Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

So if you move through space with a constant acceleration you experience longer time dilation than when you're at rest, but you also experience the same time dilation when you're under the effect of gravity like on earth, so is it possible that by standing on earth, space itself is moving relative to you at a constant acceleration, which causes the same time dilation as when you move in space with same acceleration of earth's gravity $9.8\tfrac{m}{s^2}$?

Maybe that's why under free fall you don't sense the acceleration and feel weightless, because you are moving along with space itself that is attracted to earth, neither you are moving relative to space, nor space is moving relative to you. By moving I mean accelerating and - constant speed of course.

Sorry if this question is silly or something that is maybe already suggested before.. I'm not a specialist here.

share|improve this question

4 Answers 4

up vote 2 down vote accepted

The answer is yes, and in fact I've described how this works in my answer to another question of yours: If you shoot a light beam behind the event horizon of a black hole, what happens to the light?.

I won't repeat the working from that question here, but it might be worth a comment on exactly how the idea works. When you solve the equations of GR you get an object called the metric tensor that tells you about the geometry of spacetime. To write this down you have to choose some system of coordinates i.e. the units you use to measure time and distance. The metric itself doesn't depend on what coordinates you choose, but the form you write the metric does depend on the coordinates. For a static black hole the usual way the metric is written is using the Schwarzschild coordinates, in which case it looks like:

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

In this form the coordinates $t$ and $r$ are the times and distances measured by an observer far from the black hole, so it's an obvious choice. As written in this form there is no suggestion that spacetime is flowing in any sense.

However the Schwarzschild coordinates have a singularity at the event horizon and they're difficult to use if you're trying to work out what happens at the event horizon. To get round this we use a different set of coordinates called the Gullstrand-Painlevé coordinates. Using these coordinates the metric looks like:

$$ ds^2 = -dt_{ff}^2 + \left(dr + \frac{2M}{r} dt_{ff} \right)^2 + r^2 d\Omega^2 \tag{2} $$

In these coordinates $dr$ is the distance measured by a distant observer, as in the Schwarzschild coordinates, but the time coordinate $dt_{ff}$ is the time that would be measured by an observer falling freely into the black hole. If you compare this with the metric for flat space:

$$ ds^2 = -dt^2 + dr^2 + r^2 d\Omega^2 $$

It should be obvious that this looks like flat space but where the radial distance is replaced by $dr + \tfrac{2M}{r} dt_{ff}$. In other words the distance from the black hole changes with time, and the rate of change is faster the closer you get to the centre of the black hole. The physical interpretation of this is that the spacetime is flowing inwards into the black hole, a bit like water draining in a sink. If you are falling into the black hole you are just being carried along by spacetime, just like someone being carried along in a river.

But you should be careful of attaching physical significance to this. The whole point of general relativity is that it doesn't depend on the coordinates you use. Although the two metrics (1) and (2) look different they will calculate the same value of $ds^2$ - they are just different ways of writing down the same physics. So we can write down the metric in a way that describes spacetime as flowing, just as you suggest in your question. However we can also write an equivalent equation that does not show spacetime as flowing. So you cannot say gravity is due to spacetime flowing, you can just say that the equations can be written down in that way.

share|improve this answer
    
So if we describe spacetime as flowing, can we think of spacetime as kind of being accumulated as a river accumulating water in one pond after a waterfall? And then maybe the spacetime is released back after the black holes evaporates? –  user43783 Apr 3 at 19:13
    
John Rennie: "The answer is yes [... we can think of gravity as space itself moving]" -- It is answers like these which seem to inspire (follow-up) questions like those: "Can space be accelerated and does it experience length contraction?" (or, indeed, the follow-up question asked by user43783 in the preceding comment). –  user12262 Apr 5 at 20:48
1  
@user43783: be cautious about thinking of spacetime as some kind of rubber foam that we can compress or stretch then have it spring back. Spacetime isn't a physical object, it's a manifold equipped with a metric, both of which are mathematical structures. It's really only our coordinate system that's flowing. Having said this, for a white hole the Gullstrand-Painleve radial coordinate would be flowing outwards, so this may well be the case for an evaporating BH. I'd have to sit down and think about it. –  John Rennie Apr 6 at 6:32

The time dilation is not caused by moving with respect to the space. In special relativity the time dilation is measured only by observers that move with respect to each other and it is symmetrical (both of them measure the clock of the other one slow down). This symmetry shows that they will never agree on who is actually "at rest" in space and who is "moving". They both can rightfully define a reference frame where they are at rest and it will not change the results of their measurements or calculations. You can not measure your velocity with respect to "space" - a lot of experiments tried that with negative results.

In general relativity there can be 2 observers at rest with respect to each other and yet observe different clock rates on each others clocks. This is attributed to the curvature of the spacetime, not a space "falling" towards the Earth.

The equivalence between gravity and acceleration is described by the math of the general theory of relativity. If you do not experience acceleration force, you are following the geodesic. If you do feel acceleration, you do not follow the geodesic. That "force" is described mathematically by the Christoffel coefficients. These coefficients vanish if you are in free fall in gravitational field (curved spacetime) or in inertial reference frame in flat spacetime. The Christoffel coefficients are non-zero if you are accelerating in flat spacetime (in a sense that accelerometer in your pocket is measuring non-zero acceleration) or standing still on the surface of the Earth.

share|improve this answer

As a guideline we may look to Einstein's maxim that

All our well-substantiated space-time propositions amount to the determination of space-time coincidences [such as] encounters between two or more recognizable material points.

(Instead of "recognizable material points" other authors also use equivalent terminology such as "substantial points", "principal identifiable points", "distinguishable participants" or "individual observers"; all of which may of course be actual, or be considered hypothetically, for instance in descriptions of thought experiments.)

At this foundational level there is apparently no mentioning of "space itself". Any references to "space itself" (such as the phrase "moving through space") or any attributions to "space itself" (such as "space itself moving", or "space itself expanding", or somesuch) are merely verbiage which at best disguises explicit propositions concerning "recognizable material points".

Now, much of what you wrote in the question about "space itself" may therefore be expressed explicitly by referring to concrete (actual or hypothetical) "recognizable material points". For instance, it may be said that

  • you move through an inertial system with constant acceleration, or

  • you remained rigid with respect to some other particular participants (who themselves also move through an inertial system with constant acceleration, similar to you), or

  • you remained rigid with respect to some other particular participants (such as elements of "Earth's surface"); even though some inertial system through which you were moving at some particular (e.g. constant) acceleration could not strictly be found at all, but at best only approximately, in the limit of vanishing "spatial extension".

share|improve this answer

Einstein's equivalence principle would suggest that in this case not the space, but rather Earth itself is expanding. That would be consistent with Einstein's words about the reversal of the vector of the gravitational acceleration (the famous lift thought experiment).

Obviously, in order to keep such theory coherent and things in proportion, one would need to assume also that all the matter in the universe is expanding in an accelerated manner (which would explain at the same time, why we cannot notice this expansion), as well as make room for some other type of force (electro-magnetic?) to keep planets, stars and other matter from hitting each other.

But then, if an empty space - believed not to possess any properties - can be assumed to be continuously expanding, than why can't matter?

Especially that according to this (At which point of the universe $R_{\mu \nu}=0$ if there is a source of gravitation (point mass)) conversation Einstein tensor is limited to matter only, and yet it does include the cosmological constant.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.