# Why do objects follow geodesics in spacetime?

Trying to teach myself general relativity. I sort of understand the derivation of the geodesic equation $$\frac{d^{2}x^{\alpha}}{d\tau^{2}}+\Gamma_{\gamma\beta}^{\alpha}\frac{dx^{\beta}}{d\tau}\frac{dx^{\gamma}}{d\tau}=0.$$ which describes "how" objects move through spacetime. But I've no idea "why" they move along geodesics.

Is this similar to asking why Newton's first law works? I seem to remember reading Richard Feynman saying no one knows why this is, so maybe that's the answer to my geodesic question?

• Good question. I would say "because nature is lazy", but I'm sure you'll get some better answers! – twistor59 Apr 24 '12 at 18:50
• This seems like a great question. Some will treat it as a postulate of general relativity, but I think it's more natural to look at it as a generalization as the straight-line path of a free particle. – tmac Apr 24 '12 at 19:29
• Indeed, good question. Are you looking for a mathematical explanation, or a physically intuitive one? (Or both?) – David Z Apr 24 '12 at 19:54
• @David Zaslavsky Both, but of course I can't promise I'll understand the mathematical one. – Peter4075 Apr 24 '12 at 20:02
• Newtons first is a bad example for a comparison. The law introduces "inertial frames", i.e. defines its meaning. If the law would say "An inertial frame is one in which bananas speak" the law would still "correspond to reality" (there just would be no inertial frames). An "because nature is lazy" argument is imho bad, because the fact that we can model the world using general relativity is what makes the equation possible. So in a world where we could not come up with that theory, we could neighter say "is not lazy" nor "it's lazy". And "intuitive "explainations" are merely more information. – Nikolaj-K Apr 24 '12 at 20:07

You could think of it this way:

1) Take a free particle, put it at some spacetime point, and leave it evolve.

2) Imagine the motion is not geodesic, that is $a_\mu\equiv v^\nu v_{\mu;\nu}\neq 0$, or in other words the acceleration is not zero. Note: We know that $a_\mu v ^\mu = 0$, or the 4-acceleration is normal to 4-velocity.

3) Imagine you are that very particle, that is you are in the reference frame where $v^\mu=(0,0,0,1)$. Because 4-acceleration and 4-velocity are orthogonal, you shall still "see" non-zero 3-vector of acceleration in this frame. I shall not elaborate much on this see, but if you write the equations of motion of test particles located around you, you shall see them accelerating in the direction of $\bf{a}$. I refer to the chapter on comoving reference frames.

Now the punchline. As inertial mass is equivalent to passive gravitational mass, you may never distinguish whether you are standing still or moving in a gravitational field. But if you can see an appearing 3-acceleration, then you actually can distinguish by realising that you are not standing still. Hence the contradiction.

To conclude, the fact that everything moves along the geodesics is closely related to the equivalence principle.

That this works for a test mass is essentially a postulate, which, as indicated by Alexey Bobrick's answer, is related to the equivalence principle.

On the other hand, it is hypothesized that this behaviour can actually be demonstrated to be a direct consequence of Einstein's equations for physical masses. To prove this, however, requires actually solving the full Einstein's equations, and progress in that direction is, to the best of my knowledge, incomplete. The following is a biased (and certainly incomplete) account of some which has been done.

One of the first attacks of the problem came from Einstein himself. In a joint work of Infeld and Hoffman, the point particle is treated as a point (Dirac $\delta$) singularity in space-time. Einstein's equation is written down, and expanded in the Newtonian limit. The resulting series expansion is shown to have the first term corresponding to geodesic motion, and the second term giving the first relativistic correction (which can be used to account for the perihelion precession).

The problem was raised again by Geroch and Jang in 1975. In that paper matter is treated simply as its energy momentum tensor. That is, matter is considered to be represented by a symmetric divergence-free two-tensor (the right hand side of Einstein's equation) that satisfies some energy conditions. The main result of that paper is that if $\gamma$ is a space-time curve such that for every neighborhood $U$ of $\gamma$, there exists a symmetric divergence-free two-tensor that vanishes outside of $U$ and yet is not everywhere vanishing, then $\gamma$ is a time-like geodesic. (One should also see this paper of Weatherall for some further comments.)

The Geroch-Jang theorem has been revisited and generalised by Ehlers and Geroch in 2004. It is interesting to note, as a side remark, that an analog of the Geroch-Jang theorem is also true in Newton-Cartan theories of gravity; this result is due to Weatherall.

A different approach to the problem was taken by DMA Stuart. He considered a specific matter model (in his case, the semilinear wave equation which is known to admit soliton solutions) and showed that solitons in the gravitationally-coupled theory travel along time-like geodesics. The relevant references are this paper and this other paper both from 2004. (Warning: heavy doses of PDE theory is involved in both.)

