How does reciprocal time dilation work?

[Mike_bb]

Hello!

I looked at the examples with two observers that compare time of each other.
I encountered with problem. I can't understand how is it possible that one clock shows two measurements of time at the moment.
For example, my friend's clock shows 8sec and my clock shows 10sec. But my friend says that him clock shows 10sec and my clock shows 8sec. From this friend's clock shows 8sec and 10sec. How is it possible?

Thanks.

 

[A.T.]

Look up: Relativity of Simultaneity

 

[Ibix]

Hello!

I looked at the examples with two observers that compare time of each other.
I encountered with problem. I can't understand how is it possible that one clock shows two measurements of time at the moment.
For example, my friend's clock shows 8sec and my clock shows 10sec. But my friend says that him clock shows 10sec and my clock shows 8sec. From this friend's clock shows 8sec and 10sec. How is it possible?

Thanks.

@A.T. gives the answer. In short, if the two clocks are in the same place (even if they are only passing by) then you will both agree the readings on the clocks at that instant. Those readings may be different, but you will both agree what they are. When the clocks are not in the same location, however, two frames won't generally agree on what the clocks say. This is because the notion of simultaneity depends on what frame you use, so what your friend's clock shows simultaneously with your clock showing 10s (or whatever) depends on which frame you use. Simultaneity is relative in relativity.

The underlying reason for this is that relativity models 4d spacetime, and there is more than one way to "slice" 4d spacetime into a stack of 3d slices that represent "all of space now". Depending on which "angle" you slice at, the slice that contains the event "your clock reads 10s" goes through different events on the history of your friend's clock. So your question is mis-conceived. Your friend's clock operates normally and reads 8s and 10s at different times - you just changed your mind about when on your friend's worldline you were calling "at the same time as my clock reads 10s".

 

[FactChecker]

The clocks are traveling long distances in opposite relative directions to each other.
The ticking rate of a traveling clock is most simply measured by recording it here and down the line later and comparing it to my clocks at each location, which are "synchronized" within my inertial reference frame (IRF). How I synchronized my clocks is critical. That brings in the subject of "relativity of simultaneity". It can not be avoided.

The same is true for the clocks in the moving IRF. Since the two IRFs synchronized clocks differently, we get the result that seems confusing, but is really dirt-simple.

 

[Mike_bb]

Your friend's clock operates normally and reads 8s and 10s at different times - you just changed your mind about when on your friend's worldline you were calling "at the same time as my clock reads 10s".

If it's so then what time will friend's clock show when my clock show 10s? And what time will my friend see on my clock at the same moment?

 

[robphy]

A geometric analogy may be useful here.
Consider two radius vectors to a circle, then consider their tangent-lines. (See below.)

An updated version of my desmos linked below is at
desmos.com/calculator/emqe6uyzha
- turn on the ----timeDilation folder by clicking the unfilled circle
- adjust the E-slider to go from (E=-1) special relativity, (E=0) Galilean, (E=+1) Euclidean


From my post in Why does each clock see the other's period as longer in time dilation?

My answer to https://physics.stackexchange.com/questions/383248/how-can-time-dilation-be-symmetric
uses diagrams like this to display the reciprocity

which is analogous to what happens in ordinary Euclidean geometry

(using desmos.com/calculator/wm9jmrqnw2 with E=-1 ).

 

[Dale]

If it's so then what time will friend's clock show when my clock show 10s? And what time will my friend see on my clock at the same moment?

"At the same moment" in which frame?

Whenever you mention any of the following quantities, you must also specify the frame:
1) simultaneity (e.g. at the same moment)
2) velocity
3) length or distance
4) time dilation or clock rate

 

[Ibix]

If it's so then what time will friend's clock show when my clock show 10s? And what time will my friend see on my clock at the same moment?

What do you mean by "show" here? If you're looking through a telescope at it then there's a unique answer, but we don't know what it is because we don't know how far away it is or how fast it's going relative to you, nor how it was zeroed. If you specify those things, we can calculate it.

