Question about time dilation symmetry

  • #1
Chenkel
482
109
Hello everyone,

Suppose there is an airplane that is taking off from an airport but before it takes off it synchronizes it's clock to zero with the clock at the airport.

In the rest frame of the plane the airport is moving, so you could argue ##T_{plane} = \gamma*T_{airport}##

In the rest frame of the airport the plane is moving and you could argue ##T_{airport} = \gamma*T_{plane}##

How can these two equations be true simultaneously?

If you compare the two clocks after a flight of the plane which clock will read less time and why?

Any help will be appreciated, thank you.
 
Physics news on Phys.org
  • #2
This is just your typical twin paradox setup in disguise. The rest frame of the plane is not an inertial frame and so the time dilation formula cannot be used for the full round trip.
 
  • Like
Likes vanhees71, Vanadium 50 and russ_watters
  • #3
Orodruin said:
This is just your typical twin paradox setup in disguise. The rest frame of the plane is not an inertial frame and so the time dilation formula cannot be used for the full round trip.
The plane is roughly inertial if it's flying stably and the acceleration during turn around should be a short part of the overall trip.
 
  • Skeptical
Likes jbriggs444 and Motore
  • #4
Chenkel said:
How can these two equations be true simultaneously?
They cannot. (Unless ##\gamma=1##)
 
  • #5
Dale said:
They cannot. (Unless ##\gamma=1##)
We know that ##\gamma## must be more than one because the relative velocity is non zero.
 
  • #6
Chenkel said:
We know that ##\gamma## must be more than one because the relative velocity is non zero.
Then those equations cannot both be true.
 
  • Like
Likes vanhees71
  • #7
Chenkel said:
If you compare the two clocks after a flight of the plane which clock will read less time and why?
You have, as is often the case when a relativity paradox appears, overlooked the relatively of simultaneity.
In the rest frame of the plane the airport is moving, so you could argue ##T_{plane} = \gamma*T_{airport}##
In the rest frame of the airport the plane is moving and you could argue ##T_{airport} = \gamma*T_{plane}##
The ##T_{plane}## in the first equation above is a different thing than the ##T_{plane}## in the second so it was a mistake to use the same label for both, and likewise for the two ##T_{airport}## .

Go draw a spacetime diagram of the situation and you will see what is going on here.
 
  • Like
Likes robphy, vanhees71 and Dale
  • #8
Chenkel said:
The plane is roughly inertial if it's flying stably and the acceleration during turn around should be a short part of the overall trip.
It is not the duration of acceleration that is relevant. It is the corresponding relativity of simultaneity of the turnaround and an event at the airport.
 
  • Like
Likes jbriggs444 and vanhees71
  • #9
Nugatory said:
You have, as is often the case when a relativity paradox appears, overlooked the relatively of simultaneity.
The ##T_{plane}## in the first equation above is a different thing than the ##T_{plane}## in the second so it was a mistake to use the same label for both, and likewise for the two ##T_{airport}## .

Go draw a spacetime diagram of the situation and you will see what is going on here.
Why are the two ##T_{plane}## different?

##T_{plane}## is the time on the airplane clock

##T_{airport}## is the time on the airport clock

They both synchronize to zero at the airport when they meet at the start of the journey.
 
  • #10
Chenkel said:
Why are the two ##T_{plane}## different?
Because they are based on two different simultaneity conventions. One assumes the simultaneity convention of the plane, the other that of the airport. Simultaneity is relative. Someone should put a helpful reminder in their PF signature …
 
  • Like
  • Haha
Likes Motore, vanhees71, russ_watters and 1 other person
  • #11
Chenkel said:
They both synchronize to zero at the airport when they meet at the start of the journey.
You did see the part where I suggested that you try drawing a spacetime diagram? Your reply came back so quickly that I'm pretty sure you haven't tried.

If you do not try drawing the spacetime diagram it will never make sense.
If you do try drawing the spacetime diagram eventually it will start to make sense.

Back to you....
 
  • Like
Likes robphy, Motore, vanhees71 and 1 other person
  • #12
Chenkel said:
Why are the two ##T_{plane}## different?

##T_{plane}## is the time on the airplane clock

##T_{airport}## is the time on the airport clock

They both synchronize to zero at the airport when they meet at the start of the journey.
So, I also recommend that you draw a spacetime diagram. However, for future reference, a common convention is to use ##\tau ## to denote the time on a clock and ##t## to denote a coordinate time. It helps avoid confusion to use different symbols for different kinds of quantities like that.
 
Last edited:
  • Like
Likes Chenkel and vanhees71
  • #13
Nugatory said:
You did see the part where I suggested that you try drawing a spacetime diagram? Your reply came back so quickly that I'm pretty sure you haven't tried.

If you do not try drawing the spacetime diagram it will never make sense.
If you do try drawing the spacetime diagram eventually it will start to make sense.

