Is the Duel on the Moving Train Fair in Relativistic Terms?

[atyy]

So “universal agreement” or “frame-invariance” is a characteristic that does not express well enough in what sense proper time, for instance, is qualified.

We assume that standard ideal clocks exist. These are clocks that when located at the same point in spacetime tick at the same rate. After they have been separated and read differently, we can always bring them back together to another point in spacetime and check if they still tick at the same rate. In this way we will know whether a particular real clock is standard and ideal. (This is just to start. There are some other requirements too)

The accumulated proper time along any path (no matter how short or long) in spacetime is the total number of ticks a standard ideal clock makes when it takes that path. Each tick is an event. So the proper time is related to events.

 

[Saw]

Actually, it's the other way round. The inertial Earth takes the direct "taut" route; the ship, which is not inertial for at least part of its journey, takes the indirect "slack" route. But in the topsy-turvy geometry of spacetime, straight lines are longer than bent or curved lines, unlike the normal Euclidean geometry we are familiar with.

So my analogy isn't perfect. It demonstrates how two measurements can differ, but it gets it wrong which is the longest.

The reason why in spacetime straight worldlines have longer length (i.e. more proper time τ) than bent lines is because there is minus sign in

dτ² = dt² − dx² / c²​

whereas, in Euclidean geometry, length is given by the Pythagoras formula with a plus sign

dr² = dx² + dy²​

Thanks. Well, I realize now we were talking about different things.

You were talking about a twin paradox scenario, where the Earth is inertial all the time, whereas the ship accelerates away and accelerates back to return to the Earth. Yes, in that case the Earth takes the direct "taut" route while the ship takes the indirect "slack" route through the two legs of its trip.

Instead my example was a "pure-blooded" SR case with no acceleration (and hence, I understand, analyzable with pure Euclidean geometry) :

a clock in the ship is synchronized with a clock on the Earth when they meet; the ship moves at v = 0.5 c wrt the Earth; then both frames synchronize clocks with their respective networks of assistants; when the ship reaches position x2 = 1 ls at t2 = 2 s in the Earth’s frame, will the ship’s clock read more or less time? The answer is less time and in particular t2’ = 1.732 s.

But the example is more interesting if you include a moun in it, as we were doing in the latest posts. Imagine that there is a lab in the ship and that a muon is artificially created in the ship's lab when the latter meets the Earth observer. Imagine as well that a muon's lifetime in its rest frame is, arbitrarily, for convenience, 1.8 s. Will the muon be still "alive" when it arrives at position x2 = 1 ls at t2 = 2s in the Earth frame?

Well, the detailed arguments given above, which JesseM is patiently commenting, are that:

- Given the reality under consideration (a muon is decaying), the ticks of the ship's clock accompanying such muon must directly be, if the principle of relativity is true, a "mirror" of what happens with the muon.

- Instead, the reading of the clock of the Earth frame that waits for the muon at x2 depends on two things: the Einstein convention for synchronization (which in turn also depends on the tick rate of the Earth's clock) and its own tick rate. The Einstein's convention has determined that the clock at x2 will be set at, let us say, 0 s when it received the synch light signal and later on 1 s is added to it when it is learned that such synch signal has taken 2 s to return to the Earth. This 1 s that the light signal is deemed to have taken for the trip to the clock at x2 is written as a reminder in the landmark for that clock in the Earth frame, with the label "x2 = 1 ls". When the Earth wants to solve the problem, in my opinion, the Earth combines the two values, coordinate time = dt (2 s) and rest length = dx (1 ls), in the formula sqrt (2^2 - 1^2) in order to get (this "in order to" is objected by JesseM) 1.732 s, which is the proper time of the muon. That means the muon survives. So the Earth has reached the same solution but through an indirect path that combines two simultaneity-dependent values, dt and dx. Thus the Earth values dt and dx also "mirror" what happens with the muon but only if put together in the right equation.

Geometrically speaking, I thought: if you look at the spacetime diagram of the Earth frame, there is a right angle triangle where (i) the base is the rest length, dx = 1 ls; (ii) the height is the coordinate time, dt = 2s and the hypothenuse is the wordline of the muon. If you now change the place of the dt value and you attribute this value to the hypothenuse (the muon's wordline), you can solve for the value of the height (now vacant) and you get sqrt (2^2 - 1^2) = 1.732 s. Intuitively, this looks like if the Earth were admitting that it has reached the solution by moving farther away through the hypothenuse and coming back through the base, or vice versa... But I accept in advance that this looks like an ad hoc construction and is speculation...

