Thread: Time Dilation and Length Contraction - the Long and the Short of it! Can they be rec

1. Time Dilation and Length Contraction - the Long and the Short of it! Can they be rec

Reconciling Length Contraction and Time Dilation

Using Einstein's own description

First of all we should bear in mind that Length Contraction and Time Dilation were not terms that Einstein used but have become the accepted way of referring to what he described in:

Relativity: The Special and General Theory.* 1920.* Albert Einstein* (1879–1955).
Chapter XII: * The Behaviour of Measuring-Rods and Clocks in Motion.

(Note that Einstein used the superscript ' to denote the moving Frame of Reference (K') and all associated measurements).

Let us draw diagrams to see if we can picture what Einstein was saying in each of these two cases: the measuring-rod and the clock.

He wrote:

Originally Posted by Einstein
I PLACE a metre-rod in the x'-axis of k' in such a manner that one end (the beginning) coincides with the point x' = 0, whilst the other end (the end of the rod) coincides with the point x' = 1. What is the length of the metre-rod relatively to the system K? In order to learn this, we need only ask where the beginning of the rod and the end of the rod lie with respect to K at a particular time t of the system K

the distance between the points being*

√1 – v²/c²

The rigid rod is thus shorter when in motion than when at rest.

Many will find it helpful to picture what happens to the said 1 metre rod and how well this matches Einstein's description and so I include the following diagram:

Fig. 1

This shows how the red 1 metre rod in Fig. 1 would be observed by an observer in that same K' system.

The Green 1 metre rod in Fig. 1 is that same rod rotated, as it would be seen by an observer in the stationary system K, when system K' is moving at 0.6c relative to system K.

When that green rotated rod is projected down onto the x axis of system K, as depicted by the blue rod it has the length:

√1 - v²/c² = 1/γ = 0.8

Which, being less than its original length, is quite reasonably referred to as Length Contraction and so gives us the formula:

x = x'√(1 - v²/c²) = x' / γ

And further, he continues to say:

Originally Posted by Einstein
If, on the contrary, we had considered a metre-rod at rest in the x-axis with respect to K, then we should have found that the length of the rod as judged from K' would have been*

√1 - v²/c²

this is quite in accordance with the principle of relativity which forms the basis of our considerations.
And, indeed is required, in accordance with one being the reciprocal of the other.

What then if we draw a similar diagram for the t, t' axes in Fig. 2?

The two ordinates (t axes), t and t' of systems K and K', shown in blue and red respectively, with the addition of the rotated t' axis shown in green and how they are related.

Fig. 2

Einstein described this transformation thus:

Originally Posted by Einstein
Let us now consider a seconds-clock which is permanently situated at the origin (x' = 0) of K'. t' = 0 and t' = 1 are two successive ticks of this clock.
Originally Posted by Einstein
(red rod in Fig. 2 - lying along the t' axis). The first and fourth equations of the Lorentz transformation give for these two ticks:
t = 0
and*

As judged from K, the clock is moving with the velocity v; as judged from this reference-body, the time which elapses between two strokes of the clock is not one second, but*

seconds, i.e. a somewhat larger time.
(green bar in Fig. 2 - lying along the rotated t' axis). ]As a consequence of its motion the clock goes more slowly than when at rest.
So in our diagram we take the red bar and, multiply it by the Lorentz factor, to give us 1.25 units for the green bar on that K' rotated scale. A somewhat larger measurement, which, quite justifiably, has become known as Time Dilation.

So we can see that Einstein was correct in what he wrote; but what do we conclude from it?

What lessons can we learn to resolve the mysteries, of Time dilation and Length contraction, that have so bedevilled students of Special Relativity?

Well the formula for time dilation that derives from the last equation above being:
t = γt'
leads into one of the puzzling facets of Special Relativity for if

Which flies in the face of Einstein's second postulate:

The Principle of Invariant Light Speed – "... light is always propagated in empty space with a definite velocity [speed] c which is independent of the state of motion of the emitting body."
Which demands that if c = x/t in any Inertial Frame of Reference, then in system K', c = x'/t'

But don't worry all will be reconciled in the next section.

Summary

If the speed of light, c = x/t then how can one be dilated when the other is contracted?

Only if the same thing happens to each, therefore each must be both dilated and contracted.

