B Why and how does the frequency of a moving clock affect its ticking speed?

[John Mcrain]

How moving clock ticking slower?
Let assume we have mechanical watch which moving at very high speed.
To make clock ticking slower ,frequency of balance wheel must be changed..

Why and how frequency of balance wheel will be changed when clock is moving at high speed?
I don't see any physical reason how any constant speed can changed oscillation of clock..

 

[Ibix]

Let assume we have mechanical watch which moving at very high speed.

High speed relative to what? "Moving at high speed" is physically meaningless without specifying what you consider as "at rest".

Once you've done that, analyse the motions of the balance wheel in the frame where it is at rest. Then transform all quantities into the frame where the watch is moving using the appropriate relativistic transforms. This is a somewhat non-trivial piece of work, but you will find that the physics remains consistent despite the balance wheel being non-circular and its spokes no longer being evenly spaced nor straight.

A simpler analysis is to imagine a pendulum clock whose pendulum swings across a light clock's cavity, blocking the cavity at one end of its swing. The two clocks are synchronised so that the pendulum blocks the cavity near one end while the light pulse of the light clock is at the other end, so both clocks continue to function. Now you accelerate to high speed relative to the clocks while I remain at rest with respect to them. The pendulum clock must, in your frame, slow down the same as the light clock so that they continue working. Otherwise, the pendulum would at some point absorb the light pulse and the light clock would stop - which clearly can't happen, from the analysis in my frame.

The underlying explanation in relativity is that the clock is a 4d entity. Our two frames "slice" spacetime into space and time differently, so the things you and I call the 3d clock are actually different slices through the 4d clock. We describe their operation differently because we are describing different parts of the whole. If we make our descriptions 4d (using the tensor formulation of physical laws), it becomes clear we are describing the same thing.

 

[morrobay]

The moving clock is not ticking slower. The "time between ticks" or the mechanism action is exactly same to stationary clock. The elapsed time on traveling clock is less since it traveled less distance in spacetime. See diagram of space time interval. So stationary clock elapsed time is 10 and traveling 8

 

[John Mcrain]

High speed relative to what? "Moving at high speed" is physically meaningless without specifying what you consider as "at rest".

One watch is at Earth and second is flying in rocket..
When rocket comes back to the Earth will second watch show same time?

What makes second balance wheel ticking at different rate compare to watch at earth?

 

[Ibix]

One watch is at Earth and second is flying in rocket..
When rocket comes back to the Earth will second watch show same time?

Note that this isn't quite the same as the first question you asked. A round trip is different from inertial motion.

What makes second balance wheel ticking at different rate compare to watch at earth?

Nothing. It simply took a route through spacetime that involved less elapsed time. The elapsed time measured by your wristwatch is a measure of the "length" (called interval) of the path it followed through spacetime. Just like paths through space, two different paths through spacetime are not obligated to have the same length, even if they share start and finish points.

 

[etotheipi]

The moving clock is not ticking slower. The "time between ticks" or the mechanism action is exactly same to stationary clock. The elapsed time on traveling clock is less since it traveled less distance in spacetime. See diagram of space time interval. So stationary clock elapsed time is 10 and traveling 8

I think this is wrong, or maybe I misunderstand. Suppose you have two clocks whose spacetime origins $\mathcal{O}$ coincide, moving at relative velocity $v$. Let $\mathcal{A}$ be the event where the left clock [rest frame $S$] has completed a single tick, and $\mathcal{B}$ be the event where the right clock [rest frame $S'$] has completed a single tick.

Then, $t_{\mathcal{A}} = t'_{\mathcal{B}} = 1 \text{s}$, whilst $t_{\mathcal{B}} = t'_{\mathcal{A}} = \gamma_v \text{s}$. That's to say that observers established in either frame observe the clock at rest in the other frame to tick slightly later than the one at rest w.r.t. them.

The proper time intervals $\Delta \tau_{\mathcal{O} \mathcal{A}}$ and $\Delta \tau_{\mathcal{O} \mathcal{B}}$ are most easily calculated in the respective frames where the clocks are at rest, i.e. using $c=1$, you have $\Delta \tau_{\mathcal{O} \mathcal{A}} = \sqrt{(\Delta t_{\mathcal{OA}})^2 - (\Delta x_{\mathcal{OA}})^2} = \Delta t_{\mathcal{OA}} = 1\text{s}$ and also $\Delta \tau_{\mathcal{O} \mathcal{B}} = \sqrt{(\Delta t'_{\mathcal{OB}})^2 - (\Delta x'_{\mathcal{OB}})^2} = \Delta t'_{\mathcal{OB}} = 1\text{s}$. This you expect due to symmetry from isotropy of space.

