[HydrogenBond]

Special relativity has three equations one for mass, one for distance and one for time, since these all three will change with velocity. To get past the confusing reference special effects that can occur using only distance and time, I prefer to do an energy balance by looking at mass/relativistic mass changes. Because of the conservation of energy, relativistic mass changes can not be reference dependent but need to be absolute. If it was reference dependent, we could use a reference special effects, to create or take away energy, making an universal energy balance impossible since it would be relative. We could also do perpetual motion tricks.

You simply compare SR energy/mass changes, for both references, the one with most energy, will time dilates the most.

[jakuta]

modest, on 23 October 2010 - 09:12 AM, said:

The special theory of relativity developed by Einstein in 1905. In particular, section 2 of On The Electrodynamics of Moving Bodies -- 1905 -- A. Einstein which concludes:

Yea, I understand that part.

But, since the velocity add equations are used to derive the the acceleration equations based on everything I read from you, you did agree with the SR acceleration equations.

It that true?

Quote

To accelerate is to move from one system of coordinates to another. There are two frames of reference for O'. In the first, t' = 2.5 is simultaneous with t = 2. In the second, t' = 2.5 is simultaneous with t = 3.125. Because your thought experiment relies on "when" t' = 2.5 (a matter of simultaneity) it is important not to neglect this detail.
A cynical person would think you designed this thought experiment to obfuscate the relativity of simultaneity.

May I see your equations to prove this?

And my thought epxeriment relies on the acceleration equations. I doubt you disagree with the relative motion part.

And, further given any acceleration phase "when" can be determined since that phase is absolute.

It is a common misconception that Special Relativity cannot handle accelerating objects or accelerating reference frames.
In special relativity accelerating frames are different from inertial frames. Velocities are relative but acceleration is treated as absolute.

http://math.ucr.edu/...celeration.html

So, you are trying to inject the relativity of simultaneity, but it is not applicable to acceleration under SR. Both frames will agree on the time differential between the two frames between the start of the acceleration and the end.

Quote

If that is your position, then I can only conclude that you are intentionally solving the problem in a manner inconsistent with relativity.

No, I am not using that tool to operate on this. But, can you explain the simultaneity shift when t' = t for the x and x' I gave. This is a completely issue from the twins issue.

Quote

The "plane of simultaneity shift" is explained on wiki's relativity of simultaneity page. It is not a myth. It is explained in the context of the twin paradox like so:

I will mark the shift in simultaneity on the spacetime diagram I made:

I do not have a turn around in this thought experiment, so this is not applicable.

Quote

The burn time is not absolute. You found the burn time relative to t. You could also find the burn time relative to t'. There are two equations given in that link:

t = (c/a) sh(at'/c)

t' = (c/a) sh^-1(at/c)

You can substitute t = BT or t' = BT. The two observers disagree on the burn time, but that is not at issue. The burn time is zero in the thought experiment—we are switching frames instantly.



The amount of time it takes to accelerate is not at issue.

I think I see our disagreement. The equations are based on BT in the context of the accelerating frame. So, BT is the known value. Since this is controlled in the context of the accelerating, then we can solve for t in the context of the twin that remained behind.

t = (c/a) sh(a(BT)/c)

If you disagree with this, then you disagree with the SR equations.

Therefore, if the accel frame elapses BT duing the acceleration period, then the stay at home frame elapses (c/a) sh(a(BT)/c). The frames will not disagree on this.

Since the frames do not disagree on the timing differential, then instantly acquiring v is not a problem, Einstein used this same tool in the explanation of time dilation.

From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other
http://www.fourmilab...in/specrel/www/

Quote

The velocity addition equation is not at issue. Nowhere in this thought experiment are we adding velocities.

It is used to derive the acceleration equations as I said.

You know that SR says that
"When an object P has a relative velocity u w.r.t.
to an inertial frame Q , of which the observer itself
has a relative velocity v w.r.t. an inertial observer R ,
then the object P has a relative velocity
w = (u+v)/(1+u v)
w.r.t. the observer R."

http://users.telenet...celeration.html

Quote

Rigidity is not at issue here. The thought experiment would work for point particles.

