Time Dilation = $1/\sqrt{ 1-v^2/c^2}$ but why? How do you get to that conclusion? I know you use Pythagoras'theorem and my current understand goes like this:


then you take the square root of $Ct^2$ and from there I think you divide by $c$ to get the time dilation but I'm not sure. And if this is correct what steps do you take to get to the formula used now

  • $\begingroup$ A quick search on Google returns plenty of results of the derivation, for example: archive.org/details/… $\endgroup$
    – turnip
    Commented Aug 17, 2014 at 10:39
  • $\begingroup$ @PPG I edited the question - perhaps it is more workable now. $\endgroup$
    – Danu
    Commented Aug 17, 2014 at 10:41

4 Answers 4


To me personally, the most intuitive way of understanding SRT is to always keep in mind that, in SRT, the interval $$ds^2=-c^2dt^2+dx^2=-c^2d\tau^2$$ has to be invariant. From this simple formula, everything seems to flow naturally. In particular, it is easy to see the form of the Lorentz factor arise from here, by using $\frac{dx}{dt}=v\to dx=vdt$. Using this substitution gives us

$$c^2d\tau^2=dt^2(c^2-v^2)\to d\tau^2=dt^2\left(1-\frac{v^2}{c^2}\right)\to d\tau=(\pm)dt\sqrt{1-\frac{v^2}{c^2}}=(\pm)\frac{dt}{\gamma} $$ this is the time dilation formula. The length contraction may be similarly derived. If you like this, you may be interested in this old question of mine (although you must take care not to get confused about the notational differences).


This is how you derive the equation for time dilation.

The metric used in special relativity is the Minkowski metric:

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

and the basic principle of special relativity is the the line element $ds$ is an invariant, that is all observers in all inertial frames will measure it to have the same value.

Suppose we are using the coordinates $(t, x, y, z)$ and we observe an object moving at velocity $v$ in the $x$ direction (so $dy = dz = 0)$, then:

$$ ds^2 = -c^2dt^2 + dx^2 \tag{1} $$

But we expect the position of the object in our coordinates, $x$, to be given by:

$$ x = vt + x_0 $$

and therefore:

$$ dx = vdt $$

and if we substitute this in equation (1) we get:

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

Now shift to the frame of the moving object $(t', x', y', z')$. In these coordinates the objetc is stationary so $dx' = dy' = dz' = 0$ so:

$$ ds'^2 = -c^2dt'^2 \tag{3} $$

We started out by saying that all observers will agree on the value of the line element and that means $ds = ds'$, so equating equations (2) and (3) we get:

$$ c^2dt'^2 = c^2dt^2 - v^2dt^2 $$

And dividing both sides by $c^2$ and taking the square root:

$$\begin{align} dt' &= dt \sqrt{1 - \frac{v^2}{c^2}} \\ &= \frac{dt}{\gamma} \tag{4} \end{align}$$

And this is the basis of the time dilation. If we want to find the time $t'$ corresponding to a time $t$ then we simply integrate equation (4), and because $\gamma$ is a constant this integrates to:

$$ t' = \int_0^t \frac{dt}{\gamma} = \frac{t}{\gamma} $$

which is the equation that we all know and love.

This may seem like a long winded way of deriving the result, but note that this method is applicable to situations where the velocity is not constant. In that case the relationship between $dx$ and $dt$ is not linear, and the integral will be harder, however the working is exactly the same.

  • $\begingroup$ Hah, I just wrote up my answer based on your old one - only to get another one of yours :D $\endgroup$
    – Danu
    Commented Aug 17, 2014 at 11:03
  • $\begingroup$ @Danu: damn your eyes Danu! :-) I saw a flag come up to say there was an answer while I was typing this, but I didn't bother refreshing the page to see what the answer was. Oh well, I'll leave this here for a bit as I've been more explicit about the working. Site members: please upvote Danu's answer (or both :-) as he did get in first. $\endgroup$ Commented Aug 17, 2014 at 11:05
  • $\begingroup$ On my list of things to do is write up a blog style Q/A on the twin paradox starting with this approach to time dilation and showing how acceleration affects the calculation. $\endgroup$ Commented Aug 17, 2014 at 11:06
  • $\begingroup$ Yes, yes, yes! Fits nicely within the context of my meta proposal - even if you want to do it independently. $\endgroup$
    – Danu
    Commented Aug 17, 2014 at 11:17
  • $\begingroup$ Why is it -c^2dt^2? why not just c^2dt^2? I'm 17 and don't even do physics at school so forgive me for higher-than-expected ignorance. Also how do I make my equations the way everyone else does it? When I write for example c^2 I have the '^' there while everyone else has a little 2 in the top right $\endgroup$
    – Ray Kay
    Commented Aug 18, 2014 at 0:50

Rather than rehash the equations which have been referenced already, I wanted to first cover the Michelson-Morley experiment. Michelson, Morley and even Lorentz were actually able to do a considerable amount of work on the prediction of the expected existence of the aether wind. The foundations of the underlying equations where strong by this point.

The shock of the discovery there was no aether wind was truly enormous. However, the machinery to explain the situation, the Fitzgerald-Lorentz length contraction hypothesis, quickly explained what was happening, however, without a theory like relativity there really was no foundation for how to understand why this was the case.

From a pure mathematical standpoint, the trick is to shift from circular angles to hyperbolic angles with the understanding that Lorentz contraction is understood in terms of hyperbolic functions.

Hyperbolic relationships are common in physics and occur in several places, from relativity to Heisenberg uncertainty.

For the math I would refer to the links and previous posts, but this should hopefully help your intuition on this a little.

  • 2
    $\begingroup$ nice introduction to the conceptual history of SR, i would like to add that the turning point was the understanding and formulation of E/M equations (based on newer technology) which posited the constancy of light propagation. This went into conflict with Galilean/Newtonians notions of velocity and aether was posited as the medium through which light wave is an oscilation of. Finally the MM experiment showed there is no movement of this aether. Then unification of Mechanics and EM had to be achieved somehow, and SR was one way achieving this unification at the space-time level. $\endgroup$
    – Nikos M.
    Commented Aug 17, 2014 at 15:27

i would say the intuition is the simple observation (by Einstein, Lorentz, Poincare and others) of these 2 things:

  1. Velocity of light ($c$) is $c$-onstant accross interial frames (extrapolated result from Maxwell-Lorentz equations)

  2. The velocity of light ($c$) is an upper limit on every other velocity a material body or signal can achieve (in effect $c$ assumes a role similar to infinity in mathematics)

These two derive the relativistic velocity addition formula and the Lorentz transformations which replace the Galilean transformations.

Then one, based on the Lorentz transformations, can define a Minkowski spacetime which geometrically is characterised by having the infinitesimal interval:

$$ds^2 = -(cdt)^2 + dx^2 + dy^2 + dz^2$$

an invariant of the geometry.


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