I have studied General Relativity and there is one thing that I have trouble comprehending.

What does local really mean? I will give some examples:

The Hessian

The Hessian is a way to compute the local curvature of a function. What does local mean in this case? At a small area that is not infinitesimally small or an infinitesimally small area?

The Ricci Tensor of General Relativity

It describes how much the curvature of spacetime deviates from flat space (Euclidean space) at a local point. Again, does this mean a small area/volume that is not infinitesimally small or an infinitesimally small area/volume?

In general, what does local mean in Physics and what are we really talking about whenever we say "local"? This would help me understand future concepts.


5 Answers 5


Firstly there is the mathematical understanding of locality, i.e. https://en.wikipedia.org/wiki/Local_property. Roughly "local" means "in some (sufficiently small) open set". This is also very relevant for physics, especially in GR, since the definition of a manifold (e.g. space-time) is that it looks locally like $\mathbb{R}^n$. More precisely locally here means that for every point on the manifold there exists an open neighborhood of that point which is homeomorphic to an open set in $\mathbb{R}^n$. This has to be contrasted with the term global. Very roughly this can be explained by an example, e.g. the circle $\mathbb{S}^1$, which looks locally like an interval $(0,1) \subset \mathbb{R}$ by the homeomorphism $s \mapsto (\cos 2\pi s, \sin 2\pi s)$. Globally it is however different. If you go once around the circle you end up in the same place, which you can not do in $\mathbb{R}$.

Now I agree with Vadim that as you describe it in you question "local" means "infinitesimally", since just knowing a Hessian at some point (or Gradient etc.) tells you something about the function at that point only and not in a neighborhood of that point. It tells you something about infinitesimal variations of that point. On the other hand if you know all derivatives of a function at a point, under certain assumptions, you may be able to know the function everywhere (see Taylor expansion) and knowing some derivatives gives you an approximation, that gets arbitrarily good in a neighborhood of that point as you shrink it arbitrarily close to the point. So there is some connection between the previous definition and this one.

Note also that knowing some relation of derivatives locally (i.e. on an open subset) gives you a differential equation that in combination with some conditions may or may not give you the function locally (or globally), but this is another story.

Then there is of course also the concept of a local theory or local interaction, which is correctly characterized in Vadims answer. For example in particle physics this means that interaction terms in the Lagrangian density only depend on the same space-time point. Otherwise it would lead a violation of causality. This is again another story.

  • 2
    $\begingroup$ Thank you for citing me. I readily admit that your answer is more complete than mine. Mathematical definition of locality is particularly valuable here. $\endgroup$
    – Roger V.
    Nov 20, 2020 at 15:28
  • $\begingroup$ So, does this mean that half of a circle (or even the whole circle except one point) is a sufficiently small open set of the circle to be considered the "locus" for local properties? $\endgroup$
    – Ruslan
    Nov 21, 2020 at 8:29
  • $\begingroup$ @Ruslan Exactly. $\endgroup$
    – jkb1603
    Nov 21, 2020 at 10:34
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    $\begingroup$ "..if you know the all the derivative of a function at a point you know the function everywhere..." well, not quite. It always amazes me how physicists are fine with assuming all functions are analytic and then happily take convolutions with bump functions to smoothen things... $\endgroup$ Nov 21, 2020 at 13:19
  • $\begingroup$ @DenisNardin You are of course correct. I did not want to make a precise statement, but just give some general intuiton. But I will make an edit. $\endgroup$
    – jkb1603
    Nov 21, 2020 at 18:12

Yes, local here means infinitesimally small, although it is a less well-defined term than infinitesimal. One also speaks of local theories, meaning the description of physical phenomena in terms of differential equations involving derivatives up to a finite order. Obviously, taking a derivative also means taking an infinitesimal limit. In this context non-local is associated with interaction occurring via finite distances with no continuous physical entity to mediate the interaction, famously known as the spooky action at a distance.

  • 1
    $\begingroup$ Thank you for your concise explanation. I would like to get some more answers before I pick the right one just to get a clearer picture. $\endgroup$
    – Tachyon
    Nov 20, 2020 at 13:30
  • 1
    $\begingroup$ @Tachyon You know that you can both upvote and/or accept an answer? I am not inviting you to upvote mine, just noting that you can like many answers to one of your questions (but accept only one). $\endgroup$
    – Roger V.
    Nov 20, 2020 at 13:35
  • $\begingroup$ You're right, I forgot to upvote yours. You're welcome :). $\endgroup$
    – Tachyon
    Nov 20, 2020 at 13:37

What the existing answers kinda imply but do not point out exactly is that there are two notions of locality, and one needs to exercise judgement in telling them apart.

