My question pertains to the relationship between acceleration (change in speed and/or direction of motion) and time dilation. Since acceleration (we'll consider change in speed) involves motion, and since motion itself is relative, why does time dilation due to acceleration occur for only one of the two frames we are comparing as opposed to time dilation occurring for both frames in the different case where no acceleration is involved (where the frames are moving at constant speed relative to each other)? After all, we can choose either one of the two frames to be the stationary one and the other the moving one, and with either choice, the moving frame is accelerating. It would seem that there is something other than merely a change in speed that causes time dilation, since only one of the two frames "feels" the force of resistance during this change. Also, if both frames did experience time dilation, then the "twin paradox" in the spaceship/earth scenario would have no solution. We know that a frame subjected to a gravitational field will experience time dilation whether it is in freefall (accelerating) or not (due to a force of resistance - what it "feels" as its weight - such as by contact with the mass producing the field). So, acceleration can't be the cause of the proportional increase in time dilation with increasing gravitational potential energy. Yet, despite all that I have said, acceleration is INVOLVED in the phenomenon of time dilation even though not its cause. So, precisely how is changing speed and/or direction (acceleration) connected with time dilation? When I ask "precisely how," I am referring to an intuitive visualization of what is occurring. If curvature of Space-Time is invoked in a response, again I need to "see" what's actually occurring here.

  • $\begingroup$ @FlatterMann: You're simply stating the reason why acceleration would be necessary to solve the twin paradox. But I'm not asking WHY it's necessary. Instead, I'm asking how it results in time dilation. Just because we want to solve the paradox doesn't justify creating explanations without substantiation. $\endgroup$
    – user150908
    Commented Mar 31, 2023 at 16:32
  • 1
    $\begingroup$ For visualization, it may help to draw out the planes of simultaneity for the traveller on 2-dimensional spacetime diagram. At the inversion, the traveler's plane of simultaneity changes orientations. $\endgroup$
    – g s
    Commented Mar 31, 2023 at 17:26
  • $\begingroup$ @g s: Please reread my original question. It was not about the effects of SR. It's about how acceleration (changing speed and/or direction of motion) - as opposed to constant speed and direction - is connected with time dilation (TD). Does acceleration CAUSE TD or is there something more elementary that results in both occurring together? In other words, since acceleration is also motion, what's so special about the circumstance of motion changing that additional TD occurs along with it - on top of TD due only to uniform constant motion? Also, if motion is relative, why is changing motion not? $\endgroup$
    – user150908
    Commented Apr 1, 2023 at 16:43
  • 1
    $\begingroup$ Related: the Clock Postulate. See (eg) physics.stackexchange.com/q/704658/123208 & physics.stackexchange.com/q/498799/123208 $\endgroup$
    – PM 2Ring
    Commented Apr 2, 2023 at 19:12
  • $\begingroup$ It is not involved in the slightest. Time dilation is purely a velocity effect in SR. TD happens when there is no acceleration. $\endgroup$
    – m4r35n357
    Commented Apr 9, 2023 at 8:45

5 Answers 5


and since motion itself is relative

Here is the beginning of your conceptual error. Not all motion is relative. In fact, proper acceleration is not relative, it is invariant. The traveling twin undergoes proper acceleration and by that fact the travelling twin is not symmetric with the home twin.

After all, we can choose either one of the two frames to be the stationary one and the other the moving one, and with either choice, the moving frame is accelerating.

This is not correct. Regardless of your choice of frame only one twin undergoes proper acceleration. They are not symmetrical.

NB. This is true even if you choose a non-inertial frame where the travelling twin is at rest. In that case the travelling twin has no coordinate acceleration, but still has proper acceleration. Proper acceleration is invariant under any coordinate transform.

So, precisely how is changing speed and/or direction (acceleration) connected with time dilation?

