Time dilation - what is the time on the Earth?

[Lotto]

TL;DR Summary

I have a man in a rocket with a speed relative to the Earth that is close to the speed of light. The Lorentz factor is $\gamma=2$. For the observer on the Earth, when 20 s passed, 10 s passed in the rocket according to the man on the Earth. What time passed on the Earth according to the man in the rocket?

If I understand it well, 10 s did really passed in the rocket, it is according to the observer on the Earth, but if the man in the rocket measured the time, he would measure 10 s. But when we say that the man in the rocket is in an inertial frame of reference as well, he can claim that because in his rocket passed 10 s, on the Earth passed 5 s. But that is not true. Why? Both frames of references are equal, aren't they?

 

[Ibix]

But that is not true.

Yes, it is true.

The two frames disagree on what "simultaneously" means, so they disagree what time it is on Earth "simultaneously" with the rocket clock reading 10s.

Note that this result only applies as long as both observers are travelling inertially. If one turns around and they meet again then you cannot use the naive "both frames do the same analysis" argument because one observer is not always at rest in one inertial frame.

 

[DaveC426913]

...he can claim that because in his rocket passed 10 s, on the Earth passed 5 s. But that is not true.

As Ibix points out. It is true.
But he can't go back to Earth without turning around - which breaks the symmetry of the scenario.

 

[Orodruin]

TL;DR Summary: I have a man in a rocket with a speed relative to the Earth that is close to the speed of light. The Lorentz factor is $\gamma=2$. For the observer on the Earth, when 20 s passed, 10 s passed in the rocket according to the man on the Earth. What time passed on the Earth according to the man in the rocket?

Both frames of references are equal, aren't they?

Precisely because the frames are equivalent both must observe the other’s clock rate to be slow. As has already been pointed out, the cause of this is the relativity of simultaneity. This may or may not help:
https://www.physicsforums.com/insights/geometrical-view-time-dilation-twin-paradox/

 

[Sagittarius A-Star]

TL;DR Summary: I have a man in a rocket with a speed relative to the Earth that is close to the speed of light. The Lorentz factor is $\gamma=2$. For the observer on the Earth, when 20 s passed, 10 s passed in the rocket according to the man on the Earth. What time passed on the Earth according to the man in the rocket?

If I understand it well, 10 s did really passed in the rocket, it is according to the observer on the Earth, but if the man in the rocket measured the time, he would measure 10 s. But when we say that the man in the rocket is in an inertial frame of reference as well, he can claim that because in his rocket passed 10 s, on the Earth passed 5 s.

Yes. Let the rocket frame be the primed frame $S'$ and the earth frame be the unprimed frame $S$.

Two tick-events of a clock at rest in the rocket have the same $x'$ coordinate, that means $\Delta x' = 0$. Using inverse Lorentz transformation for time:
$\Delta t = \gamma (\Delta t' + v\Delta x'/c^2)$
$\Delta t = 20s \Rightarrow$
$\Delta t' = 10s$

Please be aware, that the following deltas refer to a different pair of events.

Two tick-events of a clock at rest on earth have the same $x$ coordinate, that means $\Delta x = 0$. Lorentz transformation for time:
$\Delta t' = \gamma (\Delta t - v\Delta x/c^2)$
$\Delta t' = 10s \Rightarrow$
$\Delta t = 5s$.

But that is not true.

This statement is wrong. See calculations above.

 

[Lotto]

But then should be the man on the Earth 5 s younger than the man in the rocket, at least according to him. But for the man on the Earth, the man in the rocket is 10 s younger and only he is right. And we don't have to say that the rocket has to turn back and return to the Earth, we just talk about a short time interval, during which the rocket moves straight relative to the Earth's observer. How to explain it? This is not exactly the twin paradox.

 

[Ibix]

But then should be the man on the Earth 5 s younger than the man in the rocket, at least according to him. But for the man on the Earth, the man in the rocket is 10 s younger and only he is right.

No.