I tend to agree, but others would disagree.

Quote

Symmetry is is not achievable because the situation is non-symmetric. When O accelerates, O' is co-located with O. When O' accelerates they are separated by vt'. Clearly O does not experience the same thing as O'.

You can, however, achieve a solution in which both observers are 'at rest' through the duration of the thought experiment, which is, I think, what you are really looking for. To do this you must invoke general relativity as Einstein explains here:

http://en.wikisource...y_of_relativity

No it is not what I am looking for.

I did exactly what I wanted to do.

Here is the velocity attained by the acceleration
v = c th(a(BT)/c)

So, you have no basis to claim just because they are separated by a distance each will arrive at a different v relative to the original co-located position.

This implies the rules of physics are different basd on distance.

So, since v = c th(a(BT)/c) for both, then they are in the same frame after each accelerates because that is a v relative to the same frame that each had been a member.

[jakuta]

Proof of symmetry
Assume three clocks A, B, C are co-located, in the same frame and synchronized. Assume clock A instantly acquires v in the direction of the positive x-axis. Then, after some time on the clocks of B and C, B instantly acquires v in the direction of the positive x-axis. From the view of C both acquired v and hence are in the same frame since both are at a relative motion v from the view of C.

Now, it is shown clock A agrees B is in the same frame. Since A is in a different frame from B and C and B will move relative to C, then the velocity addition equations apply. From the view of A, B and C are in relative motion -v, i.e. in the direction of the negative x-axis. B will acquire +v relative to C. Hence, the numerator of the SR velocity addition equations is v - v and thus, from the view of A, the relative velocity of B is zero and hence they are in the same frame.

Therefore, from the view of either frame, A and B will end up in the same frame.

[modest]

jakuta, on 23 October 2010 - 11:05 AM, said:

But, since the velocity add equations are used to derive the the acceleration equations based on everything I read from you, you did agree with the SR acceleration equations.

It that true?

I have no problem with the clock postulate or the relativistic rocket equations. They aren't applicable to this thought experiment. You could, no doubt, design some variant of your thought experiment where a non-instant acceleration plays a role. If you did that, then I would note two things: First, the relativistic rocket equation (or, acceleration equation) that you’ve been considering would still be insufficient to fully describe O from the perspective of O’. Second, the result would need to be consistent with the result we already have. Both points are made here:

http://books.google....rame%22&f=false

at the end of page 168 and page 169.

There is a tool, however, that would allow you to describe things from the accelerated observer’s perspective completely (from the start of the thought experiment to the end), exactly, and easily. It is general relativity. General Relativity is a generalization of special relativity for the case of accelerated motion.

Whether you use general relativity or Rindler coordinates, the conclusion will be the same. In the line parallel to acceleration, clocks are dilated slower in the negative direction as a function of distance and faster in the positive direction as a function of distance during the acceleration. In an acceleration version of your thought experiment, while O accelerates there is no significant distance between O and O’. When O’ accelerates, O is further ahead in the direction of acceleration (higher in gravitational potential in the general relativistic sense and greater in X in Rindler coordinates). It’s clock will therefore run fast relative to O’ during the acceleration which is consistent with the result we’ve already found. The clock moves from t=2 to t=3.125 from the perspective of O’. It ran fast.

jakuta, on 23 October 2010 - 11:05 AM, said:

modest, on 23 October 2010 - 09:12 AM, said:

To accelerate is to move from one system of coordinates to another. There are two frames of reference for O'. In the first, t' = 2.5 is simultaneous with t = 2. In the second, t' = 2.5 is simultaneous with t = 3.125. Because your thought experiment relies on "when" t' = 2.5 (a matter of simultaneity) it is important not to neglect this detail.

May I see your equations to prove this?

Given on wiki here:

Quote

From the first equation of the Lorentz transformation in terms of coordinate differences



it is clear that two events that are simultaneous in frame S (satisfying Δt = 0), are not necessarily simultaneous in another inertial frame S′ (satisfying Δt′ = 0). Only if these events are colocal in frame S (satisfying Δx = 0), will they be simultaneous in another frame S′.