Time dilation is the ratio between coordinate time and proper time. Specifically, given an arbitrary Lorentzian metric $ds^2$ in coordinates which include a "time" coordinate labeled $t$, then the time dilation $\gamma$ is given by $$\frac{1}{\gamma}=\frac{d\tau}{dt}=\sqrt{-\frac{ds^2}{c^2 dt^2}}$$

I am referring to an intuitive visualization of what is occurring.

The math above is straightforward and unambiguous, but it may not be intuitive for those that are unfamiliar with it.

$ds^2$ is a sort of spacetime "length" called the spacetime interval. For a spacelike line in spacetime the interval is $\sqrt{ds^2}$ and is measured by a ruler and for a timelike line in spacetime the interval is $d\tau=\sqrt{-ds^2/c^2}$ and is measured by a clock.

The coordinates form a sort of "grid" on spacetime.

So the intuitive interpretation is that time dilation measures the length of the spacetime interval $d\tau$ between two nearby coordinate grid lines, $dt$, as measured by a clock travelling some specific path through spacetime.

A clock experiencing proper acceleration is traveling on a curved path through spacetime. Depending on the coordinate grid lines and the specific curvature, this curved path can lead to a changing angle between the path and the grid lines and hence a changing time dilation. But there is no additional time dilation due to acceleration beyond the spacetime interval between successive grid lines.

  • 1
    $\begingroup$ @Rob that is far too many follow up questions and issues for comments. This is not the place for an extended discussion. But right from the beginning, no, even in Newtonian mechanics acceleration is not a change in speed. It is a change in velocity. The distinction is important, particularly in this context, both theoretically and experimentally. $\endgroup$
    – Dale
    Commented Apr 1, 2023 at 4:15
  • 5
    $\begingroup$ @Rob I answered your question with both math and a careful intuitive explanation of the math. If you did not understand the intuitive explanation then please tell me specifically what more you want. Your mildly insulting comments are neither helpful nor true $\endgroup$
    – Dale
    Commented Apr 1, 2023 at 15:46
  • 1
    $\begingroup$ @Rob I did both. If you don’t like the math then read the following section where I explained it in intuitive language. I am also getting frustrated because I did do what you keep saying I didn’t. If my current intuitive explanation isn’t doing it for you then I need some specific feedback. Not general complaints $\endgroup$
    – Dale
    Commented Apr 2, 2023 at 0:03
  • 1
    $\begingroup$ @Rob “How is acceleration … connected with time dilation (TD)?” I already answered that. “Does acceleration of the spacehip CAUSE TD” No. I explained the more elementary thing. “what's so special … that TD occurs along with it?” I answered this already too. “Also, if speed is relative, why is changing speed not?” I did not answer this one because I don’t have an answer for it. I don’t know why. We observe that we can build self contained accelerometers but not self contained speedometers. We make our theories to reflect that observed fact, but I don’t know why the universe is that way $\endgroup$
    – Dale
    Commented Apr 2, 2023 at 0:15
  • 1
    $\begingroup$ @Rob my answer does address “the crux” of your question. Time dilation is the relationship between the interval along the clock’s worldline and the spacetime grid, as I said already. Acceleration is only relevant insofar as it is related to that. Anyway, good luck with your studies, I am done here. Clearly my answer doesn’t help you, but it appears to be helpful to others. So I am done $\endgroup$
    – Dale
    Commented Apr 4, 2023 at 2:14

The path of an object through spacetime is called its world line. A "stationary" object has a world line that points only along the time axis; a "moving" object has a world line that's at an angle in spacetime (that involves space as well as time). Acceleration is rotation in spacetime; that is, an accelerating object has a curved world-line.

The elapsed time shown on a clock between two events is the length of the clock's world-line between the events. If the clock has accelerated, its world line will be curved, and hence will be "longer" than a straight line between the events (a non-accelerated world line). I put "longer" in quotes because time is not the same as space, in fact it has an opposite sign in the spacetime metric, so a "longer" spacetime path actually corresponds to shorter time. That's why a clock which has experienced some kind of acceleration will have a shorter time showing that a clock which hasn't -- it's for the same reason that a curved line is longer in space than a straight line.

