Lag of a clock coming back to its initial position (Twins Paradox)

[Aleberto69]

Hello,
I'm a beginner on SGR and I'm struggeling with this, probably, simple problem.
I'm interested on exploring Field Theory ( relativistic) so I started reading

http://www.elegio.it/mc2/LandauLifshitz_TheClassicalTheoryOfFields_text.pdf

which was suggested and reccomended by a PF's member in another thread as being a very good book.. A sort of reference text.
At page 7-8 the matter of "proper time" is addressed and authors consider the example of a clock arbitray moving.
In order to use the invarinace of interval the authors consider that for short period of time the clock is moving uniformly and so they argument about the possibility of immagining an inertial reference system linked to the clock.

With this assumption they obtain some relevant formulas ( furthermore they also speculate on the fact that if the clock motion is uniform, it is also true that the resting clock appears to the moving clock falling behind too and this because the motion is relative and the two reference inertial, eventually argumenting that there is no contraddiction.)
In other words that means that the observer in one of the two reference knows that the moving clock falls behind, and the observer on the moving clock knows that the clock on the "resting" system is falling behind.
Eventually however they lead to the conclusion that if the clock is moving in a traiectory that after a certain time come back to the original point in the resting system reference, then the moving clock only appears to lag relative to the one at rest.
They argument that it is not possible following the converse resoning in which the movng clock would be considered at rest since the clock describying a closed trajectory does not carry out a uniform rectilinear motion and therefore the linked coordinate system will not be inertial.
MY QUESTION IS:
Why is not possible to use the same argument the authors previously used, that for short interval of time the motion of the clock is uniform and so we can consider an inertial system of coordinate linked to the clock for that short time and then integrate the following results?

In other words, why the authors argument of applying the principle of relativity to an arbitray moving clock just for short ( infinitesimal) lapse of time is not applicable for the converse reasoning?

Many thanks for your help

Aleberto69

 

[Mister T]

Why is not possible to use the same argument the authors previously used, that for short interval of time the motion of the clock is uniform and so we can consider an inertial system of coordinate linked to the clock for that short time and then integrate the following results?

You not only can do that, you must do it if you wish to know how much proper time elapses.

You are confused over their logic. The authors are not claiming that this integration process is used to show that clocks in relative uniform motion run slow. They mention that the clock is observed to run slow only when two separate K clocks are compared to the one K' clock.

This effect is known as time dilation, it's a comparison of the proper time that elapses in K' to the coordinate time that elapses in K.

The twin paradox effect is a different effect. It's a comparison of two proper times.

 

[Aleberto69]

Thank you very much Mister T for your quick reply.
I found relativity rather difficult first for the well known reason that is counter intuitive respect to every day experience and also because sometime it results a bit complicated and easy to be misundestood standing all this different system of coordinate which the brain is often not so ready to get. It happens sometime for example that authors refer to clock or event or time but is not clear which one.. or in which system.. It is often rather easy getting lost ..
I just want commenting that for a beginner it would need find lecture with an high level of rigour and especcially a step by step approch.

They mention that the clock is observed to run slow only when two separate K clocks are compared to the one K' clock.

what I have understood is that for measuring the relative rate of time in two inertial systems, one clok in one of the two system and two clocks in the second system are necessary. Then that the shorter time interval is the one measured by the clock in the system with one clock only.(since the intervals are the same but the clock is coolocated so $l_{12}'=0$ ).
The authors indeed say "moving clock goes slowly than those at rest" ..
Since motion beetwen to reference systems is relative, it is also relative which clocks are at rest and which ones are moving..however this is not causing contraddiction for inertial system becouse of the length and time dilatation. and so which clock rate is slower is not absolute but relative to the considered observever.
Differently in the case of a clock that perform a not unifor motion, they say that things are different.
Assuming that there are two clocks initially both at rest in the oringin of an inertial system of reference, consider then one of them seting off performing a circular motion with some specific velocity ( I can immagine accelerating from zero) and then slowing down and resting again in the initial position.
In this situation the clock which moved is definitively the one that marks the smaller interval and the reasoning cannot be conversed.
Hopefully I'm not still wrong.. I was convinced that this is the same situation of the two twins, one of which going for an interstellar not uniform roundtrip and then coming back to his twin on the Earth finding him older than himself.. I do not spot the difference...
Apologize and be patient as the matter is really tough for a beginner.

However the thing that is not convincing me is the following.
The principle of invariance of interval has been demostrated in the context of inertial systems and as a consequence of constancy of the speed of ligth in inertial system (I think this is SR).
How is it justified (eg in the example of the arbitrary moving clock) applying it in a differential ( punctual ) formulation and say that at each time we can consider the motion of the body uniform ( for a small amount of time ) and therefore we can associate an inertial reference linked to the body for that infinitesimal amount of time?
in other words.. is the principle true for any moving system reference provided that we consider for each time and for an infinitesimal time lapse, the proper approximating inertial reference system?

Thanks so much in advance for your answers

 

[Mister T]

Since motion beetwen to reference systems is relative, it is also relative which clocks are at rest and which ones are moving..

But as the authors point out, the situation between K and K' is not symmetrical. K has two clocks, K' has only one.

In all of the twin paradox scenarios you mention the comparison being made is fundamentally different. There are only two clocks involved, and they share two events. That is, they were co-located twice. The interval between those two events has a different value in each rest frame.

 

[stevendaryl]

There are lots of different ways to understand time dilation and the twin paradox (they are very closely related, if not exactly the same thing). In my opinion, it helps to try to understand the analogous situation for Euclidean geometry.

Suppose two roads meet at some angle, as shown in the following figure:

Let $x$ be the distance as measured along the red road, and let $x'$ be the distance as measured along the blue road. We will use $s$ (the slope) as a measure of the angle between the two roads. ($s$ is actually equal to the tangent of the angle between the roads). We can set up a correspondence between the two roads by drawing a line perpendicular to the red road at point $x$ and seeing what the corresponding value of $x'$ is. It's just elementary trigonometry, and the answer is: $x' = x \sqrt{1+s^2}$. Clearly, $x' > x$.