Of course, the proper time interval along a trajectory is frame invariant so you could also calculate e.g. $\Delta \tau_{\mathcal{O} \mathcal{A}} = \sqrt{(\Delta t'_{\mathcal{OA}})^2 - (\Delta x'_{\mathcal{OA}})^2} = \sqrt{\gamma_v^2 \text{s}^2 - v^2 \gamma_v^2 \text{s}^2} = \gamma_v \text{s} \sqrt{1-v^2} = 1\text{s}$ which is exactly what we had before.

 

[jbriggs444]

I think this is wrong, or maybe I misunderstand.

The point being made is that both clocks are ticking off one second per second of proper time. Neither is ticking incorrectly slow in that sense.

As has been pointed out, time itself does not behave as OP implicitly assumes.

 

[morrobay]

The point being made is that both clocks are ticking off one second per second of proper time. Neither is ticking incorrectly slow in that sense.

As has been pointed out, time itself does not behave as OP implicitly assumes.

Here is where there is confusion with gamma : in equation five they use it to show difference in time between ticks. But in my following write in gamma is for equating elapsed time for t = gamma t'
gamma 1.25 from post #3

 

[Dale]

How moving clock ticking slower?
Let assume we have mechanical watch which moving at very high speed.

Do you understand how a light clock time dilates?

Relativity is based on two postulates. The second postulate, the invariance of c, is used to show that a light clock dilates. Then the first postulate, the principle of relativity, is used to show that all other clocks dilate too.

Suppose that we construct two identical double clocks, each consisting of a light clock and an attached mechanical clock which tick at the same rate. Then, since the laws of physics are the same in all inertial frames, the mechanical clock must slow down to match the light clock. Otherwise you could determine an absolute velocity by looking at your double clock.

 

[jartsa]

One watch is at Earth and second is flying in rocket..
When rocket comes back to the Earth will second watch show same time?

What makes second balance wheel ticking at different rate compare to watch at earth?

Well, balance wheel of a ticking clock is more massive than balance wheel of an identical stopped clock. I mean rest mass of the wheel is larger when the wheel is turning.

Now what if we attach the ticking clock on a high speed drill? Well, the balance wheel is turning even faster now, so its mass is even larger now. I mean the mass of the wheel in the center of mass frame of the wheel is larger now . So the clock ticks slower.As you notice, I changed the method of moving the clock. Well that's because I wanted the rest mass of the balance wheel to increase. Sure I could get rockets involved in this too.

An asteroid consisting of two plate shaped rocks is spinning in space. That is now our clock. Now we use a rocket to accelerate the two plates to opposite directions. This increases the rest mass of the asteroid, and also slows down the spinning rate of the asteroid.

 

[John Mcrain]

The moving clock is not ticking slower. The "time between ticks" or the mechanism action is exactly same to stationary clock.

If not ticking slower,why clock from rocket when comes to Earth show different time compare to clock at earth?
what has speed with frequenscy of balance wheel??nothing..
Balance wheel freqency depends on material ,hairsping etc ect ,,but not speed

 

[PeroK]

If not ticking slower,why clock from rocket when comes to Earth show different time compare to clock at earth?

The clock on the rocket experienced less time than the clock on Earth. Time itself is not universal. It has nothing to do with clock mechanisms. The clocks are both working normally; but they experience different amounts of time between their two meetings.

 

[Nugatory]

If not ticking slower,why clock from rocket when comes to Earth show different time compare to clock at earth?

This is the so-called Twin Paradox - we have many threads about it and there is a good explanation at https://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

 

[DaveC426913]

If not ticking slower,why clock from rocket when comes to Earth show different time compare to clock at earth?
what has speed with frequenscy of balance wheel??nothing..
Balance wheel freqency depends on material ,hairsping etc ect ,,but not speed

1. The clock on the moving spaceship will tick at 1 second per second - just like everywhere else in the universe. The crux to understanding relativity is that time itself is relative. So, what they on the spaceship consider a day long journey might match up with what you on Earth consider a week passing - when you eventually meet up and compare clocks.