  • $\begingroup$ @ Eric Smith: Nice representation in terms of space-time grids. But, as per the twin paradox scenario, this representation still does not address the crux of my question. Again, what is so special about the circumstance of motion changing (acceleration) that TD occurs only for the spaceship (which is in addition to TD for both spaceship and earth frames and which accounts for a net TD for the spaceship clock when spaceship and Earth clocks meet up again)? After all, clocks are real things. $\endgroup$
    – user150908
    Commented Apr 5, 2023 at 22:59
  • $\begingroup$ @ Eric Smith: I thought about my own question, and the only distinguishing feature of the spaceship frame is that, unlike the Earth frame, it is subjected to a force. So both the phenomena of acceleration AND TD are the result of the force of the spaceship's jet engine acting on a mass. This would be no different than acceleration and TD for a frame positioned in a gravitational field where both would be the result of the force of gravity acting on a mass. Perhaps, here, the notion of curved space-time as an explanation for free-fall and TD is an unnecessary mathematical/geometrical construct? $\endgroup$
    – user150908
    Commented Apr 5, 2023 at 23:16
  • 1
    $\begingroup$ @Rob: Odometers are real things, too, just like clocks. You could try to make an explanation in terms of forces and accelerations for why the odometer of a car shows more if it follows a curved path from A to B instead of a straight path. But the simple, intuitive explanation is just that the odometer measures the length traveled in space, and the two paths have different lengths. Similarly, a clock is an odometer for spacetime, and if two twins follow different spacetime paths their clocks will have different elapsed times showing. Acceleration and force come into it only indirectly. $\endgroup$
    – Eric Smith
    Commented Apr 7, 2023 at 10:27
  • 1
    $\begingroup$ @Rob : No, odometers measure distance, not speed. A simple odometer can be constructed by just counting the number of revolutions of a wheel. The geometric explanation for clock differences (time dilation) is, like the geometric explanation for odometer differences, far simpler and more intuitive than any force based explanation. You do have to give up any insistence that there's one universal "true" time, but that's the case for any realistic explanation of relativity. $\endgroup$
    – Eric Smith
    Commented Apr 7, 2023 at 15:35
  • 1
    $\begingroup$ Car A drives from New York to Los Angeles in a straight line with constant velocity. Car B drives at a constant velocity, but takes a meandering path (meaning it has to accelerate laterally to change direction periodically). What is the explanation in terms of acceleration for Car B's odometer having a greater reading than that of Car A? Trying to find one will take you off into the weeds, just as trying to explain time dilation in terms of acceleration will. It's ultimately the shape/length of the paths. $\endgroup$
    – Eric Smith
    Commented Apr 7, 2023 at 21:00

Suppose an isolated locomotive L and a train T coming from opposing directions at great velocity. T has one clock in each wagon, all synchronized in its frame.

Comparing the clock's readings of the wagon temporarily in front of it, L will see bigger time intervals than its own clock according to the time dilation formula: $\Delta t_T = \gamma \Delta t_L$.

If L suffers a sudden (infinite) acceleration, and its velocity becomes equal to T, both vehicles are now in the same frame. L compares its clock with the wagon of T in front of it (and now at rest because they are in the same frame) and the relation $\Delta t_T = \gamma \Delta t_L$ is now 'frozen'. That is the effect of the acceleration: all clocks are now in the same frame, and the clock of L shows that less time passed for it. The acceleration is not the reason for the time dilation. L was observing the effect comparing the clocks of the wagons of T in front of it, while at constant velocity.

Note that it is problematic to suppose a similar sudden acceleration of T. According to Rindler analysis, the wagons should have different accelerations if the distance between them keeps constant. And during acceleration, the clocks of the wagons lose synchronism. So the situation is not symmetric. We can imagine an observer accelerating and compare its proper time with an inertial coordinate time. But not so easily to compare an accelerated frame with an inertial observer.

