Which clock was slower in special relativity?

[entropy1]

I don't know much about the math of SR, but this is what's bothering me: if a moving clock B ticks slower than the stationary one I have (A), then from the viewpoint of B, my clock (A) is ticking slower. So if we turn around and meet each other in the middle, which clock was slower than which?

Math is no problem if you want to use it in this thread.

 

[jedishrfu]

This is the twin paradox restated.

Here's a more detailed but non-math explanation:

 

[PeroK]

I don't know much about the math of SR, but this is what's bothering me: if a moving clock B ticks slower than the stationary one I have (A), then from the viewpoint of B, my clock (A) is ticking slower. So if we turn around and meet each other in the middle, which clock was slower than which?

Math is no problem if you want to use it in this thread.

There's no such thing as a "moving" clock or a "stationary" clock. Motion is relative. You can have a clock that is moving inertially: i.e. it is not accelerating. And you can have a clock that is accelerating: i.e. subject to a real force.

If you define an IRF (inertial reference frame), then you can have a clock that is stationary wrt that frame and a clock that is moving wrt that frame. But, in any other IRF the state of motion of the two clocks will be different. There is, therefore, no absolute state of motion.

If two clocks are moving inertially relative to each other, then both are measured to run slow in the inertial frame in which the other clock is at rest.

To answer your question:

Suppose the clocks are $A$ and $B$ and they start at the same location, and some time later meet up again. Choose any IRF and compute the following quantities (where the clocks meet at $t = 0$ and again at $t = T$, as measured in that IRF):
$$\tau_A = \int_0^T \sqrt{1 - v_A(t)^2/c^2} \ dt, \ \ \ \tau_B = \int_0^T \sqrt{1 - v_B(t)^2/c^2} \ dt$$
Where $v_A(t), v_B(t)$ are the speeds of the clocks as measured in the IRF. This gives the "proper" time of each clock ($\tau_A, \tau_B$) and is the time interval recorded on each clock between the two meetings.

 

[entropy1]

I am still grapling with this; to put it very simple: in my example, either A aged more than B, or B aged more than A, right? I mean, when they meet. It is not symmetrical, so there must be a variable that makes the difference!

That said, acceleration or jumping reference frames (video) could be such a variable.

 

[PeroK]

I am still grapling with this; to put it very simple: in my example, either A aged more than B, or B aged more than A, right? I mean, when they meet. It is not symmetrical, so there must be a variable that makes the difference!

That said, acceleration or jumping reference frames (video) could be such a variable.

If all you know is that they separated and came back together, then they could be the same or one could show less time than the other. The variable is the speed profile of the two clocks (as measured in any IRF).

 

[entropy1]

If all you know is that they separated and came back together, then they could be the same or one could show less time than the other. The variable is the speed profile of the two clocks (as measured in any IRF).

But if I do calculations with IRF where A is stationary and B is moving, I should get the opposite answer if I use the IRF where B is stationary and A is moving, while both are (can be) the exact same situation, right? In the first case A is older and in the second case B is older.

 

[PeroK]

But if I do calculations with IRF where A is stationary and B is moving, I should get the opposite answer if I use the IRF where B is stationary and A is moving, while both are (can be) the exact same situation, right? In the first case A is older and in the second case B is older.

If they separate and then meet again they can't both be inertial.

 

[entropy1]

If they separate and then meet again they can't both be inertial.

No, suppose only one of the two is moving, we get different results when viewed differently.

Ah, the other frame is not inertial, you mean?

Wait, now I'm really confused!

 

[PeroK]

No, suppose only one of the two is moving, we get different results when viewed differently.

Ah, the other frame is not inertial, you mean?

You must be precise about the motion of both clocks. If you establish that one clock was inertial throughout the experiment, then you could use that clock's rest frame as a convenient IRF. The inertial clock must show more time than the other.

But, if neither is inertial, then you need to specify more precisely what is happening.

They can't both be inertial. If they are both inertial then they can only meet once.

Note: we are talking SR here (flat spacetime).

