# Relation between relative simultaneity and velocity

## Main Question or Discussion Point

Hi guys, I was thinking about the relativistic effects a little bit, and I have a question regarding relative simultaneity.

Time dilation and length contraction grow as a function of the speed of the observer, and become noticeable and large on speeds close to the speed of light. By this sentence, I mean this, what's on the picture My question is: does relative simultaneity, or the time difference between two events progress by the same mathematical function, or is the time difference proportional to the velocity of the observer.
If it's not a problem, I would like to see a comparision between hyperplanes of simultaneity between different observers with different velocities so I can see how it differs between greater and slower speeds. I appreciate your help.

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Andrew Mason
Homework Helper
Hi guys, I was thinking about the relativistic effects a little bit, and I have a question regarding relative simultaneity.

Time dilation and length contraction grow as a function of the speed of the observer, and become noticeable and large on speeds close to the speed of light. By this sentence, I mean this, what's on the picture My question is: does relative simultaneity, or the time difference between two events progress by the same mathematical function, or is the time difference proportional to the velocity of the observer.
If it's not a problem, I would like to see a comparision between hyperplanes of simultaneity between different observers with different velocities so I can see how it differs between greater and slower speeds. I appreciate your help.
Events that are separated in space and simultaneous in one inertial reference frame will not be simultaneous in any other inertial reference frame (i.e. any other frame whose origin is moving relative to the origin of the first).

The time difference in frame A' between events that are simultaneous and separated by distance in frame A (A' is moving at velocity v relative to A in the x direction) is found by applying the Lorentz transformation: $t' = \gamma\left(t - \frac{vx}{c^2}\right)$ to each event.

If events at positions x1 and x2 are simultaneous in frame A, t = 0 for both events, so the time difference between those events in the moving frame will be:

$t_2'-t_1' = \Delta t' = \gamma\frac{v}{c^2}(x_1-x_2)$

AM

Nugatory
Mentor
Time dilation and length contraction grow as a function of the speed of the observer....
The time difference in frame A' between events that are simultaneous and separated by distance in frame A (A' is moving at velocity v relative to A in the x direction) is found by applying the Lorentz transformation: $t' = \gamma\left(t - \frac{vx}{c^2}\right)$ to each event.
Also, note that Andrew Mason used the Lorentz transformations, not time dilation and length contraction, to answer the question. Some of the pop-sci treatments of relativity talk too much about time dilation and length contraction because they're easy to talk about; but they're a wrong turn into a dead end for anyone who wants to understand relativity.
(Serious textbooks also discuss time dilation and length contraction, but more in the spirit of an exercise: "Here, let's test our understanding of the Lorentz transforms by using them to derive these two clever little formulas for length contraction and time dilation").

So differences in simultaneity progress just like time dilation and length contraction, because of the Lorentz-factor in the equation?

Fredrik
Staff Emeritus
Gold Member
So differences in simultaneity progress just like time dilation and length contraction, because of the Lorentz-factor in the equation?
I wouldn't say that, but then I don't know what you mean by "just like". Andrew's calculation is an accurate answer. This picture is another one: Read up on spacetime diagrams (also called Minkowski diagrams) if this doesn't make sense to you. These diagrams are by far the best way to understand special relativity.

king vitamin
Gold Member
So differences in simultaneity progress just like time dilation and length contraction, because of the Lorentz-factor in the equation?
While the presence of the $\gamma$ factor shows that the relativity of simultaneity becomes important at large velocities with the same scaling behavior as time dilation and length contraction, the presence of $x_1 - x_2$ means it is also very important on large length scales. In particular, even for very small velocities ($\gamma \approx 1$), observers on Earth must have very different notions of what events are simultaneous with themselves 10 billion light years away from Earth.

Nugatory
Mentor
So differences in simultaneity progress just like time dilation and length contraction, because of the Lorentz-factor in the equation?
That's part of it, but not all of it. Google for "Andromeda paradox".

Andrew Mason
Homework Helper
That's part of it, but not all of it. Google for "Andromeda paradox".
While it is an interesting concept, the Andromeda paradox is about more than time dilation between two different inertial frames. It involves switching between very different inertial frames of reference frames whose relative velocities approach c. The Lorentz transformations have to be reapplied when the frames of reference change. So it is rather more complicated than simple time dilation and length contraction, which I think is what the OP is asking about.

