Question about "odometer" analogy and time dilation

[FIL]

Hey again physics forums! I'm just a lug who likes watching science videos and appreciates the occasional answer to questions that pop up as I watch them.

This morning I came across a video that explained time dilation really well by using an odometer analogy, that a clock is like an odometer that measures your distance through spacetime, but the longer your "path" the less time you experience.

Well that got me thinking, and I wondered if you could use that "odometer" to settle who's moving if you've got two starships in motion drifting towards each other. Now it's my understanding that relativity makes no distinction between which starship is moving and that each can safely regard itself at rest and say it's the other moving towards it. If that's wrong the rest of this comment is gobbledygook, but I think I got that bit right!

Okay so now imagine each starship syncs their clocks with each other the moment they pass, then a few minutes later each launches out a probe synced with the starship clock. The probes are programmed to travel in an arc so that they leave their starship of origin and dock with the other starship. Now because of how the clocks were synced and the delay in launching the probes, unless both starships happened to be going the same speed, the paths through spacetime for each probe would be different. The arced path is identical, but due to the delayed launch there's a horizontal component that should be detected by the clock if the starship it originated from was moving.

If the above is accurate the probes would not have the same reading on their clocks and we could tell which starship was moving, or exactly how much each was moving.

Okay now tell me why I'm wrong

 

[PeterDonis]

a clock is like an odometer that measures your distance through spacetime

Yes, this is one way of viewing what proper time is in relativity.

but the longer your "path" the less time you experience.

No, this is backwards. The longer your path through spacetime, the more time you experience. That's what the odometer analogy means.

 

[Orodruin]

This morning I came across a video

What video? You need to link it to provide proper reference and base for discussion.

 

[Ibix]

the paths through spacetime for each probe would be different.

No, they'll be symmetric, assuming they have identical acceleration profiles.

I think the problem is that you are thinking of an odometer that measures distance travelled relative to some chosen rest frame - the surface of the Earth for example. A clock measures "distance" through spacetime, which is not the same thing.

 

[PeterDonis]

an odometer that measures distance travelled relative to some chosen rest frame - the surface of the Earth for example.

It's worth noting that the distance that an ordinary odometer measures is independent of what coordinates you choose on the Earth's surface; it is a geometric invariant, just as the distance through spacetime that a clock measures is a geometric invariant, independent of what frame you choose.

 

[FIL]

What video? You need to link it to provide proper reference and base for discussion.

I'm trying to find it again, I must not have been logged in on YouTube because I'm not seeing it in my history.

No, they'll be symmetric, assuming they have identical acceleration profiles.

I think the problem is that you are thinking of an odometer that measures distance travelled relative to some chosen rest frame - the surface of the Earth for example. A clock measures "distance" through spacetime, which is not the same thing.

Hmm, I'm afraid I don't understand why "distance through spacetime" precludes the horizonal motion in my example. Or the linear component rather. I should also have been more clear that the probes and ships all sync their clocks at the passing. Not that it probably matters, just want to make sure I'm understood.

 

[Ibix]

Hmm, I'm afraid I don't understand why "distance through spacetime" precludes the horizonal motion in my example.

I don't know what you mean by "horizontal motion".

The easiest way to see the result is to think in the frame where the ships have equal and opposite velocities. Then so do the probes. And therefore the clocks must show the same on arrival. In any other frame the paths will look asymmetric, but it's only in the sense that a square isn't left-right symmetric if you rotate it a couple of degrees. The symmetry is still there - it's just not parallel to your left-right direction any more.

I can draw a diagram later if it still doesn't make sense.

 

[FIL]

I don't know what you mean by "horizontal motion".

The easiest way to see the result is to think in the frame where the ships have equal and opposite velocities. Then so do the probes. And therefore the clocks must show the same on arrival. In any other frame the paths will look asymmetric, but it's only in the sense that a square isn't left-right symmetric if you rotate it a couple of degrees. The symmetry is still there - it's just not parallel to your left-right direction any more.

I can draw a diagram later if it still doesn't make sense.

