accelerate at a constant rate going from zero to V=c*sqrt(3)/2 at a constant rate, and the decelerates from V=c*sqrt(3)/2 back to zero at the same constant rate. Let observers in frame F0 measure that acceleration with the rocket initially having zero velocity relative to F0 and per their measurements the acceleration during the first leg of the trip is 3*sqrt(3)/2 meters per second squared and during the return leg the deceleration rate is -3*sqrt(3)/2 meters per second squared.
If F1 has an identical rocket that is initially at rest in F1, accelerates to a velocity V=c*sqrt(3)/2 and then decelerates back to zero velocity with respect to F1, will F1 measurements, based on F1's frame, identically match those taken by observers in F0 of the experiment that was conducted in F0. F1 is not measuring or looking at any events in F0 or vice versa. But do they say charts of the measurements each made in their own inertial reference frame are identical for the two identical rockets?
David Seppala
Bastrop TX

[Maciej Wozniak]

On Tuesday, 21 September 2021 at 18:20:22 UTC+2, Al Coe wrote:

> Assuming the thrust of each ship is modulated so that it maintains constant coordinate acceleration in terms of the inertial coordinates in which it was initially at rest, results (1) and (4) will both show that each rocket maintained constant coordinate acceleration in terms of the inertial coordinates in which it was initially at rest (duh). However, (1) and (2) are different, and this is the relevant point, i.e., a rocket that has constant coordinate acceleration in terms of F0 does not have constant coordinate acceleration in terms of F1. Also, (2) and (4) are different, because a rocket that has constant coordinate acceleration in terms of F1 does not have constant coordinate acceleration in terms of F0.

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

[Odd Bodkin]

No, you have not specified enough information. Remember that the rocket was
designed to accelerate at a constant rate AS MEASURED in the frame in which
the rocket was at rest pre-firing. It’s very important to state that
stipulation, because that is not the acceleration as measured in any other
frame.

So when you say the rocket is designed to decelerate at the same constant
rate, is that AS MEASURED in the frame in which the rocket was moving at
c*sqrt(3)/2 before firing? It’s very important to state that stipulation if
so, because that is not the acceleration as measured in any other frame.

> Let observers in frame F0 measure that acceleration with the rocket
> initially having zero velocity relative to F0 and per their measurements
> the acceleration during the first leg of the trip is 3*sqrt(3)/2 meters
> per second squared and during the return leg the deceleration rate is
> -3*sqrt(3)/2 meters per second squared.
> If F1 has an identical rocket that is initially at rest in F1,
> accelerates to a velocity V=c*sqrt(3)/2 and then decelerates back to zero
> velocity with respect to F1, will F1 measurements, based on F1's frame,
> identically match those taken by observers in F0 of the experiment that
> was conducted in F0. F1 is not measuring or looking at any events in F0
> or vice versa. But do they say charts of the measurements each made in
> their own inertial reference frame are identical for the two identical rockets?
> David Seppala
> Bastrop TX
>

[Al Coe]

On Tuesday, September 21, 2021 at 9:26:54 AM UTC-7, sep...@yahoo.com wrote:
> [Barnpole's text has been edited for coherence:]
> Let's say that in terms of F0 a rocket accelerates at a constant rate from 0 to V,
> and then decelerates from V back to zero at the same constant rate.... In terms of
> F1 an identical rocket is initially at rest and accelerates at constant rate to a velocity
> V and then decelerates back to zero velocity at the same constant rate. Does this
> entail any contradiction?

Of course not. The equations of physics take the same form in terms of every system of inertial coordinates, so we can have perfectly congruent scenarios in terms of each such system, but obviously this does not entail any contradiction. The relevant point is that the first rocket does not accelerate at constant rate in terms of F1, and the second rocket does not accelerate at constant rate in terms of F0. That's why your attempt to construct a contradiction fails. Understand?

[sep...@yahoo.com]

You asked:

>So when you say the rocket is designed to decelerate at the same constant
>rate, is that AS MEASURED in the frame in which the rocket was moving at
>c*sqrt(3)/2 before firing?