2. Everything on the spaceship will be seen by Earth to have experienced a one-day journey, not just the clock. The whole ship will have aged by only a day over the course of your week. Their heartbeats, plants growing, balls bouncing etc.

 

[John Mcrain]

The clock on the rocket experienced less time than the clock on Earth. Time itself is not universal. It has nothing to do with clock mechanisms. The clocks are both working normally; but they experience different amounts of time between their two meetings.

Do you understand that clock mechanism determine how much will clock show time?
So what happened with clock mechanics if clock from rocket show less time ,when come back to Earth compare to Earth clock??

 

[PeroK]

Do you understand that clock mechanism determine how much will clock show time?
So what and happened with clock mechanics if clock from rocket show less time ,when come back to Earth compare to Earth clock??

Assuming a clock accurately measures time (i.e. the mechanism is working properly) then the amount of time shown on a clock is determined by how much time has passed.

Less time passed on the rocket than on Earth between the two meetings. Both clocks show the correct time. Time is not universal.

 

[A.T.]

So what happened with clock mechanics ...

It's time dilation not clock manipulation.

 

[Janus]

Do you understand that clock mechanism determine how much will clock show time?
So what happened with clock mechanics if clock from rocket show less time ,when come back to Earth compare to Earth clock??

Relativity is about how frames frames of reference with different velocities measure time and distance differently, not about any physical mechanistic effect "motion" has on a clock.
If a clock travels out to a distance of 1 ly (as measured from the Earth) at 0.8 c and then returns at the same speed, the Earth measures the time interval of the round trip as being 2.5 yrs in duration.
The clock measures that same interval as being 1.5 yrs.
According to the Earth, this is because the clock ticks slow compared it own. But for the Clock itself, it is because it only got 0.6 ly from Earth before it turned around and thus the round trip was 1.2 ly in length and it took 1.5 yrs to travel that total distance at 0.8c.

The point is that measurements like time and distance are not absolute and universal, but depend on the inertial reference frame from which they are being measured.
Relativity involves a completely different model for time and space than the one used for Newtonian physics.

 

[jartsa]

Do you understand that clock mechanism determine how much will clock show time?
So what happened with clock mechanics if clock from rocket show less time ,when come back to Earth compare to Earth clock??

Well, it has been mentioned that the shape of the balance wheel changes. Post #2, second paragraph.

It seems that for some odd reason explanations are completely different depending on if clock moves straight ahead, or if it moves back and forth. Well I think that's ... odd.

Oh yes, I know one thing that changes: Attraction and repulsion of electric charges. I mean, moving charges are currents, and currents have magnetic field around them.

 

[Dale]

It seems that for some odd reason explanations are completely different depending on if clock moves straight ahead, or if it moves back and forth. Well I think that's ... odd.

Why would that be odd? The former depends on the reference frame and the latter is frame invariant. Why would you expect a frame variant thing and a frame invariant thing to have the same explanation?

 

[morrobay]

The first diagram shows frame variant elapsed times in S and S' with the proper times (between ticks) equal, same mechanism of action rate. In second diagram the proper times in invariant frames are different . Are there two clock mechanisms of action rates ?

 

[jbriggs444]

It seems that for some odd reason explanations are completely different depending on if clock moves straight ahead, or if it moves back and forth. Well I think that's ... odd.

We are considering a moving object which itself contains moving components.

If your expectation is that the description in the stationary frame will be the same as the description in the moving frame, just with a linearly increasing offset (+vt) added then yes, the truth will seem "odd".

 

[Herman Trivilino]

It seems that for some odd reason explanations are completely different depending on if clock moves straight ahead, or if it moves back and forth. Well I think that's ... odd.

If the clock moves back and forth then you need to look at what happens during each straight-line segment of that back and forth motion.

 

[Nugatory]

It seems that for some odd reason explanations are completely different depending on if clock moves straight ahead, or if it moves back and forth. Well I think that's ... odd.

It is odd when we first encounter it, but it will make more sense when you realize that we're considering two different phenomena - then the different explanations are not so surprising.

