How Clocks (Seemingly) Ruin Everything

[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.