Sorry I tried to clarify in an edit but you started responding before I made it. So if the ships and probes all sync at the passing, from that moment I would expect there to be a linear component of "motion through spacetime" plus the arc trajectory for the clock to read. That's why I thought you could tell which starship was moving or how much each was moving because of the linear component prior to launch.

If it's no trouble a diagram would help because I can't quite follow your explanation.

 

[PeroK]

Okay now tell me why I'm wrong

What does "a few minutes late" mean? Who is measuring these minutes?

 

[DaveC426913]

Since this:

each can safely regard itself at rest and say it's the other moving towards it.

is correct ...

... then it must follow that this:

That's why I thought you could tell which starship was moving...

must be incorrect.

It really is as if each one can consider themselves stationary, so any experiment they do will not show otherwise.

It's not like a magician's illusion where - if you look close enough, or from different angles - you'll spot the "trick".

 

[FIL]

What does "a few minutes late" mean?

(Later! Not late lol.)Well once they sync their clocks, I'm assuming as long as the probes are launched at the same clock reading it would be a symmetrical launch. The wait is to allow linear motion through spacetime to accumulate on the probe clock before sending it to the other craft. Then the ships would relay what their clocks reads and what the clock on the probe they received reads in order to make comparisons.

 

[Orodruin]

I suggest that you read this, which is kind of related:
https://www.physicsforums.com/insights/geometrical-view-time-dilation-twin-paradox/
It discusses the geometrical analogy between the twin paradox and its counterpart in Euclidean space - which is the triangle inequality.

You can determine absolute motion in special relativity as little as you can assign a preferred direction in Euclidean space, even if you can measure distances in Euclidean space with a regular odometer.

 

[phinds]

The odometer analogy does NOT explain time dilation, it explains differential ageing.

 

[PeroK]

(Later! Not late lol.)Well once they sync their clocks, I'm assuming as long as the probes are launched at the same clock reading it would be a symmetrical launch.

Okay, let's say each spaceship eaits two minutes according to its clock, then launches a probe with a clock set to zero?

Each fires the probe direct at the other spaceship at the same launch speed relative to the ship?

Then we compare the probe clock readings when they meet?

Can you do a calculation for some examples? Let's assume one spaceship is at rest and the other moving at some speed ##v## and the probes are launched at a relative speed of ##u## after time ##T##. What are the times on the probes when they meet?

By physical symmetry, they should read the same time. What do your calculations say?

 

[PeterDonis]

I don't understand why "distance through spacetime" precludes the horizonal motion in my example.

It doesn't. But "horizontal motion" is frame-dependent. "Distance through spacetime" is not. That's why you should focus on "distance through spacetime" if you want to understand what's going on.

 

[PeroK]

PS one way to calculate the answer is to use a reference frame where both spaceships are moving at the same speed in opposite directions.

In that frame, the probes are launched simultaneously with the same speed. Hence, their clocks remain synchronized in that frame and show the same time when they meet.

From that perspective, the experiment fails to detect any absolute motion of the "moving" spaceship.

 

[FIL]

I appreciate everyone trying to help me, I'm a bit overwhelmed as I'm twenty years removed from any algebra or geometry lessons and it was never my strength to begin with.

PS one way to calculate the answer is to use a reference frame where both spaceships are moving at the same speed in opposite directions.

In that frame, the probes are launched simultaneously with the same speed. Hence, their clocks remain synchronized in that frame and show the same time when they meet.

From that perspective, the experiment fails to detect any absolute motion of the "moving" spaceship.

Okay, if you'll permit to try and clarify through written explanation you may be able to help me find where I've taken a wrong turn. First let's assume all the clocks on the ships and their probes are synchronized when the two ships pass. Now let's say when each ship measures their distance from the other ship to be an agreed upon value (really any distance should do) they launch their probes, the probes are not set to zero rather keep the time they read at the moment of launch going.

Now if you can allow me to entertain the idea of absolute motion(though not really absolute motion as the whole system could be in a box going some arbitrary velocity) for a moment and say one ship is the one moving, my logic is that from the point of synchronization at the pass, that the probe in the moving ship would detect additional motion through spacetime equal to the agreed upon distance at which both probes were to be launched, and that the other ship being at rest, its probe would not detect that additional distance. This would result in the probes recording a different amount of proper time. The ships could then relay the probe reading to each other and ascertain which of them was actually moving.

