# In the theory of special relativity speed is relative so who decides which observer’s time moves slower? [duplicate]

If for example we have 2 people one moving in speed v relative to the other, according to special relativity the time passing for the moving person is slower than for the stationary person. However from the moving person’s prospective he is stationary and his friend is moving so time should move faster for him. What’s going on?

• What research did you do and what did that suggest, or rule out? Apr 25, 2021 at 22:04
• I believe the summary is "there's two timelines. The other person is slower" Apr 26, 2021 at 23:50

A core idea of special relativity is there is no right frame of reference. It doesn't matter which of the two observers you use as your point of reference, the math will work out either way. Yes, they'll disagree about whose time is what, and they're both right in their own frame of reference.

To use a simpler example, let's ignore relativistic effects for a moment. Charlie and Dave are running, but Dave's falling behind. Relative to a watching observer, Charlie is running $$5 \textrm{m/s}$$, Dave is running $$4 \textrm{m/s}$$. From Charlie's perspective, Dave is falling behind by 1 meter every second (moving $$1 \textrm{m/s}$$ backward or $$-1 \textrm{m/s}$$), the ground is moving backwards by 5 meters every second ($$-5 \textrm{m/s}$$), and relative to himself he's obviously stationary ($$0 \textrm{m/s}$$). From Dave's perspective, Charlie is running forward at $$1 \textrm{m/s}$$, the ground is falling back at $$4 \textrm{m/s}$$, and he is of course also stationary relative to himself ($$0 \textrm{m/s}$$).

So who's right, the observer, Charlie, or Dave? Well, they're all right, in their own frame of reference. They disagree, and special relativity tells us that's fine, because whenever you're running the numbers, you always pick a frame of reference for your math, and as long as you're consistent with which frame of reference you choose, you'll get the right answer from the perspective of that frame.

There are some things, like how old someone is, which order two events occur in, or even whether or not events occur at the same time, that relativity says will be inconsistent depending on which frame you choose. There is no absolute frame, there is no absolute speed, there is no absolute space, there is no absolute time. But it also tells us that there's nothing wrong with that, as long as you're consistent with frame in your math. It's just something students of physics have to learn to accept.

• This is the part of the twin paradox which seems to actually be a paradox. That is, both of the twins should say the other one is older. The solution is that only one of the twins took off in a rocket, and hence accelerated. So, if one reference frame is accelerating (relative to some third fixed body), this is the twin who will actually be younger. Apr 25, 2021 at 18:42
• @jheindel : no, that is not correct. If neither twin turns around (so they never meet up again) then it's perfectly accurate to say that they each observe the other to be aging more slowly. For example if they compare radio signals (adjusting for the finite speed of light!) they will each calculate that the other's clock is slow. If one twin does turn around then the symmetry is broken; that twin will be using different reference frames for the outbound and inbound parts of their trip. Apr 25, 2021 at 19:04
• @Eric smith I agree and what we've said are not contradictory. At least the way I was taught the twin paradox was as an out and back journey. Both start at rest, one flies away, stops, turns around and returns. My understanding is that the whole point of the twin paradox is to teach about when one can privilege a particular reference frame. In this case, the act of accelerating and decelerating breaks the symmetry, so the frame of the twin on earth should be used to correctly determine that the twin on the rocket is younger when they meet again. Apr 25, 2021 at 23:39
• The reason I brought up the twin paradox is that it's not correct to say that special relativity says there's no right frame of reference. Special relativity says there's no right inertial frame of reference. This distinction is the whole point of the twin paradox as a teaching tool. Apr 25, 2021 at 23:42
• @Vilx Velocity being non-absolute does not mean that change in velocity is non-absolute. Accelerometers can work entirely without any external reference, as opposed to speedometers. Apr 26, 2021 at 6:57

Short answer is: They both do. They both measure the same amount of time dilation. One sees the time slowed for the other observer, and the other observer sees the exact same thing happen to the original observer.

Suppose you have an observer A moving at a velocity $$v$$ relative to another observer B who is stationary (meaning you are seeing things from inside B's frame). B will measure the time lapsing in A's frame as being dilated or slowed down according to $$t' = \frac{t}{\sqrt{1-\frac{v^2}{c^2}}}$$ where $$t$$ is the time elapsed inside the frame of reference of observer B (as measured by B) and $$c$$ is the speed of light.

But from A's frame of reference, B is also moving at a velocity $$v$$ and so A will also measure the same amount of time lapsed or slowed down in B's frame. This sounds paradoxical. This paradox is the focus of what is known as the twin paradox which details this phenomena.

Note that one of the resolutions to this paradox is to consider the relativity of simultaneity

"....one must understand that in special relativity there is no concept of absolute present. For different inertial frames there are different sets of events that are simultaneous in that frame. This relativity of simultaneity means that switching from one inertial frame to another requires an adjustment in what slice through spacetime counts as the "present" ..."

They are both right.

But, then, what happens if they meet? Well, in order to meet, you have to accelerate. And if one is accelerating and the other is not, the symmetry is broken.

If they both accelerate equally, they'll end up agreeing anyhow.

