Suppose, I'm on earth and my brother is moving away from earth at a constant speed, $v=0.8c$. Now, if 5 seconds $(t_0)$ pass for me, the amount of time that will pass for my brother according to me will be $t$:




So, if $5$ decades pass for me, $8.33$ decades will pass for my brother. He will experience rapid aging according to me. So, why is everyone saying time will go slower for him when the case is the exact opposite?

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    $\begingroup$ But it is your brother's point of view that matters here. $\endgroup$
    – jamesqf
    Commented Nov 2, 2021 at 16:47
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    $\begingroup$ What matters here is which brother underwent acceleration to achieve this velocity. Nothing about velocity decides who has the "proper time". $\endgroup$ Commented Nov 2, 2021 at 20:30
  • $\begingroup$ @candied_orange does that mean that if you measure your acceleration precisely you can determine how much time dilation you are experiencing relative to the rest frame you started from? $\endgroup$
    – Michael
    Commented Nov 3, 2021 at 1:11
  • $\begingroup$ @Michael Acceleration is also observer-dependent. If you accelerate forever at a constant rate, the people on Earth will see your acceleration getting smaller as you get closer to the speed of light. Of course, just like with velocity, all you need to do is do a bit of math to get the relative time dilation between you and the Earth. The thing you need to remember is that from the PoV of Earth, your time runs slower - from your point of view, their time runs slower. That's where the (apparent) twin paradox comes from. $\endgroup$
    – Luaan
    Commented Nov 3, 2021 at 15:12
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    $\begingroup$ People who say "the faster you move through space, the slower you move through time" have it the wrong way around. It's really "the faster you move through space, the faster you move through time". Think of the classic twin paradox. Which twin spent less travel time to get to the rendezvous (assuming the twins separate and meet back up at the same space coordinates in an inertial frame, and one twin is stationary)? It's the one who moved quickly through space. Clearly, as that twin spent less time getting there, they must have moved faster through time. $\endgroup$
    – Arthur
    Commented Nov 4, 2021 at 23:32

4 Answers 4


You have applied the equation incorrectly.

This is because $t$ is the time you observe (the dilated time) on your brothers clock and $t_0$ is the proper time, or the time inside your brother’s frame of reference.

That is, if $5$ seconds elapsed on your brother's clock as measured from your frame, then the elapsed time on his clock in his frame is $t_0$ where $$5=\frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}} \\ \rightarrow t_0=5\sqrt{1-0.8^2}=3\ \ \text{seconds}$$

It's interesting to note that since your brother is also observing you to be moving away from him at $v=0.8c$, if $5$ seconds passes for you inside your frame, then your brother will observe your clock to take $$t=\frac{5}{\sqrt{1-(0.8)^2}}=8.3\ \ \text{seconds}$$ This is probably how you meant to apply the equation.

When you observe his clock you will see a dilated time and when he observes your clock, he to will also see a dilated time. So who is correct? The solution to this apparent contradiction "the twin paradox" is addressed here.

  • $\begingroup$ In your second equation you suggest that each brother, if they could observe the other brother's clock, would see it moving at a different rate from theirs, and the result would not be symmetrical. That was always my understand, except recently I have seen the opposite claim, that each would see the other's clock the same except if the other was accelerating. (e.g. see the accepted answer to the linked question) $\endgroup$
    – Michael
    Commented Nov 3, 2021 at 1:14
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    $\begingroup$ Each will see the other's clock run slower, after correcting for changing communication lag. It has nothing to do with acceleration. $\endgroup$
    – JDługosz
    Commented Nov 3, 2021 at 16:35
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    $\begingroup$ I love Relativity ... one of the greatest mind bends in physics and always a fascinating thought journey... $\endgroup$ Commented Nov 4, 2021 at 20:33

Actually, you have used the relativistic formula incorrectly, which is creating the confusion. Let's understand the equation first:

$$\Delta t=\frac{\Delta t_0}{\sqrt{1-\frac{v^2}{c^2}}}$$

Here $\Delta t_0$ is called the proper time, which is the time measured by the observer who is inside the moving frame of reference relative to the other observer. Here the person inside the spaceship measures the proper time.

$\Delta t$ is the dilated time, measured by the observer outside the moving frame of reference; or here, the person on earth. So actually, $\Delta t=5$ should yield correct answer.


Your question is as such completely unrelated to your title. However, that it seems related to the body of your question to you is at the heart of your misconception. One simply moves at $1$ second per second through time. It is a meaningless thing to ask how fast one moves through time, so it is true that it is not the case that the faster you move through space, the slower you move through time but it's not because you instead move faster through time. It's a meaningless assertion either way.

What you want to ask is whether a clock $A$, as seen by a given observer $O$, ticks at a slower or faster rate than their own clock $B$ when clock $A$ is moving faster or slower w.r.t. the said observer $O$. Now, this is a meaningful question. Given the correction to your calculation indicated in the other answers, the answer to this meaningful question is that a clock $A$ ticks at a slower rate, as observed by $O$, the faster it moves w.r.t. $O$.

However, it is crucial to understand that this does not mean that "time slows down" for clock $A$. That is, in fact, the central tenet of special relativity -- that all inertial observers observe the same physics. So, nothing unusual happens to $A$ from its own reference frame. That is the very axiom on the basis of which we derive the implication (via coupling it to the existence of a finite invariant speed) that when viewed from the reference frame of $B$, $A$ must appear to have slowed down.

  • $\begingroup$ So wait ... are you saying that it actually DOESN'T slow down for A? Because I just read about Paul Langevin dissertation where in he described the hypothetical scenario of a person traveling at 99.995% the speed of light in one direction for a year, then back in the opposite direction for another year. When that person returns, they have only aged two years while the earth would have aged 200 years. $\endgroup$ Commented Nov 4, 2021 at 20:39
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    $\begingroup$ @MichaelSims What you're describing doesn't imply what you think it implies -- precisely for the reasons I have already described in my answer. $\endgroup$
    – user87745
    Commented Nov 4, 2021 at 20:42
  • $\begingroup$ What I'm describing is a thought experiment where In a person travels at near the speed of light for two years and while he was gone, 200 years passed here on earth... is that the way it works? Or not? $\endgroup$ Commented Nov 4, 2021 at 22:27
  • $\begingroup$ @MichaelSims Yes, that is the way that works (source). However, DvijD.C. 's point is that from the perspective of the person traveling, time will go normally. He will not get any extra time than a static person. Does that make sense? $\endgroup$ Commented Nov 5, 2021 at 13:19

None of the answers point to the fact that time dilation is not an effect that happens to the clocks themselves. To explain it correctly you need to first identify what are the two spacetime event points between which you are calculating the time interval. If the two events are on the world-line of the brother on board the spaceship, then his time should be considered as the proper time and the earth-bound brother's clock will show the dilated time. For example if the spaceship-brother goes from earth to some distant star then his starting point and finishing point, both are on his worldliness. Starting the journey and finishing it are the two events. If by the spaceship clock the journey took say 5 yrs, the earth clock will show the corresponding time interval dilated by some amount depending on the velocity of the ship. It is 5 yrs times the Lorentz factor if you ignore the initial acceleration and final deceleration of the ship. If you care to include these two non-uniform motions , no big deal, you just need to use the time dilation formula with integration. Look up the 1st chapter of Landau Lifshitz Vol 3.
You may look up a few videos I made on this, here are the links: https://youtu.be/KEeEhP89SZs https://youtu.be/8iybUbKDTLs


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