Now look at the next figure: The blue road takes a turn. The way I've drawn it, the slope of the blue road switches from +s to -s. But since the relative lengths of the segments depends on $s^2$, the length of the two segments of the blue road are the same: $x' = x \sqrt{1+s^2}$. So it's no surprise that when the two roads get back together again, the blue road is longer than the red road by a factor of $\sqrt{1+s^2}$.

But slope is completely relative! If the blue road has slope $s$ relative to the red road, then the red road has slope $-s$ relative to the blue road:

So from the point of view of a traveler on the blue road, it seems that the red road is longer, by the same factor of $\sqrt{1+s^2}$. Which road is longer seems relative. But when they get back together (as shown in the middle figure), it is objectively the case that the blue road is longer than the red road.

It's just geometry: In Euclidean geometry, a bent path connecting two points is longer than a straight path connecting the same two points.

In Special Relativity, instead of paths through space, we are talking about paths through spacetime. Instead of slope, we're talking about relative velocity. Instead of talking about the length of a path, we're talking about the proper time for a path. Instead of using the formula for $x'$ in terms of $x$, in SR, we have the following formula for $t'$ in terms of $t$:

$t' = t \sqrt{1 - \beta^2}$ (where $\beta$ is the velocity in units where the speed of light has velocity 1).

Velocity is relative in exactly the same way that slope is in Euclidean geometry. But it's objectively the case that a "bent" path through spacetime (one where the velocity changes) is shorter (in terms of proper time) than the straight path. (The reason for the bent path being shorter in this case is because of the minus sign inside the square-root, while there is a plus sign in the Euclidean case).

 

[Aleberto69]

Thank you very much for your help..

In all of the twin paradox scenarios you mention the comparison being made is fundamentally different. There are only two clocks involved, and they share two events. That is, they were co-located twice. The interval between those two events has a different value in each rest frame.

and it seems to me that this is right the case of http://www.elegio.it/mc2/LandauLifshitz_TheClassicalTheoryOfFields_text.pdf at page 8 which I referred to.

If we have two clocks, one of which describes a closed path returning to the starting point (the position of the clock which remained at rest), then clearly the moving clock appears to lag relative to the one at rest. The converse reasoning, in which the moving clock would be considered to be at rest (and vice versa) is now impossible, since the clock describing a closed trajectory does not carry out a uniform rectilinear motion, so that a coordinate system linked to it will not be inertial.

My question however, which is also related to this last example, is :

Considering two system of reference, one inertial and one "not" inertial (as in the example of the clock describing a closed path quoted above)
Is the principle of invariance of intervals still valid?

I would say that ( in some way) is still applicable since:

Suppose that in a certain inertial reference system we observe clocks which are moving relative to us in an arbitrary manner. At each different moment of time this motion can be considered as uniform. Thus at each moment of time we can introduce a coordinate system rigidly linked to the moving clocks, which with the clocks constitutes an inertial reference system. In the course of an infinitesimal time interval dt (as read by a clock in our rest frame) the moving clocks go a distance $\sqrt{dx^2 + dy^2 +dz^2}$ . Let us ask what time interval dt' is indicated for this period by the moving clocks. In a system of coordinates linked to the moving clocks, the latter are at rest, i.e., dx' = dy' = dz' = 0. Because of the invariance of intervals $ds^2 = c^2dt^2 -dx^2 -dy^2 -dz^2 = c^2dt'^2$ from which $dt' = dt \sqrt {1-\frac {dx^2 + dy^2 + dz^2} {c^2dt^2}}$

The authors indeed use an auxiliary inertial reference for deducing what time interval will indicate the arbitrary moving clocks ( at rest just respect to a not inertial reference)
(furthermore it is not clear to me why they mention "clocks" and not "one clock"... Are the clocks moving in arbitrary manner but consistently between them or each of them has its own arbitrary motion?)Later on in the books, I do not understand then why the authors in the following page 8 didn't do something similar ( "using an inertial reference linked to the variable moving clock that is valid for each specific $t$ and for a small enough $dt$") for conversing the reasoning

The converse reasoning, in which the moving clock would be considered to be at rest (and vice versa) is now impossible, since the clock describing a closed trajectory does not carry out a uniform rectilinear motion, so that a coordinate system linked to it will not be inertial.

Why in the first case they claim that an accelerated motion is uniform for short enough time lapse (and so at each instant an inertial reference can be used linked to it) and in the second case they say that they can't do any reasoning since the motion of the clock is not uniform and a coordinate system linked to it won't be inertial?

In other word:
What is the right approach about of the invariance of intervals and about the speed of light when one clock is accelerated and a possible linked references not inertial?

 

[Aleberto69]

I understand that is very difficult especially for a not native writer explaining the central point of questions and that some time is better to be direct and short.

Is correct to apply concept of invariance of intervals to an arbitrary moving object (clock) using the argument that for short time and infinitesimal time lapse its motion is uniform?
If yes.. why?
If Yes, I could consider a non inertial system and ( using similar argument) approaching it as inertial for each $t$ and short enough $dt$ hence
since $ds=ds'$ integrating this will end up to $s=s'$ and therefore that the principle is true also for not inertial system.

 

[PeroK]

I understand that is very difficult especially for a not native writer explaining the central point of questions and that some time is better to be direct and short.

Is correct to apply concept of invariance of intervals to an arbitrary moving object (clock) using the argument that for short time and infinitesimal time lapse its motion is uniform?
If yes.. why?
If Yes, I could consider a non inertial system and ( using similar argument) approaching it as inertial for each $t$ and short enough $dt$ hence
since $ds=ds'$ integrating this will end up to $s=s'$ and therefore that the principle is true also for not inertial system.

