Fewer seconds or shorter seconds?

[jbriggs444]

no I don't want to, and it has nothing to do with the question discussed here

You are incorrect.

 

[PeroK]

no I don't want to, and it has nothing to do with the question discussed here

Just to clarify things: what is the outstanding question?

 

[Nugatory]

I understand the first part of your sentence :
but not the last part

.. and moving inertially”

Are you asking what "moving inertially" means or why that condition is necessary? It is necessary to exclude perverse setups such as placing the cesium atom and the cycle counting apparatus at different positions in an accelerating spaceship.

 

[Algr]

... is the travel duration shorter because: (1) it contains fewer seconds or (2) the number of seconds is the same for both brothers but the traveler's seconds are shorter?

The brothers will disagree. The brother who traveled will say that the seconds were normal, but there were fewer of them. The stationary brother will perceive the traveler as moving in slow motion, and thus say his seconds were longer.

 

[A.T.]

If you see a clock moving, it is moving relative to you... hard to argue

You can see the same clock as moving or not moving, depending on your own rotation. But that rotation won't affect the clock rates.

 

[jbriggs444]

The brothers will disagree. The brother who traveled will say that the seconds were normal, but there were fewer of them. The stationary brother will perceive the traveler as moving in slow motion, and thus say his seconds were longer.

It is a matter of perspective, of course.

As @Dale and yourself (and I) would view it, taking all of the ticks for the traveling clock and spreading them evenly on the timeline of the stay-at-home clock, there are few remote seconds, each of which is longer. The product comes out to the total stay-at-home elapsed time as it must.

Alternately, one could do the accounting from the traveller's point of view. Spreading the stay-at-home ticks evenly on the traveller's time line, one would find many remote seconds, each of which is extra short. The product comes to the traveller's elapsed time (shorter than the stay-at home elapsed time) as it must.

[You would be correct to be somewhat uncomfortable with this particular accounting from the traveling twin's point of view. The traveling twin is not inertial. There are many choices for non-inertial systems of accounting (coordinate systems or at least "foliations"). There are unexpected behaviors in all of them].

A more standard accounting from the traveller's perspective would start with a very few long ticks of the stay-at-home clock (the outbound trip), a large number of hugely short ticks (the turn-around as the definition of simultaneity sweeps forward across the stay-at-home time line) followed by a very few long ticks of the stay-at-home clock (the return voyage).

 

[Dale]

you precision about the inertial rest frame confuses me: the second is supposed to be the same irrespective of wether you are in free fall or static in a gravitational field, provided the caesium clock is present in your reference frame. That's the only important thing, isn't it?

The BIPM has published a "Mise en Pratique" for each of its units. Here is the one for the second:
https://www.bipm.org/utils/en/pdf/si-mep/SI-App2-second.pdf

It says that "The definition of the second should be understood as the definition of the unit of proper time: it applies in a small spatial domain which shares the motion of the caesium atom used to realize the definition." and further it says "the proper second is obtained after application of the special relativistic correction for the velocity of the atom in the laboratory. It is wrong to correct for the local gravitational field."

I also read that as requiring you to be at rest relative to the atom and physically close to it, but not as requiring either you or the atom to be inertial.

 

[Nugatory]

I also read that as requiring you to be at rest relative to the atom and physically close to it, but not as requiring either you or the atom to be inertial.

I agree with that reading. I also read “small spatial domain” (especially in the context of the previous sentence about negligible non-uniformity of the gravitational field) as implying that the setup can be treated as if gravity is a uniform classical force across the domain; thus there exist frames in which the setup is inertial and at rest. Practically speaking, these are the frames we use for experiments sitting on a lab bench in a lab building except when we are looking specifically for tidal effects.

 

[aperakh]

The problem with your question is the: "are".
There is no "are". That is to say, there is no universal frame of reference by which to decide how long the seconds "really are".
What would you compare a second to as a yardstick to determine how long it is?
The seconds ARE there yardstick. And nothing ever changed sizer relative to itself.
Time, in fact, never actually changes, not in any way that is meaningful to talk about.
It is merely a dimension and what changes is one's perception of the rate at which something is moving through it.

So what what you probably really mean to ask was : "Does the traveling brother perceive there being fewer seconds or does he perceive the seconds as being longer?"