 

[JesseM]

For example: a clock in the ship is synchronized with a clock on the Earth when they meet; the ship moves at v = 0.5 c wrt the Earth; then both frames synchronize clocks with their respective networks of assistants; when the ship reaches position x2 = 1 ls at t2 = 2 s in the Earth’s frame, will the ship’s clock read more or less time? The answer is less time and in particular t2’ = 1.732 s.

Well, "less time" according to the Earth frame's definition of simultaneity. Suppose there is some physical marker at rest at position x=1 in the Earth's frame. In the ship's frame, the event of the marker whizzing by the ship coincides with the event of the ship's clock reading 1.732 seconds. But in the ship's frame these events are not simultaneous with the event of the Earth's clock reading 2 s; instead, they are simultaneous with the event of the Earth's clock reading 1.5 s, so in the ship's frame the Earth's clock reads less time at the moment the ship passes this marker. Of course, if the marker has its own clock which is synchronized with the Earth's clock in the Earth's rest frame, then it's an objective fact that when the ship passes the marker, the ship's clock reads 1.732 s and the marker's clock reads 2 s (but in the ship's frame, the marker's clock already read 0.5 s at the moment the ship was passing next to Earth).

The methods for reaching this solution can be two:

- A direct or “straight path” or “taut chord” expedient is looking at the ship’s clock and comparing it with the Earth’s clock located at that place.

- An indirect or “bent path” or “slack chord” expedient is looking at the x and t measurements of the Earth frame, namely dt = 2s and dx = 1 ls, combine them in the formula based on the Pythagorean Theorem and thus guess that the proper time of the ship’s clock = sqrt(2^2 – 1^2) = sqrt(3) = 1.732 s, less than the Earth's coordinate time = 2 s.

The formula for proper time isn't really based on the Pythagorean Theorem, although it's the spacetime analogy for the Pythagorean Theorem.

But why is the Earth’s frame forced to take the indirect or “scenic” route while the ship’s frame has the privilege of using the direct route?

What do you mean by that? In the ship's frame, how would you calculate the time on the marker's clock "directly"? It seems to me you'd have to start with the fact that the marker's clock showed a reading of 0.5 s at t'=0 in the ship's frame, then calculate the proper time elapsed on the marker's worldline between t'=0 and the event of it reaching the ship, and add that to the initial 0.5 s.

It is so because the problem we have posed ourselves is related to what happens to the ship’s clock between the events of its meetings with two spatially separated Earth’s clocks. Thus the ship’s measurement does not have to rely on the concept of simultaneity, while the Earth’s does.

The ship needs to use the concept of simultaneity if it wants to predict the time elapsed on the Earth clock or the marker clock though.

In fact, if we had posed ourselves a different problem, where the question is if an Earth’s clock reads more or less time than the clock of a second ship coming behind in perfect formation with the first ship, then the roles of the ships and the Earth would be exchanged and the ships’ measurements would be the indirect route while the Earth’s measurement would be the straight route.

In this case, when calculating things in the Earth frame you'd need to know the initial time on the second ship's clock at t=0, and the elapsed time on the second ship's clock between that time and the moment the second ship reached the Earth.

 

[Saw]

JesseM, I agree with the numbers in your latest post and of course with the idea that each frame finds the other suffers TD and LC and each frame has a different view on simultaneity. But we had reached an agreement on how those concepts play in the definition and resolution of a particular practical, real-life problem, which I would not like to lose.

The problem was first described in post #27 and refined in successive posts. Unfortunately then in post #46 I placed here an answer that was intended for another thread and used for that purpose a simplified version of the problem, which (I think) has made us lose perspective. My fault.

I’ll try to summarise the conclusions reached so far, so as to introduce the reply to other posters:

- The question or problem was: will the Blue Muon, created at Event 1 (collision with the red upper atmosphere) “survive” to be “present” at Event 2 (collision with the Red Lab)?

To be noted: the definition of the question is frame-invariant, not only in the sense that its phrasing does not choose a perspective about who moves and who is stationary, but also in the sense (which is more important) that it relies on single events, on whose “happening” all frames agree, even if they assign different coordinates to them; the question is NOT whether two spatially separated events are simultaneous or not, although that plays a part in the resolution of the problem, as explained below.

- The answer or solution is: yes, the muon “survives”.

To be noted: all frames agree on this answer.

- The methods for finding the answer or solution:

* If we know that the life expectancy of a Red Muon at rest in the Red Lab is “dt” and we know that the the proper time of the Blue Muon as measured by a blue clock is less than “dt”, then we know the answer is yes.

The reason is the principle of relativity: relative motion does not change the results of experiments = if it happened to affect the muon, it would affect the clock at rest with it in the same manner, and vice versa.

