# Is this a paradox in Special Relativity?

In the Earth Frame, I am at rest with respect to my ruler, so I observe no length contraction. My clock is not at rest with respect to the muon, so I do observe time dilation.
In the Earth frame the muon takes about 30 ms to reach sea level. (Longer time interval)
In the Muon Frame, I am not at rest with respect to the ruler (Earth), so there is length contraction. My clock is at rest with respect to me, so I observe no time dilation.
In the muon frame the trip takes about 6 ms. (Shorter time interval)

The Earth frame considers the interval measured in the muon frame to be time dilated (shorter) than the interval measured in the Earth frame.

THe Earth/muon experiment can not be directly equated with the twin paradox, because if we ignore gravitational time dilation then both the Earth frame and the muon frame are essentially inertial while in the twin experiment one of the twins is non-inertial.

However, it is said they run slowly and in the case of the Twin Paradox the Moving twin is younger. Why is it said it runs slowly when it yields a larger time interval?
This is not correct. The interval measured by the moving twin is shorter.

I'm quoting the book,

In all cases, the actual time interval on a moving clock is greater than the proper time as measured on a clock at rest.
In the muon decay, the event would be the muon's journey down the arbitrary ruler and so you would assign the muon frame the proper time (shorter interval). The moving clock would be on the 0.98c Earth and would yield the larger interval.

In the Twin Paradox you would identify the event as being the journey of the spaceship and assign it the proper time or the shorter interval and the moving Earth would again get the larger interval of time since the Moving twin does actually experience the accelerations thus breaking the symmetry.

... However, it is said they run slowly and in the case of the Twin Paradox the Moving twin is younger. Why is it said it runs slowly when it yields a larger time interval? ...
I take it from your last post that you now agree that the clock of the the moving twin runs slower and yields a shorter time interval?
In the Twin Paradox you would identify the event as being the journey of the spaceship and assign it the proper time or the shorter interval and the moving Earth would again get the larger interval ...

I take it from your last post that you now agree that the clock of the the moving twin runs slower and yields a shorter time interval?
If I properly identified the correct event in the right reference frame, yes.

Hello Shackleford.

In an earlier post you describe the spaceship journey as an event, which, in normal language usage it is. However in Relativity the word event has a specific meaning. To identify the location of an object at a certain place at a certian time relative to the three spatial and one coordinate axes of sapcetime requires four numbers. This spacetime location at a certain place at a certian time determined by this set of numbers is an event. It has no spatial or temporal extension. A journey is a path in spacetime which is made up of a series of events which together constitute the wordline of an object, be it a spaceship, clock etc. The reading on a clock travelling along this worldline is a measure of the spacetime path, necessarily timelike, traversed and is called proper time. This may seem a bit long-winded but it is important for the answer to your question.

If we take the travelling twins path, as is usual in the twin scenario, as cosisting of two inertial paths out and back with instantaneous acceleration at the turn around (ignoring initial and final stages of the journey), during these two inertial phases each twin will observe (not see) the others clock running slow compared to his own. But at the turnaround, the traveller, and only the traveller, chages to a different inertial frame. This inertial frame's line of simultaneity meets the stay at home twin's coordinate time axis at a later coordinate time than the line of simultanety of the frame of the traveller before turnaround. So the traveller is now in a frame where the stay at home has aged somewhat more than before the turnaround. The rest of the journey home (ignoring the final deceleration) is again symmetrical inasmuch as each observes the other's clock to be running relatively slower than his own.

Now if we want it to be more realistic and not have an instantaneous turnaroud and also to take into account the initial and final accelerations at parting and reuniting we will have some parts of the journey following a curved path in spacetime due to accelerated, non inertial, motion. To measure proper time along this path it is neccessary to add all the infinitessimally small inertial frames that the ship is at rest in during this time. With the traveller at rest in any of these frames the situation would again be symmetrical as regards clock rates but the line of simultaneity of the traveller meets the stay at home's time axis at continually later times as he prgresses from one such small inertial portion of his worldline to the next. The addition of these small spacetime path lengths, proper time, as the path length of the individual inertial portions tend to zero, is achieved by integration of proper time along the world line.

