How Clocks (Seemingly) Ruin Everything

[striphe]

Introduction

I've been reading and thinking about special relativity recently and when I started to delve into relativity of simultaneity. I’ve seemed to of missed how relativity of simultaneity is compatible special relativity. I found it very difficult to describe this problem and it comes across rather pretentious, but if any of you can sift though it and work out what is missing from the rationale, it would be really helpful.

PART 1: The Clocks Paradox

A newly built, 1 light second long space colony begins it's journey to a distant planet, with an initial acceleration from what it considers stationary to 0.5C. As this acceleration was conducted and no problems have occurred during thrust, the captain decides that there will be a second period of acceleration, to reduce the duration of the voyage. The captain chooses to reach a velocity of 0.8C relative to the crafts pre take-off velocity or 0.5C relative to current velocity.

Before initial take off, clock at the front of the vessel was made to run 0.5 seconds behind the one at the back of the vessel so that post acceleration the clocks would be synchronised based on relativity of simultaneity calculations (XV/C^2). This process is to repeated again so that the clocks will be synchronised when the second stage of acceleration is complete. Before the captain makes the call to change the clocks, he makes an unusual announcement.

“After some thought, I’ve concluded based on the fact that we set the back clock 0.5 seconds ahead of the front clock, before the first stage of acceleration and that they are now synchronised after accelerating to 0.5C, relative of initial velocity, that our velocity is in absolute terms 0.5C or 0C. Yes, you heard correct! I've concluded that our absolute velocity is either 0.5C or 0C!”

An uproar breaks out on the bridge.

PART 2: The Captains Rationale

Before making the decision to change the clocks times, the captain had considered the fact that the two stage acceleration, accelerating to 0.5C relative to initial velocity and then another 0.5C relative the the current velocity was the same as a one stage acceleration to 0.8C relative to initial velocity.

The Captain considers if he knew that there was a good chance that the craft would be able to accelerate for a second time, he would of set the trailing clock 0.8 seconds ahead of the leading clock (based on the formula XV/C^2, which determines how ahead the leading clock, of a pair of synchronised clocks in a rest frame) so once both phases of acceleration were complete the clocks would be synchronised and he wouldn't have to adjust the clocks twice.

Considering that simultaneous events in the initial frame at each end of the space colony appear on-board to be out by 0.5 seconds, as light from the leading event reaches the centre of the craft 0.5 seconds before light from the trailing event reaches the centre. If these events were the reaching of midnight on a clock at each end. Light would be emitted by both of them simultaneously according to the the initial velocity frame with the same result of 0.5 seconds ahead for the leading. If the clocks are changed so that before take off the lead is 0.8 seconds behind, when at a velocity of 0.5C relative to initial velocity, midnight occurs according to the trailing 0.3 seconds before the leading. Based on this rationale the prescribed adjustment should be to have the leading 0.3 seconds behind the trailing so they are synchronised upon the second stage of acceleration.

Special relativity holds that any inertial frame of reference can be considered stationary. A second set of rationale, where you consider the vessel currently is at rest yields a prescribed adjustment of 0.5 seconds to the trailing clock; identical to the prescribed adjustment before the initial stage of acceleration. What is evident is that prescription will vary depending as to what reference frame is considered to be at rest and that only one of these prescription can be correct, giving substantial insight into an absolute velocity.
As the prescription for the first clock adjustment was correct, there is only two possible initial velocities. that being 0C or -0.5C as the adjustments prescribed by these initial velocities are the same as the required adjustment. Which evidently means that the current velocity of the colony is either 0.5C or 0C in absolute terms.


PART 3: Determining An Absolute Velocity

Although I am at rather odds to describe this well. The absolute velocity of an inertial reference frame, in a particular axis can be seemingly obtained, by accelerating an object twice, from relative rest with a trailing clock, a leading clock and some on-board object measurements of the time each clock displays.

This is achieved by considering V within the formula, (leading clock time – trailing clock time) = (XV)/C^2 is equivalent to S - V with V equal to the absolute velocity of the observer frame and S being equal to the absolute velocity of the object. Their for S – V = ((delta T)C^2)/X

S - V is also equal to U(1 – ((SV)/C^2)), which is obtained by rearranging the velocity subtraction formula U = (S – V)/(1 – ((SV)/C^2)). Their for ((delta T)C^2)/X = U(1 – ((SV)/C^2))

V cannot be solved from the above formula and there exists the requisite of two accelerations to effectively calculate V. The measurements required are:
(delta T) after the first phase of acceleration = Ta
(delta T) after the second phase of acceleration = Tb
The relative velocity between the observation frame and the object after the first phase of acceleration = Ua
The relative velocity between the observation frame and the object after the second phase of acceleration = Ub

these variables are then imputed into this formula:

V = ((C^2)/(Tb – Ta))((Ta/Ua) – (Tb/Ub))

Which seemingly gives you the absolute velocity of the observer (got a bit of a headache so I hope derived it properly) .

Thank you for reading. Hopefully you understood. I can elaborate upon request.

 

[DaveC426913]

All you've done is show the final speed with respect to the frame of reference of the craft its in initial state (which was defined as v=0).

But you cannot claim its initial velocity "really" was v=0.

If the entire system had been moving at .5c wrt the local star system and the crew did not know it, they would do exactly as they did above and get exactly the same results, yet, their final velocity wrt the local star system is .5c higher than they recorded.

In other words, you have demonstrated their velocity relative to a given frame of reference (whose velocity could be anything).

 

[striphe]

I'm unsure if I follow what you are saying.

But the point I'm trying to highlight is that if you set the back clock to 0.8 seconds ahead of the leading clock than accelerated to 0.8C from your initial velocity then they should be synchronised on board. Based on the same type of calculations, setting the back clock 0.8 seconds ahead and then accelerating to 0.5C the clock at the back will appear 0.3 seconds ahead. Accelerating 0.5C relative to this current frame should then even the clocks out, as this results in a velocity of 0.8C relative to the initial velocity.

The problem is that one may not have known about the initial acceleration and would think that if one was to accelerate to 0.5C that the clock at the back be 0.2 seconds trailing post acceleration.

You get different results depending on what you consider rest to be.

 

[ghwellsjr]

You also get different results depending on what you consider the acceleration to be.

You really are misapplying the Lorentz Tranform. What you should do is pick a frame of reference, like the initial rest frame of the space colony and stick with that one reference frame throughout your whole scenario. Then you have to define how this huge space colony is going to accelerate. Does it have thrusters all along its length, or just at the back end so that it gets pushed, or just at the front end so that it gets pulled? Have you taken into account the length compression that results when a body is in motion relative to a reference frame? It's just as important as the simultaneity issues. And what about the time dilation issues, have you considered how they effect the situation?

