# Relativity Acceleration Question

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### sep...@yahoo.com

Sep 18, 2021, 5:13:34 PMSep 18
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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.

Sep 18, 2021, 5:55:48 PMSep 18
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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?
Message has been deleted

### Dono.

Sep 18, 2021, 6:32:32 PMSep 18
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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

Sep 18, 2021, 6:34:10 PMSep 18
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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.

Sep 18, 2021, 6:37:38 PMSep 18
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I think that I cut him short this time. He's choking on the answer (hopefully, for good)

### sep...@yahoo.com

Sep 18, 2021, 7:01:14 PMSep 18
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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.

Sep 18, 2021, 7:18:05 PMSep 18
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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

Sep 18, 2021, 8:34:38 PMSep 18
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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

Sep 19, 2021, 12:31:13 AMSep 19
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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 19, 2021, 12:32:05 AMSep 19
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### sep...@yahoo.com

Sep 19, 2021, 10:44:11 AMSep 19
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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.

Sep 19, 2021, 10:58:09 AMSep 19
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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

Sep 19, 2021, 1:51:18 PMSep 19
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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

Sep 19, 2021, 2:12:04 PMSep 19
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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

Sep 19, 2021, 2:30:11 PMSep 19
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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

Sep 19, 2021, 2:31:55 PMSep 19
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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

Sep 19, 2021, 2:47:44 PMSep 19
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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

Sep 19, 2021, 3:07:13 PMSep 19
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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

Sep 19, 2021, 3:24:17 PMSep 19
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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

Sep 19, 2021, 3:30:45 PMSep 19
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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

Sep 19, 2021, 4:05:03 PMSep 19
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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

Sep 19, 2021, 4:44:24 PMSep 19
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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

Sep 19, 2021, 5:11:42 PMSep 19
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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

Sep 19, 2021, 5:32:33 PMSep 19
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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

Sep 19, 2021, 5:50:21 PMSep 19
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On Sunday, September 19, 2021 at 2:32:33 PM UTC-7, sep...@yahoo.com wrote:

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.

Special Relativity: 820 ... Barnpole Dave: 0

### sep...@yahoo.com

Sep 19, 2021, 6:11:36 PMSep 19
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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

Sep 19, 2021, 6:44:05 PMSep 19
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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

Sep 19, 2021, 7:10:24 PMSep 19
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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

Sep 19, 2021, 7:23:16 PMSep 19
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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

Sep 19, 2021, 7:43:21 PMSep 19
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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

Sep 19, 2021, 7:57:26 PMSep 19
to
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

Sep 19, 2021, 8:15:02 PMSep 19
to
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

Sep 19, 2021, 9:04:42 PMSep 19
to
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

Sep 19, 2021, 10:18:02 PMSep 19
to
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

Sep 19, 2021, 11:02:41 PMSep 19
to
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

Sep 20, 2021, 12:16:42 AMSep 20
to
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

Sep 20, 2021, 8:39:14 AMSep 20
to
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

Sep 20, 2021, 8:51:13 AMSep 20
to
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

Sep 20, 2021, 8:53:49 AMSep 20
to
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

Sep 20, 2021, 9:02:59 AMSep 20
to
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

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

Sep 20, 2021, 9:32:29 AMSep 20
to
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

Sep 20, 2021, 9:55:29 AMSep 20
to
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

Sep 20, 2021, 11:02:36 AMSep 20
to
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

Sep 20, 2021, 11:27:13 AMSep 20
to
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

Sep 20, 2021, 11:31:46 AMSep 20
to
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

Sep 20, 2021, 11:48:46 AMSep 20
to
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.

Sep 20, 2021, 12:05:18 PMSep 20
to
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

Sep 20, 2021, 12:09:52 PMSep 20
to
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?
David Seppala
Bastrop TX

### Al Coe

Sep 20, 2021, 2:03:21 PMSep 20
to
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

Sep 20, 2021, 2:25:33 PMSep 20
to
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

Sep 20, 2021, 4:06:40 PMSep 20
to
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

Sep 20, 2021, 4:38:38 PMSep 20
to
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.

> 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

Sep 20, 2021, 5:14:34 PMSep 20
to
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

> 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
> two frames. The two identical rockets must both reach V simultaneously.
> David Seppala
> Bastrop TX
>

### sep...@yahoo.com

Sep 20, 2021, 5:17:24 PMSep 20
to
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
Message has been deleted

### Al Coe

Sep 20, 2021, 6:23:44 PMSep 20
to
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

Sep 20, 2021, 6:48:48 PMSep 20
to
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

Sep 20, 2021, 7:14:33 PMSep 20
to
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

Sep 21, 2021, 12:29:04 AMSep 21
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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