In the round-trip twin paradox case (did you try the FAQ I linked to above?) we are counting the number of times that the clock ticks as it moves from one point in spacetime (the departure event) to another point in spacetime (the reunion event). The traveler and the stay-at-home clock follow different paths through spacetime; these paths have different lengths; and the length of a path through spacetime is measured by the amount of time a clock moving on that path measures. It's analogous to the odometer of a car, where two cars can leave point A together and arrive at point B together, yet one odometer reads less than the other if they took different routes.

Time dilation, what happens when you and I are moving relative to one another and we both find that the other clock is running slow relative to our own, is a completely different phenomenon, namely the relativity of simultaneity. If you are not famiar with that concept, stop right now, google for "Einstein train simultaneity", and be sure that you understand it - it is a complete waste of time to try to understand any of relativity until you understand relativity of simultaneity.

Say we both synchronize our clocks to read 1:00 PM as we are side by side, and then we start moving apart - maybe I'm moving relative to you, maybe you're moving relative to me, it's the same thing. We're watching each other's clocks through telescopes, and I see, after allwing for the light travel time, that at the same time that my clock reads 1:40 your clock reads 1:20. Thus, I conclude that your clock is slow by a factor of two. But you, watching my clock through your telescope, see that at the same time that your clock reads 1:20 my clock reads 1:10, so you just as correctly conclude that my clock is running slow by a factor of two.

But what's really going on is the relativity of simultaneity. Using the frame in which I am at rest, the events "my clock reads 1:40" and "your clock reads 1:20" are simultaneous; using the frame in which you are at rest they are not simultaneous but the events "my clock reads 1:10" and "your clock reads 1:20" are. Thus, the time dilation has nothing to do with the rate at which time passes - it's one second per second for both of us - and everything to do with our different definitions of "at the same time".

And at the cost of repeating myself... You must undetstand the relativity of simultaneity before you can make sense of any of the rest of relativity.

 

[jartsa]

This is the so-called Twin Paradox - we have many threads about it and there is a good explanation at https://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

Hey now I know why many people are confused:

This is actually not the Twin Paradox, because we are not interested about what opinion the traveling clock has about the Earth clock.

This is time dilation. Earth's opinion is that the spaceship clock moves, and is time dilated, like moving clocks are. And this is what we are interested about now.

 

[Nugatory]

This is actually not the Twin Paradox,

Sure it is. You may not have noticed that the poster switched from time dilation to the twin paradox (without realizing that they are different problems) partway through this thread.

The thread started with the title question “moving clock ticking slow” which is time dilation, then shifted to “One watch is at Earth and second is flying in rocket. When rocket comes back to the Earth will second watch show same time?“ which is a twin paradox situation.

 

[jartsa]

Why would that be odd? The former depends on the reference frame and the latter is frame invariant. Why would you expect a frame variant thing and a frame invariant thing to have the same explanation?

Whenever a clock is taking a different path through space-time than an inertial observer observing said clock, then the observer says that the clock is ticking slowly. Observer's wristwatch is the fastest ticking clock according to the observer, it measures observer's proper time. This is time-dilation.

Whenever a clock is taking a different path through space-time, but eventually ending at the same space-time coordinate with the observer, then the observer says that the clock was ticking slowly. Observer's wristwatch is ahead of the clock that traveled, because it was, and is, the fastest ticking clock according to the observer, it measures observer's proper time. This is differential aging.

 

[morrobay]

Whenever a clock is taking a different path through space-time than an inertial observer observing said clock, then the observer says that the clock is ticking slowly. Observer's wristwatch is the fastest ticking clock according to the observer, it measures observer's proper time. This is time-dilation.

Whenever a clock is taking a different path through space-time, but eventually ending at the same space-time coordinate with the observer, then the observer says that the clock was ticking slowly. Observer's wristwatch is ahead of the clock that traveled, because it was, and is, the fastest ticking clock according to the observer, it measures observer's proper time. This is differential aging.

For your first paragraph it's sufficient to say S and S' are moving relative to one another. As you can see from relativity of simultaneity diagram both S and S' clocks proper times , B to A,C and B to A',C' are same. It is only when events A' and C' as seen from S that S' clock is ticking slowly: t' = gamma t. For the second paragraph the traveling clocks return to Earth shows less elapsed time : t' = t/gamma .

