I Is there a solution for the behavioural contradiction in RT?

Foppe Hoekstra

Summary
A moving wagon fitted with printers at each end that simultaneously print dots on the rail makes the consequences of Lorentz transformations tangible. A lot of such wagons, forming a full circular train and a symmetrical rotating system, makes it clear that, within the scope of Lorentz transformations, the behaviour of a moving inertial system is incompatible with that of an infinitely small part of a rotating system, even though there are no differences between these systems to account for it.
In short:
Consider a train-wagon with rest-length L, having a printer on the front and the rear that can put a dot on the rail. Both printers have synchronised clocks. The wagon, being the moving frame MF, has a speed v, and at t0 both printers simultaneously put a dot on the rail. Lsf is the distance between these dots for an observer at rest along the rail, being the stationary frame SF. In accordance with Lorentz, Lsf = γL.

Next, we put a lot of the former train-wagons together, to form a full circular train of n wagons that are coupled by the printers, being the rotating system RS. The front printer of each wagon is the rear printer of the next wagon and therefore we have n printers. (To please Ehrenfest, the wagons are elastic enough to compensate for length contraction due to their velocity.)

At t0 (in RS) every printer puts a dot on the rail. Given the unbroken rotational symmetry we must assume that, in SF,
- there will be n dots on the rail,
- each dot will have the same distance to the next dot,
- the rear dot of wagon 1 will coincide with the front dot of wagon n, regardless of which wagon is numbered as number 1.
Hence, the distance between each pair of sequential dots will be L.

Now, for a close look to one unit of this rotating system. If n is taken infinitely large, pushing the radius to the limit ∞, this RS-unit has virtually the same movement as our straightforward going single wagon. The only difference is that the RS-unit is, compared to the contracted single wagon, stretched by factor γ.

So, the contradiction at hand is that in SF the stretched wagon prints its dots at a distance L, while the contracted wagon prints its dots at the larger distance γL, even though they both print the dots simultaneously in their own frame. How can this be? Somewhere the behaviour of the rotating system, or at least a very small part of it, should smoothly meet that of Special Relativity Theory in straight forward motion.

Last edited by a moderator:
Related Special and General Relativity News on Phys.org

PeroK

Homework Helper
Gold Member
2018 Award
If a sequence of wagons is moving in a circle, then they are all moving in different directions relative to each other. There is no uniform simultaneity between wagons.

This problem in various guises has been analysed on this forum before if you search for it.

Fundamentally the paradox is resolved by considering the loss of simultaneity for wagons moving in a circle, as opposed to all moving together in a straight line.

Foppe Hoekstra

If a sequence of wagons is moving in a circle, then they are all moving in different directions relative to each other. There is no uniform simultaneity between wagons.

This problem in various guises has been analysed on this forum before if you search for it.

Fundamentally the paradox is resolved by considering the loss of simultaneity for wagons moving in a circle, as opposed to all moving together in a straight line.
Whatever the notion of time on each seperate wagon may be, fact is that Lsf for the dots of a RS-wagon will be L. And that contradicts with Lsf of the single wagon.
Could you be more specific on where this problem has been analysed before?

PeroK

Homework Helper
Gold Member
2018 Award
Whatever the notion of time on each seperate wagon may be, fact is that Lsf for the dots of a RS-wagon will be L. And that contradicts with Lsf of the single wagon.
Could you be more specific on where this problem has been analysed before?
I'm on my phone. Try a search.

I did one a while back to analyze why for circular motion, as the limit of polygonal motion you get absolute time dilation in the limit but symmetric time dilation for each polygonal leg.

Ibix

So, the contradiction at hand is that in SF the stretched wagon prints its dots at a distance L, while the contracted wagon prints its dots at the larger distance γL, even though they both print the dots simultaneously in their own frame. How can this be?
The wagons are different lengths. The one on the circular track is stretched by its couplings to the adjacent wagon. The inertial wagon is not.

You keep writing as if the stretching forces (which, in an inertial frame, are due to wagons which are in motion) are irrelevant. They are not. They make the situation of the wagon on the track physically different from that of an instantaneously comoving inertial wagon. You did not seem to me to grasp this in your earlier rotating planet example and you seem to be making the same error here.