AM

DrGreg
Gold Member
While it is an interesting concept, the Andromeda paradox is about more than time dilation between two different inertial frames. It involves switching between very different inertial frames of reference frames whose relative velocities approach c. The Lorentz transformations have to be reapplied when the frames of reference change. So it is rather more complicated than simple time dilation and length contraction, which I think is what the OP is asking about.

AM
Yes, it is about more than time dilation, it's about relativity of simultaneity which is exactly what the OP asked about. And in the versions I've read, the two frames are not at high relative velocity, they are at walking speed, so γ is near-enough 1.

Andrew Mason
Homework Helper
Yes, it is about more than time dilation, it's about relativity of simultaneity which is exactly what the OP asked about. And in the versions I've read, the two frames are not at high relative velocity, they are at walking speed, so γ is near-enough 1.
Sorry about that. I was assuming that the Andromeda Paradox was another name for the Twin Paradox (which often uses a traveler going from earth to Andromeda and back).

AM

Thanks for your great replies, I think I'm starting to understand something. But what does determine what frame has straight lines of simultaneity and what doesn't? I ask this because every frame can consider to be at rest and consider the other one as moving. For instance in the twin paradox, the stay at home twin has straight lines of simultaneity, and the moving twin has different lines. Why is that, and why can't we say that from his perspective the moving twin has straight lines, while the stay at home twin which can be considered as moving from the other twin's perspective has non-straight lines? To sum-up, how can we know what frames have straight, and what frames have non-straight lines?

Fredrik
Staff Emeritus
Gold Member
In an inertial frame, a simultaneity line is always a straight line. In the case of the twin paradox, they look like this: The blue lines are simultaneity lines of the inertial coordinate system that's comoving with the astronaut twin as he's moving away from Earth. The red lines are the simultaneity lines of the inertial coordinate system that's comoving with the astronaut twin as he's coming back.

The slope of the simultaneity line is determined by the requirement that the speed of light is the same to all observers. The non-rigorous argument goes like this:

The units on the axes in a spacetime diagram are chosen so that we can draw the world line of a light ray at a 45° angle. Motion obviously tilts the time axis, because the t' axis is just the other guy's world line, and if we don't tilt his simultaneity lines by the same angle in the opposite direction, the world line of a light ray isn't going to be exactly half-way between the t' axis and the x' axis (or the x'y' plane), and would therefore have a different speed in his coordinate system. It's not super hard to make a rigorous argument, but it involves some algebra. We can't do much better than this without calculations.

@Fredrik, I understand the lines are always straight in inertial frame, it was my lapse in writing and understanding when I said straight instead of horizontal. The stay at home twin can be considered to be at rest and his lines are parallel to the x-axis of the diagram, while the moving twin has an angle between his lines and x-axis. Note that my question isn't in essence about the twin paradox, since I basically understand what causes different aging, I just want to know what determines the 'angle' (not straightness) between the x axis and the line of simultaneity. The common sense pressumption looks like 'the stay at home twin is at rest so his lines are non-diagonal', but the moving twin is also at rest in his own frame, and from his point of view the other twin is moving, so this is what's the problem.

Fredrik
Staff Emeritus
Gold Member
The second half of my previous post explains what determines the angle. If you want a more rigorous argument, we can use a Lorentz transformation. Let O and O' be two inertial observers. Let S and S' be their comoving inertial coordinate systems. Let ##\Lambda## be the Lorentz transformation from S to S'. In units such that c=1, we have
$$\Lambda=\gamma\begin{pmatrix}1 & -v\\ -v & 1\end{pmatrix},\qquad\gamma=\frac{1}{\sqrt{1-v^2}}.$$ The slope of the x' axis can be calculated by determining the coordinates in S of any point on the S' axis. If (t,x) is the coordinates of such a point, then t/x is the slope of the x' axis. Since (0,1) is the S' coordinate pair of a point on the x' axis, we have
$$\begin{pmatrix}t\\ x\end{pmatrix}=\Lambda^{-1}\begin{pmatrix}0\\ 1\end{pmatrix} =\gamma\begin{pmatrix}1 & v\\ v & 1\end{pmatrix}\begin{pmatrix}0\\ 1\end{pmatrix}=\gamma\begin{pmatrix}v\\ 1\end{pmatrix}.$$ So the slope of the x' axis is
$$\frac t x=\frac{\gamma v}{\gamma}=v.$$

