Help understanding the Relativity of Simultaneity

[Chenkel]

Hello everyone,

So I think maybe what confused me with the symmetry of time dilation was not understanding relativity of simultaneity.

After looking into it if you have two clocks P and Q that synchronize when they meet and if $\gamma$ is 2 then from P's reference frame P equaling 10 and Q equaling 5 is simultaneous from P's reference frame but P equaling 5 and Q equaling 10 is simultaneous from Q's reference frame.

Each reference frame has it's own simultaneity calculations and what's simultaneous in the rest frame of one clock is not necessarily simultaneous in the rest frame of another clock.

 

[Mister T]

After looking into it if you have two clocks P and Q that synchronize when they meet and if γ is 2 then from P's reference frame P equaling 10 and Q equaling 5 is simultaneous from P's reference frame but P equaling 5 and Q equaling 10 is simultaneous from Q's reference frame.

Assuming that when they met and synchronized clocks, they set them to zero.

I find your explanation rather clever. I've never thought about it this way and don't recall having ever seen it before.

Edit: I suggest the OP or a mentor fix the typo in the thread title.

 

[Sagittarius A-Star]

Hello everyone,

So I think maybe what confused me with the symmetry of time dilation was not understanding relativity of simultaneity.

After looking into it if you have two clocks P and Q that synchronize when they meet and if $\gamma$ is 2 then from P's reference frame P equaling 10 and Q equaling 5 is simultaneous from P's reference frame but P equaling 5 and Q equaling 10 is simultaneous from Q's reference frame.

Each reference frame has it's own simultaneity calculations and what's simultaneous in the rest frame of one clock is not necessarily simultaneous in the rest frame of another clock.

Yes. In the following Minkowski spacetime diagram with $\gamma = 2$, you could regard the vertical $ct$ axis as the world line of clock $P$ and the $ct'$ axis as the world line of clock $Q$. Please be aware, that the primed and unprimed axes have different scales. The scales of the unprimed axes can be constructed using the shown hyperbola. At event $D'$, clock $Q$ shows 5. The orange line is parallel to the $x'$ axis and connects events, that are simultaneously with reference to the restframe of $Q$.

Source:
https://www.geogebra.org/m/NnrRvA46

 

[Chenkel]

Yes. In the following Minkowski spacetime diagram with $\gamma = 2$, you could regard the vertical $ct$ axis as the world line of clock $P$ and the $ct'$ axis as the world line of clock $Q$. Please be aware, that the primed and unprimed axes have different scales. The scales of the unprimed axes can be constructed using the shown hyperbola. At event $D'$, clock $Q$ shows 5. The orange line is parallel to the $x'$ axis and connects events, that are simultaneously with reference to the restframe of $Q$.

View attachment 337770​

Source:
https://www.geogebra.org/m/NnrRvA46

The orange line looks like it's at a diagonal, how can it be parallel to the x primed axis?

 

[Sagittarius A-Star]

The orange line looks like it's at a diagonal, how can it be parallel to the x primed axis?

It's not completely diagonal. I propose that you follow the Geogebra link under the picture and play around with the angle to see it better.

 

[Chenkel]

It's not completely diagonal. I propose that you follow the Geogebra link under the picture and play around with the angle to see it better.

I'm a little unfamiliar with Geogebra but I'll try to get it working.

 

[Sagittarius A-Star]

I'm a little unfamiliar with Geogebra but I'll try to get it working.

On the top you see the green slider for the angle. You can change the angle by shifting it with a pressed mouse button.

 

[Chenkel]

It's not completely diagonal. I propose that you follow the Geogebra link under the picture and play around with the angle to see it better.

I'm not sure this app works on my phone.

 

[Sagittarius A-Star]

I'm not sure this app works on my phone.

On my phone I can use a finger to change the slider. If this does not work at your phone, it doesn't matter. If you just open the link you see the angle 21.5° and better in purple color parallel lines to the $ct'$ axis and to the $x'$ axis.

 

[Chenkel]

On my phone I can use a finger to change the slider. If this does not work at your phone, it doesn't matter. If you just open the link you see the angle 21.5° and better in purple color parallel lines to the $ct'$ axis and to the $x'$ axis.

I don't think the app is fully functional on my phone, I see a rendering with the angle you mentioned but its hard to interact.

