Proper (and coordinate) times re the Twin paradox

[Grimble]

This was straying from the point in the original thread, but I thought it made a point...

Also note that in the twin paradox we are comparing one person's proper time with another person's proper time. It is not the same thing as time dilation, and not understanding that can be a stumbling block to understanding the twin paradox. Many if not most published explanations of the twin paradox fail to mention this, and in my opinion that can cause confusion for the reader who is trying to understand it.

The stay-at -home twin is at rest in her frame and her clock must therefore measure proper time.
The traveling twin, carries his clock with him; it is therefore at rest in his frame and must also measure proper time.
As each twin is moving relative to the other, they will each measure coordinate time for their twin.
Their proper times will be identical.
Their coordinate times will be identical.
As their relative speeds are the same, their Lorentz transformations will be the same.
When the traveling twin slows on his return and comes to rest in his twin's frame they are both once again in the same frame and will have traveled exactly the same each relative to the other.

It is only if the traveller continues past at speed that there is a time difference as they pass and that will be proper time vs coordinate time...

 

[Orodruin]

This was straying from the point in the original thread, but I thought it made a point...
The stay-at -home twin is at rest in her frame and her clock must therefore measure proper time.
The traveling twin, carries his clock with him; it is therefore at rest in his frame and must also measure proper time.
As each twin is moving relative to the other, they will each measure coordinate time for their twin.
Their proper times will be identical.
Their coordinate times will be identical.
As their relative speeds are the same, their Lorentz transformations will be the same.
When the traveling twin slows on his return and comes to rest in his twin's frame they are both once again in the same frame and will have traveled exactly the same each relative to the other.

It is only if the traveller continues past at speed that there is a time difference as they pass and that will be proper time vs coordinate time...

No this is wrong. By the same reasoning in Euclidean space, you would conclude that the hypothenuse would have the same length as the sum of the length of the catheti in a right triangle.

 

[Dale]

he stay-at -home twin is at rest in her frame and her clock must therefore measure proper time

Any clock measures proper time. It doesn't matter if they are at rest or moving, if they are inertial or non inertial, in curved spacetime or flat. They always measure proper time along their worldline.

 

[Mister T]

Their proper times will be identical.

The entire point of the twin paradox is that they are not identical. One twin ages more than the other!

 

[Ibix]

I think you may be confused about "proper time". Proper is used in the Latin sense of "one's own". You keep writing as if proper time were a global thing, but it's not. Each clock measures it's own proper time (which is a tautology).

The relationship of proper time to coordinate time is like the relationship of your car's odometer reading to your OS grid northing. We can have many cars at the same grid ref with different odometer readings. This might be because they started out in different places. Or it might be because they took different routes from point A to point B. So, in a sense, there's no relationship. But, think about how the OS grid is drawn. One way to do it (a bit unlikely with real geography) would be to (a) pick a direction and call it east; (b) deploy a lot of cars one mile apart in a line in the east direction; (c) set their trip odometers to zero; (d) drive them all north; (e) add a marker every time their odometers go up one mile. Now we have a grid, built using odometers and covering the whole of the country.

Your odometer reading is like your proper time. It is the distance (SR: interval) that you personally have traveled through space (SR: spacetime). The grid northing is like coordinate time. We got an army of cars (SR: clocks) made them travel in the same direction (SR: at the same speed) and agree an odometer reading (SR: a time) to call zero. Then we build a grid from the odometer (SR: clock) readings. Then we can use our grid to identify points in space (SR: spacetime).

The important thing about proper time is its invariance. Everyone agrees the length of your path. The important thing about coordinate time is its arbitrariness. We could have chosen a different way to define our zero point, and we could have chosen a different direction to move in.

