What is time dilation really?

Question

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.

Answer 1

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 $t$ (time). Suppose your trajectory looks like this:

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

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

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 math.

And now some math

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

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:

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 $\mathrm dx$, $\mathrm dy$, $\mathrm dz$ 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 meters — so we multiply time by the speed of light $c$ so the product $ct$ has units of meters. 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 $\gamma$.

Answer 2

Appendix - accelerated motion

I started the main answer with this spacetime diagram:

Figure 1

but then switched to a simpler example when it came to churning through the math. 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 $t$ (time) axis, but 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

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:

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

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

$$ \Delta s = \sqrt{-c^2}T_{AB} $$

where $T_{AB}$ is the elapsed time recorded on your clock, and equating these gives:

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

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, $t_B - t_A$, no 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.

Answer 3

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:

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 traveled from $A$ to $B$ while I traveled 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 math 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 transformations 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:

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.

Answer 4

Frame-to-Frame Comparison of Diagrams and Intervals

This is an adjunct to John Rennie's main discussion in which we examine the classical Twin Paradox by explicitly drawing the space-time diagram of both twin's routes in two different frames and computing their intervals explicitly both ways in order to show that the outcome doesn't depend on which (single, inertial) frame the experiment is viewed from.

The scenario show here has the traveling twin (Heidi) making $0.5c$ relative the Earth on both legs of her journey, and visiting a target object one light-year from Earth with no stopover. The stay-at-home twin (Hans), of course, remains on Earth eagerly awaiting their reunion.

As a simplifying assumption here the acceleration is assumed to be fast enough that we need not bother showing it or adding it to our computations.

Earth Frame

In the Earth's frame both legs of the Journey take two years, making the diagram

The wait for Hans is \begin{align*} \tau_\text{Hans} = -\frac{\sqrt{\Delta s^2_\text{Hans}}}{c} &= \frac{\sqrt{c^2(4\,\text{years})^2 - (0\,\text{light years})^2}}{c}\\ &= \frac{\sqrt{(4\,\text{light years})^2}}{c} \\ &= 4\,\text{years}\;, \end{align*} meaning that Hans has waited 4 years. Note that $c$ is simply 1 light year per year.

For Heidi the situation is slightly more complicated, she embarks on two inertial journeys and it is easy to measure the proper time elapsed on each one, and then add them together \begin{align*} \tau_\text{Heidi} &= \tau_\text{out-bound} + \tau_\text{in-bound} \\ &= \frac{\sqrt{c^2(2\,\text{years})^2 - (+1\,\text{light years})^2}}{c} + \frac{\sqrt{c^2(2\,\text{years})^2 - (-1\,\text{light years})^2}}{c}\\ &= \frac{2\sqrt{3\,(\text{light years})^2}}{c}\\ &= 2\sqrt{3}\,\text{years} \end{align*}

At their reunion Heidi is six-and-a-half months younger than Hans.

Out-bound Frame

That's great but one of the disadvantages of using a Minkowski diagram (as opposed to a Loedel diagram) is that it seems to give a special place to the frame with the upright axes.

So let's choose a different frame of reference and re-do all the work to see if we get the same answer.

In this case I'm going to use the Frame of reference in which Heidi's out-bound leg is at rest. This means that the Earth moves backward at $0.5c$ in this frame.

To draw this figure we wouldd need to find the coordinates of Heidi's arrival at the target object and return to Earth in this frame. This can be done by direct application of the Lorentz Transform to the known coordinates of those points in the Earth-linked frame, or by knowing that a boost of $beta = v/c$ swings a line on a diagram through an angle of $\alpha = \tan \beta$ and causes the line to scale by a factor of $$ s = \frac{\sqrt{1 + \beta^2}}{\sqrt{1 - \beta^2}} \; $$ which you might recognize as the Doppler shift factor for light.

Either way, the time of arrival at the target object is $t_a = 1.73\,\text{years}$, the time of the return to Earth is $t_r = 4.62\,\text{years}$, and the location of the return to Earth is $-2.31\,\text{light years}$.

This time we get \begin{align*} \tau_\text{Hans} &= \frac{\sqrt{\Delta s^2}}{c} \\ &= \frac{\sqrt{c^2(4.62\,\text{years})^2 - (-2.31\,\text{light years})^2}}{c} \\ &= 4\,\text{years} \;. \end{align*}

Likewise, for Heidi we get \begin{align*} \tau_\text{Heidi} &= \tau_\text{out-bound} + \tau_\text{in-bound} \\ &= \frac{\sqrt{c^2(1.73\,\text{years})^2 - (0\,\text{light years})^2}}{c} + \frac{\sqrt{c^2((4.62-1.73)\,\text{years})^2 - (-2.31\,\text{light years})^2}}{c}\\ &= 3.466\,\text{years} \approx 2\sqrt{3}\,\text{years} \;, \end{align*} where the very small discrepancy at the end is just rounding error due to truncating the figures as we went along. (Did you notice that $1.73 \approx \sqrt{3}$? That's no accident, the interval of each leg of Heidi's journey has to be the same in every frame of reference.)