Drawing it

This raises the question of how we draw it! The modern way is to use the ct axis of the stationary observer to provide the time scale on the moving objects ct’ axis.

WHY?

Because then we see that time is 'Dilated' on that scale - as SR predicts. - Rubbish!
Time dilation and length contraction are two terms for the same effect! 'dilation' and 'compression' mean the same?

YES THEY DO!

How? Well they are each taking a measurement from one FoR and comparing how it is seen by each of the observers. BUT those comparison's are not performed the same way round! One compares B to A and finds B is bigger - 'Dilation'; while the other compares A to B and finds A is smaller - 'Compression'!

Back to Our diagram.
We are drawing the path of the moving object as the stationary observer sees it. When we draw the second Frame of Reference over the first, we have to use the scales as seen by that same stationary observer. (After all the diagram is from her perspective.)

The time passing, as measured within each frame, passes at the same rate - 1st Postulate - so if we draw them 1mm per unit on the first FoR we must draw them 1mm per unit for the second one too.

Measurements within each Inertial Frame of Reference will be identical in scale as each is stationary to an observer within that Frame of Reference. This is what Einstein's first Postulate is all about.

It is when measurements are transformed, that is the system of coordinates is rotated, as a function of the relative velocity, that things become interesting.

For moving objects, measurements of time and distance are transformed according to the Lorentz Transformation Equations as a function of their relative velocity. In accordance with the Lorentz Factor γ where:

and as both these transformations are applications of the Lorentz Factor, they are reciprocal, and so the Absolute Measurements (i.e. magnitude x quantity) will be unchanged.

Consider Fig. 2. The 1 second time in red (system K')rotates and dilates to the green length γ (= 1.25 seconds) when it is viewed as a moving frame. But that 1.25 seconds contracts to 1 second (1/γ) when viewed from system K, a stationary frame; (so the time in K equals the time in K'; complying with the first postulate.)

In the same way the one metre rod in K would have become γ units, = 1.25 metres in the rotated/transformed/moving system K'; and it would have then contracted back to 1 metre from the stationary observer's perspective. Remember that Einstein was placing a 1 meter rod in the moving system so, viewed from the stationary system, it contracted to 80cm.

It is all relative, reciprocal and only the observed measurements are affected.

So both time and distance are dilated and contracted, it all depends which transformation one is making.

So:
t' = γt,
t' = t/γ,
x' = γx,
x' = x/γ

are all correct! Depending on whether we are referring to the unit size or it quantity; and whether we are comparing A to B, or B to A! We must be careful to use the appropriate formulae for the calculation we are making.

This also explains the problem with the c = x'/t' for x' is referring to the unit size, while t', is referring to the number of time units; and if we change them to both refer to the same property, all is right with the world once more.

And it is worth repeating that in all cases (unit-size) x (unit-quantity) will give the same Absolute (total) Magnitude or Duration whether measuring a stationary object or a moving one.

This is not to say that the measurement of time doesn't dilate. It does. We know that because we measure it and use it in GPS calculations and innumerable experiments. Yet what we don't do is to measure the size of the units we are measuring.

Time dilates, more time passes [b] but the units of time decreases by the same factor!

And the same goes for the length. The 'metre rod' does indeed shorten, the metre unit becomes smaller, contracts, but the number of such units increase again by the same factor!

That is how a moving object can change its measured length a different amount for every observer travelling at a different relative velocity without any conflict.

I will say again, take notice of that one important phrase that is at the heart of so much:
"As a consequence of its motion …"

It is only when the system K' is observed from the system K that that extra translational vector is included in the measurements. We see this in Figs 1 & 2 where it is shown in green, rotated relative to the stationary system views; due to that rotation we see its measurements increased in quantity in comparison to the two local systems K, and K' where that relative motion is excluded. It is also where we see the measurements, projected from the rotated moving scale onto the vertical stationary scales, and the consequent contraction of those units.

It is, as was stated by Einstein himself, only when the body/Frame of Reference is moving relative to the observer, that Lorentz Transformations occur.

In Systems K, K' there are only the measurements made within each Reference Frame by the associated local observers; it is only when one is viewed by a moving observer that the appropriate vector representing that movement has to be added using the Lorentz Transformations, Transforming the Proper measurements to Coordinate measurements.

Summary

Length Contraction and Time Dilation are reconciled only when we appreciate that it is the way the measurements are calculated that changes; that all measurements of a moving Frame of Reference are both Length Contracted and Time Dilated, while the Absolute magnitude or duration remains the same.