Yes. For example, as was stated, when F0 measures the rocket to reach a velocity V=c*sqrt(3)/2 relative to F0, the rocket decelerates at a rate of 3*sqrt(3)/2 meters per second squared as measured in F0 to return to zero velocity relative to F0.
David Seppala
Bastrop TX

[Maciej Wozniak]

On Tuesday, 21 September 2021 at 18:41:38 UTC+2, bodk...@gmail.com wrote:

> No, you have not specified enough information. Remember that the rocket was
> designed to accelerate at a constant rate AS MEASURED in the frame in which
> the rocket was at rest pre-firing. It’s very important to state that
> stipulation, because that is not the acceleration as measured in any other
> frame.

[Maciej Wozniak]

On Tuesday, 21 September 2021 at 18:55:56 UTC+2, Al Coe wrote:
> On Tuesday, September 21, 2021 at 9:26:54 AM UTC-7, sep...@yahoo.com wrote:
> > [Barnpole's text has been edited for coherence:]
> > Let's say that in terms of F0 a rocket accelerates at a constant rate from 0 to V,
> > and then decelerates from V back to zero at the same constant rate.... In terms of
> > F1 an identical rocket is initially at rest and accelerates at constant rate to a velocity
> > V and then decelerates back to zero velocity at the same constant rate. Does this
> > entail any contradiction?
>
> Of course not. The equations of physics take the same form in terms of every system of inertial coordinates,

It is nowhere, but we still can make some brilliant scenarios
with this rule.

[Al Coe]

On Tuesday, September 21, 2021 at 9:57:29 AM UTC-7, sep...@yahoo.com wrote:
> > Odd Bodkin -- maker of fine toys, tools, tables

> You asked:
> >So when you say the rocket is designed to decelerate at the same constant
> >rate, is that AS MEASURED in the frame in which the rocket was moving at
> >c*sqrt(3)/2 before firing?

> Yes. For example, as was stated...

Right you already stated that in your previous message. Why are you wasting time responding to misguided replies while ignoring the clear and complete answers to your questions? Again, stated succinctly, your question was as follows: Let's say that in terms of F0 a rocket accelerates at a constant rate from 0 to V, and then decelerates from V back to zero at the same constant rate. Also, in terms of F1 an identical rocket is initially at rest and accelerates at constant rate to a velocity V and then decelerates back to zero velocity at the same constant rate. Does this entail any contradiction?

Answer: Of course not. By the principle of relativity, we can obviously have perfectly congruent scenarios in terms of each such system, and this does not entail any contradiction. The relevant point is that the first rocket does not accelerate at constant rate in terms of F1, and the second rocket does not accelerate at constant rate in terms of F0. That's why your attempt to construct a contradiction fails, because your alleged contradiction relies on your (false) belief that a trajectory with constantly coordinate acceleration in terms of F0 also has constant coordinate acceleration in terms of F1, which you have now learned is not true. That's the fallacy in your attempted contradiction. You failed to account for the relativity of simultaneity for inertial coordinate systems. Understand?

[Odd Bodkin]

OK, that’s fine. In this case, *all* measurements and behaviors you cited
are made in F0, including the start speed, the increase of speed to
c*sqrt(3)/2, the decrease of speed back to 0.

And as we mentioned before, if any observers in F1 were to measure any of
that, they would get different answers for the acceleration in both
directions than what the observers in F0 made.

And if you started an identical rocket from rest in F1, and made all the
measurements from F1, then you’d get similar measurements. Of course, any
observers in F0 would measure THAT rocket’s behavior to be different than
what the observers in F1 say.

[Odd Bodkin]

Al Coe <coea...@gmail.com> wrote:
> On Tuesday, September 21, 2021 at 9:57:29 AM UTC-7, sep...@yahoo.com wrote:
>>> Odd Bodkin -- maker of fine toys, tools, tables
>> You asked:
>>> So when you say the rocket is designed to decelerate at the same constant
>>> rate, is that AS MEASURED in the frame in which the rocket was moving at
>>> c*sqrt(3)/2 before firing?
>> Yes. For example, as was stated...
>
> Right you already stated that in your previous message. Why are you
> wasting time responding to misguided replies while ignoring the clear and
> complete answers to your questions?