And the answer is: Since there is no meaningful way for him to determine whether his seconds are shorter or longer, he experiences fewer seconds.
That is, his clock will run at such a rate as to count fewer seconds, but they would feel normal to him. While his brother's clock on Earth would seem to run too fast (count shorter seconds).
The brother on earth, watching the moving clock, would also agree that it counted fewer seconds, but to him it would appear as though the moving clock was running slower (ie - counting "longer" seconds).

 

[Nugatory]

The brother on earth, watching the moving clock, would also agree that it counted fewer seconds, but to him it would appear as though the moving clock was running slower (ie - counting "longer" seconds).

That is the conclusion earth-twin might come to (although it would be better to pursue the analogy with distance measurement described earlier in this thread) when they compare clock readings at their reunion.

It is most emphatically not what Earth twin will see if they are watching the moving clock: if the twins are watching each other’s clocks through powerful telescopes or the like, then both will see the other clock running slow on the outbound leg and fast on the inbound leg. When they allow for light travel time between the time the clock ticks and they see the tick, they both calculate the other clock is running slow on both legs.

There is a pretty good explanation in the “Doppler Shift Analysis” section of https://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_paradox.html

 

[aperakh]

It is most emphatically not what Earth twin will see if they are watching the moving clock: if the twins are watching each other’s clocks through powerful telescopes or the like, then both will see the other clock running slow on the outbound leg and fast on the inbound leg.

Opps. Misspoke. BOTH twins will perceive the other one's clock as running slower than their own.
But I don't know where you get that there would be a difference beetween the motion being inbound or outbound.
Time dilation is independent of direction. The only thing that matters are the relative speeds, not velocities. Lawrence transforms are scalar, not vectoral.

 

[PeterDonis]

I don't know where you get that there would be a difference beetween the motion being inbound or outbound.

Because the relativistic Doppler effect, which is what determines what each observer actually sees (as opposed to what they calculate after correcting for light travel time), is different for inbound vs. outbound.

Time dilation is independent of direction.

But the Doppler effect is not.

Lawrence transforms

You mean Lorentz transforms.

 

[Mister T]

That is to say, there is no universal frame of reference by which to decide how long the seconds "really are".

The entire point of SR is that you don't need a universal frame. Any inertial frame will do. As long as the clock is at rest with respect to you it can be used to define the duration of a second.

 

[aperakh]

Because the relativistic Doppler effect, which is what determines what each observer actually sees (as opposed to what they calculate after correcting for light travel time), is different for inbound vs. outbound.

I am afraid you are mistaken. Time dilation is NOT the Doppler effect.
Doppler effect does cause blue and red shifts in light from moving object, but time dilation has nothing to do with it.

You mean Lorentz transforms.

Damn you, auto-correct! lol
Thank you. yes, it is obviously what I meant.

 

[aperakh]

The entire point of SR is that you don't need a universal frame. Any inertial frame will do. As long as the clock is at rest with respect to you it can be used to define the duration of a second.

Yes, he can tell the second for himself (ie - in his frame), which will be different from one observed from another frame.
But there is no universsal frame in which one can say the "real" observation is done. Theya re all equally valid.

 

[Ibix]

Time dilation is NOT the Doppler effect.
Doppler effect does cause blue and red shifts in light from moving object, but time dilation has nothing to do with it.

We are all well aware of this (although it should be noted that the relativistic Doppler effect does factor in time dilation of the source, which is why the relativistic Doppler factor is different from the Newtonian one). But you wrote

The brother on earth, watching the moving clock,

Which implies that the stay at home is actually watching the traveller's clock through a telescope. What he sees will be dominated by the Doppler effect, and he will see the clock ticking slow on the outbound leg and fast on the inbound leg. Only if he corrects his observations to account for the changing travel time of the light will he calculate (not see) that the traveller's clock ticks slow in both directions.

 

[aperakh]

Only if he corrects his observations to account for the changing travel time of the light will he calculate (not see) that the traveller's clock ticks slow in both directions.

I see what you mean. Gotta be more careful with my language.
Always the case with this stuff.

 

[PeterDonis]

We are all well aware of this

Indeed.

