Different Clock Rates Throughout Accelerating Spaceship

[stevendaryl]

2) Actually I take it back. Since the hurling process takes time, and they are both moving wrt rest frame by the end of the boulder hurling process, I expect the forward person to finish hurling later that the back individual as seen from the rest frame.

Well, you can actually calculate it. For simplicity, let F = the rest frame before hurling the boulder. F' = the rest frame afterward. Let v = the relative velocity between the frames. Let T = the time required to hurl the boulder, as measured in frame F. Let D = the distance traveled while hurling the boulder, as measured in frame F. Let L = the distance between people, as measured in frame F.

Then identify a number of events:

1. e_1: the rear person starts to hurl the boulder.
2. e_2: the rear person finishes.
3. e_3: the front person starts to hurl the boulder.
4. e_4: the front person finishes.

Our assumptions are that the corresponding actions are synchronized in frame F. So we have the coordinates for these events in frame F:

1. x_1 = 0,\ \ t_1 = 0
2. x_2 = D,\ \ t_2 = T
3. x_3 = L, \ \ t_3 = 0
4. x_4 = L+D, \ \ t_4 = T

Now use the Lorentz transforms to see the coordinates in frame F':

1. x_1' = 0, \ \ t_1' = 0
2. x_2' = \gamma (D - vT), \ \ t_2' = \gamma (T - \dfrac{vD}{c^2})
3. x_3' = \gamma L,\ \ t_3' = -\gamma \dfrac{vL}{c^2}
4. x_4' = \gamma (L+D - vT), \ \ t_4' = \gamma (T - \dfrac{v(L+D)}{c^2})

While in frame F, t_1 = t_3 and t_2 = t_4, in frame F', the order of events is: t_3', then t_1' or t_4' and then finally t_2'. (The order of t_1' and t_4' depends on the size of L, D, and T.)

So what things look like in frame F' is this:

1. Initially, both people are traveling at speed v in the -x direction.
2. t' = t_3': the front person throws a boulder. His speed in the -x direction starts slowing down. But the rear person continues to travel at speed v in the -x direction.
3. t' = t_1': the rear person starts to throw a boulder, as well.
4. t' = t_4': the front person comes to rest.
5. t' = t_2': the rear person comes to rest.

So between times t_3' and t_2', the front person is traveling slower than the rear person, so experiences less time dilation. So when they come to rest, the front clock will have gained more time than the rear clock, and also, the distance between the rear and the front will have increased.

 

[1977ub]

So we have the coordinates for these events in frame F:

1. x_1 = 0,\ \ t_1 = 0
2. x_2 = D,\ \ t_2 = T
3. x_3 = L, \ \ t_3 = 0
4. x_4 = L+D, \ \ t_4 = T

But this is what I'm not sure I can believe anymore. Specifically that t2 and t4 are both equal to the same amount of time in the RF. Once the hurlers start pushing the boulders, the vehicles star moving, and their clocks no longer appear to be synchronized with one another from RF. Frontward clock now appears to be set later than Rearward clock. Thus RF must conclude that boulders are finished being hurled at different RF times.

 

[stevendaryl]

But this is what I'm not sure I can believe anymore. Specifically that t2 and t4 are both equal to the same amount of time in the RF. Once the hurlers start pushing the boulders, the vehicles star moving, and their clocks no longer appear to be synchronized with one another from RF.

You have to keep straight two different scenarios: (1) The front and rear accelerate so that the distance between them remains constant (as viewed by the people in the rockets). (2) The front and rear accelerate at exactly the same time (according to the initial rest frame) and in exactly the same way (according to the initial rest frame).

In scenario (1):

• The clock in the front runs faster than the clock in the rear, according to the original rest frame, and also according to those aboard the rockets.
• The distance between the front and rear contracts, according to the original rest frame.
• The distance between the front and rear remains constant, according to those aboard the rockets.
• The acceleration felt by those in the front is less than the acceleration felt by those in the rear.

In scenario (2):

• The clocks in the front and rear run at the same rate, according to the original rest frame.
• The front clock runs faster than the rear clock, according to the people in the rockets.
• The distance between the front and rear remains constant, according to the original rest frame.
• The distance between the front and rear expands, according to those aboard the rockets.
• The acceleration felt by those in the front is the same as the acceleration felt by those in the rear.

 

[1977ub]

You have to keep straight two different scenarios: (1) The front and rear accelerate so that the distance between them remains constant (as viewed by the people in the rockets). (2) The front and rear accelerate at exactly the same time (according to the initial rest frame) and in exactly the same way (according to the initial rest frame).

In my later posts, I settled on a 3rd scenario I found more fundamental, in which I do not presume to know how things will turn out, but instead start with identical programmed equipment. We've got identical springs compressed holding identical boulders. Both parties in their identical but separate vehicles activate the release on their own copy of the equipment at t0 which both parties in their vehicles and an observer in the rest frame agree is simultaneous, since at t0 all are at rest wrt one another. So how does that pan out? Presumably to the rest observer, the boulders *complete* their release at different times, since throughout the spring release, the vehicles are moving, and for him their simultaneity is lost.

 

[PAllen]

In my later posts, I settled on a 3rd scenario I found more fundamental, in which I do not presume to know how things will turn out, but instead start with identical programmed equipment. We've got identical springs compressed holding identical boulders. Both parties in their identical but separate vehicles activate the release on their own copy of the equipment at t0 which both parties in their vehicles and an observer in the rest frame agree is simultaneous, since at t0 all are at rest wrt one another. So how does that pan out? Presumably to the rest observer, the boulders *complete* their release at different times, since throughout the spring release, the vehicles are moving, and for him their simultaneity is lost.

If I understand it, your 3d scenario is identical to stevendaryl's scenario (2). In which case it doesn't correspond in a meaningful way to an idealization of real rocket or to a tall building sitting on the surface of large planet (these are scenario (1) as stevendaryl labels them).

You've got boulders expelled by identical springs at t0 per starting rest frame. This means they must be the same distance apart and moving at the same speed per this starting rest frame. If they each expel another boulder at t0+1 per their own watches (which are still in synch per the rest frame (but slow), but not per each other), again their speed and distance and clocks are in synch per the starting rest frame. Per each other, their clocks are out of synch and they have moved further apart.

 

[1977ub]

If I understand it, your 3d scenario is identical to stevendaryl's scenario (2).

I wanted to simplify the situation to a single "boost."

Either I went wrong somewhere, or else this cannot be situation #2.

1) Once both boulders *begin* to be pressed backward, both vehicles are in motion.

