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Please will someone explain what time dilation really is and how it occurs. There are lots of questions and answers going into how to calculate time dilation, but none that give an intuitive feel for how it happens.

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I'm not a physicist and can't even begin to comment on the content here, but nice job on laying it out. I have just one suggestion since I've made a post like this myself on GDSE. You can create a Contents section by hyperlinking to each answer individually. It's helpful for the obvious efficiency reason plus because votes and other factors may move the answer order around. There are 2 options for implementation, you can either add it to the question or create a new answer that's just contents, and choose it as the accepted answer so that it stays at the top. – Dom Mar 6 at 22:55
    
Also, very insightful and related: What is time, does it flow, and if so what defines its direction?. – MAFIA36790 Jul 7 at 13:59
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"I don't know what is the time" ... Here, it's about 9:20AM. – WillO Jul 7 at 14:02
    
@lucas: I can understand your predicament; at first it all looks weird and perplexing. But they are sensibly plausible and correct. Read John's answer there; it's extremely competent to give you the greatest insight on the phenomenon. Just read it. – MAFIA36790 Jul 7 at 14:14

Introduction

This answer will use ideas discussed in the answers to What is time, does it flow, and if so what defines its direction? so you really need to read the answers to that question before tackling this one.

The key concept you need in order to understand time dilation is that a clock does not measure the flow of time - time doesn’t flow in relativity (see the What is time question for more on this). A clock measures distances. To explain what I mean I’ll use the analogy of the odometer in your car. If you start at some point $A$ and drive to some point $B$ then the odometer tells you how far in space you’ve moved. So the change in the odometer reading is the distance in space $A-B$ measured along the route you took. The clock in your car measures the distance in time between the spacetime points $A$ and $B$ i.e. the change in the clock measures the number of seconds between you leaving point $A$ and arriving at point $B$, and the number of seconds is also measured along the route you took in spacetime. This last point matters, because as we’ll see the distance in time you move depends on your route, just like the distance in space moved.

The reason we have to treat time as a distance is because in relativity there isn’t a hard and fast distinction between time and space. You may split spacetime into three spatial dimensions and one time dimension, but a different observer might make this split in a different way and the two of you wouldn’t agree on what was time and what was space. In relativity we have to treat the time dimension just like the space dimensions. It is just a coordinate running from (in principle) $-\infty$ to $\infty$ just like the $x$, $y$ and $z$ coordinates run from $-\infty$ to $\infty$. See the What is time question for more on this.

The point of all this is that it gives us a very specific definition of time dilation. If two different observers measure the distance between two spacetime points $A$ and $B$ then this distance will be a four-vector with time and spatial components. Time dilation simply means that different observers will disagree on the magnitude of the time component of this distance i.e. they will observe a different amount of time between the two points.

An example of time dilation

To explain why this happens let’s take a specific example. Suppose I am watching you moving, then in my coordinates your trajectory is a line in spacetime. Because I can’t draw four dimensional graphs let’s assume you’re only moving along the $x$ axis so all I have to draw is your trajectory in $x$ and time. Suppose your trajectory looks like this:

Figure 1

Figure 1

So we both start at the point $A$. Because I am stationary in these coordinates my trajectory is straight up the time axis to $B$, while your trajectory (the red line) heads off to increasing $x$, then stops, turns round and comes back to my position. The distance I have moved in time is just the distance straight up the time axis from $A$ to $B$ - we’ll call this distance $t_{ab}$. The distance you have moved in time is, well, let’s see how to calculate that.

Figure 1 shows what happens in my coordinate system, but now let’s draw the same diagram in your coordinate system i.e. the coordinates in which you remain stationary at the origin and I move:

Figure 2

Figure 2

In your coordinates it’s me that moves (shown by the black line) and you remain stationary, so in your coordinates your trajectory (the red line) is straight up the time axis and the distance you move is just the distance in time between $A$ and $B$. We’ll call this distance $\tau_{ab}$.

