I think I understand the idea of thinking about gravity not as a force pulling an object towards another object but instead a warping of space so that an object moving in a straight line ends up following a path that brings it closer to the object, like two people at the equator both heading North and ending up at the same point even though all they did was move forward.

What I'm not following is why the speed that the object is traveling would affect the path it takes if all it is doing is moving forward and it is in fact spacetime that is bending around the planet. I can easily understand this in classical mechanics as two forces counteracting each other, but I can't visualise what is happening in a model of gravity as warped space.

Imagine a large planet and two objects passing by the planet both on the same course.

One is slower than the other. The slow object gets captured by the planet and falls into an orbit (or to the planet itself if it is too slow to make an orbit). If I understand correctly this object is simply moving forward in space but space itself bends around so that its path now takes it towards the planet. But nothing has pulled the object off its original course.

The other, a fast moving object, has its path bent slightly but flies past the planet and off into space. Same thing, it simply moves forward and again its path is bent by virtue of space itself being bent

If these two objects are both simply moving in a straight line through the same bent space time, both going only "forward" how would the speed of one object cause a path that is less bend towards the planet than the other. Surely one just travels through the same equally bent space time faster than the other.

I'm sure I'm missing something, but can't find a good explanation, most explanations I can find online about viewing gravity as curved spacetime ignore completely the speed at which the object caught by gravity is traveling.

Follow Up

Just want to say thank you to everyone who answered this question, blown away by how much people were prepared to put into formulating answers. I've not picked an acceptable answer since I don't feel qualified to know which is the best explanation, but they are all really good and have all really helped expand my understanding of this topic.


14 Answers 14


You're using the wording "curved spacetime", but you're still only thinking "curved space" with an independent, linear time.

In your curvature model, you're assuming that moving through some 3D spatial point in one spatial 3D direction will experience the same 3D path curvation independent on speed (as if you'd shoot a ball through a curved tube). You'd certainly agree that a different initial 3D direction will result in a different path.

Now we are in 4D, meaning that two different initial speeds are two different 4D directions, and as time cannot be treated as an independent component, but is curved together with space, this easily results in a different path.

| cite | improve this answer | |

Let's approach this by taking a simple analogy. Suppose you and I are in two cars at the equator and we start driving North. Even though we started off driving exactly parallel to each other we will find the distance between us decreases until when we reach the North pole we would collide. Our motion looks like this:


(this diagram is taken from my answer to When objects fall along geodesic paths of curved space-time, why is there no force acting on them?)

So the curvature of the Earth has caused us to accelerate towards each other and ultimately collide, and this acceleration depends on our speed. If we drive very slowly we would approach each other slowly, while if we drive fast we would approach each other fast. So the apparent force causing us to accelerate towards each other depends on our speed.

And this is roughly what happens in general relativity. The acceleration of an object falling in a curved spacetime is described by an equation called the geodesic equation, and the speed of the object, or more precisely the four-velocity, appears in this equation.

In my simplified analogy of the sphere the speed affects our acceleration towards each other but not the end result i.e. we'd end up colliding at the same place (the North pole).But this is an artefact of the simplified analogy I used. When we do the calculation in 4D spacetime we find the speed affects the trajectory as well. Different four velocities produce different four-accelerations and different trajectories.

| cite | improve this answer | |
  • 6
    $\begingroup$ @CormacMulhall what is hard to visualise about spacetime is that you have to include the component in the time direction. This may at first seem to be just one second per second, but it isn't because of time dilation. The way the maths works the time component of your four-velocity increases the faster you go, and this means the "direction" in 4D spacetime changes as your velocity changes. $\endgroup$ – John Rennie Oct 13 at 17:28
  • 34
    $\begingroup$ So even though you may be travelling in the same direction in space, changing your speed means you travel in different directions in spacetime. $\endgroup$ – John Rennie Oct 13 at 17:29
  • 5
    $\begingroup$ @CormacMulhall The analogy with the globe is great, but it's not perfect. It doesn't really illustrate how space time is curved very well, and as John said, you have to include the time component. In relativity, 1 millisecond of time has the same magnitude as 300 km of space, so a little bit of warped time goes a long way. ;) Also, if you have a non-zero velocity relative to me, then we disagree on how spacetime is divided into space & time, i.e., there's an angle between our time axes, as shown here. $\endgroup$ – PM 2Ring Oct 13 at 17:40
  • 3
    $\begingroup$ @CormacMulhall We can't really make nice simple diagrams that illustrate this stuff because good old-fashioned Euclidean space is governed by the Pythagorean distance formula, $d^2=x^2+y^2+z^2$, whereas the equivalent formula in Minkowski spacetime is $\tau^2=t^2-x^2-y^2-z^2$. $\endgroup$ – PM 2Ring Oct 13 at 17:50
  • 2
    $\begingroup$ I think the question makes it clear that they are aware of this analogy. $\endgroup$ – Acccumulation Oct 14 at 19:31