2) From that moment on, moving wrt the rest frame, their clocks appear out of synch in the rest frame.

3) Therefore, the final moment of boulder release from the spring will be different between the vehicles as seen by the rest frame.

I am describing the case where the apparatus, the program, the intent of the two vehicles is the same, but I think that they end up out of synch at then end of the acceleration period, as seen in the rest frame, therefore we can't in a general way say that they have "accelerated in exactly the same way in the rest frame." The intended so, they began so, but because the acceleration takes time, and they are moving during that time, they have not ended up so. They *began* accelerating in exactly the same way in the rest frame. They entire acceleration process did not take place in exactly the same way in the rest frame.

 

[PAllen]

I wanted to simplify the situation to a single "boost."

Either I went wrong somewhere, or else this cannot be situation #2.

1) Once both boulders *begin* to be pressed backward, both vehicles are in motion.

2) From that moment on, moving wrt the rest frame, their clocks appear out of synch in the rest frame.

No. They would be out of synch with starting frame if they followed a synchronization procedure while they were in motion. Since they started synchronized in the initial frame, and follow identical physical process, they remain in synch in the initial frame, and out of synch with each other.

3) Therefore, the final moment of boulder release from the spring will be different between the vehicles as seen by the rest frame.

Nope. It will be in synch from the initial frame; it will appear out of synch to each vehicle, if they were to apply a synchronization procedure.

 

[stevendaryl]

In my later posts, I settled on a 3rd scenario I found more fundamental, in which I do not presume to know how things will turn out, but instead start with identical programmed equipment. We've got identical springs compressed holding identical boulders. Both parties in their identical but separate vehicles activate the release on their own copy of the equipment at t0 which both parties in their vehicles and an observer in the rest frame agree is simultaneous, since at t0 all are at rest wrt one another. So how does that pan out?

That's not a 3rd scenario, that's scenario 2. If the front and rear follow exactly identical actions that are (initially) simultaneous in the initial rest frame, then they will always be synchronized in the initial rest frame, and the distance between them will remain constant, according to the initial rest frame.

Presumably to the rest observer, the boulders *complete* their release at different times

No. If they are doing the same actions, starting at the same time, they will finish at the same time.

...since throughout the spring release, the vehicles are moving, and for him their simultaneity is lost.

The person in the front and the rear will not view their clocks as synchronized, but they will continue to be synchronized according to the initial rest frame.

 

[stevendaryl]

I wanted to simplify the situation to a single "boost."

Either I went wrong somewhere, or else this cannot be situation #2.

1) Once both boulders *begin* to be pressed backward, both vehicles are in motion.

2) From that moment on, moving wrt the rest frame, their clocks appear out of synch in the rest frame.

Why do you think that? If the front and rear are accelerating identically, then they will always be traveling at the same velocity, according to the initial rest frame.

 

[1977ub]

ok. thanks.

 

[Austin0]

You have to keep straight two different scenarios: (1) The front and rear accelerate so that the distance between them remains constant (as viewed by the people in the rockets). (2) The front and rear accelerate at exactly the same time (according to the initial rest frame) and in exactly the same way (according to the initial rest frame).

In scenario (1):

• The clock in the front runs faster than the clock in the rear, according to the original rest frame, and also according to those aboard the rockets.
• The distance between the front and rear contracts, according to the original rest frame.
• The distance between the front and rear remains constant, according to those aboard the rockets.
• The acceleration felt by those in the front is less than the acceleration felt by those in the rear.

In scenario (2):

• The clocks in the front and rear run at the same rate, according to the original rest frame.
  [*]The front clock runs faster than the rear clock, according to the people in the rockets.
• The distance between the front and rear remains constant, according to the original rest frame.
  [*]The distance between the front and rear expands, according to those aboard the rockets.
• The acceleration felt by those in the front is the same as the acceleration felt by those in the rear.

How do they determine these effects within the frame?
Measure relative clock rates and distance??

 

[1977ub]

ok hope this is ok for this same thread. I'm moving in a more basic direction for understanding acceleration+SR:

For two intertial frames in relative motion, we can use gamma to describe how each observer measures the other's clock speed. Complete parity. Closely related to relativity of simultaneity. Fine.

Given an inertial frame RF,
and someone moving in a circle AF at velocity v with associated Lorentz gamma,

as far as I understand, RF still use gamma to determine rate of AF's clock?

what will AF use for RF's clock - 1/gamma ?

 

[PAllen]

How do they determine these effects within the frame?
Measure relative clock rates and distance??

They can determine (in principle) that the distance front to back of the rocket expanded by attaching a string that can't stretch; it will break. This is just Bell's spaceship paradox.

They can detect time difference between front and back clocks by exchanging signals.

 

[PAllen]

ok hope this is ok for this same thread. I'm moving in a more basic direction for understanding acceleration+SR:

For two intertial frames in relative motion, we can use gamma to describe how each observer measures the other's clock speed. Complete parity. Closely related to relativity of simultaneity. Fine.

Given an inertial frame RF,
and someone moving in a circle AF at velocity v with associated Lorentz gamma,

as far as I understand, RF still use gamma to determine rate of AF's clock?

what will AF use for RF's clock - 1/gamma ?

[yes, your RF can use gamma for the accelerated object.]

There are different philosophy's on this. To understand anything accelerating observers will measure (including see or photograph), it is simplest just to use any convenient inertial frame. The results of observations are invariant.

To try to come up with a frame for the accelerating observer, you run into the same issues as in GR: there is well defined local accelerated frame, just as there are well defined local frames in GR. However, there is no global frame for an accelerated observer in SR, just as there are no global frames in GR. What you can do, if you insist, is set up a coordinate system in which the accelerated observer remains at fixed coordinate position. Such a coordinate system may not be able to cover all of spacetime. Unfortunately, there are many ways to do this, none preferred. Once you have defined such coordinates (via transform from inertial frame), you can compute the metric in them. Then, using the metric, you can compute time dilation etc. per this coordinate system. It won't be as simple as a constant in place of gamma. The constant gamma results from the fact that the metric in the inertial SR frame is diag(1,-1,-1,-1). With a metric that varies by position and time, you need to integrate contraction of metric with path tangent vectors, instead of having a simple constant.

The up shot of all this is that there is no (preferred) answer to your question (what does the accelerated observer use in place of gamma?). It depends on what coordinate system you set up. On the other hand, let me stress again, if you want to know anything about what the accelerated observer measures or sees, just compute this in any convenient inertial frame.

