How Clocks (Seemingly) Ruin Everything

  • Thread starter striphe
  • Start date
  • #1
125
1

Main Question or Discussion Point

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.
 

Answers and Replies

  • #2
DaveC426913
Gold Member
18,787
2,268
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 exaclty 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).
 
  • #3
125
1
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.
 
  • #4
ghwellsjr
Science Advisor
Gold Member
5,122
146
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.
 
  • #5
125
1
Can you please explain how different means of acceleration effect relativity of simultaneity ghwellsjr?
 
  • #6
5,428
291
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.
 
  • #7
ghwellsjr
Science Advisor
Gold Member
5,122
146
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.
 
  • #8
125
1
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?
 
  • #9
ghwellsjr
Science Advisor
Gold Member
5,122
146
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.
 
Last edited:
  • #10
5,428
291
striphe said:
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.
 
  • #11
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.
 
Last edited:
  • #12
125
1
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 space craft 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.
 
  • #13
Grav-universe, My understanding of your post and ghwellsjr's post is that from which ever point along the space craft 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.
 
  • #14
125
1
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
 
  • #15
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.
 
  • #16
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.
 
  • #17
125
1
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.
 
  • #18
182
0
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 [Broken]

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

http://arxiv.org/PS_cache/physics/pdf/0411/0411233v1.pdf
 
Last edited by a moderator:
  • #19
125
1
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.
 
  • #20
182
0
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.
 
  • #21
125
1
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?
 
  • #22
182
0
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.
 
  • #23
125
1
So what will the clocks on-board read
what i mean is how do they read for someone on-board
 
  • #24
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.
 
  • #25
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.
 

Related Threads on How Clocks (Seemingly) Ruin Everything

  • Last Post
Replies
4
Views
4K
Replies
119
Views
13K
Replies
14
Views
999
Replies
5
Views
832
Replies
10
Views
968
Replies
1
Views
1K
  • Last Post
Replies
6
Views
2K
Replies
16
Views
3K
Replies
4
Views
1K
Replies
2
Views
1K
Top