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When ever i look this up all I get is sites saying how its because general relativity says "-" why does it do it though? it is because there is more motion near gravity than further away? Or is it something completely different?

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    $\begingroup$ I think Ray is asking about time dilation. GR doesn't say why. It just says what happens. $\endgroup$ Sep 5 '14 at 15:58
  • $\begingroup$ What kind of answer do you expect? Do you want to see the derivation for the time dilation (then look, for example, at the Schwarzschild metric)? Other than that, Why? is rarely a useful question - the (experimentally tested) theory predicts it, and so it (most probably) happens. $\endgroup$
    – ACuriousMind
    Sep 5 '14 at 15:58
  • $\begingroup$ Why does it predict it? whats the intuition behind this? Is my guess: "it is because there is more motion near gravity than further away?" right or wrong? $\endgroup$
    – Ray Kay
    Sep 5 '14 at 16:04
  • $\begingroup$ @RayKay Your guess is wrong. You can have more motion or less motion while near a mass, and that effect will be in addition to the gravitational effect. $\endgroup$
    – Timaeus
    Sep 5 '14 at 17:25
  • $\begingroup$ I mean motion due to gravity, now is my guess correct $\endgroup$
    – Ray Kay
    Sep 6 '14 at 0:47
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It's a fundamental principle of both special and general relativity that the line element, $ds$, given by the metric:

$$ ds^2 = g_{\mu\nu} x^\mu x^\nu \tag{1} $$

is an invariant. That is, all observers in any coordinate systems will calculate the same value for $ds$. It's this fundamental symmetry that is responsible for time dilation, along with all the other freaky effects in relativity. If you're asking why this symmetry exists then I can't answer. It just does - that's the way the universe is built. If you want to know why this symmetry causes time dilation then read on.

The metric, i.e. the form of equation(1), for a spherical mass is:

$$ ds^2 = -\left(1-\frac{2GM}{c^2r}\right)dt^2 + \left(1-\frac{2GM}{c^2r}\right)^{-1}dr^2 + r^2 d\Omega^2 $$

where $t$ and $r$ are time and distance from the centre of mass as measured by an observer at infinity (for our purposes infinity just means well outside the event horizon). Suppose you're in a spaceship hovering at some constant distance $r$ from the black hole then $dr$ and $d\Omega$ are zero so the equation simplifies to:

$$ ds^2 = -\left(1-\frac{2GM}{c^2r}\right)dt^2 \tag{2} $$

Remember, the time $t$ is measured by a clock far from the black hole.

Now consider what you measure inside your spaceship. Your local metric just looks like flat spacetime, so your line element is given by the Minkowski metric:

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

and since you're stationary $dx = dy = dz = 0$ so the above equation simplifies to:

$$ ds^2 = -d\tau^2 \tag{3} $$

I've used $\tau$ to distinguish the time you measure on your clock from the time $t$ the distant observer measures.

Anyhow, both you and the distant observer must get the same value for $ds$, because that's a fundamental symmetry. So we can just set equations (2) and (3) equal to get:

$$ d\tau^2 = \left(1-\frac{2GM}{c^2r}\right)dt^2 $$

or with a quick rearrangement:

$$ \frac{d\tau}{dt} = \sqrt{1 - \frac{2GM}{c^2r}} \tag{4} $$

And equation (4) is the equation for the time dilation because it relates the time measured on your clock, $d\tau$, to the time measured on my clock, $t$. The quantity $\sqrt{1 - 2GM/c^2r}$ is always less than one, so your time is running slowly compared to mine i.e. your time is dilated.

I've used the specific case of a black hole to do the calculation, but the principle of using the metric to calculate the time dilation is a general one. You can use exactly the same technique to calculate the time dilation due to relative velocity in special relativity, in which case you'd get the well known result:

$$ \frac{d\tau}{dt} = \sqrt{1 - \frac{v^2}{c^2}} $$

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  • $\begingroup$ Good intro paragraph. +1 $\endgroup$
    – BMS
    Sep 5 '14 at 16:43
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There is a simple thought experiment that you can use to show this. Consider a rocket accelerating in space, and consider a clock at the top of the rocket, and a clock at the bottom of the rocket. If we do this, we'll note that, if the rocket is going away from us, then the clock at the front of the rocket will have sent light to us at a time when the rocket was going slower than the light that reaches us at the same time from the bottom of the rocket. Thus, the bottom of the rocket will appear to be redshifted relative to the top of the rocket. Without going through the details, this effect will transfer to the observers inside of the rocket${}^{1}$.

Now, note that our rocket can be made to be windowless. The principle of general relativity states that if our rocket is small compared to the curvature of spacetime, and if we measure our experiments over time intervals also small compared to the curvature of spacetime, then we can't tell a difference between accelerating in space at 9.8 m/s^{2} or sitting stationary in Earth's gravitational field. Since we got a redshift between the bottom and the top of the spaceship while it was accelerating, this tells us that we're going to need to get a redshift when we're not accelerating. Then, to get the "closer to a mass" bit, just note that the acceleration effect is bigger with a bigger acceleration, and the gravitational field is bigger the closer you are to a mass.

There are more details when you actually construct models, certainly, as you can see in John Rennie's answer. But the bare physics to the situation are here.

${}^{1}$"but wait!" you say! This means that the people inside the rocket can tell that they're accelerating. "Sure," I respond. "That's true, but remember that special relativity is totally cool with experiments that can be done to prove that you're accelerating. It's absolute velocity that it does away with".

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This effect was originally predicted in special relativity, time slows for an object undergoing acceleration compared to the observer, but Einstein's big leap to general relativity was realising that gravity is an acceleration - standing on the surface of the Earth or sitting in a rocket accelerating at 9.8m/s/s are (as far as the time dilation go) the same thing.

To cut through any maths beyond primary school, think of the "speed = distance/time" equation. Add to that the observed fact that the speed of light (in a vacuum) is ALWAYS measured as the same thing, ~3x10^8m/s

So...to reconcile a constant speed of light for different observers we have to change either the distance an/or the time in "speed = distance/time" because the speed (in this case, of light) doesn't change from one observer to another. In fact, both of them warp in the same way, hence we talk about spacetime as a single entity.

WHY the universe works in this way, as pointed out above, is still a mystery. It's just a mathematical model that happens to fit reality.

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