  • $\begingroup$ Is a train that accelerates evenly along its length a "Rindler frame" (is that even a thing?)? I doubt it, but I haven't looked at special relativity in a long time, so I might be wrong. $\endgroup$ Commented Apr 1, 2023 at 17:28
  • 1
    $\begingroup$ Yes. Of course it is not relevant for the lengths, velocities and accelerations of real trains. But to be precise, an accelerated train, keeping a constant distance between wagons, has a gradient of accelerations. Each wagon is a 'Rindler observer'. $\endgroup$ Commented Apr 1, 2023 at 17:38
  • $\begingroup$ I see. I need to update my intuition. Thanks for the clarification. $\endgroup$ Commented Apr 1, 2023 at 18:43

You have asked the correct question, which is about 90% of the way to understanding!

Consider two people in spaceships, Alice and Bob. Bob is sitting "at rest" at the origin of some coordinate system, Alice is moving along its $x$ axis at, say, 0.0001 c or so. Fast, but not in the regime where we really need to fuss over length contraction or time dilation.

A bubble of light

They arrange for an “epic high-five” as they pass, but since they are both in spaceships they blast each other's shields with a short laser pulse, like any other high five, strong enough they feel it, not strong enough to do lasting damage. So they illuminate each other for a very brief instant, and instead of a loud sound wave propagating outward a very bright light wave propagates outward. If there is a very thin dust smattering through space, this light becomes visible and looks like a bubble, a thin spherical shell expanding at speed c, announcing to the world that this epic high-five has happened.

Now, for sound we know that sound propagates through air and air is at rest in some reference frame, we could clarify that this is Bob's reference frame, both Alice and Bob will agree that this bubble, as it expands, remains centered on Bob. Then Alice agrees that she is closer to one side of the bubble then to the other, not much since she is only at 0.01% c, but detectable with very fine instruments.

The problem is, in relativity the vacuum is truly a vacuum, there is no medium, no "luminiferous ether" through which light propagates. Bob thinks that he is at the center of the bubble and Alice is off-center, but Alice also thinks that she is at the center of the bubble and Bob is off-center. This is what it means for everybody to agree on the speed of light, and it is weird.

The first-order Lorentz transform

Lorentz came up with his transformation by studying the Maxwell equations, Einstein rederived it by looking at train thought experiments with clocks, and now I'm going to do it by staring at bubbles.

So a time $t$ has elapsed, freeze frame, the velocities involved are small enough that I don't need to make a distinction about who it elapsed for, Alice or Bob. They could both agree that they “see” the bubble bouncing off the dust at a distance $ct/2$ away from themselves, but that due to the reverse propagation delay of the light, the bubble is actually a distance $ct$ away from them.

First remark: they don't actually disagree in all directions. They agree that the distance to the bubble for both is $ct$ in the $y,z$ directions, to first order. There is a correction factor $y=ct-\frac{(vt)^2}{2ct}$ but at 0.01% c, this effect is 0.000001% of the distance, it is nothing compared to the 0.01% $x$-disagreement where Bob thinks Alice is 0.01% closer to one side of the bubble and 0.01% further from the other side. So we only have a true discrepancy in the $x$-direction.

We will imagine that Bob has placed a bunch of atomic clocks along the $x$-axis which were synchronized and were counting down to the high five at $t=0$ and afterwards tick up. Maybe these clocks are watching for the high-five-laser light and broadcast their timestamp when they see it. Alice and Bob disagree about who the bubble is centered on, but Bob’s clocks can attempt to furnish a proof that the bubble is centered on him. After all he has very carefully synchronized them, very carefully laid them out at constant intervals, and he has recorded their broadcasts, one “t=0” from the one at x=0, two “t=1” chirps from the clocks at x=±1, two “t=2” chirps from the clocks at x=±2, and so on. (Assume distances are measured in light seconds or so.) And Alice must agree on the messages that these broadcasts contained, and that they are laid out at these constant spacings, so surely she agrees that Bob is at the center of the bubble, right?!

