How does gravity slow a clock at a distance?

If you have two identical clocks that are synchronized and then you move one significantly closer to a black hole but still far from the black hole, the closer clock will mark time slower than the further clock. My question is, by what mechanism does the black hole slow the motion of the closer clock? Does gravity from the black hole transmit particles or waves many miles out which slow the motion of the clock? Is gravity some kind of invisible goo in another dimension that slows motion?

• The time that passes on a clock marks the distance through spacetime that the clock has traveled. Gravity is spacetime curvature. Closer to a black hole, the spacetime curvature is different so that your two clocks are traveling along paths with different lengths. – WillO May 3 '17 at 23:39
• Don, even in the context of special relativity (no gravity), two uniformly accelerated clocks that maintain constant separation (as I assume is the case after one clock moves to a closer position), do not run at the same rate. The reason in this case is that their worldlines are not congruent and so accumulate different proper time. Then, by the equivalence principle, so must be the case in your example. – Hal Hollis May 4 '17 at 1:31
• @JMac, in the context of GR, gravity is not a force, gravity is spacetime curvature. – Alfred Centauri May 4 '17 at 2:41
• You wrote: "You are only describing what the clocks do....not why. " If I drive from Pittsburgh to Chicago in a straight line, while you drive from Pittsburgh to Chicago via Los Angeles, our odometers will show different trip lengths. Why? Because an odometer, by definition, is an instrument that shows how far it has traveled along the ground, and we have traveled different distances. Likewise with the clocks. A clock is, by definition an instrument that records how far it has traveled in spacetime. One clock travels farther than the other, so it shows a different time lapse. – WillO May 4 '17 at 2:52
• @Don Tipon yes, the mechanism is well understood and has been validated experimentally for quite some time now. The mechanism is described by the Einstein Field Equations. – Dale Oct 8 '18 at 1:40

" Is gravity some kind of invisible goo in another dimension that slows motion?"

Gravity is some kind of invisible goo that slows time

• Well played sir – Punk_Physicist May 4 '17 at 1:40

Don, there is no force in gravity. Force is a Newtonian concept. General Relativity says that gravity is the curvature of spacetime.

It has been said: matter (or any form of energy) determines how spacetime curves. Spacetime (curvature) determines how matter (or light or any radiation) moves.

And you are right, curving spacetime does not happen instantaneously or at a distance. Similarly, matter and radiation move according to what the spacetime curvature is locally, where they are.

What about clocks and gravity from the earth, which seems to just be there. First, clocks. Also, I'll deal with the earth's gravity, it happens also, no need for a Black Hole. For Black Holes it is a huge effect, but the important thing is why it happens.

Clocks move like any other matter or radiation. The timing on how fast they click their seconds away, also depends on motion through spacetime. After all, time is involved. The geometry of spacetime includes the fact that proper time at one place (say the clock's timing if you're riding with it), and that at another place, will differ, because they are in different spacetime points, and times passes at different rates. Compared to an observer (or clock) far away from any gravitational sources, they will typically be slower. Motion in spacetime includes the rate at which time passes, relative to another place. The same is similarly true for spatial distance, it depends on the metRic components for the spatial coordinates. Different metric components in say x and y means that your normal motion will be curved in x and y.

What of gravity from the earth, why is it always affecting your acceleration. Gravity is equivalent to acceleration, in simple terms, from the principle of equivalence (observed, a fact), and so Newton said it was a force. But it is curvature. Earth creates the curvature of the spacetime around it (with a little contribution from the moon, Sun, etc, if we are close to the earth). How long does it take to get out there, say 300 km up. Well, when the earth was first formed, say it happened all at once, gravity's curvature propoagate to that point 300 km up (and every other point or place within 300 km, and then beyond) at the speed of light, 300 000 km/s. So it takes to go 300 km 0.001 s, or 1 ms. If something changes on earth, it propagates the equivalent gravitational perturbation at that same speed. So what you feel in space 300 km high is the curvature created by the earth 1 ms before.

It is not a gooey thing that propagates, it is curvature. If the earth vibrated in certain ways it would create a vibrating curvature out there, from gravitational waves propagating from the earth. Because the changes on the earth are so small compared to the total mass of the earth, that curvature is basically static out there, and you just calculate it once (assuming no strong gravitational wave just a msec ago, a very good assumption). We just call it a star if curvature, or static spacetime. It is mathematically represented by the Schwarzschild metric. From it we know how much a clock in the GPS satellites is faster than clocks on the earth, and have to adjust for it by actually having a slower ticking clock in the GPS satellites so that the faster effect of gravity will make it just like on earth. It's been proven and measured with the GPS satellites.