Back to you....
Thanks, I'll work on that.
 
  • #14
Dale said:
So, I also recommend that you draw a spacetime diagram. However, for future reference, the usual convention is to use ##\tau ## to denote the time on a clock and ##t## to denote a coordinate time. It helps avoid confusion to use different symbols for different kinds of quantities like that.
I was reading that article you linked me on spacetime diagrams but it was a little over my head.
 
  • #15
Chenkel said:
I was reading that article you linked me on spacetime diagrams but it was a little over my head.
Perhaps you should ask questions about that instead of this.
 
  • Like
Likes vanhees71, Vanadium 50, Nugatory and 1 other person
  • #16
Orodruin said:
This is just your typical twin paradox setup in disguise.
Not a very good disguise. Maybe a pair of Groucho glasses.
Chenkel said:
The plane is roughly inertial if it's flying stably and the acceleration during turn around should be a short part of the overall trip.
That is also true for the "regular" twin paradox.
Nugatory said:
Go draw a spacetime diagram
This is very good advice.
 
  • Like
Likes Chenkel and vanhees71
  • #17
Chenkel said:
Hello everyone,

Suppose there is an airplane that is taking off from an airport but before it takes off it synchronizes it's clock to zero with the clock at the airport.

In the rest frame of the plane the airport is moving, so you could argue ##T_{plane} = \gamma*T_{airport}##

In the rest frame of the airport the plane is moving and you could argue ##T_{airport} = \gamma*T_{plane}##

How can these two equations be true simultaneously?

If you compare the two clocks after a flight of the plane which clock will read less time and why?

Any help will be appreciated, thank you.
Consider two people standing at opposite ends of a room. One person raises their hand and notes that the size of their hand corresponds to the height of the other person. The second person does the same. How can these statements both be true?

If we let the height of the two people be ##H_A, H_b## and the size of their hands be ##h_A, h_B##, then:
$$H_A > h_A = H_B > h_b = H_A$$
 
  • Like
Likes vanhees71
  • #18
Chenkel said:
Hello everyone,

Suppose there is an airplane that is taking off from an airport but before it takes off it synchronizes it's clock to zero with the clock at the airport.
I assume you mean that the plane moves almost inertial after the start and never returns.
Let's call the airport frame ##S## and the airplane frame after a quick start ##S'##.
The clocks were set to zero approximately at the event ##x=x'=t=t'=0##.

Chenkel said:
In the rest frame of the plane the airport is moving, so you could argue ##T_{plane} = \gamma*T_{airport}##
In the rest frame of the airport the plane is moving and you could argue ##T_{airport} = \gamma*T_{plane}##
There are two concepts of time: proper time and coordinate-time. Your equations are wrong to describe time-dilation because remote clocks cannot be directly compared, only indirectly, i.e. via the coordinate-time of a reference frame.

Chenkel said:
How can these two equations be true simultaneously?
In the rest frame of the plane the airport is moving, so, because all ticks of the airport-clock happen at ##x=0##, you could argue with the Lorentz transformation:
##t'= \gamma (t-vx/c^2) = \gamma (t-0)=\gamma T_{airport}##

In the rest frame of the airport the plane is moving, so, because all ticks of the plane-clock happen at ##x'=0##, you could argue with the Lorentz transformation:
##t= \gamma (t'+vx'/c^2) = \gamma (t'+0)=\gamma T_{plane}##

Between those equations is no contradiction. Time-dilation is the ratio between the proper time of a clock and the coordinate-time of a reference-frame.

Chenkel said:
If you compare the two clocks after a flight of the plane which clock will read less time and why?
Remote clocks cannot be compared directly. But if the plane returns to the airport, then the plane-clock will read less time. Calculated with reference to the rest-frame of the airport-clock, the plane-clock has the time-dilation factor ##1/\gamma##. The plane-clock has no unique inertial rest-frame in this twin-paradox scenario. At the turnaround, it must move non-inertial.
 
  • Informative
  • Like
Likes Dale and vanhees71
  • #19
Sagittarius A-Star said:
But if the plane returns to the airport, then the plane-clock will read less time. Calculated with reference to the rest-frame of the airport-clock, the plane-clock has the time-dilation factor ##1/\gamma##. The plane-clock has no unique inertial rest-frame in this twin-paradox scenario. At the turnaround, it must move non-inertial.
The final readings on the plane and airport clocks depend on the direction in which the plane travelled - on account of the Earth's rotation. It's possible for the plane clock to show more time than the airport clock. See the results of the Hafele-Keating experiment, for example.
 
  • Like
Likes Gleb1964, vanhees71 and Sagittarius A-Star
  • #20
PeroK said:
The final readings on the plane and airport clocks depend on the direction in which the plane travelled - on account of the Earth's rotation. It's possible for the plane clock to show more time than the airport clock. See the results of the Hafele-Keating experiment, for example.
Yes, I made the idealization, that we talk about SR (local scenario), not GR, and ignoring gravitational time-dilation.
 