The bottom line is that you just can't use the Lorentz Transform to predict how an accelerated body is going to behave once it gets to a new speed. And you certainly can't use the Lorentz Transform in some convoluted scenario to figure out where absolute rest is. The purpose of the Lorentz Transform is to show how things that you have described in one frame of reference are described in a new frame of reference that has a velocity difference (and maybe a position and time offset) from your first frame of reference.

I don't think anyone is going to want to sift through your problem to help you figure out what is missing since it is based on so many misunderstandings of relativity.

 

[striphe]

Can you please explain how different means of acceleration effect relativity of simultaneity ghwellsjr?

 

[Mentz114]

striphe, if you haven't grasped that the term 'absolute velocity' is meaningless in SR, then I suggest you go back and do some more studying.

 

[ghwellsjr]

Can you please explain how different means of acceleration effect relativity of simultaneity ghwellsjr?

They don't. What I'm saying is that relativity of simultaneity is what you get when you define events in one frame and transform them into a different frame. Observers cannot tell if their remote clocks are simultaneous or not as a result of this transformation. In special relativity, you start with any inertial frame of reference and you define the simultaneity of all the clocks that are at rest in that frame. Then you use the Lorentz Transform on everything that is in that frame to see what it is like in a new frame and now what was simultaneous in the first frame is not simultaneous in the second frame. But the observers can't know what was simultaneous in one frame is not simultaneous in another frame. Nothing changes for them just because you look at them from different points of view.

 

[striphe]

Mentz114, read the rationale, i wouldn't have written this post if for one if i considered that an absolute velocity was apart of SR. I honestly haven't found anything to enlighten me on this issue. If you have some material i will defiantly study it.

ghwellsjr, i am not particularly getting "Observers cannot tell if their remote clocks are simultaneous or not as a result of this transformation" are you referring to clocks in the same frame of reference as an observer or ones with velocity? also "the observers can't know what was simultaneous in one frame is not simultaneous in another frame." It isn't too much to ask if you explain this with a hypothetical, for clarity at all is it?

 

[ghwellsjr]

OK, let's say you have an observer who is stationary in a frame. His clock is not time dilated. Now I transform him into another frame that is moving with respect to the first frame at .6c in the x direction. Now his clock is time dilated by a factor of 1.25. But does he know this? No.

Or I could pick two observers stationary in a frame and separated by 1 light second at both ends of a rigid platform with synchronized clocks. Now I view them from another frame that is moving with respect to the first one at .6c. Now the two observers are closer together by a factor of .8 and their clocks are not synchronized. However, they will think that their separation is 1 light second and that their clocks are synchronized.

This is just two different ways of viewing the same scenario.

But what you are doing is imagining that these two observers at either end of a rigid rod in the rest frame actually accelerate to a speed of .6c (or whatever) and you want to conclude that upon arrival, they will be just like as if we just viewed them from a moving frame which is not necessarily the same thing because there are so many different ways to accelerate a rigid rod as I mentioned in my first response on this thread.

 

[Mentz114]

Although I am at rather odds to describe this well. The absolute velocity of an inertial reference frame, in a particular axis can be seemingly obtained, by accelerating an object twice, from relative rest with a trailing clock, a leading clock and some on-board object measurements of the time each clock displays.

i wouldn't have written this post if for one if i considered that an absolute velocity was a part of SR. I honestly haven't found anything to enlighten me on this issue. If you have some material i will defiantly study it.

I don't see how these two quotes are compatible.

The first one is not even wrong. The only velocity that can be defined is relative to something else. So the fact that you can even use the term 'absolute velocity' means you have not understood the relativity of uniform motion.

Certainly, you could leave probe or buoy and accelerate your spaceship and work out or measure your speed relative to the buoy. But that is just a relative velocity between two objects.

 

[grav-universe]

As ghwellsjr is saying, we don't know how the colony ship accelerates, so we don't know if the front accelerates first from the rest frame and the back then accelerates faster and contracts while time dilating more or if the back accelerates first and then the front, in which case the back will still time dilate more so reads a lesser time according to the rest frame when the colony reaches the new frame. If both ends accelerated together in the same way, then there would be no time dilation or length contraction, so that wouldn't be the case with SR. You may be onto something with finding how clocks at each end must re-synch, though, depending upon how clocks at each end will time dilate while contracting toward each other, which must be in accordance to how they must re-synch after reaching each new frame. That is interesting and I have been meaning to try to work that out for a while now, so I will have to do so based upon your re-synching method between frames to see what can be determined.

In the meantime, we could have observers at each end of the colony accelerate simultanously in the same way to the new frame. That way, the colony will still measure the same distance between the observers although the distance between the observers will have elongated according to what the observers measure between themselves. The colony will also measure the same time dilation so gain the same resulting time between the observers, so we know that the colony will also not measure a simultaneity difference between the observers. We can then have the observers re-synch accordingly, although I haven't worked this out yet either.

 

[striphe]

ghwellsjr, I would consider that your opinion of what occurs in terms of relativity of simultaneity, is essentially the opposite of my opinion. Within my hypothetical, what I believe you would consider is the case, is that there is no need for the clocks to be changed at all. I think i'll start a thread based on this purely and see what it yields. If you could show some math on why this is the case then i might understand how this is so.


Mentz114, if I seemingly am able to obtain an absolute velocity, when I know that SR and absolute velocity are incompatible, then I know there is a problem that needs to be resolved; hence the thread.

Grav-universe, My understanding of your post and ghwellsjr's post is that from which ever point along the spacecraft thrust is occurring the two ends are moving towards that point. So one would consider that during acceleration the average velocity of the front is less than the back. This will only extenuate the difference that would emerge due to relativity of simultaneity rather than evening it out. In the hypothetical you require the back clock to be ahead so it evens out, but if its moving faster than the front clock then its only going to be even furthermore behind post acceleration.

 

[grav-universe]

Grav-universe, My understanding of your post and ghwellsjr's post is that from which ever point along the spacecraft thrust is occurring the two ends are moving towards that point. So one would consider that during acceleration the average velocity of the front is less than the back. This will only extenuate the difference that would emerge due to relativity of simultaneity rather than evening it out. In the hypothetical you require the back clock to be ahead so it evens out, but if its moving faster than the front clock then its only going to be even furthermore behind post acceleration.

Right. In order for the length contraction to occur, the back must travel faster than the front on average, so the back clock time dilates more and we would have to add more than just .5 or .8 seconds to compensate when reaching the new frame.

 

[striphe]

I would imagine that if it took an extreme amount of time to change velocity that this effect would become negligible, just as lost time becomes negligible over a given distance if you lower the velocity with which you cover it (I could be wrong).