Let me say that the repetition in this section is deliberate and I make no apology for that. It is a thing that I learned from the writing of Agatha Christie. Repetition drives the point home especially when read at speed; and these points are fundamental to understanding the Theory.

2. Grimble,

I am not arguing with what you are saying, (I haven't gone through the math, and in concept I very much agree), but it seems that you are only re-pointing out the interpretation of length contractions and time dilations before it was argued that there is no point in having universal time outside of local time if we can't ever observe it. No?

Regards,

JMessenger

3. Thank you, JMessenger, but I have posted this here as I thought I would be told it is non mainstream. Can you tell me more, it is the first I have heard of it and the question of universal time vs local time in that way.

4. Originally Posted by Grimble
In the same way the one metre rod in K would have become γ units, = 1.25 metres in the rotated/transformed/moving system K'; and it would have then contracted back to 1 metre from the stationary observer's perspective. Remember that Einstein was placing a 1 meter rod in the moving system so, viewed from the stationary system, it contracted to 80cm.
No, that is not what Einstein said. The second quote you provide states that a rod measuring one meter at rest will be measured to be shorter than when moving relative to an observer. This is independent of whether the frame K or K' is the moving frame.

To create your contradiction where the rod increases in length while moving, you are treating the reference frames as two distinct "spaces" that objects can move between (for lack or a better description). That is not the case, that is not what Einstein claims, and you have been told as much before. If Kenny hands a one-meter rod to Kenny', each will measure the rod to be one meter in length while they are holding it and each will measure the rod to be 80 cm in length while the other is holding it. This is what Einstein states in your quotes.

Regarding your diagrams, there are a lot of misconceptions there. Look at the Wikipedia article on Minkowski diagrams as a start to correcting those.

5. Originally Posted by Grimble
Can you tell me more, it is the first I have heard of it and the question of universal time vs local time in that way.
No. ATM threads are not for development of an idea. They are for someone to present an ATM idea, and then defend it against questioning.

6. Originally Posted by pzkpfw
No. ATM threads are not for development of an idea. They are for someone to present an ATM idea, and then defend it against questioning.
Then when someone questions it by referring to something of which the proposer has not heard (at least as referenced in that comment) is it unreasonable to ask for more information about what the questioner is referring to and how that impacts, if at all, on the proposal?
And to which part of the OP he is referring to and how he thinks it applies.

I cannot defend the idea, if the reference is unclear to me.

7. Originally Posted by Tobin Dax
No, that is not what Einstein said. The second quote you provide states that a rod measuring one meter at rest will be measured to be shorter than when moving relative to an observer. This is independent of whether the frame K or K' is the moving frame.
Yes, that a 1 metre rod will measure 1/γ metres to a moving observer. Therefore, to a stationary observer a metre rod would have had a length of γ metres in the moving frame.

To create your contradiction where the rod increases in length while moving, you are treating the reference frames as two distinct "spaces" that objects can move between (for lack or a better description). That is not the case, that is not what Einstein claims, and you have been told as much before. If Kenny hands a one-meter rod to Kenny', each will measure the rod to be one meter in length while they are holding it and each will measure the rod to be 80 cm in length while the other is holding it. This is what Einstein states in your quotes.
Exactly! If they measure a rod, held by the other to be 80cm, they would know, by calculation, that the other would measure it as 1 metre! That is exactly what I am describing.

Regarding your diagrams, there are a lot of misconceptions there. Look at the Wikipedia article on Minkowski diagrams as a start to correcting those.
And if I were to copy exactly what Wiki Says on every point, it wouldn't be in the ATM section!
Last edited by Grimble; 2012-Jun-25 at 08:25 PM. Reason: inserted missing [quote]

8. Originally Posted by Grimble
Yes, that a 1 metre rod will measure 1/γ metres to a moving observer. Therefore, to a stationary observer a metre rod would have had a length of γ metres in the moving frame.
The first statement is correct, the second is not. You will need to be more specific about to whom you are referring to as the stationary observer, which usually means an observer within a frame in which the rod is stationary, in which case the rod will measure 1 meter, which is the proper length of the rod when stationary. In any other frame where it is viewed that the rod is moving, the length of the rod will always be measured as less than the proper length, never greater, so cannot ever be measured as y meters, nor calculated to have been that.