If the two people travel inertially then they can both see the other as younger and there is mo contradiction. If one turns around then the one who turned round makes a mistake if he predicts the other to be younger, as he has used a simplified analysis in circumstances where the simplifying assumptions are violated.

And we don't have to say that the rocket has to turn back and return to the Earth,

In that case they are both inertial and cannot meet up again.

This is not exactly the twin paradox.

It appears that you want to set up a scenario where two straight lines (the observers' paths) cross twice. It can't be done, and that's why you can't make a sensible analysis of it. The solution is not to propose a self-contradictory scenario.

 

[Orodruin]

But then should be the man on the Earth 5 s younger than the man in the rocket, at least according to him. But for the man on the Earth, the man in the rocket is 10 s younger and only he is right. And we don't have to say that the rocket has to turn back and return to the Earth, we just talk about a short time interval, during which the rocket moves straight relative to the Earth's observer. How to explain it? This is not exactly the twin paradox.

Unless they meet up again there is no unique way of comparing the clocks because of the relativity of simultaneity. It does not matter how small you make the time interval.

 

[Sagittarius A-Star]

But then should be the man on the Earth 5 s younger than the man in the rocket, at least according to him. But for the man on the Earth, the man in the rocket is 10 s younger and only he is right.

That's wrong, as others have explained.

How to explain it?

The time dilation factor is the ratio between the change in proper time of a moving clock (for example between two tick-events) and the change in coordinate-time of the reference coordinate system. For a standard inertial coordinate system as reference, it is:
$\frac{d\tau}{dt}= \sqrt{1-v^2/c^2}$.

 

[Lotto]

So the man in the rocket has to return on the Earth to find out that he is not older? Otherwise he will always see the man on the Earth younger and will be true?

 

[PeterDonis]

So the man in the rocket has to return on the Earth to find out that he is not older?

Not quite. Unless the man in the rocket does return to Earth, there is no answer to the question of who is "older". The question itself is not well-defined unless the two people meet up again, because there is no invariant way of comparing the ages of the two people.

 

[obtronix]

So the man in the rocket has to return on the Earth to find out that he is not older? Otherwise he will always see the man on the Earth younger and will be true?

There's a symmetry here, the measurements are only valid in their own reference frame.

If I walk away from you you measure me smaller and I measure you smaller it seems contradictory until you realize the measurements are only valid from their perspective.

Likewise in relativity space twin measures the Earth twin younger and Earth twin measures the space twin younger but they're both valid but only in their reference frames.

Mistake people make and confuses them is they intermix measurements they take one measurement from one reference frame and think it's valid in another reference frame.

 

[Sagittarius A-Star]

Otherwise he will always see the man on the Earth younger and will be true?

No. As others mentioned, clocks can be only compared directly in an invariant way, when they meet.

Regarding time-dilation in the rest frame of the rocket:

The clock on earth ticks slower than the coordinate-time of the reference rocket-coordinate system, as long as the rocket-coordinate system is inertial.

But when the rocket makes the turn-around towards earth, the rocket-coordinate system is not inertial and there exists a coordinate time-frame, in which the clock on earth ticks faster than the coordinate time of the reference rocket-coordinate system.

 

[Mister T]

And we don't have to say that the rocket has to turn back and return to the Earth, we just talk about a short time interval, during which the rocket moves straight relative to the Earth's observer.

If they don't share the same location, they will have to do something like send signals to each other to communicate. And then they have to account for the travel time of the signal. Seems trivial until you realize there is no universal time. Say one of them sends a signal. They won't agree on what time the signal was sent or received. Thus it is not possible to compare their ages unless they share the same location.

If you really want to understand this, set up a scenario where they send each other signals and we can show you how each observes the other's clock to be running slow.

 

[Lotto]

If they don't share the same location, they will have to do something like send signals to each other to communicate. And then they have to account for the travel time of the signal. Seems trivial until you realize there is no universal time. Say one of them sends a signal. They won't agree on what time the signal was sent or received. Thus it is not possible to compare their ages unless they share the same location.

If you really want to understand this, set up a scenario where they send each other signals and we can show you how each observes the other's clock to be running slow.