I don't want to tangent off the original issue if it can be helped

 

[grav-universe]

I have been trying to work this through for a ship that accelerates to v, but we can have any speed w for an impulse wave that travels at the speed of sound of the material from any number of thrusters located at various points along the ship to other parts of the ship and many waves from a single thruster could then recoil back and forth along the length of the ship, causing it to contract and enlongate and vibrate until it finally comes to an equilibrium state in the new frame after the thrusters have been shut off. This adds too many complications and unknown variables, but I have come to realize one thing for certain, regardless of how the ship accelerates, which simplifies things immensely.

Let's say that we accelerate the front of the ship to v first and then a single impulse wave travels along the length of the ship from there, accelerating all other parts of the ship to v as it reaches them, and then the wave dissipates at the other end of the ship. According to the rest frame, then, the ship is now longer. If the ship has a ruler attached to its hull that is accelerated in the same way as the rest of the ship, then according to the ship observers, the ship will remain the same length as before it accelerated. In that case the rest frame measures a length elongation of the proper length the ship observers measure instead of a length contraction.

So in order to measure a length contraction properly as within the context of SR, the ruler must be detached from the ship itself and contract by a factor of L according to the rest frame, regardless of whatever the ship does or how it accelerates. In that case, even if the rest frame now measures the ship to be twice the original length 2d in the new frame, for instance, the ship observers, using their ruler that has contracted by a factor of L, will now measure the length of the ship to be 2d / L, so we will still have the same length contraction as the ratio of what the rest frame measures for the length of the ship as compared to the proper length the ship observers measure.

Since it doesn't matter how the ship accelerates, we can have the ship accelerate in any way we desire and the length contraction will be the same as long as the ship observers' ruler contracts independently of whatever the ship does. For simplicity, then, we can have an impulse wave that travels from the back of the ship toward the front at a speed of w = v / (1 - L), instantly accelerating all parts of the ship to v as it reaches them, until it dissipates at the front of the ship. At this speed w for the impulse wave, the new length of the ship as the rest frame measures it will be d - v t = d - v (d / w) = d - d (1 - L) = L d. Since the ship has been shortened by the same factor as the ship's ruler contracts, the ship observers will still measure the same length d of the ship in the new frame. Now I will just need to work out accordingly what the time dilations will be and how the ship observers must re-synchronize.

 

[grav-universe]

Okay well, making w = v / (1 - L) adds an additional complication, being that w works out to greater than c regardless of v, so instead of a single impulse wave that travels the length of the ship to contract it to L d, we will have thrusters placed all along the length of the ship that fire a quick burst first at the back, instantly accelerating that immediate section of the ship to v, then continuing to fire in sequence along the length of the ship to the front at an effective speed w.

 

[striphe]

grav-universe, I think I've lost you when you wrote "According to the rest frame, then, the ship is now longer" From my understanding any object that is accelerated will have contract in size according to any observer in its initial velocity.

 

[chinglu1998]

Introduction

I've been reading and thinking about special relativity recently and when I started to delve into relativity of simultaneity. I’ve seemed to of missed how relativity of simultaneity is compatible special relativity. I found it very difficult to describe this problem and it comes across rather pretentious, but if any of you can sift though it and work out what is missing from the rationale, it would be really helpful.

PART 1: The Clocks Paradox

A newly built, 1 light second long space colony begins it's journey to a distant planet, with an initial acceleration from what it considers stationary to 0.5C. As this acceleration was conducted and no problems have occurred during thrust, the captain decides that there will be a second period of acceleration, to reduce the duration of the voyage. The captain chooses to reach a velocity of 0.8C relative to the crafts pre take-off velocity or 0.5C relative to current velocity.

Before initial take off, clock at the front of the vessel was made to run 0.5 seconds behind the one at the back of the vessel so that post acceleration the clocks would be synchronised based on relativity of simultaneity calculations (XV/C^2). This process is to repeated again so that the clocks will be synchronised when the second stage of acceleration is complete. Before the captain makes the call to change the clocks, he makes an unusual announcement.

“After some thought, I’ve concluded based on the fact that we set the back clock 0.5 seconds ahead of the front clock, before the first stage of acceleration and that they are now synchronised after accelerating to 0.5C, relative of initial velocity, that our velocity is in absolute terms 0.5C or 0C. Yes, you heard correct! I've concluded that our absolute velocity is either 0.5C or 0C!”

An uproar breaks out on the bridge.

PART 2: The Captains Rationale

Before making the decision to change the clocks times, the captain had considered the fact that the two stage acceleration, accelerating to 0.5C relative to initial velocity and then another 0.5C relative the the current velocity was the same as a one stage acceleration to 0.8C relative to initial velocity.

The Captain considers if he knew that there was a good chance that the craft would be able to accelerate for a second time, he would of set the trailing clock 0.8 seconds ahead of the leading clock (based on the formula XV/C^2, which determines how ahead the leading clock, of a pair of synchronised clocks in a rest frame) so once both phases of acceleration were complete the clocks would be synchronised and he wouldn't have to adjust the clocks twice.

Considering that simultaneous events in the initial frame at each end of the space colony appear on-board to be out by 0.5 seconds, as light from the leading event reaches the centre of the craft 0.5 seconds before light from the trailing event reaches the centre. If these events were the reaching of midnight on a clock at each end. Light would be emitted by both of them simultaneously according to the the initial velocity frame with the same result of 0.5 seconds ahead for the leading. If the clocks are changed so that before take off the lead is 0.8 seconds behind, when at a velocity of 0.5C relative to initial velocity, midnight occurs according to the trailing 0.3 seconds before the leading. Based on this rationale the prescribed adjustment should be to have the leading 0.3 seconds behind the trailing so they are synchronised upon the second stage of acceleration.

Special relativity holds that any inertial frame of reference can be considered stationary. A second set of rationale, where you consider the vessel currently is at rest yields a prescribed adjustment of 0.5 seconds to the trailing clock; identical to the prescribed adjustment before the initial stage of acceleration. What is evident is that prescription will vary depending as to what reference frame is considered to be at rest and that only one of these prescription can be correct, giving substantial insight into an absolute velocity.
As the prescription for the first clock adjustment was correct, there is only two possible initial velocities. that being 0C or -0.5C as the adjustments prescribed by these initial velocities are the same as the required adjustment. Which evidently means that the current velocity of the colony is either 0.5C or 0C in absolute terms.


PART 3: Determining An Absolute Velocity

Although I am at rather odds to describe this well. The absolute velocity of an inertial reference frame, in a particular axis can be seemingly obtained, by accelerating an object twice, from relative rest with a trailing clock, a leading clock and some on-board object measurements of the time each clock displays.

