# Space-Time Curvature in General Relativity

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1. Nov 2, 2015

### klen

Suppose we are in a Minkowskian space, away from all the source of gravity, and observe an accelerated frame from this frame. Acoording to Equivalence principle, we can consider the accelerated frame to be at rest and assume we have gravity in the accelerated frame. Thus, observer in the accelerated frame will observe that his space-time is curved.
What will this observer in the accelerated frame attribute this space-time curvature to? In other words, since the curvature has to be created by a mass, what is causing the curvature in his space?

2. Nov 2, 2015

### A.T.

No. As explained in the other thread, an accelerated frame doesn't imply space time curvature. See case B in the below diagram:

3. Nov 2, 2015

### klen

Thanks A.T. and sorry I missed this idea in my previous thread.
I believe you are saying that the accelerated frame which I was talking about in the question has geometry represented by B-1 or B-2. Is that correct?
Further, I do not understand why is there no difference between B-1 and B-2 in terms of geometry, why is the space B-2 not curved?

4. Nov 2, 2015

### A.T.

Yes. It also approximates what you see locally in a hovering frame in an actual gravitational field. "Locally" means over a small space where tidal effects (gravity gradient) are negligible.

Because the distances between coordinates (measured intrinsically along the surface) are the same in both cases B1 and B2. If you can flatten out a surface, without distorting the distances, then it didn't have any intrinsic curvature in the first place. B2 merely has extrinsic curvature within the embedding space of the illustration, which doesn't have any physical significance. Only the the intrinsic geometry matters in this analogy.

Here is an animation which shows how the rolling doesn't change anything about the intrinsic geometry:

Case C has intrinsic curvature which corresponds to tidal effects (the free fallers diverge). It cannot be flatten out without distorting the intrinsic distances.

5. Nov 2, 2015

### klen

Hi A.T.,
If the space-time for the accelerating observer is flat, then does it mean that to explain the physical phenomenon around him he is constrained to use curvilinear coordinates? Because if he uses euclidean coordinates then he would observe acceleration of falling objects?
Does it mean that Equivalence Principle is just another way of looking at space-time?

6. Nov 2, 2015

### A.T.

Yes. In Newtonian terms he could also use the concept of inertial forces. But in the geometric model of GR accounting for non-inertial frames is done with coordinates.

7. Nov 2, 2015

### Staff: Mentor

He is constrained to use coordinates in which his position does not change with time. Whether they are curvilinear or not doesn't matter.

We often do choose such coordinates without even thinking about it; any time that I say that I am "standing on the earth's surface" I am assuming a coordinate system in which my position is not changing. These coordinates may be Euclidean/Cartesian ("three meters from the east wall of the room, two meters from the north wall, standing on a ladder one meter above the floor") or curvilinear ("latitude 45 degrees north, longitude 80 degrees west, two meters above the earth's surface").

But once I have adopted coordinates in which I am at rest on the surface of the earth I am faced with two anomalous observations. One is that the ground is exerting an upwards force (of $mg$ where $m$ is my mass and $g$ is 9.8 m/sec^2) on me, yet I am not accelerating; the first derivative (velocity) and the second derivative (acceleration) of my position with respect to time are both zero. Second, an object in free fall is accelerating; its position is changing and the second derivative of its position is non-zero and equal to $g$.

I can explain these anomalous accelerations by inventing a new force (called "gravity") that acts on all objects near me, always pointing down and always of magnitude $mg$. That force exactly cancels the upwards force the ground exerts on me, and it accounts for the downwards acceleration of the falling object. However, this force appears only because we chose a coordinate system in which the observer on the earth's surface is at rest - had we adopted coordinates in which the falling object were at rest (or moving in a straight line at a constant velocity) we wouldn't need this force to explain the observed accelerations.

Or as another way of looking at gravitational forces.

8. Nov 2, 2015

### klen

Thanks Nugatory for clarifying this thing....

9. Nov 2, 2015

### klen

One related question:
If we are saying that the geometry in the accelerated frame is flat, does it mean that the Euclidean geometry still holds?
If so, then consider an accelerated frame which is rotating with respect to the inertial frame, why the geometry of this frame not Euclidean?

10. Nov 2, 2015

### Staff: Mentor

Actually, even flat spacetime is not Euclidean, only its three-dimensional spatial subspace is. To see this, consider how we calculate the distance between two points at the same time: $\Delta{s}^2=\Delta{x}^2+\Delta{y}^2+\Delta{z}^2-\Delta{t}^2=\Delta{x}^2+\Delta{y}^2+\Delta{z}^2$ which is just the familiar Pythagorean theorem of Euclidean geometry. However, if we drop the "at the same time" condition so that $\Delta{t}$ is non-zero, the minus sign starts to matter and we find that the four-dimensional spacetime is Minkowski not Euclidean.

It's still the same four-dimensional spacetime described by the Minkowski metric (the "metric tensor" is the mathematical object that describes the geometry of a given spacetime in a coordinate-independent way). However, when you choose a frame, you are really choosing the coordinates you'll use to label events in spacetime.... And the coordinates that come with a rotating frame seriously obscure the underlying geometry and make easy calculations difficult:
1) Assigning time coordinates to events away from the origin is tricky; there's no non-arbitrary way to do it.
2) Differences between the values of coordinates (the $\Delta$ quantities I used above) have no natural meaning, For an example, consider that if I am spinning at the rather mild and nonrelativistic rate of once a minute, anything more than six million kilometers away from me will be moving faster than light in the rotating frame in which I am at rest - and we know that's a not a good description of what's going on.
However, after you've backed out all the stuff that comes from the choice of coordinates, we find that the spacetime itself is still flat and Minkowski.