The key difference with non-inertial motion is that it does not remain in a single inertial reference frame. If you change reference frames, then you have length contraction and relativity of simultaneity to take into account as well; not simply time dilation.

If you model accelerated motion, for example, as a sequence of inertial motions divided by instantaneous changes of velocity, then in terms of time dilation alone you can calculate a time-dilation symmetry with an inertial frame. But, each instantaneous change of velocity must come with a recalibration of lengths and synchronicity and that produces the asymmetry. For example, if we have three clocks, all relatively at rest and synchonised and one instantaneously accelerates, then the other two clocks will no longer be synchronised in the moving clock's frame. So, it's not just about time dilation.

Also, in general, you should see that there is something not right about your argument. You either do calculus properly or not at all. Your attempt to use infinitesimals as a sort of "trick" is a bad idea and likely to lead to problems and false results.

 

[stevendaryl]

Considering two system of reference, one inertial and one "not" inertial (as in the example of the clock describing a closed path quoted above)
Is the principle of invariance of intervals still valid?

Yes. But if the coordinate system is non-inertial, then the interval is computed in a more complicated way:

$\tau = \int \sqrt{g_{\mu \nu} \frac{dx^\mu}{ds} \frac{dx^\nu}{ds}} ds$

where $s$ is the parameter used to parametrize the path, and $g_{\mu \nu}$ is the matrix of components of the metric tensor in whatever coordinate system you are using, and where the indices $\mu$ and $\nu$ are summed over.

If you switch to a different coordinate system to evaluate this integral, the components $x^\mu$ change, and the metric components $g_{\mu \nu}$ change, but the value of the integral will be unchanged.

To make things simpler, let's consider only a single spatial coordinate, x, and let's parametrize our paths by the time coordinate, $t$. And let's assume that we have a coordinate system where the metric tensor is diagonal (no cross terms such as $g_{xt}$). Then this expression becomes:

$\tau = \int \sqrt{g_{tt} + g_{xx} (\frac{dx}{dt})^2} dt$

In inertial coordinates, $g_{tt} = +1$ and $g_{xx} = - \frac{1}{c^2}$, leading to the familiar expression

$\tau = \int \sqrt{1 - \frac{1}{c^2}(\frac{dx}{dt})^2} dt$

There is one problem with trying to understand the twin paradox from the point of view of a noninertial coordinate system, though. Since the traveling twin turns around instantaneously, there is no single coordinate system in which the traveling twin is always at rest. However, you can do a related experiment with a smooth turnaround.

 

[PeroK]

There is one problem with trying to understand the twin paradox from the point of view of a noninertial coordinate system, though. Since the traveling twin turns around instantaneously, there is no single coordinate system in which the traveling twin is always at rest. However, you can do a related experiment with a smooth turnaround.

One solution to that problem is to consider an inertial frame where neither twin is at rest. For example, you could consider the outbound IRF of the traveller. In that frame:

The stay-at-home twin moves inertially at a certain fixed velocity throughout.
The traveling twin is at rest for the outbound leg, then moves at a greater velocity for the inbound leg, eventually catching up with the stay-at-home.

Doing the calculation in this IRF breaks the notion that there is something special about the Earth's IRF, in particular.

@Aleberto69 I posted a solution to the above some time ago. I'll try to find it. But, as an exercise, you could study the twin paradox in this IRF and calculate what their clocks read when they meet up.

 

[Aleberto69]

Thank you PeroK.

Your attempt to use infinitesimals as a sort of "trick" is a bad idea and likely to lead to problems and false results.

I wouldn't have dare doing that..
I was just wondering if what LandauLifshitz_TheClassicalTheoryOfFields page 7 said ( which seems to me similar) is correct or not.

see my quote of the text in my previous quote ( which I'm not able to requote) partially reported below

"LandauLifshitz_TheClassicalTheoryOfFields page 7 said:,
Suppose that in a certain inertial reference system we observe clocks which are moving relative to us in an arbitrary manner. At each different moment of time this motion can be considered as uniform. Thus at each moment of time we can introduce a coordinate system rigidly linked to the moving clocks, which with the clocks constitutes an inertial reference system. In the course of an infinitesimal time interval ..."

I suppose they are right in that circumstance ( Those author are the guru of the matter.. no doubts), however from my poor PoV I do not understand
In which situation is it correct using that infinitesimal approach and when it is not as "..likely to lead to problems and false results"

What is the justification that allow them using the infinitesimal approach in their reasoning?

 

[PeroK]

What is the justification that allow them using the infinitesimal approach in their reasoning?

I don't have their book. The problem with your approach is that your calculations never took into account the changes in velocity. Effectively, you created a step function where you only considered the "flat" sections and the step discontinuities were ignored.

I have another post that I might look for where I used this technique of using a step function! But, crucially, I also considered the step discontinuities and what effect they have.

Here is is:

https://www.physicsforums.com/threa...owards-each-other-in-a-circular-orbit.896607/

It's post #14.

 

[PeroK]

@Aleberto69 here is the twin paradox analysed from an IRF where both twins move

https://www.physicsforums.com/threa...instein-or-resnick.931280/page-4#post-5882897

 

[itfitmewelltoo]

There are lots of different ways to understand time dilation and the twin paradox (they are very closely related, if not exactly the same thing). In my opinion, it helps to try to understand the analogous situation for Euclidean geometry.

Suppose two roads meet at some angle, as shown in the following figure:

View attachment 221623

Let $x$ be the distance as measured along the red road, and let $x'$ be the distance as measured along the blue road. We will use $s$ (the slope) as a measure of the angle between the two roads. ($s$ is actually equal to the tangent of the angle between the roads). We can set up a correspondence between the two roads by drawing a line perpendicular to the red road at point $x$ and seeing what the corresponding value of $x'$ is. It's just elementary trigonometry, and the answer is: $x' = x \sqrt{1+s^2}$. Clearly, $x' > x$.