If you mean "at the same time as my clock reads 10s what does my friend's clock read" then there is no unique answer. There is a range in which any reasonable answer will lie (again we need extra information to calculate this), but within that range it is a matter of personal choice. There are no physical consequences to that choice.

 

[Ibix]

And what time will my friend see on my clock at the same moment?

Forgot to answer this bit. First (again) you need to say if you mean "see" literally, and then you need to say what choice of simultaneity you're using to define "at the same moment". If you don't mean "see" literally you also have to specify what your friend chooses to mean by "at the same moment" (which need not be the same thing you mean). And then you need to specify the velocity hostory and how your clocks were zeroed before we can actually answer the question.

 

[Mike_bb]

Forgot to answer this bit. First (again) you need to say if you mean "see" literally, and then you need to say what choice of simultaneity you're using to define "at the same moment". If you don't mean "see" literally you also have to specify what your friend chooses to mean by "at the same moment" (which need not be the same thing you mean). And then you need to specify the velocity hostory and how your clocks were zeroed before we can actually answer the question.

Let ship A and ship B fly parallel to each other with velocity $v=0.6c$.
Lorentz-factor: 1.25

1) At point of view of observer A:

Time of observer's A clock: 100sec
Time of observer's B clock: 100/1.25=80sec

2) At point of view of observer B:

Time of observer's B clock: 100sec
Time of observer's A clock: 100/1.25=80sec

 

[Dale]

Yes. Assuming "point of view" means "reference frame", and assuming that $v$ is their relative velocity instead of both flying with velocity $v$ in some third frame.

 

[Ibix]

Let ship A and ship B fly parallel to each other with velocity $v=0.6c$.
Lorentz-factor: 1.25

1) At point of view of observer A:

Time of observer's A clock: 100sec
Time of observer's B clock: 100/1.25=80sec

2) At point of view of observer B:

Time of observer's B clock: 100sec
Time of observer's A clock: 100/1.25=80sec

Additional to @Dale's comments, you also assume that they zeroed their clocks when they passed one another. And you are talking about what they calculate the other's clock to read, each using the Einstein synchronisation convention associated with their rest frame, and not what they literally see on the other's clock, which would be affected by light speed delay.

 

[Mike_bb]

Ibix,

Reciprocal time dilation effect works if two observers are moving (i.e. clocks are always in motion), right?

 

[Ibix]

Ibix,

Reciprocal time dilation effect works if two observers are moving (i.e. clocks are always in motion), right?

As long as both clocks are inertial, yes. They also need to use the Einstein synchronisation convention, which they will do unless they have a really good reason not to.

 

[FactChecker]

Ibix,

Reciprocal time dilation effect works if two observers are moving (i.e. clocks are always in motion), right?

Any IRF can be considered to be "stationary". So either one of the clocks in their respective IRF can be "stationary". It is not about which one is moving (both can be moving). It is about the difference in clock synchronization within the two IRFs.

 

[Matterwave]

Let's try to get clear on what's going on. It's super easy to get a lot of misconceptions and it all has to do with lack of verbal clarity of what the mathematical objects in the theory mean.

We first start by defining what a coordinate system even is in SR.

A coordinate system for an observer $O$ is simply a set of 4 numbers $(t, x, y, z)$ that that observer assigns to any given event in spacetime. It is an idealized mathematical construction. The same way that Cartesian coordinates on a Euclidean space are an idealized construction.

How does one "visualize" how this coordinate system is constructed? One visualizes that this observer lays out an infinite grid of rigid rods, each of unit length, and at each grid point the observer attaches a clock. Every clock in this infinite set of clocks are synchronized to each other. All rulers and clocks are at rest relative to the observer. There is a point on this grid which we can define to be the spatial origin. At one time, we set every clock to 0. Now the coordinate system is constructed.

A different observer $O'$ constructs his coordinate system in the exact same way. Observers $O$ and $O'$ synchronize their origins, and we assume that the two observers align their spatial coordinate axes (i.e. we don't consider rotations) and finally that $O'$ moves with respect to $O$ with velocity $v$ along the (positive) x-direction.