Local can mean "in an open neighborhood", which is always finite.

Example: If $A$ is a closed $k$-form on a manifold $M$, there is a theorem (Poincaré's lemma) which states that then $A$ is locally exact as well. What this means is that each point $x\in M$ has an open neighborhood $U$ such that there is a $k-1$-form $B$ on $U$ satisfying $A|_U=dB$. The domain $U$ in question is finite.

There is also a notion of locality that is infinitesimal, which can be stated more rigorously using derivatives/jets. Some examples:

Example 1: It is often stated that every metric tensor is "locally flat". What this means that each point $x\in M$ has a neighborhood $U$ that is a coordinate neighborhood with some coordinate system $x^\mu$ such that at $x$ we have $g_{\mu\nu}(x)=\eta_{\mu\nu}$ and $\partial_\kappa g_{\mu\nu}(x)=0$.

Note that the neighborhood $U$ is finite, but the result is essentially valid for the "first-order infinitesimal neighborhood" of the point only. Without using some other framework such as synthetic differential geometry there is no way of stating this rigorously, but one can imagine that the first-order infinitesimal neighborhood of $x$ is the (fictious) region $U_1$ which contains $x$ and has the property that for any point $x+dx$ which is also in $U_1$ (i.e. infinitesimaly close to $x$) we have $f(x+dx)=f(x)+\partial_\mu f(x)dx^\mu$ as an exact (rather than approximative) relation for any smooth function $f$.

Example 2: Differential operators. The exterior derivative $d$, for example is a local operator in both sense. It is a local operator in the finite neighborhood sense because if $A$ and $B$ are differential forms that agree on some open neighborhood of $x\in M$, then $dA=dB$ on that neighborhood, but it is also an "infinitesimally local" operator in the sense that if $A,B$ are differential forms on $M$ such that at $x\in M$ we have $j^1_xA=j^1_xB$ (this essentially means that $A(x)=B(x)$ and in any chart they have the same first derivatives at $x$), then $dA(x)=dB(x)$.

For OP's examples, the curvature tensor is an infinitesimal measure of curvature. If the curvature tensor vanishes at a point it means that any loop in the second-order infinitesimal neighborhood of that point has integrable parallel transport.

The vanishing of the curvature at a point has no finite bearings on the manifold's geometry.

To complicate things, I am also noting that if the curvature tensor vanishes in the entire manifold, its effect on parallel transport is also only local, but now finite-local. If the entire curvature tensor vanishes, then it guarantees that parallel transport is path-independent in some open neighborhood of each point, but the corresponding global statement is not necessarily true, due to purely topological obstructions, a notion captured in the so-called null-holonomy (cf. Aharonov-Bohm effect).


Generally speaking, when a statement is said to be "locally" true, that's an epsilon-delta claim: given any $\epsilon>0$, there is some $\delta$ such that if the inputs are within $\delta$, then the outputs will be within $\epsilon$. So, for instance, if someone says that the surface of the Earth is locally equivalent to a frame of reference accelerating at 9.8 m/s^2, that means that given a point on Earth, some calculation you want to perform, and some $\epsilon$, there is some $\delta$ such that if you don't go more than $\delta$ away from that point, the calculation will be within $\epsilon$ of what you would have observed in a uniformly accelerating reference frame.


If you want a more sort of geometrical way of figuring out what "local" means, you can always compute the fermi normal coordinates for a point:


The key point here is that this coordinate system, for a given point, makes the metric tensor at that point equal to the minkowski metric, and the Christoffel symbols zero at that point, only. Then, you can pick some tolerance, and then the "local neighborhood" is the spacetime region where the largest Christoffel symbol has a value that is less than that tolerance.

A quicker procedure not involving special coordinates (but with less of a direct appeal to "similarity to flatness") is to do the same thing, but by noting that $R^{abcd}R_{abcd}$ (this is the simplest invariant that I can think of that is nonzero for every non-flat spacetime I know of) has units of inverse length to the fourth, so one over the fourth root of this will give you a rough scale for a "radius of curvature" of local spacetime, so distances less than this will be local.

  • 1
    $\begingroup$ pp-waves have not only zero Kretchmann but also the rest of scalar invariants are all zero. Look at this wiki page for references. $\endgroup$
    – A.V.S.
    Nov 23, 2020 at 17:25

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