In the twin paradox case, the two twins have different proper times when they return for the same basic reason that they would have different odometer readings if one drove direct from London to Edinburgh and the other drove from London to Edinburgh via Manchester. If they meet up in Edinburgh and continue to Aberdeen on the same road, their odometers will show the same increase in distance since Edinburgh, but will still show a different total distance since London. In this case, since Edinburgh isn't due north of London, neither of them will have odometer readings that correspond exactly to the difference in the northings of London and Edinburgh. This is analogous to a twin paradox where the traveling twin decelerates to rest at the end of the journey, but where they are working in a frame of reference where the Earth is moving slightly.

I hope that all makes sense.

 

[Grimble]

Any clock measures proper time. It doesn't matter if they are at rest or moving, if they are inertial or non inertial, in curved spacetime or flat. They always measure proper time along their worldline.

Yes, of course they do! Thank you!

 

[Grimble]

The entire point of the twin paradox is that they are not identical. One twin ages more than the other!

Yes, but any clock in its own reference frame is at rest and only measures time passing. And, in Spacial Relativity which as I understand it is where the twin paradox may be considered, identical clocks can be placed in any frame and will keep identical time - identical clocks; identical laws of science; identical conditions...
so how do they measure different times?
(and before anyone says "acceleration" I think of one twin passing the other and synchronizing with another sibling traveling in the opposite direction avoiding any acceleration)

 

[Dale]

so how do they measure different times?

Because they are traveling on different worldlines and the proper time is the spacetime interval integrated along that worldline.

Geometrically, this is like two identically constructed rulers measuring the length of two different paths. The resulting difference is not attributed to any distortion of the ruler, but to the length of the respective lines.

 

[Mister T]

When the traveling twin slows on his return and comes to rest in his twin's frame they are both once again in the same frame and will have traveled exactly the same each relative to the other.

It is only if the traveller continues past at speed that there is a time difference as they pass and that will be proper time vs coordinate time...

This is not true. The traveling twin's clock will show less elapsed time than the staying twin's, regardless.

so how do they measure different times?

How would they measure the same time? It's been predicted by theory, and confirmed by experiment, that they are different.

 

[Ibix]

identical clocks; identical laws of science; identical conditions...
so how do they measure different times?

They are not in identical conditions. There are many measurements you can make that will be different for the two clocks. But fundamentally, @Dale said it: two sides of a triangle are not the same length as the third side.

 

[Grimble]

OK. Lots of different points raised here - in what are often somewhat unclear, rather vague ways as they tend to assume that I am on exactly the same wavelength and am left trying to understand what point is being made.

So let me answer or ask for clarification point by point:

1.

No this is wrong. By the same reasoning in Euclidean space, you would conclude that the hypothenuse would have the same length as the sum of the length of the catheti in a right triangle.

Can you please explain what you mean here?
2.

The entire point of the twin paradox is that they are not identical. One twin ages more than the other!

Surely special relativity is reciprocal - 'As A is to B, so B is to A' viewed from each of the two frames of reference? So each twin measure the other to age more?
3.

I think you may be confused about "proper time". Proper is used in the Latin sense of "one's own". You keep writing as if proper time were a global thing, but it's not. Each clock measures it's own proper time (which is a tautology).

No, proper time is individual to each clock. It is the time passing on a clock measured by an observer holding that clock. And as Einstein's first postulate says that clock will measure time according to the same laws in whatever inertial frame it is placed in. It is the coordinate times that will vary according to the observer's motion relative to the clock.
"A second is a second, is a second,..." is a phrase I have often read... and any clock read in its own frame will measure the same seconds. So what I am saying is that the duration of 1 second proper time in a frame is the same as 1 second proper time, measured on another clock by that other clock's local observer in that other clock's frame.
4.

The important thing about proper time is its invariance. Everyone agrees the length of your path.

Yes, OK, but proper time can only be a time and not a distance! Yes it is the difference in time measured between two events in Spacetime, but, the time is what would be measured by the observer holding that clock and also traveling between those Spacetime events. The time measure for that clock between those events by any other observer, is by necessity coordinate time, as it includes the spatial difference between those events in that observer's frame.