If Kenny and Kenny' both carry a rod laid out along the direction of motion, each measuring a proper length of their own rod of 1 meter while it is stationary (non-moving), then Kenny will measure a length for the rod of Kenny' of 1/y and Kenny' will measure a length for the rod of Kenny of 1/y. The reason for this is hidden in relativity of simultaneity. Each measures the ends of the rod according to clocks within their own frame, simultaneously as the rod passes. Since each views a simultaneity difference upon the clocks of the other frame, a quick calculation for where each end of the rod that is stationary within their own frame will be as measured in the other frame that coincides with clocks in the other frame such that they will read the same as they pass each end of the rod, which would be considered simultaneous in the other frame although at different times in the observing frame, tells us mathematically that if both Kenny and Kenny' each carry a rod, they will each measure the proper (stationary) length within their own frame, while each measures the rod in other frame as contracted to 1/y. But you have not reached relativity of simultaneity in the other thread you have going yet, so you are not yet ready to perform this calculation to see how each frame can view the rod in the other as contracted. Once you have done so, you will see it is not quite so mind-boggling as you might now think it to be. It all has to do with how clocks are set differently within each frame and how those clocks must be used in addition to rulers to measure the length of a rod in motion.

9. Established Member
Join Date
Jun 2006
Posts
2,577

Relative to Whom?

Originally Posted by Grimble
Yes, that a 1 metre rod will measure 1/γ metres to a moving observer. Therefore, to a stationary observer a metre rod would have had a length of γ metres in the moving frame.
Stationary with respect to what? Your language most be exact, about who or what is moving relative to who or what, and you must remember that there is no absolute frame of reference.

Originally Posted by Grimble
Exactly! If they measure a rod, held by the other to be 80cm, they would know, by calculation, that the other would measure it as 1 metre! That is exactly what I am describing.
Exactly! That’s what Special Relativity is about, it allows us to calculate what the other guy sees.

Originally Posted by Grimble
And if I were to copy exactly what Wiki Says on every point, it wouldn't be in the ATM section!
The point is, your diagrams are in error.

You should keep in mind that SR is over 100 years old. Much better math and physics people than you and I have beaten on it endlessly. It holds up embarrassingly well and it works, demonstrated by observation. Nothing obvious has been missed. Contemplate the wonderful quote from korjik below. Note particularly that word "verified".

Regards, John M.
Last edited by John Mendenhall; 2012-Jun-26 at 01:14 AM. Reason: clarity; typo

10. Originally Posted by Grimble
Yes, that a 1 metre rod will measure 1/γ metres to a moving observer. Therefore, to a stationary observer a metre rod would have had a length of γ metres in the moving frame
This makes no sense as written. (See John Mendenhall's post.) This contradicts what Einstein says (if you want to appeal to authority) and this contradicts what you claim directly afterward.

Exactly! If they measure a rod, held by the other to be 80cm, they would know, by calculation, that the other would measure it as 1 metre! That is exactly what I am describing.
That's not exactly what you are describing because you still do not understand what a reference frame is. That or you are being incredibly sloppy with your descriptions. Or you do not understand that there is no difference between an observer and their reference frame.

And if I were to copy exactly what Wiki Says on every point, it wouldn't be in the ATM section!
I never suggested you copy it. I suggested that you read it and learn from it. You don't appear to understand the rotated coordinate system, where the axes are not at right angles to each other and are not parallel to the non-rotated axes. You don't appear to understand that the length intervals and time intervals between two identical points are measured differently in each coordinate system (also called a reference frame). Reading up on Minkowski diagrams may help you realize that there is no non-rotated K' frame and that, in your diagrams, your purple axis and rod are bunk.

If you were to read and understand what Wikipedia says, you may not need to use the ATM section.

11. Originally Posted by Grimble
Yes, that a 1 metre rod will measure 1/γ metres to a moving observer. Therefore, to a stationary observer a metre rod would have had a length of γ metres in the moving frame.
This seems to be confusing readers. Take it in context as a specific response to the quoted post.
Exactly! If they measure a rod, held by the other to be 80cm, they would know, by calculation, that the other would measure it as 1 metre! That is exactly what I am describing.
This restates and emphasises the previous post.[/QUOTE]

I am sorry that it was unclear. (which taken on its own it certainly was.)

