There are two kinds of time dilation:

  • One because the other clock moves fast relative to me (special relativity).

  • Another one because the other clock is in a stronger gravitational field (general relativity), or accelerating rapidly (equivalence principle).

So are these two effects are totally independent? Is it possible to derive the general relativity case from the special relativity case?

  • $\begingroup$ Would it be of interest to you to learn that both forms of time dilation are caused by the same phenomenon? $\endgroup$ Commented Jul 11, 2022 at 22:05

5 Answers 5


Let me try and expand a bit on Ben's answer.

Starting with special relativity, the key thing to understand is that all the weird stuff, and indeed the Lorentz transformations, is derived from a property called the metric. If you have two points in spacetime separated by ($\mathrm dt,~\mathrm dx,~\mathrm dy,~\mathrm dz$) then the metric tells us how to calculate the interval between them. For SR this is:

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

The interval $\mathrm ds$ is referred to as the line element and is an invariant, i.e., every observer no matter how fast they are moving, will calculate the same value for $\mathrm ds$.

The equation for the line element should remind you of Pythagoras' theorem, and indeed the only difference is that the sign of $\mathrm dt^2$ is negative not positive. It's this difference in the sign that is responsible for effects like time dilation. This is the important point to take home: this metric is all you need to calculate time dilation.

Now consider general relativity, and the effect of gravity. But first let me rewrite the special relativity equation for the line element in polar co-ordinates:

$$\mathrm ds^2 = -\mathrm dt^2 +\mathrm dr^2 + r^2 (\mathrm d\theta^2 + \sin^2\theta~\mathrm d\phi^2) $$

and now I'll write the equation for the line element near a black hole, i.e. the Schwarzschild metric:

$$ \mathrm ds^2 = -\left(1-\frac{2M}{r}\right)\mathrm dt^2 + \frac{\mathrm dr^2}{\left(1-\frac{2M}{r}\right)} + r^2 (\mathrm d\theta^2 + \sin^2\theta~\mathrm d\phi^2) $$

If you compare these two equations it should be immediately obvious that they are very similar, and indeed if you let the mass of the black hole, $M$, go to zero or if you go a long way away, so $r \rightarrow \infty$, then the two equations are the same.

This means the GR metric includes everything that the SR metric predicts, but it adds to it. So there isn't a distinction between the time dilation due to just velocity and the time dilation due to gravity. The GR metric is an extension of the SR metric and includes both. However let me reinforce Ben's cautions: it generally isn't useful to try and separate the time dilation due to velocity and the time dilation due to gravity.


It's not true that gravitational time dilation is based on the strength of the gravitational field. By the equivalence principle, the gravitational field equals zero for any inertial observer, and can have any other value you like for some other appropriately chosen observer. Gravitational time dilation is based on the gravitational potential.

Neither kinematic nor gravitational time dilation requires general relativity. You can have a gravitational field and a gravitational potential in flat spacetime, e.g., for an observer inside an accelerating elevator. There are straightforward special-relativistic arguments that derive the gravitational time dilation from thought experiments involving accelerating elevators.

Neither kinematic nor gravitational time dilation is fundamental. What's fundamental is the metric. Either effect can be calculated from the metric.

Gravitational time dilation can't be fundamental because the gravitational potential isn't even well defined unless the spacetime is static. For example, cosmological spacetimes aren't static.

A good popular-level book that explains the fundamental status of the metric very clearly is Geroch, General Relativity from A to B.

  • $\begingroup$ "By the equivalence principle, the gravitational field equals zero for any inertial observer" Locally $\endgroup$ Commented May 9, 2023 at 2:02

Yes, you can solve with a very simple formula the equivalent speed you would need to go to have the same time dilation as caused by a gravitational field.

All you need is how many Schwarzschild radii you are from the mass (this essentially gives you how deep you are in a gravitational field). Once you have that get the square root of how ever many Schwarzschild radii and you get a number that is equivalent to how much faster light is going than object moving for time dilation.

For instance:

If an object is 9 Schwarzschild radii away from Mass it will experience the same time dilation as an object going 3 times less the speed of light.