Edit: Never mind - shouldn't post when I'm in the middle of something else. See my later response.

It remains true that the circulating and inertial wagons are in different states of motion, and this is critically important. Just not quite for the reasons I stated in this post.

Last edited:

metastable

Is the measured stretching of the trains physical or an unphysical measurement effect that arises from differential velocity? If I am at rest next to a train car and I accelerate to a constant speed close to the speed of light away from the train and observe the train through a telescope, won’t I measure it’s length contracted? Has anything at all about the train actually changed or am I witnessing an optical illusion which depends on my relative velocity compared to a light emitting object?

PeroK

Homework Helper
Gold Member
2018 Award
Is the measured stretching of the trains physical or an unphysical measurement effect that arises from differential velocity? If I am at rest next to a train car and I accelerate to a constant speed close to the speed of light away from the train and observe the train through a telescope, won’t I measure it’s length contracted? Has anything at all about the train actually changed or am I witnessing an optical illusion which depends on my relative velocity compared to a light emitting object?
You should open a new thread for that.

Neither. The train doesn't change. And length contraction has nothing to do with light signals. Length can be measured without EM radiation. It's not an optical illusion.

The answer is that you need to look more closely at the definition of length measurement. A length measurement entails finding the simultaneous position of the two ends.

The flat spacetime of SR has no universal simultaneity. This leads to length contraction if you measure the length of an object that is moving in the frame of reference in which the measurement is made.

A.T.

The wagon, being the moving frame MF, has a speed v, and at t0 both printers simultaneously put a dot on the rail.
Simultaneously in what frame?

Next, we put a lot of the former train-wagons together, to form a full circular train of n wagons that are coupled by the printers, being the rotating system RS.
What simultaneity convention does the rotating system use?

FactChecker

Gold Member
2018 Award
There are common characteristics of these "paradoxes" in SR. The "paradox" will ignore some critical parts of SR, like relativity of simultaneity, length contraction in different directions of motion, etc. The setup is usually one where the correct SR calculations are complicated and intuitively difficult. Then, while the problem statement includes no proper (or any) calculations, it asks others to explain it all with valid intuition and calculations.

pervect

Staff Emeritus
Summary: A moving wagon fitted with printers at each end that simultaneously print dots on the rail makes the consequences of Lorentz transformations tangible. A lot of such wagons, forming a full circular train and a symmetrical rotating system, makes it clear that, within the scope of Lorentz transformations, the behaviour of a moving inertial system is incompatible with that of an infinitely small part of a rotating system, even though there are no differences between these systems to account for it.

In short:
Consider a train-wagon with rest-length L, having a printer on the front and the rear that can put a dot on the rail. Both printers have synchronised clocks. The wagon, being the moving frame MF, has a speed v, and at t0 both printers simultaneously put a dot on the rail.
"Simultaneous" is tricky in special relativity because of the relativity of simultaneity. Different observers have different notions of simultaneity.

For this specific example, using the standard definitions of "simultaneous", based on Einstein's simultaneity convention, when one cares out the procedure of point-wise synchronizing clocks on the train, the very last clock on the tail of the train will not be synchronized with the clock on the head of the train when they are compared directly, even though pairwise every other set of clocks in the train is syncrhonized.

If one uses a different defintion of simultaneity than Einstein's pairwise procedure, the description of the problem might change. But it's unclear what defintion you might be using and it's too much work to present a number of possible alternatives.

It's probably easier to learn about the relativity of simultaneity in non-rotating systems. The general topic is called "The relativity of simultaneity" and/or "Einsteins Train".

mfb

Mentor

In general: You can prove mathematically that special relativity is consistent, if something seems to be not working then you must have made a mistake.

metastable

"...According to special relativity an object cannot be spun up from a non-rotating state while maintaining Born rigidity..."

"...For physically reasonable materials, during the spin-up phase a real disk expands radially due to centrifugal forces; relativistic corrections partially counteract (but do not cancel) this Newtonian effect..."

"...Any rigid object made from real materials that is rotating with a transverse velocity close to the speed of sound in the material must exceed the point of rupture due to centrifugal force, because centrifugal pressure can not exceed the shear modulus of material..."