In short, the same result.

In-bound Frame or Some Frame Not Linked to Either twin.

Left as an exercise. It's worth your time to go through all the work over again and see that you continue to get the same results in other frames.

Resolving the Paradox

The paradox is fully resolved by accepting that the proper time $\tau$ is (to within a sign and a factor of $c$) the square-root of the interval. Once you accept this (both the statement and the scheme by which interval is computed) then all else is just drawing paths and adding up proper time.

Why is accepting the computational scheme important? In ordinary geometry a straight line is the shortest distance between two points. In Minkowski geometry the sign difference between $(\Delta t)^2$ and $(\Delta x)^2$ means that a straight line is the longest proper time between two events.

I should mention the role of acceleration because it is often treated as a kind of fairy dust that resolves the paradox.

What acceleration means is that the slope of a world-line is changing: that is, this world-line is not straight. And since it is not straight it is not the longest proper-time between two event. So, acceleration has that effect not because there is something magical about undergoing acceleration, but because it causes world-line to deviate. Rigging schemes in which a message is passed so that no 'thing' undergoes acceleration doesn't change the fact that your message takes a non-straight world-line between events.

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The images herein are my original work and were first prepared (in LaTeX, using TikZ) for a short note on Space-Time interval used in my Modern Physics class.

Answer 5

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:

^{public 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.

Answer 6

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.

Answer 7

We have a well defined notion of time in physics simply because the relative rates of physical processes are experimentally found to be the same always under the same conditions. Therefore, you choose one periodic physical process whose rate is influenced by factors that can be readily and repeatably controlled, and you use that as a clock. That is, you measure elapsed time by counting periods of this standard process, and compare all other physical processes to this one. See my answer here for more details.

One of the factors that is experimentally found to influence the relative rates of physical processes is the relative speed between the inertial frames that compared physical processes happen in.That is, the time dilation factor and the Lorentz transformation allow us to calculate the relative rate for two processes in different inertial frames if we know their relative rate when they happen in the same frame.

That's all time dilation is: a change in the relative rates of physical processes that is observed to arise from relative motion between different physical processes. Once you lose the cultural baggage about "Time", this difference is unsurprising: if you change a factor in an experiment, change in the experimental outcome is completely the norm, or at least extremely common.

Answer 8

Well, unfortunately I cannot comment @Rennie, but I'd like to add to the otherwise fine answer of John Rennie that it isn't right to say that:

the quantity d$s$ is called the proper distance

Because that is simply not true!

Proper distance, $σ$, in SR is analogous to proper time $\tau$. It is the invariant space-like interval between two events A and B instead of a time-like interval.

The difference is that proper distance is defined between two spatially separated events (or along a spatial path), while proper time is defined between two time-like separated events (or along a time-like path).

With d$s$ and d$s^2$ the line element and nothing else is meant in relativity. And with $Δs$ or $Δs^2$ instead of d$s$ or d$s^2$, it concerns a space-time interval.

And:

this minus sign is what explains the time dilation

Is a bit misleading imo. It doesn't explain time dilation. Not at all, physically.

The minus sign has to do with, for example, a hyperbolic geometry (such as that of Minkowski space), otherwise the mathematics will not be correct, but from a physical point of view it explains nothing at all.

Again it's just about (simple) kinematic time dilation for people who try to first understand or learn a bit about relativity. But then I think, specially for them, it, none of it should lead to possible stubborn misunderstandings.

So it's no criticism, but .. "in addition", trying to be helpful.

Answer 9

Time dilation is not a slowing of time. In SR time passes at the same rate at every point in every inertial referent frame.

However, the planes of constant time in any one reference frame are tilted relative to those of any other reference frame moving relative to the first (the degree of tilt increasing with the speed at which the two frames are moving relative to one another). This means that a horizontal time slice through one reference frame, ie a plane on which it is the same time everywhere in that frame, corresponds to a sloping plane in the other frame, upon which time is increasingly rising to the future in the direction of motion and increasingly falling back in the reverse direction.

This means that as you are moving relative to some frame you are continually passing up the slope, as it were, to points in that frame where time is progressively more advanced. Although your time passes at a normal rate, you are moving to points ahead in time in the other frame, so the time interval you move through in that frame is greater than the time you have experienced in your own.

