# Naive visualization of space-time curvature

With only a limited knowledge of general relativity, I usually explain space-time curvature (to myself and others) thus:

"If you throw a ball, it will move along a parabola. Initially its vertical speed will be high, then it will slow down, and then speed up again as it approaches the ground.

"In reality, the ball in moving in a straight line at constant velocity, but the space-time curvature created by the Earth's gravitation makes it appear as if the ball is moving in a curved line at varying velocity. Thus the curvature of space-time is very much visible."

Is this an accurate description, or is it complete nonsense?

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Depends on what you call 'straight'? –  Danu Mar 7 '14 at 8:09
people interested in this question please look at the useful link in the last answer by @answerman physics.stackexchange.com/q/102409 –  anna v Mar 11 '14 at 5:16

Yes, that's a fair description of what happens though of course from the ball's perspective it isn't moving - the rest of the universe is moving around it.

However statements like this, while true, give little feel for what's going on. Actually it's extraordinarily difficult to get an intuitive feel for the way spacetime curvature works (or at least I find it so!). The notorious rubber sheet analogy gives a fair description of the effect of spatial curvature, but neglects the curvature in the time coordinate and the time curvature is usually dominant since $dt$ gets multiplied by $c$ in the metric.

The motion of the ball is described by the geodesic equation, but a quick glance at the article I've linked will be enough to persuade you this is not an approach for the non-nerd. I have never seen an intuitive description of how the geodesic equation predicts the motion of a thrown ball.

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Many thanks to all who have answered! I appreciate your effort. I am very much aware that my initial "explanation" is a far cry from being scientifically complete. The problem of defining exactly what is meant by "reality" and "straight line" clearly illustrates that. Any popular explanation will always oversimplify things; I think that's inevitable. As the title of the question illustrates, my original explanation was intended as a very simple way to demystify the concept of curved spacetime. Something along the lines: "See, the ball moved in a curve. There's spacetime curvature for you!" –  oz1cz Mar 7 '14 at 9:30
"the time curvature is usually dominant since $dt$ gets multiplied by $c$ in the metric" has a (common) mistake: you can't compare magnitudes with different dimensions, it doesn't make sense to say "3 seconds >> 1 meter". The effects on time only are apparent when the forces are very intense. –  Davidmh Mar 7 '14 at 11:50
@Davidmh: In the parametrized post-Newtonian formalism, finding the geodesics through the action integral shows that the $\mathrm{d}t^2$ term contributes more (lower-order) to the action than any of the spatial metric coefficients precisely because of a factor of $c$. Moreover, when recovering Newtonian orbits in the low-velocity limit, the time-time term is essential but one can completely ignore deviation from flatness in the spatial terms. Thus John Rennie's statement is oversimplified but not a mistake. –  Stan Liou Mar 8 '14 at 5:55
Ok, in that sense it is correct. Then we are down the endless path of where does the physical meaning end and where the sheer mathematical derivation start. I would say that, more than oversimplified, it was imprecise because it was not clear what he meant and could be misinterpreted (as I did). –  Davidmh Mar 11 '14 at 6:07

To me, the best way of describing it is as follows (semi-stolen from Carroll's Spacetime & Geometry): we all know from high school physics that, when no force acts on an object, it should not change its velocity $v$. In other words, the velocity vector tangent to the objects trajectory through space and time (i.e. world line) undergoes parallel transport. In flat space, that's just the same as keeping its Cartesian components constant. The equation corresponding to this expresses that the acceleration is zero (i.e. the velocity is constant): $$a^\mu=\frac{d^2 x^\mu}{d\tau^2}=\frac{d}{d\tau}\frac{dx^\mu}{d\tau}=\frac{d}{d\tau}v^\mu=0$$ (the upper index $\mu$ is just there to indicate the position and other vectors have four components, three corresponding to space and one to time: $t,x,y,z$ in cartesian coordinates. This comes from special relativity)