 

[Sagittarius A-Star]

I don't think the app is fully functional on my phone, I see a rendering with the angle you mentioned but its hard to interact.

Also no problem. I marked the primed axes in orange. The angle $\alpha = \arctan {v \over c}$.

 

[A.T.]

So I think maybe what confused me with the symmetry of time dilation was not understanding relativity of simultaneity.

It can help to look at all the properties of the Lorentz-Transformation visualized together:

 

[PeroK]

I don't think the app is fully functional on my phone, I see a rendering with the angle you mentioned but its hard to interact.

I recommend focusing on Morin's book. The Relativity of Simultaneity (or Loss of Simultaneity) is the first thing he covers (1.3.1). Time dilation is the second (1.3.2). Spacetime diagrams come later (after chapter 1).

Perhaps spacetime diagrams would help at this stage - Morin obviously decided otherwise, so why not trust him? Or, perhaps, it's just not possible to learn everything at once. In any case, if you don't understand the basics (sections 1.3.1 and 1.3.2 in Morin), then I don't see how you can understand the Lorentz Transformation or Spacetime diagrams at this stage.

 

[Mister T]

The orange line looks like it's at a diagonal, how can it be parallel to the x primed axis?

Do you see why the event D' has a x' coordinate of 5?

 

[Chenkel]

Do you see why the event D' has a x' coordinate of 5?

I might be too inexperienced at Minkowski diagrams to follow, but I did try to get it, I'm going to keep reading and maybe come back to it later.

 

[PeterDonis]

I think maybe what confused me with the symmetry of time dilation was not understanding relativity of simultaneity.

Since you were explicitly told that multiple times in previous threads, yes, that is definitely something you should think. No "maybe" about it.

Each reference frame has it's own simultaneity calculations and what's simultaneous in the rest frame of one clock is not necessarily simultaneous in the rest frame of another clock.

Yes, indeed (but, as @FactChecker has now pointed out, the "not necessarily" is wrong--it should be just "not").

 

[FactChecker]

Hello everyone,

So I think maybe what confused me with the symmetry of time dilation was not understanding relativity of simultaneity.

Yes, that is the key.

Each reference frame has it's own simultaneity calculations and what's simultaneous in the rest frame of one clock is not necessarily simultaneous in the rest frame of another clock.

For clocks that are synchronized in the direction of relative motion, you can say that they can not "be simultaneous in the rest frame of another clock". "Not necessarily" is too weak. If they were the same, how could both frames measure the speed of the same light flash as c?

 

[Chenkel]

Yes, that is the key.

For clocks that are synchronized in the direction of relative motion, you can say that they can not "be simultaneous in the rest frame of another clock". "Not necessarily" is too weak. If they were the same, how could both frames measure the speed of the same light flash as c?

So if two clocks are synchronized when they meet then what's simultaneous for one clock is not simultaneous for the other clock.

Each clock will calculate the other clock is aging slowly, leading to completely different simultaneity.

 

[Chenkel]

I'm not sure if my last post was clear enough...

 

[Chenkel]

Yes, that is the key.

For clocks that are synchronized in the direction of relative motion, you can say that they can not "be simultaneous in the rest frame of another clock". "Not necessarily" is too weak. If they were the same, how could both frames measure the speed of the same light flash as c?

Maybe I should replace the phrase "not necessarily simultaneous" with "not simultaneous" in my understanding.

 

[Orodruin]

Maybe I should replace the phrase "not necessarily simultaneous" with "not simultaneous" in my understanding.

I would say no to this in a full 1+3-dimensional spacetime. Assuming the boost is in the x-direction, two simultaneous events in S will still be simultaneous in S’ as long as they have the same x-coordinate.

They will of course not be simultaneous in some other frame (eg after boosting in y-direction) but that is not how you worded it.

 

[robphy]

To help develop geometric reasoning, I think it is helpful to first appreciate the Euclidean analog.

For arithmetic simplicity in special relativity, let's work with a velocity of (3/5)c, where $\gamma=5/4$.

Visit my visualization
"robphy SIMULTANEITY spacetime diagrammer for relativity (simplified from v8c) - 2023"
https://www.desmos.com/calculator/ajktes8bp5
which produced the graphics shown below.