Well I know that when the rocket goes away from the Earth and sends signals, then because of the relativistic Doppler effect the frequences of the signals are lower on the Earth than the original ones. So we on the Earth see the man on the rocket age slower? But how is it related to the twin paradox?

Moreover, as far as I know, we do not need to think about an acceleration to explain the paradox, but [U]Sagittarius A-Star[/U] used this argument to exlain it, and this explanation I understand, but the explanation without it is confusing for me.

 

[PeterDonis]

how is it related to the twin paradox?

The scenario you are discussing in this thread is not related to the twin paradox, because the two observers only meet once. A "twin paradox" scenario requires the two twins to meet twice.

as far as I know, we do not need to think about an acceleration to explain the paradox, but @Sagittarius A-Star used this argument to exlain it

Do you mean here?

when the rocket makes the turn-around towards earth

If so, this statement is irrelevant to this thread, as in this thread, per what I said above, there is no turn-around.

If you want to discuss the twin paradox, you should start a separate thread; but first you should read the many previous threads we have had on that topic.

 

[Sagittarius A-Star]

If so, this statement is irrelevant to this thread, as in this thread, per what I said above, there is no turn-around.

If you want to discuss the twin paradox, you should start a separate thread; but first you should read the many previous threads we have had on that topic.

The OP referred to the twin paradox in the posting, I answered to:

So the man in the rocket has to return on the Earth to find out that he is not older? Otherwise he will always see the man on the Earth younger and will be true?

 

[Ibix]

The OP referred to the twin paradox in the posting, I answered to:

Yes. I think it might be more accurate to say that the OP is confused about why the "one way" scenario posited here is different from the "two way" twin paradox.

 

[Ibix]

...so I'll try to unconfuse.

Let's draw a simple diagram called a displacement-time diagram. You may have come across these in physics at school - you simply plot the distance of an object from the origin at a given time:

That's the situation you were talking about - the blue line represents the stay-at-home, who stays in the same place as time goes from zero to ten seconds, and the red line represents your traveller who is doing about 0.87 times the speed of light, so travels about 8.7 light seconds in those ten seconds. Additionally, I've marked crosses on each line when their clocks tick - the stay-at-home's clock ticks once per second while (in this frame) the traveller's clock ticks every two seconds.

In high school physics these diagrams are just that - a sometimes useful record of data. But in relativity, we think about spacetime, which is a four dimensional entity. In this context, these diagrams become something rather more: they are a map of spacetime - or two dimensions of it anyway (and they are usually called Minkowski diagrams). The "time" direction is up the page and the x direction in space is across the page (you can add a second spatial dimension, but we don't need to here because all our motion is in 1d).

Let's add a feature to our diagram:

This version includes horizontal lines, each one of which marks "all of space at one time". For example the line going through the $t=4\mathrm{s}$ mark is the whole of space at the time the (blue) stay-at-home calls four seconds after the start of the experiment. You can see that the (red) traveller is about 3.46 light seconds away at that time.

In pre-relativistic physics that's all there is to it. But Einstein made a key observation: two observers in relative motion do not agree about what "at one time" means. He also wrote down equations that let us draw lines showing what that the (red) traveller would call "all of space at one time":

And this is the answer to your original question. The traveller's last "now" line, what the traveller calls "all of space 5s after I passed the stay-at-home", passes through the stay-at-home's (blue) line half way between the stay-at-home's 2s and 3s tick. That is, according to the traveller, the stay-at-home's clock ticked 2.5 times "during" five ticks of his own clock, while according to the stay-at-home their clock ticked ten times while the traveller's ticked five times. This is not a paradox: they just mean different things by the word "during", and neither is wrong.