We would like to generalize this notion to curved spaces. Our idea is to have the velocity vector (the derivative of the position with respect to proper time) undergo parallel transport. However, parallel transport turns out to be trickier when spacetime is curved. We can no longer use 'simple' derivatives like we do in flat space, and have to emply what is called the covariant derivative, an introduction to which is probably not very enlightening if you're not familiar with the framework. Suffice it to say that this covariant derivative is the generalization of a normal derivative in curved spacetime. The more general notion of parallel transport (i.e. the curved spacetime version of 'keeping the velocity constant') which uses this covariant derivative gives rise to the equation $$\frac{d^2x^\mu}{d\tau^2}+\Gamma^\mu_{\rho\sigma}\frac{d x^\rho}{d\tau}\frac{d x^\sigma}{d\tau}=\frac{d}{d\tau}v^\mu+\Gamma^\mu_{\rho\sigma}v^\sigma v^\rho=0$$ The exact meaning of the second term is not so important if you're just trying to get a feel of what is going on. It is instructive, however, to compare the first equation with the second on. You see that the spacetime curvature gives rise to a second term which changes the way an object moves through spacetime. In flat space, coordinates can be chosen such that the second term reduces to zero everywhere in spacetime, recovering our original equation. The full expression is known as the geodesic equation, by the way, and objects with a worldline that obeys it are following a geodesic through spacetime.

Note that, during this explanation, we never referred to gravity as a force, but rather just assumed spacetime wasn't flat. This is, of course, a huge conceptual leap (and probably the source of many misconceptions and mistakes). In a very crooked sense, the motion of a test particle under gravity is 'straight', in that its 'velocity vector' is parallel transported, a notion which reduces to our intuitive idea of a straight line if spacetime is flat.

P.S. To experts appalled at the omission of essential details etc: Any comments and suggestions for improvement are very welcome.

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Full marks for effort :-) However this is still far from a non-nerds explanation of why we see geodesics as curved. Maybe an intuitive explanation (intuitive for non-nerds that is) is just not possible. –  John Rennie Mar 7 '14 at 8:42
@JohnRennie The covariant derivative is really what does it :( –  Danu Mar 7 '14 at 8:50
$\Gamma$'s don't necessarily reduce to zero when space(time) is flat. The advice to compare the two equations doesn't make sense, because nonvanishing $\Gamma^\mu_{\rho\sigma}$ has just about nothing to do with curvature--they can be nonzero even in flat spacetime, depending on coordinates. I suppose one way to fix this without getting bogged down in details is to note that the $\Gamma$'s are quantitatively different in curved spacetime without bothering overmuch in the specific way that they are. –  Stan Liou Mar 7 '14 at 8:51
@StanLiou would it be acceptable to add the caveat that coordinates can always be chosen such that they vanish when spacetime is flat? Or is this also a gross misrepresentation? –  Danu Mar 7 '14 at 8:54
@Danu: if you add an additional qualifier like 'globally' (over entire spacetime), that would work. Otherwise, coordinates can always be chosen so that they vanish along any given geodesic even in curved spacetime (more than just at a point). If they're made to vanish, the information about curvature is entirely in their derivatives/rates of change. –  Stan Liou Mar 7 '14 at 9:02

I'll raise some issues. Firstly you say

In reality...

Do you contrast something against something else here? It implies you say there the preceding sentence

If you throw a ball, it will move along a parabola. Initially its vertical speed will be high, then it will slow down, and then speed up again...

wasn't right, but that sentence seems pretty resonable. The problem is that there are many perspectives/frame to describe the world and all are valid. In particularly you then say

In reality, the ball in moving in a straight line at constant velocity

but that's of course just as relative. If that's true, how is the above sentence not true? In curved spacetime, you can't pic a global inertial spacetime, so I'd not refer to right and wrong velocities in the elaborations.

Now in general relativity, and that's what Danus answer is in particular about, you use another language to specify what straightness is, the mathematical language of Riemannian geometry, and the expressive power of the formulas doesn't really fit into two englisch sentences. The

straight line

sections of your explanation is problematic. Because, of course, if you already know Riemannian geometry, you read "straight line" as "solution of the Geodesic equation", but the people who you explain relativity to will not. It's a little like if you don't catch sarcasm. While hearing it, you might understand the words perfectly well but when you come back to think about, what has been said doesn't quite add up.