Nugatory
Mentor
@Fredrik, I understand the lines are always straight in inertial frame, it was my lapse in writing and understanding when I said straight instead of horizontal. The stay at home twin can be considered to be at rest and his lines are parallel to the x-axis of the diagram, while the moving twin has an angle between his lines and x-axis. Note that my question isn't in essence about the twin paradox, since I basically understand what causes different aging, I just want to know what determines the 'angle' (not straightness) between the x axis and the line of simultaneity. The common sense pressumption looks like 'the stay at home twin is at rest so his lines are non-diagonal', but the moving twin is also at rest in his own frame, and from his point of view the other twin is moving, so this is what's the problem.
A similar question came up in another thread: https://www.physicsforums.com/showpost.php?p=4711070&postcount=13

Everyone's lines of simultaneity are parallel to their x-axis, and we could jus as easily have drawn the diagram with the traveling twin's x-axis and t-axis intersecting at right angles.

Well then, I hope it's not a problem if someone could demonstrate to me with diagrams the difference between the diagrams of the stationary twin when his line of simultaneity is parallel with the x-axis, and the diagram of the moving twin when his line is parallel with the x-axis (only the outbund trip is enough to grasp this). This animation shows pefectly well the differences, let's say that the earth twin has a velocity of v=0. Now, what would it look like if we take the moving observer and say that he has the velocity of v=0, and how would the second twin's line look like on the diagram?

jtbell
Mentor
If events at positions x1 and x2 are simultaneous in frame A, t = 0 for both events, so the time difference between those events in the moving frame will be:

$t_2'-t_1' = \Delta t' = \gamma\frac{v}{c^2}(x_1-x_2)$
To put it in perhaps more concrete terms: If in frame S, two clocks are at rest, a distance Δx apart, and synchronized with each other, then in frame S' moving at velocity v along the line joining the two clocks, the clocks are out of sync by the amount Δt' = γvΔx/c2. That is, their readings (which show values of t) differ by that amount.

Fredrik
Staff Emeritus
Gold Member
Now, what would it look like if we take the moving observer and say that he has the velocity of v=0, and how would the second twin's line look like on the diagram?
Freeze your animation when the ct axis is tilted to the left. The ct axis is how the (outbound) astronaut twin draws the world line of the Earth twin, and the x axis is how the astronaut twin draws a simultaneity line of the Earth twin.

Freeze your animation when the ct axis is tilted to the right. The ct axis is how the Earth twin draws the world line of the (outbound) astronaut twin, and the x axis is how the Earth twin draws a simultaneity line of the astronaut twin.

Nugatory
Mentor
Well then, I hope it's not a problem if someone could demonstrate to me with diagrams the difference between the diagrams of the stationary twin when his line of simultaneity is parallel with the x-axis, and the diagram of the moving twin when his line is parallel with the x-axis (only the outbund trip is enough to grasp this).
Everyone's lines of simultaneity are parallel to the x-axis in their coordinate system, so the traveling twin's lines of simultaneity are always parallel to the x' axis (because he uses the primed coordinates) and the earth-bound twins lines of simultaneity are always parallel to the x axis (because he uses the unprimed coordinates).

Your animation shows what the x and t axes of the earthbound twin look like to an observer moving at various speed relative to the earthbound twin. If you freeze your animation at the v=0 point, you'll get the picture in which the earthbound twin is at rest. If you freeze it at the v=-.3c point (Note the negative sign, the earthbound twin is moving at -.3c relative to the spaceship twin) and you'll see what the earthbound twin's axes and lines of simultaneity look like when we draw the picture with the traveling twin at rest.

We could use the same animation for the traveling twin's x' and t' coordinates; then if you freeze the animation at v=.3c (note the positive sign) you'd see what the x' and t' axes look like to the earthbound twin.

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Andrew Mason
Homework Helper
To put it in perhaps more concrete terms: If in frame S, two clocks are at rest, a distance Δx apart, and synchronized with each other, then in frame S' moving at velocity v along the line joining the two clocks, the clocks are out of sync by the amount Δt' = γvΔx/c2. That is, their readings (which show values of t) differ by that amount.
The Andromeda Paradox is not really a paradox as much as it is an illustration that the concept of 'simultanaeity' of two events separated by a great distance does not have much meaning.

If two events can be causally related (Δx/Δt<c), the order of the events in all inertial reference frames is the same, which means that they cannot be simultaneous in any reference frame. If the events cannot be causally related, the order of events is not the same in all reference frames. This means that there is an inertial reference frame in which the events will appear to be simultaneous.

So, if the order of non-causally related events is relative (i.e. it is meaningless to speak of the order in which such events occur), then the concept of absolute simultaneity of such events is lost.

AM

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