Now this is the point where things get strange, but actually it’s the only point where things get strange so if you can get past this point you’re home. The distance $\tau_{ab}$ in figure 2 has a special significance in relativity. It’s called the proper time, and it’s a fundamental principle in relativity that the proper time is an invariant. This means the proper time is the same for all observers, and specifically it the same for both you and me. This means that - and here’s the key point:

The length of the red line is the same in both figure 1 and figure 2

Let’s go back to figure 1 for a moment and see why this means there must be time dilation:

Figure 3

Figure 3

The length of my line from $A$ to $B$, $t_{ab}$, is obviously different from the length of the red line from $A$ to $B$, $\tau_{ab}$. But we’ve already agreed that the length of the red line is the time you measure between the two points, and that means the time I measure between $A$ and $B$ is different from the time you measure between $A$ and $B$:

$$ t_{ab} \ne \tau_{ab} $$

And that’s what we mean by time dilation.

If my aim was to give an intuitive idea of how time dilation arises then I’ve probably failed because it is far from intuitively obvious why the length of the red line should be the same in figure 1 and figure 2. But at least I’ve narrowed it down to one unintuitive step, and if you’re prepared to accept this then the rest follows in a straightforward way. To make this quantitative, and explain exactly what I mean by the length of the red line, we need to get stuck into some maths.

And now some maths

The situation I’ve drawn in figures 1 and 2 is actually somewhat complicated because it involves acceleration i.e. you speed away from me, decelerate to a halt then accelerate back towards me. To get started we’ll use the simpler case where you just head off at constant velocity and don’t accelerate. Our two spacetime diagrams look like this:

Figure 4

Figure 4

In my frame you are travelling at velocity $v$, so after some time $t$ measured on my clock your position is $(t, vt)$. In your frame your are stationary, so after some time $T$ measured on your clock your position is $(T, 0)$. And remember we said that the length of the red line must be the same for both you and me.

To calculate the length of the red line we use a function called the metric. You probably remember being taught Pythagoras’ theorem at school. Which tells you for the right angled triangle:

Triangle

the length of the hypotenuse is given by:

$$ s^2 = a^2 + b^2 $$

This equation tells one how to measure total (that is, in this case diagonal) distances, given the displacements in each coordinate direction. That is precisely the information contained in a metric: It tells you how to measure distances. The above equation does this by giving an explicit formula for the length of a line, resulting from coordinate displacements in the horizontal and vertical directions (let's call those $x$ and $y$). Now, one can of course also think about infinitesimal (infinitely small, in a limiting sense) distances. The formula then simply becomes

$$ \mathrm ds^2=\mathrm dx^2+\mathrm dy^2$$

This is called the line element for two-dimensional Euclidean space, and it encodes the corresponding (Euclidean) metric. For special relativity we need to extend this idea to include all three spatial dimensions plus time. There are various ways to write the line element for special relativity and for the purposes of this article I’m going to write it as:

$$ \mathrm ds^2 = - c^2\mathrm dt^2 + \mathrm dx^2 + \mathrm dy^2 +\mathrm dz^2 $$

where $\mathrm dt$ is the distance moved in time and $\mathrm dx$ etc are the distances moved in space.

This equation encodes the Minkowski metric and the quantity $\mathrm ds$ is called the proper distance. It looks a bit like Pythagoras’ theorem but note that we can’t just add time to distance because they have different units - seconds and metres - so we multiply time by the speed of light $c$ so the product $ct$ has units of metres. Also note that we give $ct$ a minus sign in the equation - as you’ll see, this minus sign is what explains the time dilation. Since we are only considering two dimensions our equation becomes:

$$ \mathrm ds^2 = -c^2\mathrm dt^2 + \mathrm dx^2 $$

OK, let’s do the calculation. Since all motion is in a straight line we don't need the infinitesimal line element and instead we can use:

$$ \Delta s^2 = -c^2\Delta t^2 + \Delta x^2 $$

Start in your frame - you don’t move in space so $\Delta x = 0$ and you move a distance $\tau$ in time, so $\Delta t = \tau$, giving us:

$$ \Delta s^2 = -c^2 \tau^2 $$

Now let’s do the calculation in my frame. In my frame you move a distance in space $\Delta x=vt$ and a distance in time $\Delta t = t$ so the equation for the length of the red line is:

$$ \Delta s^2 = -c^2t^2 + (vt)^2 = -t^2c^2\left(1 - \frac{v^2}{c^2}\right) $$

Since the lengths $\Delta s$ are equal in both frames we combine the two equations to get:

$$ -c^2 \tau^2 = -t^2c^2\left(1 - \frac{v^2}{c^2}\right) $$

And rearranging gives:

$$ \tau = t\sqrt{1 - \frac{v^2}{c^2}} = \frac{t}{\gamma} $$

where $\gamma$ is the Lorentz factor:

$$ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} $$

And that’s the result we need showing the time dilation. The distance you have moved in time $\tau$ is less than the distance I have moved in time $t$ by a factor of $\gamma$.

