Two clocks along different worldlines

Question

I have been reading and watching videos about this subject for a while now. I just can not seem to grasp the idea. Let's say we have two clocks. I leave one at home and keep one in my pocket. Then, I started running at speed that is close to speed of light to my school then come back to my house. If I compare those two clocks how would they differ in time?

Answer 1

The world lines of both clocks pass through two particular events ('points' in spacetime), the event of your leaving the home and the event of your returning.

The worldline of the clock at home is straight while the worldline of the clock in your pocket must be curved due to the acceleration you undergo during your near light speed trip out and back.

A result in special relativity is that the elapsed time is longest along the straight (inertial) worldline; all other worldlines have less elapsed time.

According to the Wikipedia article "Proper time":

In relativity, proper time is the elapsed time between two events as measured by a clock that passes through both events. The proper time depends not only on the events but also on the motion of the clock between the events. An accelerated clock will measure a smaller elapsed time between two events than that measured by a non-accelerated (inertial) clock between the same two events. The twin paradox is an example of this effect.

Thus, the clock in your pocket will show less elapsed time that the clock at home.

Answer 2

Let's say we have two clocks.

Let's call the (coordinate) times of these two clocks $t_A$, and $t_B$, respectively.

I leave one at home and keep one in my pocket. Then, I started running [...] then come back to my house. If I compare those two clocks how would they differ in time?

If it is also given (corresponding to the comment by the OP above: "they increased at same rate") that each of these two clocks had been "good" (i.e. "running at a definite, constant rate") during the separation , i.e.

$\frac{d}{d \tau_A}[t_A] = \frac{d}{d \tau_B}[t_B] \text{=(abbreviated as)=} \text{rate}$

then the "difference in time" between these two clocks, at the return, is

$t_A[ \text{at return to B} ] - t_B[ \text{at return to A} ] :=$

$t_A[ \text{at separation from B} ] - t_B[ \text{at separation from A} ] +$ $\text{rate} * \Delta\tau_A[ \text{from separation from B, until return to B} ] * $ $\left(1 - \frac{\Delta\tau_B[ \text{from separation from A, until return to A} ] }{ \Delta\tau_A[ \text{from separation from B, until return to B} ] } \right) =$

$t_A[ \text{at separation from B} ] - t_B[ \text{at separation from A} ] +$ $\left( t_A[ \text{at return to B} ] - t_A[ \text{at separation from B} ] \right) * $ $\left(1 - \frac{\Delta\tau_B[ \text{from separation from A, until return to A} ] }{ \Delta\tau_A[ \text{from separation from B, until return to B} ] } \right),$

where $\Delta\tau_A$ is the duration (a.k.a. "proper time") of one clock, and $\Delta\tau_B$ the duration of the other, respectively, between their indications as specified.

If it is also given (corresponding to the comment by the OP above: "[...] started out at same time") that

$t_A[ \text{at separation from B} ] = t_B[ \text{at separation from A} ],$

then this "difference in time" between these two clocks, at the return, simplifies to

$t_A[ \text{at return to B} ] - t_B[ \text{at return to A} ] =$

$\left( t_A[ \text{at return to B} ] - t_A[ \text{at separation from B} ] \right) * $ $\left(1 - \frac{\Delta\tau_B[ \text{from separation from A, until return to A} ] }{ \Delta\tau_A[ \text{from separation from B, until return to B} ] } \right).$

Finally, in order to evaluate the number

$\frac{\Delta\tau_B[ \text{from separation from A, until return to A} ] }{ \Delta\tau_A[ \text{from separation from B, until return to B} ] }$

it would of course be necessary to measure (or to assume) the geometric-kinematic relations between these two clocks during their being separated from each other more specificly. (It might for instance be relevant which, if any, of these two clocks had remained a member of an intertial system throughout the separation; at least within some suitable accuracy.)

Answer 3

Without getting to mathy, the reason is because you are dilating time and what you would see is that the clock at home was much ahead of your clock you held when you ran near the speed of light. In other words, you would be in the future.

Answer 4

Okay, let's say you have a friend staying at home measuring the time and observing you with a telescope - the whole situation will be described from his point of view. Namely, we will use his measure of distance $x$ and the time $t$ he measures with the stopwatch in his hand.

