# On the question of experimental evdidence for 'real' SR time dilation?

1. Jul 26, 2014

### objecta99

Do experiments like 1971's Hafele & Keating show that SR relativistic effects of time dilation are not mere products of measuring symmetry? Does such an experiment show that for the jets it is their clock that has kinematic time dilation in relation to the other clock that 'remains' stationary'. I want to say no but I don't know completely why. is this just the ladder thought experiment all over again and an issue of simultaneity? I realize we can say that some object is time dilated greater than another object but is this merely a factor that's agreed upon and not an 'absolute' understanding of which frames is actually undergoing the greater time dilation due to their speed. if speed is relative to an IRF and there is no way to tell who is moving or who is stationary then it seems that the factor of time dilation doesn't ascribe time dilation to any one frame of the other.

It seems to me that the only non-controversially non-symmetrical time dilation effect is the gravitational effect where two frames of reference or more can agree on who's clock is ticking faster or slower relative to some altitude on a mass. Is there reason to believe that time dilation and length contraction as treated in a non-gravitational deep space SR scenario could have effects such that between two or more frames of a reference an agreed upon ordered set of greater than less than time dilations and or length contractions can be obtained?

Or even if its a periodic motion aka an orbital motion around a planet where a stationary clock and a meeting up clock on a plane whos clock ticks slower after a 'round trip' relative to a stationary clock, does this tell us anything interesting about frame dependent and symmetrical time dilation or off set why there is a difference in clock meet-up. Is the issue cashed out in terms of the angular velocity (gravitational acceleration or centrifugal effect though this makes no sense since gravity slows down clocks (or is it a lack of constant speed on the jet clocks part) that accounts for a difference in clock times.

Last edited: Jul 26, 2014
2. Jul 26, 2014

### Staff: Mentor

You might also want to consider measurements of the lifetime of unstable particles; the faster they're moving relative to an observer the longer they live according to that observer's clock.

3. Jul 26, 2014

### Staff: Mentor

Yes. But bear in mind that the reason the experiment shows this is that all of the clocks involved started out at the same place, and were brought back to the same place at the end of the experiment to have their readings compared. *That* is what makes the time dilation "real"--the fact that at the end of the experiment there is an invariant (i.e., not frame-dependent) comparison between the clocks, so all observers agree on the comparison between their elapsed times.

No, because, as above, all the clocks are brought back together again at the end of the experiment to have their readings compared. Simultaneity is only an issue when you're trying to compare clock readings at events which are spatially separated.

Yes, because this (periodic motion) is the other case in which an invariant, frame-independent comparison can be made between the clocks: you just compare how much elapsed time there is on each clock for one period of the motion (for example, one orbit of a satellite around a planet). For example, GPS clocks "run faster" than clocks on the ground in an invariant sense because they have more elapsed time per orbit than ground clocks do.

The measurements of lifetimes of unstable particles that Nugatory mentioned, at least when done in the laboratory, use a similar comparison: the unstable particles go around and around in an accelerator ring, so their motion is periodic, while the observer in the lab stays stationary.

Last edited: Jul 26, 2014
4. Jul 29, 2014

### accel2know

An example of asymmetric SR time dilation

Let me jump here and give an example of asymmetric time dilation. This is not done by two observers looking at each other's clocks, but by comparing their clocks to a common event. A little background.

Muons are created when cosmic rays strike molecules in the upper atmosphere, and they constantly rain down on Earth. These muons are traveling at near the speed of light and experience time dilation effects that allows them to reach Earth's surface. A muon's life is only 2.2 µs, but, because of their speed, it is dilated to about 500 µs to travel 90 miles. From the muon's perspective, Earth is length contracted and the atmosphere is thin enough to go thru in 2.2 µs.

Let's say that a muon is falling straight down and that there is a pulsar some 1000 light-years away (common event) with a period of 100 µs and in a direction perpendicular to the relative motion of the muon and Earth. Common Doppler effects.

See the animation in this post for clarity. (Click on it to see animation if not moving)

From the Earth observer point of view, the pulsar waves are moving from right to left and are 100 µs apart. The muon (blue dot) is moving straight down for 500 µs. The Earth observer is at the point where the muon decays and disappears.

The Earth's observers sees 5 pulses from the pulsar in the time the muon falls to Earth. These pulses are 100 µs apart as stated earlier.

The Muon also sees 5 pulses in its 2.2 µs life. For the muon, these pulses have a period of 0.44 µs.