Einstein’s radical idea: maybe she disagrees specifically because she does not think that Bob’s clocks are synchronized. If she thinks that they have a delay which depends on their $x$-coordinate, then she might say “your clocks DID chirp t=1.0000, but when I reconstruct what they were saying when the high five happened based on all of my observations from before the high fives, your clock at $x=1$ actually said $t=0.0001$. Therefore only 0.9999 seconds had actually elapsed between the high five and when it broadcasted that message, and if your clock had been properly synchronized it would have said 0.9999, and if you roll the tape forward by that last 10ms you will find that I was at the center of the bubble, not you. You just chose the offsets to make it look like you were at the center, but you weren't actually because I am at the center.” Einstein starts off his relativity works by saying “let's get into the nitty gritty about how we synchronize clocks”, he recognized that this was the fundamental point.

The conventional Galilean coordinate transform to a moving reference frame was $$\begin{align} t'&=t,\\ x'&=x-vt,\\ y'&=y,\\ z'&=z\end{align}$$ but now we need instead to change the time coordinate just a little, let $w=ct$ and $\beta=v/c$ and we can write the needed transform as $$\begin{align} w'&=w-\beta x,\\ x'&=x-\beta w,\\ y'&=y,\\ z'&=z.\end{align}$$ This is the first-order Lorentz transformation. If we use this to get Alice's coordinates we find that she is at the center of something that's approximately a sphere, maybe it's a little ellipsoidal to second order in $\beta$, it will become a sphere once we correct it in the next section. I claim that there is no other fundamental physics. Everything you need to understand about special relativity is right here.

To all orders!

Now, you asked about time dilation, this equation does not contain time dilation. Or does it? Remember that we can use calculus ideas to build something big out of something small. For this it helps to define some new coordinates, $p=w+x$ and $q=w-x$, these are based on the “eigenvectors” for the transform, so that in terms of these the first-order transform is ( $$\begin{align} p'&=(1-\beta) p,\\ q'&=(1+\beta) q,\\ y'&=y,\\ z'&=z.\end{align}$$ Now let us build up a bigger change $\phi$ out of $N$ smaller changes $\beta=\phi/N$, this leads to the overall change $$\begin{align} p'&=\left(1-\frac\phi N\right)^N p = e^{-\phi} p,\\ q'&=\left(1+\frac\phi N\right)^N q =e^\phi\\ y'&=y,\\ z'&=z,\end{align}$$ where these limits for large $N$ are classics of precalculus. Unfold using the hyperbolic sine and cosine to get the all-orders Lorentz transformation, $$\begin{align} w'&=w\cosh\phi - x\sinh\phi\\ x'&=x\cosh\phi - w\sinh\phi\\ y'&=y,\\ z'&=z,\end{align}$$ This angle $\phi$ is a rapidity, you have to work through a couple of examples to find the actual speed of the transformation is $\beta=\tanh\phi$ and therefore from hyperbolic trig identities that $\cosh\phi=\gamma=1/\sqrt{1-\beta^2}$.

But the point is that these effects that concern $\gamma$ like time dilation and length contraction are nonphysical. They are the aggregated consequence of how space and time axes twist into each other under the Lorentz boost, it is a mathematical consequence of the actual physical thing, which is the relativity of simultaneity. Those are second order effects, the relativity of simultaneity is a first order effect in $\beta$.

How this answers your question

Now let's go back to Alice and Bob. Imagine in the distant past, Alice is stationary relative to Bob, she's on the Moon or so, and she watches him set up all of these clocks in preparation for the high-five and agrees that they are all in-sync. They're in sync now, they are out of sync by the time of the high five, and the only thing that has happened is that Alice has accelerated towards Bob, but magically these clocks pick up an offset from her perspective?