LOL. I’m not biting.

> Again, stated succinctly, your question was as follows: Let's say that
> in terms of F0 a rocket accelerates at a constant rate from 0 to V, and
> then decelerates from V back to zero at the same constant rate. Also, in
> terms of F1 an identical rocket is initially at rest and accelerates at
> constant rate to a velocity V and then decelerates back to zero velocity
> at the same constant rate. Does this entail any contradiction?
>
> Answer: Of course not. By the principle of relativity, we can obviously
> have perfectly congruent scenarios in terms of each such system, and this
> does not entail any contradiction. The relevant point is that the first
> rocket does not accelerate at constant rate in terms of F1, and the
> second rocket does not accelerate at constant rate in terms of F0. That's
> why your attempt to construct a contradiction fails, because your alleged
> contradiction relies on your (false) belief that a trajectory with
> constantly coordinate acceleration in terms of F0 also has constant
> coordinate acceleration in terms of F1, which you have now learned is not
> true. That's the fallacy in your attempted contradiction. You failed to
> account for the relativity of simultaneity for inertial coordinate systems. Understand?
>

--

[sep...@yahoo.com]

Now let's say the F0 rocket reached a velocity V=c*sqrt(3)/2 with respect to F0, and at that same time and x coordinate, frame F1 starts his identical rocket accelerating toward F0, just as the F0 rocket starts accelerating back toward F0. F0 and F1 have a relative velocity along the x-axis of |V|=c*sqrt(3)/2. The F1 and F0 rockets are identical, they both accelerate side by side, going the same direction along the x-axis, they both have the identical velocity at that point in time and space, why do you say the two identical rockets do not accelerate at the same rate? What is PHYSICALLY different between the two rockets that makes them accelerate at different rates compared to each other?
David Seppala
Bastrop TX

[Maciej Wozniak]

On Tuesday, 21 September 2021 at 19:53:23 UTC+2, bodk...@gmail.com wrote:

> OK, that’s fine. In this case, *all* measurements and behaviors you cited
> are made in F0, including the start speed, the increase of speed to
> c*sqrt(3)/2, the decrease of speed back to 0.
>
> And as we mentioned before, if any observers in F1 were to measure any of
> that, they would get different answers for the acceleration in both
> directions than what the observers in F0 made.
>
> And if you started an identical rocket from rest in F1, and made all the
> measurements from F1, then you’d get similar measurements. Of course, any
> observers in F0 would measure THAT rocket’s behavior to be different than
> what the observers in F1 say.

[Al Coe]

On Tuesday, September 21, 2021 at 11:26:07 AM UTC-7, sep...@yahoo.com wrote:
> Let's say the F0 rocket reached a velocity V with respect to F0, and at that same

> time and x coordinate, frame F1 starts his identical rocket accelerating toward F0,
> just as the F0 rocket starts accelerating back toward F0. F0 and F1 have a relative
> velocity along the x-axis of |V|=c*sqrt(3)/2.

> The F1 and F0 rockets ... both accelerate side by side...

Nope, you specified that what you call the F0 rocket accelerates at constant rate in terms of F0, and what you call the F1 rocket accelerates at constant rate in terms of F1, so they do not accelerate along the same trajectories and their paths diverge. You can easily plot their paths (in terms of either coordinate system), and show exactly how they diverge.

> they both have the identical [initial] velocity [and position] at that point in time and space...

Right, they have the same initial conditions, but they accelerate differently, one at constant rate in terms of F0 and the other at constant rate in terms of F1. That's why their trajectories diverge.

> why do you say the two identical rockets do not accelerate at the same rate?

Because in the situation you specified they do not accelerate at the same rate. One of them must modulate its thrust to maintain constant acceleration in terms of F0, and the other modulates its thrust to maintain constant acceleration in terms of F1.

> What is PHYSICALLY different between the two rockets that makes them accelerate at
> different rates compared to each other?