To be noted: “proper time” is self-sufficient in the resolution of the problem and proper time is a simultaneity-free concept.

* If we know the red dt (coordinate time difference between Event 1 at the birth place of the Blue Muon, which is simultaneous with Event 0 at the Red Lab, and Event 2) and the red dx (the rest length of the red atmosphere), then we also know the answer. We can explain why in two ways: either that (i) by combining these two values in the appropriate formula, we get the proper time of the Blue Muon (my view) or that (ii) we conclude that the Blue Muon’s coordinate time is less than the life expectancy of a muon moving at the given velocity in the Red Frame (an alternative view that for you stands on equal footing).

To be noted: The red dt (coordinate time), which is simultaneity-dependent, is not self-sufficient in the resolution of the problem. Neither is the red dx (rest length). You have to combine the two values to find the answer.

Open points:

Is rest length also a simultaneity-dependent concept?

Is one way of explaining the solution more meaningful in some sense than the other? For example, how do you link your explanation (ii) with the principle of relativity?

How to label “proper time” in view of its capacity for self-sufficiently solving problems like this?

* There are other routes but the same reasoning applies to them.

Well, given this basic agreement (if I summarised correctly), I draw conclusions as to the significance of SR, which in my opinion would avoid many misunderstandings and objections that one often finds. (At least this has served myself to overcome my initial reluctance to accept it.) I have already advanced them, but will try to put them in a simpler manner.

The accumulated proper time along any path (no matter how short or long) in spacetime is the total number of ticks a standard ideal clock makes when it takes that path. Each tick is an event. So the proper time is related to events.

Yes, I agree with that. That is why, since the problem is here about “local events” in JesseM’s terminology, I suggested that the proper time reading of a clock is different (it self-sufficiently solves the problem) because it directly “mirrors” those events, in accordance with the principle of relativity.

An analagous geometrical question is this: There are two rulers at an angle to each other, whose zero marks coincide. Is the 10cm marker on one ruler at a greater vertical height than the other? I hope you will agree that there is insufficient information to answer this question, specifically I haven't said which way is "up".

I agree with that, of course.

Similarly in your question you haven't specified in which frame you want your answer, the earth's, the ship's or someone else's.

That is the question. After some mutual efforts, in this thread we have defined “answer” as the ultimate practical solution to a practical problem. I understand that this “answer” cannot be frame-dependent: it must be the same in the earth’s, in the ship’s or in anybody’s frame. During the calculation process of the answer, certainly, different frames may play with different measurements about what is “relatively up” or what is “relatively simultaneous” or whose clock “relatively dilates”, but this does not lead to disagreement on the final answer to the problem.

 

[DrGreg]

That is the question. After some mutual efforts, in this thread we have defined “answer” as the ultimate practical solution to a practical problem. I understand that this “answer” cannot be frame-dependent: it must be the same in the earth’s, in the ship’s or in anybody’s frame. During the calculation process of the answer, certainly, different frames may play with different measurements about what is “relatively up” or what is “relatively simultaneous” or whose clock “relatively dilates”, but this does not lead to disagreement on the final answer to the problem.

Sorry, I must have been half asleep yesterday. The question I answered was not the question you asked -- I misread it. I withdraw my last post, it is not applicable to what you really said! Sorry.

 

[JesseM]

Sorry to take a few days to reply again, but in this case I see my response can be fairly short since there doesn't seem to be any real disagreement on the main issues:

I’ll try to summarise the conclusions reached so far, so as to introduce the reply to other posters:

- The question or problem was: will the Blue Muon, created at Event 1 (collision with the red upper atmosphere) “survive” to be “present” at Event 2 (collision with the Red Lab)?

To be noted: the definition of the question is frame-invariant, not only in the sense that its phrasing does not choose a perspective about who moves and who is stationary, but also in the sense (which is more important) that it relies on single events, on whose “happening” all frames agree, even if they assign different coordinates to them; the question is NOT whether two spatially separated events are simultaneous or not, although that plays a part in the resolution of the problem, as explained below.

- The answer or solution is: yes, the muon “survives”.

To be noted: all frames agree on this answer.

- The methods for finding the answer or solution:

* If we know that the life expectancy of a Red Muon at rest in the Red Lab is “dt” and we know that the the proper time of the Blue Muon as measured by a blue clock is less than “dt”, then we know the answer is yes.

The reason is the principle of relativity: relative motion does not change the results of experiments = if it happened to affect the muon, it would affect the clock at rest with it in the same manner, and vice versa.

To be noted: “proper time” is self-sufficient in the resolution of the problem and proper time is a simultaneity-free concept.