Now look at the next figure: The blue road takes a turn. The way I've drawn it, the slope of the blue road switches from +s to -s. But since the relative lengths of the segments depends on $s^2$, the length of the two segments of the blue road are the same: $x' = x \sqrt{1+s^2}$. So it's no surprise that when the two roads get back together again, the blue road is longer than the red road by a factor of $\sqrt{1+s^2}$.

View attachment 221624

But slope is completely relative! If the blue road has slope $s$ relative to the red road, then the red road has slope $-s$ relative to the blue road:

View attachment 221626

So from the point of view of a traveler on the blue road, it seems that the red road is longer, by the same factor of $\sqrt{1+s^2}$. Which road is longer seems relative. But when they get back together (as shown in the middle figure), it is objectively the case that the blue road is longer than the red road.

It's just geometry: In Euclidean geometry, a bent path connecting two points is longer than a straight path connecting the same two points.

In Special Relativity, instead of paths through space, we are talking about paths through spacetime. Instead of slope, we're talking about relative velocity. Instead of talking about the length of a path, we're talking about the proper time for a path. Instead of using the formula for $x'$ in terms of $x$, in SR, we have the following formula for $t'$ in terms of $t$:

$t' = t \sqrt{1 - \beta^2}$ (where $\beta$ is the velocity in units where the speed of light has velocity 1).

Velocity is relative in exactly the same way that slope is in Euclidean geometry. But it's objectively the case that a "bent" path through spacetime (one where the velocity changes) is shorter (in terms of proper time) than the straight path. (The reason for the bent path being shorter in this case is because of the minus sign inside the square-root, while there is a plus sign in the Euclidean case).

. "But when they get back together (as shown in the middle figure), it is objectively the case that the blue road is longer than the red road"

How so? It appear to me that all you've done is present the view from that of the red road then done some hand waving to ignore the symmetry. "Objectively the case"? I see nothing in the explanation that negates the objectively obvious problem that neither reference frame is preferred over the other. Given this obvious symmetry, the identical analysis can be performed the other way around to get the same result for both reference frames.

Indeed, a careful reading of Einstein's paper, "On the Electeodynamics of Moving Bodies" will show Einstein clearly comments on this unavoidable conclusion.

The argument you've presented provides no explanation of how there is an asymmetry between the two frames. Simply stated, in terms of your red and blue line, including the projection of one vector upon the other, all of it is equally applicable by simply swapping the red for blue in assignment to the "stationary" and "moving frames".

 

[PeroK]

. "But when they get back together (as shown in the middle figure), it is objectively the case that the blue road is longer than the red road"

How so? It appear to me that all you've done is present the view from that of the red road then done some hand waving to ignore the symmetry. "Objectively the case"? I see nothing in the explanation that negates the objectively obvious problem that neither reference frame is preferred over the other. Given this obvious symmetry, the identical analysis can be performed the other way around to get the same result for both reference frames.

Indeed, a careful reading of Einstein's paper, "On the Electeodynamics of Moving Bodies" will show Einstein clearly comments on this unavoidable conclusion.

The argument you've presented provides no explanation of how there is an asymmetry between the two frames. Simply stated, in terms of your red and blue line, including the projection of one vector upon the other, all of it is equally applicable by simply swapping the red for blue in assignment to the "stationary" and "moving frames".

This is wrong.

 

[Aleberto69]

I don't have their book. The problem with your approach is that your calculations never took into account the changes in velocity. Effectively, you created a step function where you only considered the "flat" sections and the step discontinuities were ignored.

I have another post that I might look for where I used this technique of using a step function! But, crucially, I also considered the step discontinuities and what effect they have.

Here is is:

https://www.physicsforums.com/threa...owards-each-other-in-a-circular-orbit.896607/

It's post #14.

TY very much. I'll more carefully study your answer and the linked post tonight...

find here http://www.elegio.it/mc2/LandauLifshitz_TheClassicalTheoryOfFields_text.pdf
an old version of the books. Hopefully the book is copyright free as the link was posted in another thread from another PF members ..
The matter is at page 7.

 

[itfitmewelltoo]

The key difference with non-inertial motion is that it does not remain in a single inertial reference frame. If you change reference frames, then you have length contraction and relativity of simultaneity to take into account as well; not simply time dilation.

If you model accelerated motion, for example, as a sequence of inertial motions divided by instantaneous changes of velocity, then in terms of time dilation alone you can calculate a time-dilation symmetry with an inertial frame. But, each instantaneous change of velocity must come with a recalibration of lengths and synchronicity and that produces the asymmetry. For example, if we have three clocks, all relatively at rest and synchonised and one instantaneously accelerates, then the other two clocks will no longer be synchronised in the moving clock's frame. So, it's not just about time dilation.

Also, in general, you should see that there is something not right about your argument. You either do calculus properly or not at all. Your attempt to use infinitesimals as a sort of "trick" is a bad idea and likely to lead to problems and false results.

"Yes. But if the coordinate system is non-inertial, then the interval is computed in a more complicated way"

Thank you. Exactly. The resolution is in that one frame is non-inertial. The analysis of special relativity is strictly applicable to inertial reference frames. And that leads to the paradox. A clock at the equator compared to one at the pole, any clock moving in a closed loop, any reference frame undergoing a change in velocity is not inertial and the difference is observable within that frame of reference. Acceleration can be measured by the body under acceleration. That's the requirement to resolve the paradox. The twin leaving Earth behind and returning is different in that her motion is not uniform.

 

[PeroK]

TY very much. I'll more carefully study your answer and the linked post tonight...

find here http://www.elegio.it/mc2/LandauLifshitz_TheClassicalTheoryOfFields_text.pdf
an old version of the books. Hopefully the book is copyright free as the link was posted in another thread from another PF members ..
The matter is at page 7.

