# How can a point-like particle “feel” gravity, if locally the curvature of spacetime is always flat?

I imagine a point-like particle can only experience the local properties of spacetime. But locally there is no curvature and no gravity, as it is often stated that

Locally, as expressed in the equivalence principle, spacetime is Minkowskian, and the laws of physics exhibit local Lorentz invariance. (Wikipedia:General relativity)

But if this is the case how does a point-like particle know, which direction it should follow in a gravitational field?

Can this be a seen as a hint that point-like particles cannot be used as underlying concept for a field theory like general relativity?

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Different regions of a general spacetime that are Minkowskian to $\mathcal{O}(\Delta x^2)$ can have light cones which have null rays that point in different directions. A particle in this spacetime moves from one such region to another by connection coefficients, sometimes called the Christoffel symbols, which patch together these different locally flat regions. This patching together is what defines the geodesic equation, which in effect cancels out any measurable effect of passing from one region of flatness to another. In that way a particle falling in a gravity field is equivalent to a local frame that is globally flat. This is the equivalence principle. One can patch up a spacetime into locally flat spacetime regions any possible way, which is a matter of choosing a coordinate system to work in. This results in different connection terms which define the motion through this different coordinate choice.

An extended body “feels” gravity because different points, or small masses which compose it at various points, will tend to move on different geodesics. This is actually more physical. The deviation between two geodesics is a measure of curvature $$\frac{dU^a}{ds}~=~R^a_{bcd}V^bU^cV^d,$$ where we think of $V^b$ as tangent vectors to two geodesics and $U^a$ as a vector between them at each point. This will cause a body do distend and is what causes tides.

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+1 for writing the geodesic tide equation – lurscher May 15 '11 at 15:03
@Lawrence, Since gravity is nothing but space time curvature. and force on any object depend upon this curvature. so under zero curvature no force should be exerted on object. Please correct me if i am wrong. – Rahul kumar walia Jun 6 '14 at 12:56

"How can a point-like particle “feel” gravity"

It can't. That's in fact true not only for perfect point like particles, but also for sufficiently small objects, like... Can you feel gravity? I don't think so. What you can feel is the ground pushing against your feet, but there is no way of knowing whether this is because of gravity or because the earth accellerates.

The point is that an object does not need to know about gravity in order to obey it. It follows some path which is indeed locally indistinguishable from a zero-curvature path, but if you integrate over time the locally neglectable $\mathcal{O}(\Delta x^2)$ components will sum up to something finite, thus yielding a nonzero-curvature path.

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