Special Relativity

In step 3:

modest, on 22 October 2010 - 11:20 PM, said:

jakuta, on 19 October 2010 - 03:33 PM, said:

3. After time t', O' will acquire v in precisely the same way as O in precisely the same direction.

From the perspective of O', t jumps from 2 to 3.125 (because the plane of simultaneity shifts)

set Δt' = 0, v = -.6 (negative because the change in velocity is toward the other frame) and Δx = 1.5 (the distance in the first frame) and you should get the correct shift of 1.125. In fact, let's do that now,







There you have it. By the Lorentz transformations, when O' changes frames t changes by 1.125. It also should be clear from the properties of Minkowski spacetime. The diagram I made is perhaps helpful in that regard.

Here, also, is a very good wikibook on the topic:

Special Relativity/Simultaneity, time dilation and length contraction

The first section on phase shifting describes the part you are doing wrong well. In particular, that switching frames of reference moves the present instant along the time axis in a manner that depends on distance. The Andromeda paradox is a good example of this. Over very large distances (to the Andromeda galaxy) the present instant shifts quite a lot even with a small velocity change.

I think your question has been answered rather well and any inconsistencies are now shown to result from your refusal to apply relativity properly. Perhaps someone else has a different perspective that you would find helpful.

~modest

[jakuta]

modest, on 24 October 2010 - 02:58 PM, said:

I have no problem with the clock postulate or the relativistic rocket equations. They aren't applicable to this thought experiment. You could, no doubt, design some variant of your thought experiment where a non-instant acceleration plays a role. If you did that, then I would note two things: First, the relativistic rocket equation (or, acceleration equation) that you’ve been considering would still be insufficient to fully describe O from the perspective of O’. Second, the result would need to be consistent with the result we already have. Both points are made here:

http://books.google....rame%22&f=false

at the end of page 168 and page 169.

There is a tool, however, that would allow you to describe things from the accelerated observer’s perspective completely (from the start of the thought experiment to the end), exactly, and easily. It is general relativity. General Relativity is a generalization of special relativity for the case of accelerated motion.

Whether you use general relativity or Rindler coordinates, the conclusion will be the same. In the line parallel to acceleration, clocks are dilated slower in the negative direction as a function of distance and faster in the positive direction as a function of distance during the acceleration. In an acceleration version of your thought experiment, while O accelerates there is no significant distance between O and O’. When O’ accelerates, O is further ahead in the direction of acceleration (higher in gravitational potential in the general relativistic sense and greater in X in Rindler coordinates). It’s clock will therefore run fast relative to O’ during the acceleration which is consistent with the result we’ve already found. The clock moves from t=2 to t=3.125 from the perspective of O’. It ran fast.



Given on wiki here:


Special Relativity

In step 3:



set Δt' = 0, v = -.6 (negative because the change in velocity is toward the other frame) and Δx = 1.5 (the distance in the first frame) and you should get the correct shift of 1.125. In fact, let's do that now,







There you have it. By the Lorentz transformations, when O' changes frames t changes by 1.125. It also should be clear from the properties of Minkowski spacetime. The diagram I made is perhaps helpful in that regard.

Here, also, is a very good wikibook on the topic:

Special Relativity/Simultaneity, time dilation and length contraction

The first section on phase shifting describes the part you are doing wrong well. In particular, that switching frames of reference moves the present instant along the time axis in a manner that depends on distance. The Andromeda paradox is a good example of this. Over very large distances (to the Andromeda galaxy) the present instant shifts quite a lot even with a small velocity change.

I think your question has been answered rather well and any inconsistencies are now shown to result from your refusal to apply relativity properly. Perhaps someone else has a different perspective that you would find helpful.

~modest

You have proven nothing above. Your objective was to prove they did not end up in the same frame since that is what you claimed was false. That was your claim. You claim distance operates on whether A and B will end up in the same frame.
The SR acceleration equation suggest the v relative to the original frame is the same. Therefore, you would have to prove the SR acceleration equations are false and that is actually what I asked you to prove because that is your assertion.