---

In the Euclidean case ($E=-1$), this is a slope of (3/5) with respect to the vertical (anticipating the standard convention of time running vertically upwards).
The angle with the vertical is ${\rm arctan}(3/5) \approx 40.84^\circ$.
So, the unit vector along the sloped line has coordinates
adjacent-side $\cos(\arctan(3/5))=\frac{5}{\sqrt{5^2+3^2}} \approx \cos(40.84^\circ)\approx 0.85749$
opposite-side $\sin(\arctan(3/5))=\frac{3}{\sqrt{5^2+3^2}} \approx \sin(40.84^\circ)\approx 0.51449$
Imagine each [geodesic-]surveyor has an odometer and a long ruler
that he holds "perpendicular" to his path,
where "perpendicular" is along tangent to the circle [with center on the surveyor path].
Think "tangent is perpendicular to radius".
Note the symmetry that each surveyor's long ruler intercepts the other surveyor path at location $1.1662=\frac{1}{\cos(\arctan(3/5))}$ along that other surveyor path.
Indeed, the associated right-triangles with adjacent-side 1 and hypotenuse 1.1662 are
isometric triangles.

Thus, each surveyor can define a coordinate-system grid with lines parallel and perpendicular to his path.

---

For special relativity ($E=+1$), move the E-slider to +1. The "future circle" in the "geometry of Minkowski spacetime" ("Minkowski-circle") is a hyperbola, whose asymptotes are along sloped lines that are parallel to the spacetime-paths of light-rays.
Rather than Euclidean-angles involving the circular-functions,
it is more natural to work with "rapidities" (Minkowski-angles) involving the hyperbolic-functions.
[One can alternatively think in terms of areas of sectors... but I won't use this idea here.]

("Time runs upwards".)

So, the unit vector along the sloped line has coordinates
adjacent-side $\cosh({\rm arctanh}(3/5))=\frac{5}{\sqrt{5^2-3^2}} =\frac{5}{4} =1.25=\gamma$ (the time-dilation factor)
opposite-side $\sinh({\rm arctanh}(3/5))=\frac{3}{\sqrt{5^2-3^2}} =\frac{3}{4} = 0.75 = \frac{3}{5}\frac{5}{4} =v\gamma$
Imagine each [inertial-]observer has a wristwatch and a long ruler
that he holds "Minkowski-perpendicular" to his path,
where "Minkowski-perpendicular" is along tangent to the "Minkowski-circle" [with center on the observer worldline].
Think "tangent is perpendicular to radius".
(The long ruler Minkowski-perpendicular to his worldline indicates the notion of simultaneity for that observer.)
Note the symmetry that each observer's long ruler intercepts the other observer's worldline at wristwatch-time $0.8=\frac{4}{5}=\frac{1}{\frac{5}{4}}= \frac{1}{\cosh({\rm arctan}(3/5))}$ along that other observer worldline.
(This is the symmetry of time-dilation.)
Indeed, the associated Minkowski-right-triangles with adjacent-side 1 and Minkowski-hypotenuse 0.8 are
Minkowski-isometric triangles.

Thus, each observer can define a coordinate-system grid with lines parallel and Minkowski-perpendicular to his worldline.

For the case of $\gamma=2$, use $v_2=0.866\approx \tanh( {\rm arccosh}(2))=\frac{\sqrt{3}}{2}$.

It is important to note that the implications of this Minkowski-spacetime-geometry
lead to predictions that agree spectacularly with experiment (better than the E=0 Galilean case below).
https://en.wikipedia.org/wiki/Tests_of_special_relativity

---

The "($E=0$)" case corresponds to the Galilean-spacetime-geometry (the geometry of the ordinary position-vs-time graph in PHY 101... but with time running upwards). The "future Galilean-circle" is a horizontal line (more generally, a "hyperplane" ruled by segments corresponding to infinite-speeds). (Imagine opening up the hyperbolas (and the light-cone, a hyperbola of radius zero) so that the asymptotes correspond to an infinite signal-speed.)

There are analogous Galilean-trig functions (as defined by the mathematician I.M. Yaglom).
(Details are in my "Spacetime Trigonometry" poster https://www.aapt.org/doorway/Posters/SalgadoPoster/SalgadoPoster.htm
as part of https://www.aapt.org/doorway/TGRU/ ).
In the E=0 Galilean case, the tangent-lines determined by the various inertial observers in this diagram coincide. This implies "absolute simultaneity in Galilean relativity".... and "no time-dilation",
In the scheme, E=0 Galilean-spacetime-geometry is the very special case (...not the general case).