This is achieved by considering V within the formula, (leading clock time – trailing clock time) = (XV)/C^2 is equivalent to S - V with V equal to the absolute velocity of the observer frame and S being equal to the absolute velocity of the object. Their for S – V = ((delta T)C^2)/X

S - V is also equal to U(1 – ((SV)/C^2)), which is obtained by rearranging the velocity subtraction formula U = (S – V)/(1 – ((SV)/C^2)). Their for ((delta T)C^2)/X = U(1 – ((SV)/C^2))

V cannot be solved from the above formula and there exists the requisite of two accelerations to effectively calculate V. The measurements required are:
(delta T) after the first phase of acceleration = Ta
(delta T) after the second phase of acceleration = Tb
The relative velocity between the observation frame and the object after the first phase of acceleration = Ua
The relative velocity between the observation frame and the object after the second phase of acceleration = Ub

these variables are then imputed into this formula:

V = ((C^2)/(Tb – Ta))((Ta/Ua) – (Tb/Ub))

Which seemingly gives you the absolute velocity of the observer (got a bit of a headache so I hope derived it properly) .

Thank you for reading. Hopefully you understood. I can elaborate upon request.

A newly built, 1 light second long space colony begins it's journey to a distant planet, with an initial acceleration from what it considers stationary to 0.5C. As this acceleration was conducted and no problems have occurred during thrust, the captain decides that there will be a second period of acceleration, to reduce the duration of the voyage. The captain chooses to reach a velocity of 0.8C relative to the crafts pre take-off velocity or 0.5C relative to current velocity.

Before initial take off, clock at the front of the vessel was made to run 0.5 seconds behind the one at the back of the vessel so that post acceleration the clocks would be synchronised based on relativity of simultaneity calculations (XV/C^2). This process is to repeated again so that the clocks will be synchronised when the second stage of acceleration is complete. Before the captain makes the call to change the clocks, he makes an unusual announcement.

Your theory does not conform to the SR uninform acceleration equations.

You must first calculate your t and then t' based on acceleration.

I will leave that to you since this is your thread. Make sure you conform to the below links.


http://www.ias.ac.in/currsci/feb252007/416.pdf

http://www.ejournal.unam.mx/rmf/no521/RMF52110.pdf

http://users.telenet.be/vdmoortel/dirk/Physics/Acceleration.html

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

http://en.wikipedia.org/wiki/Twins_paradox

http://arxiv.org/PS_cache/physics/pdf/0411/0411233v1.pdf

 

[striphe]

chinglu1998 I'm taking a very 'applied' approach with these threads, so it difficult to take any utility with ease from your posts in relation to the issues. In the beginning post I avoid describing acceleration (where thrust occurs and the duration of acceleration) as i lack any clear understanding of how these variables would effect the vessels clocks.

 

[chinglu1998]

chinglu1998 I'm taking a very 'applied' approach with these threads, so it difficult to take any utility with ease from your posts in relation to the issues. In the beginning post I avoid describing acceleration (where thrust occurs and the duration of acceleration) as i lack any clear understanding of how these variables would effect the vessels clocks.

I am OK with your reasoning. But, you included acceleration and the consequences therefrom. I simply made note.

If you choose to use instantaneous v, I can know that as Einstein made that claim.

But, I can warn you, your reasoning will not stand this test.

Please proceed.

 

[striphe]

So what will the clocks on-board read if I accelerate the 1 lightsecond long vessel to 0.5C instantaneously if the clocks on-board were synchronised before acceleration?

 

[chinglu1998]

So what will the clocks on-board read if I accelerate the 1 lightsecond long vessel to 0.5C instantaneously if the clocks on-board were synchronised before acceleration?

They will all read the same time at that instant. It is after that you will see the relativity of simultaneity and time dilation.

So after the instantaneous acquiring of v = .5c, what do you want to do?

I will need a stationary frame of reference origin.

 

[striphe]

So what will the clocks on-board read

what i mean is how do they read for someone on-board

 

[grav-universe]

grav-universe, I think I've lost you when you wrote "According to the rest frame, then, the ship is now longer" From my understanding any object that is accelerated will have contract in size according to any observer in its initial velocity.

If the front of the ship accelerates first, then the ship could end up elongated according to the rest frame, depending specifically upon how it was accelerated. It will appear to be stretched according to the ship observers as well. But it doesn't matter what actually happens to the ship when it comes to the length contraction. As long as the ship's ruler contracts in the line of motion, the ratio of the coordinate length that the rest frame measures to the proper length the ship observers measure when using their own ruler, which is what the length contraction really is, will still be equal to the length contraction of the ruler itself as the rest frame measures it.

I'm starting to wonder, however, why the ship's ruler should contract naturally when not attached to the ship itself while the resulting length of the ship depends upon the specific conditions about how it is accelerated, when the ruler is still likewise accelerated to the same speed. It may be that rulers do not naturally contract according to SR either, but must be re-constructed in the new frame in a similar way that clocks must be re-synchronized in order to measure the same speed c isotropically. The real physics of SR, then, it seems, might only be that rigid bodies will contract in the line of motion when rotated within the same frame. For instance, if the rest frame measures some length of a rigid body when tangent to the line of motion, when positioned along the y-axis when the line of motion is along the x axis, the body will be measured as contracted to sqrt[1 - (v/c)^2] when turned to line up along the line of motion with the x axis. Indeed, that is all the MMX really demonstrates.

 

[grav-universe]

Okay, now for the simultaneity differences along the length of a ship when accelerated. The ship accelerates to v from the rest frame A by firing thrusters in sequence from the back of the ship to the front at an effective speed of w = v / (1 - L), so the back of the ship instantly accelerates to a speed of v while the front accelerates to v a time of d / w later, and the resulting length of the ship is

d - v (d / w)

= d (1 - v / w), where w = v / (1 - L), so

= d (1 - (1 - L))

= L d

L is the length contraction as measured by the rest frame while the ship observers measure the same proper length of the ship. Since the back accelerated to v first, a clock at the back will time dilate by a factor of z for a time of d / w while the front clock continues to tick at the previous rate, so the resulting difference in readings between the clocks is with the back clock reading a time of (1 - z) (d / w) less than the front clock. In order for the ship observers to synchronize to their new frame, the back clock must be tl = d v / c^2 ahead of the front clock according to the rest frame, while it is currently (1 - z) (d / w) behind, so it must be set forward by a time of t = tl + (1 - z) (d / w). Now let's say we do the same thing from the new rest frame, accelerating again to v from the new frame. The the physics is the same in every frame, and the thrusters can be fired in the same sequence with the same results. Then again the ship observers must set the back clock forward again by the same amount, so since departing the original rest frame, the back clock has been set forward a total time of 2 t.

