Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: thedoc on 04/12/2014 15:30:01
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MALCOLM FOWLER asked the Naked Scientists:
I've heard of the experiment where two atomic clocks were synchronized, one was taken on a high speed trip on a plane and when the plane returned the clock that took the plane ride recorded a slower time. I can just about grasp the principle that as speed increases time slows down relative to the frame of reference of the stationary clock.
However what puzzles me is that if you take your frame of reference as being that of the 'moving clock' then it is the supposedly stationary clock that is moving at high speed and that should be showing a slower time?
The following example might better explain what I am getting at;
Imagine two space rockets docked together in geostationary orbit both carrying an atomic clock. One of the rockets remains in geostationary orbit but the other one ignites it's engines and does an extra orbit of the earth before docking with the other rocket again. Which clock would show the slower time?
From all the examples I have seen it would be the clock on the rocket that undocked and completed an extra orbit of the earth, but surely when considered from the frame of reference of the rocket that did the extra orbit, the other rocket that remained in geostationary orbit has travelled away from it and completed exactly the same orbit in reverse?
Hopefully I've explained it adequately because it puzzles me and I haven't seen the answer in any textbooks.
Best Regards
Malcolm
PS. It's over 30 years since I did my physics A level. I wasn't intelligent enough to make it to university but I do enjoy listening to your podcasts (http://www.thenakedscientists.com/HTML/podcasts/).
What do you think?
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I've heard of the experiment where two atomic clocks were synchronized, one was taken on a high speed trip on a plane and when the plane returned the clock that took the plane ride recorded a slower time. I can just about grasp the principle that as speed increases time slows down relative to the frame of reference of the stationary clock.
It isn't really "time slows down". An atomic clock uses microwaves and other electromagnetic phenomena, which move at the speed of light c. When the clock moves on its high-speed trip at speed v, the electromagnetic propagation inside it isn't moving at a speed c + v. It's still moving at speed c, so the effective speed of electromagnetic propagation inside the clock is less than c. Have a look at the simple inference of time dilation due to relative velocity (http://en.wikipedia.org/wiki/Time_dilation#Simple_inference_of_time_dilation_due_to_relative_velocity). This uses an idealised parallel-mirror light clock, wherein the Lorentz factor used to calculate time dilation is derived from Pythagoras's theorem. The hypotenuse of a right-angled triangle is the light path, and the light moves at c. The base represents speed v through space, and the height gives the Lorentz factor. This simplified scenario "works" because of the wave nature of matter, wherein light and matter are "made of the same essence" (http://en.wikipedia.org/wiki/Gravitational_time_dilation#Important_features_of_gravitational_time_dilation).
(https://www.thenakedscientists.com/forum/proxy.php?request=http%3A%2F%2Fupload.wikimedia.org%2Fwikipedia%2Fcommons%2Fthumb%2Fa%2Fa5%2FTime-dilation-002.svg%2F400px-Time-dilation-002.svg.png&hash=c7512e7864501d98abc9f6143bbd2e8d)
However what puzzles me is that if you take your frame of reference as being that of the 'moving clock' then it is the supposedly stationary clock that is moving at high speed and that should be showing a slower time?
That's the famous "twins paradox". It's not really a paradox. When you and I are separated by distance we don't express amazement that you look smaller than me and I look smaller then you. In similar vein when we are separated by relative motion we shouldn't be amazed that my clock looks slower than yours and yours looks slower than mine. Note though that the symmetry of the situation is broken when one of the clocks turns round and comes back. Then we can say that the other clock was stationary.
The following example might better explain what I am getting at; Imagine two space rockets docked together in geostationary orbit both carrying an atomic clock. One of the rockets remains in geostationary orbit but the other one ignites it's engines and does an extra orbit of the earth before docking with the other rocket again. Which clock would show the slower time?
The clock that went on the round trip. You could replace the orbit of the Earth with a fast out-and-back trip to Neptune.
From all the examples I have seen it would be the clock on the rocket that undocked and completed an extra orbit of the earth, but surely when considered from the frame of reference of the rocket that did the extra orbit, the other rocket that remained in geostationary orbit has travelled away from it and completed exactly the same orbit in reverse?
No, there's a difference. One of the rockets fired its engines. Simplify the situation to two rockets motionless in space, then one fires its engines and goes off on a return trip. The clock on that rocket shows the lower time.
Hopefully I've explained it adequately because it puzzles me and I haven't seen the answer in any textbooks.
