Fewer seconds or shorter seconds?

[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?

 

[Ibix]

As the 2 is correct, then your calculation is not an average period as I imagined

It is. It's just not the clearest form. The point is that $2T$ is the total time experienced by the traveller and $2\gamma T$ is the total time experienced by the stay at home. Thus the stay at home argues that one tick of the traveller's clock is $\gamma$ times longer than his own.

Similarly, the traveller can argue that the stay at home's clock must, on average, have ticked once every $1/\gamma$ ticks of his own. Learning how to reconcile that with the fact that the teaveller also says that due to time dilation the stay at home's clock ticks once every $\gamma$ ticks of his own clock is the point of the twin paradox.

 

[Ibix]

Is this duration measured by one of the brothers or do they both find a different one?

Who is moving and who is measuring?

 

[Kairos]

Each brother (the stay at home and the traveler) can measure the duration of the trip (between the separation and the reunion of the brothers) using his own clock. Does one of them verify that this measured duration corresponds to 2D/v? where v is the speed of the rocket (on which they agree), and D is the distance between two distant planets (on which they also agree according to astronomical measurements).

 

[Ibix]

D is the distance between two distant planets (on which they also agree according to astronomical measurements).

I think you've forgotten that the planets are moving in one frame and the distance between them is length contracted by a factor of $\gamma$ compared to the other frame.

 

[Nugatory]

where v is the speed of the rocket (on which they agree)

Do they? According to the traveling twin the speed of the rocket is zero at all times, while the speed of the Earth is $v$, first receding and then approaching.

 

[Kairos]

I think you've forgotten that the planets are moving in one frame and the distance between them is length contracted by a factor of $\gamma$ compared to the other frame.

You mean that the contraction of lengths is not only valid for the rocket but also for space. So for the traveler, the distance traveled is in fact shorter than the one predicted by the terrestrial astronomers ? this would indeed answer my question.

 

[Ibix]

So for the traveler, the distance traveled is in fact shorter than the one predicted by the terrestrial astronomers ?

Well, as Nugatory notes, for the traveller the planets are moving. But yes, the inbound and outbound inertial frames regard the distance between the planets as length contracted. Imagine actually laying a physical ruler that just touches each planet. Since the ruler is moving in the traveller's frames it must be length contracted, but both frames must agree whether or not the ruler touches the planets. So the distance between the planets must be contracted too.

 

[Kairos]

Do they? According to the traveling twin the speed of the rocket is zero at all times, while the speed of the Earth is $v$, first receding and then approaching.

One can certainly consider that the whole universe shifts with respect to the rocket seen as immobile, but in the case of a terrestrial takeoff, it is the cosmonaut which experienced an acceleration.

 

[Nugatory]

but in the case of a terrestrial takeoff, it is the cosmonaut which experienced an acceleration.

This is true, but irrelevant to distance measurements made using either the inertial frame in which the cosmonaut is at rest on the outbound leg or the different inertial frame in which the cosmonaut is at rest during the inbound leg.

 

[Kairos]

Indeed, I agree that when a constant velocity is reached, one goes into a perfectly symmetrical system, although it is generally easier for an astronaut (who remembers his acceleration) to think that it is he who is flying rather than the earth.
But your remark makes me ask another question: when the system is symmetrical, why decide that it is the astronaut's path that contracts and not the Earth's one? If the two contract symmetrically, Ibix's argument is no longer valid and the 2D/v duration is indeed reciprocal

 

[Dale]

when the system is symmetrical, why decide that it is the astronaut's path that contracts and not the Earth's one?

When the system is symmetrical then each contracts with respect to the other equally, preserving the symmetry. However, that is not relevant here since the situation is not symmetrical.

If the two contract symmetrically, Ibix's argument is no longer valid and the 2D/v duration is indeed reciprocal

His argument is valid for the asymmetrical situation under discussion.

 

[jbriggs444]

both frames must agree whether or not the ruler touches the planets

To be clear, both frames agree that the front end of the ruler touches the front planet at some time and that the back end of the ruler touches the back planet at some time. They disagree on whether these two events are simultaneous.

 

[Ibix]

when the system is symmetrical

It isn't. The destination planet is at rest in the stay at home frame, and there is no analogous object at rest in the traveller's frame.

 

[Ibix]

To be clear, both frames agree that the front end of the ruler touches the front planet at some time and that the back end of the ruler touches the back planet at some time. They disagree on whether these two events are simultaneous.

I was imagining a ruler at rest with respect to the planets, so that it was length contracted in the traveller's frame.

 

[Kairos]

When the system is symmetrical then each contracts with respect to the other equally, preserving the symmetry. However, that is not relevant here since the situation is not symmetrical.

His argument is valid for the asymmetrical situation under discussion.

That's what I thought but the discussion drifted from above remarks. It is indeed because the astronaut's experience is non-inertial that there is asymmetry. This is precisely why I would have been interested in a demonstration in the style of the post #61 but from the point of view of the other brother, to show the asymmetry.

 

[Kairos]

When the system is symmetrical then each contracts with respect to the other equally, preserving the symmetry. However, that is not relevant here since the situation is not symmetrical.
His argument is valid for the asymmetrical situation under discussion.

is 2D/v the duration of the trip for the stay at home? and introducing length contraction 2D*sqrt(1-(v/c)^2)/ v for the traveler?

 

[Dale]

is 2D/v the duration of the trip for the stay at home? and introducing length contraction 2D*sqrt(1-(v/c)^2)/ v for the traveler?

Yes. Although for the traveler there is no standard reference frame since they are non-inertial. So the number for the non-inertial frame is dependent on how things are defined.

 

[Kairos]

Thank you very much. I suppose that this result holds for a simple U-turn like a rebound without velocity change. I think I will contact you later for a new thread about the twin experiment. This simple demonstration poorly convinces me because the only parameter that distinguishes the twins: the U-turn, is not introduced in the reasoning and the difference of travel duration between the brothers is proportional to D, whereas things are perfectly symmetrical between them during the phases of inertial motion.

 

[lightarrow]

In the twin experiment, the travel time is shorter for the traveling brother than for the static brother. Since the unit of time in physics is the second, 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?

(1) it "contains fewer seconds": the traveller's clock doesn't "walk slower", it travels a shorter path in spacetime.
Sorry, I haven't read all the answers, probably this has already been written more times.

--
lightarrow

 

[A.T.]

the U-turn, is not introduced in the reasoning and the difference of travel duration between the brothers is proportional to D

As an analogy: Turning with your car doesn't affect the rate of the odometer, but the total distance traveled between A and B is affected by the turns you make.