 

[1977ub]

what does the accelerated observer use in place of gamma?

I'm asking a simpler question than how to define gamma if one is AF.

What RF calls gamma can be used to determine the click speed on AF and its inverse can be used to determine RF's click speed measured from AF.

I mean if RF measures one second while AF is making a revolution, and RF find AF's clock to have moved forward by .5 second, then it's a given that AF will find RFs clock to have clicked twice as fast as his own. I just want to verify it is this simple. I don't see how this can't be true.

 

[PAllen]

I'm asking a simpler question than how to define gamma if one is AF.

What RF calls gamma can be used to determine the click speed on AF and its inverse can be used to determine RF's click speed measured from AF.

I mean if RF measures one second while AF is making a revolution, and RF find AF's clock to have moved forward by .5 second, then it's a given that AF will find RFs clock to have clicked twice as fast as his own. I just want to verify it is this simple. I don't see how this can't be true.

I use AO (accelerated observer) rather than AF, below, because there is no such thing as an accelerated frame (only choices of many possible coordinate systems).

For AO, the behavior they see on clocks in the inertial frame depend on where in the inertial frame they are, and the visual rates vary in time. That is, the observed behavior of inertial clocks will be both position and time dependent. The rates on these clocks averaged over time will show them (per the AO) to fast compared to AO clock. It is true that for pure circular motion at constant speed, the averaged rate seen on the inertial clocks will be gamma (as determined by RF) times faster than AO's clock.

 

[1977ub]

It is true that for pure circular motion at constant speed, the averaged rate seen on the inertial clocks will be gamma (as determined by RF) times faster than AO's clock.

And inescapably, if we simply use a single clock in RF, and AO measures it once per revolution (passing right by it, say) then we can use 1/RF's-gamma to determine the time anti-dilation of RF as perceived by AO. I do understand that this will not apply to all of RF's clocks throughout a revolution. For one thing, AO is moving away from some of them while moving toward others, etc. "On average" AO must be able to use 1/RF's-gamma to determine the average speed of RF's clocks. Every time AO scrapes by clock-RF-0, RF finds AO's clock to have ticked slower by gamma, thus AO must find clock-RF-0 to have ticked faster by 1/RF's-gamma.

 

[PAllen]

And inescapably, if we simply use a single clock in RF, and AO measures it once per revolution (passing right by it, say) then we can use 1/RF's-gamma to determine the time anti-dilation of RF as perceived by AO. I do understand that this will not apply to all of RF's clocks throughout a revolution. For one thing, AO is moving away from some of them while moving toward others, etc. "On average" AO must be able to use 1/RF's-gamma to determine the average speed of RF's clocks. Every time AO scrapes by clock-RF-0, RF finds AO's clock to have ticked slower by gamma, thus AO must find clock-RF-0 to have ticked faster by 1/RF's-gamma.

Agreed. Just don't try to generalize this to other situations without understanding the complexities I described.

 

[1977ub]

Agreed. Just don't try to generalize this to other situations without understanding the complexities I described.

Sure.

1) for uniform linear motion, gamma is purely "subjective" - nobody's clock is moving slower than anyone else's in a meaningful way. it is all wrapped up with planes of simultaneity.

2) for uniform circular motion, gamma gains a clear "objective" meaning - RF can use it to describe *all* of AO's time dilation, and AO can use the inverse to describe *average* time anti-dilation of RF. It is similar to the gravity situation where there are agreed differences in time dilations.

3) The next part is harder for me to work out -
the linearly accelerating observer - how RF and AO can measure their relative clock speeds.

I gather than RF can calculate AO's relative time dilation by simply integrating ever changing gamma with ever changing velocity?

But to ask how AO determines RF's clock speed, this situation is neither the simple 'subjective' or 'objective' case above.

 

[PAllen]

3) The next part is harder for me to work out -
the linearly accelerating observer - how RF and AO can measure their relative clock speeds.

I gather than RF can calculate AO's relative time dilation by simply integrating ever changing gamma with ever changing velocity?

But to ask how AO determines RF's clock speed, this situation is neither the simple 'subjective' or 'objective' case above.

Yes, RF could just integrate gamma(t) over the accelerating path. RF can do this for any path.

For AO, this simple example raises one of the SR concepts difficult to grasp on first encounter. This is the so called Rindler horizon. If you try to ask about how AO would model RF clocks by factoring out light delay using some simultaneity convention, you face the following observation:

RF clocks that AO is accelerating away from appear to red shift until they disappear, at a fixed distance behind the AO. Any RF clock further away cannot be seen or communicate in any way with the AO, so mutual clock comparison is impossible. Note that there is a one way aspect to this (as for all horizons): any RF clock can eventually receive a signal from any point in AO's history; however, there is a last time, for every clock in RF, after which it cannot send any signals to AO.

Visually, you can certainly say the case is more like your (1): any clock at rest in RF and AO's clock each eventually see the other clock freezing and red shifting to infinity.

 

[1977ub]

For normal SR we can use gamma for both observers to determine that they see each other's clocks as ticking slower by gamma. This works for clocks they are passing or any other clocks in either frame.

I understand that it becomes more complex doing this between RF and AO for remote clocks, so I would just wish to focus on RF clocks being passed by the AO observer, all of which are deemed simultaneous and synchronized by RF.

RF uses integrating gamma to watch AO's single clock continually slow down relative to the RF network of clocks as v and gamma increase.

What does AO decide passing these RF clocks. How fast does this network of RF clocks tick? The same method?

 

[PAllen]

For normal SR we can use gamma for both observers to determine that they see each other's clocks as ticking slower by gamma. This works for clocks they are passing or any other clocks in either frame.

I understand that it becomes more complex doing this between RF and AO for remote clocks, so I would just wish to focus on RF clocks being passed by the AO observer, all of which are deemed simultaneous and synchronized by RF.

RF uses integrating gamma to watch AO's single clock continually slow down relative to the RF network of clocks as v and gamma increase.

What does AO decide passing these RF clocks. How fast does this network of RF clocks tick? The same method?

A direct clock comparison is invariant. If, using RF, you compute (correctly) that AO's clock will be further and further behind each RF clock it passes, then, ipso facto, AO will find each passing RF clock further and further ahead. But please note, the same would be true if AO were moving uniformly (a inertially moving clock passing this sequence of clocks would interpret that each is ticking slow, but they are increasing out of synch with each other per the moving clock).