Now all this window dressing can be avoided by simply stating that the clock accompanying the twin who traverses the longest spacetime path length records the shorter proper time. The inertial spacetime path between two events is always the shortest spacetime path and a clock travelling along it will record more proper time than a clock traversing any other spacetime path. The stay at home twin follows such an inertial spacetime path. No matter how you pose the scenario, both twins travelling etc. the shortest spacetime path will give the longest proper time. Clock rates can be, in some cases, just an unnecessary complication and the cause of much confusion.

As regards biological ageing, it is only another physical process subject, as any other, to the laws of nature and so what can be said about clocks can be said about biological mechanisms.

There are many finer points and loose ends to be tied up, as always, but the above, as I see it, are the basics.

Matheinste.

jtbell
Mentor
Okay, I busted out the textbook and perused the SR chapter a bit, particularly the muon experiment.

In the Earth frame the muon takes about 30 ms to reach sea level. (Longer time interval)

In the muon frame the trip takes about 6 ms. (Shorter time interval)
Let's apply relativity of simultaneity here. Suppose there are two clocks at rest in the Earth's reference frame, one at the muon's starting point and one on the ground; and a clock that rides along with the muon (and a hypothetical observer who rides along with them). We set the clocks so that in the Earth's reference frame, all three read zero when the muon passes its starting point.

In the Earth's reference frame, the muon takes 30 ms to reach the ground, so both the clock at the starting point and the clock on the ground read 30 ms when the muon reaches the ground. Because of time dilation, the clock on the muon reads 6 ms when it passes the clock on the ground.

In the muon's reference frame, 6 ms elapse on its clock during the trip, of course. In this frame the other two clocks are time-dilated, so only about [STRIKE]0.17 ms[/STRIKE] 1.2 ms elapse on those clocks. (The time-dilation ratio is the same in both cases: 1/5 = 6/30 = [STRIKE]0.17/6[/STRIKE] 1.2/6).

In the muon's reference frame, the two "Earth frame clocks" are also not synchronized. At the start of the trip, the clock at the starting point reads 0, and the clock on the ground reads about [STRIKE]21.83 ms[/STRIKE] 20.8 ms. During the trip, [STRIKE]0.17 ms [/STRIKE] 1.2 ms elapse on both of these clocks. At the end of the trip, the clock at the starting point reads [STRIKE]0.17 ms[/STRIKE] 1.2 ms, and the clock on the ground reads 22 ms.

So both an "Earth observer" and the "muon observer" agree that when the trip starts, the clock at the starting point and the clock on the muon both read 0; and when the trip ends, the clock on the ground reads 22 ms and the clock on the muon reads 6 ms. However, they disagree about whether the clock at the starting point and the clock on the ground are synchronized.

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Hello Shackleford.

In an earlier post you describe the spaceship journey as an event, which, in normal language usage it is. However in Relativity the word event has a specific meaning. To identify the location of an object at a certain place at a certian time relative to the three spatial and one coordinate axes of sapcetime requires four numbers. This spacetime location at a certain place at a certian time determined by this set of numbers is an event. It has no spatial or temporal extension. A journey is a path in spacetime which is made up of a series of events which together constitute the wordline of an object, be it a spaceship, clock etc. The reading on a clock travelling along this worldline is a measure of the spacetime path, necessarily timelike, traversed and is called proper time. This may seem a bit long-winded but it is important for the answer to your question.
Thanks for clearing up the vernacular. An event is basically a well-defined four-"dimensional" point. A journey is a path comprised of all these space-time "points" from one event to the next.

If we take the travelling twins path, as is usual in the twin scenario, as cosisting of two inertial paths out and back with instantaneous acceleration at the turn around (ignoring initial and final stages of the journey), during these two inertial phases each twin will observe (not see) the others clock running slow compared to his own. But at the turnaround, the traveller, and only the traveller, chages to a different inertial frame. This inertial frame's line of simultaneity meets the stay at home twin's coordinate time axis at a later coordinate time than the line of simultanety of the frame of the traveller before turnaround. So the traveller is now in a frame where the stay at home has aged somewhat more than before the turnaround. The rest of the journey home (ignoring the final deceleration) is again symmetrical inasmuch as each observes the other's clock to be running relatively slower than his own.
So, my original assertion was partly correct insofar as I was looking only at the comparison of the inertial frames (inertial phase). However, my analysis was incomplete. I didn't take into account the acceleration, or non-inertial phase, of the moving frame which breaks any symmetry in the simplified case. Why does the line of simultaneity change when the traveler changes to another inertial frame? Is this where the diagram is helpful?