This is also the resolution to the twin paradox, with which you are confusing the above scenario. Let's add a return leg to the above diagram to turn it into the twin paradox:

The stay at home says that "during" the outbound leg the traveller's clock ticked five times; the traveller says that "during" the outbound leg the stay-at-home's clock ticked 2.5 times. And we can draw the "now" lines for the traveller on the return leg too:

Again, you can see that the traveller says that there were 2.5 ticks of the stay-at-home's clock "during" the return leg. But you can also see the error the traveller makes if they simply add those two 2.5s: they failed to remember that they'd changed direction and therefore changed their definition of "all of space at one time", so they failed to account for the 15 ticks of the stay-at-home's clock between "the end of the outbound leg" and "the beginning of the inbound leg". If they remember that they've changed their definition they can correct for that mistake and will correctly predict that the stay-at-home's clock should read 20s at return.

There are more sophisticated ways of defining "all of space at one time" so that you don't get funny time skips like this (indeed, any approach that yields time skips is basically a mistake). I'm not going to go into them because the maths is a lot messier and this post is long enough already, but there are several ways to do that.

I hope that helps a bit.

 

[Mister T]

Well I know that when the rocket goes away from the Earth and sends signals,

When the rocket sends his signal he can look at the rocket clock and see what it reads. But on Earth they have to account for the travel time of the signal to figure out what time it was on Earth clocks when the signal was sent.

And when they do that they will not agree that the rocket clock reading was correct. This is what I mean when I say there is no universal time. You will have to understand this point to understand the answer to your original question. If not you will be left confused by the twin paradox.

 

[Lotto]

When the rocket sends his signal he can look at the rocket clock and see what it reads. But on Earth they have to account for the travel time of the signal to figure out what time it was on Earth clocks when the signal was sent.

And when they do that they will not agree that the rocket clock reading was correct. This is what I mean when I say there is no universal time. You will have to understand this point to understand the answer to your original question. If not you will be left confused by the twin paradox.

Yes, and they won't agree because of the relativistic Doppler effect, will they?

 

[PeroK]

Yes, and they won't agree because of the relativistic Doppler effect, will they?

Two observers both moving inertially and moving relative to each other will measure the same, invariant speed of light in a vacuum. This is one of the postulates of SR and is incompatible with the classical notion of absolute space and time. The relativistic Doppler effect is another consequence of the postulates of SR.

 

[Mister T]

Yes, and they won't agree because of the relativistic Doppler effect, will they?

You miss the point. They won't agree. Certainly you can explain this observation using the Doppler effect, but my point concerns the relativity of simultaneity. And more specifically, how it explains the symmetry of time dilation.

 

[pervect]

Possibly of interest, an old thread. "Symmetrical time dilation implies the relativity of simultaneity". https://www.physicsforums.com/threa...on-implies-relativity-of-simultaneity.805210/

"Symmetrical time diation" is how I refer to the idea that if A and B are moving, both think the other's clocks are running slow.

Possibly the OP is under the impression that this is a new thread. It may be new to them, but to the rest of us, it's not so new, and I'll point to my post addressing the issue as a specific example of how it's not new.

There's at least one paper out there about how hard it is to get students to appreciate the relativity of simultaneity, "The challenge of changing deeply held student beliefs about the relativity of simultaneity", bh Scheer, et al. https://arxiv.org/abs/physics/0207081

So, basically - I'd encourage the original poster to read about the relativity of simultaneity, of which there has been much written about.

From experience, as well as the cited paper by Scheer, it can be difficult to get people to understand the issue. But it is the issue at the heart of this thread, and it is a correct observation that special relativity is not compatible with the natural-seeming assumption that different frames of reference share the same idea of "now", the idea of what it means for events to be "at the same time". This is an important realization, but it's only the first step. Relativity is internally self consistent, it's just not consistent with the external idea of simultaneity being the same for all observers.

The main point I can really make is that on my end (and, I would imagine, many other PF posters), this is a very old issue. We've written about it before. I can't make people read what was written before, much less force them to understand it. I can't even say that what I've written is the best on the topic - there's just so many authors who have tackled it. But I can try to point out that it's a well-known issue with a lot of history, and a lot written about it. Pointing this out and encouraging people to read about it is really all I can do, in the end.

