Why Does the Spaceship Record the Proper Time in Relativity?

[Undoubtedly0]

On several homework questions I have encountered the following type of scenario:

Two clocks are perfectly synchronized. One is placed in a spaceship, and the other is left on the ground. The spaceship flies to some distant location at some high speed, and then returns to earth. When the clocks are compared, the spaceship has recorded that less time has passed. Mathematically, this is because the spaceship's clock records the "proper time".

Why though, is the spaceship the one with the proper time? Proper time is defined to be the time interval between two events that occur at the same point. Does not the observer on Earth record both events from the same location in his inertial reference frame? The reference frame of the spaceship does not even stay inertial during the turn-around.

Thanks all!

 

[HallsofIvy]

There are several things at work here. First, the clock that stayed on Earth was always subject to Earth's gravity and the one that went to the "distant location" was not. (Actually, that would tend to make the clock on Earth run slower and so is NOT the main cause of the other clock running slower.)

More importantly, the clock that went to the "distant location" was subject to several accelerations- going up to the "high speed", turning around at the "distant location", and then decelerating when it gets back to earth. Each acceleration "breaks" the symmetry and will make that clock run slower.

 

[Undoubtedly0]

Thanks HallsofIvy. I agree with all this, except, what rules out the clock at rest (on Earth) from "carrying" the proper time? Consider this in deep space so as to ignore effects of gravity.

 

[Passionflower]

I think you misunderstand what proper time stands for. A proper time interval is simply that what is measured by a clock between two events.

 

[Undoubtedly0]

According this textbook, however, these two events must occur at the same point. This is why I am confused.

 

[Passionflower]

Take event A and event B, different travelers can take different paths between these events and as a consequence they may all record a different elapsed time between them all these recorded times are proper times.

An event is a 4-dimensional point so two events occurring at the same points implies they are actually the same event.

 

[morrobay]

On several homework questions I have encountered the following type of scenario:
Why though, is the spaceship the one with the proper time? Proper time is defined to be the time interval between two events that occur at the same point.

Thanks all!

No - proper time is the time interval between two points as recorded on ship.
Actually in this case the proper time is equal to the time-like spacetime interval .
s = ( ct²-x²)^1/2

And you can see why proper time recorded by traveling clock is less than proper time recorded by Earth clock ( add after yuiop )

see also : https://www.physicsforums.com/showthread.php?t=354622@page=3

post # 36 must be logged into see graph by Dr Gregg

 

[yuiop]

On several homework questions I have encountered the following type of scenario:

Two clocks are perfectly synchronized. One is placed in a spaceship, and the other is left on the ground. The spaceship flies to some distant location at some high speed, and then returns to earth. When the clocks are compared, the spaceship has recorded that less time has passed. Mathematically, this is because the spaceship's clock records the "proper time".

Why though, is the spaceship the one with the proper time?

You have been mislead by your sources. Both the clock on the spaceship and on the Earth record proper time. The reason the spaceship records less time is not because it is "the one" that records "proper time", but because it records less proper time than the clock that remains on Earth.

Proper time is defined to be the time interval between two events that occur at the same point.

Proper time is the time recorded by a clock that remains stationary with respect to the observer. The clock on the Earth remains stationary with respect to the Earth observer and the clock on the spaceship remains stationary with respect to the spaceship observer. Therefore both clocks record proper time.

Does not the observer on Earth record both events from the same location in his inertial reference frame?

Yes.

The reference frame of the spaceship does not even stay inertial during the turn-around.

Correct. It is precisely because the spaceship clock is not inertial and the Earth clock is inertial that the difference in recorded proper times occurs. In nit picking detail the clock on the Earth is not inertial either, due to the acceleration of gravity, but for the sake of the thought experiment the relativistic speed of the spaceship is great enough that the time dilation due to the Earth's gravity can be thought of as insignificant.

 

[yuiop]

According this textbook, however, these two events must occur at the same point. This is why I am confused.

That is correct in the casual sense that a point is spatial rather a 4 dimensional combination of time and space. Here is an example that makes things clearer. Imagine we have two synchronised clocks, A at location A and B at location B. These clocks remain stationary in reference frame S. A third clock C, travels from A to B in 1 second at 0.8c as measured in S. An observer moving with clock C (call this reference frame S') considers clock C to be stationary in his reference frame. The time measured by clock C as it travels from A to B is 0.6 seconds. This is the proper time between the events "Clock C leaving A and clock C arriving at B" and it is measured by clock C because C is present at both events. Clocks A and B do not record the proper time between those two events because neither one is present at both events. In reference frame S the events "clock leaving A" and "clock arriving at B" do not occur at the same place, but in reference frame S' they do, because the observer at rest with clock C considers himself to be stationary.