Basically, I offered a very simple proof that both O and O' would end up in the same frame.

However, let's look at the words of Einstein.

From this there ensues the following peculiar consequence. If at the points A and B of K there are stationary clocks which, viewed in the stationary system, are synchronous; and if the clock at A is moved with the velocity v along the line AB to B, then on its arrival at B the two clocks no longer synchronize, but the clock moved from A to B lags behind the other which has remained at B by

It is at once apparent that this result still holds good if the clock moves from A to B in any polygonal line, and also when the points A and B coincide.

http://www.fourmilab...in/specrel/www/

We can note that Einstein used this concept of instantaneous v. Note always he used an arbitrary length AB. He also said A and B can be co-located.

So, if this procedure is applied twice, then we must conclude each acquires v relative to the stationary B which is what this twins experiment does. If this is false, then it is impossible to ever claim you can acquire v relative to to the rest frame and therefore, Einstein is wrong.

On the other hand, I took the view of A and proved using two observers B and C in the original frame using the velocity add equations, the result is the same and A concludes B will end up in the same frame as A. Threfore, they end up in the same frame regardless.

Hence, the Einstein instant v method ends up with both in the same frame.

The SR acceleration equations cause both to end up in the same frame.

Are these false?

If so, then SR is false.

[Kharakov]

Hey Jakuta, consider this analogy to SR that I just read (I think on wikipedia, not sure which article, but I like it):

2 people who are the same size are some distance from one another. From A's perspective, B looks smaller, and vice versa.

I think you need to think about the pi-meson decay experiments. Read the top of page 60 in: Relativity: an introduction to space-time physics By Steve Adams.

Basically, when we observe that something's clock is moving slower, it is. The meson decays slower because it observes less time passing, at least when it "looks" at us. Tangent: Do mesons have eyes? I know freemasons do.

back to topic: Anyways, so since it sees time as passing slower in everything around it, it thinks less time has passed, spaces out for a bit, and then checks its watch and says "Crap, it's time to decay! I'm f-in late. I'm always fricken late!". Now, because it sees time as moving slower around it and spaces out for a bit, thinking it's going to be on time, then it ends up late (decaying farther down the road then it would have otherwise).

Now to us, it looks like the particle went through less time, because more of our time passed before it decayed (which made it look like it went through less time), but really, it thought less time had passed so it did stuff slower.


Now for the kicker.. the main point of relativity, that everyone just plain forgets until some joker mentions it:

Matter moves towards light at the speed of light, from light's perspective.

[jakuta]

Kharakov, on 04 November 2010 - 08:04 PM, said:

Hey Jakuta, consider this analogy to SR that I just read (I think on wikipedia, not sure which article, but I like it):

2 people who are the same size are some distance from one another. From A's perspective, B looks smaller, and vice versa.

I think you need to think about the pi-meson decay experiments. Read the top of page 60 in: Relativity: an introduction to space-time physics By Steve Adams.

Basically, when we observe that something's clock is moving slower, it is. The meson decays slower because it observes less time passing, at least when it "looks" at us. Tangent: Do mesons have eyes? I know freemasons do.

back to topic: Anyways, so since it sees time as passing slower in everything around it, it thinks less time has passed, spaces out for a bit, and then checks its watch and says "Crap, it's time to decay! I'm f-in late. I'm always fricken late!". Now, because it sees time as moving slower around it and spaces out for a bit, thinking it's going to be on time, then it ends up late (decaying farther down the road then it would have otherwise).

Now to us, it looks like the particle went through less time, because more of our time passed before it decayed (which made it look like it went through less time), but really, it thought less time had passed so it did stuff slower.


Now for the kicker.. the main point of relativity, that everyone just plain forgets until some joker mentions it:

Matter moves towards light at the speed of light, from light's perspective.

Yea, the issue under consideration is whether the simultaneity shift is necessary under uniform acceleration . I produced several links and mainstream papers that said it is not part of deciding the timing differentials between the frames. Sure under SR, time is relative, but that is not the case under acceleration. For example, when accelerating, you can feel it.

Here is a wiki calculation that demonstrates the symmetry of the acceleration equations regardless of the distance, regardless of the direction and regardless of negative of negative or positive acceleration relative to a frame.

http://en.wikipedia....spacetime_paths

This calculation uses asinh as they calculate from the view of the stay at home frame. I used the view of the acceleration frame and hence sinh.
Either way, symmetry is conclusive with regard to timing differentials regardless of direction, distance or negative/positive acceleration..

[Kharakov]

jakuta, on 05 November 2010 - 03:29 PM, said:

Yea, the issue under consideration is whether the simultaneity shift is necessary under uniform acceleration .

Ohh, I read differently. I thought you were attacking SR specifically with an example of "instantaneous acceleration" to eliminate any GR effects (due to the acceleration).

Quote

Sure under SR, time is relative, but that is not the case under acceleration. ...

Isn't that (acceleration) covered by GR, rather than SR? In other words, treat acceleration as gravitation, such as in gravitational time dilation.

Quote

Either way, symmetry is conclusive with regard to timing differentials regardless of direction, distance or negative/positive acceleration..

I still don't see what you're getting at? Experimental evidence supports SR time dilation, unless the evidence is falsified. Is that what you are saying, or are you simply attacking the (semantic and syntactic) formulation of the theory, and not the evidence that supports it?

In any event, the evidence at least supports that the object associated with (in a similar reference frame to) a greater group (or mass) of local objects, such as the earth or the sun, is the preferred frame (or more influential frame) than that of a space traveler with much less mass (time, by some accounts) to pull reality along with it.

At the very least, it would seem likely that the mass of a frame of reference directly impacts upon its share** of time dilation. So if you have an atomic clock moving through space and an atomic clock in the reference frame of the earth, the one associated with the earth's reference frame would have the preferred reference frame.

** And now, for our exciting offer. You too can have your own time share, in your very own reference frame! It's only 50000 down, and a few incidentals down the road! Own your very own share of the space time continuum, right now, for a one TIME payment!

[jakuta]

Kharakov, on 08 November 2010 - 10:40 PM, said:

Ohh, I read differently. I thought you were attacking SR specifically with an example of "instantaneous acceleration" to eliminate any GR effects (due to the acceleration).

No, i used instantaneous acceleration only for simplicity. Originally, I used the SR uniform acceleration equations.

Quote

Isn't that (acceleration) covered by GR, rather than SR? In other words, treat acceleration as gravitation, such as in gravitational time dilation.

It can be, but on e need to be carful in that one is not accelerating and one is. Certainly, GPS satellites are programmed prior to launch to slow down the clock since the ground and the GPS orbital path exist in different gravity potentials. The observers in this example are in open space. Also, since both observers are not accelerating at the same time, SR can be used.
http://math.ucr.edu/.../SR/rocket.html

Kharakov, on 08 November 2010 - 10:40 PM, said:

I still don't see what you're getting at? Experimental evidence supports SR time dilation, unless the evidence is falsified. Is that what you are saying, or are you simply attacking the (semantic and syntactic) formulation of the theory, and not the evidence that supports it?

In any event, the evidence at least supports that the object associated with (in a similar reference frame to) a greater group (or mass) of local objects, such as the earth or the sun, is the preferred frame (or more influential frame) than that of a space traveler with much less mass (time, by some accounts) to pull reality along with it.

At the very least, it would seem likely that the mass of a frame of reference directly impacts upon its share** of time dilation. So if you have an atomic clock moving through space and an atomic clock in the reference frame of the earth, the one associated with the earth's reference frame would have the preferred reference frame.

** And now, for our exciting offer. You too can have your own time share, in your very own reference frame! It's only 50000 down, and a few incidentals down the road! Own your very own share of the space time continuum, right now, for a one TIME payment!

I am just showing that the observers would believe time dilation applies but since they are in the same frame, they can perform Einstein's clock sync method which can only give one answer.

Since frequency has been demonstrated to change based on the strength of a gravitational field, it is not surprising a device based on frequency (clock) would also be altered.