To try to create an symmetric situation for AO, we need a configuration of co-accelerating clocks, with one RF clock going past them. For this, we have to decide the acceleration profile of each clock, and also how to synchronize them. For the latter, unfortunately, there is no preferred approach (what is special about inertially comoving clocks is that any reasonable synchronization procedure produces the same result; this is not true for a family of accelerating clocks. In particular, Einstein clock synch using light signals, and Born rigidity based simultaneity, disagree.)

I urge you to focus on questions about what AO observes, and stop trying to treat AO as defining a frame of reference.

 

[1977ub]

I urge you to focus on questions about what AO observes, and stop trying to treat AO as defining a frame of reference.

AO has a clock. For me that's most important thing. I'm trying to build up my understanding of the principles necessary for understanding that the two identical accelerating craft along an axis judge themselves to have differently ticking clocks according to both of them.

But now I realize I need to look back at the SR issue you brought up.

inertially moving clock passing this sequence of clocks would interpret that each is moving slow, but they are increasing out of synch with each other per the moving clock.

As a particular clock in a moving frame passes me, and continues along, I decide that it ticks at a steady rate, slower than mine by gamma, a determination which relies on my network of "simultaneous" clocks.

However since I also determine at any moment using my network that gamma-adjusted clocks placed further back on along the moving frame are set later than clocks forward along the frame, that is a different effect. Do they cancel each other out? As a moving frame passes me, and I watch the counts of these different clocks in the frame, passing me, a different clock each time, do I see this time change at my own rate?

 

[PAllen]

As a particular clock in a moving frame passes me, and continues along, I decide that it ticks at a steady rate, slower than mine by gamma, a determination which relies on my network of "simultaneous" clocks.

However since I also determine at any moment using my network that gamma-adjusted clocks placed further back on along the moving frame are set later than clocks forward along the frame, that is a different effect. Do they cancel each other out? As a moving frame passes me, and I watch the counts of these different clocks in the frame, passing me, a different clock each time, do I see this time change at my own rate?

If a line of inertial clocks synchronized with each other by Einstein convention, passed you, and the first one matched your clock, each successive clock would be further ahead of yours. The clock synch issue dominates the slower rate on each clock. Think of a muon generated in the upper atmosphere passing Earth frame clocks on the way to the ground. It sees the last of these clocks, say, a millisecond ahead of the first clock it passed; meanwhile, for the muon less than 2 microseconds have passed.

 

[1977ub]

If a line of inertial clocks synchronized with each other by Einstein convention, passed you, and the first one matched your clock, each successive clock would be further ahead of yours.

I see. There must be a simple expression for how quickly that succession of clocks appears to tick to me - an expression involving gamma - not sure if you can easily find that... ?

 

[PAllen]

I see. There must be a simple expression for how quickly that succession of clocks appears to tick to me - an expression involving gamma - not sure if you can easily find that... ?

There is. It is gamma. You are 'seeing' why the other observer thinks you are slower by gamma.

 

[PAllen]

Quote by Austin0

Assuming they are physicists:
Distance
Wouldn't they assume that the string broke because of Lorentz contraction?
And so opt for a different method to determine distance , like radar ranging?

Assuming constant proper acceleration and the initial synchronization of the launch frame: 1) it appears to me that reflected distances and times would be equivalent at both ends.
2) I see no obvious reason to infer that the measured distances would increase with time. In fact I would think that the raw times would decrease due to the dilation factor due to acceleration between transmission and return and the increased velocity.
This would seem to be the case if carted from the initial frame and if we assume frame agreement between observation events it seems like it should hold in the accelerated frame.

Time
Wouldn't they ,knowing they are accelerating , attribute the difference in received signals to Doppler due to the relative motion from acceleration during transit??.

Couldn't they integrate the dilation factor during transit and the effect of relative motion to extract a value solely related to the clock rates at transmission?

I find this post an argumentative distraction in a thread where the OP is genuinely trying to learn. I will not answer in this thread.

More to the point, the OP asked about direct observations, not interpretations. If front and back experience same g force, then string between front and back will break. This is a fact. Similarly, if they compare clock rates by exchanging signals, the front clocks will be observed to be going faster, by both front and back rocket passengers. Again a simple fact.

That is all that was asked - direct observations.

 

[Austin0]

I find this post an argumentative distraction in a thread where the OP is genuinely trying to learn. I will not answer in this thread.

More to the point, the OP asked about direct observations, not interpretations. If front and back experience same g force, then string between front and back will break. This is a fact. Similarly, if they compare clock rates by exchanging signals, the front clocks will be observed to be going faster, by both front and back rocket passengers. Again a simple fact.

That is all that was asked - direct observations.

I am sorry if you interpreted my query as argumentative. it was out of pure interest in the subject and directly related to stevedaryl's post. Certainly I understand if you want to end it here or move it to another thread.

 

[stevendaryl]

How do they determine these effects within the frame?
Measure relative clock rates and distance??

In the same way you might measure the height of a building on Earth. For instance, if you take a long rope, and make a mark every meter or so, and let the rope dangle from the front rocket to the back, that would show the distance constant (according to scenario 1) and increasing (according to scenario 2). Alternatively, you could measure the distance by bouncing a laser off the front rocket and back to the rear rocket, and measure the round trip time (or you could look at interference patterns).

It really is the case that aboard an accelerating rocket, things don't seem much different from a rocket hovering over the Earth, except that the variation of "gravity" with height is different in the two cases (although you have to have a really huge rocket to notice the difference in either case).

 

[1977ub]

There is. It is gamma. You are 'seeing' why the other observer thinks you are slower by gamma.

Interesting.

ok so the problem of AO's "frame"...

RF can easily decide there is a grid of evenly-spaced meters and synchronized clocks using exchanges of light pulses.

we can say that RF has no trouble assigning AO an x coordinate and measuring his even acceleration 'a', and clock speed related to AO's current gamma.

what happens when AO attempts this... Will AO encounter any ambiguity determining an x coordinate of the RF observer at the origin?

If AO can do this, what will he think of RF-origin's clock speed and 'acceleration' with regard to him?

 

[PAllen]

Interesting.

ok so the problem of AO's "frame"...

RF can easily decide there is a grid of evenly-spaced meters and synchronized clocks using exchanges of light pulses.

we can say that RF has no trouble assigning AO an x coordinate and measuring his even acceleration 'a', and clock speed related to AO's current gamma.

what happens when AO attempts this... Will AO encounter any ambiguity determining an x coordinate of the RF observer at the origin?

If AO can do this, what will he think of RF-origin's clock speed and 'acceleration' with regard to him?