 

[elfmotat]

They BOTH record proper time. Proper time is defined as the time measured on clocks. This does not prevent the two clocks from experiencing a different amount of time.

For example, assuming the Earth is stationary (i.e. dx/dt=0), the proper time of Earth clock is given by the path integral (in one dimension):

$$\Delta \tau = \int \sqrt{1-\frac{1}{c^2}\left (\frac{dx}{dt}\right )^2}dt = \int dt=\Delta t$$

For the spaceship, since it is accelerating/decelerating over some nontrivial path, it is easy to see that the integral will result in less proper time.

 

[Michael C]

According this textbook, however, these two events must occur at the same point. This is why I am confused.

The problem here is the phrase "the same point". This phrase only makes sense when defined with respect to a particular frame of reference.

At the moment when the clocks are synchronised, it's clear that they are at the same point for all observers. After the spaceship has left the Earth, for the observer on the spaceship his clock is still at the same point (it hasn't moved relative to him), while the clock on the Earth is no longer at the same point. For the observer on the Earth, it's the spaceship clock that has moved: his Earth clock is still at the same point relative to him.

 

[morrobay]

They BOTH record proper time. Proper time is defined as the time measured on clocks. This does not prevent the two clocks from experiencing a different amount of time.

For example, assuming the Earth is stationary (i.e. dx/dt=0), the proper time of Earth clock is given by the path integral (in one dimension):

$$\Delta \tau = \int \sqrt{1-\frac{1}{c^2}\left (\frac{dx}{dt}\right )^2}dt = \int dt=\Delta t$$

For the spaceship, since it is accelerating/decelerating over some nontrivial path, it is easy to see that the integral will result in less proper time.

Nice equation , would you refer to the thread reference in my post #7
In that referenced thread, post #36 by yuiop ( Nov 16, 09 ) and the graph by Dr. Gregg.
The graph indicates that the recorded proper time for rocket is function of velocity only not acceleration .
So is proper time determined by both velocity and acceleration ?
It looks like the evaluation of this integral involves velocity.
"The Integrator " produced a very complicated function.

 

[yuiop]

Nice equation , would you refer to the thread reference in my post #7
In that referenced thread, post #36 by yuiop ( Nov 16, 09 ) and the graph by Dr. Gregg.
The graph indicates that the recorded proper time for rocket is function of velocity only not acceleration .
So is proper time determined by both velocity and acceleration ?
It looks like the evaluation of this integral involves velocity.
"The Integrator " produced a very complicated function.

I once read a nice analogy between wind chill factor and time dilation. Consider a bike moving at 5 mph relative to the air on a cold day. When the bike accelerates to a new constant velocity of 10 mph relative to the air the wind chill factor increases but this increased chill factor is due to the increased velocity rather than due to the acceleration in itself.

Here is another example. Consider two inertial clocks far away from any gravitational sources. One clock is accelerated to a high relativistic speed and travels in a large circle and returns to the un-accelerated clock. In this case, the clock that travels in large circle experiences continuous proper acceleration and experiences the greatest time dilation. Now consider of the case of two clocks in a gravitational field. One clock is hovering at radius r from a massive gravitational body and the other clock is orbiting at radius r around the gravitational body. For each complete orbit when the clocks pass each other the accelerating clock (the hovering one) experiences the least time dilation. In this case the orbiting clock is inertial and feels no proper acceleration and yet this is the clock that experiences the greatest time dilation and shows the least elapsed proper time on each passing. Clearly in this case, acceleration is not the cause of time dilation. However I should point out that this example includes gravitational time dilation and Special Relativity specifically excludes effects due to gravity, but I hope this example clearly illustrates that you can not universally assume acceleration equates to greater time dilation.

Now let's return to the case of the traveling twins. Ignore the acceleration due to gravity on the twin that stays on the Earth. Plot the paths of the twins on a spacetime diagram. In the Earth reference frame the path of the twin that remains on the Earth is a vertical straight line. The path of the traveling twin is a sloped line on the outward journey and another sloped line in the opposite direction on the return journey. It is obvious that the path through spacetime for the the traveling twin is greater than the spacetime path of the stay at home twin, so the traveling twin experiences the greatest time dilation. The important concept is that we can transform the paths to any reference frame and the spacetime path of the traveling twin is always longer than the path of the Earthbound twin so there is never any ambiguity about which experiences the greater time dilation.

 

[morrobay]

Well thanks but given the endless discussions on proper time in this section
and even the different evaluations of this:
Integral t_o> t₁ sqrt[1-v(t)²/c²]dt
I posted a numerical problem in the Homework Intro Physics section on proper time : velocity and acceleration. On page two now
That hopefully will get a numerical answer for this flexible term.
yuiop, elfmotat ,Matterwave, Pervect , Dr Greg, DaleSpam Janus - thank you