* If we know the red dt (coordinate time difference between Event 1 at the birth place of the Blue Muon, which is simultaneous with Event 0 at the Red Lab, and Event 2) and the red dx (the rest length of the red atmosphere), then we also know the answer. We can explain why in two ways: either that (i) by combining these two values in the appropriate formula, we get the proper time of the Blue Muon (my view) or that (ii) we conclude that the Blue Muon’s coordinate time is less than the life expectancy of a muon moving at the given velocity in the Red Frame (an alternative view that for you stands on equal footing).

To be noted: The red dt (coordinate time), which is simultaneity-dependent, is not self-sufficient in the resolution of the problem. Neither is the red dx (rest length). You have to combine the two values to find the answer.

Yes, agree with all of this.

Open points:

Is rest length also a simultaneity-dependent concept?

Well, what do you think about my proposal of defining "rest length" in terms of the two-way time for light to get from one end to the other and back, as measured by a clock at the end the light departs from and returns to?

Is one way of explaining the solution more meaningful in some sense than the other? For example, how do you link your explanation (ii) with the principle of relativity?

I guess I would just say that the principle of relativity tells us that the laws of physics work the same way when expressed in the coordinates of any inertial frame, and that one of those laws is that the time on a clock moving at v in a given frame is dilated by the gamma factor, so we can dilate the blue muon's lifetime by the gamma factor and see whether it's longer or shorter than the coordinate time needed for it to reach the surface.

 

[Saw]

As to rest length

Well, what do you think about my proposal of defining "rest length" in terms of the two-way time for light to get from one end to the other and back, as measured by a clock at the end the light departs from and returns to?

This method is the Einstein convention, that is to say, precisely the method used for synchronizing two distant clocks. If you use the same signal to both (i) measure rest length and (ii) synchronize the clock located at the other end of the rod and the signal returns after 2 s as measured by the origin clock, you will conclude that the distant clock is separated from you by 1 ls and that it should read 1 s at the time when it received the signal…

That is why I said that the x value, the separation in length units of a distant clock from the origin of a coordinate system, is like a reminder of the time it took for the synch signal to get there, which is in turn a component of the t value that such distant clock will show later on.

Does all that justify saying that rest length is a “simultaneity-dependent” value? Well, I am not very happy with the expression, because, as noted in another post, the concept of simultaneity is also intertwined with clock rate … In the end, too, I do not see a clear utility for this distinction. It might have one if, as I proposed in other threads, it were true that the length of a given rod, after acceleration (the pure change of frame, leaving aside the temporary physical effect of the acceleration process) had to be “recalibrated” = a new length should be attributed to the rod, just as you would also have to re-synchronize clocks in that frame. The latter (clock re-synchronization) was accepted in one thread, but the former (length re-calibration) seems to be rejected (another thread). If you confirm that is wrong, I’ll leave the idea, then.

As to how to link the principle of relativity with the idea that from the Red Frame you need to combine two values to get the answer to the problem:

I guess I would just say that the principle of relativity tells us that the laws of physics work the same way when expressed in the coordinates of any inertial frame, and that one of those laws is that the time on a clock moving at v in a given frame is dilated by the gamma factor, so we can dilate the blue muon's lifetime by the gamma factor and see whether it's longer or shorter than the coordinate time needed for it to reach the surface.

Yes, that must be the way to put it in technical terms. I was just trying to get a more didactic expression for a non-specialized audience. Do you think the following would be correct?

- A measurement is something that happens in an “instrument”, which mirrors what happens in a “real-life” object we’re interested in (though in standard units).
- The ticks of a clock accompanying the Blue Muon mirror the decay ticks of the same. This would be so both for classical and special relativity.
- But SR notes that relative motion brings about a number of discrepancies between the measurements of different frames (I don't need to repeat them). Thus, for example, the difference between the readings of the two Red clocks that the Blue Muon passes by do not directly mirror what happens to the Blue Muon.
- However, the combined red readings of dt and dx do mirror the combined blue readings of dt and dx, through the right formula. For example, sqrt(red dt^2 – red dx^2) = sqrt(blue dt^2 – blue dx^2) = 1.732 s in this example.
- This is because SR uses a sort of MM-experiment instrument, where two light pulses (or two tennis balls, although that would be less precise and less workable), leaving simultaneously from the origin and traveling perpendicular to each other, always reunite at the origin. That is to say, even if two instruments of this sort give out different values for length/distance (in the X axis, where there is relative motion) and time, they have something in common: same Y length and the light pulse returns to the origin simultaneously in all cases…

I admit all this is very rough, but I would welcome criticism... if it is not so rough that it can't even be criticized!