Get rid of the planet in your scenario. Just have two objects at the same place and same time in (1+1D) flat spacetime. Let's build our reference frame so that they both start at the origin $(t,x)=(0,0)$, with one moving at $1\,\mathrm{m}/\mathrm{s}$ in the $+x$ direction and one moving at $2\,\mathrm{m}/\mathrm{s}$ in the $+x$ direction. In spacetime, are these objects moving on the same path? I think you might say yes, because both are following the spatial path $t = 0$, but the answer is emphatically no! The path of an object through spacetime is just that: the path through space and time. Our "slow" object follows the path $x=t\cdot1\,\mathrm{m}/\mathrm{s},$ and our fast one $x=t\cdot2\,\mathrm{m}/\mathrm{s}.$

time-space diagram of objects' paths through spacetime; they are lines both passing the origin, but with different slopes

What you're thinking of as the "path" is the "shadow" of the full spacetime paths onto the "spatial hyperplane" (in this case, on the x-axis; in your question that would be 3-dimensional "space"). But this is SR/GR: the whole point is that looking at just space is not enough. In any case, now that we've established that objects with different velocities already follow different paths through spacetime, even if spacetime is flat and even if they start at the same point. All that I really need to say is that a curved spacetime can allow this difference, which looks "temporal" right now, to bleed over and become "spatial".

Now, I'm not going to go all in on the GR, but for low-mass objects like Earth, most of the gravitational attraction comes from the curvature of time, not space. All objects naturally move towards the future, and the gravity of Earth means the futureward direction acquires an inward radial component near its surface (as compared to an observer in free-fall "far away"). Falling towards the Earth is as inevitable as moving through time... which is, as shown above, pretty "evitable" if you go fast enough. In the case of us evidently not falling through the floor, this is because the repulsion between our atoms and the Earth's constantly accelerates us at $1 g$ upwards, as long as we are mechanically connected to the surface.

Now, I said I wouldn't go full GR. Instead, I'll say this: even here, on the surface of the Earth, we can approximate spacetime as flat (so we're in SR land), and things appear to accelerate under gravity simply because we're in a non-inertial frame constantly accelerating upwards under the normal force of the ground. As an SR trick, we should use Rindler coordinates. Rindler coordinates in SR are the coordinates of a non-inertial frame of reference that has constant proper acceleration. Seen from an inertial frame, the Rindler coordinate axes are curved. Seen from the Rindler frame, the Cartesian axes of the inertial frame are curved. Assuming we're accelerating at $a=9.8\,\mathrm{m}/\mathrm{s}^2$ along the $+y$ direction and we let the origin be shared, the transformation from inertial $(t, x, y)$ coordinates to Rindler $(T, X, Y)$ coordinates is $$T=\frac{c}{a}\operatorname{arctanh}\left(\frac{tc}{y+\frac{c^2}{a}}\right),\quad X=x,\quad Y=\sqrt{\left(y+\frac{c^2}{a}\right)^2-c^2t^2}-\frac{c^2}{a}.$$ If we extend our above graph with a $y$-axis, which pokes into/out of your screen, then the $t$-axis is defined by $x=y=0.$ We can graph this in the $T-Y$ plane of our new coordinates:

Graph of t-axis in Rindler coordinates, which looks like a parabola

(Note: this is almost but not quite (a few parts in [insert-big-power-of-10-here] off) a parabola). In the above graph, the $X$-/$x$-axis is going into/out of the screen. If you imagine taking our graph from above, aligning its $x$- and $y$-axes with the $X$- and $Y$-axes here, and then bending the $y$- and $t$-axes so they line up with the $Y$- and $T$-axes, then the worldlines of the two objects also bend to give the path as we see them from our reference frame attached to the "ground". Since the objects had no $y$-component to their motion, their worldlines are actually "on top of" the $t$-axis curve above, so the above plot also serves to show the (almost)-quadratic relationship between height and time elapsed for the objects as they fall under gravity. Note that their apparent acceleration and subsequent displacement in the $Y$-direction (which you might consider the "spatial" direction "height") comes purely from the bending of the time axis.