Earth's clock and the muon's clock are not running at the same rate since they are measuring different period for the pulsar.

Muons are going right thru you as you read this post.

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5. Jul 30, 2014

### pervect

Staff Emeritus
My \$.02. The question of what is "real" is not really one that can be answered by science. The results of specific experiments are things that can be answered directly, the experiments can be performed. The question of "reality" is a question of how to interpret those results.

It may be helpful to discuss briefly the underlying philosophy when trying to come to grip with the experimental results, as long as one avoids the "never-ending arguments" trap common to a lot of philosophy.

Requiring "real" objects to have some degree of observer independence, I would say that proper time is real, and that coordinate time, in isolation, isn't real, because it doesn't meet this requirement. So we need to distinguish proper time and coordinate time when talking about "reality" of time.

"Time dilation", considered in isolation, can be defined as the ratio of coordinate time to proper time. Because it coordinate time is observer dependent, time dilation is also observer dependent, and thus "not real" by the particular requirement of observer independence I'm imposing.

This is good so far, but a bit over-restrictive.

To go on a bit, I need to get technical enough to introduce the idea of the 4-vector and tensors. If they aren't familiar concepts, I apologize for being too technical, but I don't see how I can say what I want to say without them. I'll briefly cite the wiki for those totally unfamiliar http://en.wikipedia.org/wiki/Tensor. http://en.wikipedia.org/wiki/Four-vector may also be helpful.

Let us consider, as a specific example of a tensor, the 4-vector. Now, the status of the "reality" of a 4-vector, which includes time and space coordinates both, is an interesting question. I regard the entity of the 4-vector as real, because it is a tensor. The individual components (for instance the time coordinate of the 4-vector which is just the coordinate time we were talking about earlier) are observer dependent, and hence "not real" in and of themselves. The group behavior of the complete 4-vector taken as a whole (the coordinate time, plus the space coordinates) allows the 4-vector itself to be regarded as "real". Here we replace the concept of "observer independence" for a simple scalar quantity with the less restrictive "transforms in a standard manner" for a vector or tensor quantity.

I believe this philosophy (of regarding tensors as real) is reasonably common, but it's hard to document how common it is. The purists will probably not stress the issue of whether or not tensors are "real", it's possible to do the math without regarding them as real. I believe it helps a lot in understanding the math to regard tensors components (which individually are observer dependent) to be regarded as a description of something that is itself fundamentally observer independent and "real", while the tensor itself as an abstract object (as opposed to its particular representation by a specific observer) is real.

So in a nutshell:

proper time is one of the simplest "real" (obsever independent) quantities in relativity.
coordinate time isn't in and of itself "real", being observer dependent, but it is part of a larger abstract structure (the space-time 4-vector) which can be regarded as "real". It's not precisely observer independent, but the component description of one observer can be translated in a standard manner to the component description of another observer via standarized transformation laws.

6. Jul 30, 2014

### Staff: Mentor

And from the muon's perspective, the Earth is also time dilated. And you also have to include relativity of simultaneity to fully understand what's going on. See further comments below.

But they also have different senses of simultaneity, so you can't use the pulsar pulses as a common standard the way you can for a periodic orbit. It's worth working through the math in some detail to see how this goes.

Let's say that the pulsar pulses, which we will idealize as plane wave fronts, are moving in the positive $x$ direction, and the muons are moving in the negative $y$ direction. The Earth is stationary, and we'll call the point where the muon hits the Earth the spatial origin, $(x, y) = (0, 0)$. We'll call the spacing of the pulsar pulses $d$ (more precisely, this is the spacing of events along the muon's worldline where it crosses the wave fronts of the pulsar pulses), and we'll say that the muon starts falling from a height $y = 5 d$, so that, as you have stipulated, the muon crosses 5 pulsar pulses as it falls--for definiteness, we'll say that the pulses are at heights $y = 0.5 + k d$, where $k$ runs from $0$ to $4$. The muon falls at speed $v$ relative to the Earth, so the time it takes to fall, relative to the Earth, is $T = 5d / v$.