Now if we want it to be more realistic and not have an instantaneous turnaroud and also to take into account the initial and final accelerations at parting and reuniting we will have some parts of the journey following a curved path in spacetime due to accelerated, non inertial, motion. To measure proper time along this path it is neccessary to add all the infinitessimally small inertial frames that the ship is at rest in during this time. With the traveller at rest in any of these frames the situation would again be symmetrical as regards clock rates but the line of simultaneity of the traveller meets the stay at home's time axis at continually later times as he prgresses from one such small inertial portion of his worldline to the next. The addition of these small spacetime path lengths, proper time, as the path length of the individual inertial portions tend to zero, is achieved by integration of proper time along the world line.

Now all this window dressing can be avoided by simply stating that the clock accompanying the twin who traverses the longest spacetime path length records the shorter proper time. The inertial spacetime path between two events is always the shortest spacetime path and a clock travelling along it will record more proper time than a clock traversing any other spacetime path. The stay at home twin follows such an inertial spacetime path. No matter how you pose the scenario, both twins travelling etc. the shortest spacetime path will give the longest proper time. Clock rates can be, in some cases, just an unnecessary complication and the cause of much confusion.

As regards biological ageing, it is only another physical process subject, as any other, to the laws of nature and so what can be said about clocks can be said about biological mechanisms.

There are many finer points and loose ends to be tied up, as always, but the above, as I see it, are the basics.

Matheinste.
That last part seems counter-intuitive. The twin with the longer space-time path yields a shorter proper time, and the twin with the inertial space-time path, which is the shorter space-time path, yields a larger proper time.

Does this even make sense? Let's say you're comparing a stationary frame and a moving frame. A slower-moving object will "accumulate" more time at each event along a space-time path, but a faster-moving object will "accumulate" less time at each event along a space-time path.

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That last part seems counter-intuitive. The twin with the longer space-time path yields a shorter proper time, and the twin with the inertial space-time path, which is the shorter space-time path, yields a larger proper time.
You are right to think that it seems counter-intuitive, but the statement by matheinst is correct. Consider two cars moving travelling from A to B. Both start at A at the same time and both arrive at B at the same time, but one takes the straight path and the other takes a long winding path. The car that takes the long winding route has to travel faster and as a consequaence experiences more time dialation and less proper time. This is not not an exact analogue. The cars in this example are travelling a path through 3D space rather than 4D spacetime, but the idea is similar.

In the muon's reference frame, 6 ms elapse on its clock during the trip, of course. In this frame the other two clocks are time-dilated, so only about 0.17 ms elapse on those clocks. (The time-dilation ratio is the same in both cases: 1/5 = 6/30 = 0.17/6).
Did you mean 1.2 ms?

You are right to think that it seems counter-intuitive, but the statement by matheinst is correct. Consider two cars moving travelling from A to B. Both start at A at the same time and both arrive at B at the same time, but one takes the straight path and the other takes a long winding path. The car that takes the long winding route has to travel faster and as a consequaence experiences more time dialation and less proper time.
Hmm. So, is my "accumulation" analogy kind of close?

Fredrik
Staff Emeritus
Gold Member
Thanks for clearing up the vernacular. An event is basically a well-defined four-"dimensional" point. A journey is a path comprised of all these space-time "points" from one event to the next.
I don't see the point of using the word "journey" like this. I would call that curve the object's "world line", and use words like "journey" only when I'm talking about the corresponding sequence of events in the real world.

Why does the line of simultaneity change when the traveler changes to another inertial frame? Is this where the diagram is helpful?
It changes in that particular way only when we have chosen to use the comoving inertial frames. (That's not the only possible choice, but we don't need to get into that now). The comoving inertial frame is constructed by applying the synchronization procedure that I described briefly in my previous post, to the tangent of your world line at some point on it. How to do this is one of the first things you learn when you study spacetime diagrams, which by the way is by far the best and easiest way to learn SR.

See this post for my standard reply (including a spacetime diagram) to questions about the twin paradox. There are good contributions from other posters as well in that thread.