So, I know your question was a bit of jest, but this is the real answer. In physics stack exchange we can take a joke, but generally people want to understand things.

And hope this more serious answer wasn't too boring.

• Thanks for your comments, Bob. I am not a physicist and I am trying to understand the curvature of space-time. I absolutely believe that space-time is curved by a large mass but I do not know if anyone understands how a large mass does this. One possible explanation is that large masses radiate gravitons which would travel out to a clock and the gravitons would stick to the sub-atomic particles like glue and slow their movement. I just do not know if this is a proposed physics theory. Do gravitons do such a thing? – Don Tipon May 4 '17 at 5:20

The important thing to understand before you start talking about black holes is that these things are universal to special relativity. General relativity didn't invent them, it just understood gravity as a special case of the same rules.

What led up to relativity

Define $w=ct$ and $\beta = v/c$ as a "spatial" time coordinate and a unitless speed. The classical-mechanics rules for switching to new coordinates $(w', x', y', z')$ moving in the $x$ direction with speed $\beta$ relative to an old one $(w,x,y,z)$ are:\begin{align}w'&=w,\\ x'&=x-\beta~w,\\ y'&=y,\\ z'&=z.\end{align} We have a not-really-shorthand way to write these with matrices, $$\vec r' = \begin{bmatrix}w'\\x'\\y'\\z'\end{bmatrix} = \begin{bmatrix}1&&&\\-\beta&1&&\\&&1&\\&&&1\end{bmatrix}\begin{bmatrix}w\\x\\y\\z\end{bmatrix} = G_\beta~\vec r,$$ where blank entries can be interpreted as zeros. One can then use matrix mathematics to confirm that $G_{\alpha}~G_{\beta} =G_{\alpha + \beta}$ and then confirm that this theory is completely consistent by proving that you get back the identity matrix, $G_{-\beta}~G_{\beta} = I.$

Now Maxwell discovered that his very successful equations predicted that there were electromagnetic waves which would move with a fixed speed $c$, which he could calculate and found to be approximately the speed of light. (He possibly noticed the speed first--it had appeared in other publications if I recall correctly--and then looked for the wave theory after.) The most common interpretation was that this speed was measured in a "luminiferous ether," which we were probably moving relative to, but there were some interesting intimations for example from Lorentz, who calculated that since the electromagnetic force makes up rulers, and it predicts a length contraction in a certain distance when you're moving relative to that luminiferous ether, you might not be able to use a moving ruler in order to measure this length contraction anywhere else -- or for that matter, the different speed of light.

Einstein, a great pragmatist and very physical, geometric thinker, took this idea much more seriously, indeed wanted to know what a world would look like if everyone had to agree on the speed of a beam of light moving away from them, even if they were moving relative to each other. It turns out that it is sufficient to work out the right theory to first order in $\beta$ and then build out from there. So: an event happens and a short time $\delta w$ later, a "bubble" of light which announces that event in all directions has moved a distance $\delta w$ away. Let us suppose that there were two moving observers who were at the event when it happened: one, we'll say, stays at rest, the other moves with speed $\beta$ and is now at $x = \beta~\delta w.$ So surely they think that the distance to the light "behind" them is $(1+\beta)~\delta w$, and the distance to the light "ahead" is $(1-\beta)~\delta w.$

The relativity perspective says: there is no contradiction, if they think that the time we're talking about, ahead of them, is the time $(1-\beta)~\delta w$ while the time we're talking about, behind them, is the time $(1+\beta)~\delta w.$ Einstein immediately realized that both reference frames disagree about what it means for two clocks to be synchronized, and the correct first-order transformation is not $G_\beta$ but is actually, $$\ell_\beta = \begin{bmatrix}1&-\beta&&\\-\beta&1&&\\&&1&\\&&&1\end{bmatrix}.$$ Now this is only consistent to first order because $\ell_{-\beta}~\ell_\beta = \operatorname{diag}(1-\beta^2,1-\beta^2,1,1) \ne I$. That will not matter for this comment. But I will say that there is an "easy" way to resolve the consistency (just to divide that 2x2 block by $1/\sqrt{1-\beta^2}$ to make this term vanish) and a "correct" way to resolve the consistency (since it's consistent to first order, solve $L_\beta = \lim_{N\to\infty} (\ell_{\beta/N})^N$ and they happen to both come to the same conclusion, they just somewhat disagree on what $\beta$ means (for the former approach it is an actual speed; for the latter it turns out to be a rapidity).