Now we can start labeling events in the Earth frame, which is the one we've been using so far. We'll call the time when the muon starts to fall time $t = 0$, so the event of the muon starting to fall, which we'll call event S, has coordinates $(t, x, y) = (0, 0, 5d)$. The event of the muon reaching the Earth, which we'll call event E, has coordinates $(5d / v, 0, 0)$. The five events at which the muon crosses a pulsar wave front are: event M1, $(0.5d / v, 0, 4.5d)$, event M2, $(1.5d / v, 0, 3.5d)$, event M3, $(2.5 d / v, 0, 2.5 d)$, event M4, $(3.5d / v, 0, 1.5 d)$, and event M5, $(4.5 d / v, 0, 0.5 d)$. There are also five events at which a pulsar wave front passes the point on the Earth's surface where the muon will hit; these are: event P1, $(0.5 d / v, 0, 0)$, event P2, $(1.5 d / v, 0, 0)$, event P3, $(2.5 d / v, 0, 0)$, event P4, $(3.5 d / v, 0, 0)$, and event P5, $(4.5 d / v, 0, 0)$. Finally, we need to pick out the event on the Earth's worldline that is simultaneous, in the Earth frame, with the muon starting its fall, which is simply the origin O of the frame: $(0, 0, 0)$.

The key things to note about these events, for the comparison we'll be doing in a moment, are these: events S and O are simultaneous; events M3 and P3 are simultaneous; event E is where the two worldlines cross; and the time from event O to event E (which is also the time from event S to event E) is just $T = 5 d / v$, as expected.

Next, we need to do a Lorentz transformation into the muon's rest frame to find the event coordinates there, so we can compare the times between the events of interest. The Lorentz transformation to the primed frame, $(t', x', y')$, is a boost in the $y$ direction with velocity $- v$ (because the muon is moving in the negative $y$ direction), so the transformation equations are:

$$t' = \gamma \left( t + v y \right)$$
$$x' = x$$
$$y' = \gamma \left( y + v t \right)$$

where $\gamma = 1 / \sqrt{1 - v^2}$. So some representative event coordinates in the primed frame are:

S: $(5 d \gamma v, 0, 5 d \gamma)$

E: $(5 d \gamma / v, 0, 5 d \gamma)$

M3: $(2.5 d \gamma ( v + 1 / v ), 0, 5 d \gamma)$

P3: $(2.5 d \gamma / v, 0, 2.5 d \gamma)$

P5: $(4.5 d \gamma / v, 0, 4.5 d \gamma)$

O: $(0, 0, 0)$.

Now, we can see that the muon does indeed have a shorter elapsed time between events S and E, than the Earth does between events O and E; that time is $T' = 5 d \gamma \left[ 1 / v - v \right] = 5 d / \gamma v = T / \gamma$, as expected. But, in this frame, event S happens *after* event O. In fact, in this frame, for any $v > 0.95$ or so (and the actual muons in the Earth's atmosphere are certainly traveling faster than this), event S happens after event P5. That means that, from the muon's perspective, when it starts its fall, the Earth has *already* passed *all five* of the pulsar pulses that the muon will pass on its way down.

So from the muon's perspective, the only reason the Earth has a longer elapsed time for the five pulsar pulses is that it has a long head start; it starts moving past the pulses way before the muon does. This is why you can't use the pulsar pulses as a common time standard in this case; the muon's motion is not periodic, so relativity of simultaneity comes into play. (Note also that the elapsed time for Earth during the muon's fall, from the muon's perspective, is *shorter* than the muon's elapsed time, by a factor $\gamma$. You can see this by finding the event on the Earth's worldline that is simultaneous with event S, in the muon's frame.)

7. Jul 30, 2014

### HALON

Sorry, I can’t see how the Earth is time dilated in this situation.

The earth is rotating on its axis. Therefore the muon’s path from its initial position relative to any point on the Earth’s surface is slightly accelerated. The muon would ‘feel’ accelerated due to gravity. This suggests non-symmetry, meaning the Earth would be time contracted relative to the muon. The same thing happens when a particle is accelerated in a synchotron. An observer is time contracted relative to a time dilated accelerating particle.

8. Jul 30, 2014

### Staff: Mentor

The effects of both the earth's gravitation and its rotation are negligibly small across the time that the muon is in flight (a few hundred microseconds) so we can analyze this as a pure special relativity problem in which we consider the muon to be approaching the observer on the surface of the earth at a constant velocity close to $c$.

In this case, as in all constant relative velocity situations, the time dilation is symmetrical. Both observers claim to be at rest while the other is moving relative to them, and both correctly consider the other clock to be running slow.

Last edited: Jul 31, 2014
9. Jul 30, 2014

### Staff: Mentor

It's time dilated in the muon's rest frame. That's obvious from the math I posted.

