Just curious...

Is there a certain acceleration where the time and length distortion is increasing at a rate equal to the objects increase in velocity such that from an external frame of reference the object appears to not be moving at all?

No. From an external reference frame, an accelerating object always appears to be moving.

No. From an external reference frame, an accelerating object always appears to be moving.


Perhaps I am missing something. What do you call an external reference frame.? All frames except the observer considered moving would seem to be external.

In that respect we have shown that it is possible that at sufficiently high velocity, acceleration causes direction to appear to reverse. That is moving AWAY we appear to be getting CLOSER.

I infact mentioned the affect as being possible for accelerated recession to appear to being stationary.

***************************************

OK forget that you are taking the one view point again.

The rocket could see himself as standing still however.

A rocket in free space, firing its engine in a constant direction, will appear to all external observers in inertial frames to be continuously increasing its speed in the opposite direction (opposite to the direction of ejected fuel), although no one will see it reach the speed of light.

All observers will see the rocket undergoing continuing acceleration in one direction, although in some frames, its acceleration might be in a direction opposite to its veloctity.

Every observer is at rest in his own reference frame, by definition.

A rocket in free space, firing its engine in a constant direction, will appear to all external observers in inertial frames to be continuously increasing its speed in the opposite direction (opposite to the direction of ejected fuel), although no one will see it reach the speed of light.

All observers will see the rocket undergoing continuing acceleration in one direction, although in some frames, its acceleration might be in a direction opposite to its veloctity.

Every observer is at rest in his own reference frame, by definition.

Unfortunately, I think you are wrong on this. In the original "Lorentz Contraction Paradox" thread it was determined that in fact as one accelerates away at sufficient velocity and from a sufficient distance that Lorentz Contraction causes one to see themselves getting closer instead of further away. It follows that at the appropriate speed and distance therefore that a point of reversal takes place and at that point a state of rest would appear to have occured.

Further the rate of distance change over time indicates such closlure to occur at super FTL speeds. i.e. - many millions times the speed of 'c'.

Also, and I was going to ask James R., about this too, it has been determined in another thread that ALL relavistic functions are mutual and reversable but may be shifted by simultaneity.

My question becomes how does that apply in this case?

Unfortunately, I think you are wrong on this....causes one to see themselves getting closer instead of further away....
The question was what would an external observer see, not what would someone in the accelerated frame see. An external observer in an inertial frame would see continuing acceleration in one direction only.

The question was what would an external observer see, not what would someone in the accelerated frame see. An external observer in an inertial frame would see continuing acceleration in one direction only.

That is where we disagree. relative velocity is a recipocal function. That is if 'A' (rocket) has a velocity relative to 'B' (Earth) then 'B' must also have an equal velocity relative to 'A'. Otherwise it is not relative.

At given different velocities V1 and V2 during the acceleration of 'A', 'B' sees the same relative velocity (Perhaps shifted by time dilation and simulteneity) but ultimately any distance seen by the rocket between them must also be seen by Earth.

Therefore the changing distance must also be seen. It seems you agree that the rocket can see the reversal of distance affect and hence could see a pseudo-rest velocity while actually accelerating away.

But the question remains as to when and how that view can be made recipocal to Earth. Relavistic affects are based on the relavitive velocity which is and must be recipocal.

Now the gray area in my mind is that period of acceleration. While we can state that the rocket is accelerating, the fact is the rocket can by the equivelence principle simply argue I am in a gravity field and at rest, it is Earth that is accelerating away.

Because of the equivelence principle the forces of acceleration are the same affect as the gravity on Earth. There would appear to be no actual difference in their condition, particularily if the rocket is maintaining the same 1 g acceleration.

Further the rate of distance change over time indicates such closlure to occur at super FTL speeds. i.e. - many millions times the speed of 'c'.


This is full of crap, isn't it? Your MacM's velocity is a crap velocity, which you have no way to compare it to the real velocity.

This is full of crap, isn't it? Your MacM's velocity is a crap velocity, which you have no way to compare it to the real velocity.

NOTICE TO READERS: Paul T., seems to have a problem with basic physics. He wants to claim that V' = Vo + at and Vavg = (V1 + V2)/2 or that Velocity = Change in distance/Change of time; are not valid at cosmological scales or at relavistic speeds.