AO will not find an unambiguous answer to assigning x and t coordinates. One issue is that there is no such thing as rigid accelerated motion in SR, while there is no problem with rigid inertial motion. You can choose a way to work around this by assuming a mathematical idealization that works for simple acceleration profiles (Born rigid acceleration). The next issue is that different clock synch methods that produce the same result for an inertial observe will produce different results for AO.

While AO would have no difficulty making some choices to set up some coordinate system, the ambiguity means you can't give a single preferred answer to RF origin's clock speed - it depends on which choices you make for setting up the coordinate system.

Note that any choice made by AO will have a different metric that RF. This means that gamma will not apply (gamma as a time dilation factor is a direct consequence of the inertial metric). Further, the standard SR Doppler formula will not apply.

As an interesting aside, a common idealization of the 'rigid framework' coordinates (Fermi-Normal coordinates), and building simultaneity using Einstein's clock synch convention, produce quite different answers for the rate of RF clock (actually, they agree when the RF origin clock is coincident with AO; the further away this clock is, the more they disagree).

 

[1977ub]

AO will not find an unambiguous answer to assigning x and t coordinates. One issue is that there is no such thing as rigid accelerated motion in SR, while there is no problem with rigid inertial motion.

I'm imagining AO as an idealized point (large enough for a "clock" I guess), thus no issues of rigidity... I presume the limit of a point AO will have a sense of the passing of time, and some form of clock, without rigidity questions? So pulses from the origin will take longer and longer to reach him, just as in unaccelerated motion but more so - but unlike the unaccelerated motion case, he will not have a simple way to decide how far they traveled to reach him, I take it.

 

[PAllen]

I'm imagining AO as an idealized point (large enough for a "clock" I guess), thus no issues of rigidity... I presume the limit of a point AO will have a sense of the passing of time, and some form of clock, without rigidity questions? So pulses from the origin will take longer and longer to reach him, just as in unaccelerated motion but more so - but unlike the unaccelerated motion case, he will not have a simple way to decide how far they traveled to reach him, I take it.

Correct, on all counts. Viewed from AO as a single world line, the difficulty all comes to simultaneity, and distance is defined in terms of a simultaneity convention. Why I brought up rigidity is that rigid, comoving rulers, to the extent you can idealize them, provide one possible answer to distance; as such they also define a simultaneity, because you can't have distance without a simultaneity convention. As such, it is interesting that this ruler based simultaneity will disagree with Einstein clock synch.

In no way am I saying you can't set up coordinates - just that you can't go from there to talking about 'the' rate of a distant clock 'now' for an accelerated observer. You also can't use inertial frame formulas (like gamma).

The approach I think you have in mind leads to AO coordinates with some nice properties for non-inertial motion with rapidly changing accelerations. Using two way light signals you simply define that if event e1 is reached by a signal you sent at t1 on your clock, and you got a return signal at t2, then you define that e1 is simultaneous to (1/2)(t2+t1) on your clock. Then define radial coordinate in polar type coordinates by c(t2-t1). Such coordinates have the following nice properties:

- coordinate speed of light is c for radial paths from the origin
- these coordinates cover larger regions of spacetime than many other accelerated coordinates
- they behave 'smoothly' around sudden changes in acceleration (other common coordinates for accelerated observers do not).

However, you also have to accept that coordinate distance fairly quickly diverges from ruler distance (either idealized, or approximate real rulers), where both are possible. For example, as defined above, someone at the bottom of the (accelerating) rocket using the above coordinates would get a slightly different result than using a tape measure along the length of the rocket. [If the rocket was coasting, these would always agree].

 

[PAllen]

Using two way light signals you simply define that if event e1 is reached by a signal you sent at t1 on your clock, and you got a return signal at t2, then you define that e1 is simultaneous to (1/2)(t2+t1) on your clock. Then define radial coordinate in polar type coordinates by c(t2-t1).

Of course, the radial coordinate should be (c/2)(t2-t1).

 

[1977ub]

Back to the usenet rocket example, and related discussions, let's take the two rockets along x which RF finds to be identical in acceleration, velocity, click-rate.

It is said that the tail and head rocket observers can agree that their clocks are moving at different rates.

This has some implication of a 'frame' that each has...

So, there is some ambiguity regarding 'distance' and 'simultaneity' between them...

At the least, each person/clock can see the other as subtending the same or similar visual arc, even if they cannot agree on the distance between them exactly...

 

[PAllen]

Back to the usenet rocket example, and related discussions, let's take the two rockets along x which RF finds to be identical in acceleration, velocity, click-rate.

It is said that the tail and head rocket observers can agree that their clocks are moving at different rates.

This has some implication of a 'frame' that each has...

So, there is some ambiguity regarding 'distance' and 'simultaneity' between them...

At the least, each person/clock can see the other as subtending the same or similar visual arc, even if they cannot agree on the distance between them exactly...

The person in the back rocket would see the the front rocket subtending smaller angle over time. Yet they would see their clock running faster and light be blueshifted.

 

[Wes Tausend]

...

I have a couple of questions if no one minds.

I have a simplistic way of imagining the accelerating rocket scenario. Perhaps it is not correct.

We first imagine Captain Kirk and Scotty on Earth planning an experiment regarding the Equivalence Principle. Both weigh exactly the same, side-by-side, on Earth before departure. They synchronise their absolutely indentical watches which thereby keep identical time when held side-by-side on Earth during weigh-in.

The rocket is long and narrow and Kirk rides in the nose to steer while Scotty rides in the rear near the propulsion engine. They depart Earth to an imaginary region free of external gravitational influence and began an extended acceleration at 32 feet per second per second.

Since the rocket is initially now moving at some forward velocity, it will have experienced a foreshortening in the direction of travel, in relation to the rest of the universe which may consider itself, at average, to be at rest. Both watches will slow down as the rocket gains speed relative to the rest of the universe. In addition, the rocket will experience a continuously greater foreshortening, again in comparison to the rest of the universe, because the rocket is gaining speed in approach to the universal speed of light. We may regard this rocket to have an unlimited supply of fuel.

Because the rocket must continually foreshorten more and more, Scotty is always moving faster than Kirk. It does not matter if one regards the rear of the rocket moving forward faster, or the front of the rocket moving slower to achieve foreshortening, rationally Scotty must move faster than Kirk. The rocket must continuously get shorter at least until it reaches the comparative speed of light where it will have no dimension whatsoever in the direction of motion. Yet also because of Special Relativity and endless fuel, the trip may go on forever which insures the different experiences must be continuous.