...but the space-time curvature created by the Earth's gravitation...

Do you have a working definition of "Earth's gravitation"? Because to me it's the space-time curvature around it.

...makes it appear as if the ball is moving in a curved line at varying velocity.

Same velocity-problem as above. Keep in mind that there are frames where your left eyeball is rotating around your nose, and there are frames where your nose is rotating around your mouth.

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In flat space(time), straight lines keep the same direction and is the length/inverval-extremizing path between two points. In curved space(time), both are still correct if the qualifier "locally" is added; the connection is specifically chosen to make the two notions equivalent. So I don't agree that there is any "sarcasm" involved in saying that geodesics are straight. –  Stan Liou Mar 7 '14 at 9:23
@StanLiou: Then you are assuming you can explain the notion of a local interial frame to a layman. –  NikolajK Mar 7 '14 at 9:41
I am implying that neither derision nor scorn is appropriate for the label "straight", because it is correct. –  Stan Liou Mar 7 '14 at 9:50
@StanLiou The sarcasm he mentioned in his answer is meant as an analogy about only 'insiders' understanding something. This was obscured by his poor grammar; the relevant sentences should read something like: 'It's similar to when a very naive person hears a sarcastic remark. He/she may understand the literal meaning of each word, but is oblivious to the real significance of what's being said'. This analogy is (not very) relevant because to call a geodesic 'straight' assumes an understanding of non-Euclidean geometry, and thus not very helpful to a layperson. –  Danu Mar 7 '14 at 10:07

You have the right basic idea. But it gets simpler to visualize if you just drop the ball, or throw it vertically. Then there is just one spatial dimension to consider, and you can directly compare the paths in space and in space-time, like shown here:

But note that this doesn't involve any intrinsic space-time curvature. Such curvature is related to tidal effects (geodesic deviation), which are negligible over small distances (like the height of an apple tree or a ball throw). However, over larger distances intrinsic curvature is inevitable, as shown in the last link of the video description:

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nice, useful visualization and link . It is worth bringing it up front again. –  anna v Mar 11 '14 at 5:06
@annav: I don't agree. That visualization is deeply flawed, because it shows no intrinsic curvature. The illustration is a cylinder for Newton and a conical frustum for Einstein, but both surfaces have zero curvature. –  Stan Liou Mar 12 '14 at 9:41
@StanLiou look what happens within the plane, one x and t dimension forget about the cone and cylinder which are just the last frames. –  anna v Mar 12 '14 at 12:24
@annav: It doesn't matter. Morphing a flat sheet into a differently-sized flat sheet doesn't demonstrate curvature. –  Stan Liou Mar 12 '14 at 22:04
@StanLiou: I have updated the answer to clarify this issue. –  answerman Mar 14 '14 at 0:26

Is this an accurate description, or is it complete nonsense? "If you throw a ball, it will move along a parabola. Initially its vertical speed will be high, then it will slow down, and then speed up again as it approaches the ground.

More accurately:
The ground (as well as anything rigidly "connected to the ground") would move along a parabola wrt. the "thrown ball", or rather: wrt. the entire "freely moving inertial system" (such as a "falling ruler") of which the ball is one member.

The difference is that the members of a (suitable) inertial system are (at least to some approximation) at rest to each other; accordingly they are capable of determining distances as well as simultaneity relations between each other, and in these terms they may describe the trajectories of other participants who the met in passing (such as elements of "the ground", or for instance the hands/fingers which are typically shown in the pictures alongside the "falling rulers", moving along parabolas).

In contrast, the geometric relations of elements of "the ground" and other participants (especially: "above" or "below") are more complicated; they are (at best) rigid to each other, i.e. their mutual relations are not characterized by distances but by quasi-distances, and they fail at determining simultaneity of their indications. Accordingly, it would be inaccurate to say that they could determine and express the trajectories of those participants who they met in passing (such as a "thrown ball", or the ends of a "falling ruler") as a definite parabola.

"In reality, the ball in moving in a straight line at constant velocity,

The "thrown ball" is at rest wrt. suitable other participants;

• either exactly, in a flat region,

• or at least to some accuracy (and: "the closer, the more accurate") otherwise.

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