Now let's look at what the rest frame observes for both parts of the trip. After accelerating to the second frame, the ship observers had to set the back clock forward by t so that it now reads tl = d v / c^2 ahead of the front clock. When accelerating from the second frame, the effective speed for the sequence of the thrusters bacomes w2 = (w + v) / (1 + w v / c^2) and the new relative speed of the ship to the original rest frame becomes v2 = 2 v / (1 + v^2 / c^2). The back of the ship will accelerate a time of L d / (w2 - v) before the front of the ship accelerates, time dilating by a factor of z2 while the front clock continues to time dilate by a factor of z, so the difference in readings between the clocks now becomes tl + (z2 - z) L d / (w2 - v). In order to synchronize to the final frame, the difference in readings between the clocks should be tl2 = d v2 / c^2. The ship observers add the same time to the back clock as before of t = tl + (1 - z) (d / w), so the resulting difference in readings is

tl2 = tl + (z2 - z) L d / (w2 - v) + t

= 2 tl + (z2 - z) L d / (w2 - v) + (1 - z) (d / w)

d = 1 ls
v = .5 c
z = L = sqrt[1 - (v/c)^2] = .866025403
v2 = 2v / (1 + v^2 / c^2) = .8 c
z2 = sqrt[1 - (v2 / c)^2] = .6
w = v / (1 - L) = 3.732050808 c
w2 = (w + v) / (1 + w v / c^2) = 1.476627109 c
tl = d v / c^2 = .5 sec

tl2 = .5 sec + (-.266025406) (.866025403) (1 ls) / (.976627109) + (.133974597) (.035898385)

= .8 sec, which of course must also match tl2 = d v2 / c^2, which it does.

Okay, now let's say that the ship observers go ahead and set the back clock forward by 2t = 2 tl + 2 (1 - z) (d / w) = 1.07179677 sec beforehand, before the ship accelerates from the original rest frame, but instead of accelerating first to v and then again to v2, we will just have them accelerate directly to v2 by firing the thrusters at an effective speed of w3 = v2 / (1 - L2) = (.8 c) / (1 - .6) = 2 c, so that the contracted length the rest frame measures of the ship in the final frame is L2. In this case, the back of the ship accelerates first to v2 for a time of d / w3 before the front of the ship accelerates, so the difference in readings between the clocks after accelerating directly to the final frame becomes

2t + (z2 - 1) (d / w3) = 1.07179677 sec + (-.4) (.5 sec) = .87179677 sec

It appears that if the ship accelerates directly to the final frame after going ahead and adding 2t to the back clock as they did before when accelerating to one frame and then the other, the back clock will now read too far ahead by .07179677 sec. Interestingly enough, if they had only added 2 tl = 1 sec beforehand instead, however, then it would have worked out to .8 sec as it should be, while the additional amount the back clock falls behind comes from accelerating to v twice. This does not show any discrepency with simultaneity shifts in SR, however, but only indicates that accelerating a ship to v and then again to v2 in the manner described does not result with the same simultaneity shifts as accelerating directly to v2.

 

[striphe]

Does the effect of length contraction having the clocks operating at different speeds, become less if decrease the rate of acceleration? e.g. instead of instantaneous reaching of 0.5C of the back end of the vessel, it takes say 100 years.

 

[grav-universe]

Okay, I want to try a different acceleration profile by which to compare to the previous result. This time, instead of accelerating the entire ship, we will just have two disconnected observers that accelerate from each end of the ship, with the ship remaining in the rest frame. Of course, we can make what we did last time amount to the same thing by having the back observer accelerate a time of d / w = (d / v) (1 - L) before the front observer and find their time dilations and simultaneity differences accordingly, so this time I want to have each of the observers accelerate simultaneously. We could do the same thing to the ship as well, by having all parts of it accelerate simultaneously, so remaining the same length according to the rest frame but elongating according to the ship observers, so it is really the same difference, but two disconnected observers might be easier to visualize.

Two observers have a distance of d between them in the rest frame and then they instantly and simultaneously accelerate to a speed v. The rest frame measures the same distance between them as before with no simultaneity difference between the readings upon their clocks. With rulers by a factor of L, however, the observers measure a distance of d / L between themselves. They want to synchronize to the new frame, so the back observer sets his clock forward by tl = (d / L) v / c^2. Clocks within the new frame are synchronized in the same way. Then the observers accelerate again in the same way, simultaneously according to the clocks of the new frame to a speed of v, so to a speed of v2 = 2v / (1 + v^2 / c^2) to the original frame. The distance the observers measure between themselves again elongates in the same way as before by another factor of L, so the back observer now sets his clock forward by tl' = (d / L^2) v / c^2, so has set it forward a time of (1 + 1 / L) tl in all.

According to the original frame, if the observers accelerate simultaneously from the new frame according to their own clocks, then since the back observer's clock is set tl ahead of the front observer's clock, the back observer will accelerate first and then the front observer a time of tl / z later, while the back observer's clock time dilates by a factor of z2 while the front observer's clock continues to time dilate at a factor of z during this time. The resulting difference in the readings of the clocks will be tl + (z2 - z) (tl / z) = (z2 / z) tl, but then the back observer sets his clock forward by an additional tl' = tl / L when reaching the final frame in order to synchronize to the final frame, so the back clock now reads tl2 = tl' + (z2 / z) tl = (1 / L + z2 / z) tl ahead of the front clock.

Since the back observer accelerates first from the second frame, the distance the rest frame measures between the observers in the final frame is d - (v2 - v) (tl / z), so the observers measure (d - (v2 - v) (tl / z)) / L2 between themselves. Once they have synchronized to the final frame, the simultaneity difference that the rest frame measures between their clocks should be tl2 = [(d - (v2 - v) (tl / z)) / L2] v2 / c^2, while before we found tl2 = (1 / L + z2 / z) tl, so these must be equal. We have

[(d - (v2 - v) (tl / z)) / L2] v2 / c^2 = (1 / L+ z2 / z) tl

d = 1 ls
v = .5 c
L = z = sqrt[1 - (v/c)^2] = .866025403
v2 = 2v / (1 + v^2 / c^2) = .8
L2 = z2 = sqrt[1 - (v2 / c)^2] = .6
tl = (d / L) v / c^2 = .577350269 sec

[(1 ls - (.3 c) (.666666667 sec)) / (.6)] (.8 c) / c^2 = (1.847520863) (.577350269 sec)

1.066666667 sec = 1.066666667 sec, so that works out.