Since Scotty is moving faster than Kirk, Scotty's watch must now generally run slower than Kirks watch, according to Special Relativity. Since, by deduction, Kirk is accelerating at a slightly slower rate, he should now "weigh" less than Scotty in the artificially induced gravity of inertial acceleration.

After weigh-in on earth, when Kirk and Scotty first took their separate stations in the vertical rocket in preparation for take-off, Kirk already once weighed less than Scotty because he was at the top of the tall rocket, whereas Scotty was closer to Earth near the rear mounted engines. Kirks watch also ran faster because he resided in slightly less gravity because of his higher altitude and distance from Earth's center.

Because of the Equivalence Principle, which I assume here is absolutely equivalent, the difference in clock rates and apparent weights are identical whether Kirk and Scotty are in test mode at 32 feet per second per second in an accelerating rocket far from earth, or split-up on the 32 feet per second per second launchpad shortly before take-off right on earth. Is this essentially correct?

===

The other question is about a simplified scenario I once read where Einstein had imagined two scientists in an elevator being drawn up by a very long cable in an accelerated manner. It's been about 30 years, so I am not certain I read this was Einsteins personal thought experiment.

Anyway, the two scientists were supposed to be drawn up equivalent to the acceleration of gravity. I believe Einstein surmised that the scientists would not be able to distinguish between being in a gravitational field or that acceleration provided by such a cable.

Einstein went on to imagine that if there were a hole in the elevator sidewall, and a lightbeam entered the elevator in a perfectly perpendicular manner to that wall, the lightbeam would strike the opposing wall slightly below the height of the hole because of the elevator's acceleration. The beam would do so because the beam would seem to be bent by the acceleration of the elevator. Supposedly, it was these proposed results (which became the Equivalence Principle) that dictated curved space and the rest of the development of General Relativity.

So one might further assume that if the two scientists had a stepladder and the elevator were a bit taller, another hole even higher up on the same sidewall would permit another perpendicular lightbeam to enter and curve similarly to strike the opposing wall at a corresponding greater height. But the two observations would only be similar because the top beam would not bend quite as much. This would occur exactly like the elevator were sitting still on Earth's surface and the bottom beam was bent more than the other because of it's greater nearness to Earth center. The gravitational field becomes weaker at greater altitudes, so the top beam would not bend as much. Is this two-beam Equivalence assumption correct?

Thanks,
Wes
...

 

[stevendaryl]

Since Scotty is moving faster than Kirk, Scotty's watch must now generally run slower than Kirks watch, according to Special Relativity. Since, by deduction, Kirk is accelerating at a slightly slower rate, he should now "weigh" less than Scotty in the artificially induced gravity of inertial acceleration.

After weigh-in on earth, when Kirk and Scotty first took their separate stations in the vertical rocket in preparation for take-off, Kirk already once weighed less than Scotty because he was at the top of the tall rocket, whereas Scotty was closer to Earth near the rear mounted engines. Kirks watch also ran faster because he resided in slightly less gravity because of his higher altitude and distance from Earth's center.

Because of the Equivalence Principle, which I assume here is absolutely equivalent...

Just to interrupt here, the equivalence principle is only an approximate statement. If you ignore the variation of g with location, then there is no difference between being at rest on Earth with a gravitational field g, and being in a rocket with acceleration g. But the two cases are different, if you perform precise measurements of g. The gravitational field of the Earth obeys an inverse-square law: g \propto \dfrac{1}{r^2} while the pseudo-gravitational field of an accelerating rocket is inverse-linear: g \propto \dfrac{1}{r}.

 

[stevendaryl]

Just to interrupt here, the equivalence principle is only an approximate statement. If you ignore the variation of g with location, then there is no difference between being at rest on Earth with a gravitational field g, and being in a rocket with acceleration g. But the two cases are different, if you perform precise measurements of g. The gravitational field of the Earth obeys an inverse-square law: g \propto \dfrac{1}{r^2} while the pseudo-gravitational field of an accelerating rocket is inverse-linear: g \propto \dfrac{1}{r}.

I should add that in the case of gravity on the Earth, r measures distance from the center of the Earth. In the case of an accelerating rocket, r is measured so that the rear of the rocket is at r=\dfrac{c^2}{g}, where g is the acceleration felt by someone in the rear. So if the rocket has length L, then the gravity felt by someone in the front of the rocket will be

g' = \dfrac{g}{1+\dfrac{gL}{c^2}}

 

[jartsa]

...

I have a couple of questions if no one minds.

I have a simplistic way of imagining the accelerating rocket scenario. Perhaps it is not correct.

I think you have the correct idea about the bending of light.

An inertial observer sees the rear of the rocket accelerating faster and moving faster, which both things explain why light seems to bend more at the rear.

The people inside the rocket explain the bending of light by the faster motion of photons nearer to the nose, and the fact that faster moving objects tend to move more straight.

(An observer staying in one position inside the rocket will say that "in this homogeneous gravity field the acceleration of falling objects seems to be the same everywhere, although the readings of accelerometers are larger closer to the rear")

And an observer climbing a ladder inside the rocket might say "the gravity field feels weaker here closer to the nose, that must be the reason why the bending of light seems to be decreasing while I'm climbing"

 

[1977ub]

Because the rocket must continually foreshorten more and more, Scotty is always moving faster than Kirk. It does not matter if one regards the rear of the rocket moving forward faster, or the front of the rocket moving slower to achieve foreshortening, rationally Scotty must move faster than Kirk.

This is important thing to remember and which I ran into earlier: "foreshortening" is in the eye of the beholder. There is no shortening for Kirk & Scotty the way there would be for an intertial observer nearby. And furthermore any foreshortening is only found after comparing notes with other observers in one's own frame. It could be argued to be "illusory" (not to put too fine a point on it.) Any shortening experienced by Kirk & Scotty is due to materials compressing I believe. This is one reason why I modified my model to cover two identically accelerating rockets one in front of another. This allows issues of how front / back see / measure the other without issues of materials, etc.

Even "universal speed of light" is relative. Sitting here, we are moving very close to "the universal speed of light" relative to some body somewhere (with its putative observers), without doing anything to make that happen.

 

[1977ub]

Related:
Two rockets differently placed along x.
Both accelerating identically according to RF.
AO1 (rear) 'sees' AO2's clock ticking faster and receding.
Presumably if AO1 puts a 2nd clock, initially synched with his own, on a stick and sends up forward,
attached to AO2 for a bit, he will see it ticking faster.
When he gets it back later, it will have ticked ahead of his own? Or returned to synch with his?