Most of our confusion with special relativity stems from
our difficulty with letting go of our "common sense intuition" suggested by Galilean relativity.

 

[Sagittarius A-Star]

After looking into it if you have two clocks P and Q that synchronize when they meet and if $\gamma$ is 2 then from P's reference frame P equaling 10 and Q equaling 5 is simultaneous from P's reference frame but P equaling 5 and Q equaling 10 is simultaneous from Q's reference frame.

The following illustration shows the rest frame of $P$ at 3 different instances of coordinate-time (P-time 0, P-time 5, P-time 10). The coordinate-time is represented by a grid of clocks.

• At P-time 0, the "moving" clock $Q$ gets directly compared to clock $P$.
• At P-time 10, the "moving" clock $Q$ gets directly compared to another clock $P^*$, which is synchronous to clock $P$ with reference to the common rest frame of $P$ and $P^*$.

Source (Einstein, Infeld "The Evolution of Physics", page 193, electronic page number: 212):
https://archive.org/details/B-001-002-263/page/n211/mode/2up

 

[Sagittarius A-Star]

I might be too inexperienced at Minkowski diagrams to follow, but I did try to get it, I'm going to keep reading and maybe come back to it later.

Maybe the following can help understanding Minkowski diagrams and the relativity of simultaneity.

Leonard Susskind constructs in his book "Special Relativity and Classical Field Theory" the $x'$ axis of a Minkowski diagram.

The chosen unit system is based on $c=1$. The dashed 45°-lines are worldlines of light.

Art, Maggie and Lenny are all at rest in the primed frame. Maggie is located in the middle between Art and Lenny. Art and Lenny both send light pulses to Maggie. The common receiving-event is $a$. Art and Lenny must have sent the light pulses at the "same time" $t'=0$. Therefore, event $b$ must be on the $x'$ axis, as the origin is.

Simultaneity is relative between the unprimed and primed frame, because the $x'$ axis is not horizontally.

Related lecture video of Leonard Susskind:
https://theoreticalminimum.com/cour...-classical-field-theory/2012/spring/lecture-1

 

[Orodruin]

One can alternatively think in terms of areas of sectors... but I won't use this idea here.

This is true also for the Euclidean case!

 

[robphy]

One can alternatively think in terms of areas of sectors... but I won't use this idea here.

This is true also for the Euclidean case!

Related:

"On the definitions of the trigonometric functions" (1894)
Alexander Macfarlane (1851-1913)
https://archive.org/details/ondefinitionstr00macfgoog
tries to develop circular and hyperbolic trigonometry using directed areas.

See also
"On the introduction of the notion of hyperbolic functions" (1895)
M.W. Haskell
Bull. Amer. Math. Soc. 1 (1895), 155-159
https://www.ams.org/journals/bull/1895-01-06/S0002-9904-1895-00266-9/

 

[Renato Iraldi]

Hello everyone,

So I think maybe what confused me with the symmetry of time dilation was not understanding relativity of simultaneity.

After looking into it if you have two clocks P and Q that synchronize when they meet and if $\gamma$ is 2 then from P's reference frame P equaling 10 and Q equaling 5 is simultaneous from P's reference frame but P equaling 5 and Q equaling 10 is simultaneous from Q's reference frame.

Each reference frame has it's own simultaneity calculations and what's simultaneous in the rest frame of one clock is not necessarily simultaneous in the rest frame of another clock.

To reason with time dilation we must take into account that in relativity you can only compare clocks that are in the same place, you cannot compare clocks from one system with those from another system.
We had two possible cases, the first looking from a reference system where we have clocks synchronized along the spatial axis and we look at a single given traveling clock, and we follow this same clock with the synchronized ones from the stationary system. In this case We will observe that the traveling clock goes slower compared to the synchronized stationary ones. put x'=0 in the Lorentz equation.
This is reciprocal because we can compare with synchronized clocks in the s or s' system, what I have explained is symmetrical.
The second possibility is to stop with a fixed clock and compare it with the clocks that pass from the moving system, in this case you will see that the moving clocks seem go faster. but this is not an indicative measurement because they are being compared with clocks that were not synchronized.

 

[Mister T]

The second possibility is to stop with a fixed clock and compare it with the clocks that pass from the moving system, in this case you will see that the moving clocks seem go faster. but this is not an indicative measurement because they are being compared with clocks that were not synchronized.