 

[phinds]

No, suppose only one of the two is moving,

As has already been pointed out (see post #3), this is a meaningless statement. Motion is relative.

What is NOT relative is acceleration, so any differences in their clocks when they meet back up will be based on which one accelerated.

EDIT: If both accelerated, then it gets slightly more complicated but that's just math.

 

[Janus]

But if I do calculations with IRF where A is stationary and B is moving, I should get the opposite answer if I use the IRF where B is stationary and A is moving, while both are (can be) the exact same situation, right? In the first case A is older and in the second case B is older.

In order for A and B to meet up again (so that they can directly compare clocks), one or the other has to have changed velocity. Let's say it's B.
As far as A is concerned, B travels away at some speed, turns around and returns at that same speed. All A needs to know to work out how much time passes for B is what speed B was traveling and for how long. (the fact that B spent some time slowing down and then speeding up again in the reverse direction will have some small effect on the total time, but as far as A is concerned, the rate at which B's clock ticks only depends on B's relative speed with respect to A at any given moment. In other words, other than the change in B's speed, the acceleration B is undergoing adds no additional effect).
For B, the above applies during the two legs of his trip. When he and A are separating at a constant speed or approaching at a constant speed, he would measure A's clock running slow at a rate dependent on their relative speed.*
Where B's observations differ from A's is during that period when B is reversing direction and thus changing his own velocity. This is when B is non-inertial. And measurements made from non-inertial frames are not as simple as those made from inertial ones.
During this period, it is not enough for B to know the relative speed between A and himself to determine how fast A's clock is ticking. He also has to factor in the distance to A and how he is accelerating with respect to A.
By transitioning from going away from A to approaching A, he is accelerating towards A, And this causes him to determine that A's clock runs fast by a factor that depends on the magnitude of the acceleration and the distance between A and himself. In other words, B's acceleration does effect how B measures A's clock.
The result is the B would measure A's clock running slow on the outbound trip, running very fast during B's turn-around phase, and then runs slow during the return leg. The end result after returning to A is that more time has accumulated on A's clock than B's clock.
So, while during different points of the trip, A and B will disagree as to what their respect clocks are doing at any moment, when they meet up again, they agree as to how much time has accumulated on each of their clocks.

* And by "measure", I mean what they would determine after accounting to light propagation delay.

 

[PeterDonis]

if I do calculations with IRF where A is stationary and B is moving, I should get the opposite answer if I use the IRF where B is stationary and A is moving

If A and B separate and then meet up again, and spacetime is flat (so SR applies), it is impossible for there to be a single IRF in which A is always stationary and B is moving, and also a (different) single IRF in which B is always stationary and A is moving. Only one of them can be stationary in the same single IRF the whole time. And that one will be the one who ages the most.

 

[entropy1]

If A and B separate and then meet up again, and spacetime is flat (so SR applies), it is impossible for there to be a single IRF in which A is always stationary and B is moving, and also a (different) single IRF in which B is always stationary and A is moving. Only one of them can be stationary in the same single IRF the whole time. And that one will be the one who ages the most.

But you can apply both of those examples alternately to a single real-life situation, right? Only from a different vantage point (IRF). In real life A is younger than B or the other way round.

 

[Ibix]

But you can apply both of those examples alternately to a single real-life situation, right?

I don't quite understand what you mean. In a real life situation, at least one of A and B is not moving inertially. Thus you can have a situation where A is inertial and B is not - in that case, there's an inertial frame where A is always at rest but no inertial frame in which B is always at rest. Or you can have a different situation where B is inertial but A is not - in that case, there's an inertial frame where B is always at rest but no inertial frame in which A is always at rest.

But these are two different scenarios.

 

[PeterDonis]

you can apply both of those examples alternately to a single real-life situation, right?

I don't understand what you mean. It is impossible to have a single real-life situation where both A and B are stationary in a single IRF the whole time. Only one of them can be.

In real life A is younger than B or the other way round.