Given this set up, the Lorentz transformation gives us how the coordinates of events transform between the two coordinate systems. I.e.
$$
\begin{aligned}
t' &= \gamma \left( t - \frac{v x}{c^2} \right) \\
x' &= \gamma (x - v t) \\[1.5ex]
\end{aligned}
$$
And:
$$
\begin{aligned}
t &= \gamma \left( t' + \frac{v x'}{c^2} \right) \\
x &= \gamma (x' + v t')
\end{aligned}
$$

Finally the pay off. What does "time dilation" mean? Suppose two events occur, $A$ and $B$. Suppose further that both events happen at observer $O$'s spatial origin, but some time $(\Delta t)_{AB}$ apart, with event $A$ happening at $t_A=0$. Then if you take the Lorentz transformation, you will find:

$$(\Delta t')_{AB} = t'_B-t'_A = \gamma (\Delta t)_{AB}$$

It's easy to see this since $t'_A=t_A= 0$ and $x_A=x_B=0$. Voila, you have time dilation!

Now, suppose that there are events $C$ and $D$ which both happen at $O'$'s spatial origin and some time $(\Delta t')_{CD}$ apart with $C$ happening at $t'_C=0$. Use the latter set of Lorentz transformations to see:

$$(\Delta t)_{CD} = t_D-t_C = \gamma (\Delta t')_{CD}$$

This is your "reciprocal time dilation" as best as I can understand it. The key things to take away:

1. Coordinate systems are idealized. Things like "how long does it take for the light from clocks in $O$'s grid far away from $O$ to reach $O$?" are not considered. In particular, events which happen spatially separated from each other in a particular frame are given time-coordinates (by that same frame) based off of different, though synchronized, clocks.
2. "Reciprocal" time dilation is a very loose term. Note how in the construction, we kept pairs of events $A,B$ and $C,D$ distinct from each other. That's because they are all different events! (Technically, given our set up, events $A$ and $C$ could be the same event since they both happen at $x=t=0$ and $x'=t'=0$ but that's just an artifact of how we set up and synchronized our coordinate systems and events). If you keep things separated like this in your head, you will avoid a lot of confusion.
3. In relativity, is best to talk about invariants. Events $A$, $B$, $C$, $D$ are physically meaningful events. The coordinates of those events in any particular frame are just a set of numbers assigned to them. The spacetime interval between them are invariants and you can always ground your reasoning on invariants. Invariants are the geometrically and physically meaningful objects of the theory. This will be especially true once you transition from SR to GR.

 

[robphy]

This comment complements the recent comment by @Matterwave ,
as well as directly connect to your recent calculation.

Let ship A and ship B fly parallel to each other with velocity $v=0.6c$.
Lorentz-factor: 1.25

1) At point of view of observer A:

Time of observer's A clock: 100sec
Time of observer's B clock: 100/1.25=80sec

2) At point of view of observer B:

Time of observer's B clock: 100sec
Time of observer's A clock: 100/1.25=80sec

This is exactly the situation in my first diagram above,
where each diamond refers to a 10-sec interval of proper-time.

$V_{Bob\ wrt\ Alice}=\frac{PP'}{OP}=\frac{6}{10}$,
where $OP$ is along Alice's worldline (maybe "Alice's timeline" is a better term),
meeting the hyperbola "Minkowski-circle" at event $P$,
and $PP'$ is parallel to the tangent to that circle at $P$.
Geometrically, following Minkowski (1908),
$PP'$ is Minkowski-orthogonal [or "-perpendicular", "-normal"] to $OP$.
Physically, $PP'$ trace out events that are simultaneous according to Alice.
($PP'$ could be called "a spaceline-segment according to Alice".)

