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

[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? *

 

[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.