Purely kinematically, of course, there is nothing to distinguish the two clocks. The argument can be turned round unless you know which clock is accelerating (changing its IRF). The analysis works in the IRF, because there is nothing more to consider in an IRF. But, it can't be turned round and used in a non-inertial RF because of the added requirement to recalibrate lengths and synchronicity.

Your question is good, because as you can see from this thread, some people never actually manage to understand this point about the lack of symmetry between the IRF and the non-IRF. They believe that there is a paradox here, unresolved in SR!

 

[Mister T]

Considering two system of reference, one inertial and one "not" inertial (as in the example of the clock describing a closed path quoted above)
Is the principle of invariance of intervals still valid?

No. Each twin has a different value for the size of the interval between the same two events.

Later on in the books, I do not understand then why the authors in the following page 8 didn't do something similar ( "using an inertial reference linked to the variable moving clock that is valid for each specific $t$ and for a small enough $dt$") for conversing the reasoning

They never used that idea in the first place to do that.

 

[Aleberto69]

@Aleberto69 here is the twin paradox analysed from an IRF where both twins move

https://www.physicsforums.com/threa...instein-or-resnick.931280/page-4#post-5882897

TY PeroK,
I regret to have put in the title of the thread the mention to the Twins Paradox which is interesting as well but it lead away from my questions that are related mostly to an argument that LandauLifshitz use at page 7 of their book where they use infinitesimal to approach the problem of estimating proper time for clocks that move of arbitrary motion.
To do that they say that for any $t$ and small $dt$ ( the relate to a reference system were the observer is) the motion can be considered uniform and therefore for just that small lapse of time, a inertial reference can be linked to the moving clocks.
I do not know why this approach is right and however why they instead don't do the same on the next page 8 considering the example of a clock that goes in a circular trajectory ( which is not in any case a uniform motion) coming back to its initial position.

Effectively, you created a step function where you only considered the "flat" sections and the step discontinuities were ignored.

By the way I do not understand to what "step" you are referring... Are you referring to the abrupt change on velocity ( big acceleration) that is often considered in the example of the twins whre the spaceship is traveling uniformly then turn instantaneously and then come back with the same velocity but in the opposite direction?
If yes .. I understood this is adding one more complication which is not in my example were the variable motion is at least considered smooth..
In the example at page 8 of the book the second clock could be imagined doing a circular ( maybe several light years) path and coming back to the original position ( likeli starting and arriving with finite and maybe small accelerations)

 

[stevendaryl]

. "But when they get back together (as shown in the middle figure), it is objectively the case that the blue road is longer than the red road"

How so? It appear to me that all you've done is present the view from that of the red road then done some hand waving to ignore the symmetry. "Objectively the case"?

I'm not exactly sure I understand the point of view behind your question. If you have two roads as shown in the middle figure, you agree that the total length of the blue road is greater than the total length of the red road, right? You can measure the two roads.

So I'm assuming you're not asking whether the blue road is longer, but how to understand this fact, from the point of view of a traveler on the blue road.

• Let's assume that on each road, every meter, there is a marker beside the road with an integer on it. The first marker reads 0, the second ,aler reads 1, etc.
• Furthermore, let's assume that when the roads cross the first time (the left end of the middle figure), the markers on both roads read 0.
• Just to make it definite, let's assume that the slope is equal to 0.75.
• Let's also assume that the midpoint of the red road is at marker number 800.

With those assumptions, geometry tells us that:

• At the midpoint of the blue road (where it turns the corner), the marker on the blue road reads 1000.
• When the roads get back together, the red road will be at marker number 1600, an the blue road will be at marker 2000.

It's clear how to interpret these facts from the point of view of the traveler along the red road. He uses this formula to compute the length of a road:

$L = \int \sqrt{1+s^2} dx$

The starting point is at $x=0$ the end point is at $x=1600$. For the red road, the slope is 0, so he computes:
$L_{red} = \int dx = 1600$

For the blue road, the slope is 0.75, so he computes:

$L_{blue} = \int \sqrt{1+(0.75)^2} dx = \int 1.25 dx = 1.25 \cdot 1600 = 2000$

So there's no problem, conceptually, from the point of view of the traveler along the red road. How can we understand things from the point of view of the traveler along the blue road, though? Well, here's his experience of the trip:

1. He starts off with the red road at marker 0, and the blue road at marker 0.
2. He travels along until he gets to marker number 100 on the blue road. He looks along a line perpendicular to the blue road to see what the corresponding marker is for the red road. It's number 125.
3. He keeps going until he gets to the turn-around. The marker on the blue road reads 1000.
4. A second before the turn, he looks along a line perpendicular to the blue road, and he sees the corresponding marker on the red road is 1250.
5. Now he makes the turn, and again looks along the line perpendicular to the blue road. The corresponding marker on the red road is marker number 350! The act of turning around has apparently caused a "leap" in the numbering on the red road. A leap backwards by 900 meters, in this case.
6. He continues for another 1000 meters. The markers on the blue road increase by 1000, so that the final marker is number 2000.
7. The markers on the red road increase by 1250, so that the final marker is number 1600.

This is illustrated by the figure below. When the traveler along the blue road makes the turn at marker 1000, his notion of which red marker corresponds to which blue marker changes. Before turning, he has a correspondence where the blue marker number 1000 corresponds to the red marker number 1250. Immediately after turning, blue marker number 1000 corresponds to the red marker number 350. So if he just recorded how the corresponding red markers changed along the trip, then he would say that the markers increased smoothly from 0 to 1250, then jumped back to 350, and then increased smoothly from 350 to 1600.

The corresponding accounting of ages for a traveling twin would similarly have a "jump" in the age of the stay-at-home twin at the turnaround. This jump isn't physical, but just indicates that the traveling twin has changed his notion of which ages of the traveling twin correspond to which ages of the stay-at-home twin.

