You have seen this demonstration. Somebody stretches a rubber sheet over a frame, drops a bowling ball in the middle, and the sheet sags. Then they roll a marble across it and the marble curves toward the bowling ball, maybe even loops around it once or twice before spiralling in. And the narrator says: that is gravity. Mass curves spacetime, and curved spacetime tells matter how to move. Everyone nods. It is on the cover of textbooks. It is in every documentary. I have used it myself when someone asks me at a party what general relativity is about, because it takes ten seconds and it gets the shape of the idea across.
But it is wrong. Not wrong in the sense of being a simplification that leaves out details, which would be fine, but wrong in a specific and interesting way: it is an accurate picture of something that is not the thing causing gravity. And once you see exactly what it has left out, you understand general relativity considerably better than the demonstration was ever going to teach you.
The objection you usually hear is that the demonstration is circular. The marble rolls toward the bowling ball because real gravity is pulling it down into the dip, so you are using gravity to explain gravity. This is true, and it is the weakest objection available. It is an inelegance in the staging, not an error in the physics. You could imagine running the whole thing in free fall with the sheet under tension and a marble constrained to its surface, and you would still have a geometry, and the marble would still follow the shortest path across it. The circularity is a presentational problem. I want to make a harder objection.
Here it is. Take an apple and hold it at rest above the ground. Let go. It falls. Now go back to the rubber sheet and put a marble on it, at rest, on a flat part of the sheet far from the bowling ball. It does not move. To make the marble move on the sheet, you have to give it a velocity, or you have to put it where the sheet is already sloped, at which point real gravity does the work. But the apple was at rest. It had no velocity. It was not sitting on a slope in space. So what curved geometry acted on an object that was not moving?
The rubber sheet cannot answer this, and the reason is that it has deleted the axis along which the apple was moving. An apple at rest in space is not at rest in spacetime. It is moving through the time direction, and moving through it fast. That motion is what the curvature acts on. The rubber sheet shows you two space directions and no time direction at all, which means it has thrown away the entire mechanism.1Sometimes people defend the demonstration by saying the sheet is meant to represent spacetime, not space, with time somehow implicit. It is not. The surface drawn in every version of this demonstration is a two dimensional surface, and both of its dimensions are spatial. There is no axis on the sheet along which a stationary object advances. If there were, a stationary marble would move along it, which is exactly what does not happen.
Let me make that precise, because “the time direction matters more” is the kind of claim that should come with an equation attached. Take a weak, static gravitational field, meaning a Newtonian potential $\Phi$ with $|\Phi|/c^2 \ll 1$, which covers the Earth, the Sun, and essentially every situation in which anyone has ever demonstrated a rubber sheet. To first order in $\Phi/c^2$, the metric of spacetime is $$ds^2 = -\left(1 + \frac{2\Phi}{c^2}\right)c^2 dt^2 + \left(1 – \frac{2\Phi}{c^2}\right)\left(dx^2 + dy^2 + dz^2\right).$$ There are two corrections to flat spacetime here. One sits on the time part, $g_{00}$, and one sits on the space part, $g_{ij}$. They are the same size, both of order $\Phi/c^2$. The rubber sheet is a picture of the second one. Let us find out what the first one does.2This form of the metric is written in what is called the Newtonian gauge or the conformal Newtonian gauge. The two potentials appearing in $g_{00}$ and $g_{ij}$ are in general independent functions, often written $\Phi$ and $\Psi$, and they are only forced to be equal when the stress energy tensor has no anisotropic stress. For ordinary matter, dust and stars and planets, this condition holds, so a single $\Phi$ does the job.