Well, slow down for just a moment, all of Bob's clocks appear to be ticking fast for Alice if she just looks at them! This is true without relativity too, where it is called the Doppler shift. There it has magnitude $$\frac{1}{1-\beta } = 1 + \beta +\beta^2+\dots,$$ whereas in relativity it has magnitude $$e^\phi=\sqrt{\frac{1+\beta}{1-\beta}} = 1 + \beta +\frac12\beta^2+\dots.$$ Either way, you move towards a clock and it appears to tick fast. But special relativity’s clocks don't quite tick as fast as the Doppler shift would normally account for. This tiny little $\frac12 \beta^2$ discrepancy is what we are calling time dilation, it is always overwhelmed by the Doppler effect itself, which is first-order.

With that qualifier, the statement is yes, that must be how it happens. The central claim of relativity is precisely that, it must be a universal property of acceleration that, in addition to other Doppler shifts that you can understand classically, there is a weird shift that you can't because it is tied to both your acceleration and the coordinate distance along x to the clock in question.

In other words, you were expecting after correcting for any Doppler shifts that you know about, that the clocks would tick at a rate of one second per second. Alice, while accelerating, actually observes that a clock at coordinate $x$ ticks at a rate of $1+\frac{ax}{c^2}$ seconds per second, where $a$ is her acceleration. This is the ultimate reason why the clock at x=1 was showing the time t=0.0001 when the high-five happened, it was ticking very subtly faster for Alice when she accelerated.

Now you have all of the pieces.

  • The acceleration causes someone to perceive this subtle weird vaguely-Doppler-like shift in the definition of “now” as it applies to distant location.

  • The shift is always buried by other Doppler effects in terms of what you literally see, to perceive the shift you must always correct for the overwhelming classical shifts that also take place.

  • When you do this you find, most dominantly, that the definition of “now” at remote locations has changed, their clocks appear to have ticked fast during your acceleration.

  • The aggregated result of those coordinate redefinitions for time, twisting into the Galilean coordinate redefinitions for space, ultimately cause a sense of time dilation where a clock moving towards you is seen to be ticking fast (Doppler) but not as fast as it should be ($\beta^2/2$) given the relative velocity.

To combine the stories: Alice says the clock ticks at some position $x'$ at some time $t'$ and then ticks again at some point $x'-v~\delta t'$ at the following time $t'+\delta t'$, and she is scandalized that $\delta t'$ is longer than one second. This is based on her sense of “now”. But the clock’s sense of “now” is at an angle relative to that, and she sees it in a sort of “projection,” and this is time dilation. According to the Bob-clock, this new point in spacetime Alice has described is at, $$\begin{align} \delta x &= \gamma~(\delta x' + v~\delta t') = 0,\\ \delta t &= \gamma~ \left(\delta t' +\frac{v}{c^2} \delta x'\right) = \delta t'~\sqrt{1-\beta^2}\end{align}$$ According to the Bob clock, Alice's implicit clocks that she measures this clock by, are also all ticking slow, but they are also offset from “now” across space in an effect that is about twice as strong as this $\gamma$ effect, leading her to think a much longer time $\delta t'$ has elapsed when only $\delta t$ has elapsed.

And if we Taylor expand of course we find $\frac12\beta^2,$ just the necessary Doppler correction.