Proper time is the time measured on a clock between two events - on a clock that passes through both events, where the two events are on that clock's world line and, measured on that clock, in that clock's frame of reference, where both events occur at location (0,0,0). Where there is a distance measured between those events it they have to be measured by an observer in another frame and it must therefore be coordinate time.
5.

Because they are traveling on different worldlines and the proper time is the spacetime interval integrated along that worldline.

Worldlines. An easy concept to grasp. The path of a particle through spacetime - but in what reference frame? For surely any worldline measured from the frame of that particle (or clock if you prefer) is a fixed location (t,0,0,0) moving through time... and it will have a different path for every other observer... ?

Geometrically, this is like two identically constructed rulers measuring the length of two different paths. The resulting difference is not attributed to any distortion of the ruler, but to the length of the respective lines.

Where the two different paths are due to the relative movements of the observers.
The worldline measured by a remote observer from another frame, is in effect the coordinate interval, comprising the spacetime interval and the travel time measured from the observer's frame (added vectorially). This is after all the basis of the Lorentz transformation that is, at its core no more than vectorial addition by the courtesy of Pythagoras.

 

[Mister T]

OK. Lots of different points raised here - in what are often somewhat unclear, rather vague ways as they tend to assume that I am on exactly the same wavelength and am left trying to understand what point is being made.

Surely special relativity is reciprocal - 'As A is to B, so B is to A' viewed from each of the two frames of reference? So each twin measure the other to age more?

There's nothing vague or unclear in this statement:

The traveling twin's clock will show less elapsed time than the staying twin's, regardless.

Or this one.

They are not in identical conditions. There are many measurements you can make that will be different for the two clocks.

The traveling twin is the one who returns, the staying twin is the one who stays. Draw the worldlines of the two twins. The worldline of the staying twin is a straight line, the worldline of the traveling twin is not.

 

[Nugatory]

So let me answer or ask for clarification point by point:

You have two basic misunderstandings here. You'll want to work through them first, then take on the other questions.

Proper time is the time measured on a clock between two events - on a clock that passes through both events, where the two events are on that clock's world line and, measured on that clock, in that clock's frame of reference, where both events occur at location (0,0,0). Where there is a distance measured between those events it they have to be measured by an observer in another frame and it must therefore be coordinate time.

Think of proper time as something that we observe: Say we design our clock so that every time it ticks it punches a hole in a piece of paper somewhere inside; we start with a fresh piece of paper at event A and remove it at event B. How many holes are there in the piece of paper? That's a simple direct observation; all observers everywhere will agree about the answer without any rigamarole about reference frames or relative velocity or time dilation. We call the number of holes in the piece of paper "the proper time along the path from A to B", and it is a fact that has nothing to do with any other observers and their notions of time, distance and speed.

(The twin paradox can be summarized by saying that when the twins rejoin they compare their pieces of paper and find that there are more holes punched in stay-at-home's than in traveller's. This is because they took different paths through spacetime between the departure and reunion events).

Worldlines. An easy concept to grasp. The path of a particle through spacetime - but in what reference frame? For surely any worldline measured from the frame of that particle (or clock if you prefer) is a fixed location (t,0,0,0) moving through time... and it will have a different path for every other observer... ?
Where the two different paths are due to the relative movements of the two observers.

The worldline is the same in all frames and for all observers. Consider some object; in some frames it is moving fast, in others it is moving slowly or not at all, even the direction of motion is frame-dependent. But for any given point in spacetime, there are exactly two possibilities: either the worldline of the object passes through that point or it doesn't.

If we assign coordinates using a frame in which the object is at rest we'll label the points that the worldline passes through (t,0,0,0) and if we use a frame in which the object is moving these points might be labeled (t',vt',0,0), but they're the same points and the same worldline either way. When we change frames we're changing the axes of the coordinate system we're using to assign coordinates to points, but this doesn't change the points themselves.