A free particle in general relativity moves along a geodesic, which is the curve satisfying $$\frac{d^2 x^\mu}{d\tau^2} + \Gamma^\mu{}_{\alpha\beta}\, \frac{dx^\alpha}{d\tau}\frac{dx^\beta}{d\tau} = 0,$$ where $\tau$ is proper time along the path. This is just the statement that the particle goes as straight as the geometry allows. Now take our apple, which is moving slowly, so that its spatial velocity is negligible compared to $c$. Then $dx^i/d\tau \approx 0$ while $dx^0/d\tau = c\, dt/d\tau \approx c$, and the double sum over $\alpha$ and $\beta$ collapses to the single term with $\alpha = \beta = 0$: $$\frac{d^2 x^i}{d\tau^2} \approx -\Gamma^i{}_{00}\, c^2.$$ Everything now depends on one Christoffel symbol. For a static metric, $$\Gamma^i{}_{00} = -\tfrac{1}{2} g^{ij} \partial_j g_{00},$$ and substituting $g_{00} = -(1 + 2\Phi/c^2)$ and $g^{ij} \approx \delta^{ij}$ gives $\Gamma^i{}_{00} = \partial_i \Phi / c^2$. Therefore $$\frac{d^2 x^i}{dt^2} = -\partial_i \Phi, \qquad \text{that is,} \qquad \vec{a} = -\nabla\Phi.$$ That is Newton’s law of gravitation, recovered exactly, and I want you to look carefully at where it came from. It came from $g_{00}$. The spatial part of the metric, the part the rubber sheet is drawing, never appeared in the calculation. Not once. It contributed nothing.
You can see why from the structure of the geodesic equation. The spatial components of the metric couple to $dx^i/d\tau$, the rate at which the particle moves through space. For an apple, that rate is tiny. The time component couples to $dx^0/d\tau$, the rate at which the particle moves through time, and that rate is enormous. The two corrections to the metric are the same size, but they get multiplied by wildly different things. The relative contribution of the spatial curvature to the motion of a particle moving at speed $v$ is suppressed by a factor of order $(v/c)^2$. An apple that has fallen a metre is moving at about $4.4$ metres per second, so $(v/c)^2 \approx 2 \times 10^{-16}$. The rubber sheet is showing you the part of the effect that contributes roughly one part in ten thousand million million.
If you want independent confirmation that this accounting is right, look at the bending of starlight, which is the one case where the spatial part genuinely matters. A photon has $v = c$, so the suppression factor is gone and both parts of the metric contribute equally. Compute the deflection using only the time curvature, which is what you get from a naive Newtonian calculation treating light as a fast particle, and you find $\Delta\theta = 2GM/(c^2 b)$ for impact parameter $b$. Compute it in full general relativity and you find $$\Delta\theta = \frac{4GM}{c^2 b},$$ exactly twice as much. That missing factor of two is the spatial curvature, contributing its equal half. For the Sun this is $1.75$ arcseconds against a Newtonian $0.87$, and Eddington’s 1919 expedition measured the larger number, which is what made Einstein famous.3The 1919 measurement was considerably less precise than the confident retellings suggest, with error bars large enough that the two candidate values were not separated as cleanly as one would like. The result has since been confirmed to far better precision by radio interferometry of quasars occulted by the Sun, which now agrees with the general relativistic prediction to better than one part in ten thousand. The factor of two is not in doubt. So the spatial curvature is real, and it is measurable, and for light it accounts for half of everything. For an apple it accounts for essentially nothing, and an apple is what the demonstration claims to be explaining.
None of this means the rubber sheet is a fantasy. It is a picture of a real geometric object, and here it is worth being precise about which one. Take the Schwarzschild solution, freeze the time coordinate at some value $t = \text{const}$, and look at the equatorial plane $\theta = \pi/2$. The geometry of that two dimensional slice is $$dl^2 = \left(1 – \frac{r_S}{r}\right)^{-1} dr^2 + r^2 d\phi^2,$$ and if you ask what surface embedded in ordinary flat three dimensional space has this same intrinsic geometry, you get $z(r) = 2\sqrt{r_S(r – r_S)}$, a paraboloid opening outward. That is the funnel. That is the rubber sheet, and it is not an approximation or an artist’s impression: it is the exact embedding diagram of a spatial slice of Schwarzschild spacetime, computed by Flamm in 1916. The demonstration is drawing a correct picture. It is a correct picture of the wrong thing.
So what is the right thing? The curvature that makes apples fall lives in $g_{00}$, and $g_{00}$ is the thing that tells you how fast clocks run. Write out the proper time for a stationary observer at position $x$: setting $dx = dy = dz = 0$ in the metric gives $$d\tau = \sqrt{1 + \frac{2\Phi}{c^2}}\; dt \approx \left(1 + \frac{\Phi}{c^2}\right) dt.$$ A clock deeper in the potential well, where $\Phi$ is more negative, ticks slower. Near the Earth’s surface, $\Phi = gh$ for height $h$, and the fractional difference in tick rate between two clocks separated vertically by $h$ is $gh/c^2$. This is not a metaphor and it is not a small print correction. Pound and Rebka measured it in 1959 using a $22.5$ metre tower at Harvard, where the predicted fractional shift is $$\frac{gh}{c^2} = \frac{9.81 \times 22.5}{8.99 \times 10^{16}} = 2.46 \times 10^{-15},$$ and they confirmed it to about one percent. Fifty years later, optical lattice clocks at NIST resolved the same effect over a height difference of $33$ centimetres. Your head ages faster than your feet by about half a microsecond over an eighty year life, and that is a real, measurable, in principle observable fact about you.4The half microsecond figure comes from $gh/c^2 \approx 1.9 \times 10^{-16}$ for a height difference of $1.7$ metres, multiplied by roughly $2.5 \times 10^9$ seconds in eighty years. Whether this is worth mentioning at parties depends heavily on the party.
Now, why should a gradient in clock rate make anything move? Here is the principle, and it is one of the most beautiful statements in physics: a free particle follows the path through spacetime that maximises the proper time elapsed along it. Not minimises. Maximises. Of all the ways of getting from one event to another, an unforced particle takes the one on which its own watch reads the longest. Let me show you that this single statement contains Newtonian mechanics whole. For a slow particle in a weak field, $$d\tau = dt\sqrt{1 + \frac{2\Phi}{c^2} – \frac{v^2}{c^2}} \approx dt\left(1 + \frac{\Phi}{c^2} – \frac{v^2}{2c^2}\right),$$ so the total proper time along a path is $$\int d\tau = \int dt + \frac{1}{c^2}\int\left(\Phi – \frac{v^2}{2}\right) dt.$$ The first term is the same for every path between the same two events, so maximising $\int d\tau$ means maximising the second integral, which means minimising $$\int \left(\frac{v^2}{2} – \Phi\right) dt.$$ That integrand is the kinetic energy minus the potential energy, per unit mass. It is the Lagrangian. Extremising proper time is the principle of least action, and running it through the Euler-Lagrange equation gives back $\ddot{x}^i = -\partial_i \Phi$, which is where we came in.5The sign flip between maximising proper time and minimising the action is not a slip. It comes from the overall minus sign in the timelike part of the metric signature, and it means that free particles in relativity extremise a quantity that reduces to the classical action with the conventional sign.
So here is the picture that actually explains gravity. Everything in the universe is moving through spacetime at speed $c$, always. An object sitting still on your desk is not stationary; it is travelling through the time direction at the full speed, and through the space directions at nothing. Gravity does not push on it. What gravity does is make the time direction run at different rates in different places, and a path that keeps going straight through a region where time runs at a gradient does not stay parallel to where it started. Some of the object’s motion gets rotated out of the time direction and into a space direction. That rotation is what you call falling.
And that gives us the analogy I actually want, which is not a sheet at all. It is a cart with two wheels on a common axle, rolling forward. The cart has no steering. Nobody pushes it sideways. If both wheels turn at the same rate, it goes dead straight. Now let the ground under the right wheel be slightly slower, so that the right wheel covers a little less distance per turn than the left one. The cart veers to the right. No sideways force acted on it. The cart was going straight the entire time. It curved because “straight” in a medium with a speed gradient means “bending toward the slow side.”
The translation is direct. The cart’s forward motion is the object’s motion through time, which never stops and never slows. The speed gradient across the axle is the gradient in clock rate, $gh/c^2$. The turning is the worldline bending out of the pure time direction and acquiring a spatial component, which is to say the object accelerating downward. And notice what this analogy explains that the rubber sheet cannot. It explains why a stationary object falls, because the cart was never stationary, it was rolling forward the whole time. It explains why gravity is universal and independent of mass or composition, because the turning depends only on the ground, not on what the cart is made of or how heavy it is. And it needs no external gravity to run, because nothing in the setup falls.
It is worth getting a feel for how gentle this bending is. Take the apple falling under $g = 9.81$ and draw its worldline on a spacetime diagram with the time axis measured in metres, so that one second of elapsed time is $c \times 1\,\text{s} = 3 \times 10^8$ metres. In that second the apple moves $4.9$ metres sideways. Its worldline is a parabola $x = gT^2/2c^2$ where $T = ct$, and the radius of curvature of that parabola at its vertex is $$R = \frac{c^2}{g} = \frac{8.99 \times 10^{16}}{9.81} \approx 9.2 \times 10^{15}\,\text{m},$$ which is $0.97$ light years. The worldline of a falling apple is bent on a scale of about one light year. It is almost perfectly straight. Gravity feels overwhelming to us not because the curvature is large but because we persist through enormous stretches of the time direction, and a tiny angle applied over a light year of lever arm still knocks you off a ladder.
Now let me be honest about where my cart analogy breaks, because an analogy you cannot break is an analogy you do not understand. A uniform speed gradient across the axle is not curvature. It is the flat spacetime of an accelerating observer, which is what the equivalence principle tells you a uniform gravitational field is indistinguishable from. Genuine curvature is what you get when the gradient itself varies from place to place, so that two carts started off parallel do not merely both turn but turn by different amounts and converge. That is the real signature, and it is called geodesic deviation. If $\xi^\mu$ is the separation between two neighbouring geodesics with four velocity $u^\mu$, then $$\frac{D^2 \xi^\mu}{d\tau^2} = -R^\mu{}_{\alpha\nu\beta}\, u^\alpha \xi^\nu u^\beta,$$ where $R^\mu{}_{\alpha\nu\beta}$ is the Riemann tensor. This equation is the honest definition of gravity. In the weak field limit its leading component reduces to $R^i{}_{0j0} \approx \partial_i \partial_j \Phi / c^2$, the Newtonian tidal tensor, which is the thing that stretches you along the radial direction and squeezes you sideways as you fall toward a planet. A uniform field can be transformed away by jumping. Tides cannot, and that is how you know the curvature is real.
So the summary is this. The rubber sheet draws a real object, the spatial geometry of a slice of Schwarzschild spacetime, and it draws it accurately. But that object contributes essentially nothing to why things fall, suppressed relative to the effect that matters by a factor of order $(v/c)^2$, which for anything slower than light is a very small number indeed. The gravity you feel comes from the curvature of the time direction, from the plain fact that clocks run at different rates at different heights, and from the principle that a free object takes the path through spacetime along which its own clock reads the most. Nothing pulls the apple. The apple goes straight. Straight, in a place where time runs at a gradient, means down.
I still use the rubber sheet at parties. It takes ten seconds and it gets across the idea that geometry is doing the work, which is the part most people have never heard. But if the conversation lasts longer than ten seconds, I put the sheet away and start talking about clocks, because that is where the physics actually is. The demonstration answers the question “what does curved space look like.” The question we asked was “why do things fall,” and those turn out to be almost entirely different questions.
References and Footnotes
- 1Sometimes people defend the demonstration by saying the sheet is meant to represent spacetime, not space, with time somehow implicit. It is not. The surface drawn in every version of this demonstration is a two dimensional surface, and both of its dimensions are spatial. There is no axis on the sheet along which a stationary object advances. If there were, a stationary marble would move along it, which is exactly what does not happen. ↩︎
- 2This form of the metric is written in what is called the Newtonian gauge or the conformal Newtonian gauge. The two potentials appearing in $g_{00}$ and $g_{ij}$ are in general independent functions, often written $\Phi$ and $\Psi$, and they are only forced to be equal when the stress energy tensor has no anisotropic stress. For ordinary matter, dust and stars and planets, this condition holds, so a single $\Phi$ does the job. ↩︎
- 3The 1919 measurement was considerably less precise than the confident retellings suggest, with error bars large enough that the two candidate values were not separated as cleanly as one would like. The result has since been confirmed to far better precision by radio interferometry of quasars occulted by the Sun, which now agrees with the general relativistic prediction to better than one part in ten thousand. The factor of two is not in doubt. ↩︎
- 4The half microsecond figure comes from $gh/c^2 \approx 1.9 \times 10^{-16}$ for a height difference of $1.7$ metres, multiplied by roughly $2.5 \times 10^9$ seconds in eighty years. Whether this is worth mentioning at parties depends heavily on the party. ↩︎
- 5The sign flip between maximising proper time and minimising the action is not a slip. It comes from the overall minus sign in the timelike part of the metric signature, and it means that free particles in relativity extremise a quantity that reduces to the classical action with the conventional sign. ↩︎