In the stationary frame the moving clocks run SLOWER. And they don't just seem to do that, they actually do.

The fact that they may not be synchronized doesn't change this. In other words, they run slower in the stationary frame, whether they are synchronized or not.

 

[jbriggs444]

In the stationary frame the moving clocks run SLOWER. And they don't just seem to do that, they actually do.

This statement seems overly strong. The sense in which the moving clocks run slower is relative to a chosen coordinate system. It is a coordinate-relative fact. It does not rise to the level of what I would call "actually".

If you compare a moving clock against a series of clocks that are stationary in a selected inertial reference frame and which are synchronized in that selected inertial reference frame then the comparison will reveal that the proper time of the moving clock advances more slowly than the sequence of displayed times on the momentarily co-located rest clocks. That part is certainly "actual". It is an invariant fact about the described experiment.

I try to reserve the word "actual" to describe invariant facts. A clock which is at rest in one frame and is moving in another has no invariant fact about its "actual" rate of progress.

 

[Ibix]

In the stationary frame the moving clocks run SLOWER. And they don't just seem to do that, they actually do.

I suspect @Renato Iraldi is imagining looking at a stream of clocks moving past his location, ignoring all but the one right in front of him. The slow ticking plus the changing zeroing conspires to make "the clock in front of him" tick fast, if you ignore the fact that it's not one clock.

 

[Renato Iraldi]

In the stationary frame the moving clocks run SLOWER. And they don't just seem to do that, they actually do.

The fact that they may not be synchronized doesn't change this. In other words, they run slower in the stationary frame, whether they are synchronized or not.

I think it should be specified what it means that one clock runs slower than another. The Lorentz transformation gives us t' =t*gamma for x =0 Time on the train seems to go faster. And t' = t/ gamma for x'=0. Time on the train runs slower
If you don't understand this you have to study.

I suspect @Renato Iraldi is imagining looking at a stream of clocks moving past his location, ignoring all but the one right in front of him. The slow ticking plus the changing zeroing conspires to make "the clock in front of him" tick fast, if you ignore the fact that it's not one clock.

 

[FactChecker]

The IRF that is assumed to be "stationary" is only observing one moving clock to determine how fast it is running. But in order to do that, it compares that one moving clock to at least two of it's own clocks along the moving clock's path. For any of that to make sense, the two "stationary" clocks should be synchronized. The synchronization is where the "stationary" and "moving" IRFs disagree. They can not agree on what events along the relative path are simultaneous. That is the "relativity of simultaneity". The disagreement allows either IRF to consider itself to be "stationary" and the other to be the "moving" IRF with slow clocks.

 

[FactChecker]

I suspect @Renato Iraldi is imagining looking at a stream of clocks moving past his location, ignoring all but the one right in front of him. The slow ticking plus the changing zeroing conspires to make "the clock in front of him" tick fast, if you ignore the fact that it's not one clock.

In that sense, the observer is not observing the ticking speed of one moving clock. He is observing the synchronization of multiple moving clocks as they pass one "stationary" location. And he would say that those moving clocks are not synchronized correctly.

 

[Ibix]

In that sense, the observer is not observing the ticking speed of one moving clock

Agreed. @Renato Iraldi is correct that the reading on the ever-changing clock in front of him is advancing faster than the one in his rest frame, but is wrong to call this "the time" in any frame, since no clock is measuring it, and no one clock can ever do so.

 

[jbriggs444]

I think it should be specified what it means that one clock runs slower than another. The Lorentz transformation gives us t' =t*gamma for x =0 Time on the train seems to go faster. And t' = t/ gamma for x'=0.

That is not a comparison between two clocks, exactly. Instead, it is a comparison between two coordinate systems -- observing the relationship between the two time coordinates ($t$ and $t'$) while agreeing to hold a particular position coordinate (either $x$ or $x'$) fixed.

Alternately, since the coordinate time delta between two events along a geodesic with constant spatial coordinate is constructed to be identical to the proper time along that trajectory, the coordinate comparison is functionally identical to a comparison between one clock and one coordinate system. Not really a comparison between two clocks.

One cannot compare two relatively moving clocks for rate without invoking a synchronization convention. A choice of coordinate system is one way of adopting a synchronization convention -- two events are synchronous if they share the same time coordinate.