Yes, and the one that ages more will be the one who is stationary in a single IRF the whole time. Only one of them can be.

 

[Mister T]

But you can apply both of those examples alternately to a single real-life situation, right?

No, you cannot. One clock changes direction and one clock doesn't.

To fully understand this you need to grasp the relativity of simultaneity.

 

[entropy1]

So one of the frames is not inertial. But isn't that dependent on which of the frames (A/B) is considered inertial?

 

[Vanadium 50]

You don't get to pick who is inertial.

 

[entropy1]

You don't get to pick who is inertial.

So where does that depend on? (I guess no acceleration)

 

[PeroK]

So where does that depend on?

Real forces! Newton's laws.

 

[PeterDonis]

where does that depend on? (I guess no acceleration)

No proper acceleration--the one who never feels any force (no rocket engine firing) is the one who is inertial the whole time.

 

[entropy1]

No proper acceleration--the one who never feels any force (no rocket engine firing) is the one who is inertial the whole time.

But then we get to my point: SR clock differences should be the result of velocity, but actually in this twin paradox, it seems to me dependent (also) on acceleration! (who is accelerating delivers an asymmetry)

 

[PeterDonis]

SR clock differences should be the result of speed, but actually in this twin paradox, it seems to me dependent (also) on acceleration!

Neither of these is correct.

The result of relative speed is time dilation, but time dilation is not an invariant. It's frame-dependent.

The result of acceleration is that two observers who already met once in flat spacetime can meet again; in flat spacetime that is impossible unless one of them accelerates. But the acceleration itself does not affect the rate at which their clocks tick.

The difference in the elapsed time for the two observers when they meet up again is due to the difference in lengths of their paths through spacetime. In other words, it's geometry. It's not that one clock ticked slower than the other: both clocks tick at one second per second. But the path through spacetime that one clock takes is fewer seconds long.

It's the same as if two cars set out from city #1 to city #2 taking two different routes that are different lengths. The elapsed distance on their odometers will be different when they meet up again at city #2, but that's not because either odometer was "ticking" distance at a different rate. It's because the paths they took have different lengths.

 

[PeroK]

But then we get to my point: SR clock differences should be the result of velocity, but actually in this twin paradox, it seems to me dependent (also) on acceleration! (who is accelerating delivers an asymmetry)

Ultimately, if you have a real physical clock, then the speed (relative to a given IRF) is determined by the initial velocity and the subsequent acceleration profile. The calculation simply uses the speed - that is all you need to calculate $\tau$.

 

[Ibix]

But then we get to my point: SR clock differences should be the result of velocity, but actually in this twin paradox, it seems to me dependent (also) on acceleration! (who is accelerating delivers an asymmetry)

No. SR clock differences are dependent on the "lengths" of the worldline the clocks followed. This is directly dependent on the velocity only, at least in flat spacetime. Acceleration is only necessary in order for the clocks to meet up for a second time - so it's necessary but not explanatory. It's perfectly possible to construct scenarios where the twins undergo different accelerations but end up the same age, and scenarios where they undergo the same accelerations (at slightly different times) and end up different ages.

 

[entropy1]

Thanks guys! I feel I understand this, what was always kind of a mystery to me, at least a little better. So I don't really know much about relativity, but am I correct that SR operates on/in a flat spacetime whereas GR does not?

 

[Ibix]

General relativity is a generalisation of special relativity. It includes special relativity as the case where you consider flat spacetime.

 

[entropy1]

General relativity is a generalisation of special relativity. It includes special relativity as the case where you consider flat spacetime.

So is it easy to explain what is the difference between flat spacetime and non-flat spacetime? (I guess curvature of spacetime?)

Why do we have to consider this example in flat spacetime?

 

[PeterDonis]

is it easy to explain what is the difference between flat spacetime and non-flat spacetime?

In flat spacetime, there is no tidal gravity: two objects that are moving inertially and are at rest relative to each other, will stay at rest relative to each other.