This is a consequence of my original "Lorentz Contraction Paradox" thread which shows that not only when you accelerate AWAY from an observer at relavistic speeds, do you actually get CLOSER but you can do so at many millions of times the speed of light.!

See: http://www.sciforums.com/showthread.php?p=597677#post597677
Starting 6/1 @ 8:28 PM.

After running his own calculation to find that indeed his results yielded a ds/dt in excess of 1 million 'c' he then started calling this MacM's velocity and claims it doesn't happen. :bugeye:

He can't believe that relativity actually mandates such results. James R., and others have verified the phenomena but Paul T knows more that the rest of us and knows that it just can't be. :D

MacM,

NOTICE TO READERS: Paul T., seems to have a problem with basic physics. He wants to claim that V' = Vo + at and Vavg = (V1 + V2)/2 or that Velocity = Change in distance/Change of time; are not valid at cosmological scales or at relavistic speeds.


MacM, you have problem. You applied Vo + at in V1 reference frame and expect the result (V') to be velocity relative to earth. For non relativistic case, that's okay. But, for relativistic case, that's not correct. Hahaha, MacM casting BS.

READERS:

Afraid the BS is flowing from elsewhere. Mr Paul T would have you agree that the relavistic distances are real.. The time of distance change is real but that ds/dt = v is not real.

Why because Relativity creates FTL changes and he doesn't like that. :D

MacM,

READERS:

Afraid the BD is flowing from elsewhere. Mr Paul T would have you agree that the relavistic distances are real.. The time of distance change is real but that ds/dt = v is not real.

Why because Relativity creates FTL changes and he doesn't like that. :D

Hahahaha. MacM, your physics understanding is so...so...soo...sloopy. And, it seem to be widespread. Don't even need to look for your misconception in relativity. You even thought velocity equation under acceleration was Vo + at/2, hahaha. God knows, what else your misconception are.

That's why I have so much to say about you. Everytime you tell us something, you reveal more misconception.

Now, you wanted to convince us that equation V = Vo + at is applicable in the relativistic condition as good as it is on non-relativistic condition. Hahahaha. What's a joke. You even discovered that FTL is possible within SR. Hahaha.

What are you trying to say by: "The time of distance change is real but that ds/dt = v is not real." ??? Haven't you taken my advice to read some books about velocity? May be a bit too late, but trust me, you need that.

READERS:

MacM,

Hahahaha. MacM, your physics understanding is so...so...soo...sloopy. And, it seem to be widespread. Don't even need to look for your misconception in relativity. You even thought velocity equation under acceleration was Vo + at/2, hahaha. God knows, what else your misconception are.

That's why I have so much to say about you. Everytime you tell us something, you reveal more misconception.

Now, you wanted to convince us that equation V = Vo + at is applicable in the relativistic condition as good as it is on non-relativistic condition. Hahahaha. What's a joke. You even discovered that FTL is possible within SR. Hahaha.

What are you trying to say by: "The time of distance change is real but that ds/dt = v is not real." ??? Haven't you taken my advice to read some books about velocity? May be a bit too late, but trust me, you need that.

First my apology to members here for Paul T's conduct. He will persist at drawing this discussion off topic with his unwarranted personal attacks.

Before ignoring him however, I do want to post this from another thread which he started for the sole purpose of challenging me (actually 1 of 3 threads he has started to do so and now comes here with the same ill advised BS).

************************************************** *****
:“ If your Vo + at/2 was a typo, why did you do it multiple times at different occasions? The possibilities are: 1) your computer changed your 'at' to at/2 without you knowing it....very unlikely, or 2) key board problem...also very unlikely since on your QUERTY keyboard '/' and '2' are located relative far away from 't', or 3) you thought Vo + at/2 was correct...this is the most likely posibility. ”


:Fine then I suggest you must believe that the speed of light was 3E6 m/s since you posted it multiple times as such. You have absolutely no understanding of my thoughts, no more than I yours when you claimed the speed of light was only 1% its generally accepted value.

:“ Also, when you didn't type the equation, you computed using Vo + at/2 too, such as your acceleration of 54m/s2 in one second that you thought would produce additional velocity of 27 m/s. MacM, it's not a typo...it was in your brain. ”


:Just why do you choose to ignore (for about the 10th time) the fact that I had already stated time dilation would affect the computed acceleration rate long before you complained about it? I posted the closure rate between observers for 1 sec or 7,071 seconds in my prior example to show that considering actual acceleration based on time dilation had no affect on the conclusion. Hence there was no necessity to actually compute theoretical acceleration as part of the point of the issue which was the reversal of the distance and the FTL conclusion. (Being distance change over time) was FTL with or without specific time dilated acceleration calculations.