Alice says the distant-event $P'$ on Bob's worldline is simultaneous with her local-event $P$,
which is (by counting) after an elapsed time of 10 of Alice's ticks (= 100 sec).
We can count off that 8 of Bob's ticks (= 80 sec) elapses from the meeting event $O$ to event $P'$.
So, as you say,

1) At point of view of observer A:

Time of observer's A clock: 100sec
Time of observer's B clock: 100/1.25=80sec

Geometrically, note the 6-10-8 Minkowski-right-triangle (similar to a 3-5-4 one),
where $OP$ and $PP'$ are the orthogonal legs
(whose ratio $(PP')/(OP)$ gives the velocity of Bob according to Alice),
and $OP'$ is the hypotenuse (opposite the "Minkowski-right-angle at $P$").
The time-dilation factor is $\gamma=\cosh\theta=\frac{\rm ADJACENT}{\rm HYPOPTENUSE}=\frac{OP}{PP'}=\frac{10}{8}$,
where $\theta$ is the relative-rapidity between $OP$ and $OP'$
and $\tanh\theta=\frac{\rm OPPOSITE}{\rm ADJACENT}=\frac{PP'}{OP}=\frac{6}{10}$ is the relative-[dimensionless-]velocity.

Bob (also being an inertial observer from event $O$) meets the same "circle" at $Q$,
which is also after an elapsed time of 10 of Bob's ticks (= 100 sec).
That's why this is a "circle" in Minkowski spacetime.
Bob can do a (literally) similar construction.
$QQ'$ is tangent at $Q$.
Distant-event $Q'$ on Alice's worldline is simultaneous-according-to-Bob with his local-event $Q$.
etc... etc...
So, as you say,

2) At point of view of observer B:

Time of observer's B clock: 100sec
Time of observer's A clock: 100/1.25=80sec

Time-dilation qualitatively:
With 10 ticks to get to the "circle",

• at Alice's 10th-tick, Alice's tangent line to the "circle" meets Bob's worldline
  before Bob reaches the "circle" (before Bob's clock elapses 10 ticks).
• at Bob's 10th-tick, Bob's tangent line to the "circle" meets Alice's worldline
  before Alice reaches the "circle" (before Alice's clock elapses 10 ticks).

For the Euclidean geometric analogue (the E= -1 case) , look back (above) at the ordinary circle and its tangent-lines, each meets the other radial-axis at a distance along that axis greater than the circle-radius.

In these two cases, along different radii meeting the circle, the tangents are in different directions.
This is the relativity-of-simultaneity.

For the Galilean analogue (the E=0 case), look at the desmos.
The "circle" in Galilean-spacetime is the hyperplane (in 1+1, a spatial straight-line),
whose tangent-[hyper]planes (tangent-lines) coincide.
This leads to "absolute time", with no-time-dilation.
(In Galilean spacetime geometry, the hypotenuse has the same size as the adjacent side. The underlying geometry of the PHY101 position-vs-time graph is non-euclidean (a flat non-euclidean one).)

 

[Sagittarius A-Star]

Let ship A and ship B fly parallel to each other with velocity $v=0.6c$.
Lorentz-factor: 1.25

You describe 4 events (E1 ... E4).

1) At point of view of observer A:

Time of observer's A clock: 100sec
Time of observer's B clock: 100/1.25=80sec

E1: clock A shows 100s
E2: clock B shows 80s

2) At point of view of observer B:

Time of observer's B clock: 100sec
Time of observer's A clock: 100/1.25=80sec

E3: clock B shows 100s
E4: clock A shows 80s

I call the coordinates of frame A (x, y, z, t) and the coordinates of frame B (x', y', z', t').

1) => Events E1 and E2 happen simultaneous for observer A: $\Delta t_{21} = t_2 - t_1 = 0s$.
But events E1 and E2 do not happen simultaneous for observer B, if they happen at different $x$ coordinates, according to the Lorentz-transformation: $\Delta t'_{21} \neq 0s$.

2) => Events E3 and E4 happen simultaneous for observer B: $\Delta t'_{43} = t'_4 - t'_3 = 0s$.
But events E3 and E4 do not happen simultaneous for observer A, if they happen at different $x'$ coordinates, according to the Lorentz-transformation: $\Delta t_{43} \neq 0s$.

