# Curvature gravity and a falling apple? [duplicate]

I know very little of physics after Einstein.

I am aware of that Einstein's gravity theory says that the existence of matters creates curvature of a space-time, so that our Earth orbits our Sun. I can grasp this idea.

But I do not see how to use the language of general relativity theory to justify why an apple can fall?

## marked as duplicate by Pranav Hosangadi, fibonatic, Kyle Kanos, JamalS, ACuriousMind♦Jan 21 '15 at 17:59

• Why couldn't it? – Demosthene Jan 21 '15 at 15:15
• the key point is that it's not space that's curved, it's spacetime. The apple falls because the accelerating downward path through spacetime is the geodesic through curved space, rather than the "straight" hovering path through spacetime. – Jerry Schirmer Jan 21 '15 at 15:16
• Thanks for concern. But that is my problem. I do not see the reason. @Demosthene – Megadeth Jan 21 '15 at 15:16
• @JerrySchirmer: If you would like to make your comment an answer, then I will immediately accept it :) – Megadeth Jan 21 '15 at 15:18
• possible duplicate of How does the curvature of spacetime induce gravitational attraction? – Pranav Hosangadi Jan 21 '15 at 15:23

I don't have time to write this up properly, but I was asked to make this an answer. (I also feel that John Rennie is about to post something much better)

The key point is that it's not space that's curved, it's spacetime. The apple falls because the accelerating downward path through spacetime is the geodesic through curved space, rather than the "straight" hovering path through spacetime.

• @Chou I've always liked this visualization by Rickard Jonsson – garyp Jan 21 '15 at 21:30

The path in spacetime followed by a freely moving object is given by the geodesic equation:

$${d^2 x^\mu \over d\tau^2} + \Gamma^\mu_{\alpha\beta} {dx^\alpha \over d\tau} {dx^\beta \over d\tau} = 0 \tag{1}$$

Now this looks absolutely horrible, but it's a lot simpler than it looks.

The variable $x$ is a vector that gives a position in spacetime. So if we use $x$, $y$ and $z$ for the position in space and $t$ for the position in time then the vector $\vec{x}$ has components $(t, x, y, z)$. The superscript, e.g. $x^\mu$, isn't a power but instead tells us which component it is. So $x^0$ would be $t$, $x^1$ would be $x$, $x^2$ would be $y$ and $x^3$ would be $z$.

The variable $\tau$ is called the proper time, and is the time shown on a clock carried by the moving object. That is, if you were the one moving through spacetime $\tau$ is just the time shown on your wristwatch.

So the expression $d^2 x^\mu/d\tau^2$ is just an acceleration e.g. when $\mu=2$ it's just $d^2y/d\tau^2$ or the acceleration in the $y$ direction. Similarly the expression $dx^\mu/d\tau$ is a velocity, so when $\mu=3$ it's just $dz/d\tau$ or the velocity in the $z$ direction.

The symbols $\Gamma^\mu_{\alpha\beta}$ are called Christoffel symbols, and they basically describe how spacetime is curved. In flat spacetime, i.e. no gravity, the Christoffel symbols are all zero$^1$. In curved spacetimes the Christoffel symbols are non-zero.

And now we're ready to go. Let's see what happens if you release your apple in flat spacetime. For flat spacetime the Christoffel symbols are zero and equation (1) becomes:

$${d^2 x^\mu \over d\tau^2} = 0$$

and this is really four equations, one for each value of $\mu$:

\begin{align} {d^2 t \over d\tau^2} &= 0 \\ {d^2 x \over d\tau^2} &= 0 \\ {d^2 y \over d\tau^2} &= 0 \\ {d^2 z \over d\tau^2} &= 0 \end{align}

So what this tells us is that the acceleration in all directions is zero. In other words the apple stays where it is. And of course that's quite correct. If you're floating in space far from any masses and you let go of the apple it will stay where it is.

A quick aside before I get on to what happens in curved spacetime. You may be wondering what on Earth $d^2t/d\tau^2$ is. Well the time coordinate $t$ is what we measure while we're watching the apple and the time coordinate $\tau$ is what the apple measures on its wristwatch. If the apple is stationary relative to us the $t$ and $\tau$ are the same. But if the apple is moving there will be time dilation and $t$ and $\tau$ will not be the same. The velocity $dt/d\tau$ is just the time dilation, and the acceleration $d^2t/d\tau^2$ is the rate of change of time dilation.

Anyhow let's go back to your apple and see why equation (1) tells us it will fall.

Actually calculating the trajectory is quite hard. If you really want to see how it's done this article has the gory details. I'm not going to do the calculation, I'm just going to show you why GR tells us the apple must move. To do this I'll use the geodesic equation to calculate the acceleration at the instant we release the apple, and to show it is non-zero.

We start with the apple stationary so the velocities $dx/d\tau$, $dy/d\tau$ and $dz/d\tau$ are all zero. Though the apple isn't moving in space it is moving in time at one second per second, so at the moment the apple is released $dt/d\tau = 1$ (actually it's $c$, but we tend to use units in which $c = 1$.

So looking back at equation (1), the velocities $dx^\alpha/d\tau$ and $dx^\beta/d\tau$ are zero except when $\alpha = \beta = 0$. So equation (1) simplifies to (remember this is only at the moment the apple is released):

$${d^2 x^\mu \over d\tau^2} + \Gamma^\mu_{00} = 0 \tag{2}$$

The trouble is that the coordinates I've chosen, $(t, x, y, z)$, aren't ideal for (roughly) spherically symmetric gravitational fields like the one round the Earth. So I'm going to cheat a bit and assume I can use the Rindler metric to describe the curvature in the small region around the apple and us. We'll take $z$ to be the height, so $x$ and $y$ will be the position on the ground.

With these coordinates the only non-zero Christoffel symbol $\Gamma^\mu_{00}$ is when $\mu = 3$, i.e. $\Gamma^3_{00}$, and this has the value (at the moment the apple is released):

$$\Gamma^3_{00} = \frac{g}{c^2}$$

where $g$ is the gravitational acceleration (and remember we're using units where $c = 1$). Plug this into equation (2) and expand it into the four separate equations and we get:

\begin{align} {d^2 t \over d\tau^2} &= 0 \\ {d^2 x \over d\tau^2} &= 0 \\ {d^2 y \over d\tau^2} &= 0 \\ {d^2 z \over d\tau^2} &= -g \end{align}

And there's our result. So the apple doesn't accelerate sideways, i.e. in the $x$ or $y$ directions, but it does accelerate downwards in the $-z$ direction. And the acceleration downwards is just $g$ i.e. the gravitational acceleration.

$^1$ even in flat spacetime the Christoffel symbols will be non-zero if you're using curved coordinates e.g. polar coordinates. We'll gloss over that for the purposes of this discussion.