 

[vanhees71]

Though it's not entirely wrong to say that time dilation is also part of the Doppler effect, particularly for light. The "transverse Doppler effect", i.e., observerving the em. wave emitted perpendicular to the velocity of the light source (relative to the observer), is entirely due to "time dilation": $\omega=\omega_0/\gamma$, where $\omega_0$ is the frequency of the em. wave in the rest frame of the source.

 

[Speady]

We are all well aware of this (although it should be noted that the relativistic Doppler effect does factor in time dilation of the source, which is why the relativistic Doppler factor is different from the Newtonian one). But you wrote

Which implies that the stay at home is actually watching the traveller's clock through a telescope. What he sees will be dominated by the Doppler effect, and he will see the clock ticking slow on the outbound leg and fast on the inbound leg. Only if he corrects his observations to account for the changing travel time of the light will he calculate (not see) that the traveller's clock ticks slow in both directions.

Slower in both directions?
If the traveler looks at my clock with a telescope, then, looking through his telescope, he can just as well count how many circles the Earth has revolved around the sun. After all, the movements of the clock and of the orbiting Earth are in a fixed relationship to each other, aren't they? No matter how far and how fast the traveler goes: it cannot be the case that he counted more or fewer laps than the person who stayed at home? Surely it cannot be that one says "3 rounds" and the other says "5 rounds"? During the experiment, the Earth could only have made one and the same number of circles.
Okay: I can understand that the traveler counts fewer rounds on the way there, because of the delay, because of the time it took for the light from my surroundings to reach him. But on the way back it seems logical to me that he sees the orbiting Earth as an accelerated image, due to the ever shorter time it takes for the light from my surroundings to reach him. So at the end the number of laps is the same again. We call turning one circle "the passing of a year". So, isn't it the case that for the traveler and me the same number of years have passed?
I hope I don't get blocked right away after asking this question.

 

[jbriggs444]

Okay: I can understand that the traveler counts fewer rounds on the way there, because of the delay, because of the time it took for the light from my surroundings to reach him. But on the way back it seems logical to me that he sees the orbiting Earth as an accelerated image, due to the ever shorter time it takes for the light from my surroundings to reach him. So at the end the number of laps is the same again. We call turning one circle "the passing of a year". So, isn't it the case that for the traveler and me the same number of years have passed?

You are correct that the number of trips made around the sun by the Earth during the trip is an invariant. It is the same number according both stay-at-home observer and traveling observer.

You are correct that the telescopic image during the outbound trip will show an Earth making lazy circles around the sun. Much longer than 365 * 24 * 60 * 60 seconds per year according to the traveler's wristwatch. That is a combination of the ordinary Doppler effect and relativistic time dilation. The combination of the two is the relativistic Doppler effect.

You are correct that the telescope imaging during the return trip will show an Earth making rapid circles around the sun. Somewhat faster than 365 * 24 * 60 * 60 seconds per year according to the traveler's wristwatch. There is a speed-up due to the ordinary Doppler effect and a slow-down due to relativistic time dilation.

So yes, again, the number of orbits during the trip comes out right. But the elapsed time on the traveler's wristwatch is less than the number of years times 365 * 24 * 60 * 60 seconds.

 

[Dale]

Slower in both directions?

Yes

After all, the movements of the clock and of the orbiting Earth are in a fixed relationship to each other, aren't they?

Yes. Both slow down equally.

So, isn't it the case that for the traveler and me the same number of years have passed?

The orbiting of the Earth around the sun does not mark a year of the traveler’s time.

 

[Speady]

With these short answers, I assume that I am meant to believe you at your word.

I don't quite understand yet what the definition of a year (the Earth around the sun once?) should be for the traveler.

I also have the following problem: just as the traveler looks at my watch on his way back, looking at the traveler's watch, I also see an accelerated recording of his watch, due to the increasingly shorter time that the light from his watch needs to reach me. How can I reconcile this fast-paced display of his watch with his claim that his watch actually passed less time on the way back than mine?

Does it just follow from a calculation and don't I have to worry about what I see?

 

[Dale]

I assume that I am meant to believe you at your word.

You are welcome to consult the professional scientific literature instead.

I don't quite understand yet what the definition of a year (the Earth around the sun once?) should be for the traveler.

The year (measured by orbits of the Earth around the sun) is a different number of seconds (measured by atomic clocks) for the traveler and the stay at home twin.