So there is no contradiction.

The Lorentz transformation for time depends on $\Delta x$:
$\Delta t '= \gamma (\Delta t - v \Delta x / c^2)$.
If $\Delta t=0$, then $\Delta t '= -\gamma v \Delta x / c^2$.

The inverse Lorentz transformation for time depends on $\Delta x'$:
$\Delta t = \gamma (\Delta t' + v \Delta x' / c^2)$.
If $\Delta t'=0$, then $\Delta t = \gamma v \Delta x' / c^2$.

 

[Mike_bb]

1) => Events E1 and E2 happen simultaneous for observer A: $\Delta t_{21} = t_2 - t_1 = 0s$.
But events E1 and E2 do not happen simultaneous for observer B, if they happen at different $x$ coordinates, according to the Lorentz-transformation: $\Delta t'_{21} \neq 0s$.

2) => Events E3 and E4 happen simultaneous for observer B: ##\Delta t'_{43} = t'_4 - t'_3 = 0s ##.
But events E3 and E4 do not happen simultaneous for observer A, if they happen at different $x'$ coordinates, according to the Lorentz-transformation: $\Delta t_{43} \neq 0s$.

Big thanks! But I don't understand why " events E1 and E2 do not happen simultaneous for observer B," and "events E3 and E4 do not happen simultaneous for observer A, "?

If I understand correct Frame A and Frame B move then they don't happen simultaneous, right?

 

[A.T.]

Big thanks! But I don't understand why

Because that's how the Lorentz Transformation works.

 

[Mike_bb]

Sagittarius A-Star,

$\Delta t$ is slowed time ?

Could you explain what time do clocks show in our example with ship A and ship B? Thx.

 

[Sagittarius A-Star]

Big thanks! But I don't understand why " events E1 and E2 do not happen simultaneous for observer B," and "events E3 and E4 do not happen simultaneous for observer A, "?

If I understand correct Frame A and Frame B move then they don't happen simultaneous, right?

You wrote in posting #10 under 1) that for observer A, events E1 and E2 happen simultaneously.

• Newtonian physics postulates (wrongly) that an absolute true time exists and that, if E1 and E2 are simultaneous for one observer, they must be simultaneous for all observers.
• Einsteinean physics requires (correctly), that no absolute true time exists. Each observer has his own time. If E1 and E2 are simultaneous for one observer, they are not simultaneous for all observers.

 

[Mike_bb]

You wrote in posting #10 under 1) that for observer A, events E1 and E2 happen simultaneously.

• Newtonian physics postulates (wrongly) that an absolute true time exists and that, if E1 and E2 are simultaneous for one observer, they must be simultaneous for all observers.
• Einsteinean physics requires (correctly), that no absolute true time exists. Each observer has his own time. If E1 and E2 are simultaneous for one observer, they are not simultaneous for all observers.

I understand it. Thx! Please response to my post#27.

 

[Sagittarius A-Star]

Sagittarius A-Star,

$\Delta t$ is slowed time ?

$\Delta t$ is the time interval between two events with respect to frame (x, y, z, t).

Could you explain what time do clocks show in our example with ship A and ship B? Thx.

You did not specify the problem completely. Therefore I adopt the following assumption:

Additional to @Dale's comments, you also assume that they zeroed their clocks when they passed one another.

I call the event, at which both clocks met and were zeroed: E0.

Case 1)
E1: clock A shows 100s
E2: clock B shows 80s

The time-interval between E0 and E2 with respect to frame A (x, y, z, t) is given as $\Delta t_{20} = t_2 - t_0 = 100s.$
The distance between E0 and E2 with respect to frame B (x', y', z', t') is $\Delta x'_{20} = 0$ (clock B is at rest in frame B).

Inverse LT:
$\Delta t_{20} = \gamma (\Delta t'_{20} + v \Delta x'_{20} /c^2) = \gamma (\Delta t'_{20} + 0)$
$100s = \gamma \Delta t'_{20}$
Clock B shows: $\Delta t'_{20} = 100s / \gamma.$

Case 2) can be calculated in an analog way.