Again, each rocket must set and adjust its thrust level to achieve and maintain the desired acceleration, and they desire two different acceleration profiles, one that is constant in terms of F0 and the other that is constant in terms of F1. They are not the same. One rocket is trying to accelerate at the constant rate A in terms of the inertial coordinates in which it is starting out at rest, and the other is trying to accelerate at the constant rate A in terms of inertial coordinates in which it is starting out moving at speed V. Those are different acceleration rates.

Is there something about this you don't understand?

[sep...@yahoo.com]

You state:
>Again, each rocket must set and adjust its thrust level to achieve and maintain the desired acceleration.
If the F0 rocket goes from zero relative to F0 to V=c*sqrt(3)/2 at a constant rate as measured in F0 and then decelerates back to zero velocity relative to F0 at the same constant rate as measured in F0 during the first leg and then accelerates back to V=c*sqrt(3)/2 at the same rate as before and then back to zero relative to F0 at the same rate as before, with each leg of the journey taking the same amount of time as measured by observers in F0, and the F1 rocket does the analogous accelerations in F1 as measured by the observers in F1, then what happens when the F1 rocket starts its acceleration at the same x coordinate and time as when the F0 rocket has zero velocity relative to F1. F1 and F0 both measure that the time of each of the four cycles are identical during each of the accelerations and decelerations, the accelerations and decelerations are equal (as measured in each of their respective frames). F1 does not measure that the acceleration rate of his rocket changes during the second cycle, so he has no reason that the two rockets being side by side would accelerate at different rates, neither do observers in F0.
Explain what physically is different between the first cycle and the second cycle where observers in each frame say there is nothing different happening with their rocket during the first and second cycle, but something changes with the identical rocket in F1 during its first cycle compared to the F0 rocket's second cycle.
David Seppala
Bastrop TX

[Al Coe]

On Tuesday, September 21, 2021 at 12:26:54 PM UTC-7, sep...@yahoo.com wrote:
> What happens when the F1 rocket starts its acceleration at the same x coordinate
> and time as when the F0 rocket has zero velocity relative to F1?

That is exactly what I explained in the previous message. Again, when you have two rockets initially at the same location with the same state of motion, e.g., at rest in terms of F1, and you want to accelerate these rockets, one at the constant rate A in terms of F1 (in which both rockets are initially at rest), and the other at constant rate A in terms of F0 (in which the rockets are initially moving at speed V), then you need to set their throttles differently, because those are two different amounts of proper acceleration needing two different amounts of thrust.

Remember, a given amount of thrust produced by the rocket engines gives a certain amount of proper acceleration (for a given mass), but the amount of proper acceleration the rockets need to give the desired coordinate accelerations is different, because the coordinate acceleration equals the proper acceleration for the coordinates in which the rockets are at rest, but not for the coordinates in which the rockets are moving.

> F1 does not measure that the acceleration rate of his rocket changes...

Be careful. The proper acceleration of each rocket changes from the beginning of the leg to the end, so you need to modulate the rocket thrust to give the varying proper acceleration in order to maintain the constant coordinate acceleration. When the rocket is at rest in a coordinate system, the proper acceleration and coordinate acceleration are equal, but when the rock is moving with speed V in terms of that coordinate system, the proper and coordinate accelerations are different. (The faster a rocket movers, the more thrust is required to maintain a constant coordinate acceleration... and the required thrust goes to infinity as the speed approaches the speed of light.) Inside each rocket, you are setting proper acceleration, so you need to adjust the thrust as your speed changes to maintain constant coordinate acceleration.

> ... he has no reason that the two rockets being side by side would accelerate at
> different rates...

No, for each rocket, a constant coordinate acceleration in terms of a given system of coordinates is achieved by a certain amount of proper acceleration that varies depending on the speed, meaning it is different when speed is zero versus when speed is V. So, when you have two rockets initially at rest in F1, and one of them wants to accelerate at coordinate rate A in terms of F1 in which it is moving at speed 0 while the other wants to accelerate at coordinate rate A in terms of F0 in which it is moving at speed V, they require different proper accelerations, and hence they must set their thrust levels differently and hence their trajectories diverge.

> Explain what physically is different between the first cycle and the second cycle...