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Appendix - accelerated motion

I started the main answer with this spacetime diagram:

Figure 1

Figure 1

but then switched to a simpler example when it came to churning through the maths. This is because I didn’t want to distract from the main message in my answer, however if anyone is interested I’ll explain how we deal with accelerated motion now.

Incidentally, you’ll hear people claim that special relativity can’t deal with accelerated motion but as you’re about to see this is simply not true. The basic principle is the same - the length of the trajectory is the same for all observers. It’s just that calculating the length of the trajectory is a bit harder.

The calculation we’re going to do is the same as before i.e. I’ll calculate the distance from $A$ to $B$ along my trajectory then calculate the distance along your trajectory, and the time dilation will be the difference between them. The distance along my trajectory is obviously just the distance up the time axis, but to for you we have to calculate the length of the red curve.

We do this by splitting up the curve into infinitesimal straight lines:

Figure 2

Figure 2

If we approximate the red curve by a series of straight lines of length $\mathrm ds$ then the total length of the curve, $\Delta s$, will just be the sum of the lengths of all these straight lines. We let the lengths $\mathrm ds$ go to zero and replace the sum by an integral:

$$ \Delta s = \int_A^B \,\mathrm ds \tag{1} $$

And the length $\mathrm ds$ is given by the same equation that we used in the main answer:

$$ ds^2 = -c^2\mathrm dt^2 + \mathrm dx^2 \tag{2} $$

The trick we use is to note that if you move a distance $dx$ in a time $dt$ then your velocity is $v = {\mathrm dx}/{\mathrm dt}$, because that’s exactly how we define velocity. Rearranging this gives:

$$ \mathrm dx = v\,\mathrm dt $$

And we can substitute this into equation (2) to get:

$$\mathrm ds^2 = -c^2\mathrm dt^2 + v^2(t) \mathrm dt^2 $$

where $v(t)$ is your velocity as a function of time measured in my frame. Now put this into equation (1) and we get:

$$ \Delta s = -c \int_A^B \, \left(1 - \frac{v(t)^2}{c^2}\right)\, \mathrm dt \tag{3} $$

Finally we note that in your frame the distance you move is still given by the same equation as before:

$$ \mathrm ds^2 = -c^2T^2 $$

And we get:

$$ T_{AB} = \int_A^B \, \sqrt{1 - \frac{v(t)^2}{c^2}}\,\mathrm dt $$

where $T_{AB}$ is the elapsed time measured by your clock.

To do the calculation we need to know the equation for your velocity as a function of time, and this depends on how you accelerate away. Actually doing the sums gets quite complicated quite quickly so I won’t go through the detail. However we can see immediately that there is time dilation and you measure less elapsed time than I do.

Whether your velocity $v(t)$ is positive or negative the square, $v^2(t)$ is always positive, and that means the factor in the square root is always less than 1:

$$ 1 - \frac{v(t)^2}{c^2} \lt 1 $$

So we are integrating a function that is always less than one from $t = t_A$ to $t = t_B$ and that means the result must be less than $t_B - t_A$, that is:

$$ T_{AB} \lt t_B - t_A $$

So your elapsed time, $T_{AB}$ is always less than my elapsed time, $t_B - t_A$, not matter how you change your velocity during your round trip.

And by now you should have spotted that this is just the twin paradox in disguise. This shows that the elapsed time for the accelerating twin is always less than the elapsed time for the stationary twin, though there are more details that will have to wait for another post on another day.

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Appendix - what did the twin observe?

The more attentive of you might have noticed something I left out of my calculation in the last section of the main answer. I gave this figure showing the spacetime diagrams:

Figure 4

Then I did the calculation of the length of the red line in my frame, and I showed that your elapsed time is less than my elapsed time. All quite correct of course, but hang on, isn’t time dilation symmetric? Shouldn’t you observe my time to be dilated? Yes indeed, and the purpose of this appendix is to explain what’s going on.