:“ I don't need crystal ball to see that it wasn't a typo. See my above arguement. ”


:Nor do I to know you don't even know the correct speed of light.

:“ I have seen your equation: V' = Vo + at/2, for quite a few times. I first thought, it was just a typo, but since you repeatedly write it that way, I am now convince you don't even know what the correct equation for velocity is. May be you will tell me: "Well, it just a minor error. ”

:Actually you are correct. I did miss type it, it is not at/2 which has to do with distance (actually "at^2/2" but it has been corrected above before I got to this paragraph. Need I remind you of your repeated c = 3E6 m/s posts, which I pointed out to you.
************************************************** *******

His deliberate repeating of this ...at/2 typo error in my case (now at least 12 times) and his repeated ignoring that he has commited simular error by posting c = 3E6 m/s repeatedly but then wants to claim his was just a typo.

His error post when starting another of his multiple attack threads on the same topic:

:Sorry guys, Lorentz's Contraction "Paradox" again!

There is a ship moving away from earth at velocity V1=0.6c relative to earth. In V1 reference frame, the ship accelerates at acceleration a=1E6 m/s2 for an elapse time of Dt = 1 second and reaches velocity V2 relative to earth.

Velocity u gained in V1 reference frame is approximately 1E6*1=1E6 m/s and V2:

V2 = (0.6c+1E6)/(1+0.6c*1E6/c2) = 180,638,722 m/s (based on c = 3E8 m/s)

The "red" 8 was his correction after I pointed out his repeated error in the value of 'c'.

:Reposted by MacM:“ (based on c = 3E6 m/s) ”


:What is this?


:I'll note that you are calculating V2, where in my example I approximated an acceleration and specified V2. This is not a complaint, just a note about the changing of criteria of the example. I also note the continued value of c = 3E6 m/s. It is repeated throughout so I don't see that as a typo.
”

:It was just a typo, Mr MacM. It didn't affect the whole calculation.

Nor did my discontinuance of calculating time dilated acceleration (which I had pointed out in the beginning) alter the conclusion of the issue of the topic.

This shows his lack of character or should I more correctly say shows his extremly poor character.

If he continues to distrupt this thread I just hope you ALL tell him to kiss your ass.

Because of the equivelence principle the forces of acceleration are the same affect as the gravity on Earth. There would appear to be no actual difference in their condition, particularily if the rocket is maintaining the same 1 g acceleration.

Wrong. The gravity field of Earth falls off by the square of the distance from the center of the Earth. The gravity field that is equivalent to the forces of accleration of the Rocket extends uniformly in strength all the way from ship to Earth.

Since Relativistic effects due to gravitation depend on the difference in gravitational potential not the local measured strength, you do not see the same Relativistic effects sitting on the Earth as you do in the rocket while it is under accleration. The relative potential as seen from the Earth is proportional to GM/r-GM/d. As seen from the rocket it is proportional a*(d-r).

where M is the mass of the Earth, r the radius of the Earth, d the distance from the rocket to the surface of the Earth (as measured by each observer), and a the acceleration of the Rocket.

These lead to two different Relativistic results, especially when you consider that the Earth's field exists for both observers, (the rocket has to take its potential within the Earth's gravitational field in account also.) while the equivalent gravotational field to the rocket's accleration only exists in the rocket;s frame.

That is where we disagree. relative velocity is a recipocal function. That is if 'A' (rocket) has a velocity relative to 'B' (Earth) then 'B' must also have an equal velocity relative to 'A'. Otherwise it is not relative.

At given different velocities V1 and V2 during the acceleration of 'A', 'B' sees the same relative velocity (Perhaps shifted by time dilation and simulteneity) but ultimately any distance seen by the rocket between them must also be seen by Earth.

Therefore the changing distance must also be seen. It seems you agree that the rocket can see the reversal of distance affect and hence could see a pseudo-rest velocity while actually accelerating away.

But the question remains as to when and how that view can be made recipocal to Earth. Relavistic affects are based on the relavitive velocity which is and must be recipocal.

Now the gray area in my mind is that period of acceleration. While we can state that the rocket is accelerating, the fact is the rocket can by the equivelence principle simply argue I am in a gravity field and at rest, it is Earth that is accelerating away.

