# Why does time dilation only affect one of the twins?

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## Main Question or Discussion Point

I am more confused by the theory of relativity as I start thinking about it. I have a question and it might sound silly but please, correct me if I am wrong.
Suppose, A and B are twins where A is at the Earth, and B is moving on a spacecraft at a speed near to the speed of light. In this scenario, since there is a relative motion between two, time moves slower for the other observer with respect to own frame(i.e. For A, time slows down for B and vice-versa). But the twin paradox says that the twin who makes a trip with high speed and returns back to the Earth seems younger than the twin who is at the Earth. If A and B both see time slowing for each other, why is that only B is younger? Since B sees that time is slowing down for A, shouldn't A be the one who is younger(in B's frame of reference)?
Also, does the direction of speed have any effect on the time dilation(i.e. if moving towards or away from the observer)?

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Orodruin
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See this Insight. The main thing you are missing is the relativity of simultaneity (see my signature). B is not an inertial observer and changes rest frames mid-way. In each of the two different rest frames for B (outbound and inbound, respectively), A is indeed time dilated. However, simultaneity in the outbound frame is not the same as simultaneity in the inbound frame, making you miss a large part of A's world line if you just try to blindly apply time dilation.

Also, does the direction of speed have any effect on the time dilation(i.e. if moving towards or away from the observer)?
No.

ersa17, Ibix and vanhees71
Dale
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Since B sees that time is slowing down for A, shouldn't A be the one who is younger(in B's frame of reference)?
B’s frame of reference is non-inertial, so the rules for inertial frames do not apply. Things look very different in non inertial frames

ersa17 and Ibix
Ibix
But the twin paradox says that the twin who makes a trip with high speed and returns back to the Earth seems younger than the twin who is at the Earth.
Not quite. "High speed" is never a meaningful phrase unless you specify soeed with respect to something. The twin paradox actually says that the twin who moves non-inertially will be (not seem, actually be) younger than the twin who moves inertially.
If A and B both see time slowing for each other
It's not that simple. They don't see the other's clock tick slowly directly. They will see the other's clock appear to tick fast or slow due to the changing light speed delay from the changing distance. Only when they correct for that will they calculate that the other's clock is ticking slowly. This is a point that often isn't made clearly.

The problem is that the calculations that lead to "the other guy's clock is ticking slow" assume that you are both moving inertially, so do not apply to the twin paradox where one of you isn't. You can naively apply the calculation to the outbound and inbound legs of the journey separately but, as Orodruin says, you need to look up the relativity of simultaneity in order to join the results together correctly.

ersa17
phinds
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Why does time dilation affects only one of the twins?
It doesn't. Time dilation is symmetrical and an observational effect. You are confusing it with differential aging, which is NOT symmetrical and is a real effect, not an observational effect. The traveling twin takes a different path through space-time so even though his clock ticks away at one second per second, just as does the clock of his Earthbound twin, his clock gets fewer tics because of the different path through spacetime, thus he ends up younger.

ersa17 and Ibix
Now I believe that I have somewhat understood it. Because of the different worldline, they're traveling in and t ≠ t', even though both the observers see the other observer's time slowing, B is still younger than A. If t = t' then this would have contradicted(or hadn't been possible).

I want to expand on what Ibix said about actual appearances, and discuss what they'll see and not what they compute:
It's not that simple. They don't see the other's clock tick slowly directly. They will see the other's clock appear to tick fast or slow due to the changing light speed delay from the changing distance.
Because of the different worldline, they're traveling in and t ≠ t', even though both the observers see the other observer's time slowing, B is still younger than A.
So about what they actually see, suppose there is a rod of length 1.732 light hours and A is stationed at one end of it, and B is going to travel at 0.866c to the other end and come back. A should age 4 hours during this and B just 2, but I want to describe it in terms of what they see when they watch each other's clocks, in order to show the complete symmetry between the two.
Relativistic dilation factor (gamma) at that speed is 2. Doppler effect is much stronger, so while receding, each will see the other's clock run at about 0.268 the normal rate, and a clock approaching at that speed will appear to run about 3.732 the normal rate.

So B sets out. In his frame, B is stationary and the fast moving rod length-contracts to 0.866 light hours, and it takes the contracted rod one hour to pass by him at that speed, and another hour to pass by in the other direction. Total duration of 2 hours. In the first hour, A's clock appears to log 0.268 hours, and during B's second hour, A's approaching clock appears to log 3.732 hours, for a total of 4 hours .

In A's frame, only B moves. It takes B two hours to go the length of the 1.732 light-hour rod before turning around, and it takes 1.732 hours for the light from the distant turnaround event to reach him, so B's clock will appear to run at 0.268 for 3.732 of A's hours at which point the B clock will appear to read exactly one hour. For the next 0.268 hours A will see B returning with a clock running at 3.732 which accumulates one more hour on B's clock for a total of 2 when they are again in each other's presence.