 

[darth boozer]

The moving clock is not ticking slower. The "time between ticks" or the mechanism action is exactly same to stationary clock. The elapsed time on traveling clock is less since it traveled less distance in spacetime. See diagram of space time interval. So stationary clock elapsed time is 10 and traveling 8View attachment 271974

It looks like you have transposed hypotenuse and opposite side! The hypotenuse cannot be shorter than the other two sides.

 

[Ibix]

It looks like you have transposed hypotenuse and opposite side! The hypotenuse cannot be shorter than the other two sides.

In Euclidean geometry you would be correct. But this is Minkowski geometry - a straight line is the longest distance between two points.

 

[Dale]

It looks like you have transposed hypotenuse and opposite side! The hypotenuse cannot be shorter than the other two sides.

It can when your metric is $ds^2=-c^2 dt^2 + dx^2 + dy^2 + dz^2$

 

[danb]

It can when your metric is $ds^2=-c^2 dt^2 + dx^2 + dy^2 + dz^2$

What is the justification for classifying the 4-interval as a measure of "distance" or "path length", as opposed to being a function with some other physical significance? It makes no sense to me at all.

 

[Dale]

What is the justification for classifying the 4-interval as a measure of "distance" or "path length", as opposed to being a function with some other physical significance?

Because mathematically the path length is $ds^2=dx^2+dy^2+dz^2$. So when $dt=0$ the spacetime interval reduces directly to the standard distance formula. Then the spacetime interval formula is a straightforward generalization of the distance formula.

The justification for it being a generalization distance isn’t terribly interesting. The interesting thing is that it is experimentally validated.

 

[A.T.]

It looks like you have transposed hypotenuse and opposite side! The hypotenuse cannot be shorter than the other two sides.

There is an Euclidean version of space-time diagrams, where the coordinate-time is the path length (hypotenuse ), while the proper-time is the axis. It is more intuitive for a single object, because you see the time-dilation and length contraction directly as projections onto the axes:

http://www.adamtoons.de/physics/relativity.html

But it's less useful for multiple objects, because you don't see their meetings as crossings of the world lines. Here a comparison of the two diagrams:

http://www.adamtoons.de/physics/twins.html

Note that modern browsers block flash-animations by default. You have to explicitly allow it.

 

[Ibix]

What is the justification for classifying the 4-interval as a measure of "distance" or "path length", as opposed to being a function with some other physical significance? It makes no sense to me at all.

Mathematicians and some physicists will shout at you if you call an interval a length. But it occupies basically the same position in the maths - it's the simplest thing you can build from coordinate differentials that is invariant under coordinate transform. In Cartesian coordinates in Euclidean space, the length associated with a coordinate difference $d\vec x$ is $dl$, which is given by $(dl^2)=d\vec x^T\mathbf{I}d\vec x$ where $\mathbf{I}$ is the identity matrix. In Minkowski spacetime, $(ds)^2=d\vec x^T\mathbf{\eta}d\vec x$, where $\mathbf{\eta}=\mathrm{diag}(-1,1,1,1)$ is the metric of Minkowski spacetime. Since the metric is what defines the properties of spacetime and the only difference between these two expressions is the choice of metric ($\mathbf{I}$ for Euclidean space and $\mathbf{\eta}$ for Minkowski) they are clearly analogous. You probably shouldn't, strictly, call the interval a length (its square can be negative, and this has Implications) but it's in the realm of "typical physicist sloppy attitude to fine mathematical distinction".

You might ask why I can't write $d\vec x^Td\vec x$. That's a good question but the answer would involve explaining more about tensors than I'm willing to explain in a post. Suffice to say that it doesn't make sense, and it only works in the Euclidean+Cartesian case because the metric tensor happens to be the identity matrix which let's you get away with stuff.

 

[vanhees71]

Well, the Minkowski pseudometric is describing special-relativistic spacetime, and all high-precision tests about the validity of special relativity show that it is a very good spacetime model underlying all physical phenomena except gravity, for which you have to extent the spacetime model to a Lorentzian differentiable manifold, and also there this fundamental bilinear form plays the same role for local phenomena. The specific signature (3,1) (which is the east-coast convention; equivalently you can use (1,3), the west-coast convention) particularly enables to establish a causality structure for these spacetime models.

 

[danb]

The justification for [the relativistic 4-interval] being a generalization distance isn’t terribly interesting.

The justification goes to the heart of the OP's question: What makes moving clocks tick more slowly? Why does the clock from the rocket in the twin paradox show a different elapsed time compared to clocks on Earth when it returns to Earth? The only explanation I've seen on this website is that the spaceship takes a "shorter route through spacetime", and the only way that answer can be a scientific explanation is if it has a scientific justification. But when I ask what that justification is, you tell me it's "not interesting".

There's no evidence that the 4-interval measures anything that can reasonably be called "distance" or "length", so use of those terms in "explaining" relativistic phenomena is a misleading abuse of classical concepts. No matter what reference frame you use to analyze the twin paradox, less time elapses on the spaceship than on Earth, and the principle of relativity is inconsistent with any deterministic mechanism to make that happen. What is consistent with determinism is the interpretation proposed by HA Lorentz: that there's some kind of trade-off between center-of-mass motion through the vacuum and other behaviors of matter, including the internal motions of ticking clocks.

 

[Ibix]

There's no evidence that the 4-interval measures anything that can reasonably be called "distance" or "length"

Apart from the maths I did in my last post. And that it actually is literally distance for a spacelike path. And that it's literally elapsed time for any timelike path.

No matter what reference frame you use to analyze the twin paradox, less time elapses on the spaceship than on Earth, and the principle of relativity is inconsistent with any deterministic mechanism to make that happen.

Um... you can use the principle of relativity to derive the maths used in special relativity. So the results you say are inconsistent with the principle of relativity actually follow from it.

the interpretation proposed by HA Lorentz

Please don't confuse "you don't like the geometrical interpretation of relativity" with "it is wrong". Lorentz ether theory gives the same results as the geometrical interpretation, but requires the addition of an undetectable "preferred frame".

 

[danb]

Since the metric is what defines the properties of spacetime and the only difference between these two expressions is the choice of metric ($\mathbf{I}$ for Euclidean space and $\mathbf{\eta}$ for Minkowski) they are clearly analogous.

Great, they're analogous. But analogy is a similarity of form, not of content or meaning. What is the physical significance of the negative sign in the t term? It's okay to say you don't know. There's nothing wrong with not knowing something that has no experimental evidence to confirm or deny it.

You probably shouldn't, strictly, call the interval a length (its square can be negative, and this has Implications) but it's in the realm of "typical physicist sloppy attitude to fine mathematical distinction".

And yet that's what regular posters on this website keep saying: The traveling twin takes a "shorter path through spacetime" than the earthbound twin. There's no experimental evidence to support that interpretation of the Lorentz transformations over the interpretation preferred by Lorentz himself.

 

[Ibix]

What is the physical significance of the negative sign in the t term?

It makes the timelike direction distinct from the spacelike ones and, coupled with the fact that there's only one negative sign, it leads to the causal structure of spacetime - like lightcones.

There's nothing wrong with not knowing something that has no experimental evidence to confirm or deny it.

We've literally a hundred years of evidence of the accuracy of relativity. Again, you are failing to distinguish between "I don't like Minkowski's interpretation" and "Minkowski's interpretation is wrong". The former is fine (a rare opinion to say the least, but fine). The latter would be an unevidenced and unevidenceable claim since both interpretations give the same physical results.

And yet that's what regular posters on this website keep saying: The traveling twin takes a "shorter path through spacetime" than the earthbound twin.

Because it's far and away the simplest explanation and leads towards the more sophisticated differential geometry of general relativity. I, at least, usually point out that interval and distance are not quite the same thing. Most of us do.

There's no experimental evidence to support that interpretation of the Lorentz transformations over the interpretation of Lorentz himself.

Of course not. It's an interpretation. If there could even be experimental evidence favouring one interpretation over the other then they'd be different theories, not interpretations. But Minkowski's explanation doesn't involve an arbitrary and undetectable parameter (the choice of preferred frame). And it gels nicely with the maths of general relativity. And it's easily the most popular and the one you will encounter in most textbooks.

If someone asks about Lorentz ether theory we're happy to say what it is. But cod-philosophical navel gazing over which interpretation is "better" for its own sake is pointless. The geometrical interpretation is vastly more widely used (can you even find one textbook that teaches special relativity using Lorentz ether theory?) so is the better choice to teach.

 

[russ_watters]

Thread locked pending moderation.

 

[Dale]

Several anti-relativity posts and responses have been removed and the thread is reopened.

 

[Dr Wu]

It would be interesting to know if the poster believes he has had his questions answered.

 

[Frodo]

How moving clock ticking slower?

I really dislike the phrase "a moving clock runs slow" because I find it ambiguous and, arguable, incorrect.

Why? Because you need to specify who is observing the clock so as to remove the ambiguity. I prefer the phrasing "If a clock is moving relative to an observer that observer measures it to run slow even though an observer moving with the clock sees it to run at its proper rate".

Say Observer_2 is "stationary" and the clock is moving relative to her.

Observer_1 is moving with the clock. Observer_1 observes that the clock running at the correct speed.

Because the clock is moving relative to her, Observer_2 observes that the clock running slow.

So the clock is running at the correct speed for Observer_1 and, at the same time, the clock is running slow for Observer_2.

Now have a third Observer_3 who is moving relative to both Observer_1 and Observer_2. He observes the clock to be running at a third, different rate.

So, if you have a million observers, each moving relative to each other, they will between them observe the same clock to be running at one million different rates. But the observer moving with the clock always sees it running at its proper rate.

 

[PeterDonis]

if you have a million observers, each moving relative to each other, they will between them observe the same clock to be running at one million different rates. But the observer moving with the clock always sees it running at its proper rate

It's also worth noting that the words "observe" and "see" do not have their usual intuitive meanings here. They do not refer to what the observers actually see with their eyes or with some device that records incoming light signals. They refer to what the observers calculate to be the case with reference to an inertial frame in which they are at rest. What the observers actually see depends on the direction of the relative motion and the relativistic Doppler effect.

 

[Frodo]

OK - let's remove that potential confusion.

Imagine there are 1 million observers who are all traveling at different speeds. They arrange things so that they all arrive together at the clock's location when the clock shows 12 o'clock.

Every observer is at the same place at the same time, albeit they are moving. They all observe the clock to display 12 o'clock.

But each traveller will measure that the clock is running at a different rate. So, among the 1 million travellers, they measure 1 million different clock rates.

I am stationary, holding the clock and I measure its proper rate. It will be the fastest rate - all others will measure it to be running more slowly. Someone traveling close to c will measure the clock running so slowly it has virtually stopped.

 

[PeroK]

OK - let's remove that potential confusion.

Imagine there are 1 million observers who are all traveling at different speeds. They arrange things so that they all arrive together at the clock's location when the clock shows 12 o'clock.

Every observer is at the same place at the same time, albeit they are moving. They all observe the clock to display 12 o'clock.

But each traveller will measure that the clock is running at a different rate. So, among the 1 million travellers, they measure 1 million different clock rates.

I am stationary, holding the clock and I measure its proper rate. It will be the fastest rate - all others will measure it to be running more slowly. Someone traveling close to c will measure the clock running so slowly it has virtually stopped.

The Lorentz Transformation is a transformation of coordinates from one inertial reference frame to other. That's a better way to remove any confusion.

 

[jbriggs444]

But each traveller will measure that the clock is running at a different rate.

The devil is in the details of that measurement. Normally it involves three clocks and a simultaneity convention.

 

[Nugatory]

But each traveller will measure that the clock is running at a different rate. So, among the 1 million travellers, they measure 1 million different clock rates.

To be precise, they will calculate that the clock is a running at a different rate. What they measure is the time, according to their clock, between the "all clocks at the same place at the same time" event and the arrival at their eyes of light carrying an image of the clock at some later time. This, the indicated time in the image, and the light travel time is sufficient to calculate the rate at which the remote clock ticks.

 

[morrobay]

. Case 1.In this diagram the traveler has .6c velocity . Then Lorentz Factor = .8 So the elapsed time for traveler round trip is( Lorentz Factor ) (elapsed time for stationary observer) And is so since less spacetime path for traveler as shown in proper times. Ie. both clocks had the same mechanism of action rate. Case 2. Now also t' = gamma t where t is the time between two events in frame where ∆x = 0 And t' is the time between same two events where they happen in different locations. (gamma) (one year) or (gamma ( one clock tick) is 1.25 time dialated . So to re ask the OP question: In case 1. the mechanism of action is the same , the proper times. In case 2. it appears as if the mechanism of actions or proper times is different for the two clocks ? Is this the relativity of simultaneity *or actually the traveling clock mechanism time dialated? *