 

[Ibix]

Surely special relativity is reciprocal - 'As A is to B, so B is to A' viewed from each of the two frames of reference? So each twin measure the other to age more?

While both twins are under the same conditions, you can make this argument. But one of the twins turns around. Why should there be reciprocity when the twins are doing different things?

Yes, OK, but proper time can only be a time and not a distance! Yes it is the difference in time measured between two events in Spacetime, but, the time is what would be measured by the observer holding that clock and also traveling between those Spacetime events.

You misunderstand me. Proper time (multiplied by c if you insist) is the spacetime equivalent of distance in space (it's called the "interval"). If I drop a ruler on a piece of graph paper its length will always be 30cm, but it's x-extent and its y-extent may vary. Similarly, proper time is the "length" of a clock's path through spacetime - whether that path is aligned with the time axis of the imaginary graph paper we call a reference frame or not.

That is, at the core, the resolution of the twin paradox. The twins took different routes through spacetime. They can have different intervals, just like two different routes through space can have different lengths.

 

[Orodruin]

And this

That is, at the core, the resolution of the twin paradox. The twins took different routes through spacetime. They can have different intervals, just like two different routes through space can have different lengths.

Is essentially the basic point of this

No this is wrong. By the same reasoning in Euclidean space, you would conclude that the hypothenuse would have the same length as the sum of the length of the catheti in a right triangle.

 

[Dale]

The path of a particle through spacetime - but in what reference frame?

In any reference frame. It is invariant.

For surely any worldline measured from the frame of that particle (or clock if you prefer) is a fixed location (t,0,0,0) moving through time... and it will have a different path for every other observer... ?

The worldline is the geometric figure itself, irrespective of the coordinate system that you might use for describing it. One worldline will have different coordinates in different coordinate systems but it is the same worldline.

If you have experience with hiking with a map and a compass (pre-GPS), then this might make sense: a mountain peak might have different coordinates using magnetic north and true north, but it is the same point on the map and the same mountain peak in reality.

The reference frame is the coordinates and is just a convenient way to label points on the map.

The worldline measured by a remote observer from another frame, is in effect the coordinate interval,

No, the worldline is more basic than a reference frame. It is not necessary to define a reference frame in order to have a valid worldline.

You can draw a line on a blank piece of paper. The paper doesn't have to have a grid on it. And even without a grid on the paper, you can tell if a line is straight or bent.

 

[Arkalius]

Another way to look at it is by the principle of maximal aging. We have 2 events in spacetime, the departure from Earth, and the return to Earth. Both twins are present at each event. The staying twin's worldline through spacetime between these two events is inertial-- he never accelerates. Thus, his path between these two events is the shortest possible path in spacetime (a single straight line on a Minkowski diagram). The traveler, on the other hand, leaves and comes back. His worldline isn't a straight line, and thus its overall length in spacetime is longer. It includes at least one instance of acceleration. The principle of maximal aging says that the shortest (inertial) path between the two events is the path of maximal aging (it's the path upon which the most proper time passes). This means that the longer your path through spacetime between two events, the less proper time is experienced along that path (approaching a light-like path, where the proper time elapsed approaches 0). Since the traveling twin's path through spacetime is longer than the staying twin's, it involves less proper time.

 

[PeterDonis]

his path between these two events is the shortest possible path in spacetime (a single straight line on a Minkowski diagram)

No, it's the longest. Proper time is the path length through spacetime, and the stay at home twin has longer proper time.

The traveler, on the other hand, leaves and comes back. His worldline isn't a straight line, and thus its overall length in spacetime is longer.

No, shorter. See above.

The principle of maximal aging says that the shortest (inertial) path between the two events is the path of maximal aging

The inertial path is the longest; proper time = path length. See above.

 

[Arkalius]

No, it's the longest. Proper time is the path length through spacetime, and the stay at home twin has longer proper time.