 

[anuttarasammyak]

For example, my friend's clock shows 8sec and my clock shows 10sec. But my friend says that him clock shows 10sec and my clock shows 8sec. From this friend's clock shows 8sec and 10sec. How is it possible?

I try to keep reciprocality in explanation.

A and B agree that their watches show the same time when they pass by.

A has resident agents everywhere. A ordered them to observe watches of B and their own when B and they pass by. Integrated report reveals that watch of B goes 8 sec for every 10 sec.

B has resident agents everywhere. B ordered them to observe watches of A and their own when A and they pass by. Integrated report reveals that watch of A goes 8 sec for every 10 sec.

Question to both A and B : How do you think of other's observation ?
Answers : I do right way and he/she does wrong because his/her agents have wrongly synchronized watches.

 

[Mike_bb]

Case 2) can be calculated in an analog way.

Yes. But how is it possible that clock B shows 80s (case 1) and at the same moment clock B shows 100s(case 2)?

 

[Dale]

at the same moment

Whenever you say “at the same moment” you need to specify the reference frame. I have told you this before.

Undergrad Post in thread 'How does reciprocal time dilation work?'

Yesterday, 11:06 AM

If it's so then what time will friend's clock show when my clock show 10s? And what time will my friend see on my clock at the same moment?

"At the same moment" in which frame?

Whenever you mention any of the following quantities, you must also specify the frame:
1) simultaneity (e.g. at the same moment)
2) velocity
3) length or distance
4) time dilation or clock rate

• Dale

• 

 

[Mike_bb]

Whenever you say “at the same moment” you need to specify the reference frame. I have told you this before.

I see on the clock B from the frame A and clock B shows 80s. My clock in frame A shows 100s. Observer from the frame B see on my clock in the frame A that shows 80s. Observer's clock in frame B shows 100s.

How is it possible?

 

[Nugatory]

Yes. But how is it possible that clock B shows 80s (case 1) and at the same moment clock B shows 100s(case 2)?

Because of that phrase "at the same moment". They are different "same moments", because if A and B are moving relative to one another they have different definitions of "same moment". That's relativity of simultaneity.

And how can this be?
Well, let's go back to a more intuitive situation. It is 2026 and you are looking at the sky; you see a flash of light from a star ten light-years away exploding. When did the star explode? In 2016, of course, and then it took ten years for the light to to get to your eyes in 2026.

And (maybe you're just lucky spotting astronomical phenomena) it so happens that in 2021 you saw another star explode, this one just five light-years away. When did that star explode? Also in 2016, of course, so that five years of light travel time got the light to your eyes in 2021.

So both stars exploded at the same time, in 2016. You should be able to convince yourself that not only is this a sensible way of establishing that they both exploded at the same time, it is the only sensible way.

Now consider this situation as understood by someone moving relative to you and the stars - and remember that the speed of light is the same for everyone. They will go through the same "allowing for light travel time" calculation because it is the only sensible way of determining when something "really happened"... And they will conclude that the two explosions did not happen at the same time. That's relativity of simultaneity: the "same instant" is different in different frames.
(I won't step through the arithmetic myself because you really should try it for yourself but I will give you one hint - it's easier if we consider the "someone else" to be at rest while you and the stars are moving towards them, instead of you and stars at rest while they're moving. But if you try it and get stuck, I or someone else here can show you).

Any time that you say something like "At time X their clock reads Y" you are really saying "their clock reads Y at the same time, using the frame in which I am at rest, that my clock reads X" and clearly that depends on what we mean by "at the same time".

 

[Matterwave]

I see on the clock B from the frame A and clock B shows 80s. My clock in frame A shows 100s. Observer from the frame B see on my clock in the frame A that shows 80s. Observer's clock in frame B shows 100s.

How is it possible?

These are not the same moment, there is no universal same moment. That is exactly the point of the relativity of simultaneity.

Please see any of our replies above. You have everything you need to answer this question, sometimes you just need to sit and think on it.