 

[PeroK]

@

TY PeroK,
I regret to have put in the title of the thread the mention to the Twins Paradox which is interesting as well but it lead away from my questions that are related mostly to an argument that LandauLifshitz use at page 7 of their book where they use infinitesimal to approach the problem of estimating proper time for clocks that move of arbitrary motion.
To do that they say that for any $t$ and small $dt$ ( the relate to a reference system were the observer is) the motion can be considered uniform and therefore for just that small lapse of time, a inertial reference can be linked to the moving clocks.
I do not know why this approach is right and however why they instead don't do the same on the next page 8 considering the example of a clock that goes in a circular trajectory ( which is not in any case a uniform motion) coming back to its initial position.By the way I do not understand to what "step" you are referring... Are you referring to the abrupt change on velocity ( big acceleration) that is often considered in the example of the twins whre the spaceship is traveling uniformly then turn instantaneously and then come back with the same velocity but in the opposite direction?
If yes .. I understood this is adding one more complication which is not in my example were the variable motion is at least considered smooth..
In the example at page 8 of the book the second clock could be imagined doing a circular ( maybe several light years) path and coming back to the original position ( likeli starting and arriving with finite and maybe small accelerations)

Okay. If you look at what L&L are doing on page 7, it boils down to this.

First, they ought to be using $\Delta t$ instead of $dt$, in my opinion. And then use a limiting argument to get from to the integral, with $dt$. That's what I'll do.

First, split the time into a sequence of small intervals of length $\Delta t$. Then, approximate the non-uniform motion by a (different) constant velocity on each interval. In each interval, we have:

$\Delta t' \approx \Delta t \sqrt{1 - \frac{v^2}{c^2}}$

That gives:

$t'_2 - t'_1 = \Sigma \Delta t' \approx \Sigma \Delta t \sqrt{1 - \frac{v^2}{c^2}} \ \$ (*)

And, taking the limit as $\Delta t \rightarrow 0$ gives an integral:

$t'_2 - t'_1 = \int dt' = \int_{t_1}^{t_2} \sqrt{1 - \frac{v^2}{c^2}} dt$

Effectively, the velocity has been modeled as a step function, with the limit being taken.

But, if you try to turn the argument round, the problem is that all the steps you are adding in line (*) above are in different IRF's. There is more than just time dilation to consider. (Go back to the example of the three clocks.)

That's why the reverse argument breaks down: it would ignore the relativity of simultaneity in adding together time intervals from different IRF's.

 

[Aleberto69]

Purely kinematically, of course, there is nothing to distinguish the two clocks. The argument can be turned round unless you know which clock is accelerating (changing its IRF). The analysis works in the IRF, because there is nothing more to consider in an IRF. But, it can't be turned round and used in a non-inertial RF because of the added requirement to recalibrate lengths and synchronicity.

Your question is good, because as you can see from this thread, some people never actually manage to understand this point about the lack of symmetry between the IRF and the non-IRF. They believe that there is a paradox here, unresolved in SR!

Hi Perok.
I understood ( apologize if I'm wrong) that you had the chance to looka at the referenced page 7 .

for me it is not really convincing that they can calculate the elapsed time of clocks which, at the end, are in general at rest with NIRFs.
If I'm not wrong the calculated time lapse would be time in a NIRF (linked to the specific arbitrary moving clock) between two collocated event.. isn't it?
And this seems to me that also NIRF are easily approachable using "infinitesimal" concepts approach...I do not also understand why they talk about "clocks" ; I mean using the plural...
It may mean that the formula applies to each of them once replaced the specific expression of $v(t)$.
I suffer from a a bit of language barrier as well especially for relativity which is really difficult itself for a beginner...
However I consider myself not a believer ... so I do not say I have understood if I'have not...

 

[Aleberto69]

Hi Perok.
I understood ( apologize if I'm wrong) that you had the chance to looka at the referenced page 7 .

View attachment 221678
View attachment 221679

for me it is not really convincing that they can calculate the elapsed time of clocks which, at the end, are in general at rest with NIRFs.
If I'm not wrong the calculated time lapse would be time in a NIRF (linked to the specific arbitrary moving clock) between two collocated event.. isn't it?
And this seems to me that also NIRF are easily approachable using "infinitesimal" concepts approach...I do not also understand why they talk about "clocks" ; I mean using the plural...
It may mean that the formula applies to each of them once replaced the specific expression of $v(t)$.
I suffer from a a bit of language barrier as well especially for relativity which is really difficult itself for a beginner...
However I consider myself not a believer ... so I do not say I have understood if I'have not...

Sorry PeroK..
I think you already answered in the last post..
Ill study it and comment ...
TY very much

 

[PeroK]

It may mean that the formula applies to each of them once replaced the specific expression of $v(t)$.
I suffer from a a bit of language barrier as well especially for relativity which is really difficult itself for a beginner...
However I consider myself not a believer ... so I do not say I have understood if I'have not...

This is all good. You are doing very well in English. See my post above. Remember that the simultaneity of relativity (SoR) is just as important as time dilation. It's easy to go wrong if you just think about time dilation. In fact, symmetric time dilation without SoR is an obvious paradox. It's really important to understand that. Time Dilation "is not the only game in town"!

 

[Aleberto69]

This is all good. You are doing very well in English. See my post above. Remember that the simultaneity of relativity (SoR) is just as important as time dilation. It's easy to go wrong if you just think about time dilation. In fact, symmetric time dilation without SoR is an obvious paradox. It's really important to understand that. Time Dilation "is not the only game in town"!

TY for your help..
I didn't write down carfully everithing but I will..
I think I understand what you meant ..
I did some weeks ago some example ( starship and interstellar trip) on myself using Lorenz transformation ( before discovering this books that is nicely starting from this concept of "intervals")
And I saw how things are consistent looking from different observer point of view because of time dilation, space dilation and simultaneity...