In curved spacetime, that is no longer the case: two objects that are moving inertially and start out at rest relative to each other, won't stay at rest relative to each other. For example, consider two objects that are momentarily at rest in space above the Earth, at different altitudes, and both moving inertially (zero proper acceleration). They will not stay at rest relative to each other. In Newtonian terms, this is because the gravity of the Earth pulls them with slightly different accelerations; the lower one gets pulled a little more, so it falls faster than the higher one and the two separate. But in GR terms, there is no force of gravity; the two objects separate because of spacetime curvature.

 

[Ibix]

I guess curvature of spacetime?

Yes. The presence of curvature is another name for gravity. Edit: I see Peter beat me to it with a longer and more precise reply.

 

[entropy1]

So does my example have to be considered in flat spacetime, and why?

 

[Ibix]

Why do we have to consider this example in flat spacetime?

I think you edited this in - didn't see it before.

We consider flat spacetime because the maths is simpler (you don't need integrals right out of the gate) and the concepts are not so complicated. They remain complicated enough with just SR.

 

[PeterDonis]

does my example have to be considered in flat spacetime, and why?

If you're using SR, you're in flat spacetime. SR is only valid in flat spacetime.

 

[Dale]

So does my example have to be considered in flat spacetime, and why?

Your example was given in flat spacetime, but the answer can be generalized. In general, given an arbitrary spacetime equipped with metric $g$, expressed in coordinates $x$, the proper time $\tau$, along a given object’s worldline $P$, is given by
$$\tau_P =\frac{1}{c} \int_P \sqrt{g_{\mu \nu} \ dx^{\mu} \ dx^{\nu}}$$
This is true in any reference frame (inertial or not) in any spacetime (curved or not) for any massive object’s worldline.

So simply calculate this for both and the answer is obtained. It is invariant. This is the generalization of the formula posted by @PeroK in post 3.

 

[pervect]

I see this thread has already gotten very long, and I haven't had time to read it, alas. But there is a fundamentally simple point that I'd like to make, though others may have made it before.

That point is that the process of comparing clocks in an inertial frame is frame dependent.

It is convenient to compare clocks using the Einstein synchronization convention, as it is very standard, and most schemes are equivalent. Operationally, one way of describing this is that a light signal is emitted from the midpoint in the inertial frame, then the light signal is received "at the same time" , when it reaches its destination.

As is explained by the "Einstein's Train" thought experiment, this process is frame dependent. Google for it, or for "the relativity of simultaneity". This is one of the trickiest parts of SR to get across, by the way.

This is a necessary insight to understand how A can think B's clock is slow, and B can think A's clock is slow.

Since it was mentioned that the OP is not afraid of math, I'll go through an abstract argument about why symmetrical time dilation implies simultaneity must be relative. It's not much math - it's just the abstract notion of an invertible map, a 1:1 correspondence between sets, also known as a bijection. It may be overkill, but there's a wiki article on this at https://en.wikipedia.org/wiki/Bijection. For every element in one set (say A) there is one unique corresponding element in set B. And since the mapping is invertible, for every element in set B, there is one, and only one, corresponding element in set A.

The necessary assumption is that comparing times is done by such a one-one map, by a bijection.

Let's take a specific example. Let's say, for simplicity, that A thinks B's clock is running at half seed, and B thinks A's clock is running at half speed.

If there is only one mapping from A's time to B's time, this is logically impossible. With only one mapping, if B's clock is running at half A's rate, then A's clock is running twice as fast as B's clock.

A=1 corresponds to B=2. Logically, B=2 must correspond to A=1, as the map is unique and invertible - it's unique in both directions.

However, if the mapping from A's time to B's time is different from the mapping from B's time to A's time, this is perfectly possible.

In special relativity, this is the case. The mappings are done by clock syncrhonizattion convetions. A is in A's frame, and A uses the convention for this frame to make A's map. B is in B's frame, and B uses frame B's convention to make B's map. The important thing to realize is that A's map is NOT THE SAME as B's map.

If necessary, we can talk more about why we use Einstein's synchronization convention, but at this point it would distract from the main point, I think.