Because of the equivelence principle the forces of acceleration are the same affect as the gravity on Earth. There would appear to be no actual difference in their condition, particularily if the rocket is maintaining the same 1 g acceleration.
Since what I said applies to any observer of the rocket in an inertial reference frame, then we can talk about an Earth bound observer, if you like, although we need not. Any external inertial frame would work. Earth, though, is close enough to inertial for our purposes.

You are making this seem much more complicated than it actually is. If a rocket takes off from the earth and continues to fire its engines with no change in direction (just to keep it simple), then an Earth bound observer will see the distance to the rocket continue to increase, according to the formula Distance = speed x time, using, of course, the speed of the rocket as seen by the Earth bound observer. For a changing speed, just integrate speed with respect to time.

He will also see the velocity of the rocket continue to increase (that is its acceleration will continue in the same direction). To make this plausible, the Earth bound observer will see the rocket engine push the rocket away from the Earth with a force greater than zero, which will always cause an acceleration in the direction of the force, even relativistically. In order for him to see the rocket slow down, he would have to observe some force exerted on the rocket in a direction opposite to the direction its engine is acting on it, causing a net force pointing back towards the Earth. Clearly no force to counterbalance the rocket engine exists. Therefore, it will continue to accelerate in the same direction. Although F = MA is not true relativistically, F = dP/dt is.

MacM,


First my apology to members here for Paul T's conduct. He will persist at drawing this discussion off topic with his unwarranted personal attacks.


Hehehe, you don't have to apologize. Or, if you wanted to anyway, please apologize for all the BS that you have said since I was just commenting your BS. :D


If he continues to distrupt this thread I just hope you ALL tell him to kiss your ass.

Good idea. How about your telling BS. Should we keep quite? :D

Paul T, I see that you have made 158 posts, and that nearly all of them have been in conflict with MacM in some sort of personal vendetter. Personally I find this whole bun fight to me a total waste of your time and my reading efforts.
MacM has always taken the position of "radical exposer" and his posts are always provocative. This I think the forum has generally got used to.

However I would look forward to seeing something more productive of your posts instead of repeating the same as I feel you have something to offer to all readers and not just an example of a failed strategy to teach an "ole dog " to lay down, because I can tell you, MacM does not and wil not lay down.

Maybe you should use your intelligence a bit more how can I say "cleverlly" in proving MacM's point as wrong......for this repetition of "quote " posting is getting no where.

For example this thread I think is about the the conundrum of length contraction to do with a rigid object, although by now I am not sure what it's about. as the last 10 posts have been about the BS btween two posters and not much more.

Janus58,

:Because of the equivelence principle the forces of acceleration are the same affect as the gravity on Earth. There would appear to be no actual difference in their condition, particularily if the rocket is maintaining the same 1 g acceleration. ”

Wrong. The gravity field of Earth falls off by the square of the distance from the center of the Earth. The gravity field that is equivalent to the forces of accleration of the Rocket extends uniformly in strength all the way from ship to Earth.

1- I truely hope you know that I (we non-professionals) out here know that gravity is inverse square (sort of). Particularily since you know I have written extensively about gravity issues.

2 - You mis-lead others by your statement since inferring an error which doesn't exist. We are comparing only surface gravity of 1 g to the force of acceleration at 1 g. They according to Einstein are equivelent in every respect. (This assumes of course that the rocket is sufficiently fr away that its postion in Earth's grvity field is simply neglected).

:Since Relativistic effects due to gravitation depend on the difference in gravitational potential not the local measured strength, you do not see the same Relativistic effects sitting on the Earth as you do in the rocket while it is under accleration. The relative potential as seen from the Earth is proportional to GM/r-GM/d. As seen from the rocket it is proportional a*(d-r).

OK. Then translate those formulas into hard numbers. i.e. Earth's gravity at 1g on the surface and the rocket accelerating with a force of 1g. Show us the difference. Should there actually be a difference then it doesn't take a genius to sacle one against the other to set them equivelent in any case. Point being this issue does not trump the arguement.

:These lead to two different Relativistic results, especially when you consider that the Earth's field exists for both observers, (the rocket has to take its potential within the Earth's gravitational field in account also.)

Just what value shift do you suggest we give the earth's gravity field to the rocket 10 B Ly away?