Now I'll study better ( honestly It is not clear yet ) why in one case is correct and in the other way round is not .. But I need time for first spending a bit of time instead of keep bothering you...
The mathematics you presented is easy and was clear ..
what I wasnt' sure of is the legitimation of saing that since infinitesimally ( for Δt tending to zero) teh motion is uniform I can model the situation with a IRF and using the same result that apply for IRF ( invarinace of $ds$).
Then eventually each $dt'$ belong to a different IRF linked to the moving clock and for each $t+dt$ we have a different IRF..
Is difficult to justificate ( so far, but I will) why i can then sum ( integrate ) all the dt' in this example .. why I cannot do similarly in the other way round .
I repaet it ..
I need a bit of time to think a bit focused on th problem

TY again

 

[PeroK]

T

Then eventually each $dt'$ belong to a different IRF linked to the moving clock and for each $t+dt$ we have a different IRF..
Is difficult to justificate ( so far, but I will) why i can then sum ( integrate ) all the dt' in this example .. why I cannot do similarly in the other way round

One last point:

Clocks A, B together and clock C some distance away. All at rest initially and synchronised.

Clock B accelerates. Using only time dilation arguments, clocks A and C must read the same (in B's frame). It doesn't matter whether you integrate using $dt$ or sum using $\Delta t$. After B accelerates, using only time dilation, clocks A and C must remain synchronised in B's frame.

But, clocks A & C are not synchronised in B's frame. So, if you change your IRF you cannot simply think about time dilation.

 

[Aleberto69]

@Okay. If you look at what L&L are doing on page 7, it boils down to this.

First, they ought to be using $\Delta t$ instead of $dt$, in my opinion. And then use a limiting argument to get from to the integral, with $dt$. That's what I'll do.

First, split the time into a sequence of small intervals of length $\Delta t$. Then, approximate the non-uniform motion by a (different) constant velocity on each interval. In each interval, we have:

$\Delta t' \approx \Delta t \sqrt{1 - \frac{v^2}{c^2}}$

That gives:

$t'_2 - t'_1 = \Sigma \Delta t' \approx \Sigma \Delta t \sqrt{1 - \frac{v^2}{c^2}} \ \$ (*)

And, taking the limit as $\Delta t \rightarrow 0$ gives an integral:

$t'_2 - t'_1 = \int dt' = \int_{t_1}^{t_2} \sqrt{1 - \frac{v^2}{c^2}} dt$

Effectively, the velocity has been modeled as a step function, with the limit being taken.

But, if you try to turn the argument round, the problem is that all the steps you are adding in line (*) above are in different IRF's. There is more than just time dilation to consider. (Go back to the example of the three clocks.)

That's why the reverse argument breaks down: it would ignore the relativity of simultaneity in adding together time intervals from different IRF's.

Hello PeroK.
I spend a bit of time this week end try to be eventually convinced but I didn't succeeded.
What I do not understand is how con be justified extending the result of SR which applies only to ISRs to a NISR. ( Hopefully I'm not wrong saying that SR applies only to IRSs)
Concerning your explanation for example, why it is acceptable to sum together all the $Δt'$ ( or if you like the $dt'$ in the limit that that became an integral)?
Each $Δt'$ indeed belongs to a different IRFs ( the one that at each time approximate the motion, so I'm not sure that it is correct to sum them together.

It is not clear why in your explanation , in one way is correct, while wouldn't be correct when conversing the reasoning...?A second point is :
In a lot of text I read about the twin paradox that is not solvable with SR as there is the acceleration ( in the time the traveling twin revert is motion) and that for solving it GR is needed. However in the same paper/text, the authors progress anyway saying that "however" the twin paradox can be explained and solved with SR as well...
So I'm confused.. Or SR do not explain the paradox and we need GR, or SR explains the paradox and then GR is not needed...
Or a theory cover a case or it does not... It can't be both.#
What is your opinion in the specific matter ( Twins paradox need GR to be explained or can be explained rigorously with SR only) ?

And the same question for calculating the time marked by the two clocks above: one at rest ( let say in a ISR) and the other one moving of arbitrary motion...3rd point
Can I ask you if you can kindly suggest good books on SR first and on GR as well?
I noticed some of them are terribly difficult for the math they use, while some of them are just qualitative text ( I meant the one without formula that are for everybody want to know the conclusion only). I'm not interested on the latest .. but also not used to understand the first ones...
I wonder if SR at least ( maybe for GR is different ) can be treated with rigour without using strange "mathematical theories of fields" and using just concept as differential geometry ( vector and tensor calculus)...?

Some books use indeed concept that are known to engineers ( L&L for example ), but, on the other side, I find some other authors that from the beginning start quoting about starnge words as Homomorphism , starange field theory, topological concepts...
Does this means that that such a mathematical tough approach is necessary or that there are different methods approaching SR ( and GR)?
It maybe that some approaches are more recent, and although complex they are giving more elegance or insight to the matter...? I'm not sure.
What was historically the maths used by Einstein and by the first insighter?
In other world... I what I'm looking for is finding the simpler approach ( in therm of advanced maths) provided that still rigorous...

Dear PeroK, Thanks in advance, for your help

 

[PeterDonis]

Hopefully I'm not wrong saying that SR applies only to IRS

You are. SR can be done perfectly well in non-inertial frames. The key requirement for SR to be accurate is that spacetime is flat.

 

[Aleberto69]

You are. SR can be done perfectly well in non-inertial frames. The key requirement for SR to be accurate is that spacetime is flat.

TY very much PeterDonis
So when spacetime is flat?
Is space time flat for a clock or for an observer at rest respect a NIRF (Not Inertial Reference Frame e.g a reference system locked to a spacecraft / or a lifter that is moving with accelerated motion)?