Well okay... yeah I guess its reversed. Hyperbolic geometry is a little funky in that way. The shortest path seems "longer" in a certain sense.

 

[Orodruin]

Well okay... yeah I guess its reversed. Hyperbolic geometry is a little funky in that way. The shortest path seems "longer" in a certain sense.

There is no "shortest path". You are trying to impose your sense of a Euclidean metric on Minkowski space.

 

[Arkalius]

There is no "shortest path". You are trying to impose your sense of a Euclidean metric on Minkowski space.

Shortest worldline then. My point is, if you look at a minkowski diagram, the lengths of worldlines with lower amounts of proper time get longer (on the diagram) as proper time along them gets shorter between two intersections. I know this apparent length difference doesn't represent any physical reality, but I find it a useful concept to consider because it makes it easier (for me) to quickly estimate which worldlines represent shorter proper times when looking at such diagrams.

 

[Orodruin]

Shortest worldline then.

No, the "length" of the world-line is the proper time. There is no other concept of "length" in Minkowski space (or a general space-time). That you are drawing your Minkowski diagram on a piece of paper with a natural Euclidean metric is irrelevant. Furthermore, in GR (or even just curvilinear coordinates on Minkowski space) this completely depends on your choice of coordinates and what form the metric takes in those coordinates. Given your avatar, you must be aware of this already.

 

[robphy]

Another way to look at it is by the principle of maximal aging. We have 2 events in spacetime, the departure from Earth, and the return to Earth. Both twins are present at each event. The staying twin's worldline through spacetime between these two events is inertial-- he never accelerates. Thus, his path between these two events is the shortest possible path in spacetime (a single straight line on a Minkowski diagram). The traveler, on the other hand, leaves and comes back. His worldline isn't a straight line, and thus its overall length in spacetime is longer. It includes at least one instance of acceleration. The principle of maximal aging says that the shortest (inertial) path between the two events is the path of maximal aging (it's the path upon which the most proper time passes). This means that the longer your path through spacetime between two events, the less proper time is experienced along that path (approaching a light-like path, where the proper time elapsed approaches 0). Since the traveling twin's path through spacetime is longer than the staying twin's, it involves less proper time.

Shortest worldline then. My point is, if you look at a minkowski diagram, the lengths of worldlines with lower amounts of proper time get longer (on the diagram) as proper time along them gets shorter between two intersections. I know this apparent length difference doesn't represent any physical reality, but I find it a useful concept to consider because it makes it easier (for me) to quickly estimate which worldlines represent shorter proper times when looking at such diagrams.

Aside from the error of using "shortest [Euclidean-] length" instead of "longest [Minkowski-]length" (as others have mentioned), the main idea is okay.

A spacetime diagram on rotated graph paper may help quantify the metrical relations on a spacetime diagram because it is easy to draw the ticks of light-clocks carried by various observers.

Here is the Twin Paradox/Clock Effect for two non-inertial travelers (with there-and-back speeds of 3/5 and 4/5) and the inertial stay-at-home twin.
You can verify that the time-dilation factors $\gamma=\frac{1}{\sqrt{1-v^2}}$ are 5/4 and 5/3, respectively, which could be read off the diagram.
(Note that there are segments of 5 inertial reference-frames shown on this diagram:
Alice, outgoing-Bob, incoming-Bob, outgoing-Carol, incoming-Carol.
You can't easily draw this on two-observer graph paper or hyperbolic graph paper.)

 

[Battlemage!]

This was straying from the point in the original thread, but I thought it made a point...
The stay-at -home twin is at rest in her frame and her clock must therefore measure proper time.
The traveling twin, carries his clock with him; it is therefore at rest in his frame and must also measure proper time.
As each twin is moving relative to the other, they will each measure coordinate time for their twin.
Their proper times will be identical.
Their coordinate times will be identical.
As their relative speeds are the same, their Lorentz transformations will be the same.
When the traveling twin slows on his return and comes to rest in his twin's frame they are both once again in the same frame and will have traveled exactly the same each relative to the other.