Brandon9000,

We actually agree on this issue. My posts however attempt to address the reciprocity of relative velocity. That is the fact that relative velocity must be equal to both observers and hence relavistic affects must be the same.

Paul T, I see that you have made 158 posts, and that nearly all of them have been in conflict with MacM in some sort of personal vendetter. Personally I find this whole bun fight to me a total waste of your time and my reading efforts.
MacM has always taken the position of "radical exposer" and his posts are always provocative. This I think the forum has generally got used to.

However I would look forward to seeing something more productive of your posts instead of repeating the same as I feel you have something to offer to all readers and not just an example of a failed strategy to teach an "ole dog " to lay down, because I can tell you, MacM does not and wil not lay down.

Maybe you should use your intelligence a bit more how can I say "cleverlly" in proving MacM's point as wrong......for this repetition of "quote " posting is getting no where.

For example this thread I think is about the the conundrum of length contraction to do with a rigid object, although by now I am not sure what it's about. as the last 10 posts have been about the BS btween two posters and not much more.

Thank you QQ. I agree with your post in every detail; including the fact that Paul T seems to likely to be able to actually contribute. But unfortunately he seems to like these pissing contests and it isn't 158 post (unless you actually counted all five (5) threads where he is spreading his same repeated bile over and over.

Janus58,
2 - You mis-lead others by your statement since inferring an error which doesn't exist. We are comparing only surface gravity of 1 g to the force of acceleration at 1 g. They according to Einstein are equivelent in every respect. (This assumes of course that the rocket is sufficiently fr away that its postion in Earth's grvity field is simply neglected).

The only one mis-leading here is you. Einstein very carefully said that the Relativistic effects of gravity were due to difference in potential not difference in local force (either by gravity or acceleration.) Surely someone who has written so much about gravity understands the difference between gravitational force and gravitational potential.




OK. Then translate those formulas into hard numbers. i.e. Earth's gravity at 1g on the surface and the rocket accelerating with a force of 1g. Show us the difference. Should there actually be a difference then it doesn't take a genius to sacle one against the other to set them equivelent in any case. Point being this issue does not trump the arguement.



Okay, let's put the rocket at 2,000,000,000 m from the Earth. Thus the difference in potential would be a factor of 62574974 due the gravity of the Earth and 19600000000 as seen from the accelerating rocket.



Just what value shift do you suggest we give the earth's gravity field to the rocket 10 B Ly away?

The difference in potential for an object at ten lightyears distance with respect to the Earth's gravity field compared to an object sitting on the surface of the Earth would be a tad under 62775164.628410

Mac, I wouldn't advise arguing with Janus58 about special relativity. He is one of the
most knowledgable on this forum about SR. Even I can see your lack of understanding,
not meant as a put-down, just a fact.

Mac, I wouldn't advise arguing with Janus58 about special relativity. He is one of the
most knowledgable on this forum about SR. Even I can see your lack of understanding,
not meant as a put-down, just a fact.

Not so taken. Also if my post seemed to be an arguement it wasn't so intended. I asked that he post his math which he has done.

I'll only note that this doesn't address the point I raised about scaleing and creating the equality if not by force then by potential. In other words we can still have the circumstances as described being a case of recipocal relavistic function by adjusting distance from earth and/or acceleration.

I guess what I see here then is that the equivelence principle isn't really equivelent. To say that 1g acceleration is equivelent to 1g gravity should mean everything is the same. But according to this then that isn't true.

Personally I find this whole bun fight to me a total waste of your time and my reading efforts. MacM has always taken the position of "radical exposer" and his posts are always provocative. This I think the forum has generally got used to.

Don't read what you don't like. And to be honest, to call math the "radical exposer" is hilarious and a mockery to all those who have attempted to educate Mac on the basics of some physical theories.

But sure he'll reply that he knows what he is talking about... we've been there done that a dozen times already. Oh f*ck this...

Janus58,

Surely someone who has written so much about gravity understands the difference between gravitational force and gravitational potential.

It is clear this is meant as a put down. Let me simply point out that my gravity work does not include Relativity. Which is a rather nice result I think.

Don't read what you don't like. And to be honest, to call math the "radical exposer" is hilarious and a mockery to all those who have attempted to educate Mac on the basics of some physical theories.

But sure he'll reply that he knows what he is talking about... we've been there done that a dozen times already. Oh f*ck this...