Either way, how does RF explain this?
I can see during clock's travel forward it is accelerating faster than AO1 or AO2.
Thus gamma is larger, clock moves slower than AO1 or AO2 clocks, as seen by RF.
It would arrive to AO2 reading slower than his own, then while set earlier than his own, begin ticking at same rate as his own.
Then when he sends it back, RF will see it accelerating more slowly than AO1 or AO2, lower gamma, faster ticking, and arrives back to AO1 back in synch with his?

Oh:
Since AO1, AO2 and 3rd clock appear to be continually slowing down throughout this exercise, presumably when clock3 is being sent back, this is during a period of greater gammas all around, and thus slower ticking all around, and so the ticks 'lost' during the travel forward are not all 'regained' on the trip back?

 

[jartsa]

Related:
Two rockets differently placed along x.
Both accelerating identically according to RF.
AO1 (rear) 'sees' AO2's clock ticking faster and receding.
Presumably if AO1 puts a 2nd clock, initially synched with his own, on a stick and sends up forward,
attached to AO2 for a bit, he will see it ticking faster.
When he gets it back later, it will have ticked ahead of his own? Or returned to synch with his?

Either way, how does RF explain this?
I can see during clock's travel forward it is accelerating faster than AO1 or AO2.
Thus gamma is larger, clock moves slower than AO1 or AO2 clocks, as seen by RF.
It would arrive to AO2 reading slower than his own, then while set earlier than his own, begin ticking at same rate as his own.
Then when he sends it back, RF will see it accelerating more slowly than AO1 or AO2, lower gamma, faster ticking, and arrives back to AO1 back in synch with his?

Oh:
Since AO1, AO2 and 3rd clock appear to be continually slowing down throughout this exercise, presumably when clock3 is being sent back, this is during a period of greater gammas all around, and thus slower ticking all around, and so the ticks 'lost' during the travel forward are not all 'regained' on the trip back?

A trip to the upper regions and then back at moderate velocity will cause a clock to gain ticks. This is why:

(I'm not going to consider 3 clocks, 2 should be plenty enough.)

RF sees the clock that moves upwards to run slower than a clock that stays at the rear.
RF sees the clock that moves downwards to run faster than a clock that stays at the rear.

The latter effect wins, because relativistic effects become larger as velocity increases.
(that's why not immediately starting the return trip causes even more ticks to be gained.)

 

[Wes Tausend]

...

I apologise. It will take a while to come back with fully intelligent questions and answers. I've recently retired and hoped that I will finally have time to pursue some basic layman questions about the Equivalence Principle that arose during science books I read long ago, before the internet. There is also a bit of domestic pressure to tackle home projects that have accumulated over the years rather than spend time on the internet. I will have to be devious.

Lacking advanced formal education, I think I may be in over my head here for the time being. The first post was simple enough for me to understand in terms of geometry. But I don't have enough of an advanced education to quickly understand some of the math language and acronyms used as this thread proceeded, so there is a language barrier for me to overcome that will take more time.

===

In reply to stevendaryl,

Apparently I may have been mistaken in my belief that the Equivalent Principle is fully equivalent. I took info such as this rendition of Einsteins statement, and subsequent development, quite literally.

I don't entirely understand what you meant by,"But the two cases are different, if you perform precise measurements of g." Does that mean that the loss of apparent "weight" by Kirk will be present in both cases, but different, when the rocket is sitting on the ground as opposed to an inertial reaction in an accelerating rocket? What about the clock variation? Will time differences also be different when sitting on the ground as opposed to accelerating in the rocket? I am unable to see this.

Not knowing any better, regarding the Equivalence Principle not being equivalent, I think I worked out a simple relativity model where tidal forces are seemingly also included on the rocket because of Special Relativity, but have since found that that is a highly unusual perspective. It is easier to envision with a different, rather odd thought experiment. There is even a paragraph to this conventional "tidal shortcoming" towards the end of the wiki development section mentioned earlier.

===

In reply to jartsa,

Yes, the bending of the light should occur in both cases, whether the elevator is sitting still on the ground or being drawn up by cable. I am delighted we see it the same. But again, would it be a varied curve difference between the two?

===

In reply to 1977ub,

It is you I must apologise to most profusely for dragging your thread down to my level and slightly diverting it. Thank you for your tolerance.

I tend to regard foreshortening as a real phenomena that may even be easily observable in a very strange way. My thought was that regardless of motions throughout the endless universe, there is a theoretical average that is most at rest. Since the average rest might be any position and light must be at a relative constant speed to include all other observers, that is the philosophical reason any external observer (beholder) may regard themselves at rest.

To Kirk and Scotty, the rest of the universe is really foreshortening while they observe themselves to remain "uncompressed" in assumed current local time. They actually observe the external universe in an earlier, shorter condition. I know this may not make sense to you. It too, is hard for me to explain without a suitable visual descriptive thought experiment.

I do regard the speed of light as universal in a vacuum. Whereas "nothing may go faster than the speed of light in a vacuum", my thought is nothing can be observed to go faster than the speed of light in a vacuum. Perhaps most importantly, light cannot go slower in a vacuum.

In my opinion, the time-dilation/reverse-time-dilation in an accelerating rocket can best be explained by considering that all indicators of time involve additional motion to create an event (tick-tock). Since a vibrating atom, a pendulum, a reciprocating flywheel and all other event-creating clocks must have their own inherent motion, the time mechanism must have additional motion in addition to the forward motion of the spacecraft . But the closer to lightspeed the addition of motions become, the slower the periodic clock events must progress so they may theoretically gradually arrive at zero, the time-stop instant when all the total available "speed" is entirely used up by forward motion at "C", lightspeed. In the above case of Kirk vs Scotty at acceleration, one gentleman is traveling faster than the other, and their ever-slowing respective clock event motions cannot proceed at equal ticks until they both arrive at zero. I hope this makes sense.

...
Wes

 

[stevendaryl]

I don't entirely understand what you meant by,"But the two cases are different, if you perform precise measurements of g." Does that mean that the loss of apparent "weight" by Kirk will be present in both cases, but different, when the rocket is sitting on the ground as opposed to an inertial reaction in an accelerating rocket? What about the clock variation? Will time differences also be different when sitting on the ground as opposed to accelerating in the rocket? I am unable to see this.

Yes, the variation of "g" with height is different for a rocket sitting on the Earth and for a rocket accelerating through space. The variation of "clock rate" with height is different for a rocket sitting on the Earth and for a rocket accelerating through space. In both cases, "g" gets weaker as you go "higher", and in both cases, clock rates go up as you go higher. But the precise rate at which "g" changes and "clock rates" change with height is different in the two cases.