This has nothing to do with cycles. On each leg, each rocket needs to have a varying level of proper acceleration (and thrust) to maintain constant coordinate acceleration. The proper acceleration equals A when the rocket is at rest in the coordinates in which the coordinate acceleration is A, and when the rocket has speed V in terms of those coordinates the proper acceleration must be greater in order to maintain the same coordinate acceleration. So, when you have two rockets initially side by side, both initially at rest in F1, the one that wants to accelerate at A in terms of F1 just needs to set hus thrust for proper acceleration A, but the one that wants to accelerate at A in terms of F0 (in which it is moving at speed V) needs the increased proper acceleration (and thrust). Of course, by the end of the leg, the situation is reversed, meaning the rocket accelerating at constant rate in F1 is now moving at speed V is F1, so it needs proper acceleration greater than A, whereas the other rocket now has speed 0 in terms of F0, so it just needs proper acceleration A to maintain that coordinate acceleration in F0.

Do you understand this? If not, what part is unclear?

[Odd Bodkin]

No, they will not be accelerating side by side. You’ve lost track of which
frames these accelerations are measured from. Remember that the F0 rocket’s
acceleration from that common time and location in F1 is AS MEASURED by
observers in F0. It has that value of 3*sqrt(3)/2 as measured in F0, but
that’s not the value of that rocket’s acceleration as measured in F1. On
the other hand, the rocket firing from rest in F1 DOES have the
acceleration 3*sqrt(3)/2 in the frame F1. These two rockets do not have the
same acceleration profile in F1 and so they will not stay side-by-side.
Likewise, the acceleration of the F1 rocket will not be 3*sqrt(3)/2 as
measured in F0, while the F0 rocket’s acceleration will be that value in
F0, so they will not travel side by side according to F0 either.

Recall what I recapped above. When you have the F0 rocket turning on its
engine to have the deceleration 3*sqrt(3)/2, that was AS MEASURED in F0,
not as measured in F1. That is measured from the reference frame in which
the F0 rocket is at rest POST-firing. This is a completely different
acceleration profile than what you get for the case where the F0 (or F1)
rocket is at rest PRE-firing.

[sep...@yahoo.com]

If the F0 rocket is shipped to F1 and the two rockets are side by side, accelerating at the same start time, no change in thrusters is needed so that the two rockets travel side by side with zero relative velocity in their journey. However, you say if the F0 rocket instead of being shipped to the starting location in F1 travels there on its own its thrusters need to be adjusted so the two rockets travel side by side with zero relative velocity in their journey. Explain what changes in the F0 rocket to make that necessary if the rocket is shipped versus traveling on its own.
David Seppala
Bastrop TX

[sep...@yahoo.com]

There are two identical rockets, the F0 rocket and the F1 rocket. If in F0 the F0 rocket accelerates from 0 to V, and then decelerates to 0, and then accelerates back to V and then back to 0, and the acceleration rate and decelerate rate are a constant |3*sqrt(3)/2| meters per second squared in F0, and the analogous measurements are made by observers in F1 for the F1 rocket, explain what happens when the F0 rocket decelerates (or in other words accelerates toward F0) and the identical F1 rocket accelerates toward F0, with both rockets starting that part of their accelerations simultaneously and at the same x coordinate. Why does one rocket accelerate at a different rate?
David Seppala
Bastrop TX

[Odd Bodkin]

You keep saying, “everything is the same” when it isn’t. The nature of the
burn to decelerate it is different than the nature of the burn to
accelerate it. One is measuring the acceleration from a frame where the
rocket is at rest PRE-burn. The other is measuring the acceleration from a
frame where the rocket is at rest POST-burn. For it to have constant
acceleration as measured in the F0 frame in the two cases, the burn
profiles have to be different. (Again, I refer to the car accelerating from
0 to v in one frame and from v to 2v in another frame and the increase in
kinetic energy is completely different between the two, and that’s a
*classical* result.)