Last edited:

Dale

Mentor
the behaviour of a moving inertial system is incompatible with that of an infinitely small part of a rotating system, even though there are no differences between these systems to account for it.
First, it is incorrect to claim that "there are no differences between these systems to account for it". The angular velocity $\omega$ is physically measurable and accounts for the difference between the systems. See also below:

Now, for a close look to one unit of this rotating system. If n is taken infinitely large, pushing the radius to the limit ∞, this RS-unit has virtually the same movement as our straightforward going single wagon. The only difference is that the RS-unit is, compared to the contracted single wagon, stretched by factor γ.
For any rotating ring $\omega=v/R$ where $v$ is the tangential velocity and $R$ is the radius. Since $v<c$ then we have $$\lim_{R \to \infty} \omega = 0$$. So the angular velocity becomes zero and all angular effects disappear. Thus the angular and linear motion do smoothly transform to each other in that limit.

Ibix

If the length of the wagons is much less than the radius of the track then one can reasonably Einstein-synchronise the clocks at each end of a wagon. But if one synchronises each front-of-wagon clock to its back-of-wagon clock this way, the cumulative effect of the "leading clocks lag" rule leads to the final clock (the back clock of the wagon behind the wagon where we started) showing a different time from the front-of-wagon clock. This tells us that either (a) we break circular symmetry somewhere, or (b) we do not use Einstein synchronisation. Either way gets us out of any paradox.

pervect

Staff Emeritus
If the length of the wagons is much less than the radius of the track then one can reasonably Einstein-synchronise the clocks at each end of a wagon. But if one synchronises each front-of-wagon clock to its back-of-wagon clock this way, the cumulative effect of the "leading clocks lag" rule leads to the final clock (the back clock of the wagon behind the wagon where we started) showing a different time from the front-of-wagon clock. This tells us that either (a) we break circular symmetry somewhere, or (b) we do not use Einstein synchronisation. Either way gets us out of any paradox.
Yes. People often don't appreciate the need for Einstein clock synchronization, or properly deal with the consequences if they decides not to use it. It is possible not to use Einstein clock synchronziation, but it's tricky and opens up a whole diffeent can of worms. Many laws of physics that one learns in high school or college have been formulated assuming a standard, conventional, clock synchronization scheme, so they need to be modified if one chooses to abandon the standard scheme.

Most commonly, people assume that simultaneity is not dependent on the observer, and possibly even go further and think that if Einstien clock synchronization is observer dependent, it must be wrong, because they "know in their hearts" that the "one true simultaneity" is universal.

Unfortunately, this belief that simultaneity is universal is inconsistent with special relativity. I've seen a lot of people blame the theory for faults that trace back to a conflict between their beliefs about simultaneity being universal and what the theory of relativity demand.

A good clue for whether someone is falling into this trap is to note that they specify "clocks are synchronized" without adding the necessary details about what frame they are synchronized in.

Foppe Hoekstra

Simultaneously in what frame?

What simultaneity convention does the rotating system use?
Simultaneously in the moving frame (MF).
RS: From the full story: "Every printer has a clock that is synchronized when the train is at rest and thereafter the train is accelerated to a constant velocity v. Thus all clocks are accelerated in the same way and so all clocks will develop the same deviation in relation to SF. So all clocks will still be synchronized in RS when the train is at speed."

Last edited:

Foppe Hoekstra

There are common characteristics of these "paradoxes" in SR. The "paradox" will ignore some critical parts of SR, like relativity of simultaneity, length contraction in different directions of motion, etc. The setup is usually one where the correct SR calculations are complicated and intuitively difficult. Then, while the problem statement includes no proper (or any) calculations, it asks others to explain it all with valid intuition and calculations.
For "proper calculations" and the notion of relativity of simultaneity see the full text.

Foppe Hoekstra

"Simultaneous" is tricky in special relativity because of the relativity of simultaneity. Different observers have different notions of simultaneity.

For this specific example, using the standard definitions of "simultaneous", based on Einstein's simultaneity convention, when one cares out the procedure of point-wise synchronizing clocks on the train, the very last clock on the tail of the train will not be synchronized with the clock on the head of the train when they are compared directly, even though pairwise every other set of clocks in the train is syncrhonized.