Now, if we rotate the superimposed graphs so the $X$- and $Y$-axes are visible but the $T$-axis disappears, we finally recover your spatial paths. While in the inertial frame, the spatial paths of the two objects coincided, the curvature of the Rindler coordinates has turned the temporal separation between them (due to their different velocities) into a spatial one. My demonstration is purely mathematical—the spacetime described by Rindler coordinates is still flat, even if the coordinates are curved—but I hope you can see that in GR, where spacetime really does curve, that curvature can "detect" the difference between objects moving at different velocities, because the objects simply are going in different spacetime directions.

(Almost)-parabolic "spatial" trajectories of the objects, in the X-Y plane

| cite | improve this answer | |
  • $\begingroup$ I.e. You should contemplate the relativistic view in flat spacetime first. $\endgroup$ – Keith Oct 14 at 23:34

Different initial speeds determine different initial directions for the geodesic through spacetime. For example, think of a light cone in simple flat spacetime. The worldline for an object with zero speed is along the axis of the cone. The worldline for an object moving at the speed of light is along the surface of the cone. Other worldlines for various speeds lie at various angles between these.

| cite | improve this answer | |
  • $\begingroup$ "Different initial speeds determine different initial directions for the geodesic through spacetime". I guess I'm trying to visualize why this is the case. Take something like Kerbal Space program (I know it doesn't model space time but reasonably good modeling of realistic orbital physics). I'm in a stable orbit and I slightly decelerate then my path relative to Earth will change dramatically. I slightly accelerate and I fly off into space instead. What is actually changing here if the spacetime I'm moving through is essentially the same either way given the warping of the Earth? $\endgroup$ – Cormac Mulhall Oct 13 at 17:19
  • 2
    $\begingroup$ I'm trying to visualize why this is the case. As I mentioned, light cones make it easy to visualize this. The point is to visualize spacetime. $\endgroup$ – G. Smith Oct 13 at 17:22
  • $\begingroup$ Ok, I'm not really following how that comes into play but thanks for answering $\endgroup$ – Cormac Mulhall Oct 13 at 17:29
  • $\begingroup$ @CormacMulhall take one spatial dimension away from the KSP and stack the resulting planes at different times one on top of each other to get a 3D visualization of a 2+1 dimensional spacetime (2 spatial, 1 temporal dimension). Orbits will be helicoidal paths through spacetime and different velocities will correspond to different angles of these paths like in the answer. $\endgroup$ – Danijel Oct 14 at 12:35
  • 1
    $\begingroup$ @CormacMulhall - imagine a horiz. 2D plane, representing our 3D space. The way you are currently thinking about it is "if I go faster in some direction in this plane, I'm still going in the same direction along the same line". But now take that plane and add a 3rd, vertical axis to represent time. In this picture, if you're just standing still, you're moving "upwards" (because your space-plane moves through time). If you go at different speeds, you go at different angles through spacetime. If you now add curvature, you'll see that these paths can project differently onto your space-sheet. $\endgroup$ – Filip Milovanović Oct 15 at 8:32

As others have explained, the main point is that curvature is in 4D, not just 3D. In fact, the main "warping" happens in the time direction.

I just want to help your imagination with two pictures.

Consider a 2D space (horizontal) + time (vertical) spacetime, and a reference frame with the Earth at rest, as in the first picture below. The Earth is a 2D disk; its world-tube (thin blue line) in this spacetime is a 3D cylinder.

Flat 2+1 spacetime

Take three projectiles starting a tangential motion above the Earth's surface (thick red lines). The first has zero initial velocity with respect to the Earth, so its worldline starts vertically. The second has a non-vanishing tangential velocity, so its worldline starts at some angle with a horizontal plane. The third has a higher initial tangential velocity than the third, so its worldline starts at a smaller angle with a horizontal plane (same space = horizontal span in less time = vertical span).

If this spacetime were flat, as in the picture above, the three worldlines would lie within a plane (green) parallel to the earth's worldtube. The first projectile would stay still, not falling, with a straight vertical worldline. The other two would also have straight worldlines continuing away from the earth's worldtube.