Then, there is the muon which appears time dilated and also GPS satellites.

These cases however, are examples of absolute time dilation. Under SR, time dilation is reciprocol. This means when taking a frame as stationary, it believes the other frames clocks beat slower. Then, when taking the other frame as stationary, that frame will also claim the other frame clocks are beating slower. This is a deduction of the relativity postulate in which the rules of physics are the same for all frames. Consequently, absolute time dilation is not compatible with SR's reciprocol time dilation.

Hence, absolute time dilation does not prove reciprocol time dilation which is the correct version under SR.

[Kharakov]

jakuta, on 10 November 2010 - 07:56 AM, said:

Originally, I used the SR uniform acceleration equations.

I read differently, it may be the way you worded your original post (the instantaneous velocity acquisition is without acceleration because acceleration requires time- acceleration is change in velocity over time).

Quote

Then, there is the muon which appears time dilated and also GPS satellites.

These cases however, are examples of absolute time dilation. Under SR, time dilation is reciprocol. This means when taking a frame as stationary, it believes the other frames clocks beat slower. Then, when taking the other frame as stationary, that frame will also claim the other frame clocks are beating slower. This is a deduction of the relativity postulate in which the rules of physics are the same for all frames. Consequently, absolute time dilation is not compatible with SR's reciprocol time dilation.

I basically said as much in the last post I made by saying something along the lines of: there have to be preferred reference frames or SR time dilation wouldn't work "In any event, the evidence at least supports that the object associated with (in a similar reference frame to) a greater group (or mass) of local objects, such as the earth or the sun, is the preferred frame (or more influential frame) than that of a space traveler with much less mass ...gobbledygook "

I've seen some very sloppy math and/or logical justification for "why" SR time dilation works without preferred reference frames, but it just doesn't work mathematically or logically (which you are apparently stating).

The problem being that if you say "SR is incorrect", people will think you are a kook, unless you explain that it is only a specific portion of the theory that is incorrect, as SR's predictions are more or less experimentally confirmed (at least in the sense of a preferred reference frame).

In the muon decay scenario: If the earth's reference frame was not "preferred" a muon would decay at a rate quicker than average as more time would pass for it than did for earth from the muon's reference frame. The "picking" of frames could be a statistical (quantum) process, or it could be a simple process related to mass of reference frames as I stated earlier (which might make more sense in the light of mass and SR/GR).

LOL.. I almost forgot: the multi-frames interpretation of SR (which is quite the bit of BS, or load of crap as those in the other reference frame say, but...):

There are an undefined amount of frames, each of which has it's own preferred reference frame (it's own) in which it advances forward in time. Every action in one frame has a "reciprocal action" in another frame, although the frame that one is in is always the preferred frame for all effects. This means that in the muon's frame, it decays a lot sooner than in the earth's frame: there are 2 (well, infinite, but in the case of these 2 frames, 2) separate universes that exist, one in which the muon decayed quicker, one in which it didn't (the earth's frame universe).

[zendagimigzara]

I'm new to this...


but from what I understand from what Kharakov said is that while O moves, in O's frame/universe, O is younger than O', but in O''s frame/universe, O' is younger than O?

So both universes can occur at once?

[jakuta]

zendagimigzara, on 10 November 2010 - 04:55 PM, said:

I'm new to this...


but from what I understand from what Kharakov said is that while O moves, in O's frame/universe, O is younger than O', but in O''s frame/universe, O' is younger than O?

So both universes can occur at once?

Yes, that is the theory of relativity.

But, with this experiment, I forced the two universes to come back into the same one.

In that same universe, Einstein provides a tool called the Einstein clock synchronization method.

With this, you are able to determine timing differentials between two clocks in the same universe/frame.

Einstein said about his clock synch method,

"We assume that this definition of synchronism is free from contradictions, and possible for any number of points; and that the following relations are universally valid"
http://www.fourmilab...in/specrel/www/

Hence, one clock is older, they are equal or the other clock is older, and those are the only choices which refutes the reciprocal time dilation of relativity.