If we look at my spacetime diagram we note that you and I didn’t end up at the same points. You travelled from $A$ to $B$ while I travelled from $A$ to $C$. In my frame the points $B$ and $C$ are simultaneous i.e. they have the same time coordinate, $t_B = t_C$, and that’s why I can claim there is time dilation. My claim is that we both started at the same time $t=t_A$ and we both ended at the same time $t=t_B=t_C$ but our clocks measured different elapsed times while we did it. Hence there must be time dilation.

But my claim that the points $B$ and $C$ are simultaneous is only true in my frame, and in all other frames $B$ and $C$ are not simultaneous. This means different observers will disagree with my calculation of the time dilation, and that’s why you and I can both think the other person’s time is dilated. Let’s see how this works.

I’m going to shortcut a lot of maths and simply tell you that to find where spacetime points are in different frames we use a couple of equations called the Lorentz transformations. These are:

$$\begin{align} t’ &= \gamma\left(t - \frac{vx}{c^2}\right) \\ x’ &= \gamma\left(x - vt \right) \end{align}$$

Take the point $B$, which in my coordinates is $(t,vt)$. To find the corresponding point $B’$ in your coordinates just plug $t = t$ and $x = vt$ into the equations to get:

$$\begin{align} t’ &= \gamma\left(t - \frac{v(vt)}{c^2}\right) = \gamma t \left(1 - \frac{v^2}{c^2}\right) = \frac{t}{\gamma} \\ x’ &= \gamma\left(vt - vt \right) = 0 \end{align}$$

So in your frame the point $B = (t/\gamma, 0)$. But we already knew this. In your frame you are stationary at the origin so your position $x$ is always zero, and we have already worked out that your elapsed time is $T = t/\gamma$. So the Lorentz tranformations tell us what we already knew, which is just as well really!

But now take the point $C$, which is $(t, 0)$ in my frame, and let’s see where it is in your frame. Again, just bung these values for $t$ and $x$ into the Lorentz transformations and we get:

$$\begin{align} t’ &= \gamma\left(t - \frac{v\,0}{c^2}\right) = \gamma t \\ x’ &= \gamma\left(0 - vt \right) = -\gamma vt \end{align}$$

Let’s draw our frames with all these points on them:

Figure 8

So in in my frame the time interval measured on my clock while I move from $A$ to $C$ is $t$, but in your frame the time interval while I move from $A$ to $C$ is the distance $AD$ i.e. it is $\gamma t$. And since $\gamma t \gt t$ you observe my time to be dilated in the same way as I observe your time to be dilated. It’s just that we disagree about our start and end points.

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Asked by lucas:

I know nothing about relativity but I cannot accept that there is a phenomenon called time dilation. However I have no problem with it because of mathematics behind it. I have no problem if time is dilated, because I don't know what time is. But I wonder when they say a clock will work slowly with respect to the other same clock if its speed is higher.

  1. Which kind of clocks they mean? Analog clock, digital clock, etc.

  2. As far as I know some mechanical clocks work by a torsion spring inside them. So, how does the material of the spring know that it must unroll slowly at higher speed? Does higher speed change chemical structure or physical properties of the material of the spring?

Answered by Gennaro Tedesco:

The clock obviously neither slows nor speeds. That is only unfortunate terminology to mean that time intervals depend on the reference frame and different observers in different reference frames may measure different time intervals if in relative motion with respect to each other.

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Sorry but time is composed of time intervals and time is measured by clocks, by some periodic motion etc. The frequency of ticking of clocks and watches really slows down relatively to an observer that moves relatively to the clocks or watches. This means that time as such is dilated. It applies to time as measured by all kinds of clocks. There is nothing unfortunate about the term "dilatation of time" or claims that "time slows down" etc. I think that by far the most likely "lesson" one may take from your vague criticism of the terminology is a full denial of dilation of time. – Luboš Motl Jul 8 at 16:34
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@LubošMotl What one can prove in special relativity is that time intervals depend on the reference frame, no more, no less; whether this is due to the mechanics of the ticking of the clocks isn't the point (as you may always invent some other type of clocks, theoretically, and the results in physics must not depend on how you construct your apparatus). Moreover, I don't see how one can imply the statement "the most likely "lesson" one may take from your vague criticism of the terminology is a full denial of dilation of time". – Gennaro Tedesco Jul 8 at 16:53
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Many comments deleted after they veered off into a "Is not!"/"Is so." argument. – dmckee Jul 9 at 17:35
    