Same for 4 Schwarzschild radii and the speed of light going 2 times faster.

it boils down to $$x=y^2$$

To get to this formula, first you start with the gravitational time dilation formula where:

$$ T_1=T\sqrt{1-\frac{2GM}{rc^2}} $$

and rather than entering $r$ for the radius we replace $r$ with the Schwarzschild radius formula $(2GM/c^2)x$ with an $x$ at the end representing how many Schwarzschild radii you are away from the center. This brings the formula to look like:

$$ T_1=T\sqrt{1-\frac{2GM}{\frac{2GM}{c^2}xc^2)}} $$

Which when simplified breaks down to:

$$ T_1=T\sqrt{1-\frac{1}{x}} $$

and if you make $T=1$ then you just get

$$ T_1=\sqrt{1-\frac{1}{x}} $$

This is very similar to the one in many physics books $=\sqrt{1-r_0/r}$, where $r_0$ is equal to the Schwarzschild radius and then $r$ equals the radius from the center. The formula above it just makes it slightly simpler due to making $r_0$ equal to 1 and $x$ equal to how many radii a point you are observing is from the center of the mass.

That is the gravitational time dilation side portion of this relationship. Now for the velocity time dilation side we use a similar methodology and start with:

$$ T_0=T\sqrt{1-\frac{v^2}{c^2}} $$

Now we make $T$ equal to 1, $v$ equal to one, and $c$ to $y$ because now we are going to make $c$ a variable.

$$ T_0=\sqrt{1-\frac{1}{y^2}} $$

What you see now "$1/y^2$" is showing the velocity as a constant 1 and $y$ represents how much faster light is going than the velocity constant of 1. If the above were to show the fraction as $1/5^2$ then this would be the same as saying an object is going at a velocity 1/5th the velocity of light. So now if we solve the velocity and gravitational time dilation formulas so that we can see how they dilate time to come up with the same result:

$$ \sqrt{1-\frac{1}{x}}=\sqrt{1-\frac{1}{y^2}} $$

We can simplify this to

$$ x=y^2 $$

  • 1
    $\begingroup$ More details about how you arrive at this result would be nice. $\endgroup$ Commented Jan 4, 2015 at 22:34
  • $\begingroup$ I just edited my answer to include how you get to the simple x=y^2 $\endgroup$
    – Joe
    Commented Jan 4, 2015 at 23:21

From an engineer: I interpreted the answer from the practical point of view of how to combine effects from velocity-related time dilation and gravity-related time dilation. They can be treated independently, and then either multiple the factors together or, since the factors are usually very close to 1, you can add (1-factors). For example, for GPS satellites, the velocity is high so the clock runs a little slower via the equation $t/t_0 = 1/\sqrt{1-v^2\over c^2}$ (about 7 microseconds per day). But the gravity is lower so the clock runs about 45 microsecs/day faster: You need to solve $t/t_0=1/\sqrt{1-2GM\over rc^2}$ for the surface and the orbit and subtract the difference. The net result is we build clocks that would run 45 - 7 = 38 microseconds a day slow here on earth, but they work perfectly in the GPS orbit.

As to how the two factors are related, I think of it conceptually as the gravity on a planet's surface is the energy equivalent of the kinetic energy to escape earth. Kinetic energy is $1/2 mv^2$ and gravitational potential energy of an object far from Earth is $GMm\over r$. Set them equal and you get the escape velocity, or the value of $v^2 = {2GM\over r}$. That is, instead of the kinetic energy causing the object's internal clock to slow down, the gravitational potential energy is causing the clock to slow down. When the energies are equal, the impact on the internal clock is the same.


In the calculation of the relativistic precession of Mercury there is a factor of 3. This includes 1 part kinematic time dilation, and 2 parts gravitational time dilation. They are not mutually exclusive.

It's quite easy to get the necessary precession for Mercury if you take gravitational and kinematic time dilation / length contraction into account.

The Newtonian acceleration vector is: \begin{equation} \vec{g}_n = \frac{\hat{d} G M}{{\lvert\lvert \vec{d} \rvert\rvert}^2}. \end{equation}

One important value is closely related to the kinematic time dilation: \begin{equation} \label{eq_kinematic} \alpha = 2 - \sqrt{1 - \frac{\lvert\lvert \vec{v}_{o}\rvert\rvert^2}{c^2}}. \end{equation} Another important value is the gravitational time dilation: \begin{equation} \beta = \sqrt{1 - \frac{R_{s}}{\lvert \lvert \vec{d} \rvert \rvert}}. \end{equation}

Finally, the semi-implicit Euler integration is: \begin{align} \vec{v}_{o}(t + \delta_t) &= \vec{v}_{o}(t) + \delta_{t} \alpha \vec{g}_n, \\ \ell_{o}(t + \delta_t) &= \ell_{o}(t) + \delta_{t} \beta \vec{v}_{o}(t + \delta_t). \end{align}

I won't pass on my code quite yet, but here's a beginner's C++ source code that calculates strictly Newtonian orbit. Adding in the required code isn't all that tough, but I just don't know if my code has bugs, so I need independent validation:


Let me know if you have any questions!


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