Somewhere I'm sure there's a relativistic effect that eats up this possibility, I just don't know where.

I hope that my new explanation is clearer.

 

[Ibix]

Somewhere I'm sure there's a relativistic effect that eats up this possibility, I just don't know where.

The lack of an absolute frame, probably. If you insist there is one then you'll probably conclude that experiment will detect it...

 

[PeroK]

I hope that my new explanation is clearer.

What you need is a calculation to show that the times are different. Or the same. Unless you have a calculation, you have nothing.

 

[FIL]

The lack of an absolute frame, probably. If you insist there is one then you'll probably conclude that experiment will detect it...

I'm not insisting on anything, this is just the scenario that popped in my head after learning of the odometer analogy.

What you need is a calculation to show that the times are different. Or the same. Unless you have a calculation, you have nothing.

I get that, and I'm not trying to be obstinate or anything. Nor am I trying to say relativity is wrong, this is just what I thought about after the odometer thing.

 

[PeroK]

PS I've given a calculation that the times are always the same. Unless that calculation is wrong, there is nothing more to discuss.

Note that it makes no difference whether the probes are fired out showing a time of zero. The current ship time works just as well.

 

[DaveC426913]

Now if you can allow me to entertain the idea of absolute motion(though not really absolute motion as the whole system could be in a box going some arbitrary velocity) for a moment and say one ship is the one moving, my logic is that from the point of synchronization at the pass, that the probe in the moving ship would detect additional motion through spacetime equal to the agreed upon distance at which both probes were to be launched, and that the other ship being at rest, its probe would not detect that additional distance. This would result in the probes recording a different amount of proper time. The ships could then relay the probe reading to each other and ascertain which of them was actually moving.

Right, well, what you've got here is a tautology.

It boils down to:

1. If absolute motion exists then it is detectable. (logically valid)
2. I posit that absolute motion exists, therefore I should detect it with my experiment (logically valid)

None of which actually leads to the conclusion that it does, in fact, exist.

At the risk of seeming glib, the logic above can just as easily have the word "unicorns" substitute for "absolute motion" and "I could ride one" substitute for "should be detectable".

 

[PeroK]

I'm not insisting on anything, this is just the scenario that popped in my head after learning of the odometer analogy.

It can be hard to spot an error when you describe a scenario and give a conclusion without calculation.

There might be no error. Just a false conclusion.

 

[pervect]

Hey again physics forums! I'm just a lug who likes watching science videos and appreciates the occasional answer to questions that pop up as I watch them.

This morning I came across a video that explained time dilation really well by using an odometer analogy, that a clock is like an odometer that measures your distance through spacetime,

This is a valid (and IMO good) way of looking at things, but a reference to your source would be helpful to settle any fine details. Such as:

but the longer your "path" the less time you experience.

That part is probably a misunderstanding of what was said, though it's possible the video itself was wrong, which is why it would be helpful to cite it.

Well that got me thinking, and I wondered if you could use that "odometer" to settle who's moving if you've got two starships in motion drifting towards each other. Now it's my understanding that relativity makes no distinction between which starship is moving and that each can safely regard itself at rest and say it's the other moving towards it. If that's wrong the rest of this comment is gobbledygook, but I think I got that bit right!

Yes, that sounds right.

Okay so now imagine each starship syncs their clocks with each other

The issue with this is called "the relativity of simultaneity". Any given inertial frame of reference does have a notion of how to synchronize clocks, due to Einstein.

However, the notions of different frames of how to synchronize clocks, are different for each different frame, there is not a common, shared notion of simultaneity.

This causes a lot of confusion. There are numerous references about "relativity of simultaneity" on the web, some due to Einstein. The applicable thought experiment is often referenced to as "Einstein's train". One source of this, possibly not the best, is in one of Einstein's popularizations, see https://www.bartleby.com/lit-hub/re...ral-theory/ix-the-relativity-of-simultaneity/.