I think the edit was highly inappropriate. An answer shouldn't answer a different question, it should answer the question it is posted on. If the answer in its previous form didn't satisfy that criterion, the questions shouldn't have been merged in the first place - but if that is the case, now that they are merged, the appropriate solution would be to either edit this answer to address the question it is currently on (which would probably have to be done by Gennaro), or delete the answer outright. – David Z Jul 9 at 22:58
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I also suppose an edit would be inappropriate, since the original question is pretty different from the one I address. I would say we either keep it with the original text quoted in marks (as it is now) or we move it back elsewhere (or anything similar). – Gennaro Tedesco Jul 10 at 11:18

What is time dilation really?

A reduced rate of local motion. See What is time, does it flow, and if so what defines its direction? As Einstein said, time is what clocks measure. And if you take a scientific empirical look at what a clock really does, you will see that it doesn't actually measure the distance in time between the spacetime points A and B. It merely features a vibrating crystal or a rocker or a pendulum, and some kind of gears or electronics to count or translate this regular cyclical local motion to provide some kind of cumulative display. A clock "clocks up" local motion, that's all. And when the clock goes slower it's because that local motion is going slower.

Please will someone explain what time dilation really is and how it occurs.

As above, time dilation is a reduced rate of local motion. See On the Electrodynamics of Moving Bodies where Einstein talked about time:

Now we must bear carefully in mind that a mathematical description of this kind has no physical meaning unless we are quite clear as to what we understand by "time". We have to take into account that all our judgments in which time plays a part are always judgments of simultaneous events. If, for instance, I say, “That train arrives here at 7 o'clock,” I mean something like this: “The pointing of the small hand of my watch to 7 and the arrival of the train are simultaneous events”.

This operational definition of time is nothing more than the position of the hands, which is just a cumulative version of all the regular cyclical local motion inside the clock. The internal mechanism of a clock isn't called a movement for nothing. Einstein later talked about the “time” required by light to travel from A to B, which nicely relates to the simple inference of time dilation on Wikipedia:

enter image description herepublic domain image by Mdd4696

This features light moving in a parallel-mirror light clock. The time is nothing more than the number of times the light has reflected off the mirrors. Time dilation occurs when the ensemble moves fast because the light takes a zigzag path rather than a straight up-and-down path. But if it zoomed across the clear night sky and you could watch it through your gedanken telescope, you'd have to pan to keep it in your field of view. And in that field of view the light beam would appear to move straight up and down, at a slower rate than normal. That's special-relativity time dilation. That's all it is. It's that simple. The Lorentz factor $$\Delta t' = \frac{\Delta t}{\sqrt{1-\frac{v^2}{c^2}}}$$ is simply derived from Pythagoras's theorem, wherein the hypotenuse is the light path, and the base is the speed as a fraction of c. The height gives the Lorentz factor, and we employ a reciprocal to distinguish time dilation from length contraction.

There are lots of questions and answers going into how to calculate time dilation, but none that give an intuitive feel for how it happens.

I think the Wikipedia article is good enough for special relativity. It's very simple. The rate of local motion is of necessity reduced by the macroscopic motion through space because the maximum rate of motion is c. This time dilation applies not just to light, but to all material things too, because of the wave nature of matter.

It's also very simple for general relativity. A concentration of energy alters the surrounding space, this effect diminishing with distance, such that parallel-mirror light clocks run slower when they're lower, along with all material processes, again because of the wave nature of matter.

enter image description here

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Time dilation is simply the slowing of clocks. It is really the change in time at a velocity relative to a clock at rest as demonstrated in the lorentz transformation. As a simple illustration, imagine that all clocks work on some mass related mainspring. As you go faster and faster the effective mass increases and therefore the period of the clock tick slows down. This not a perfect model but helps draw a more sophisticate image.

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Nice idea. This is the closest to "intuitive" that I see on this page. The rest are the standard responses that you find anywhere. goo.gl/fz4R0I – Sean Jul 24 at 16:54

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