Could you explain how to treat NIRFs using SR or in other words what are the relevant result of SR applied to NIRFs?
I have some difficulties on understanding that since most of the results and demonstration I read textbooks ( in-variance of intervals respect to IRFs , Lorentz transformation between IRFs) were obtained for IRFs starting from SR postulates ( c= constant and invariance of physical laws in any IRF) which are enunciated to apply for IRFs?

Could you suggest a good text on SR that does examples of problem involving NISRs?

e.g The twin paradox can be explained rigorously using SR or , as some authors tell us, it need using GR?

Thanks in advance for your answers

 

[PeterDonis]

Is space time flat for a clock or for an observer at rest respect a NIRF

Spacetime being flat is not observer-dependent or frame-dependent. Spacetime being flat means the absence of tidal gravity.

Could you suggest a good text on SR that does examples of problem involving NISRs?

I can't remember if Taylor & Wheeler discuss this, it's been quite a while since I looked through that textbook. Misner, Thorne, & Wheeler discuss it, but that's a textbook on GR and only its first few chapters discuss SR.

The twin paradox can be explained rigorously using SR

Yes. In the standard formulation of the twin paradox, spacetime is flat, so SR works just fine.

 

[Aleberto69]

Spacetime being flat is not observer-dependent or frame-dependent. Spacetime being flat means the absence of tidal gravity.
I can't remember if Taylor & Wheeler discuss this, it's been quite a while since I looked through that textbook. Misner, Thorne, & Wheeler discuss it, but that's a textbook on GR and only its first few chapters discuss SR.
Yes. In the standard formulation of the twin paradox, spacetime is flat, so SR works just fine.

TY very much ,
Now I know that SR works also on NIRF if the space time is flat ( i.e. no tidal gravity), however I do not know how to use SR in that case .
For example if a RF is accelerated respect to a IRF:
1) What do Lorentz transformations become?
2)Do Lorentz transformation still have the same expression but with $v=v(t)$, $v(t)$ being the instantaneous velocity of the origin of the NIRF measured from the observer on the IRF ?
3)Is velocity of light constant (and the same constant) once measured in that NIRF?
3)Are intervals (distance in the ST between events) invariant against considered RF (assuming one IRF and one NIRF) ?

TY in advance for answers

 

[PeterDonis]

What do Lorentz transformations become?

Lorentz transformations only work between inertial frames. For non-inertial frames you have to use more general coordinate transformations, and the exact transformation equations will depend on the specific non-inertial coordinates you are using.

Is velocity of light constant (and the same constant) once measured in that NIRF?

The coordinate velocity of light might not be $c$ in a non-inertial frame. However, it is still true that nothing can go faster than light--the coordinate velocity of any object other than light will be less than the coordinate velocity of light at the same point.

Are intervals (distance in the ST between events) invariant

Yes, but it might be much harder to calculate them in a non-inertial frame since the metric tensor will look different.

 

[Mentz114]

TY very much ,
Now I know that SR works also on NIRF if the space time is flat ( i.e. no tidal gravity), however I do not know how to use SR in that case .
For example if a RF is accelerated respect to a IRF:
1) What do Lorentz transformations become?
2)Do Lorentz transformation still have the same expression but with $v=v(t)$, $v(t)$ being the instantaneous velocity of the origin of the NIRF measured from the observer on the IRF ?
3)Is velocity of light constant (and the same constant) once measured in that NIRF?
3)Are intervals (distance in the ST between events) invariant against considered RF (assuming one IRF and one NIRF) ?

TY in advance for answers

Objects with constant acceleration can be described by the Rindler coordinates. Objects in a circular motion which experience centripetal acceleration are described by the Born coordinates. Both of these are coordinate charts on Minkowski spacetime.

 

[stevendaryl]

Could you explain how to treat NIRFs using SR or in other words what are the relevant result of SR applied to NIRFs?

In any coordinate system, you can compute elapsed time by using the expression:

$\tau = \int \sqrt{\sum_{\mu \nu} g_{\mu \nu} \frac{dx^\mu}{ds} \frac{dx^\nu}{ds}} ds$

where $\mu$ and $\nu$ range over the indices for the different coordinates, and where $g_{\mu \nu}$ are the components of the metric tensor in whatever coordinate system you are using.

An example of a noninertial coordinate system where it is possible to do the calculation is the "Rindler coordinates" $X$ and $T$, which are related to the usual $x$ and $t$ through:

$x = X cosh(gT/c)$
$t = \frac{X}{c} sinh(gT/c)$

or the inverse:

$X = \sqrt{x^2 - c^2 t^2}$
$T = \frac{c}{g} tanh^{-1}(\frac{ct}{x})$

The significance of this coordinate system is that it is the natural coordinate system to use for someone traveling in a rocket that is accelerating so as to give an apparent gravity of $g$ at all times (in the rear of the rocket). Objects at "rest" aboard the rocket will have a constant value of $X$. The coordinate $T$ is the time as measured by a clock in the rear of the rocket.

For these coordinates, $g_{TT} = \frac{g^2 X^2}{c^4}$, $g_{XX} = -\frac{1}{c^2}$. So the proper time for a path is given by:

$\tau = \int \sqrt{\frac{g^2 X^2}{c^4} (\frac{dT}{ds})^2 - \frac{1}{c^2} (\frac{dX}{ds})^2} ds$

For the particularly simple case of $X =$ constant (a clock that is at "rest" on board the accelerating rocket), this simplifies enormously to:

$\tau =\frac{g X}{c^2} T$

which shows that the proper time for a clock is proportional to its location $X$ (higher clocks run faster).

The odd thing about the Rindler coordinate system is that the point $X=0$ isn't actually the rear of the rocket. Instead, the rear of the rocket is at $X = \frac{c^2}{g}$, which means that for a clock in the rear, $\tau = T$, as I said.