How can I reconcile this fast-paced display of his watch with his claim that his watch actually passed less time on the way back than mine?

He visually sees the watch running fast. Since the distance is changing the light travel delay is also changing. When he corrects for the light travel delay he finds that the watch is running slow.

 

[PeterDonis]

just as the traveler looks at my watch on his way back, looking at the traveler's watch, I also see an accelerated recording of his watch, due to the increasingly shorter time that the light from his watch needs to reach me. How can I reconcile this fast-paced display of his watch with his claim that his watch actually passed less time on the way back than mine?

See here:

https://math.ucr.edu/home/baez/physics/Relativity/SR/TwinParadox/twin_doppler.html

The whole FAQ is worth reading, but I've linked to the particular page that addresses the specific issue you raise in the above quote.

Does it just follow from a calculation and don't I have to worry about what I see?

As the page linked to above makes clear, you can explain the difference in elapsed times entirely in terms of what each twin directly observes through their telescope, without having to calculate at all.

 

[Ibix]

We call turning one circle "the passing of a year".

A long time ago when we were not so well aware of how time works we did, yes. Now, we make a distinction between the orbital period of the Earth as measured in an arbitrary frame where the Sun is moving, and the orbital period of the Earth in the Sun's rest frame. Only in the latter does the period match what we call a year, since a year is a number of seconds and the second is defined in terms of experiments done with kit at rest with respect to you.

Does it just follow from a calculation and don't I have to worry about what I see?

Let's say the traveller's watch ticks $T$ times on the outbound leg and $T$ times on the inbound leg. The stay-at-home must also see $2T$ ticks, but the time taken for them to arrive will be larger/smaller by the Doppler factor. Thus the time the stay-at-home waits to see all the ticks is$$
\begin{eqnarray*}
&&\sqrt{\frac{c+v}{c-v}}T+\sqrt{\frac{c-v}{c+v}}T\\
&=&\left(\sqrt{\frac{c+v}{c-v}}+\sqrt{\frac{c-v}{c+v}}\right)T\\
&=&\left(\sqrt{\frac{c+v}{c-v}\left(\frac{c+v}{c+v}\right)}+\sqrt{\frac{c-v}{c+v}\left(\frac{c-v}{c-v}\right)}\right)T\\
&=&\left(\frac{c+v}{\sqrt{c^2-v^2}}+\frac{c-v}{\sqrt{c^2-v^2}}\right)T\\
&=&\frac{2c}{\sqrt{c^2-v^2}}T\\
&=&\frac{2}{\sqrt{1-v^2/c^2}}T\\
&=&2\gamma T
\end{eqnarray*}$$Thus, if the stay-at-home is to see $2T$ ticks of the traveller's watch between departure and return, he must experience his own watch ticking $2\gamma T$ times. This is exactly consistent with the time dilation based calculation.

A similar calculation can be carried out from the traveller's perspective, yielding the reciprocal factor. Care must be taken with how many ticks are received before and after the turnaround, since the traveller isn't always inertial. Why don't you have a go? Post your working if you get stuck.

 

[vanhees71]

Slower in both directions?
If the traveler looks at my clock with a telescope, then, looking through his telescope, he can just as well count how many circles the Earth has revolved around the sun. After all, the movements of the clock and of the orbiting Earth are in a fixed relationship to each other, aren't they? No matter how far and how fast the traveler goes: it cannot be the case that he counted more or fewer laps than the person who stayed at home? Surely it cannot be that one says "3 rounds" and the other says "5 rounds"? During the experiment, the Earth could only have made one and the same number of circles.
Okay: I can understand that the traveler counts fewer rounds on the way there, because of the delay, because of the time it took for the light from my surroundings to reach him. But on the way back it seems logical to me that he sees the orbiting Earth as an accelerated image, due to the ever shorter time it takes for the light from my surroundings to reach him. So at the end the number of laps is the same again. We call turning one circle "the passing of a year". So, isn't it the case that for the traveler and me the same number of years have passed?
I hope I don't get blocked right away after asking this question.

Usually you get this confusion about space-time measurements solved when remembering that what's simultaneous in one inertial reference frame is not simultaneous in another.

