# Confusion in General Relativity

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1. Oct 31, 2015

### klen

According to equivalence principle, gravity can be treated like acceleration "locally". Based on this principle we can treat a non-inertial frame at rest and explain the fictitious forces (of Newton's Laws) as gravity. From this we can prove that time elapses at different rates at different positions in gravity and that the space is curved.
But Einstein also said that the curved space-time can be treated "locally" as Minkowski space which is flat.
Doesn't these two ideas contradict each other? From the former reason we are saying that space is curved locally because of gravity and latter is saying that it is flat.
Could anyone please explain this reasoning.

2. Nov 1, 2015

### Staff: Mentor

Yes.

No. To show that spacetime (not space) is curved, we need to look beyond a single local inertial frame, i.e., beyond the scope of the equivalence principle. The EP alone does not show that spacetime is curved; curvature is a non-local effect, and as you say, the EP is local.

3. Nov 1, 2015

### Staff: Mentor

The equivalence principle says that gravitational effects behave locally like acceleration in flat spacetime. The curvature becomes apparent only when you take a non-local view.

An observer inside a sufficiently small closed room will not be able to tell whether the room is resting in the curved spacetime at the surface of the earth or being accelerated through flat spacetime at 1g - the spacetime curvature within the room is too small to detect. That's the part about treating curved space-time as locally flat and treating gravity as equivalent to acceleration in flat spacetime.

However, if we make the room large enough, non-local effects that can only be attributed to curvature will start to appear - for example, two objects dropped at the same time will follow very slightly non-parallel paths as their trajectories converge to intersect at the center of the earth.

4. Nov 1, 2015

### sweet springs

Hi. Let me tell you some cases.

1. Suppose you are in the accerelating rocket in the space. You feel gravity. You interprete there is constant gravitational field in the whole universe and it is true in your frame of reference.
Suppose you are on the earth. You feel same gravity as you are in the rocket. Your previous interpretaitin on gravity becomes wrong because the gravity has center at the center of the Earth.
"Locally" the gravitational effect on the earth is same as that of the accerelating rocket.

2. Suppose your brother floats in space and the accerelatin rokcekt passes him.
You and your brother share the same space time point. You feel gravity but your brother feel no gravity. His spacetime is Minkowsky.
Suppose you are on the Earth and your brother is free falling nearby. You and your brother share the same spacetime point again. You feel gravity but your brother does not. His "Local" spacetime is Minkowsky.

5. Nov 1, 2015

### klen

Hi, I agree if my brother is falling then his local spacetime is Minkowski but when I am standing on earth will my local spacetime be Minkowski? Because, by above the argument by Einstein, my spacetime appears to be curved.

6. Nov 1, 2015

### sweet springs

His spacetime is "Locally" Minkowsky because gravity disappears only where he is.

No. your spacetime is curved thus you get gravity. If you want to straighten the spacetime around you jump off following your brother.

7. Nov 1, 2015

### Staff: Mentor

Either spacetime is curved at a given point or it isn't; it can't be curved for one person and not curved for another. A person standing on the surface of the earth is accelerating at 1g (we measure this by asking him to hold an accelerometer, or to stand on a spring scale so that we can directly observe the force that is accelerating him) through a locally flat spacetime. The nearby falling person is moving inertially (no force acting on him as long as he remains in free fall) through the same locally flat spacetime.

Of course this local region of spacetime is not exactly flat - it is very slightly curved, and with sufficiently sensitive instruments both observers would be able to measure this very slight curvature. Their measurements would agree, as the curvature is what it is no matter how the person doing the measurement is moving.
This is incorrect. It's the same slightly curved locally flat spacetime for both.

Last edited: Nov 1, 2015
8. Nov 1, 2015

### sweet springs

Yes. The free faller should be infinitesimal small dwarf to escape from your criticism.

9. Nov 1, 2015

### A.T.

Gravitational acceleration and different clock rates doesn't imply space time curvature. See the intrinsically flat cone the first figure here:

http://www.physics.ucla.edu/demoweb..._and_general_relativity/curved_spacetime.html

Also this great post:

10. Nov 1, 2015

### loislane

Riemannian curvature is mathematically a local property. So what is non-local about it according to you? If possible give some reference.
The EP must limit arbitrarily to first derivatives of the metric, but how dos this make curvature something non-local?