 

[nitsuj]

It could be argued to be "illusory" (not to put too fine a point on it.)

To your previous point it is "after comparing notes" & not optical observation (Penrose-Terrell rotation) that length contraction is observed.

The "ladder paradox" is imo great because it demonstrates the "strength" of that Principal of Relativity postulate (which seems to hold via RoS). How it "smushes" the two FoRs together; unlike that nasty twin paradox.

Here, clearly the ruler is "ticking more slowly"... Or is "shorter" in simultaneous terms, and of course the concept of "simultaneous" is the illusion; not the fact the ladder is actually shorter and fit(s) in the barn; such a fine line. But maybe made more clear by remembering we live in a continuum, and not a sequence of "simultaneity frames" of a mathematical nature.

 

[1977ub]

Here, clearly the ruler is "ticking more slowly"... Or is "shorter" in simultaneous terms, and of course the concept of "simultaneous" is the illusion; not the fact the ladder is actually shorter and fit(s) in the barn; such a fine line. But maybe made more clear by remembering we live in a continuum, and not a sequence of "simultaneity frames" of a mathematical nature.

"The ladder is actually shorter and fits in the barn" - this highlights the way we see an object has having permanent and synchronized areas or parts. The sleight of hand performed by the moving ladder is to have the back of the ladder arrive "soon" and the front of the ladder arrive "late" compared with how we think the whole of the ladder is traveling as a singular object.

The "gee whiz" nature of all many beginner-level SR descriptions leaves out some of the complexities that would remind one of the simultaneity issues.

The "shortened" train with "slower" clocks for instance - if we think of all the things on a moving train that are not static, then the simple idea of things simply being "shorter" gets muddled - after you really work out the ladder paradox you see this - for instance if there is a ceiling fan, the blades are no longer coordinated and opposite, but flop around like psychedelic beagle ears [edit: and this is not an optical illusion, if resting observer could poke fingers down from above at the right times he'd "feel" the misshapen fan]. If there is an axle running under the train, it is twisted like a peppermint stick. If instead of a "ruler" traveling at a high rate, we were to think of a movie screen, then in addition to noting it seems shortened, we'd have to note that the action at the trailing end is from a part of the film a bit later than the leading end - by seconds or millennia or whatever.

 

[nitsuj]

"The ladder is actually shorter and fits in the barn" - this highlights the way we see an object has having permanent and synchronized areas or parts. The sleight of hand performed by the moving ladder is to have the back of the ladder arrive "soon" and the front of the ladder arrive "late" compared with how we think the whole of the ladder is traveling as a singular object.

The "gee whiz" nature of all many beginner-level SR descriptions leaves out some of the complexities that would remind one of the simultaneity issues.

The "shortened" train with "slower" clocks for instance - if we think of all the things on a moving train that are not static, then the simple idea of things simply being "shorter" gets muddled - after you really work out the ladder paradox you see this - for instance if there is a ceiling fan, the blades are no longer coordinated and opposite, but flop around like psychedelic beagle ears [edit: and this is not an optical illusion, if resting observer could poke fingers down from above at the right times he'd "feel" the misshapen fan]. If there is an axle running under the train, it is twisted like a peppermint stick. If instead of a "ruler" traveling at a high rate, we were to think of a movie screen, then in addition to noting it seems shortened, we'd have to note that the action at the trailing end is from a part of the film a bit later than the leading end - by seconds or millennia or whatever.

Should have also mentioned I like the Wikipedia Ladder Paradox description, specifically the diagram that reads like a Sunday comic strip.

I don't understand how comparative motion is "slight of hand". In what sense does the "back of the ladder arrive "soon" and the front of the ladder arrive "late""? I can only assume you mean the ladders FoR. The one that's equally valid as the barn FoR.


this highlights the way we see an object has having permanent and synchronized areas or parts.
Is that length? In either case yes, the barn doors closing is a "permanent"/simultaneous "area or "parts". And of specific coordinates, coordinates unlike the ladders. both "fall under" principle of relativity.

I can't follow that last paragraph, more comic strip, less big block of words.

 

[jartsa]

...
Yes, the bending of the light should occur in both cases, whether the elevator is sitting still on the ground or being drawn up by cable. I am delighted we see it the same. But again, would it be a varied curve difference between the two?

A scientist in a closed accelerating elevator can not be sure that he is in an accelerating elevator, and not standing still on the ground on some planet that has very homogeneous gravity field. That's what the equivalence principle says.

Clocks and accelerometers behave so that it is possible that there is either a real gravity field, and no acceleration, or acceleration and no gravity field.BUT an inertial observer will observe that clocks inside an accelerating rocket do NOT behave like clocks a in a gravity field. The time dilation difference between clocks at different positions is "too small".

For example the case of the non-contracting rocket: No time dilation difference at all between clocks at different positions.Addition: An accelerometer in a rocket with constant proper acceleration looks the same from any frame: It looks like a stopped clock.

Addition2: Oh yes the question was about the bending of light. Well a bending light beam is an accelerometer. So light beams bend the same way in an accelerating rocket and in a gravity field.

 

[1977ub]

I can't follow that last paragraph, more comic strip, less big block of words.

What do you get for a ceiling fan hanging/rotating in a "train" traveling at relativistic speeds?

Do you see that it is "distorted" and changing its shape? not merely "shortened" ?

[edit/added]

Let's say there is an apparatus set up to make sure that both barn doors close simultaneously as seen by barn frame. There is a single post which is held above the barn and it rotates regularly [on its axis, like an axle]. Large planks are attached at either end and the whole thing rotates so that the two planks cover the barn doors together. Since the ladder perspective sees the two doors covered at different times, this suggests that the post which rotates overhead is measure in ladder frame to be grotesquely twisted like a peppermint stick. This same desynch along a relatively moving frame applies to anything that might exist in the frame, such as a ceiling fan or a TV screen. For those moving relative to this frame, things nearer the direction of motion are set "later" than those toward the rear. Anything like a ceiling fan will be distorted and the blades will no longer be opposite.

[more]

What if a ruler were more like a candle? If it existed for a few minutes, started small, then grew in width as it changed color, then disappeared.

In a moving frame, it would appear to start as a spec and grow out to its full length. but the time it fully existed in the front, the color would be very different in the back, and the width would be greater. finally it would disappear from the back to the front.

We'd certainly be reminded of the simultaneity issues which accompany the shortening. That's all I meant.