I was trying to learn General Relativity. While I was learning about curved geometry, I found out this post. As said by the author, those lines are geodesics on a 2-manifold. I think I don't quite understand it. What I interpret is that those lines are straight lines that curve because of the intrinsic curvature of the manifold, no? Or are they how the lenghts (you measure) are increasing due to curvature?

They have something to do with coordinates or with the basis vectors you have chosen? What exactly those lines means and what they have to do with the metric tensor?


1 Answer 1


Newton's first and second laws of motion respectively state momentum is unchanged without a net force and the rate of momentum is equal to net force. Every student naturally wonders what the point is of the first law, as surely it's a special case of the second. But whatever Newton's original motive for such delineation, it's helpful here to understand what geodesics do in general relativity.

Let's first phrase the first law (for a constant mass) another way: without a net force, velocity is fixed, so displacement is a linear function of time. Or, if we construe such a path as through spacetime, we can see its rate of change with respect to time as constant. In other words, it's a straight line through spacetime.

Now, when people say of the GR it implies "gravity isn't a force", what they mean is its equivalent of Newton's first law is a specification of the paths followed in the absence of non-gravitational forces. These paths, which gravity determines, are the geodesics, the equivalent of the aforementioned straight lines. But they're only straight with a suitable parameterization that bears in mind both time and space in the right way, and the metric tensor encodes which way is right.

Explicitly, the first law for a constant mass is upgraded from $\frac{d^2}{dt^2}x^i=0$ to$$\frac{d^2}{d\tau^2}x^\mu+\Gamma^\mu_{\nu\rho}\frac{d}{d\tau}x^\nu\frac{d}{d\tau}x^\rho=0,$$with $\tau$ the proper time. (The metric determines the Christoffel symbols $\Gamma^\mu_{\nu\rho}$.) Either way, the $=0$ assumes no force, and is replaced with $=f^i$ or $=f^\mu$, a force or $4$-force, when we generalize to the second law. But GR's key insight into gravity is that there's another term at work here. Rearranging,$$\ddot{x}^\mu=f^\mu-\Gamma^\mu_{\nu\rho}\dot{x}^\nu\dot{x}^\rho$$(with dots indicating $\tau$-derivatives). The term $-\Gamma^\mu_{\nu\rho}\dot{x}^\nu\dot{x}^\rho$ looks like a $4$-force, which is why objects seem to "fall" and "orbit" and so on. But its origin is fundamentally different.

People always offer a qualitative analogy whereby the spacetime geometry is like bends in a rubber sheet. But I'd like to offer another one. We see from the above that geometry-induced gravity adds something quadratic in $\dot{x}$ to $\ddot{x}$. On at least one model, that's reminiscent of the drag force in a fluid. In this analogy, a spacetime geometry further from "flat" spacetime (i.e. stronger gravity) is like a stronger drag force, while a geodesic is like the path a body with no propulsion (and, ironically, no gravity!) acting on it will follow due to drag. I'm sure this analogy has all sorts of drawbacks on further scrutiny, but it hopefully makes GR's account of gravity seem less bizarre.

  • $\begingroup$ Amazing explanation. But I have two questions: Those lines (on the post), represent geodesics, that, at the same time, represents how objects naturally would move in this surface, if I didn't understand it wrong. But my concern is, if you were a 2D living being on the surface and trace out an straight line, those lines are what you would see? And lastly. You said that Newton's Second Law "updates" to this new equation. In an external frame of reference, that's the acceleration someone would have, but from your own frame of reference you will measure that: $\ddot{x}^\mu = 0$, no? $\endgroup$ Jun 26 at 22:02
  • $\begingroup$ @ÁlvaroRodrigo (i) The straight lines you'd trace have $\ddot{x}^\mu=0$, whereas geodesics instead have $\ddot{x}^\mu=-\Gamma^\mu_{\nu\rho}\dot{x}^\nu\dot{x}^\rho$, which is why gravity appears to cause curved orbits. Geodesics are "straight" in a more fundamental sense than the lines we see as straight. (ii) If you adopt a coordinate system that says you're always stationary, it would say of your position $\frac{d}{dt}x^i=0$ so $\frac{d^2}{dt^2}x^i=0$, but that's very different from the question of $\frac{d^2}{d\tau^2}x^\mu$. $\endgroup$
    – J.G.
    Jun 27 at 7:38

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