11. Nov 1, 2015

### Staff: Mentor

When we say that spacetime is locally flat and that curvature is non-local, we're saying that the approximation "it is flat across this region" can be made arbitrarily good by choosing the region in question to be sufficiently small. No matter how small a curvature effect you can detect, there is a region small enough that you can't detect it inside that region.

Conversely, to observe any curvature effect we must consider the trajectories of two test particles that are not colocated, and then we can interpret "non-local" to mean "at least as separated as our two non-colocated particles".

12. Nov 1, 2015

### Staff: Mentor

I think you may still be missing the point. Yes, you are right that if we were to shrink both observers down to infinitesimal dwarfs, the "it's locally flat" approximation would become exact in the infinitesimally small locality around each observer. But I was objecting to the suggestion that "If you want to straighten the spacetime around you jump off following your brother", because that's just plain wrong. The spacetime is what it is whether you're accelerating through it or in free fall through it.

13. Nov 1, 2015

### Staff: Mentor

That is explained in detail in chapters 3 and 4 here
http://www.preposterousuniverse.com/grnotes/

14. Nov 1, 2015

### sweet springs

Thanks. Will restating "If you want to change your frame of reference so that it is locally flat where you (an infinitesimal dwarf) are, " satisfy you ?

15. Nov 2, 2015

### A.T.

@loislane
As an analogy: The Riemannian curvature of a spherical planet surface is the same at each point. But the effect it has on the accuracy of a flat map depends on the size of the mapped area. Over small areas (locally) the distortion due to curvature is negligible.

Last edited: Nov 2, 2015
16. Nov 2, 2015

### loislane

You are restating that the EP arbitrarily restricts itself to the first derivatives of the metric(non-tensorial coordinate-dependent Christoffel coefficients) for that approximation. It might be clarifying to know the reason why a physical principle is restricted to a coordinate approximation when physicists are so concerned about considering physical only what is coordinate-independent(invariants).

I guess you are referring here to geodesic deviation that uses the Riemannian tensor in its formulation and therefore a local notion o curvature. Riemannian curvature is preserved by local isometries.

17. Nov 2, 2015

### Staff: Mentor

The curvature tensor is covariant, so scalars derived from it that measure "the amount of spacetime curvature" (the Ricci scalar is the simplest one) are coordinate-independent. The EP can just as easily be stated as applying only over length and time scales that are small compared to the radius of curvature derived from such scalars. That is a coordinate-independent statement.

18. Nov 2, 2015

### loislane

Sure.

But here you are introducing the second derivatives of the metric(derivatives of the Christoffels) since you are introducing curvature . These are explicitly left out in the EP, that is only concerned with the first derivatives of the metric, that vnish at the point by using Riemann normal coordinates(and you need this local vanishing or you don't have any equivalence principle). So you can't use a local measure of curvature here to state a principle about equivalence to local flatness by arguing that curvature is coordinate-independent. Curvature can never be equivalent to flatness, that's precisely the reason the EP must be stated in terms of only first derivtives and Riemann normal coordinates.

Last edited: Nov 2, 2015
19. Nov 2, 2015

### loislane

page 108 of chapter 4 indeed explains what I'm saying about the EP starting by:"
In fact, let us be honest about the principle of equivalence: it serves as a useful guideline,
but it does not deserve to be treated as a fundamental principle of nature. From the modern
point of view, we do not expect the EEP to be rigorously true".
But it doesn't give any alternative other than wait and see if the principle is empiracally ruled out in practice(explaining why this would be quite hard in fact).

20. Nov 2, 2015

### Staff: Mentor

If what you are saying is consistent with what Carroll is saying then I don't think that there is any real disagreement here.

I am not sure why you think some alternative is needed. We have a theory which makes testable predictions. The theory reflects the principle meaning that it follows the "useful guideline". So what more is needed?