Question about time dilation symmetry

[Chenkel]

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.

 

[Orodruin]

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.

 

[Chenkel]

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.

 

[Dale]

How can these two equations be true simultaneously?

They cannot. (Unless $\gamma=1$)

 

[Chenkel]

They cannot. (Unless $\gamma=1$)

We know that $\gamma$ must be more than one because the relative velocity is non zero.

 

[Dale]

We know that $\gamma$ must be more than one because the relative velocity is non zero.

Then those equations cannot both be true.

 

[Nugatory]

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.

 

[Orodruin]

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.

 

[Chenkel]

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.

 

[Orodruin]

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 …

 

[Nugatory]

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....

 

[Dale]

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.

 

[Chenkel]

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.

 

[Chenkel]

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.

 

[Dale]

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.

 

[Vanadium 50]

This is just your typical twin paradox setup in disguise.

Not a very good disguise. Maybe a pair of Groucho glasses.

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.

Go draw a spacetime diagram

This is very good advice.

 

[PeroK]

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:
$$H_A > h_A = H_B > h_b = H_A$$

 

[Sagittarius A-Star]

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 $x=x'=t=t'=0$.

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.

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.

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.

 

[PeroK]

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.

 

[Sagittarius A-Star]

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.

 

[PeroK]

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.

 

[Sagittarius A-Star]

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?

 

[PeroK]

I don't assume this. Which statement in my post do you mean specifically?

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.

 

[Sagittarius A-Star]

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.

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.

 

[PeroK]

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.

 

[Sagittarius A-Star]

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".

 

[Sagittarius A-Star]

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.

 

[PeroK]

Then you disagree with Lufthansa. They offer air travel from Munich to Frankfurt.

Perhaps more on environmental than relativistic grounds.