Time dilation and understanding which is $\Delta t$ and which is proper time $\tau$

Question

In the textbook that we are using, the definition for proper time $\tau$ is the interval between two events, as measured by an observer who is at rest with the two events. The definition for $\Delta t$ is the time interval between two events measured by an observer who is in motion with respect to the events. Using this information our teacher asked us to figure out the following question

Superman leaves Lois in Metropolis to rescue a malfunctioning space probe sent up from Earth. Flying at a speed of $0.7c,$ superman reaches the probe in 20 hours according to his wrist watch. How long would the trip take according to Lois's clock on Earth?

The answer is the textbook is $\Delta t$ is equal to 28 hours.

I do not understand however why the time on Lois clock is $\Delta t$; based on the definition above, should we not be solving for proper time $\tau$, because superman is not at rest.

Answer 1

superman is not at rest

There is no preferential frame in relativity. The problem supposes both Louis frame and Superman frame as inertial, with a relative velocity of $0.7$c.

The two events (say good bye to Louis and save the space probe) happens at the same location for Superman. It is like to get the meal at your seat in an airplane, and afterwards the empty packaging be collected by the flight attendant. Both events happens at your seat in the airplane frame. For people on the earth, the first event may happen over Málaga (Spain), and the other over Bordeaux (France). In this case, the time interval of your cell phone (in flight mode) between the 2 events is proper time. The time interval measured by the Málaga and Bordeaux airports is not.

Answer 2

I've always found talk of an observer confusing in Special Relativity. Much clearer, in my opinion, to use the idea of an inertial frame of reference (frame, for short).

The proper time between two events is the time between the events as measured in a frame in which the events occur at the same place. In your example, superman's frame (the frame in which he is at rest) is moving at 0.7 $c$ with respect to the Earth and the events (superman leaving Earth and superman reaching the probe) occur in the same place.

In the Earth's frame the two events occur in quite different places, so the time between them is an improper time.

Answer 3

According to superman's wrist watch, superman is not moving, and instead Earth and space probe are moving.

Answer 4

Superman is at rest, in Superman's frame ($S'$).

It's good to assign a frame name to each observer, btw. It increases clarity.

In $S'$, Lois's frame ($S$) is moving at $\beta=-0.7$.

Use events:

$$ E_i = (t', x')_{\rm frame} $$

Events are points in spacetime and do not change with Lorentz transforms, but their coordinates do change. Physics is independent of coordinates.

Basically, there are 3 events: $E_0$, the start, with:

$$ E_0 = (0, 0)_S = (0, 0)_{S'} $$

Superman reaches the probe:

$$ E_1 = (t_1, x_1)_S = (20h, 0)_{S'}$$

And then Lois's coordinates when Superman reaches the probe:

$$ E_2 = (t_2=t_1, 0)_S $$

Note that $t_2=t_1$ because the events are simultaneous in $S$. Ofc, $t'_2 \ne t'_1$, because $E_2$ is in the future for $E_1$, according to $S'$.

Answer 5

Whatever Superman does, he does right where Superman is. Superman kisses Lois goodbye right where Superman is. Twenty hours later in Superman's proper time $\tau$, Superman fixes the space probe right where Superman is. He comes to a complete stop with respect to Lois and measures that she is $x$ distance away. He sends Lois a message letting her know that he's on the way home, and what time it is on his clock.

Whatever Lois does, she does right where Lois is. Lois kisses Superman goodbye right where Lois is. $\Delta t + x/c$ time later in Lois's proper time, she receives Superman's message on her radio (which is right where Lois is). She triangulates and calculates that Superman was $x$ away from her when he sent it. She subtracts $x/c$ from the time her clock shows and determines that Superman fixed the space probe $\Delta t$ time after he kissed her goodbye, and that his time dilation factor was $\tau / \Delta t$, which matches her measurement of his velocity, since $20/28 = 1/\gamma(0.7c)$ as predicted by relativity.

Answer 6

Superman's wrist watch is at rest with respect to superman. So the time range measured by superman's wrist watch is smaller (i.e. the time runs slower for superman and for his watch) than the corresponding time range measured by Lois on earth. In particular Lois sees superman moving at 0.7c.

There is the argumentation that a watch on earth is at rest with respect to the earth, so that the roles of superman and Lois(earth) are swapped (Superman sees Lois moving). That would lead to the conclusion that the time on earth runs slower than the time measured by superman's wrist watch. It seems actually to be a paradoxon.

However, the situation of superman and Lois is not completely symmetrical, which means that a swap of roles between superman and Lois with its projected consequences on the time advance respectively delay is in fact impossible.

Actually superman not only travels to the space probe, but he also travels back to earth (This fact is extremely important, because otherwise the watch of superman and Lois' watch on earth cannot be compared). Therefore superman's invariant line element is shorter (due to the hyperbolic geometry of the Minkowski space) than the corresponding line element on earth. This cannot be changed by a change of reference system, because the line element is invariant under change of reference system.

The last paragraph is explained now in more detail:

The line element is defined as

$$ (\Delta s)^2 = c^2 (\Delta t)^2 - (\Delta x)^2 -(\Delta y)^2 - (\Delta z)^2$$

in the most general case.

In a reference system on earth Lois has coordinates x=y=z=0, i.e.

$$ (\Delta s)_{earth}^2 = c^2 (\Delta t)^2\quad \text{therefore}\quad (\Delta t)_{earth} = (\Delta s)_{earth}/c = (\Delta t) $$

whereas at superman because he is moving $z\neq 0$, $y\neq 0$ and $z\neq 0$ it is:

$$ (\Delta s)_{superman}^2 = c^2 (\Delta t)^2 - (\Delta x)^2 -(\Delta y)^2 - (\Delta z)^2 < c^2 (\Delta t)_{earth}^2 $$

So the proper time $\Delta\tau = (\Delta s)_{superman}/c$ at superman is shorter, i.e. its time runs slower.

One could now try to change the coordinate system in order to swap the role of superman and Lois. However, the computed line elements for both persons do not change, because the line element is invariant with respect to changes of reference systems.