Subject: | OT: amazing, 1/4 of Americans think sun orbits the earth Date: | 08 Mar 2014 Message-ID: | carlip-502C66.17124408032014@news.eternal-september.org

Here
Richard Norman states:

>>> Gravity is a fictitious force dependent on the coordinate system. That >>> is an important central point of general relativity.

A disoriented participant in the discussion requests:

>> Cite?

Resulting in a quote from wikipedia:

> http://en.wikipedia.org/wiki/Fictitious_force#Gravity_as_a_fictitious_force
which then said:

The notion of "fictitious force" comes up in general relativity. All fictitious forces are proportional to the mass of the object upon which they act, which is also true for gravity. This led Albert Einstein to wonder whether gravity was a fictitious force as well. He noted that a freefalling observer in a closed box would not be able to detect the force of gravity; hence, freefalling reference frames are equivalent to an inertial reference frame (the equivalence principle). Following up on this insight, Einstein was able to formulate a theory with gravity as a fictitious force; attributing the apparent acceleration of gravity to the curvature of spacetime. This idea underlies Einstein's theory of general relativity.

Here Dr Carlip's Post of the Month begins:

This is not quite right, and I think it may be causing confusion. There are really two
separate issues -- is gravity a fictitious force, and is gravity a force at all?
First: the standard definition of a fictitious force is one that can be eliminated by choosing a different coordinate system. The principle of equivalence implies that gravity can be considered *locally* to be a fictitious force. "Locally" has a technical meaning, but roughly, it means, "If you give me a measuring method with a given precision, I can find a region small enough in space and time that the effects of gravity can be removed to that precision by a coordinate choice." That choice is, basically, a freely falling reference frame; since locally everything falls at the same rate, you can't detect any acceleration in that frame.

The "locally" is important, though. At larger scales, gravity is certainly not a fictitious force. As soon as you are looking at a region big enough that two nearby objects experience different accelerations, there is no single freely falling reference frame, and you can't treat gravity as a fictitious force.

(Think about holding two rocks above the ground, one higher than the other, and letting go of both at once. The lower rock is closer to the center of the Earth than the higher, so it will accelerate slightly more, and the distance between the rocks will increase. Or start with two rocks at the same height, but not too close together. Both will fall toward the center of the Earth, and since the Earth is round, the distance between the rocks will decrease. In both cases, this relative change in distance won't go away with any choice of coordinates.)

These relative changes are called "tidal gravity," and they are definitely not fictitious forces. Technically, these relative accelerations are determined by the curvature tensor, which cannot be transformed to zero in any coordinate system.

Second, the question of whether gravity is a "force" at all is at least partly a matter of semantics. In general relativity, mass (and energy) causes spacetime to curve, and objects under the influence of gravity follow the straightest possible paths in this curved spacetime. In this description, it's natural to say that there is no force acting on an object in free fall -- it's following the straightest path it can -- and that it's only when something is stopped from falling that there is a force.

(Note that the "time" in "spacetime" matters here. Any object, even if it's at rest in space, is following a path in spacetime. For a rock that's held above the ground and then dropped, the straightest path in spacetime is not the one where it only moves in time -- i.e., floats above the ground -- but rather the one where it moves in space as well as time -- i.e., falls. If you're sitting in a chair, the straightest path in spacetime is one in which you fall toward the center of the Earth. The chair, and the floor and the ground beneath it, prevent you from following this path, causing you to accelerate, where "accelerate" just means "not move along the straightest path." So when you're sitting still, you are actually accelerating, which is why your butt eventually gets sore.)

But you can also ask about relative accelerations of nearby objects, the accelerations described by tidal gravity. The equation that describes these -- called the geodesic deviation equation -- looks very much like F=ma, where the "force" is a quantity that depends on the spacetime curvature. It's not quite as simple as a Newtonian force (technically, it's a sort of velocity-dependent force, and it also depends on the relative separation), but it's very close. So depending on exactly how you phrase the question, it might be perfectly fine to describe gravity as a force.

I think what most people who work in this field would say is that if you understand general relativity, it doesn't matter too much what terminology you use; and if you don't, I guess it doesn't matter too much, either, except that various words have extra connotations that may not apply.

Steve Carlip

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