The answer is around in various places. But maybe it isn't explained simply enough. IMHO the simplest way to look at it is to say that a clock "clocks up" some kind of regular cyclical local motion, and when the clock moves fast it has to tick at an effectively slower rate because the local motion plus the macroscopic motion through space cannot exceed c.
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First, let's get a nasty complication out of the way. When you take a clock up in a plane and fly round the world, that clock is slowed by the faster movement, but the plane will typically be at a sufficiently high altitude for the clock to run faster rather than slower during the trip because gravity also slows a clock, and raising a clock to high altitude has a stronger effect than the high speed travel of the plane. I mention this because you may at some point see results from real experiments and be puzzled by the moving clock recording more time rather than less. With that out of the way though, we can now ignore the gravity effect and pretend that a clock on a plane runs slower, as it would do if there was no gravity involved. Alternatively we can imagine that the plane flies at the same altitude as the stationary clock throughout its trip, and that will eliminate the gravity effect from the results.
It's easier to answer your question by putting two spaceships in deep space. One of them, ship A, will stay still (well, it thinks it's stationary), and the other, ship B, will initially sit next to ship A. Ship B fires its rockets and moves away from ship A at speed for a while. It then fires its rockets the other way so as to stop and then to accelerate back the other way. It eventually fires them the first way again to slow down and stop beside ship A. When the clocks on the two ships are compared, B's clocks have run slow. We have looked at this from the perspective of ship A's frame of reference, a frame in which ship A did not move at all. Ship B doesn't have a constant frame of reference in the say that ship A does because it accelerates several times, and each acceleration changes the frame of reference it is sitting in. This means we can't tie a single frame of reference to B and assert that B didn't move at all relative to that frame during the trip. What we have to do is pick a single frame and analyse the whole set of events from that one frame of reference. We have already done it for the frame of reference of ship A. We can now do it for the frame of reference in which ship B is stationary while it is travelling away from ship A, and when we do this, we find that for part of the time ship B is indeed stationary within this frame, but on the second part of its journey when it's heading back towards ship A, ship B is now moving through that frame of reference at very high speed, so its clocks will be slowed dramatically during this second part of its trip and lead to less time passing over all than for the clocks in ship A. Whichever frame of reference you choose for your analysis of this thought experiment, you must stick with it throughout and not change it in the middle, and whichever frame you use you will get the same end result with ship B recording less time than ship A, and it will be exactly the same amount less too.
Your idea of two ships orbiting the Earth with one doing an extra orbit adds complexity, but you are not allowed to analyse events from a shifting frame of reference, and both ships are now moving relative to any frame that you pick as a base for your analysis because the ships are continually changing direction. Any valid frame of reference which you choose will show that the ship which did an extra orbit will have recorded the passage of less time.
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The simplest definition of whose clock will 'differ', is to ask yourself which one of them are accelerating. The other point is the one John mention, in where it have to return, to make a final comparison with its 'stationary' sibling possible.
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also called being in a same frame of reference btw.
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MALCOLM FOWLER asked the Naked Scientists:
I've heard of the experiment where two atomic clocks were synchronized, one was taken on a high speed trip on a plane and when the plane returned the clock that took the plane ride recorded a slower time. I can just about grasp the principle that as speed increases time slows down relative to the frame of reference of the stationary clock.
However what puzzles me is that if you take your frame of reference as being that of the 'moving clock' then it is the supposedly stationary clock that is moving at high speed and that should be showing a slower time?
The following example might better explain what I am getting at;
Imagine two space rockets docked together in geostationary orbit both carrying an atomic clock. One of the rockets remains in geostationary orbit but the other one ignites it's engines and does an extra orbit of the earth before docking with the other rocket again. Which clock would show the slower time?
From all the examples I have seen it would be the clock on the rocket that undocked and completed an extra orbit of the earth, but surely when considered from the frame of reference of the rocket that did the extra orbit, the other rocket that remained in geostationary orbit has travelled away from it and completed exactly the same orbit in reverse?
Malcom, the physical situation is not simmetric for the two rockets because they make different paths on space-time. The rocket which switch on its engine, makes a longer path on space and then switch-on again its engines to decelerate and come again near the other rocket, is the one for which time interval is lower, after computing the "space-time interval" between the two events:
1) the second rockets starts its engines and goes away
2) the second rocket decelerates and meet again the first.
After the computation it comes out that, when the rockets meets again, the astronauts/the clocks in the second rocket are younger than the others in the first (which stayed in the geostationary orbit) *if the gravitational effects of the Earth's field on the rockets can be neglected with respect to the effects due to their speed differences*.