But the period of this rotation is so long compared to the muon's fall time that it's negligible for this discussion. Also, the key thing is not whether the Earth is rotating, but whether the muon's motion is periodic; and it isn't. See further comments below.

Only in the sense of coordinate acceleration (Coriolis acceleration, to be specific); the muon is in free fall.

No, it wouldn't. Coordinate acceleration is not the same as proper (i.e., felt) acceleration.

If this were true, it would be true even though the muon is in free fall, because the Earth observer in this scenario is certainly *not* in free fall; he feels a 1 g acceleration. However, acceleration, in and of itself, does not affect "rate of time flow"; it only affects it to the extent that it causes a change in relative velocity. For the scenario under discussion here, the effect of the proper acceleration of the Earth observer on this is negligible because, once again, the time for the muons to fall is so short.

That has nothing to do with acceleration; it's because the particle's motion in the synchrotron is periodic; it is spatially co-located with the observer at rest in the lab once per revolution. That enables a direct comparison of the two "clocks". No such periodicity is present in the motion of the muons relative to the Earth in the scenario we're discussing in this thread.

(To see that it's periodicity of motion, not acceleration, that makes the difference, consider a satellite in a free-fall orbit about the Earth; it experiences the same time dilation due to motion that the particle in the synchrotron does, relative to an observer at rest on the Earth's surface. In the case of an actual Earth satellite, there are other effects involved, which is why the Hafele-Keating experiment had to be carefully analyzed to separate out those effects; a better idealized case is a satellite in orbit just above the surface of a non-rotating Earth-like planet with no atmosphere. In that case, the satellite is time dilated relative to the Earth observer for the same reason that the particle in the synchrotron is time dilated relative to the lab observer.)

10. Jul 31, 2014

### HALON

I understand this now.I also understood Peter's point about the coordinate acceleration vs proper acceleration, and the negligibility of the earth’s rotation.

Regarding "assymetry"...

Good point. It is symmetric after all.

Yes, and if the observer’s position in the lab was at the centre of the synchrotron, then in this situation I understand co-location to be continuous, not just once per revolution of the particle. Nevertheless, periodicity exists, so I gather the “clocks” can be compared despite the lack of a common reference frame, and in a continuity due to the continuous co-location, to coin a phrase.

11. Jul 31, 2014

### e.bar.goum

What Nugatory said. Measurements of half-lives of states that are measured using Ion Storage Rings (e.g. ESR at GSI) need to be lorentz corrected.

e.g. http://arxiv.org/pdf/1306.0475.pdf

And you can go the other way: The ESR at GSI has been used to test time dilation.
Nature paper: http://www.nature.com/nphys/journal/v3/n12/abs/nphys778.html

12. Jul 31, 2014

### Staff: Mentor

No; in this case the two are never co-located. If the observer is at the center of the synchrotron, then in order to compare the observer's and the particle's clock rates, you have to pick a particular direction--say the direction the observer has to look to see the door to the lab--and say that one "period" of the particle's motion elapses every time the particle passes the observer's line of sight in that direction.

There's no way for an observer at rest and a moving particle to be "continuously co-located"; that doesn't make sense. Relative motion is relative motion, even if it's periodic.

13. Jul 31, 2014

### HALON

OK, so the term co-location might mean something else in relativity. (I’ve only done first year university physics). I imagined it to mean something like “co-proximity”. Specifically, due to the fixed (negligible) separation between observer and particle then any number of “agreed” directions is valid. And as the motion is uniformly periodic around a central point, the observer and particle could just compare each other’s “clocks” continuously without waiting for a revolution. The clocks would be asymmetrically synchronized, if such a phrase is used.

(I have to admit sneaking this concept in from my last post in my own thread)

14. Jul 31, 2014

### Staff: Mentor

You can pick any direction you like, yes, but you have to pick one specific one in order to define a periodicity for the motion of the particle in the synchrotron.

Only in the sense that you can pick any direction, as above. If you don't pick a direction at all, then you have no way of defining what constitutes one "period" of the motion, and therefore you have no way of invariantly comparing the two clocks.

They can continuously exchange light signals showing their clock readings, sure. But again, without defining some specific direction to mark one "period" of the particle's motion, how do either of them know what the standard of comparison is? In other words, without defining some specific line of sight such that, each time the particle crosses it, one "period" has elapsed, for all the particle knows, its clock could be running faster and the observer's clock could be running slower, since there's no common standard for them to refer to.