Take a deep breath Crisp, breath. :D I really think you have mis-read QQ post. I don't see math any place. He is basically simply saying I have been a lightening rod. Which I have.

But in self defense I think the Lorentz Contraction Paradox thread went quite well. You all sure taught me a lot there huh?

Be honest now. Had you ever for a single moment realized the consequences of relavistic acceleration with vast distances between observers? Did you ever think you would see James agree to an FTL calculation?

the 158 figure shows as a mumber of posts as per sci forums stats....and whilst I am happy you agree I can only wish that Paul also agrees and we can get down some quality time

But sure he'll reply that he knows what he is talking about... we've been there done that a dozen times already. Oh f*ck this...
Crisp, I share your hmmmmm.....frustration but from another perspective.

The thought that came to mind is that you can't support a flawed theory with a flawed theory. Now before you scream foul!!....it works both ways.

To me as an observer I see certain problems from both sides and all I see is a chasing of the wind......going around in circles due to weaknesses in both cases. Both sides considering the other with contempt, when it is the weaknesses of both that prompted the discussion in the first place.
There are arguements raized that suggest contradiction, a FTL result....for example.....to me I think ...so what?

A mathematical model that has limitations in how it can be applied is all I see, and that aint so bad...is it?

MacM has allowed a "paradox" to co-exist and has stated that this doesn't invalidate relativity but exposes a possible area of concern.

And since all the heat that has been generated it has turned into a relativity vs MacM contest to see what should be thrown our and what should be kept....which I happen to think is totally unecessary.

But the thing that really gets my goat is the Utter crap that has filled this forum and all the others instead of dealing with errors we all end up defending the errors. And to take any issues personally I find as trivial and unproductive.

hmmmmmm.....sorry about that,,,,a little hot in here.....at the moment

hmmmmmm.....sorry about that,,,,a little hot in here.....at the moment

Here, here!. Not to say I accept your assumption of mine being a weak challenge. :D

You are correct however in pointing out I have not and do not claim these results in any way invalidate Relativity. That was never the intent of the thread but only to point out the "Paradox" (they don't have to accept the Webster definition, they make up their own definitons all the time which is part of the problem.)

Just as V' = ds/dt isn't valid at relavistic speeds. That is nonsense and is simply an arguement to provide cover for the result which they don't like.

Not so taken. Also if my post seemed to be an arguement it wasn't so intended. I asked that he post his math which he has done.

I'll only note that this doesn't address the point I raised about scaleing and creating the equality if not by force then by potential. In other words we can still have the circumstances as described being a case of recipocal relavistic function by adjusting distance from earth and/or acceleration.

No, you can't. For one thing, relative potential depends on the respective direction of the objects with respect to the gravity field/acceleration. In your case where the rocket is accelerating away from the Earth, the Earth is at a lower potential with respect to the rocket. The rocket however is at a higher potential with respect to the earth's gravitational field. For them to be recipocal, Both would have to be at higher potentials with respect to each other. For another, the Earth's gravitational field exists for both rocket and Earth, while the equivalent gravity field for the rocket due to its acceleration only exist for the rocket.



I guess what I see here then is that the equivelence principle isn't really equivelent. To say that 1g acceleration is equivelent to 1g gravity should mean everything is the same. But according to this then that isn't true.

No, the equivalence principle holds, it is just that you have, once again, misinterpreted what it means. Einstein was very specific by what he meant by equivalence. When it come to discussing Relativity it doesn't matter one whit what you think equivalence should mean; All that matters is what Einstein meant by equivalence.

No, you can't. For one thing, relative potential depends on the respective direction of the objects with respect to the gravity field/acceleration. In your case where the rocket is accelerating away from the Earth, the Earth is at a lower potential with respect to the rocket. The rocket however is at a higher potential with respect to the earth's gravitational field. For them to be recipocal, Both would have to be at higher potentials with respect to each other. For another, the Earth's gravitational field exists for both rocket and Earth, while the equivalent gravity field for the rocket due to its acceleration only exist for the rocket.



No, the equivalence principle holds, it is just that you have, once again, misinterpreted what it means. Einstein was very specific by what he meant by equivalence. When it come to discussing Relativity it doesn't matter one whit what you think equivalence should mean; All that matters is what Einstein meant by equivalence.

OK, I'll back peddle on this point for now. Thanks for clarifying the gravity equivelence issue.