This already starts with the clock synchronization, which by definition is made only between clocks all at rest wrt. one inertial reference frame. By definition you think about one standard clock at the origin of this frame and any other standard clock at another arbitrary point at rest relative to the clock at the origin at a distance $r$ from the origin. Then you can synchronize both clocks by sending a light signal to the other clock being reflected there. Now you can measure the time needed from sending the signal to receiving the reflected one. This is a measurement at one position, i.e., the origin of the frame and thus really feasible in the lab. By definition of the time units (like the second in the SI units) this time is $2 \Delta r/c$, where $c$ is the (arbitrarily defined as in the SI!) speed of light. Now by definition it's assumed that the signal takes as long to go from the origin to the clock to be synchronized as the reflected signal needs to be reflected back (again note that this is assumed in an inertial reference frame and for clocks being both at rest in this frame!). Thus to synchronize the distant clock with the clock at the origin. You have to preset the distant clock to a time $r/c$ and send the signal from the origin at $t=0$, and start the distant clock as soon as the light signal arrives at it. That you do with all (fictitious) clocks at rest wrt. this frame at any position. In this way you can define locally what "simultaneity" of two events means within this inertial reference frame: Two events at different places are by definition simultaneous when the two synchronized clocks at each of these places show the same time $t$.

Now since the speed of light by Einstein's 2nd postulate should be the same, independent of the velocity of the source, you get the Lorentz transformations between the space-time coordinates of two different inertial frames, i.e., the frame $\Sigma'$ moving with constant velocity $\vec{v}=\beta c \vec{e}_1$, which immediately tells you that the synchronized clocks defining the time coordinate $t$ of $\Sigma$ are not synchronized with the synchronized clocks defining the time coordinate $t'$ of $\Sigma'$. Thus two events being simultaneous wrt. $\Sigma$ are not simultaneous anymore wrt. $\Sigma'$ and vice versa. Since the Lorentz transformations form a group, there can never be contradictions between the description of physical events within either frame of reference. The "coordinate times" $t$ and $t'$ refer to different sets of synchronized clocks.

You can always describe any physical situation in terms of invariant quantities, i.e., scalars, vectors, and (most generally) tensors, which shows that the physics does not depend on the choice of any inertial reference frame.

Usually that's done by choosing some convenient inertial reference frame to define quantities in tensor form. E.g., take relativistic fluid dynamics. There all the needed quantities characterizing the material properties of the fluid are always defined in an inertial frame, where the fluid element is at rest at the time under consideration, like the number density, the temperature, density of thermodynamical potentials, etc. In this way all these densities get scalar fields. In addition you need the four-velocity field $u^{\mu}$ (with $u_{\mu} u^{\mu}=1$) to express all quantities in an easy way as invariant/covariant quantities. E.g., the four-current of some conserved charge is
$$J^{\mu}(x)=n(x) u^{\mu}(x),$$
where $n$ is the number density as measured in the momentaneous rest-frame of the fluid cell located at $x$, or the energy-momentum tensor in the case of an ideal fluid
$$T^{\mu \nu}=(e+P) u^{\mu} u^{\nu}-P g^{\mu \nu},$$
where $e$ is the internal-energy density and $P$ the pressure (both as measured in the momentaneous rest frame of the fluid cell). In this way everything is neatly expressed in explicitly invariant tensor quantities (or, as written here, in terms of the corresponding components of all these quantities wrt. to one fixed "observational inertial reference frame", aka the "lab frame").

 

[Kairos]

The problem with your question is the: "are".
There is no "are". That is to say, there is no universal frame of reference by which to decide how long the seconds "really are".

Your view seems to contradict the posts #42 and #48. There is no need for the brothers to compare their seconds to measure THE second, provided they carry close to them cesium atoms at rest relative to them. No?

 

[Kairos]

A long time ago when we were not so well aware of how time works we did, yes. Now, we make a distinction between the orbital period of the Earth as measured in an arbitrary frame where the Sun is moving, and the orbital period of the Earth in the Sun's rest frame. Only in the latter does the period match what we call a year, since a year is a number of seconds and the second is defined in terms of experiments done with kit at rest with respect to you.