12. Originally Posted by grav
If Kenny and Kenny' both carry a rod laid out along the direction of motion, each measuring a proper length of their own rod of 1 meter while it is stationary (non-moving), then Kenny will measure a length for the rod of Kenny' of 1/y and Kenny' will measure a length for the rod of Kenny of 1/y. The reason for this is hidden in relativity of simultaneity. Each measures the ends of the rod according to clocks within their own frame, simultaneously as the rod passes. Since each views a simultaneity difference upon the clocks of the other frame, a quick calculation for where each end of the rod that is stationary within their own frame will be as measured in the other frame that coincides with clocks in the other frame such that they will read the same as they pass each end of the rod, which would be considered simultaneous in the other frame although at different times in the observing frame, tells us mathematically that if both Kenny and Kenny' each carry a rod, they will each measure the proper (stationary) length within their own frame, while each measures the rod in other frame as contracted to 1/y. But you have not reached relativity of simultaneity in the other thread you have going yet, so you are not yet ready to perform this calculation to see how each frame can view the rod in the other as contracted. Once you have done so, you will see it is not quite so mind-boggling as you might now think it to be. It all has to do with how clocks are set differently within each frame and how those clocks must be used in addition to rulers to measure the length of a rod in motion.
Well you certainly make it mind boggling with that description!
But all three effects, time dilation, length contraction and relativity of simultaneity are really quite straightforward, as described in my diagrams.
Once you tie them together, with the concept of rotation, they all fall together seamlessly.

13. Originally Posted by John Mendenhall
The point is, your diagrams are in error.
How so?

It would be helpful if you could be specific, how do they not illustrate what Einstein was describing?

You should keep in mind that SR is over 100 years old. Much better math and physics people than you and I have beaten on it endlessly. It holds up embarrassingly well and it works, demonstrated by observation. Nothing obvious has been missed. Contemplate the wonderful quote from korjik below. Note particularly that word "verified".
Very True! And I am not making any assertion that that is not the case. Rather, all I am making is a different view of how it works, one that eliminates the apparent difficulties that new students find so confusing. Simplifying the understanding of how it works, not trying to say that it doesn't!
With some scientific theories there can be endless arguments from those who do not agree, but no one can argue with the success of GPS, based on SR, which must be one of the best validations of any theory.

Regards JM (my initials too!)

14. Originally Posted by Grimble
This seems to be confusing readers. Take it in context as a specific response to the quoted post.
It's confusing because it's not true.

In the same way the one metre rod in K would have become γ units, = 1.25 metres in the rotated/transformed/moving system K'; and it would have then contracted back to 1 metre from the stationary observer's perspective. Remember that Einstein was placing a 1 meter rod in the moving system so, viewed from the stationary system, it contracted to 80cm.
Is the one meter rod in frame K measured to be one meter in length when stationary in that frame? If not, under what conditions is it measured to be one meter in length by an observer in frame K?

Is the rod that is 80 cm in length in frame K measured to bewhen stationary in that frame? If not, under what conditions is it measured to be 80 cm in length by an observer in frame K?

15. Originally Posted by Tobin Dax
It's confusing because it's not true.
Yet when restated, as
Originally Posted by Grimble
If they measure a rod, held by the other to be 80cm, they would know, by calculation, that the other would measure it as 1 metre! That is exactly what I am describing.
it was accepted as a valid statement. Can we forget about the phrasing in the first case which has been confusing my readers?

Is the one meter rod in frame K measured to be one meter in length when stationary in that frame? If not, under what conditions is it measured to be one meter in length by an observer in frame K?

Is the rod that is 80 cm in length in frame K measured to bewhen stationary in that frame? If not, under what conditions is it measured to be 80 cm in length by an observer in frame K?
A one metre rod, (that is a rod measured to be 1 metre by an observer in the same frame), will, in a moving frame (i.e. moving with respect to an observer), will be measured as only 80cm by that moving observer. (That is by the observer relative to whom the rod is moving)

Let me say that again.

An observer in the same frame as the rod will measure that rod's true, proper length. (1metre)
An observer, relative to whom the rod is moving will measure length contraction. (80cm)
An observer, relative to whom the rod is moving will be able to calculate the proper length, that an observer moving with (i.e. in the same frame as), moving with the rod will measure. (1metre)

Is that clearer?