Considering gravitational effects too, the answer could be the opposite; however I'm not able to make a computation, probably others are.
To avoid considering the gravitational effects, you should consider two close rockets which are both still with respect to an inertial frame; then one of the two switches on its engines, goes apart from the other and then comes back and meets again the first after some time.
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lightarrow
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Increasing the amount of space/time you occupy with a increase of speed does not slow time but it increases the length of your present moment thereby slowing the tick rate.
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Increasing the amount of space/time you occupy with a increase of speed does not slow time but it increases the length of your present moment thereby slowing the tick rate.
That statement makes no sense to me.
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There is another definition that I hope will make sense to you :)
Make 'c' equivalent to your local clock. then use it to define all other 'clocks'. at no time you will find light to go 'backwards'. It may be defined as 'stopping' from your frame of reference, but that's probably because you don't have the 'time' to measure it correctly.
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MALCOLM FOWLER asked the Naked Scientists:
However what puzzles me is that if you take your frame of reference as being that of the 'moving clock' then it is the supposedly stationary clock that is moving at high speed and that should be showing a slower time?
What do you think?
For inertial motion, when they observe each other while moving, they each describe the same scenario. Only when they make comparisons together, can they make an assessment of elapsed time. The explanation of asymmetry due to acceleration is misleading, since both clocks can be accelerated, and time dilation is a function of speed v/c. The true explanation is the longer speed profile as shown in a Minkowski drawing. Since time is an inverse function of speed, t=x/v, the faster the clock moves for the same distance, the less time it shows. This is true even before adding the relativistic effects. It is confusing for those new to Minkowski diagrams, since they want to interpret them as 2D road maps, i.e. x=vt. The simplest “twin” case is true because any path that departs and rejoins an inertial motion path, will lose more time. Obviously you have to accelerate to change speed. In a more complex case of both changing speeds at random, the paths/profiles would have to be integrated for total elapsed time. You could also bring the clocks together and compare as done in experiments, since the clock acts as a mechanical integrator.
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Hi everybody, I have a question.
For two clocks in relative movement, is the frequency adjustment between them instantaneous? For instance, if one of them changes its speed, will its frequency change immediately with regard to the other? And if so, how will it know that the other has not changed its speed also?
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Hi everybody, I have a question.
For two clocks in relative movement, is the frequency adjustment between them instantaneous? For instance, if one of them changes its speed, will its frequency change immediately with regard to the other? And if so, how will it know that the other has not changed its speed also?
Well, the clock can quess that quite many clocks in the universe did change their speed, and also quite many clocks in the universe did not change their speed.
One more thing that the clock does not know, is its own speed, and that's quite a problem, if you are supposed to slow down your ticking when gaining speed.
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MALCOLM FOWLER asked the Naked Scientists:
However what puzzles me is that if you take your frame of reference as being that of the 'moving clock' then it is the supposedly stationary clock that is moving at high speed and that should be showing a slower time?
Yes this seems quite paradoxial.
The "moving" clock should see a slowed down other clock. (This is true all the time, if "moving" clock is circling the other clock)
The "moving" clock should see a speeded up other clock, in order to later meet that clock and see that the other clock has run fast. (If moving clock circles the other clock it should see the other clock having a constant ticking rate)
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(If moving clock circles the other clock it should see the other clock having a constant ticking rate) Toffo?
exchange it for gps, and those satellites orbiting Earth. The information from those has to be corrected.
http://www.astronomy.ohio-state.edu/~pogge/Ast162/Unit5/gps.html
And yes, both will find a time dilation, it also depends on how they define their relative speeds. A rocket in a constant uniform motion, faster than some 'exactly equivalent siblings', will be able to prove a time dilation though, assuming they meet, being at rest with each other. Also assuming no accelerations involved at all, so it's a pure thought example, having little to do with our reality in where a acceleration always seem to be involved. The only exception to this should be the idea of a accelerating expansion of space.
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forgot to wonder here. In the twin experiment they refer to the biological age of identical twins. Any other way one think we could measure this change from, as they meet up?
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Increasing the amount of space/time you occupy with a increase of speed does not slow time but it increases the length of your present moment thereby slowing the tick rate.
That statement makes no sense to me.
I will always see my light clock a counting my present, John's diagram line L, others may see my light clock as forming a triangle, john's diagram as in the d lines. The difference is the area inside of the triangle, a graph of space/time..