15. Aug 2, 2014

### accel2know

Let me try to get the discussion back to the question that Objecta99 had asked. I want to make a few points regarding my previous post.

Let me first thanks to PeterDonis for his explanation. I was trying to indicate that observers in different inertial frames of reference will measure different periods for the pulsar using their own individual clocks, without regards for how they see each other. The beauty of pulsars is that they transmit a continuous train of pulses equally separated in time and traveling at a constant speed that all agree. The period may not be the same for all observers, but as long as the observer is in an inertial state (non accelerated), the period should remain constant (as long as the pulsar is also not moving, which is very unlikely)

Point #1: Out of the billions and billions of muons falling to Earth every second, lets say that there is another muon (#2) that is falling on the other side of Earth 180° from the other and that is created in the same way as the first (#1). Going back to the animation of my previous post, flip the animation vertically and now Earth is on the top, and the muon is moving from the bottom upwards; waves are still moving from right to left.

We now have three observers, two of them (the two muons) are in very different Inertial Frame of Reference (IFR) see very short pulsar periods; the other observer (Earth's) has an IFR that is somewhere in between the two muons, but this observer reports pulsar periods that are much larger.

Point #2: Let's say that the Earth observer is in a "transformer" planet (like the movie), and when he sees muon #1 go by, the planet is transformed into a small ship, fires up the engine and accelerates along the trajectory of the muon. Keep in mind that the acceleration is in a direction that is perpendicular to the direction of the pulsar. It may have to accelerate for a very long time to catch up, but as it gets closer and closer to the same IFR of muon #1, the pulsar periods will get smaller and smaller, just at the muon saw short pulsar periods.

Things are looking weird here and if this observer gets a case of the ‘willies’ and decides to turn around (180°), the pulsar periods will start getting larger as it approaches his original IFR and then at some point the pulsar periods will start getting shorter again as it approaches the IFR of muon #2.

Point #3: The unit of length (meter in SI) is now defined in terms of the distance traveled by light in a certain amount of time.

For the Earth observer, at 100 µs (Te) apart in time, the pulsar pulses are 30 km (Te x c) apart from each other in space.

For the muons, the pulsars pulses are only 0.44 µs (Tm) apart in time, which translates into only 132 meters apart in space. The muons see not only Earth length contracted in the direction of its motion, but they also see length contraction 90° from Earth motion.

Last edited: Aug 2, 2014
16. Aug 2, 2014

### Staff: Mentor

Fair enough. But note that the reason the Earth observer measures a longer period is that you set up the scenario so that the Earth was at rest relative to the pulsar, while the muon is moving at right angles to the pulsar wave fronts (as seen from the Earth rest frame). The muon seeing a shorter period is just an example of the relativistic transverse Doppler effect. (So is the associated "length contraction" of the pulsar wavelength, btw, which is indeed present, as you correctly observe later in your post.)

In other words, what determines the observed period (and wavelength) of the pulsar is the relative motion of pulsar and observer. There's nothing special about the Earth with respect to spacetime itself; it's just that in this scenario the Earth happens to be at rest relative to the pulsar. Similar remarks apply to your other scenarios.

17. Aug 4, 2014

### pervect

Staff Emeritus
This may be sort of an aside, but the variation in quasar periods as seen from the Earth has been of concern to astronomers for some time, and written about in detail, For an early paper, see for instance Annual Variation of an Atomic Clock.

Compensating for relativistic effects has become a matter of routine, basically one has to pick one of the IAU recomended coordinate systems, and stick with it. (The most recent IAU resolutions as of this post are as of 2006, some 40 years after the above paper was written, so don't expect modern terminology from this older paper0.

To avoid being confused by the jumble of acronyms associated with the fairly large number of coordinate systems that are or have historically been defined, it serves well to keep one basic principle in mind.

The observer independent notions of time and space are "proper time" and "proper distance". The rest of the alphabet soup is just a coordinate choice, it's funamentally just a manner of assigning labels ("numerical coordinates") to events in space-time in a consistent manner.

Each coordinate system has a tool, known as a metric, that can be used to convert the observer dependent coordinate notions of location into the observer independent notions of proper time, proper distance, or Lorentz interval.

People who aren't familiar with relativity tend to adopt an observer-dependent notion for time and distance, and win up getting snowed under by the mass of details as a result. It may take a little thought to understand the observer-independent notions of proper time and proper distance, but it will save an immense amount of confusion once one does so.