If it helps at all, remind yourself from children’s books about relativity
that applying a constant force to an object will not result in the same
increase in speed for every second the force is applied. As the object gets
faster and faster in some frame of reference, the increment in speed will
slowly start to drop. (This is where the unfortunate “relativistic mass”
gets brought up in children’s books.) So think about what that means in
this case. For the acceleration to be CONSTANT as measured in F0, then for
the acceleration phase, you’re actually have to apply more and more thrust
as time goes on. But for the deceleration to be constant as measured in F0,
then the force to be applies is going to get less and less as time goes on.
The point is, the burn profile is different for acceleration than for
deceleration, even though the acceleration is held constant as measured in
F0. DO NOT ASSUME that because the acceleration is held constant, then the
burn is constant and everything is the same. Remember the children’s books
about relativity and the fact that a constant force does NOT produce a
constant increment in speed each successive second.

Now it should be obvious why the F0 rocket decelerating from being at rest
in F1 to at rest in F0 is different than the F1 accelerating from being at
rest in F1 to at rest in F0. The burn profiles are completely different for
acceleration and deceleration. But you’ve insisted the rockets are
identical, and so you are racing one that is on an acceleration burn
against one that is on a deceleration burn. They won’t be side by side.

[Al Coe]

On Tuesday, September 21, 2021 at 2:00:31 PM UTC-7, sep...@yahoo.com wrote:
> If the F0 rocket is shipped to F1 and the two rockets are side by side,

> accelerating at the same start time, no change in thrusters is needed...

That's incorrect. If you want one rocket to accelerate at constant rate A in terms of F0 and the other to accelerate at constant rate A in terms of F1, they must set their throttles differently, to deliver the different amounts of proper acceleration that correspond to those two different coordinate accelerations.

> so that the two rockets travel side by side with zero relative velocity in their journey.

No, see above.

> However, you say if the F0 rocket instead of being shipped to the starting location

> in F1 travels there on its own...

No, this has nothing to do with how the rockets are placed into the specified initial condition. It is stipulated that the two rockets are initially at rest in terms of F1, and we want them to undergo constant coordinate accelerations in terms of F1 and F0 respectively. This requires that the rockets set their throttle differently to produce the different proper accelerations corresponding to the specified coordinate accelerations.

> its thrusters need to be adjusted so the two rockets travel side by side with
> zero relative velocity in their journey.

Well, if you want them to remain side by side, you would need to apply equal accelerations to them (meaning equal poroper accelerations or, equivalently, equal coordinate accelerations in terms of the same coordinate system), which you are not doing.

> Explain what changes in the F0 rocket to make that necessary if the rocket
> is shipped versus traveling on its own.

Non-sequitur. See above.

> ... with both rockets starting their accelerations simultaneously and at the same
> x coordinate, why does one rocket accelerate at a different rate?

They accelerate at different rates because the scenario you described entails burning fuel at different rates to deliver the different amounts of thrust and proper acceleration corresponding to the coordinate acceleration rates that you specified. For example, if you begin with two rockets initially co-located and at rest in terms of F0, and you want one rocket to accelerate at the rate A in terms of F0 and the other to accelerate at rate A in terms of F1, the former rocket needs initial proper acceleration A, and the latter needs the greater initial proper acceleration A+. Then, at the end of that leg, when the speed of the former is V in terms of F0 and the speed of the latter is 0 in terms of F1, the proper acceleration requirements are reversed, i.e., the former needs A+ and the latter needs A.

In other words, the proper acceleration for the first rocket increases from A to A+, and the proper acceleration for the second rocket decreases from A+ to A. Understand?

[Python]

You're not going well Maciej. You're know posting this blatant lie
hundreds of times completely out of context as there is not a single
word about GPS in the post you are answering to.

What about asking for some medical help, Maciej? Think about it.

[sep...@yahoo.com]

In the scenario, I say one rocket (F0 rocket) is going from v to 0 (F0) rocket and the other rocket (F1 rocket) is traveling in the negative x direction so it is also going from V (relative to F0) to 0 (or in terms of F1, from 0 to -V).

> Now it should be obvious why the F0 rocket decelerating from being at rest
> in F1 to at rest in F0 is different than the F1 accelerating from being at
> rest in F1 to at rest in F0. The burn profiles are completely different for
> acceleration and deceleration. But you’ve insisted the rockets are
> identical, and so you are racing one that is on an acceleration burn
> against one that is on a deceleration burn. They won’t be side by side.