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(If moving clock circles the other clock it should see the other clock having a constant ticking rate) Toffo?
A clock riding on a carousell sees a clock at the midde of the carousell to run at constant fast rate.
The clock at the middle of the carousell sees the clock riding on a carousell to run at constant slow rate.
Why does a clock riding on a carousell see a clock at the midde of the carousell to run at constant fast rate?
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Well, first of all the clock doing that orbit around the other clock should be defined as accelerating. There are no geodesics, that i know of at least? That will keep a perfect orbit. Maybe Pete has some other definition of it. If you on the other hand define it as one frame consisting of a rotating carousel with two clocks being at rest (also in terms of 'resting' on it) with it, and each other? Is that what you mean?
(there is a addendum to this, it takes the form of, if you want something to be defined as rotating it seems, to me at least, that you also will need a frame from which it can be found rotating.) although you can apparently argue that a universe can 'rotate' as described the only way it exist. As measured from its 'inside' https://en.wikipedia.org/wiki/Mach%27s_principle (and that's the opposite to Mach principle, The Gödel rotating universe)
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I don't think it matter though, what definition you find appropriate, as long as all observers agree on the carousel 'rotating' for this, the further out you go from this carousel center the greater your displacements, as expressed, and measured, in (your local) time, which you then can translate to a speed, relative the speed of the clock at the center. That will give you a time dilation too.
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(there is actually one more point to it, that I choose to avoid. In Newtonian terms any circular motion consist of a acceleration. If you think of yourself swinging a staff, you rotating around your center holding it out, as soon as you let it go it will break this rotation, to go at a straight angle from you. That's another way to define it. And in relativistic terms the path that staff, once released, will take is a geodesic.)
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ah, ok. Not about clocks at all is it? It's about what makes it happens, right? One answer is that we have found 'c' to be correct. And that is doesn't matter how uniformly 'fast' you find yourself to move relatively others, you will still measure 'c' everywhere, and so will those others. It's a 'jagged' universe in a way, and what communicates our perception of it is light. that is a mystery though, Einstein had no explanation to why 'c' is 'c' as far as I know. And it is also so that you either can think of it, as if only when proved, did that 'time dilation' really exist, which then always will crave some sort of twin experiment for it to be proved. Or you can do as I do and accept time dilations on their face values, that means that what you measure is what you also will find to exist, for you, at that place and time you measured. It simplifies it.
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And that is to the heart of the mystery, for me :)
Because as soon as you accept a local definition as the one correct, the one we normally use, in where we constantly search for an explanation covering us all must become the one you need to question. The local definition is the one defining repeatable experiments, and it is first when several people can repeat it at other locations, and times, we define it as repeatable. so now you have two mysteries, one is why 'c' is 'c' always? The other is how this universe can 'coexist'. You have one common proposition for both in light, as that seems to be what communicates, both quantum mechanically and relatively. The other is the idea of causality.
Causality is what glues this universe together.
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Take away causality, and you will question the logic(s) that have made us what we are, scientifically. Myself I doubt we ever will be able to violate causality, because without that, this universe would be magic.
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So.
Causality
Your local arrow, or 'clock'
'c'
they go together. 'c' and your local clock, they are equivalent and, only locally definable. they are what makes a repeatable experiment exist, as those always is locally made. Then we have this 'common mind space' from where we look out, finding a universe in where we 'coexist'. To make that have a meaning you need to introduce causality. Without causality this universe wouldn't exist, the local arrow would have no meaning, and as this is equivalent to 'c', neither would relativity. It would mean that what communicates had no logic.
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which actually bring us back to that link I gave on Gödels rotating universe. This I find to be a excellent reason why I don't expect this notion to exist. It violates causality in my eyes. Mach universe is different to me, and possible. But both of those ideas goes out from a 'global perspective', presuming this 'common universal container', that I seriously wonder if existing. By that I mean existing in the old Victorian terms of existing, a hard thing to define actually, that one is about preconceptions and archetypes to me. You can't doubt causality though, unless you're prepared to invalidate your own existence.
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[diagram=741_0]
How do those two clocks see each other?
Answer: They see each other as approaching, blueshifted, fast ticking clocks.
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Would that be how you imagine them on your carousel? Or are you now describing two clocks in motion, going in opposite directions Toffo?