A yet different point-of-view was given by Gralla and Wald. In that paper they considered a point-particle as a scaling limit of a family of metrics corresponding to solutions of Einstein's equations possessing a coherent body or a black hole, and derived an equation of motion for the limiting particle. The point of view was also taken up by Iva Stavrov where initial data sets generating such a family were constructed. In some sense this method is the rigorous counterpart to the original work of Einstein-Infeld-Hoffman mentioned above.

Remark: Any omissions from the above represent the limits of my own knowledge; it is quite likely that there are other large bodies of work regarding the geodesic hypothesis I am not familiar with. Unfortunately the phrase "geodesic hypothesis" has two distinct meanings in theoretical physics that I am aware of. One is the above in general relativity. The other refers to a hypothesis (due to M.F. Atiyah and N.S. Manton) in high energy physics that the dynamics of solitons can be described by geodesics on a certain moduli space of solutions. So it can be a bit confusing for doing literature search.

• Thanks for the replies, though I'm afraid they are way over my head. I still don't get it. – Peter4075 Apr 25 '12 at 15:32
• The short version is: mathematically the fact that "extended bodies" travel along geodesics can be seen as a direct consequence of Einstein's equations of general relativity, at least in certain cases where the maths have been worked out; for the other cases it is a long-standing conjecture. For the case of test-particles and for the physical interpretation, you'll have to refer to the other answers/comments on this question. – Willie Wong Apr 25 '12 at 15:56
• Charged particles don't move along geodesics, unless spacetime is reinterpreted as something with an increased number of dimensions such as in Kaluza-Klein theory. But, if every particle follows a geodesic in such an expanded spacetime, it may not be unreasonable to equate "particle" with "geodesic": to say that a particle is, per se, an aspect of topological features that can only occur along geodesics. In that case, it would be better to say that a particle is a sort of geodesic, not that it follows a geodesic. The above is not mainstream, and I don't necessarily believe it. – S. McGrew Sep 21 '20 at 3:21
• @S.McGrew: charged particles are rather obviously not within the spirit of the question being asked. – Willie Wong Sep 21 '20 at 11:32
• True, but the same argument would apply to non-charged particles. I just tossed that in as something to think about. The fact that a set of equations demands that a particle move along a geodesic doesn't satisfactorily answer the question of "why" the particle does that. It just translates the question to "why do the equations have that form?" – S. McGrew Sep 21 '20 at 13:41

The result that a freely falling test particle(a particle whose effect on space-time can be neglected) in a gravitational field moves along a geodesic can be deduced(just like a theorem of maths) from the equivalence principle, which is a hypothesis of the general theory of relativity.

It is just like asking why Newton's first law works. First we must define inertial reference frames, which are the frames in which Newton's first law works (text from The Large and the Small).

• An inertial reference frame is one in which inertial bodies remain at rest or in uniform motion.

It is implicit in this definition that inertial reference frames are local. That is to say, an inertial reference frame describes only a finite region of spacetime in which deviation from rest or uniform motion is not measurable and can be neglected. The size of this region depends upon the accuracy of measurement, but it is worth noting that a second in time corresponds to a light-second in distance. In terms of normal timescales, local refers to quite short intervals of time.

An inertial object, that is one in free fall, follows a locally straight path in an inertial reference frame, along the direction of its 4-velocity in spacetime.

A geodesic is defined by parallel transport of a vector in the direction which it specifies. Thus, the 4-velocity determines a geodesic in spacetime. Geodesic motion is simply a statement of Newton’s first law, that in an inertial reference frame, inertial objects remain at rest or in uniform motion in a straight line. Curvature of spacetime means that, on larger distance scales, inertial objects follow curved paths, like the orbit of the moon about the Earth.

It is related to what Einstein called "the happiest thought of my life", that for an observer falling freely from the roof of a house, the gravitational field does not exist.

If we could choose a system of coordinates and a suitable definition of derivative, so that the acceleration (the derivative of velocity) were zero for that case, the feeling of the person in free fall would match the maths.

GR does that job. The system of coordinates is defined by the metric tensor, and the geodesic equation is nothing more than the result of setting the covariant derivative of the 4-velocity to zero.

If we make an analogy from the 4-dimensional spacetime to a 2-dimensional surface of the earth, the change of perspective is similar to realize that the airplane journey from Tokyo to Paris is a curve if observed in a common world atlas of the airline magazine. But it is "straight" if we join the 2 cities by a string in a world globe.

Simplicity is the is the root of the explanation: The shortest distance between two points (relatively noted) in our inclusive spherical system, is a curve and the function of the geodesic design. The reason is that the path of least resistance reduces the waste of total energy to a finite number, and thus the total contained energy can be distributed equally within the system...

N/0= f(n){1=1}

Because everything in space moves in straight lines until acted by an outside force, (gravity) ,where it still moves in straight lines around the mass of the objects in which it orbits that's been brought in to its gravitational pull. Simple answer i apologize for such