Under relativity, events can be separated by spacelike or timelike curves. Any two world lines will agree that things on timelike curves occur in a specific order, and for any two things seperated by spacelike curves, you'll find wordlines that consider them one to be before the other, vice versa, or "at the same time".

That is, barring exotic geometry. Relativity ends up depending on the geometry of your universe; if the universe is flat and boring, the above statements work. If your universe's geometry contains wormholes, or wraps around, things get more interesting.

• "if one is accelerating and the other is not" Well, each of them will see other is accelerating. So acceleration will not help explaining this.
– Atul
Feb 26, 2022 at 0:50

One of the keys to understanding SR is to remember that all the effects it predicts are reciprocal. If Observer A thinks that Observer B's spaceship has contracted in length, Observer B will consider that Observer A's spaceship has contracted by the same proportion. There are many analogous cases in everyday life. If we stand some distance apart, I will seem smaller to you and you will seem smaller to me. There is no contradiction- we just have different perspectives on the same reality.

There is a simple picture for intuitively explaining the concept without maths. It's not an exact analogy, because the geometry of spacetime is not the same as the geometry of space, but the principle by which it works is the correct one.

Imagine that you and a friend are walking together across a field that that represents space and time. The distance you walk is the same as the time you experience, so for each of you 'forwards' is the time axis and 'sideways' is the space axis. You each walk in a straight line in a slightly different direction. As you proceed, you see your friend drift to the side (moving in space) and fall behind you (progressing slower through time). Your friend has a different definition of 'forwards' and 'sideways', and thus sees you gradually drift sideways in space in the other direction, and fall behind her (progress slower through time). You can each fall behind the other because you are facing in different directions, you each have a slightly different version of 'forwards'. One time axis shows you going faster than your friend, the other time axis shows your friend going faster than you. Similarly with length contraction - each of you sees the other's ruler shrinking compared to your own. There is thus no inconsistency.

The actual geometry of spacetime is different to this, in that for Euclidean space the Pythagoras Theorem saying that the length of a diagonal line squared is the distance forward squared plus the distance sideways squared, with spacetime it's the distance forward squared minus the distance sideways squared (according to one possible convention). The minus sign radically changes the behaviour of lengths. But the reason that both clocks can run slower than the other in special relativity is exactly as described above. It's just the changing of coordinate values looked at in different coordinate systems.

• For a Nullus in Verba, you have a good way with words! ;-) Apr 26, 2021 at 20:54

I'll expand on the mathematics in @josephh's answer. This will involve some calculus, but drawing the diagram @Bob's comment recommended will make the same points visually obvious.

Your question is a common one, and it boils down to this: if $$\partial t^\prime/\partial t=\gamma:=(1-v^2/c^2)^{-1/2}>1$$, and in another equally valid reference frame $$\partial t/\partial t^\prime=\gamma>1$$, whose time is passing faster? Even if we ignore the inequalities' directions, how can $$\partial t^\prime/\partial t\cdot\partial t/\partial t^\prime$$ be $$\gamma^2$$ instead of $$1$$?

The answer lies in why these are partial, not total, derivatives; some variables have been forgotten. From $$t^\prime=\gamma(t-vx/c^2),\,x^\prime=\gamma(x-vt)$$ we deduce that $$\partial t^\prime/\partial t=\gamma$$ assumes constant $$x$$, but $$\partial t/\partial t=\gamma$$ assumes constant $$x^\prime$$. Far from being reciprocals, these partial derivatives are equal. (To take a "pure mathematics" example of this phenomenon, note in $$2$$ dimensions $$x/r=\cos\theta$$ is the constant-$$\theta$$ value of $$\partial x/\partial r$$ and the constant-$$y$$ value of $$\partial r/\partial x$$.) And length contraction paradoxes admit the same solution.

As a sanity check, $$t=\frac{t^\prime}{\gamma}+\frac{vx}{c^2}$$, so the constant-$$x$$ value of $$\partial t/\partial t^\prime$$ is actually $$1/\gamma$$.

There is no universal reference frame as per special relativity. Both observers have their own reference frame, and looking from there, the other one's time slows. This is why we say that speed is symmetrically relative.

We happen to live in a universe where speed is relative, but acceleration is absolute.

Let's compare this to a simpler example from normal (Galilean) relativity. We have two observers $$A$$ and $$B$$. They move with a certain velocity with respect to eachother.

 A's frame   | B's frame
|
A   B-->    | <--A   B


Observer $$A$$ claims that he is standing still but observer $$B$$ claims that he is the one standing still. So who is right? They are somehow both right but only in their respective frames. The same holds true for time dilation in special relavity. Each observer sees the clock of the other observer moving slower. They are both right but only in their respective frames. This might seem like a stupid analogy because time is a measurable property and more fundamental than the notion of 'who is standing still' but this analogy is how you should think about it.

To make meaningful statements about the time dilation between the two observers they have to meet to compare clocks. For inertial observers they will only meet once so to compare clocks between two instants in time one of the observers has to accelerate towards the other, giving the twin paradox. At this point we are answering a whole different question thought.