Now let's find out what the back observer must add to his clock when the observers accelerate directly to the final frame, after setting the back observer's clock ahead by (1 + 1 / L) tl beforehand, which was the total time added to the back clock as found earlier. The final distance between the observers as measured by the rest frame was d - (v2 - v) (tl / z) and we want the result for the final frame to be identical to what we had before, so this time we must accelerate the back observer a time of [(v2 - v) (tl / z)] / v2 before the front observer to gain the same resulting distance between them. The back observer's clock, then, will time dilate by a factor of z2 during this time while the front observer's clock continues to tick at the same rate as the rest frame, resulting in a simultaneity difference between the clocks of

(1 + 1 / L) tl + (z2 - 1) (v2 - v) (tl / z) / v2

= (2.154700539) (.577350269 sec) + (-.4) (.3 c) (.666666667 sec) / (.8 c)

= 1.144016936 sec, so the back observer's clock is greater than it should be by a time of .07735027 sec. Apparently, accelerating from one frame to another and then to a third frame is not the same as accelerating directly from the first to the third frame.

 

[grav-universe]

Does the effect of length contraction having the clocks operating at different speeds, become less if decrease the rate of acceleration? e.g. instead of instantaneous reaching of 0.5C of the back end of the vessel, it takes say 100 years.

I suppose it would depend upon precisely how the acceleration takes place, but I will go ahead and work on an example for that to find out what happens.

 

[striphe]

I speculate that the effect may be less, but i was unsure how i would approach the calculations and be confident in the results. I look forward to your next reply grav-universe

 

[Mentz114]

Striphe,
can you define what you mean by 'absolute velocity'. In what sense is this expression

V = ((C^2)/(Tb – Ta))((Ta/Ua) – (Tb/Ub))

an 'absolute velocity' ?

 

[striphe]

Mentz114, let's focus on part 2 first, so everyone gets on the same page.

 

[Mentz114]

Mentz114, let's focus on part 2 first, so everyone gets on the same page.

You are avoiding this question because you don't have an answer.

 

[striphe]

no, its more i CBF writing some massive explanation on my iPhone.

 

[Mentz114]

no, its more i CBF writing some massive explanation on my iPhone.

So, you are saying that the postulate of relativity "it is not possible to distinguish a state of rest from a state of uniform motion" is wrong ?

'Yes' or 'no' will suffice.

 

[grav-universe]

I'm still working on the acceleration profile that will bring this all together better, but I have been thinking about a couple of things in the meantime. Before I mentioned that if the thrusters are placed at the front of the ship, the ship might be stretched according to both the ship observers and the rest frame, and likewise if placed at the back, the ship might contract more. I'm thinking now that is only the initial conditions, but after some time has passed, the molecular binds that hold the ship together should compensate for the stresses involved. That is to say, as long as the ship doesn't break apart when the thrusters are initially fired, then it shouldn't any time after that (at least until the materials wear down from the stresses), so from the frame of the ship, the ship will be compressed or stretched after the thrusters are initially fired, but then the whole of the ship will establish equilibrium and things will thereafter remain the same as long as the thrusters continue firing for any duration of time. This means that instead of comparing the length of the ship before the thrusters fire to the length afterward, we should compare the lengths from some time after the thrusters are fired and the ship has established equilibrium. It also means that after equilibrium is maintained, the ship should also then contract at the same rate as the ship's ruler.

I'm also wondering why accelerating to one frame, synchronizing to that frame along the length of a ship or between two observers, then accelerating in the same way to another frame and synchronizing again to that frame should have more time added to the back clock than accelerating directly to the final frame. It may have something to do with the acceleration profile I gave. There may be a "natural" way that ships must be accelerated such that the same time will be added to the back clock, which might give some insight into how accelerations take place naturally, so I will work on that also.

 

[chinglu1998]

what i mean is how do they read for someone on-board

Sorry, not clear from me.

Here is Einstein which is answer to your question.

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

http://www.fourmilab.ch/etexts/einstein/specrel/www/

Einstein used instantaneous v and did not consider acceleration effects on time.

At instant v is acquired all clocks read same in both frame but only for that instant.

Then from rest frame (not ship frame), front clock ship back clock ship become unsynchronized after that instant.

In ship frame, both clocks remain at same time.

Now assume uniform acceleration equations of SR. This is done as integral with instanteous co-moving frame as slice and so both clocks at any instant are in same frame.

So, even with uniform acceleration equations clocks in ship frame remain synchronized.

 

[grav-universe]

Does the effect of length contraction having the clocks operating at different speeds, become less if decrease the rate of acceleration? e.g. instead of instantaneous reaching of 0.5C of the back end of the vessel, it takes say 100 years.

If we decrease the acceleration of the ship, then the time dilations at either end will be closer to that of the rest frame at any given time, so the tick rate at either end is closer to unity, and each end's rate of ticking will be closer to the other's after some amount of time passes for the rest frame with a lesser acceleration, although that just means it will take more time to reach the final frame, so eventually the difference between the front and back will increase in the same way, but it will just take longer. But I think what you're really asking is does the overall difference in readings between the front and back of a ship decrease upon reaching the final frame if it takes longer to accelerate to that frame, right? Let's find out.

The usual equations for acceleration in SR are local, used primarily for how a single observer accelerates, and I could not find a way to apply constant accelerations to the front and back of a ship in such a way that it is contracted by sqrt[1 - (v/c)^2] when it reaches the final frame and shuts off the thrusters that is consistent for any final speed v as compared to continuing to accelerate in the same way until it reaches a different frame u, for instance, shuts off the thrusters, and is now contracted by sqrt[1 - (u/c)^2]. I'm sure that's how you wanted the ship to contract, such that when the reaches the final frame, regardless of whatever speed is achieved, it will be then be contracted such that the ship observers measure the same length of the ship as before, but let's do the next best thing.

We will have both ends of the ship accelerate simultaneously from the rest frame, the front at some constant acceleration a1 and the back at some constant acceleration a2 in the ship frame. We can see from one of the relativistic equations for acceleration, v / c = (a t / c) / sqrt[1 + (a t / c)^2], that if both ends of the ship are to accelerate to the same frame so that they have the same speed v after shutting off the engines, then (a t / c) must be a constant for both ends of the ship. That is, if half the acceleration is applied to the back of the ship than to the front, then according to the rest frame, the back must continue to accelerate for twice the time to achieve the same speed. Next, we can see from the equation t' = (c / a) ln[(a t / c) + sqrt(1 + (a t / c)^2)], where t' is the time that passes upon a clock during acceleration, that since (a t /c) is a constant, then the amount of time that passes upon an accelerating clock is inversely proportional to the acceleration applied, so directly proportional to the time it takes for the clock to accelerate to the final frame according to the rest frame.