16. Originally Posted by Grimble
An observer in the same frame as the rod will measure that rod's true, proper length. (1metre)
An observer, relative to whom the rod is moving will measure length contraction. (80cm)
An observer, relative to whom the rod is moving will be able to calculate the proper length, that an observer moving with (i.e. in the same frame as), moving with the rod will measure. (1metre)
Careful! do not confuse "proper" with "true." In relativity "proper" refers to measurements made in the same frame as the object being measured. It does not mean "true" or "correct"::

An observer in the same frame as the rod will measure a length (referred to as proper because it is measured in the same frame as the bar). (1 meter)
An observer, moving relative to the bar will measure the bar to be shorter in length than what was measured in the same frame with the bar. (80cm)
Any observer (knowing the relative speeds) will be able to calculate what another observer would measure. (1 m or 80 cm)

Measurements made in any frame are equally valid and equally correct for the frame in which they are made.

17. Originally Posted by Grimble
Yet when restated, as it was accepted as a valid statement. Can we forget about the phrasing in the first case which has been confusing my readers?
No, we can't. That's where your error lies.
A one metre rod, (that is a rod measured to be 1 metre by an observer in the same frame), will, in a moving frame (i.e. moving with respect to an observer), will be measured as only 80cm by that moving observer. (That is by the observer relative to whom the rod is moving)

Let me say that again.

An observer in the same frame as the rod will measure that rod's true, proper length. (1metre)
An observer, relative to whom the rod is moving will measure length contraction. (80cm)
An observer, relative to whom the rod is moving will be able to calculate the proper length, that an observer moving with (i.e. in the same frame as), moving with the rod will measure. (1metre)

Is that clearer?
Okay, so if the one meter rod is in frame K and the observer is in frame K', moving at 0.6c relative to frame K, the observer in K' measures the rod to be what length as he moves by it?

If the observer in K' grabbed the rod as he moved by it, what length would he measure the rod to be in his own frame? (Again, the rod measures one meter when at rest in frame K.)

18. Originally Posted by Tobin Dax
Okay, so if the one meter rod is in frame K and the observer is in frame K', moving at 0.6c relative to frame K, the observer in K' measures the rod to be what length as he moves by it?
An observer, relative to whom the rod is moving will measure length contraction. (80cm)

If the observer in K' grabbed the rod as he moved by it, what length would he measure the rod to be in his own frame? (Again, the rod measures one meter when at rest in frame K.)
An observer in the same frame as the rod will measure that rod's true, proper length. (1metre) - 1st postulate, '... whether these changes of state be referred to the one or the other of two systems in uniform translatory motion relative to each other.'

19. Originally Posted by Grimble
An observer, relative to whom the rod is moving will measure length contraction. (80cm)

An observer in the same frame as the rod will measure that rod's true, proper length. (1metre) - 1st postulate, '... whether these changes of state be referred to the one or the other of two systems in uniform translatory motion relative to each other.'
Seems okay. When you say "the same frame," do you mean "stationary with respect to"? I just want to be clear there.

Originally Posted by Grimble
In the same way the one metre rod in K would have become γ units, = 1.25 metres in the rotated/transformed/moving system K'; and it would have then contracted back to 1 metre from the stationary observer's perspective.
Is the "one meter rod in K" stationary in that frame when it is measured to be one meter?

Considering what you have told me so far, do you still stand by the quoted statement? If so, please explain why.

And have you read up on Minkowski diagrams yet?

20. Originally Posted by Tobin Dax
Seems okay. When you say "the same frame," do you mean "stationary with respect to"? I just want to be clear there.
Yes that is so. If they are in the same frame they are stationary one to the other. When in different frames they are moving.

In the same way the one metre rod in K would have become γ units, = 1.25 metres in the rotated/transformed/moving system K'; and it would have then contracted back to 1 metre from the stationary observer's perspective.
Is the "one meter rod in K" stationary in that frame when it is measured to be one meter?
Yes,I made a mistake. I meant the one metre rod in K', in Fig 1, the one metre rod shown in red.
Considering what you have told me so far, do you still stand by the quoted statement? If so, please explain why.
The red rod is stationary in K', and therefore moving in K.
Stationary in K' it measures 1 metre.
That rod rotated would still be 1 metre in K', but rotated (shown in green) it would measure 1.25 metres as a moving rod in K.
Projected then down from the rotated axis of K' (green) to the normal axis of K (blue) it would then have contracted to 1 metre once again.