What relativity implies

So now we have the first-order theory, what does it really say? It says that in the process of the one frame going at speed $\beta$ relative to the other, all of the clocks "ahead" of it now seem to tick some time $w$ at a true time $w' = w - \beta x$, in other words, they are ticking "early" from clocks which would truly be in-sync in that reference frame. The clocks behind that frame (negative $x$) are ticking "late".

Well, how did it get to be moving at speed $\beta$? Assuming that they were both stationary in the past, one of them accelerated. And this change must have happened during acceleration. Suppose you are accelerating at some rate $a$ (measured let's say conventionally, in length / time^2 rather than our trick with $\beta$ which would make it 1/length). This says that, to first order, some clock which is ahead of you a distance $L$ must appear to tick faster, at where the multiplicative factor is $1 + aL/c^2.$ A clock behind you must appear to tick slower by the multiplicative factor $1-aL/c^2$.

If you're paying careful attention you'll notice that the latter expression can and will hit zero: is that correct? Yes! Light emitted from a certain distance behind an accelerating observer will never reach that observer, and in fact that accelerating observer must, by virtue of that acceleration, see all of the objects that pass them slow down and approach this "wall of death" a distance $c^2/a$ behind them! (The all-order theory does not change the existence of this wall of death, the shape of this wall of death, or even the distance to it.)

How any of this relates to your question

What you may not realize is that this "wall of death" is actually an event horizon, and this clocks-ahead-of-you-tick-faster phenomenon is actually gravitational time dilation. They are not something which general relativity adds, they are something present in special relativity.

The only thing that general relativity adds is the idea that the "natural states" of things -- the inertial reference frames of special relativity -- are in free-fall. You think that sitting at your desk you are stationary, but you are wrong. You are actually constantly accelerating upward, away from the center of the Earth. You are doing so because there is an uncompensated force acting upon you: the force of your chair. "But that is compensated by gravity!" you wish to day: balderdash! Baloney. Gravity is a fictitious force like the centrifugal or Coriolis forces, or like the sense that you are being pushed forward when you have to suddenly brake in your car. When you see the centrifugal and Coriolis terms you imagine that you stand outside the Earth and watch it spin, maybe even outside of the Solar system and watch the planets orbit: when you see the gravitational force, if you want to "stand outside" it, go to an amusement park and do one of those rides that puts you into free-fall. Those are the inertial reference frames.

So it's a combination of two things. First there is an effect that you might not have been expecting--might not even have wanted!--which states that when you accelerate in some direction the clocks in front of you tick faster. Second there is the effect that, in order to remain a fixed distance from the center of the Earth, you actually need to be constantly accelerating upwards, which causes clocks far above you, like the ones on satellites, to be ticking faster than clocks near you. This effect does not exist in free-fall.

In fact it is now well known that if you want to, you can fall past an event horizon in finite time. It's no different than anything we've already discussed. Imagine an accelerating observer and a star crossing the "wall of death." Then think about it from the star's reference frame: that "wall of death" still has meaning to the star, but only as a boundary across space, moving in time: anything which happens on one side, that light will reach this accelerating observer; anything on the other side, will not. The boundary passes over the star at some time, let's say $w=10$: the star doesn't care, the star's clock ticks just fine, $w=11,12,13,\dots$. But it can no longer shine any light on the accelerating observer, the accelerating observer can never see it tick $10$ because that ray of light announcing that tick is the very first one that cannot quite reach the observer in any finite time. The observer, on the other side, appears to observe this star's clock ticking slower and slower -- red-shifting and becoming dimmer -- as it gets closer and closer to the wall and closer and closer to saying $w=10.$ But it never gets all the way there; it appears to "freeze" in time.

Similarly you drop something into a black hole, it appears to slow down more and more, as it arrives on the event horizon. But actually in the free-fall frame it does pass through, it is just no longer able to send any more light out to you living outside of the black hole.

All aspects of time dilation have been based on the workings of atomic clocks. Atomic clocks contain the Caesium Atom and the clocks work by using the vibrations of the caesium Atom.

As you move away from gravity or towards the atom vibrates changes.

So regardless if it is alterations in Space-Time or gravity or by any other scientific concept, the end result is the mechanism within the clocks is what slows down as you get closer to a black hole.