  • $\begingroup$ Wow, that was the longest explanation for a trivial delay effect that I have ever seen. $\endgroup$ Commented Apr 2, 2023 at 18:46
  • $\begingroup$ If you think that the twin paradox is a trivial delay effect, you are trivially wrong. If you tried to reproduce the twin paradox without special relativity, both friends would have the same age when they finally meet. $\endgroup$
    – CR Drost
    Commented Apr 2, 2023 at 19:06
  • $\begingroup$ The delay only occurs in a relativistic world. Signal propagation in a non-relativistic world is instantaneous. Why is this supposed to be a problem? The delay explanation gives the correct result both in Newtonian mechanics (there is no time dilation) and in a relativistic scenario. OTOH, yesterday was the first of April. Did I miss the joke? $\endgroup$ Commented Apr 2, 2023 at 22:39
  • $\begingroup$ So, signal propagation does not have to be instantaneous in non-relativistic physics; you can model light as mediated by a luminiferous ether at rest relative to the rest twin, at a finite speed $c$, and the Doppler effect exists but the two twins are necessarily the same age when they meet up. This proves that you have to actually engage with the time dilation that relativity gives you after you correct for the Doppler shift, where both twins think the other twin must have legit aged less. I think this will come out of your calculation if you are careful to do it out in full detail. $\endgroup$
    – CR Drost
    Commented Apr 2, 2023 at 23:04
  • $\begingroup$ That Newtonian gravity seemed to be instantaneous bothered even the folks in Newton's time. The insertion of media is not the Newtonian explanation for non-contact forces. It only crops up as an aether when it became clear that electromagnetism didn't play the Newtonian game. In any case, the point is that the asymmetry of the twin paradox is simply Doppler effect plus signal delay. Nothing else is needed for an intuitive understanding of what is happening there. There is not even a need to understand the exact relativistic Doppler formula because the Doppler shift is symmetric. $\endgroup$ Commented Apr 2, 2023 at 23:14

The reason why acceleration enters into the twin paradox is because the traveling twin can neither leave nor return without accelerating. This is no different from a classical version, though.

That the total time for the traveler is less than for the stationary observer is not caused by acceleration, however. It's caused by signal delay. If we evaluate the Doppler shift of the clock tick signals of the other twin, then it turns out that the Doppler of the traveler turns from slow (red shifted) to fast (blue shifted) immediately when he turns around. The signal delay between the traveler, however, delays this change for the stationary twin for an amount of time, so he will count fewer clock ticks from the traveling twin than the traveling twin will count from the stationary observer.

In a more theoretical treatment we would have to apply the Poincare transformation that includes signal delay, rather than the Lorentz transformation, which does not, if I am not mistaken. So the "paradox" is simply explained by the wrong mental model and calculation procedure of this scenario. It does not require some complex physics.

  • $\begingroup$ But Doppler shift is observer dependent. Clocks are real things. When the spaceship clock returns to Earth and is compared with the non traveling Earth clock, the spaceship clock reads an earlier time. $\endgroup$
    – user150908
    Commented Mar 31, 2023 at 16:45
  • $\begingroup$ @Rob The amount of Doppler shift is the same for both observers because the relative velocity is the same. When that Doppler shift kicks in is different. It's immediate for the observer who changes velocity. It's delayed by the signal propagation for the stationary observer. To me that's the most simple and most intuitive explanation for the twin paradox. $\endgroup$ Commented Mar 31, 2023 at 16:59
  • $\begingroup$ But when the spaceship returns to Earth and their clocks are compared, only the spaceship clock is behind. What each observer sees during the spaceship's travel due to Doppler shift cannot alter the reality of what the clocks actually read when they are finally compared. $\endgroup$
    – user150908
    Commented Apr 1, 2023 at 3:57
  • $\begingroup$ The clock ticks that each observer receives from the other observer's clock are exactly the amount of time that has passed for the other observer. When the traveler returns, his clock sends out the last clock tick and since the signal delay is now zero the count is correct. The same is true for the resting observer. The problem with using only the Lorentz transformation is that it's only correct when both observers are in the same place. That's simply not the case for the twin paradox. Counting clock ticks with delay, however, keeps the correct time. $\endgroup$ Commented Apr 1, 2023 at 4:44
  • 3
    $\begingroup$ @Rob If you are hoping for some sort of validation, then I must disappoint you. I am here to explain physics in the most intuitive way I can think about, not to validate other people's ideas about it. $\endgroup$ Commented Apr 1, 2023 at 16:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.