So let's say we accelerate the front and back of the ship at a1 and a2 for times t1 and t2. Since (a t / c) is a constant for both ends to achieve the same speed v in the final frame after the engines have been cut off, then a1 t1 = a2 t2. After the ship has been accelerated in this way to the new frame, the difference in readings between the clocks is tx. Now let's say we start over and accelerate the ship again, but this time we will apply only half of the original accelerations to the front and back of the ship. With half the accelerations for both, the ship will now take twice as long to reach the final frame according to the rest frame. Since the time that passes upon the clocks at the front and back are directly proportional to the time that they are accelerated, then each of the clocks will read twice the time that they did when accelerated before with twice the accelerations. Everything will be as it was before after the ship reaches the final frame and cuts off the engines, but now the clocks read twice the times, so they now have twice the difference between them also. So it appears that decreasing the acceleration will actually increase the difference in times between the clocks in inverse proportion and in direct proportion to the time of acceleration according to the rest frame, not decrease it, at least when both ends constantly accelerate and are accelerated simultaneously from the rest frame, but otherwise still in general.

 

[striphe]

grav-universe, have a look at the other thread I posted up
https://www.physicsforums.com/showthread.php?t=455853
in it a user called DaleSpam calculates the effects on the clocks using based on Rindler coordinates http://en.wikipedia.org/wiki/Rindler_coordinates

 

[grav-universe]

grav-universe, have a look at the other thread I posted up
https://www.physicsforums.com/showthread.php?t=455853
in it a user called DaleSpam calculates the effects on the clocks using based on Rindler coordinates http://en.wikipedia.org/wiki/Rindler_coordinates

Aha, yes, I've got it now, thank you. I would have to see it for myself, but the links motivated me to run back through my own calculations. It turns I forgot to carry the d, sort of speak , which is why I couldn't find a constant proper acceleration for both the front and for the back that would contract the ship to sqrt(1 - (v/c)^2) d when accelerating to a speed v or to sqrt(1 - (u/c)^2) d when continuing to accelerate in the same way for a longer duration of time to another speed u, but it turns out they can. Here's how.

Let's say we have a ship that is originally stationary to the rest frame with a length d. The back of the ship then accelerates from the rest frame with a constant acceleration aB and the front accelerates with a constant proper acceleration aF a time of tx later. The back of the ship accelerates for a time of tB, then becomes inertial at a speed v in the new frame. The front of the ship remains stationary for a time of tx, then accelerates for a time of tF, then becomes inertial at a speed v as well. Let's say the rest frame measures the distance the front and back have traveled after a duration of t3, some time after both the front and back of the ship have become inertial. According to the relativistic acceleration formula for distance, the rest frame would then find the distances to be

dF = d + (c^2 / aF) [sqrt(1 + (aF tF / c)^2) - 1] + v (t3 - tx - tF)

dB = (c^2 / aB) [sqrt(1 + (aB tB / c)^2) - 1] + v (t3 - tB)

The relativistic acceleration formula for speed is

(v / c) = (a t / c) / sqrt[1 + (a t / c)^2], which becomes

(v / c)^2 = (a t / c)^2 / [1 + (a t / c)^2]

(v / c)^2 + (v / c)^2 (a t / c)^2 = (a t / c)^2

(a t / c)^2 [1 - (v / c)^2] = (v / c)^2

(a t / c) = (v / c) / sqrt[1 - (v / c)^2]

You can see from this that if both the front and back are to accelerate to the same speed v, then (aF tF / c) = (aB tB / c) = (v / c) / sqrt[1 - (v / c)^2]. Substituting this into our formulas for distance gives

dF = d + (c^2 / aF) [sqrt(1 + (v / c)^2 / (1 - (v / c)^2)) - 1] + v (t3 - tx - tF)

= d + (c^2 / aF) [1 / sqrt(1 - (v / c)^2) - 1] + v (t3 - tx - tF)

dB = (c^2 / aB) [sqrt(1 + (v / c)^2 / (1 - (v / c)^2)) - 1] + v (t3 - tB)

= (c^2 / aB) [1 / sqrt(1 - (v / c)^2) - 1] + v (t3 - tB)

The difference in distance the rest frame will measure between the front and the back after the ship has become inertial in the new frame at speed v, then, is

dF - dB = d + c^2 (1 / aF - 1 / aB) (1 / sqrt(1 - (v / c)^2) - 1) + v (tB - tF - tx)

= d + c^2 (1 / aF - 1 / aB) (1 / sqrt(1 - (v / c)^2) - 1) - v tx - v (tF - tB), where (aF tF / c) = (aB tB / c) = (v / c) / sqrt(1 - (v / c)^2), so we have

= d + c^2 (1 / aF - 1 / aB) (1 / sqrt(1 - (v / c)^2) - 1) - v tx - v [v / (aF sqrt(1 - (v / c)^2)) - v / (aB sqrt(1 - (v / c)^2))]

= d + c^ (1 / aF - 1 / aB) (1 / sqrt(1 - (v / c)^2) - 1) - v tx - (v^2 / sqrt(1 - (v / c)^2)) (1 / aF - 1 / aB)

= d + [c^2 (1 / sqrt(1 - (v / c)^2) - 1) - v^2 / sqrt(1 - (v / c)^2)] (1/ aF - 1 / aB) - v tx

= d + [(c^2 - v^2) / sqrt(1 - (v / c)^2) - c^2] (1 / aF - 1 / aB) - v tx

= d + c^2 [sqrt(1 - (v / c)^2) - 1] (1 / aF - 1 / aB) - v tx

Now, we want the distance that the rest frame measures between the front and the back when the ship accelerates to v to be equal to the contracted distance sqrt(1 - (v / c)^2) d, but it should also equal sqrt(1 - (u / c)^2) if the ship continues to accelerate in the same way for an extended period of time to a speed u, so we have

d + c^2 (sqrt(1 - (v / c)^2) - 1) (1 / aF - 1 / aB) - v tx = sqrt(1 - (v / c)^2) d

c^2 (sqrt(1 - (v / c)^2) - 1) (1 / aF - 1 / aB) = d (sqrt(1 - (v / c)^2) - 1) + v tx

c^2 (1 / aF - 1 / aB) = d + v tx / (sqrt(1 - (v / c)^2) - 1)

But also similarly for u with

d + c^2 (sqrt(1 - (u / c)^2) - 1) (1 / aF - 1 / aB) - u tx = sqrt(1 - (u / c)^2) d

c^2 (sqrt(1 - (u / c)^2) - 1) (1 / aF - 1 / aB) = d (sqrt(1 - (u / c)^2) - 1) + u tx

c^2 (1 / aF - 1 / aB) = d + u tx / (sqrt(1 - (u / c)^2) - 1)

We can see first of all from these that either v / (sqrt(1 - (v / c)^2) - 1) = u / (sqrt(1 - (u / c)^2 - 1) = constant, which it is not, or else tx is zero. Before I mentioned that depending upon how the ship accelerates, front first or back first, the ship might be stretched or compressed, but this is only an initial condition that the stress of acceleration will apply to the ship. Unless the ship breaks up during the initial acceleration, the hull will eventually compensate until the ship is in equilibrium. tx is essentially tx = d / s where s is the speed of sound of the ship with a thruster fired at either the front or the back. We can consider that any initial compression or expansion will even itself back out during the initial acceleration for the most part, as the impulse waves pass through the ship back and forth, so we can use the stationary length of the ship d before acceleration is applied and consider that any left over stretching or compression that takes place will be minimal, so that tx should become insignificant in the long run after the initial accelerations have taken place and the ship works toward equilibrium. In this case, we are left with just

c^2 (1 / aF - 1 / aB) = d

which is consistent for the constant proper accelerations at the front and back of the ship, providing the correct contracted length between the front and back for any frame the ship might accelerate to. I will work out the difference in proper times accordingly next.