This may not be the best reference to learn from, Sherr, et al, argue for different presentations in their paper https://arxiv.org/abs/physics/0207081 entitled "The challenge of changing deeply-held student beliefs about the relativity of simultaneity", and in other papers based on their teaching experience. This paper (and most of their others that I am aware of) are aimed at teachers rather than students, I do not have a particular student-friendly reference on this point, just an observation that many, many people struggle with it, including people who have had a formal course with a live teacher explaining the topic :(.

My main position here is that I can try to point out some resources which may help interested people learn about the topic, but I'm unlikely to write anything better myself than the slew of things that have already been written about it.

I'll snip the rest of what you wrote, as there isn't any obvious problem with it.

I will add one thing. Remember when I said that the odometer was a good way of looking at things? The reason the odometer is a good way of looking things is precisely the fact that odometers have no notion of "synchronization". They just measure distances. But using odometers, we can make statements such as "the shortest distance between two points is a straight line", without invoking any notion of "syncrhonizing odometers". We can make similar statements about clocks, such as the "principle of maximal aging".

The point of the odometer analogy can be regarded as using a notion of "what clocks measure", a notion formally known as "proper time", which is basic enough that it is not frame-dependent because it omits the frame dependent parts of notions of time, notions that include the additional idea that "clocks can be synchronized".

 

[Ibix]

Here's the promised diagram. It's a Minkowski diagram, also called a spacetime diagram. You may have come across displacement-time graphs in physics in school, where you just plot the distance of some object from its starting point as a function of time. This is exactly the same, except that we take them a bit more literally in relativity - since spacetime is a 4d structure, this is a map of that 4d structure (or two dimensions of it anyway). It's also conventional to draw time up the page and space horizontally.

What's going on? The two thick lines are the positions of the ships. They meet at the origin at time zero, and the red one moves to the right and the blue one to the left, both at 0.4c. Five minutes later they launch probes, represented by the thinner lines moving at 0.95c. They catch up with the opposite spaceship at 12.27 minutes after the first meeting.

Those times, 5 and 12.27 minutes, are noted using clocks at rest in this frame. The clock ticks of the ships are marked by little crosses on the lines - note that due to time dilation the probes launch at times the ships call about 4.58 minutes after launch. The ticks of the clocks on the probes are also marked by little crosses. The probes are moving much faster, so the ticks are much further apart - so they reach the opposite spaceship about 2.27 minutes by their own clocks after launch, although this frame's clocks show 7.27 minutes elapsing.

In this frame it is clear that the two probes' clocks will read the same when they reach their destination ships, because the situation is manifestly symmetrical when drawn in this frame. The odometer analogy comes in because the clock ticks are evenly spaced along the worldlines of the ships and probes. The clocks measure the "distance" along the lines.

Now let's draw that same diagram, but this time we'll choose to treat the red rocket as stationary. The blue rocket moves faster to the left, the red probe moves even faster to the left, and the blue probe moves slower to the right.

If you care to count the crosses and compare this diagram to the previous one you'll find that all the junctions lie in the same place in the diagram. In fact, we've just done something very like rotating a map - none of the distances change when you turn a map round, but lines on it that were slanted may now be straight up. In fact, the transformation here is a Lorentz boost rather than a rotation, which is a little more complicated (and I genuinely mean "a little") than a rotation but is closely analogous. Because of this boost, things that were at the same level on the first diagram ("happened at the same time") now aren't at the same level ("didn't happen at the same time") and clocks tick at different rates. So the symmetry is hidden, but it's still there. You can see that the probes still launch 4.58 ticks (by the ships' clocks) after meeting and still meet 2.27 ticks (by the probes' clocks) after they're launched.

Just like rotating a road map doesn't change the distance from your house to the shops, boosting a map of spacetime doesn't change the times for the ships. And the original diagram shows you that the times must be the same for the two halves of the experiment because there they're mirrors of one another.

 

[FIL]

Thanks pervect, I'll read up about relative simultaneity some more. I don't think the video was wrong I probably just misunderstood it. I'll have another go at trying to find it. It was depicting a minkowski diagram of the twin paradox and showed the traveling twin moving away on an angle then back on an angle whereas as the stay at home twin was a straight line. I probably got confused by that and drew the wrong conclusion about path length.