I did that to show that what they see is entirely symmetrical. Both will see the other clock running at the same slow or fast rate depending on if they're receding or approaching, but due to the asymmetry of their actions, they see these rates for different amounts of time

ersa17
For what it's worth, when I first puzzled over the twins paradox, I found it helped to think about the about the problem without non-inertial frames. It highlights that the reason for what appears to be non-symmetrical time dilation is actually a change of frames, as stated in #2 above.

Here's a quick description of the inertial version:
• Spaceship B, moving at a constant velocity, passes adjacent to A.
• At that moment, both A and B set their clock to 0.
• B continues on, eventually meeting C, who is traveling at a constant velocity toward A.
• At the moment B and C are adjacent, C sets his clock to match B's.
• When C passes by A, they compare times.
• C's clock time will always read less than A's.
It's not necessary that B's and C's velocities match.

ersa17 and PeroK
robphy
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To drive home that the non-inertial traveler is not equivalent to the inertial twin,
I think it's useful to consider the traveler with an asymmetric trip (with numbers chosen to allow calculations with fractions).

While this diagram for inertial twin OPZ
can be thought of as "a splicing together
of the diagrams for OP (using simultaneity according to OP)
and for PZ (using simultaneity according to OP)",
the same can't be said for OQZ.

In accord with what @Orodruin and @Dale said, the non-inertial traveler OQZ
has a very different looking "[attempted] spacetime diagram". A portion of OPZ is missing in OQZ's diagram.
• a portion of OPZ's worldline is missing and OPZ is discontinuous.
In fact, event P and other events in the white region are missing.
The attempted spacetime diagram is incomplete... events are missing.
(This doesn't happen for inertial observers.)

• Event X and other events in the green region appear twice.
(This doesn't happen for inertial observers.)
No Lorentz transformation will straighten out the kink at Q.

The takeaway message is
"Being able-to-be-at-rest" $\neq$ "Being inertial".

ersa17
To drive home that the non-inertial traveler is not equivalent to the inertial twin,
I think it's useful to consider the traveler with an asymmetric trip (with numbers chosen to allow calculations with fractions).
This looks like an artificial example, one that assumes instantaneous acceleration. This is unphysical.

If you add a short period of acceleration, the moving observer's frame changes with time. At any instant, there is no discontinuity and X has a single set of coordinates.

If you allow for instantaneous acceleration, then even a symmetric trip will have missing events and events with multiple coordinates. I don't think the asymmetry is necessary to make that point.

I suspect none of this is going to help the OP as it may require a foundation of knowledge he may not have (if one can make sense of your diagrams, one probably doesn't need help with the twins paradox).

robphy
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This looks like an artificial example, one that assumes instantaneous acceleration. This is unphysical.
That critique applies to any piecewise-inertial motion, including the standard twin paradox/clock effect in special relativity and piecewise-constantVelocity motion in Galilean kinematics.

Ibix
If you add a short period of acceleration, the moving observer's frame changes with time. At any instant, there is no discontinuity and X has a single set of coordinates.
If by this you mean to patch together the instantaneous inertial rest frames of the accelerating twin, this is not correct. The process inevitably assigns multiple coordinates to single events on the outside of the corner in the worldline as depicted on a Minkowski diagram.

Orodruin
That critique applies to any piecewise-inertial motion, including the standard twin paradox/clock effect in special relativity and piecewise-constantVelocity motion in Galilean kinematics.
The twins paradox is vritually unchanged if you allow for a very quick (but not instantaneous) acceleration--and all the unphysical stuff goes away. I always think of it as: "we're not going to look super-closely at what happens at turn-around". If you focus on this and insist on the turn-around being instantaneous, you create a lot of artificial warts.

I proposed an alternate version of the twins paradox in which there is no acceleration back at #9 that also removes a lot of these problems. And the velocities do not need to be symmetrical.

If by this you mean to patch together the instantaneous inertial rest frames of the accelerating twin, this is not correct. The process inevitably assigns multiple coordinates to single events on the outside of the corner in the worldline as depicted on a Minkowski diagram.
I'll go out on a limb and claim that, for the moving observer, there is no single instant in which the process assigns multiple coordinates to a single event. At different instants, an event will have different coordinates, but never simultaneously for the moving observer (unless you allow for instantaneous acceleration).

But perhaps you can show me what you mean with a diagram. I can't picture it.

Ibix
But perhaps you can show me what you mean with a diagram. I can't picture it.
Draw a curved worldline on a Minkowski diagram. Draw a line perpendicular to that line at some event (in the Minkowski sense of perpendicular), and that line is the locus of events sharing a t-coordinate value using your methodology. Now draw another perpendicular line at another event on the worldline. That line is also the locus of events sharing a t-coordinate, but a different value from the first line.

Unless the velocity of the worldline is equal at the two events from which you drew perpendiculars, those lines are not parallel. Thus they intersect at some event - and you have assigned two different t-coordinates to this event. For the special case of constant acceleration this event is the intersection of the Rindler horizon and anti-horizon.