That last part seems counter-intuitive. The twin with the longer space-time path yields a shorter proper time, and the twin with the inertial space-time path, which is the shorter space-time path, yields a larger proper time.
Don't forget that the "length of the spacetime path" is defined as the proper time of the curve. It's actually a useless concept. It's better to just use the term proper time.

Edit: Now I see that you actually said "shorter space-time path" = "larger proper time". I guess what you (and matheinsteine) meant by "length of the spacetime path" is just the length of the curve in a diagram, using the Euclidean concept of distance and length. (I recommend that you don't use such terminology. People like me tend to assume that we're talking about Lorentzian geometry, not Euclidean, when "spacetime" is mentioned).

Does this even make sense? [...] A slower-moving object will "accumulate" more time at each event along a space-time path, but a faster-moving object will "accumulate" less time at each event along a space-time path.
That's not the prettiest way of saying it, but yeah, it makes sense, because expressed in the coordinates of an inrtial frame, the proper time is the integral of $$\sqrt{dt^2-dx^2-dy^2-dz^2}$$ along the curve. A line parallel to the time axis (which represents zero velocity) has dx=dy=dz=0 everywhere, so every other curve between two points on that line that's parallel with the time axis has to have a shorter proper time, because of the negative contribution from every little movement in space. (Higher speed=bigger dx,dy,dz for a given dt).

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So, it's critical to make the observation that the observer in the Fixed Frame never leaves his respective inertial frame, but the observer in the Moving frame does in fact change inertial frames. Because of those non-inertial frames/phases, it's valid to put the relativistic effect/corrections on the Moving frame and look at it from the "fixed" Fixed frame, i.e. the Moving frame is the moving clock which runs slowly and yields the shorter time interval. You then also have to make corrections for length contraction.

I don't see the point of using the word "journey" like this. I would call that curve the object's "world line", and use words like "journey" only when I'm talking about the corresponding sequence of events in the real world.
Is that not what a journey is? That's how I'm looking at it - a series of infinitesimally-small events that constitute the curve. Any point on the world line or curve is an event at a specific point in space and time. There's just a bunch of them depending on how you sub-divide the time interval along that curve.

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Fredrik
Staff Emeritus
Gold Member
So, it's critical to make the observation that the observer in the Fixed Frame never leaves his respective inertial frame, but the observer in the Moving frame does in fact change inertial frames.
I'm not a fan of phrases like "leaves his inertial frame". I recommend that you never say that an object is "in" an inertial frame or that it "changes" inertial frames. Instead, say that it's stationary in an inertial frame and that it changes velocity in that same inertial frame. If it changes velocity, it now has a different comoving inertial frame.

Inertial frames are just global coordinate systems. A global coordinate system is a function from spacetime into $$\mathbb R^4$$. So the inertial frames are just functions that assign coordinates to events. You can't be "in" one, but you have a velocity in all of them.

Is that not what a journey is? That's how I'm looking at it - a series of infinitesimally-small events that constitute the curve.
I'm just saying that if you travel to South Africa, that would be a journey (unless you happen to live there). I don't see a reason to use the word "journey" for the mathematical representation of that journey when the we already have a standard term for it: "world line".

jtbell
Mentor
Did you mean 1.2 ms?
Ugh, you're right. I don't remember how I could have possibly come up with 0.17 ms this morning. I should have waited until after I had my quota of coffee. I've corrected it and the other related numbers. Thanks for catching that.

Start: Two clocks sitting next to each other on earth are synchronized.
Next: One clock is put into a rocket and launched on a high-speed trip to Alpha Centuari and back.
Finish: The two clocks are sitting next to each other again on earth.

Which is ahead and which is behind at the finish?
The rocket clock is behind (less time has elapsed on it) upon its return. On the way out to Alpha Centauri the rocket occupant can measure the rate of the earth clock (using the clocks distributed in his own rest frame) and will deduce that the earth clock runs more slowly than those at rest in his own rest frame. But when he changes rest frames upon reaching Alpha Centauri, he will admit that the clocks in his initial rest frame are actually not synchronized with one another, etc., and that it is actually his clock that has been running slowly during the trip. The same situation repeats on his way back to Earth. By way of a real example, short half-life mesons last much longer (don't decay) when shot around the loop of an accelerator.