To take a concrete example, suppose you move at some speed, starting at 12:00 both in your frame and in the frame through which you are moving. After five minutes on your watch you have risen up the sloping plane of time in the other frame to a point which is ahead by one minute compared with the point at which you started, so your watch will show 12:05 and the time in the stationary frame where you now are is 12:06. After another five minutes on your watch you have moved up the sloping plane of time in the other frame to a point that is advanced by yet another minute, so you now see 12:10 on your watch and 12:12 on an adjacent clock in the stationary frame. And so on. Time passes on your watch at the normal rate, but you are moving in the other frame from regions of earlier time to regions of later time, so the clocks you pass are progressively showing a later time than your watch does.

The effect is entirely symmetrical. From the perspective of a person in the frame through which you are moving, their plane of time is level and it is yours that is sloping. If that person had been with you at 12:00 when you started to move, they will see, after five minutes on their watch, that an adjacent passing clock in your frame will read 12:06, and five minutes later, at 12:10 on their watch, they will see another adjacent clock in your frame reads 12:12, so it is their watch which is progressively falling behind the time in your frame.

So, time dilation is not a result of time slowing down- time for both of the people in the scenario I just described continues at the same rate- it is a result of the fact that planes of constant time are tilted, so as one moves through a frame one progressively moves to regions of later time in that frame so that the gain in time in that frame seems greater than the time that has elapsed where you are in your own frame.

The common phrase 'moving clocks run slow' is a misleading one if taken out of context, and a frequent source of confusion.

Answer 10

Time doesn't itself slow down or seed up. We slow down or speed up in time by accelerating (or decelerating) in space.

If we ever reach the speed of light in space, our speed in time will reach zero, and time dilation will be infinite.

If we consider ourselves to be stationary in space, our speed in space will then be considered as zero, thus our speed in time will be equal to the speed of light. Therefore we can consider that we age at maximum speed.

Answer 11

I read through the answers here, and I think I could provide another perspective.

Let us go back to the end of 19th century, when out of all fundamental forces we knew about were only electromagnetism and gravity. Let us forget about gravity and just look at Maxwell's equations, and imagine that the whole world is governed by them. All structures that one sees in such world are just some manifestations of solutions to the Maxwell's equations.

Let us consider a clock C in such world. The clock is some kind of a particular localized structure, probably based on an oscillator, satisfying Maxwell's equations. We will assume that we know how the clock looks when it does not move in space, and we will try to "derive" how a moving clock looks. The game is, having a non-moving (on average) configuration of Maxwell's equations, construct a corresponding moving configuration. In general case, for an arbitrary partial differential equation, it is a tricky job. For instance, you cannot construct a "moving" solution to a diffusion equation. You can construct a moving solution to a Navier-Stokes equations (fluid dynamics) though, by a change of coordinates $$ x' =x-vt, $$ but such change does not work for Maxwell's equations!

Lorentz and Larmore discovered a crucial fact (see A Note on Relativity Before Einstein) that the form of Maxwell's equations remains invariant under Lorentz transformations : $$ \begin{array}[c]\\ x' =(x-vt)\gamma,\\ t' =(t-vx)\gamma, \end{array}\tag{1} $$ (here speed of light here assumed to be equal 1)

Why the invariance important? Because it allows us to make a moving clock from a stationary clock. Imagine, you made a stationary clock and you apply Lorentz transformation (1) to all of its parts. What are you going to get? You are going to get a moving clock that is squeezed along $x$ and that ticks slower. Let is sink: (1) Maxwell's equations are still valid in the new coordinates; (2) we have a clock that works by the same principles as the original ones but it (a) moves, (b) squeezed along $x$ (c) ticks slower. And that's exactly what we wanted! If you know how a stationary clock looks like - you know how a corresponding moving clock is going to look! And this is, of course, true not only for clocks, but for any constructions. Having a stationary configuration of the field, We can make a moving configuration by applying Lorentz transformation.

Now, real clocks are not made of solutions to Maxwell's equations, but involve lots of other stuff. The crucial idea of Einstein was that Lorentz transformations not only leave Maxwell's equations invariant, but they leave all laws of the universe invariant, which means that you can make a moving clock from any kind of stationary clock by applying transformation (1).

To summarize, clocks slow down because the laws of the universe are invariant under Lorentz transformations, which tells us how to make a moving clock out of a stationary one.

Answer 12

Before you can understand what time dilation is, you need to understand what time is. The word time is a term describing temporal motion. It's the motion of physical things, through the temporal dimension. In layman's terms, we move through the spatial dimension and we time through the temporal dimension.

Because these two dimensions are interlinked, our temporal velocity is inversely proportional to our spatial velocity, and this gives rise to time dilation. Time dilation is not time speeding up or slowing down. Time dilation is when someone or something times slower or faster.

Unfortunately, because our temporal velocity determines the rate of atomic, biological and mechanical processes, we never perceive any change locally. Our rate of perception slows down exactly the same amount as our clocks. To us, everything appears normal. It's only when we compare our clock with someone who has a different temporal velocity, that we see a difference.

Gravity also affects the rate at which we time, but I'll get into that another time.