While I believe the primary issue is one of motivation - as I said, there's a lot written about the issue - there ARE some tools that can be helpful to illustrate the underlying concepts, such as space-time diagrams. I could write more this if there is some interest, but there's not a lot of sense in writing more if the target audience is in what I call "denial mode".

 

[Lotto]

Yes. Let the rocket frame be the primed frame $S'$ and the earth frame be the unprimed frame $S$.

Two tick-events of a clock at rest in the rocket have the same $x'$ coordinate, that means $\Delta x' = 0$. Using inverse Lorentz transformation for time:
$\Delta t = \gamma (\Delta t' + v\Delta x'/c^2)$
$\Delta t = 20s \Rightarrow$
$\Delta t' = 10s$

Please be aware, that the following deltas refer to a different pair of events.

Two tick-events of a clock at rest on earth have the same $x$ coordinate, that means $\Delta x = 0$. Lorentz transformation for time:
$\Delta t' = \gamma (\Delta t - v\Delta x/c^2)$
$\Delta t' = 10s \Rightarrow$
$\Delta t = 5s$.
This statement is wrong. See calculations above.

So if I understand it well, if I use $\Delta t = \gamma (\Delta t' + v\Delta x'/c^2)$, that is when I take datas from S' into S. And when I use $\Delta t' = \gamma (\Delta t - v\Delta x/c^2)$, I take (the same?) datas from S into S'. When I calculate time $t'$ from the first equation, it is the acual time that passed in the rocket. And when I want to calculate time $t$ from the second equation, I cannot suddenly say that this time really passed on Earth. According to the man in the rocket it passed, that I understand, but why cannot I say it is the real time that passed on Earth? In the first case I can say that it is the real time in the rocket.

 

[Ibix]

According to the man in the rocket it passed, that I understand, but why cannot I say it is the real time that passed on Earth?

It is the proper time on Earth between the start of the experiment and the time the rocket's simultaneity convention says the experiment ended. But that simultaneity convention is just a convention, and you can pick others. So you can calculate the elapsed time for the Earth between the start of the experiment and the end, but you have considerable latitude in defining what "the end of the experiment" actually means.

 

[PeterDonis]

@Lotto, there are three things in relativity that always go together: time dilation, length contraction, and relativity of simultaneity. You can never just look at one in isolation. You always have to look at them all together. That means you can never just look at one Lorentz transformation equation in isolation. You always have to look at the full transformation, including all coordinates. Otherwise you will be leaving out crucial information.

 

[Lotto]

@Lotto, there are three things in relativity that always go together: time dilation, length contraction, and relativity of simultaneity. You can never just look at one in isolation. You always have to look at them all together. That means you can never just look at one Lorentz transformation equation in isolation. You always have to look at the full transformation, including all coordinates. Otherwise you will be leaving out crucial information.

Observer S is stationary and throws a ball up and measures time $t$ until the ball falls down on earth at the coordinate $x=0$. Observer S' is moving relative to him in a rocket with a velocity $v$. His coordinate of the whole event is thus $x'=-vt'$.

So I can write $t=\gamma(t'+\frac{x'v}{c^2})=\frac{t'}{\gamma}$. And also $t'=\gamma (t-\frac{xv}{c^2})=\gamma t$. So it means that for the observer S' the event takes longer according to his clock. But when I am considering only time that elapses in his rocket, it means that $x'=0$, I can see it as an "event" with this coordinate. Thus for S is $t=\gamma t'$ and for S' $t'=\gamma t$, because I consider an event that have different locations in different frame of references. So time elapses slower for him, but he will see the ball to fall down in longer time and that will also "see" the observer S.

Do I understand it well?

 

[PeroK]

It's simpler perhaps to give the coordinates of both events in both frames. In the $S$ frame:

Ball is thrown up: $t = 0, x = 0$
Ball is caught: $t = T, x = 0$.

Note that it's better to leave $t$ as the coordinate label and use, for example, $T$ for the specific coordinate value.

Then, in frame $S'$:

Ball is thrown up: $t' = 0, x' = 0$.
(Note tthat we've taken this event as the common origin.)