If one uses a different defintion of simultaneity than Einstein's pairwise procedure, the description of the problem might change. But it's unclear what defintion you might be using and it's too much work to present a number of possible alternatives.

It's probably easier to learn about the relativity of simultaneity in non-rotating systems. The general topic is called "The relativity of simultaneity" and/or "Einsteins Train".

m4r35n357

And by the way, the mathematical 'proof' of SRT is a circulair reasoning.
There are no proofs of SR, only deductions from some or other postulates, and that is linear reasoning.

Foppe Hoekstra

First, it is incorrect to claim that "there are no differences between these systems to account for it". The angular velocity $\omega$ is physically measurable and accounts for the difference between the systems. See also below:

For any rotating ring $\omega=v/R$ where $v$ is the tangential velocity and $R$ is the radius. Since $v<c$ then we have $$\lim_{R \to \infty} \omega = 0$$. So the angular velocity becomes zero and all angular effects disappear. Thus the angular and linear motion do smoothly transform to each other in that limit.
When the angular and linear motion smoothly transform to each other in the limit, should not the distances between the dots do as well?

Foppe Hoekstra

If the length of the wagons is much less than the radius of the track then one can reasonably Einstein-synchronise the clocks at each end of a wagon. But if one synchronises each front-of-wagon clock to its back-of-wagon clock this way, the cumulative effect of the "leading clocks lag" rule leads to the final clock (the back clock of the wagon behind the wagon where we started) showing a different time from the front-of-wagon clock. This tells us that either (a) we break circular symmetry somewhere, or (b) we do not use Einstein synchronisation. Either way gets us out of any paradox.
We use the synchronisation as described (see also reply to A.T.) and we certainly do not break circular symmetry! How on earth would 'breaking the circular symmetry' look like?
A 'time gap' perhaps? Wouldn't that mean that if you travel around the world's axis (which we do at a daily basis by just staying at home) you would not exist for this time-gap-moment?

Foppe Hoekstra

There are no proofs of SR, only deductions from some or other postulates, and that is linear reasoning.
The circular reasoning I am refering to is the one in which it is claimed that 'the observed speed of light is independent of the speed of the observer' follows from RT. Where at the same time it is the main initial hypothesis, so of course it does. Which brings us to another point: namely that a consistent theory does not necessarily need to be true.

jbriggs444

Homework Helper
Simultaneously in the moving frame (MF).
RS: From the full story: "Every printer has a clock that is synchronized when the train is at rest and thereafter the train is accelerated to a constant velocity v. Thus all clocks are accelerated in the same way and so all clocks will develop the same deviation in relation to SF. So all clocks will still be synchronized in RS when the train is at speed."
So you have adopted a coordinate system in which the time coordinate is referenced against a stationary clock in the center of the circular train tracks. That's fine. (The GPS system runs that way). But since you have discarded Einstein clock synchronization, the coordinate speed of light is no longer isotropic in your coordinate system. It is no longer equal to c in every direction.

Ibix

We use the synchronisation as described (see also reply to A.T.) and we certainly do not break circular symmetry!
Clearly you break circular symmetry if you Einstein synchronise your wagon clocks. From the track's rest frame, the front clock of any wagon lags behind the back clock if you Einstein synchronise them. So the front clock of the wagon infront is even further behind. What happens when you get all the way round your circle to the first wagon that you synchronised again? There's your break in symmetry.
A 'time gap' perhaps? Wouldn't that mean that if you travel around the world's axis (which we do at a daily basis by just staying at home) you would not exist for this time-gap-moment?
Nothing is happening to time. It's just that you cannot set clocks consistently in the way you want. The break in clock synchronisation behaves like crossing from one time zone to the next - nothing interesting happens, except that your watch suddenly doesn't agree with the wall clocks.

The actual cause of the synchronisation issue is very different, of course, but it's just a problem with experimental design.

Last edited:

Ibix

But since you have discarded Einstein clock synchronization, the coordinate speed of light is no longer isotropic in your coordinate system. It is no longer equal to c in every direction.
...and, additionally, the wagon clocks are not synchronised to the clocks on the instantaneously co-moving inertial wagon if you use this method.

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving