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

[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? *

 

[Ibix]

Clocks always measure their own proper time. In the case you illustrated, one clock travels along thediagonal paths and one along the straight line AC. Fine.

I'm not clear exactly what your other case, case 2, is. Are you trying to ask about the time between events A and B as determined by the frame you illustrated and by the frame of the moving clock? There's no physical mechanism at work here, they're just measuring different things. Clocks at rest in your diagram's frame measure the vertical distance between A and B. The clock moving from A to B measures the diagonal.

 

[morrobay]

This is about the "time between ticks" in the proper times of both clocks. In case 1. those units in both clocks are equal. Less elapsed time by traveler caused by lesser spacetime path as measured by traveling clock . t (tick) = t' (tick) .The case 2. is a definition : Here time between ticks is not equal, t' = gamma t.

 

[Ibix]

The proper time between ticks of identical clocks is always the same. I suspect you are not clear on the distinction between coordinate time and proper time. Perhaps you could draw a diagram of case 2, since it doesn't appear to derive from part of the spacetime diagram you posted, as I initially thought.

 

[morrobay]

In this diagram t' = gamma t , as viewed from S .

 

[Ibix]

There is no $t'$, $S$, or $S'$ in those diagrams. The only times marked are the $t$ axis and the coordinate times, $t_1$ and $t_2$, of the events $A_1$ and $C_1$. There are no proper times marked. So I'm still confused about what you are asking.

 

[Ibix]

Just to add: if that is meant to be two views of the same scenario from two frames then it's very poorly labelled. The time and space axes are both labelled $x$ and $t$, although they're not the same thing. The positions of the events on the axes are labelled $A$ and $C$ in both diagrams, although they're different. And the events themselves (which are genuinely invariant things) are labelled differently. Also, although this may just be a perspective effect from your camera, the diagonal lines look below 45° to me, which would normally imply spacelike lines.

Can I ask where you found these diagrams?

 

[morrobay]

View attachment 272876 In this diagram t' = gamma t , as viewed from S .

S is left side S' is right side. t' would be t(2) -t(1) * t or t'= gamma t . Edit typo: t'=gamma t

 

[Ibix]

S is left side S' is right side. t' would be t(2) -t(1) * t or t' gamma t

I'm sorry, this makes no sense. Presuming you mean $t_1$ and $t_2$ by t(1) and t(2) then you seem to have written $t'=t_2-t_1t=t'\gamma t$, which is nonsense. You've got terms with units of time and others with units of time squared. Perhaps if you used LaTeX to typeset your maths it would be easier to follow.

 

[morrobay]

I'm sorry, this makes no sense. Presuming you mean $t_1$ and $t_2$ by t(1) and t(2) then you seem to have written $t'=t_2-t_1t=t'\gamma t$, which is nonsense. You've got terms with units of time and others with units of time squared. Perhaps if you used LaTeX to typeset your maths it would be easier to follow.

Given that t' = gamma t .What this means is that from the diagram ,rs, t2 -t1 is numerically equivalent to gamma .And I'll take the task of showing this tomorrow morning.

 

[Ibix]

Defining $\Delta x$ as the distance from A to B and B to C in the left hand diagram:$$\begin{eqnarray*}
t_2&=&\gamma(t+v\Delta x/c^2) \\
t_1&=&\gamma(t-v\Delta x/c^2)\\
t_2-t_1&=&2\gamma\frac{v}{c^2}\Delta x
\end{eqnarray*}$$I do not think this is "numerically equivalent to gamma". And I don't understand how any of this relates to this thread. It's just messing around with coordinates.

 

[PeterDonis]

In case 2. it appears as if the mechanism of actions or proper times is different for the two clocks ?

What makes you say that? In both cases, the proper time is the length along each worldline. The $t$ coordinates in your case 2 are not proper times, so why should you expect them to behave like proper times?

 

[morrobay]

What makes you say that? In both cases, the proper time is the length along each worldline. The $t$ coordinates in your case 2 are not proper times, so why should you expect them to behave like proper times?

Since the graph depicts the relativity of simultaneity . And the coordinates are measuring light flashes from B to A and C then it would seem the proper times can be inferred. Ok good , then can you graph the proper times/worldlines on two identical clocks in S and S' and show that the units (time between ticks) are also identical ? Because this is what the OP is questioning.

 

[robphy]

Possibly useful...

From a little sleuthing, I think @morrobay is using diagrams from

Sanjoy Mahajan (http://web.mit.edu/sanjoy/www/ ) http://www.inference.org.uk/sanjoy/teaching/corpus/IA/2002/relativity-notes.pdf
( http://www.inference.org.uk/sanjoy/teaching/corpus/IA/2002/ )

A.P. French
https://books.google.com/books?id=udA0Gjq8vHIC&q=simultaneity#v=snippet&q=simultaneity&f=false (p 71. Fig 3-2)

...from the caption, we learn that A, B, C are not events but “stations” (parallel inertial worldlines) arranged as described. Looking further on the page

we see that $A_1$ and $C_1$ are events, as are $A_1{}'$ and $C_1{}'$.and Wikipedia
https://books.google.com/books?id=J4-30n591U8C&pg=PA170&lpg=PA170&dq="writing+the+Lorentz+Transformation+and+its+inverse+in+terms+of+coordinate+differences+we+get"&source=bl&ots=Q19MgnBfGz&sig=ACfU3U076-NBpbsruEpqZcltSrcc9MPQ5A&hl=en&sa=X&ved=2ahUKEwiz4ZTEopTtAhUxtDEKHe97CXsQ6AEwBHoECAEQAg#v=onepage&q="writing the Lorentz Transformation and its inverse in terms of coordinate differences we get"&f=false

 

[Nugatory]

Ok good , then can you graph the proper times/worldlines on two identical clocks in S and S' and show that the units (time between ticks) are also identical ? Because this is what the OP is questioning.

It is difficult to do that with a graph, because the surface we're drawig on is Euclidean, not Minkowskian. Thus the length of a line segment in such a diagram is generally not equal to the proper time between the two events at the ends of the segment. The twin paradox diagram from @robphy's post directly above is a good example: If you hold a ruler up to your screen and measure, you will find that the distance AC is less than sum of the distance AB and the distance BC - yet more proper time elapses on the AC path and it is "longer".

 

[Frodo]

The point I was trying to make with my 1 million observers is that it isn't the clock itself which ticks slower. It is the moving observer(s) who measure the clock to tick slower. If 1 million observers each measure the clock to tick at 1 million different rates then it cannot be the clock itself causing the effect - it must be the observers.

My explanation has the clock stationary and the multiple observers moving as it makes the point so clear.

The original poster asked

To make clock ticking slower, frequency of balance wheel must be changed..

That is a fundamental misunderstanding of what is happening and I was trying to illuminate it.

The clock doesn't tick slower because it gets a weaker balance spring, or it gets a larger balance wheel, causing the clock physically to tick slower. The clock always ticks at its proper speed as observed by someone stationary with the clock. There is nothing "funny" about the clock and nothing about the clock changes. The clock has no knowledge that 1 million traveling observers are checking to see how fast it is running. It carries on blissfully unaware, ticking at its normal rate.

It is the observers, who are moving relative to the clock and are traveling across spacetime, who are causing the effect. (Perhaps "experiencing" would be better than "causing" because, of course, it is 'how spacetime works' which is the real cause, and the travellers are experiencing the changes spacetime forces when they travel through it.) As they travel, their view of what is simultaneous at locations separated by distance in space changes. It is these changes which cause them to measure the stationary clock to run slow. And it is why different observers traveling at different speeds all measure different clock rates.

For me, the most interesting item in the Lorentz transformation is the " x " in the time transformation equation. By varying " x ", the distance separating the events, we can make t' as different from t as we want, from microseconds to billions of years. And we can put " x " as positive or negative so t' can go into the future or into the past, relative to t, as we wish.

 

[PeterDonis]

can you graph the proper times/worldlines on two identical clocks in S and S'

No. You can't draw space vs. proper time diagrams that include different clocks in relative motion; they don't work.

 

[robphy]

Here is a more accurate drawing of the spacetime diagram from French's textbook.
(The original diagram doesn't include length contraction...
possibly because that is developed in a later chapter.)

It should be noted that the areas of triangles $OA_1C_1$ and $OA_1{}′C_1{}′$ are equal,
since the Lorentz Boost has determinant 1.

Note that Alice measures the time-difference $t_{C_1{}′}−t_{A_1{}′}$,
however, this does not easily tell us the time-dilation factor $\gamma$.

To see what is going on,
consider a scaled up version of the "light-clock diamond with spacelike-diagonal $A_1{}'C_1{}'$ ",
namely, the causal-diamond with timelike-diagonal OP,
as part of the clock effect diagram based on the diagram by Sanjay Mahajan.

The spacetime diagram below shows the ticking light-clocks belonging to Alice (20 ticks along OZ)
and to non-inertial Bob (8 ticks along OP, followed by 8 ticks along PZ).
Note that all clock diamonds have the same area.

Note that OP has velocity (6)/(10) and
that the area of the causal diamond* with timelike-diagonal OP is (16)(4)=64,
which is equal to the square of the number of clock-diamonds along OP,
which are similar to the causal diamond with diagonal OP. Hence, OP=(8).

(*The causal diamond of OP is the intersection of the causal future of O with the causal past of P.)

By the way, from causal diamond edges (16) and (4),
the square of the doppler factor $k^2=(16)/(4)=4$...
so, $k=2$, which is the eigenvalue of this transformation...
so the clock diamond for Bob is stretched in the forward direction by a factor of 2
and shrunk in the opposite direction by a factor of 2.
Then since $k=\displaystyle\sqrt{ \frac{1+v}{1−v}}$,
we have $v_{Bob}=\displaystyle\frac{k^2−1}{k^2+1}=\frac{4−1}{4+1}=\frac{3}{5}$.

Similarly, PZ=(8) and OZ=(20).

Note the time-dilation factor is (by counting) $\gamma=(10)/(8)$.

The left corner of the causal diamond has $t_{M{}′}=2$ and the right has $t_{N{}′}=8$,
so $t_{N{}′}−t_{M{}′}=6$, which doesn't easily lead to $\gamma=(10)/(8)$.

(For more information on these diagrams, look at my Insights.)UPDATE:

Why is the area of the causal diamond
equal to the square-interval of its timelike diagonal?
Look at this diagram...

• take the Minkowski-right-triangle with hypotenuse OP
  and slide the diamonds along the legs in the lightlike direction of the black arrow to determine
  the width of the causal diamond of OP as the sum $\Delta t +\Delta x$ in clock diamond edges
• then slide along the other lightlike direction to determine
  the edge-height of the causal diamond of OP as the difference $\Delta t -\Delta x$ in clock diamond edges
• the area of the causal diamond of OP is the product,
  which equals the square-interval $(\Delta t)^2 - (\Delta x)^2$
  in clock diamond areas.

 

[vanhees71]

It is difficult to that with a graph, because the surface we're drawig on is Euclidean, not Minkowskian. Thus the length of a line segment in such a diagram is generally not equal to the proper time between the two events at the ends of the segment. The twin paradox diagram from @robphy's post directly above is a good example: If you hold a ruler up to your screen and measure, you will find that the distance AC is less than sum of the distance AB and the distance BC - yet more proper time elapses on the AC path and it is "longer".

I think in Minkowski diagrams one should always give the unit-lengths tix on the axes on the two systems, exactly for the reason that you must forget about Euclidean geometry of the plane and think in terms of the Minkowskian/Lorentzian geometry. The unit tix have to be constructed with help of hyperbolae (Lorentzian) rather than circles (Euclidean).

It's also important to emphasize the indefiniteness of the fundamental form, establishing the causality structure of spacetime.

 

[robphy]

I think in Minkowski diagrams one should always give the unit-lengths tix on the axes on the two systems, exactly for the reason that you must forget about Euclidean geometry of the plane and think in terms of the Minkowskian/Lorentzian geometry. The unit tix have to be constructed with help of hyperbolae (Lorentzian) rather than circles (Euclidean).

It's also important to emphasize the indefiniteness of the fundamental form, establishing the causality structure of spacetime.

And this is exactly what the "rotated graph paper" does for a Minkowski spacetime diagram.

• In fact, it can easily show unit lengths in multiple frames of reference systems
  (the above clock-effect diagram posted earlier shows three systems;
  see the elastic collision diagram in my PF Insight https://www.physicsforums.com/insights/relativity-rotated-graph-paper/ which shows unit lengths for five systems).
• The underlying hyperbolas at each event are encoded by
  "the family of equal-area causal diamonds with a common corner event",
  where the area of a causal diamond is equal to the square-interval.
  (See the hyperbola in the above Insight, which shows light-clock diamonds for nine systems.)
• The causal diamond encodes the causal structure,
  since its edges are lightlike and determined by the light cones,
  and its diagonals (one timelike and the other spacelike) are Minkowski-orthogonal
  and have the same absolute magnitude.
  (The timelike diagonal is essentially the radius vector,
  and the spacelike diagonal is along the tangent line to a hyperbola.
  Thus, the "tangent line is orthogonal to the radius".)
• The light-clock diamonds (the basic unit of the rotated graph paper) are determined
  by the light-clocks of inertial observers.
  The reshaping of the clock diamond
  (stretching by k in one direction and shrinking by k in the other direction)
  is a Lorentz boost transformation,
  where k (the Doppler factor, which is exp(rapidity) )
  is the larger eigenvalue of the particular boost (the smaller eigenvalue is its reciprocal 1/k)...
  and this k encodes the velocity and the time-dilation factor.
• So, from counting clock diamonds, one can do quantitative calculations
  [especially when k is rational, leading to simple arithmetic involving Pythagorean triples].
  (You are secretly working in light-cone coordinates, in the spirit of the Bondi k-calculus.)
• and it's easy to get rotated graph paper
  [no need for (just-)two-observer diagrams,
  no need for hyperbolic-graph-paper (which distinguishes an event)]
• and one can (and should) emphasize the physical interpretation to the measurements read off from the diagram, and demote the algebraic Lorentz-transformation formulas [until needed].
• ...and you don't have to forget everything about Euclidean geometry...
  You can keep the "affine structure" (parallel lines, addition of vectors, invariance of planar areas).
  You just have to get the invariant square-magnitudes
  by counting up the number of light-clock diamonds in a causal diamond.

 

[vanhees71]

Sure, that's using light-cone coordinates implicitly. Here the hyperbolae have the equations x^+ x^-=\text{const}. and this leads to the equal-area prescription of the diamonds. It's a pretty elegant method.