 

[grav-universe]

Okay, now to find the simultaneity difference between the front of the ship and the back after the ship has reached its new frame and travels inertially at v. We are considering tx to be insignificant, or working itself back out toward zero immediately after the initial acceleration occurs, so we are left with just c^2 (1 / aF - 1 / aB) = d for the rest of the duration for the acceleration of the ship. Considering both ends essentially begin acceleration simultaneously, then, when some time t3 passes in the rest frame, after both ends of the ship have become inertial and now travel at v in the new frame, we have

tF' = (c / aF) ln[(aF tF / c) + sqrt(1 + (aF tF / c)^2)] + z (t3 - tF)

= (c / aF) ln[(v / c) / sqrt(1 - (v / c)^2) + sqrt(1 + (v / c)^2 / (1 - (v / c)^2))] + z (t3 - tF)

= (c / aF) ln[(v / c) / sqrt(1 - (v / c)^2 + 1 / sqrt(1 - (v / c)^2)] + z (t3 - tF)

= (c / aF) ln[sqrt((1 + v / c) / (1 - v / c))] + z (t3 - tF)

where z is the time dilation of the ship's clocks when traveling at v in the new frame, whereas z = sqrt(1 - (v / c)^2), so we have

tF' = (c / aF) ln[sqrt((1 + v / c) / (1 - v / c))] + sqrt(1 - (v / c)^2) (t3 - tF)

and similarly for tB', giving

tB' = (c / aB) ln[sqrt((1 + v / c) / (1 - v / c))] + sqrt(1 - (v / c)^2) (t3 - tB)

The difference between their times as viewed from the rest frame, then, is

tF' - tB' = c (1 / aF - 1 / aB) ln[sqrt((1 + v / c) / (1 - v / c))] - sqrt(1 - (v / c)^2) (tF - tB)

= c (1 / aF - 1 / aB) ln[sqrt((1 + v / c) / (1 - v / c)] - sqrt(1 - (v / c)^2) [v / (aF sqrt(1 - (v / c)^2)) - v / (aB sqrt(1 - (v / c)^2))]

= (1 / aF - 1 / aB) [c ln[sqrt((1 + v / c) / 1 - v / c))] - v]

Since before we had c^2 (1 / aF - 1 / aB) = d, this now gives

tF' - tB' = (d / c^2) [c ln[sqrt((1 + v / c) / (1 - v / c))] - v]

= (d / c) [ln[sqrt((1 + v / c) / (1 - v / c))] - v / c]

and there we have it.

Well, oops. A couple of posts back in post #37, I wasn't able to work it completely out as I just have, so compared the equations instead. I figured that since aB tB = aF tF, we could just cut both accelerations in half and have the front and back accelerate to the new frame in twice the time, giving twice the difference in times between the front and back. I will have to remember now never to do that again. The problem is that according to the equation we came up with c^2 (1 / aF - 1 / aB) = d, we cannot simply cut both accelerations in half, since in order to do that, we must then have 2d, not d. The proper accelerations of the front and back are related in a particular way, so we cannot just cut both of them in half, but rather, we can cut one in half and then determine what the other must be and go from there, so the conclusions of that post are incorrect, sorry about that. We can now see from the last equation,

tF' - tB' = (d / c) [ln[sqrt((1 + v / c) / (1 - v / c))] - v / c]

that regardless of the amount of constant proper acceleration that is applied as the ship accelerates to the new frame, the resulting difference in times between the front and back as the rest frames views them will be the same. That is very interesting; I didn't expect that. The resulting difference in times between the front and back that the rest frame measures depends only upon the proper length of the ship and the final speed v. Comparing some values, if v is small, we have

tF' - tB' is approximately (d / c) (v / c)^3 / 3

For v = .5 c and d = 1 ls as in the examples earlier, we get tF' - tB' = .049306144 sec.

 

[grav-universe]

Okay, so after the ship reaches the new frame and travels inertially at v, the rest frame views the front clock as having a greater reading than the back clock by

tF' - tB' = (d / c) [ln[sqrt((1 + v / c) / (1 - v / c))] - v / c]

In order to synchronize to the new frame, the reading of the back clock must be greater than that of the front clock by d v / c^2 according to the rest frame, but instead it is behind, so to properly synchronize between the back and front of the ship in the new frame, the back clock would have to be set forward a time of (tF' - tB') + d v / c^2 as the rest frame views it, which of course would mean that the back clock must be set forward by the same amount according to the ship's frame also, so as far as the ship observers are concerned, the back clock lags behind the front clock by a total time of

tF - tB = (tF' - tB') + d v / c^2

= (d / c)[ln[sqrt((1 + v / c) / (1 - v / c))] - v / c] + d v / c^2

= (d / c) ln[sqrt((1 + v / c) / (1 - v / c))]

= (d / c) ln[(1 + v / c) / (1 - v / c)] / 2

= (d / c) arctanh(v / c)

So for v = .5 c and d = 1 ls, from the perspective of the ship's frame, the back clock lags behind the front clock by .549306144 sec.

 

[grav-universe]

Now we can also go back and re-work the earlier examples to find out what occurs with length contraction and simultaneity shifts by applying this natural relation for the constant proper acceleration between the front and back of the ship when accelerating to a second frame with v = .5 c and then again to a final frame with u = .8 c. After accelerating to v = .5 c, the ship observers must add .549306144 sec to the back clock to be properly synchronized. Then considering themselves to be at rest in that frame, they accelerate again to v = .5 c, so to u = 2 v / (1 + v^2 / c^2) = .8 c according to the original rest frame, and add another .549306144 sec to the back clock to synchronize to the final frame, so 1.098612289 sec has been added to the back clock in all.

This time the ship observers want to accelerate directly to u = .8 c, so they add 1.098612289 sec to the back clock beforehand. If they don't add to the back clock, then after accelerating directly to the final frame, they will find the back clock to be behind by a time of

tF - tB = (d / c) ln[(1 + u / c) / (1 - u / c)] / 2 = 1.098612289 sec

but since they have already added this same amount of time to the back clock beforehand, they now find the clocks to be perfectly synchronized.