Let's say the traveller's watch ticks $T$ times on the outbound leg and $T$ times on the inbound leg. The stay-at-home must also see $2T$ ticks, but the time taken for them to arrive will be larger/smaller by the Doppler factor. Thus the time the stay-at-home waits to see all the ticks is$$
\begin{eqnarray*}
&&\sqrt{\frac{c+v}{c-v}}T+\sqrt{\frac{c-v}{c+v}}T\\
&=&\left(\sqrt{\frac{c+v}{c-v}}+\sqrt{\frac{c-v}{c+v}}\right)T\\
&=&\left(\sqrt{\frac{c+v}{c-v}\left(\frac{c+v}{c+v}\right)}+\sqrt{\frac{c-v}{c+v}\left(\frac{c-v}{c-v}\right)}\right)T\\
&=&\left(\frac{c+v}{\sqrt{c^2-v^2}}+\frac{c-v}{\sqrt{c^2-v^2}}\right)T\\
&=&\frac{2c}{\sqrt{c^2-v^2}}T\\
&=&\frac{2}{\sqrt{1-v^2/c^2}}T\\
&=&2\gamma T
\end{eqnarray*}$$Thus, if the stay-at-home is to see $2T$ ticks of the traveller's watch between departure and return, he must experience his own watch ticking $2\gamma T$ times. This is exactly consistent with the time dilation based calculation.

A similar calculation can be carried out from the traveller's perspective, yielding the reciprocal factor. Care must be taken with how many ticks are received before and after the turnaround, since the traveller isn't always inertial. Why don't you have a go? Post your working if you get stuck.

To understand your demonstration, what is T? Is it, as suggested by your text, a number of ticks (unitless integer) or is it a period (time)? To my knowledge Doppler effects modify apparent wavelengths but not numbers of ticks.

 

[Ibix]

To understand your demonstration, what is T? Is it, as suggested by your text, a number of ticks (unitless integer) or is it a period (time)?

It doesn't matter. Say you are traveling away from me at $v$ and you emit radiation for $T$ ticks of your watch, which is a duration of $T×1\mathrm{s}$. Say that I am illuminated by the radiation for $T'$ ticks of my watch, which is a duration of $T'×1\mathrm{s}$. The two durations are related by the Doppler factor, so$$T'×1\mathrm{s}=\sqrt{\frac{c+v}{c-v}}T×1\mathrm{s}$$The units cancel out, so you are free to interpret $T$ and $T'$ as a number of ticks, as I wrote, or to mentally multiply them by a second/month/year to get a duration.

To my knowledge Doppler effects modify apparent wavelengths but not numbers of ticks.

It doesn't modify the number of ticks of your watch that I see. It does give the number of ticks of my watch that I have to wait to see the given number of ticks of your watch.

 

[Kairos]

OK, so in your demonstration you determine a mean period over the whole trip?

 

[Ibix]

OK, so in your demonstration you determine a mean period over the whole trip?

I was showing that the total time for the stay-at-home to see $2T$ ticks of the traveller's watch, taking into account the Doppler effect, was consistent with a simple minded time dilation calculation. You can interpret that as giving an average tick rate of a twin's watch as measured by the other, yes.

 

[Kairos]

Thank you. I find it hard to accept that a number of periods and a period size are the same thing, but I will think about it... If your formulas describe an average period T' of the outbound and inbound journey, the 2 should be removed. I would be curious to see the calculation from the traveler's point of view.

 

[Ibix]

I find it hard to accept that a number of periods and a period size are the same thing

They aren't. But in this case we're calculating a ratio of two time periods, which is equal to the ratio of the number of seconds in those periods.

If your formulas describe an average period T' of the outbound and inbound journey, the 2 should be removed.

The 2 is correct. The formula is for the ratio of the total time for the stay at home to the total time ($2T$) for the traveller.

I would be curious to see the calculation from the traveler's point of view.

Let's see if @Speady can figure it out before I give the answer. The final answer is obvious, of course - since you know the stay-at-home's duration in terms of the traveller's you can just invert the equation.

 

[Kairos]

As the 2 is correct, then your calculation is not an average period as I imagined, so I fall back on a number of ticks ..
I have a question that deviates a bit from the original one: in my (simplistic) physics, the total travel time is 2D/v, where D is the length of the path before the turn. Is this duration measured by one of the brothers or do they both find a different one?