It is only if the traveller continues past at speed that there is a time difference as they pass and that will be proper time vs coordinate time...

I found this article an interesting example showing why one twin must age less.

https://www.scientificamerican.com/article/how-does-relativity-theor/

Edit- it is an algebraic way of showing what robphy's post shows. In other words, actually working out the light signals/ticks shows it pretty clearly.

 

[Arkalius]

That you are drawing your Minkowski diagram on a piece of paper with a natural Euclidean metric is irrelevant.

It's not irrelevant to me. I have to look at the Euclidian representation of the graph. Knowing that longer Euclidian length = shorter worldline length is useful to me. In fact, what I was trying to point out is how the Euclidian length on a minkowski diagram has, in effect, the opposite meaning to what one's intuition would suggest.

 

[PeterDonis]

what I was trying to point out is how the Euclidian length on a minkowski diagram has, in effect, the opposite meaning to what one's intuition would suggest

As you have hopefully realized by now, using "length" unqualified, in this context, to mean "Euclidean length as it appears on the diagram", rather than "Minkowski length, i.e., the length that is physically real", does not work well. Whatever your personal heuristics might be, the word "length" in this context (along with all related words like "longest", "shortest", etc.) has the latter meaning, and you need to use it with that meaning if you want to be understood.

 

[Arkalius]

Whatever your personal heuristics might be, the word "length" in this context (along with all related words like "longest", "shortest", etc.) has the latter meaning, and you need to use it with that meaning if you want to be understood.

If I was talking to people who already knew the subject, sure. The way I'm approaching it here reflects how I moved from being confused to understanding it, which I'm hoping will help someone else who is confused. I'm a firm believer in teaching by exploring the misconceptions first to understand why they're wrong. People tend not to fully retain merely correct explanations that don't address their currently flawed (unbeknownst to them) misunderstandings.

 

[Orodruin]

People tend not to fully retain merely correct explanations that don't address their currently flawed (unbeknownst to them) misunderstandings.

So you want to remove one misconception by introducing another. I assume we have already agreed that your heuristic only works in the very special case of Minkowski coordinates and that in other coordinates (such as the Penrose coordinates in your avatar) this will lead people astray? In Minkowski coordinates, the point is not that it is the shortest Euclidean distance when you draw it on a paper, the point is that it is a straight line and a straight line satisfies the geodesic equations both in the Euclidean and the Minkowski case.

 

[PeterDonis]

I'm a firm believer in teaching by exploring the misconceptions first to understand why they're wrong.

You're assuming that everyone who isn't already familiar with the subject will share your own misconception. That's not a good assumption. In this thread, in particular, I see no evidence that the OP's misconception was the one you describe. He was confused about simultaneity, not about Euclidean vs. Minkowski geometry.

Also, the way you started out trying to "correct" the misconception was itself incorrect. In your first post in this thread, you used the phrase "the shortest path through spacetime". That is not correcting a misconception about Euclidean vs. Minkowski geometry. It is repeating the misconception. So even if the OP had been confused about that point, you would not have unconfused him; you would only have reinforced his confusion. Even for your next few posts, you continued to state things in a way that would reinforce rather than dispel the particular confusion you say you were trying to fix.

Finally, Orodruin's point is quite valid: your heuristic only works for the specific case of Minkowski coordinates in flat spacetime. (And only for timelike worldlines at that.) So you'll have to unlearn it anyway as soon as you go to either curvilinear coordinates or curved spacetime.

 

[Arkalius]

I see no evidence that the OP's misconception was the one you describe. He was confused about simultaneity, not about Euclidean vs. Minkowski geometry.

He also seemed to have some misconceptions about the concept of proper time. That's kind of what I was trying to address.