Ball is caught: $t' = \gamma T, x' = \gamma(0- vT) = -\gamma vT$
(This agrees with the general coordinates of the ball thrower: $x' = -vt'$)

 

[Lotto]

It's simpler perhaps to give the coordinates of both events in both frames. In the $S$ frame:

Ball is thrown up: $t = 0, x = 0$
Ball is caught: $t = T, x = 0$.

Note that it's better to leave $t$ as the coordinate label and use, for example, $T$ for the specific coordinate value.

Then, in frame $S'$:

Ball is thrown up: $t' = 0, x' = 0$.
(Note tthat we've taken this event as the common origin.)

Ball is caught: $t' = \gamma T, x' = \gamma(0- vT) = -\gamma vT$
(This agrees with the general coordinates of the ball thrower: $x' = -vt'$)

So they both have to agree on times when the particular event happend, for S it was $t$ and for S' it was $t'$, if I use the Lorentz transformation for $t$ or $t'$, I have to get the same relation between these two times. And each stationary observer thinks that in the moving frame of reference time passes more slowly.

 

[PeroK]

So they both have to agree on times when the particular event happend, for S it was $t$ and for S' it was $t'$, if I use the Lorentz transformation for $t$ or $t'$, I have to get the same relation between these two times. And each stationary observer thinks that in the moving frame of reference time passes more slowly.

You have a special case where the two events both occur at $x = 0$. That means that throwing the ball was an unnecessary detail. You really just took two times at the spatial origin of $S$. And have essentially an example of time dilation.

The Lorentz Transformation (LT) can transform any two events. In general, as I think has been mentioned before in this thread, the LT encapsulates time dilation, length contraction and relativity of simultaneity.

Ultimately, the Lorentz Transformation is a pair of invertible equations and you can't boil them down to "each observer thinks time passes more slowly".

 

[Sagittarius A-Star]

So if I understand it well, if I use $\Delta t = \gamma (\Delta t' + v\Delta x'/c^2)$, that is when I take datas from S' into S. And when I use $\Delta t' = \gamma (\Delta t - v\Delta x/c^2)$, I take (the same?) datas from S into S'.

As I wrote, these are different data:

Please be aware, that the following deltas refer to a different pair of events.

When I calculate time $t'$ from the first equation, it is the acual time that passed in the rocket.

Yes, it is the proper time passed in the rocket while the coordinate-time interval $\Delta t$ with reference to the rest-frame of the earth. Time-dilation is in this case the ratio between the proper time of a moving clock and the coordinate-time of the inertial reference frame.

And when I want to calculate time $t$ from the second equation, I cannot suddenly say that this time really passed on Earth. According to the man in the rocket it passed, that I understand, but why cannot I say it is the real time that passed on Earth? In the first case I can say that it is the real time in the rocket.

That's wrong. The situation is reciprocal. To understand time-dilation, you need to understand the definition of a standard inertial 4D-coordinate system:

The basic principle of clock synchronization is to ensure that the coordinate description of physics is as symmetric as the physics itself. For example, bullets shot off by the same gun at any point and in any direction should always have the same coordinate velocity dr/dt .
...
We should, strictly speaking, differentiate between an inertial frame and an inertial coordinate system, although in sloppy practice one usually calls both IFs. An inertial frame is simply an infinite set of point particles sitting still in space relative to each other. For stability they could be connected by a lattice of rigid rods, but free-floating particles are preferable, since keeping constant distances from each other is also a criterion of the non-rotation of the frame. A standard inertial coordinate system is any set of Cartesian x,y,z axes laid over such an inertial frame, plus synchronized clocks sitting on all the particles, as described above. Standard coordinates always use identical units, say centimeters and seconds.

Source:
http://www.scholarpedia.org/article...nematics#Galilean_and_Lorentz_transformations

 

[Mister T]

According to the man in the rocket it passed, that I understand, but why cannot I say it is the real time that passed on Earth? In the first case I can say that it is the real time in the rocket.

Because in this context there is no such thing as a real time. Your question is prompted by a misunderstanding of the way time passes.