The energy-momentum-stress of the earth curves spacetime instead, as shown in the second picture below. The worldline of the projectile with initial zero velocity is bent toward the Earth's worldtube – thus acquiring a radial velocity and eventually touching the Earth's surface. The worldline of the second projectile is bent around the Earth's worldtube; this is seen as an orbital motion. The worldline of the third projectile is also bent toward the Earth's worldtube, but not as much as the second. It eventually continues far away from the Earth (and becomes "straighter", as the curvature diminishes); this is seen as an escape from the Earth's gravitation.

Curved 2+1 spacetime

So the spacetime curvature bends worldlines with different "inclinations" in different ways. Hence the dependence of velocity, which is how we see such inclination.

The fact that most of the curvature is in the time direction becomes clear if you take natural units for space distance and time lapse (1 s = 300000 km). The worldlines of ordinary projectiles are almost "vertical", and their bending only occurs over huge "vertical" distances in this example spacetime.

The moon, for example, has a velocity of roughly 1 km/s. In natural units this would be a worldline with an angle of 89.9998° from the horizontal plane. And the spiral of its worldline would form one coil only after a vertical distance of roughly $56\,000\,000$ times the diameter of the Earth's worldtube depicted here – you'd need roughly $56\,000\,000$ screens on top of each other to see one coil, if the picture here respected natural units.

An even simpler intuitive picture is obtained considering a ball thrown vertically, with different initial speeds. I invite you to draw a 1+1 spacetime picture of the worldlines of the ball with different initial velocities (they'll look like parabolae) – you'll see the effect of curvature, and its dependence on velocity, directly in front of you. Also check out how these parabolic worldlines would look like, using natural units.

(Please note that the pictures above only have an illustrative purpose, they aren't plots of solutions of 2+1 Einstein equations or anything of the kind; and sorry for the poor draughtsmanship!)

| cite | improve this answer | |

On how to take existing velocity into account:

Start with the standard thought demonstration of the Principle of Equivalence: a spacecraft is accelerating in uncurved spacetime. The spacecraft is accelerating; it is pulling G's. By the principle of equivalence: all motion of objects in the spacecraft can be treated as motion subject to gravitational acceleration.

Next step: a projectile is launched from one side of the spacecraft, the initial velocity of the projectile is perpendicular to the G-load.

When that projectile arrives at the other side of the spacecraft it is no longer moving exactly perpendicular. Over the course of its flight the projectile has dropped.

The amount of drop that you expect is dependent on two factors:
The magnitude of the G-load
The velocity of the projectile

In terms of relativistic physics any projectile is negotiating spacetime.

The factor time cannot in any circumstance be omitted from the picture; if it is omitted then the very picture is gone.

Going back to the projectile in the spacecraft: the faster the projectile is moving, the less time there is available for the G-load to cause the projectile to drop.

Now to the example in your question: satellites are delivered to orbit by giving them sufficient velocity (in the direction perpendicular to the Earth's gravity).

Just as in the spacecraft: the amount of drop per unit of time is the same for any object. But when the object has a large perpendicular velocity the amount of drop per unit of distance traveled is comparitively small.

More generally, you shouldn't think of curved spacetime as some sort of conduit. The expression 'curved spacetime' expresses that an object that is negotiating that region of spacetime will undergo change of velocity. This change of velocity comes in addition to existing velocity, if there is any.

I agree with you: a proper presentation of what curved spacetime is should have the capacity to accomodate that a different starting velocity will lead to a different outcome. Conversely: if a presentation cannot accommodate that then it is fatally deficient.

Additional remarks:
Even for the celestial bodies of the solar system the spatial non-straightness is still very small. In the case of the Sun and the orbit of Mercury: the curvature of spacetime as a whole gives rise to the orbit of Mercury, the precession of the perihelion of Mercury's orbit correlates with the degree of spatial non-straightness.

For non-relativistic velocities the contribution of the spatial non-straightness to the total effect is very small, exemplified by the orbit of Mercury.

On the other hand, light moves so fast that there is very little time for spacetime curvature to have effect. Because of that very-little-time the spatial effect is a larger proportion of the total effect. (The effect of the spatial non-straightness does not depend on how much time is available; it is a spatial effect.)

There is the bending of light by the curvature of spacetime around a star. The 1919 Eddington experiment sought to measure the amount of deflection of light that grazes the Sun. The GR-prediction for that is 1.75 seconds of arc. (Half of that 1.75 arc seconds is attributed to the spatial non-straightness of spacetime). This underlines again that the spatial non-straightness of the space around the Sun is very, very small.

| cite | improve this answer | |
  • 2
    $\begingroup$ " you shouldn't think of curved spacetime as some sort of conduit" This might be where I'm going wrong. I visualize warped space time as sort of like a track or grid and obviously if you curve the track or grid anything traveling 'forward' along the track/grid bend around to the degree that the track/grid is bent. But these answers are making it clear I'm probably using a very simplified notion of what is actually 'warped' about warped spacetime $\endgroup$ – Cormac Mulhall Oct 13 at 18:17

The expression curved spacetime can lead to this type of association of ideas. It is better to think that the effect of gravity is to impose some type of curvilinear coordinates.

What happens in GR is that the typical accelerated movement followed by orbiting bodies, turns to be non accelerated if:

  1. the curvilinear coordinates of the metric are used,

  2. the calculus of the acceleration are corrected for the fact that the coordinates are curvilinear.

While I can't imagine how to visualize 4D, it is possible to explain how curvilinear coordinates are tricky in a 2D example.

A plane chooses the shortest path between 2 cities, unless there is some other reason not to do that. Because the longitudes and latitudes are curvilinear coordinates, a fly between 2 points at almost the same latitude (say San Francisco to Washington DC) doesn't take a route constant to east. If you see in a fly magazine chart, the fly seems to be a curve, with the plane having some velocity component to the north in the first half, and to south in the second half of the trip.

But if you see the route in a globe, it is easy to see it is indeed the shortest path. The compass is always showing that the direction of the velocity is changing, but it isn't really. There is a mathematical heavy machinery called covariant derivative that corrects the inputs of the compass, resulting in a constant velocity.

It is similar for 4D spacetime. Our coordinates show an accelerated movement. But when corrected by the covariant derivative, it becomes a movement with a constant velocity.

| cite | improve this answer | |

According to your reasoning, if a particle is not moving wrt to curved space, it would stay up in space.
But if space is curved, time (being an integral part of spacetime, in contrast to the Newtonian view where they are considered as being separate and absolute) is curved too. The connected space and time are the absolute entity in relativity instead of the separate absolute space and absolute time in Newtonian mechanics.

The reason why clocks run at different speeds at different places in spacetime is nicely explained by Feynman in his little book "Six not so easy pieces" (you can look at this nice book here; the rocket part is centered around page 162), where he discusses what happens to the pace of two clocks placed on the top and the bottom of a rocket (in outer space) when the rocket is accelerated (which according to Einstein's equivalence principle means that we just as well can say the rocket finds itself in a gravity field).

Because of this intimate connection between space and time, if you travel on a curved space (as you assume in your question) you ignore the accompanying curved time.
The reason I fall to the Earth is the (curved) time component of the curved spacetime.

There are three regimes:

  1. I move very slow through curved spacetime. In that case, curved space has the biggest grip on me in making me freely move.
  2. I move with a speed that has a value somewhere in the middle of zero and lightspeed. In that case, both the curvature of time and that of space have a comparable influence on my trajectory.
  3. Not I, but photons that always, from whatever frame of reference they are observed, travel at the speed of light. The curvature of space only grips the photons (as time stands still for photons, the curvature of time doesn't have a grip on them). They get deflected by Earth (though very slightly) because of the space curvature part of the connected curved spacetime.

That's why the speed with which an object travels gives different results for the trajectory in space, as you stated.

If you're interested, in this article (which you can do download) the "famous" factor 2 in the deflection of light by a spherically mass is discussed:

The problem of the deflection of light in a medium with varying refractive index is applied to the motion of light in a weak Schwarzschild gravitational field. In contrast to the standard derivation, the present method is physically transparent, providing a clear reason for the factor-of-2 deviation of the general relativistic result from that of the Newtonian theory without any detailed calculation.

| cite | improve this answer | |

For the sake of argument, let's assume that the two small objects have exactly the same amount of stress-energy and they are relatively small (little stress-energy) compared to the planet, and let's assume the planet is Earth.

Now the answer to your question is:

  1. the slower object spends more time inside Earth's gravitational field

  2. GR time dilation

  3. the four velocity vector's magnitude has to stay constant

  4. 1,2,3 will cause the slower object to deflect more on its trajectory

Now 1,2 are pretty clear, GR time dilation is an effect caused by Earth's gravitational field, causing the object being inside the gravitational field to slow down (relatively) in time.

Now what needs more explanation is the connection between GR time dilation and the four velocity vector and this causing the slower object to deflect more on its path.

Gravitational time dilation is a form of time dilation, an actual difference of elapsed time between two events as measured by observers situated at varying distances from a gravitating mass. The lower the gravitational potential (the closer the clock is to the source of gravitation), the slower time passes, speeding up as the gravitational potential increases (the clock getting away from the source of gravitation).


If you accept that the universe is set up so and the four velocity vector is set up so, that the magnitude of the four velocity vector has to stay constant, then it is very important to understand that GR time dilation causes the object's four velocity vector's temporal component to change. This is what we mean when we say that the object slows down (relatively) in time.

in short, the magnitude of the four-velocity for any object is always a fixed constant:


Now remember, the magnitude of the four velocity vector has to stay constant. If its temporal component changes, the spatial components have to compensate. This is very important. This means that the object will deflect on its path towards the center of Earth.

The more time (bigger period of time relative to the faster object) the slow object spends under the influence of Earth gravitational field, the more its four velocity vector's temporal component (the more it will slow down in time relatively) will change. The more its temporal component changes, the more the spatial component has to compensate (the more it will deflect from its path towards Earth's center).

Please note, this is one of the very reasons why we say that space and time are interconnected.

| cite | improve this answer | |
  • $\begingroup$ yes, in general terms, but I do not know if the OP will get it. $\endgroup$ – anna v Oct 14 at 17:15
  • $\begingroup$ @annav thank you so much! $\endgroup$ – Árpád Szendrei Oct 14 at 19:31

This is not a complete answer to your question, it's more of a supplement to the existing answers, and a response to some of the comments you've made.

In a comment, you said:

I visualize warped space time as sort of like a track or grid and obviously if you curve the track or grid anything traveling 'forward' along the track/grid bend around to the degree that the track/grid is bent.

That's perfectly fine, as long as you bear in mind that while travelling through space you cannot avoid moving forwards in time at 1 second per second according to a clock you carry with you. The time measured by that clock is called your proper time, and we generally use the Greek letter $\tau$ (tau) to represent proper time.

In flat spacetime, if you're moving with a constant velocity relative to me (so we measure each other to have constant speed & to be moving in a constant spatial direction), you can consider yourself to be at rest, so your space coordinates are constant, but of course your proper time keeps on ticking ahead, as usual. As I said earlier in a comment, we will divide spacetime up into space & time slightly differently, and there will be an angle between our time axes.

A point in spacetime is called an event. Let's say that you travel from some event A to some other event B. You're at rest in your frame, so in your frame A & B have the same space coordinates, but B will have a later proper time.

In my frame, the spacetime "track" from event A to event B has a nonzero spatial component, as well as its temporal component. So while you say that the time "distance" between A & B is $\tau$ and the space distance is 0, I measure that the space distance between A & B is $s$ and the time distance is $t$ (according to my proper time), and there's a simple formula connecting those numbers, the Minkowski version of the Pythagorean formula: $$\tau^2 = t^2 - s^2$$ where we use compatible units for our space & time measurements, eg, light-seconds and seconds.

Now in General Relativity we can cut a chunk of curved spacetime into small chunks of spacetime, where the curvature of each small chunk is negligible. If the big chunk is highly curved, then we just need to make those small chunks very small. (This is exactly the same process that we use to make an atlas of flat maps of the curved surface of the Earth. On each page of the atlas we can ignore the curvature and use simple 2D flat geometry, and the errors from ignoring the curvature are negligible). So in each of those small chunks of spacetime we can ignore the spacetime curvature and do our calculations using the equations of flat spacetime from Special Relativity. The mathematics of General Relativity is essentially the machinery required to slice spacetime up into small chunks using standard calculus techniques, and to keep track of how all the chunks connect with each other.

As I mentioned in a previous comment, it's not easy to visualise 4D spacetime, with its Minkowski distance formula replacing the standard Pythagorean distance formula. We can simplify things a little by dropping one spatial dimension. For instance, if we use a frame where the Sun is at rest, the Earth's orbit around the Sun is pretty much in a plane. So we can use that plane for our two spatial dimensions, and we can use the vertical direction to represent time (but bearing in mind that the time direction is a bit weird because of the $\tau^2 = t^2 - s^2$ distance formula). To make things even simpler, let's pretend that the Earth's orbit is a perfect circle, so it orbits the Sun at a constant distance of approximately 499 light-seconds with a constant speed of $10^{-4}\,c$, that is $10^{-4}$ light-seconds per second, or 30 km/s in more conventional units.

Such a circle has a rather small spatial curvature relative to typical human scales. A 55 km long arc of that circle deviates from a perfectly straight line by just over 1 cm. (That is, if you draw a chord from one end of the 55 km arc to the other, the distance between the arc & the chord at their midpoints is around 1 cm). However, that spatial curvature is huge compared to the spacetime curvature.

A path in spacetime is called a worldline. In our frame where the Sun is at rest, the Sun's worldline is a vertical line. The worldline of the Earth is then a helix, with one turn of the helix per year. Now one year is about 31,557,000 seconds, so the pitch of the helix (the vertical distance between turns) is about 63,240 times its radius.

In units of reciprocal light-seconds, the curvature of the orbit circle is $1 / 499 \approx 0.002$. In contrast, the curvature of the orbit helix is $$\frac{499}{(3155700/2\pi)^2 + 499^2}\\ \approx 1.978\times 10^{-11}$$

which is a lot smaller. So it doesn't take much spacetime curvature to keep a planet in orbit.

Actually, I probably should use a minus sign in the denominator of that helix curvature calculation, to respect the Minkowski metric. However, that doesn't affect the numerical result at this level of precision, it's still $\approx 1.978\times 10^{-11}$.

| cite | improve this answer | |

In a much simplified form, what you are thinking of only occurs at a black hole, where spacetime actually forms a complete circle around the body so even light will simply travel in a complete circle. Otherwise, with anything less than a black hole, the curvature you are thinking of is only partial.

You probably are aware that if you drop a bullet from your hand, and if you shoot a bullet horizontally, they will both hit the ground at the same time. (Lets say 1 second). This is because they are both impacted by gravity (the curvature of spacetime) at the same rate. But what if you shoot a very high speed bullet horizontally. Although it will fall at the same rate as the other bullets, it travels much further in 1 second, so the curvature of the earth becomes a factor. So it won't hit the ground in 1 second because the ground has moved away from it. The same thing happens if you move to even faster objects, like a fighter jet traveling at mach 3, it will take even longer because the curvature of the earth and the ground has moved even further away. This becomes very apparent with a beam of light. The light will be bent in the same manner, but in 1 second it will have traveled far past the earth and is no longer subject to the earth's gravity. But, if you had a planet that was very very big, and very very flat, then indeed, the dropped bullet, the shot bullet, and the beam of light would all hit the ground at exactly the same time.

| cite | improve this answer | |

Exactly this question was addressed on pp. 32-33 of Gravitation by Misner, Thorne, and Wheeler:

How can the tracks of a ball and of a bullet be curved so differently if that curvature arises from the geometry of space? ... Depicted in spacetime (C), the tracks of ball and bullet appear to have comparable curvature.

| cite | improve this answer | |

In 4 dimensions a different speed is already a different path.

Given that spacetime is warped it should be unremarkable that the part of the path in the three spacial dimensions can also differ depediong on the steepness in the fourth.

| cite | improve this answer | |

Your perfectly right when you said the reason things fall is because they take straight paths in space time and it is space time that's curving... I have two arguments...the second one is a guess(but comes from special relativity, Its up to you to accept the answer that you think is more right)-

the first argument-

The reason something has to take a straight path is because its moving in time. For example the reason something falls when you drop something is because it has to move forwards in time and space time is curved around and in to the earth. It follows a straight coordinate space time line (that line of course being just going to infinity in the time direction like in the graph y=1 where x is the time coordinate)...Now think about the dimensions of space time the objects are moving through...they are moving through space and time, both...unlike the last example where an object just moved through time...so it doesn't really (at all) have to follow a straight path. It could deviate from the space time coordinate line ...And if it deviates enough then it doesn't get caught by the gravity... The reason the smaller object falls is because its not deviating enough...

Heres my second argument(my favourite)-

In special relativity an object moving would dilate time i.e. time would flow slower for it ... now the time flow rate for the faster moving object is less than that of the other slower moving one...Like I said before the reason an object falls is because it has to follow a straight path through space time since its moving through time. But here time is slowed down so the time component is not as strong as the space component for the faster moving object. So it moves slowly in the time coordinate line and faster in the space coordinate line. As for the other object the opposite happens (wasn't that neat? connecting einsteins main theories...)

I hope you got the answer to your question

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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