Special relativity tells us that he has a means of calculating your "proper time", i.e. the time that runs on the stopwatch in your pocket while running, from observing your velocity. The first statement is trivial: "the total proper time is the sum of the proper times during each phase of motion. This is formally written as: $$\tau_{1,2} = \sum_{\dots} \Delta \tau = \int_1^2 d \tau$$ Where the second relation describes the situation when you continuously change your velocity so that every infinitesimal piece of your time changes it's rate towards your friend at home and the sum has to be done through an integral. The trick is now to find out the rate of change of your proper time $ \tau$ with respect to the home time $t$. Relativity tells us that for a fixed running velocity $v$ we have $$\Delta \tau = \sqrt{1-v^2/c^2} \Delta t\,\to\,d \tau = \sqrt{1-v(t)^2/c^2} d t $$ Hence, the only thing to find out how much time the stopwatch in your pocket will measure is to compute the following $$\tau_{1,2} = \int_1^2 \sqrt{1-v(t)^2/c^2} d t$$ But this is highly nontrivial for reasonable velocity histories $v(t)$, so I am going to use a model for a qualitative understanding.

Why a model? Because the description will contain infinite accelerations, where we would have to include general-relativistic effects for consistency, because horrible energy transfers would be happening. Thus this description gives us a qualitative understanding of the resulting time-difference only, and should be understood as a very rough approximation to a process with small accelerations (smoothed out slowing down etc.).

Ok, so you start at rest at home - $v=0,x=0$. From $t_1$ to $t_1 + (t_1-t_2)/2$ you are moving with a velocity $0.5 c$ and from $t_1 + (t_1-t_2)/2$ you move with a velocity $-0.5 c$ until $t_2$ when you arrive home again and stop to compare your times. Your proper running time will be $$\tau_{1,2} = \sqrt{1 - (0.5 c)^2/c^2}(t_1-t_2)/2 + \sqrt{1 - (-0.5 c)^2/c^2}(t_1-t_2)/2 = \sqrt{0.75} (t_1 - t_2)$$ That is, to you, your stopwatch would show $0.87$ times less runtime than the stopwatch left at home. But since relativity is relative and does not discriminate, how is it decided that you are the one who has a shorter time? Because you are the one who starts and stops. That is, you go back into the home frame of reference to compare your times, so the home frame of reference is suddenly "privileged".

Remember that this description is highly unrealistic, but a realistic descriptions always lead to shorter time of the "runner" and longer time of the "home" at least until you are in a complicated gravitational situation. An experiment has actually shown this effect, the Hafele-Keating experiment.

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Gravitation? - We can neglect it as long as you move strictly at the same altitude $\to$ same gravitational potential and acceleration. There would be some technical complications with moving around a considerable part of the Earth because it's spherical shape would require a change in our description, but no new qualitative effects would appear.

However, if for a considerable time you would move with a speed smaller than the Earth's rotation and orbit, you would actually gain a time difference not from your movement with respect to the Earth's surface but from a smaller or larger speed with respect to the "background space". You can see in the already linked Hafele-Keating experiment that in that case the relative speeds were so small that the westward motion prolonged the time of the "runner" clock with respect to the "home". The intuitive reason is that the westward clock was actually counter-rotating against the faster Earth rotation and thus moving slower. But a non hand-waivy description would require general relativity.

Answer 5

Technically they wouldn't differ in time because to come back to your house you would have to decelerate and accelerate, which is under the realms of general relativity or as seen in comments below (thanks to @dgh) we can use comoving frames. You could use two frames one moving towards school and one away from school and Lorentz transform between the two, or you could use an infinite number of comoving frames in both directions and use the infinitesimal Lorentz transformations between each of the frames, hence integrate them to get total transformation. However, in simple special relativity your clock would be at say 2 mins past 12 and the clock at home would be at 30 mins past 12.This is because when you travel "fast" (relative to some observer) your time slows down even though you wouldn't notice it. What you would notice is that the distance you travel is decreased by

$$L=L_{0}\sqrt{1-\left(\tfrac{v}{c}\right)^2}$$

Where $L$ is the distance you would see, and $L_{0}$ is the distance your mom (waiting for you at home) would see you travel. This is similar to the Twins paradox which you can learn about here http://www.phys.unsw.edu.au/einsteinlight/jw/module4_twin_paradox.htm

or you can google twin paradox.

You can find how much the time differs by using time dilation equation (or Lorentz transformations):

$$\Delta t=\frac{\Delta t_{0}}{\sqrt{1-\left(\tfrac{v}{c}\right)^2}}$$

Where $\Delta t$ is the time elapsed for the clock in your pocket and $\Delta t_{0}$ is the "stationary" clock left at home.