Unfortunately Einstein placed the one metre rod in the rotated K' so it contracted to just 80cm.

point noted.
And have you read up on Minkowski diagrams yet?
Yes indeed, many times and I can see that they should really be in 3D as then the diamond shape on rotation is a natural consequence. Rotating in the same plane is mixing up the Galileian rotation for the path tan β = v/c with the hyperbolic rotation where it is Sin β = v/c

viz: 3D Minkowski diagram.

21. Originally Posted by Grimble
Yes that is so. If they are in the same frame they are stationary one to the other. When in different frames they are moving.

Yes,I made a mistake. I meant the one metre rod in K', in Fig 1, the one metre rod shown in red.

The red rod is stationary in K', and therefore moving in K.
Stationary in K' it measures 1 metre.
That rod rotated would still be 1 metre in K', but rotated (shown in green) it would measure 1.25 metres as a moving rod in K.
Projected then down from the rotated axis of K' (green) to the normal axis of K (blue) it would then have contracted to 1 metre once again.

Unfortunately Einstein placed the one metre rod in the rotated K' so it contracted to just 80cm.
If K' is considered to be moving (and so K is stationary), K' is the rotated frame. There is no "non-rotated K' frame." K' is the green grid and K is the red grid in your "3D" diagram.

If K is considered to be moving, K becomes the rotated frame and K' is stationary. There is no "non-rotated K frame." K is the green grid and K' is the red grid in your "3D" diagram. The frames switch roles.

A moving rod will never be measured to be longer than its proper length because the moving frame is always the rotated frame, and the projected length of the rod will always be less than the proper length.

The labels K and K' are only for convenience because each frame is stationary in relation to itself and moving in relation to the other. The physics is identical in both, and Einstein says as much. You have refused to understand this for two months, at least, so I'm not sure why I should continue in this thread.

Yes indeed, many times and I can see that they should really be in 3D as then the diamond shape on rotation is a natural consequence. Rotating in the same plane is mixing up the Galileian rotation for the path tan β = v/c with the hyperbolic rotation where it is Sin β = v/c
That's not much better than your earlier diagrams. I think it's better to stick to the two overlapped coordinate systems describing the same plane, since that is the reality of the situation. The third dimension is just confusing.

On the plus side, you did get rid of your non-physical "non-rotated moving frame." Keep that change.

22. It does take some thought to work out just what is happening in length contraction, doesn't it?

Take this simple diagram showing the vector addition of the relative velocity of 0.6c vector to that of the light in the light clock.

This is how the moving clock (system K') will be observed by the stationary observer (system K); Einstein's second postulate the limiting speed of light, means that the time has to increase to 1.25 seconds.

That is the rotated magnitude.

Now the sum of those vectors has a magnitude of 1.25ct. Given that ct represents time we see the lights journey time dilate from 1 in the stationary system K, to 1.25 seconds in the moving system K'. Time Dilation.

Our observer in K though, measures sees this projected onto the ordinate where it is only 1ct.

So taking the 1.25ct as the length of the new path the light follows in system K', as viewed from system K, we see the length of 1.25ct, contract to become 1ct in system K. Length contraction.

Note that this is what Einstein described when he stipulated that the 1 metre rod be placed in the moving system; it was a one metre rod, where we have a length of 1.25metres, which then contracted to 80cm. Whereas if he had placed it into system K', [b]as observed by an observer in system K', i.e. an observer, who was stationary with respect to that rod, would have measured it to be one metre but it would have been dilated to 1.25metres, which, as a moving rod, would then have contracted back to 1 metre, as observed by the observer in system K.
So the same rod viewed by a local observer, a stationary observer in each system (K and K') would have the same length of 1 metre, complying with the first postulate.

Note time dilation from K to K'
Length contraction from K' to K

The same process viewed from opposite sides.

Not too that this is in the vertical, the y axis; because time is non directional c = x/t can only work if both these effects happen in all three dimensions. Can c = x'/t' = x/γt ?

There is no need to limit it to the x axis. It is not required by any other part of the theory, nor by any experiment for length contraction, to only happen in one direction.
Last edited by Grimble; 2012-Jun-29 at 10:23 AM.

23. Originally Posted by Grimble
It does take some thought to work out just what is happening in length contraction, doesn't it?
No, not really.

Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•