Last edited:
  • Like
Likes vanhees71
  • #21
Sagittarius A-Star said:
Yes, I made the idealization, that we talk about SR (local scenario), not GR, and ignoring gravitational time-dilation.
It is not dependent on gravity. The plane could be replaced by ground transportation, such as a high-speed train. The problem is the false assumption that "acceleration causes real time dilation", which is implicit in your post.
 
  • #22
PeroK said:
The problem is the false assumption that "acceleration causes real time dilation", which is implicit in your post.
I don't assume this. Which statement in my post do you mean specifically?
 
  • #23
Sagittarius A-Star said:
I don't assume this. Which statement in my post do you mean specifically?
Sagittarius A-Star said:
The plane-clock has no unique inertial rest-frame in this twin-paradox scenario. At the turnaround, it must move non-inertial.
It's just part of the same overall error in attributing differential ageing to some characteristic of a trajectory, as opposed to its overall length. The problem here is that the Earth is a rotating reference frame. Hafele-Keating did, indeed, have to take account of the direction of travel. And, the plane that flew westwards showed more time than the airport clock.
 
  • Like
Likes vanhees71
  • #24
PeroK said:
It's just part of the same overall error in attributing differential ageing to some characteristic of a trajectory, as opposed to its overall length.
No. It's just an argument, why it is easier to calculate in the rest frame of the airport, given a local SR scenario, the OP asked for.

PeroK said:
The problem here is that the Earth is a rotating reference frame. Hafele-Keating did, indeed, have to take account of the direction of travel. And, the plane that flew westwards showed more time than the airport clock.
The Sagnac effect needs only to be taken into account at a full travel around the earth. I am talking about a local SR scenario.
 
  • #25
Sagittarius A-Star said:
No. It's just an argument, why it is easier to calculate in the rest frame of the airport, given a local SR scenario, the OP asked for.
It's an interesting confusion of ideas to invoke airline travel within a local experiment! The purpose of invoking air travel would be to make the experiment non-local, I would have thought.

Ultimately, the simple answer is to use spacetime geometry to measure the length of each path. That's a simple, universal answer that puts differential ageing on a firm foundation. Everything else involves ad hoc assumptions about what is and what isn't taken into account.
 
  • Like
Likes Sagittarius A-Star
  • #26
PeroK said:
Ultimately, the simple answer is to use spacetime geometry to measure the length of each path. That's a simple, universal answer that puts differential ageing on a firm foundation. Everything else involves ad hoc assumptions about what is and what isn't taken into account.
I agree. But the OP is not yet there. According to other recent threads, he is reading Morin's book and is somewhere before chapter "2.1 Lorentz transformations" and has not yet reached "2.3 The invariant interval".
 
  • Like
Likes PeroK
  • #27
PeroK said:
The purpose of invoking air travel would be to make the experiment non-local, I would have thought.
Then you disagree with Lufthansa. They offer air travel from Munich to Frankfurt.
 
  • #28
Sagittarius A-Star said:
Then you disagree with Lufthansa. They offer air travel from Munich to Frankfurt.
Perhaps more on environmental than relativistic grounds.
 
  • Like
Likes Sagittarius A-Star

FAQ: Question about time dilation symmetry

What is time dilation symmetry?

Time dilation symmetry refers to the principle in special relativity that time dilation is reciprocal between two observers in relative motion. Each observer perceives the other’s clock as running slower than their own, demonstrating a symmetric relationship in their observations.

How does time dilation symmetry arise from special relativity?

Time dilation symmetry arises from the Lorentz transformations in special relativity, which describe how measurements of time and space change for observers moving relative to each other. These transformations ensure that the laws of physics, including the speed of light, are the same for all observers, leading to symmetric time dilation effects.

Can both observers be right in claiming the other's clock is slower?

Yes, both observers can be correct in claiming the other's clock is slower. This seeming paradox is resolved by understanding that each observer's measurements are valid in their own inertial frame of reference. The symmetry comes from the fact that the situation is entirely reciprocal.

Does time dilation symmetry apply in all situations?

Time dilation symmetry specifically applies to inertial frames of reference, where observers are moving at constant velocities relative to each other. In non-inertial frames, such as those involving acceleration, general relativity must be used, and the symmetry can be more complex.

How is time dilation symmetry experimentally verified?

Time dilation symmetry has been experimentally verified through various tests, including observations of particles moving at high speeds, such as muons created by cosmic rays, and precise measurements using atomic clocks on fast-moving airplanes or satellites. These experiments consistently confirm the predictions of special relativity, including symmetric time dilation effects.

Similar threads

Replies
16
Views
1K
Replies
88
Views
5K
Replies
36
Views
2K
Replies
54
Views
2K
Replies
45
Views
4K
Replies
34
Views
1K
Replies
55
Views
3K
Back
Top