When you say the "burn profiles are different" for accelerating versus decelerating, I said that in F0 the acceleration of the F0 rocket is |3*sqrt(3)/2 meters per second squared| whether the rocket is accelerating or decelerating, just as they are as measured in F1 for the F1 rocket. What do you mean that they are different?
David Seppala
Bastrop TX

[Al Coe]

On Tuesday, September 21, 2021 at 3:38:44 PM UTC-7, sep...@yahoo.com wrote:
> When you say the "burn profiles are different" for accelerating versus decelerating,

> I said that in F0 the acceleration of the F0 rocket is [A] whether the rocket is

> accelerating or decelerating, just as they are as measured in F1 for the F1 rocket.
> What do you mean that they are different?

Again, the "rocket burn profile" corresponds to the *proper* acceleration. For each rocket to have constant *coordinate* acceleration A in terms of the respective coordinate systems during this leg, the proper acceleration of one rocket must be varied from A1 (=A) to A2 (=A+delA), and the proper acceleration of the other rocket must be varied from A2 to A1. On the reverse legs, the first rocket's proper acceleration must be varied from A2 back to A1, and the second rocket's proper acceleration must be varied from A1 back to A2. If you still don't understand this, can you point out what you think is unclear?

[Maciej Wozniak]

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

[Odd Bodkin]

But with different acceleration profiles. For the F0 rocket, the velocity
is going from v to 0 (relative to F0) at a constant acceleration as
measured from the frame in which the rocket is at rest POST-burn (F0). For
the F1 rocket, the the velocity is going from v to 0 (relative to F0) at a
constant acceleration as measured from the frame in which the rocket is at
rest PRE-burn (F1). The latter is because you specified that the F1 rocket
is designed the same way as the F0 rocket, which means that it is designed
to burn in such a way that it has a constant acceleration from the frame in
which it was at rest PRE-burn.

Since the burns cannot be constant (and in fact, one applies more and more
force to the rocket, and the other applies less and less force to the
rocket), they cannot accelerate the same way.

>
>> Now it should be obvious why the F0 rocket decelerating from being at rest
>> in F1 to at rest in F0 is different than the F1 accelerating from being at
>> rest in F1 to at rest in F0. The burn profiles are completely different for
>> acceleration and deceleration. But you’ve insisted the rockets are
>> identical, and so you are racing one that is on an acceleration burn
>> against one that is on a deceleration burn. They won’t be side by side.
>
> When you say the "burn profiles are different" for accelerating versus
> decelerating, I said that in F0 the acceleration of the F0 rocket is
> |3*sqrt(3)/2 meters per second squared| whether the rocket is
> accelerating or decelerating,

That’s right. But I explained to you that this demands different burn
profiles. In the case of accelerating, you have to apply more and more
force with time to maintain the same acceleration as viewed from the F0
frame. In the case of decelerating, you have to apply less and less force
with time to maintain the same acceleration as viewed from the F0 frame.
This is what you learn from a children’s book about relativity — that a
constant applied force does not produce a constant acceleration.

Now, from the F1 frame, if there were a rocket fired from rest in F0 and
more and more force were applied with time, the acceleration of that rocket
as viewed from F1 will SURELY not be constant, even it’s constant with
respect to F0. In fact, F1 will see that rocket’s acceleration (from -v to
0, with respect to F1) get higher and higher with time.

You are mistakenly thinking “F=ma. Constant acceleration means constant
force.” But F=ma is not the right rule. F=dp/dt is the right rule, and in
this rule, a constant force produces a constant change in momentum, but not
a constant acceleration. To produce a constant acceleration, you have to
increase the force with time, as viewed in a frame in which the object is
initially at rest and speeding to moving; and to decrease the force with
time, as viewed in a frame in which the object is initially moving and is
slowing to rest.

> just as they are as measured in F1 for the F1 rocket. What do you mean
> that they are different?
> David Seppala
> Bastrop TX
>