For illustrations see figure 1 (instantaneous turnaround) and 2 (smooth turnaround) in Dolby and Gull.

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Orodruin
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I'll go out on a limb and claim that, for the moving observer, there is no single instant in which the process assigns multiple coordinates to a single event.
You can define a coordinate system where the non-inertial twin is at rest, but this coordinate system is not going to be an inertial frame. If you want to use this, you need to dig a lot deeper than describing things in terms of inertial frames.

Ibix
Draw a curved worldline on a Minkowski diagram. Draw a line perpendicular to that line at some event (in the Minkowski sense of perpendicular), and that line is the locus of events sharing a t-coordinate value using your methodology. Now draw another perpendicular line at another event on the worldline. That line is also the locus of events sharing a t-coordinate, but a different value from the first line.
Sure, that's obvious. But those are two different points in time for the observer whose worldline you are drawing tangents to. There is no single instant for that observer at which an event has two coordinates.

On the other hand, if you allow for instantaneous acceleration, there is one instant in which the observer would have two valid and different sets of coordinates for some event.

In any case, I feel I have introduced a discussion that threatens to hijack the OP's thread and I'd rather not go further down that path.

Relevant to the OP are my thoughts that one can view the problem using only inertial frames and one can view the problem with a very short and dramatic velocity change at the turn-around point.

weirdoguy
robphy
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There is no single instant for that observer at which an event has two coordinates.
Why raise this issue?

The point concerning my diagram is that the "attempted spacetime diagram" is not a complete map of spacetime and is not obtained as a one-to-one map from the spacetime diagram for an inertial observer.

I made no claim concerning "a single instant... at which an event has two coordinates."

I claimed that there are events, like event X, that have two sets of coordinates.
Said another way, the same event X (like the particular intersection of two worldlines) occurs at different times according the non-inertial observer.

Orodruin
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There is no single instant for that observer at which an event has two coordinates.

On the other hand, if you allow for instantaneous acceleration, there is one instant in which the observer would have two valid and different sets of coordinates for some event.
This is a misconception and confuses observers with inertial frames.

Relevant to the OP are my thoughts that one can view the problem using only inertial frames and one can view the problem with a very short and dramatic velocity change at the turn-around point.
Your misconceptions only serve the purpose of confusing the OP. The only relevant thing to this problem in reality is piecewise smooth curves and the geometry of Minkowski space. Whether the turnaround is smooth or not is not that relevant.

Even in the case of continuous acceleration the entire process can be described using a single inertial frame, two inertial frames, three inertial frames, or whatever coordinates you want to impose. The physics will remain the same though.

Ibix, vanhees71 and PeterDonis
vanhees71
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The twin paradox is only invented to confuse students ;-). It's easily resolved when just expressing the relevant quantities in a manifestly covariant form and the conjectore that a moving proper clock shows proper time,
$$\tau=\int_{\lambda_1}^{\lambda_2} \mathrm{d} \lambda \sqrt{g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}},$$
where ##x^{\mu}(\lambda)## is the parametrization of the time-like worldline of the clock. In this form it's completely independent of the choice of space-time coordinates and the parametrization. You can calculate it in any frame of reference (inertial or not), and it's even valid in general relativity with gravitation (or spacetime curvature) present.

That this conjecture is correct has been experimentally demonstrated in many high-precision experiments (for both situations that gravitation is necessary to be taken into account or not).

I claimed that there are events, like event X, that have two sets of coordinates.
That shouldn't be a big surprise with two different frames of reference. With non-instantaneous acceleration X has even infinite different sets of coordinates. Yes, this is just an artefact resulting from the use of inertial rest frames for a non-inertial observer. But there is nothing wong with it.

vanhees71
robphy
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That shouldn't be a big surprise with two different frames of reference. With non-instantaneous acceleration X has even infinite different sets of coordinates. Yes, this is just an artefact resulting from the use of inertial rest frames for a non-inertial observer. But there is nothing wong with it.
It is this feature (yes, which is to be expected, but only if one looks hard enough)
that I use to distinguish a non-inertial observer from an inertial one.
That was the point of my initial posting in this thread.
Straightening out the non-inertial worldline (which is implied by "doing time-dilation calculations in the traveler-leg frames") is not enough
to justify an attempted equivalence of a non-inertial observer with an inertial one.
One has to transform [map] the spacetime as well.

"Being able-to-be-at-rest" ≠ "Being inertial".

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PeterDonis
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I'll go out on a limb and claim that, for the moving observer, there is no single instant in which the process assigns multiple coordinates to a single event.
You are wrong. The moving observer's lines of constant time, according to your definition, will inevitably cross at some distance "below" him (i.e., in the direction opposite to the direction in which he is accelerating), which means his coordinate chart will assign multiple values of his ##t## coordinate to the same event. This is a well-known limitation of the method you are describing.

vanhees71