Also, the way you started out trying to "correct" the misconception was itself incorrect. In your first post in this thread, you used the phrase "the shortest path through spacetime". That is not correcting a misconception about Euclidean vs. Minkowski geometry. It is repeating the misconception. So even if the OP had been confused about that point, you would not have unconfused him; you would only have reinforced his confusion. Even for your next few posts, you continued to state things in a way that would reinforce rather than dispel the particular confusion you say you were trying to fix.

Yes, I will concede my initial post wasn't worded well for its purpose, and perhaps too narrowly focused in its context. My subsequent posts were to explain where I was coming from, but I certainly could have done better in the initial post.

Finally, Orodruin's point is quite valid: your heuristic only works for the specific case of Minkowski coordinates in flat spacetime. (And only for timelike worldlines at that.) So you'll have to unlearn it anyway as soon as you go to either curvilinear coordinates or curved spacetime.

Sure, but the original post was about the twin paradox which is typically a scenario that assumes a flat spacetime.

 

[Battlemage!]

No, the "length" of the world-line is the proper time. There is no other concept of "length" in Minkowski space (or a general space-time). That you are drawing your Minkowski diagram on a piece of paper with a natural Euclidean metric is irrelevant. Furthermore, in GR (or even just curvilinear coordinates on Minkowski space) this completely depends on your choice of coordinates and what form the metric takes in those coordinates. Given your avatar, you must be aware of this already.

On a slightly off topic tangent, isn't mc² also the "length" of the energy/momentum equation? I.e.

$mc^2 = \sqrt {E^2 - (pc)^2}$

I wonder because I'm pretty sure mc² shares the same property of proper time, that it is universally agreed upon by all inertial observers, and time and energy tend to have a very intense romantic relationship.

 

[robphy]

On a slightly off topic tangent, isn't mc² also the "length" of the energy/momentum equation? I.e.

$mc^2 = \sqrt {E^2 - (pc)^2}$

I wonder because I'm pretty sure mc² shares the same property of proper time, that it is universally agreed upon by all inertial observers, and time and energy tend to have a very intense romantic relationship.

It is.
So, in fact, one can draw energy-momentum diagrams of collisions (analogous to spacetime-diagrams).
From my Insight, https://www.physicsforums.com/insights/relativity-rotated-graph-paper/ , here is an elastic collision of two particles.
[edit: added italicized words to following sentence]
Note the analogy of a totally-inelastic collision to the clock-effect... although I didn't draw in the "mass-diamonds" to describe the invariant-mass of the system for the center-of-momentum frame--that is, suppose instead of the elastic collision, we had the incident particles collide totally-inelastically to form a single particle in the center-of-momentum frame. That single particle has rest-mass $\sqrt{800}\approx 28.28 > 12+8$.

 

[Grimble]

Think of proper time as something that we observe: Say we design our clock so that every time it ticks it punches a hole in a piece of paper somewhere inside; we start with a fresh piece of paper at event A and remove it at event B. How many holes are there in the piece of paper? That's a simple direct observation; all observers everywhere will agree about the answer without any rigamarole about reference frames or relative velocity or time dilation. We call the number of holes in the piece of paper "the proper time along the path from A to B", and it is a fact that has nothing to do with any other observers and their notions of time, distance and speed.

Why not just count the 'ticks' of the cloak: the time read from the clock face or display?

 

[Nugatory]

Why not just count the 'ticks' of the cloak: the time read from the clock face or display?

Of course that works too.

 

[Grimble]

Perhaps, gentle folk, it would be useful to correct my 'misunderstanding' of proper time?

As I have understood it, proper time is the time measured by a clock between two events on that clock's world line.
Suppose another clock takes a different path between those same two events, then the second clock (that of the traveller for example) would read a different time at event 2. Clock 2's world line also passes through events 1 & 2. What then of clock 2's proper time? Can it also be the time read on its clock? For then we have two different proper times between the same two events and I understood that proper time was invariant...?

Is the proper time interval not the same as the Spacetime interval?