 

[Ibix]

It was depicting a minkowski diagram of the twin paradox and showed the traveling twin moving away on an angle then back on an angle whereas as the stay at home twin was a straight line. I probably got confused by that and drew the wrong conclusion about path length.

The important thing to remember about Minkowski diagrams is that the "length" you're measuring is ##\sqrt{c^2\Delta t^2-\Delta x^2}##, rather than the more familiar ##\sqrt{\Delta x^2 +\Delta y^2}##. That usually means that lines that are longer on the diagram are shorter in reality. You just can't draw Minkowski spacetime completely accurately on a Euclidean plane.

Edit: I'm probably over-generalising there, and I really need some absolute value signs in the Minkowski metric to avoid complex numbers, but hopefully you see the point.

 

[FIL]

The important thing to remember about Minkowski diagrams is that the "length" you're measuring is ##\sqrt{c^2\Delta t^2-\Delta x^2}##, rather than the more familiar ##\sqrt{\Delta x^2 +\Delta y^2}##. That usually means that lines that are longer on the diagram are shorter in reality. You just can't draw Minkowski spacetime completely accurately on a Euclidean plane.

Edit: I'm probably over-generalising there, and I really need some absolute value signs in the Minkowski metric to avoid complex numbers, but hopefully you see the point.

Thanks, and the diagram with accompanying explanation of it you provided is really helpful. I've downloaded a screenshot of your comment for future reference.

Edit: am I reading it wrong, or did you accidentally put the probe return at 11 instead of 12? Edit: nevermind it's my bad eyes!

 

[Ibix]

nevermind it's my bad eyes!

The diagram is generated by a python script that does most of the maths for me. It's always possible that I made a mistake in the code, but random arithmetic errors are fairly unlikely.

 

[Mister T]

Well once they sync their clocks, I'm assuming as long as the probes are launched at the same clock reading it would be a symmetrical launch

By symmetrical do you mean simultaneous? Recall that simultaneity is relative. Two events that are simultaneous in the rest frame of one of the ships will not be simultaneous in the rest frame of the other ship.

Edit: By this I mean that if the two launches are simultaneous in the rest frame of one of the ships, they will not be simultaneous in the rest frame of the other ship.

 

[Orodruin]

The odometer analogy does NOT explain time dilation, it explains differential ageing.

This is incorrect. It explains both if applied correctly.

 

[Orodruin]

I don't think the video was wrong I probably just misunderstood it.

But this is the point. There are several members here who are qualified to determine whether the video was wrong or if you misunderstood. Both happen frequently.

 

[FIL]

By symmetrical do you mean simultaneous? Recall that simultaneity is relative. Two events that are simultaneous in the rest frame of one of the ships will not be simultaneous in the rest frame of the other ship.

Edit: By this I mean that if the two launches are simultaneous in the rest frame of one of the ships, they will not be simultaneous in the rest frame of the other ship.

Later in the thread I amended the scenario so that the probe launches occur when the two ships measure an agreed upon distance between themselves after they pass.

After seeing the spacetime diagram Ibix provided I understand it doesn't matter, if motion is regarded as relative the proper time each probe accumulates will always be identical. I was supposing some version of absolute motion without entirely realizing it, I thought it was more of a case of "objectively, there is a true velocity each ship carries with respect to the other, but there's no way to detect it" as opposed to it truly being relative.

 

[robphy]

both at 0.4c. Five minutes later they launch probes, represented by the thinner lines moving at 0.95c. They catch up with the opposite spaceship at 12.27 minutes after the first meeting.

Why were these values chosen?
Of course, they are correct,
and a computer wouldn't have any problem with them,
but the arithmetic seems (in my opinion) unnecessarily complicated to make your point.
A return leg of 0.8c would have been better.
(An outgoing leg of 0.6 would also help simplify the arithmetic.
I have a fondness for 3/5, 4/5, 15/17... https://en.wikipedia.org/wiki/Pythagorean_triple .)

 

[Ibix]

Why were these values chosen?

0.6c and 0.8c made for a very long chase phase (5 mins to launch, then 30 mins to overtake) and I wanted the diagram to be reasonably wide and short so you can see the ticks clearly. 0.4 and 0.95 were pretty arbitrary, I admit.