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If they are moving in a opposite direction to each other, measuring light from the other must become red shifted, not blue shifted. If you are defining it as a carousel one of the clocks should be in the center of that carousel, according to your own earlier description? In that case the one at the edge/rim will have a faster motion, as in covering more ground per 'time', than the one at the center, as I wrote before.
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In the previous picture the clock on the left is moving straight up, and the clock on the right is moving straight down.
In the picture below on the left a upwards moving solar panel is not receiving many photons from a light bulb, because it's pointed to the wrong direction. The photons are blueshifted.
On the right a upwards moving solar panel is receiving many photons from a light bulb, because it's pointed to the right direction. The photons are blueshifted.
[diagram=744_0]
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I'm not sure how you think there? It's motion and mass that create a blue or redshift, not the way you lean whatever it is that moves, or being 'still'. Things moving away from each other, finding the other to have a opposite direction will describe each other as 'red shifted'. Things mowing towards each other will define the opposite, a blue shift. With mass one can use the expression of 'climbing out of a gravity well'. Climbing out of it implies a red shift for the 'inertial observer' on the ground observing it. The opposite happens when it falls into a gravity well, also called a 'gravitational acceleration', as defined from the same observer on the ground. This 'gravitational acceleration' is not the same as a rocket expending energy to accelerate though, as the object falling in expends no energy doing so, it's just following a geodesic.
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Yes. I changed "redshift" to "blueshift". And now it's absolutely correct.
Hey here's an idea: Electric motor rotates a hot flat iron very fast ... lighting with good efficiency.
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If we use light as the object infalling or climbing out of a gravity well, you can think of it as a result of conservation laws acting on it. A light quanta is a light quanta, on its own it does not change energy as far as I understand, it's always in the interaction with something this happen, in this case gravity. actually not so unlike a time dilation which also demands someone measuring, comparing his local clock and ruler to someone elses. The difference there being the twin experiment, in where you will find a different biological aging after they join up, being in a same frame of reference as it is called. https://en.wikipedia.org/wiki/Blueshift#Gravitational_blueshift
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Conservation laws are one strange pimpernel though, similar to the idea of 'c'. We know they exist, but why? It would be easier to understand if we assumed that the universe was 'contained' in something that kept all energy inside it, only allowing transformations. but defining it that way would also define a outside, and there is none. Without that outside you have to look elsewhere for a explanation to why they exist. We have them and can use them to explain some stuff, but just as 'c', I don't see any easy explanation to why they exist.
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ah, ok. Not about clocks at all is it? It's about what makes it happens, right? One answer is that we have found 'c' to be correct. And that is doesn't matter how uniformly 'fast' you find yourself to move relatively others, you will still measure 'c' everywhere, and so will those others. It's a 'jagged' universe in a way, and what communicates our perception of it is light. that is a mystery though, Einstein had no explanation to why 'c' is 'c' as far as I know. And it is also so that you either can think of it, as if only when proved, did that 'time dilation' really exist, which then always will crave some sort of twin experiment for it to be proved. Or you can do as I do and accept time dilations on their face values, that means that what you measure is what you also will find to exist, for you, at that place and time you measured. It simplifies it.
Outward from a focal point is the motion of energy, dilation is the motion of time, both are the same direction. Think of big bang as a white hole with an outward flow of energy from a focal point that today we measure as the force of gravity, 9.8 meters per second per second. Every where we look we see the same motion and it is all traveling in the same direction, in time. Why is the speed of light always "c"? IMME it is because we are not colliding into light we are always catching it in time, as a static energy level in space.
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hmm, you're thinking of gravity as coming out of a fountain, if I read you right? And the Big bang as the fountains origin with the flow of the fountain being time? Difficult concept to defend as there is no 'point' from where that Big Bang originated, it's not of a localized origin according to inflation. And the proof of that is when you look out at the starry sky finding it to look much the same in any direction. A homogeneous and isotropic universe. So you will need a lot of fountains as each one of us then can be considered a 'origin' of a Big Bang, alternatively use particles to define 'origins', including bosons. Each one of them are a 'center' of a Big Bang as I read it, alternatively you need more dimensions, or fewer :)
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Professor Brian Fox did a programme on BBC he nearly got it right. The delay in the measurement produces a decrease in the moving objects clock (back and forth) but the clock on the moving object is going actually faster. If you think about this you may come up with a different answer.
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Doesn't matter. Logically a "delayment" means nothing. Think of it as a movie that someone sits on, he will only leave you it after some time, defined by him. Doesn't matter for the movie. And I'm severly disapointed in English not having 'delayment'. Make it happen :)