Expert: Philip Stahl - 10/22/2007

Question

Hi Philip
Need few clarification about gravity. If we assume to be in a freely falling elevator, we experience weightlessness ie, zero gravity. Now assume that we are travelling in a spaceship, obviously there is no gravity there and we will be expriencing weightlessness. If the spaceship is accelarated at 9.8m/s^2, we will experience the same
effect as on earth, i mean the gravitation. My question is, Is the gravity a result of accelaration. Our universe is constantly expanding and is gravity a result of this acccelarating universe. If the expansion of our universe stops, what will happen to the graviational effect on a planet ? Einstein defined gravity as a result of warping of Space-Time continum, ie, the heavy mass object such as SUN tends to warp the Space-time and the smaller mass object will revolve around the heavier objects in a geodesic curve(ie, Orbit). How the elevator example (free fall) differs from this Warping of Space-Time.
Rather how is Gravity defined ?

Thanking you in expectation.

Answer
Hello,

Your questions are as follows:

i) Is the gravity a result of accelaration?

The gravity experienced in the spaceship is a result of the reaction to a pseudo-force generated (in this case, perhaps a rotation of the ship in order to emulate Earth gravity)

More generally, astronauts in any craft orbiting Earth say, can detect any general gravitational field F_G by the tidal effects it produces.

Thus for a liquid drop in space (or absence of external forces) a "double bulge" indicates the existence of a gravity field. (One "bulge" toward the Earth, the other away from it)

In certain cases, clearly, the (inertial) force generated by a local gravity field, say, will be indistinguishable from a force generated via acceleration. We refer here to the "Principle of Equivalence" postulated by Einstein in conjunction with his General Relativity theory.

So, whether an astronaut is being accelerated at some rate, a, or is in a planet’s gravitational field with some rate g, is neither here nor there - assuming a = g. He feels the same “weight” in each case.

To illustrate:  an astronaut of mass 100 kg will have a mass of 109 kg, when traveling at 40% of the speed of light. If he finds himself in a gravity field with rate of acceleration g (= 9.8 m/S^2), then his weight will be:

W = mg = (109 kg) (9.8 m/s^2) =  1068 N (newtons)

If, on the other hand, he’s in a tall rocket traveling at the same speed, in NO gravity field, but inside an elevator (in the rocket)  moving at an acceleration upwards of a = g, then:

W = ma = mg = 1068 N  (e.g. the same weight).


What all of this means is that in any unaccelerated ship and in the absence of gravity fields, the astronaut would be unable to tell if his weight was changing. However, in a ship undergoing acceleration (or deceleration) from a gravity field – OR from some external force – he would.



(ii) If the expansion of our universe stops, what will happen to the graviational effect on a planet ?

Nothing. Because the planet's gravitational effects are predicated on a local INERTIAL reference frame which the universe at large doesn't share. Hence, for all intents and purposes - you - on some planet or other -would never know IF or WHEN the expansion of the universe halted, since you are not in that inertial reference frame. Rather you are occupying a much smaller, more local one.


iii) Einstein defined gravity as a result of warping of Space-Time continum, ie, the heavy mass object such as SUN tends to warp the Space-time and the smaller mass object will revolve around the heavier objects in a geodesic curve(ie, Orbit). How the elevator example (free fall) differs from this Warping of Space-Time.
Rather how is Gravity defined ?

You are talking chalk and cheese here. Einstein's geodesic (least-time path in a Riemannian space-time) pertains to what occurs in the vicinity of massive gravitating bodies. Thus, light from a distant star passing near the Sun (in the line of sight will curve (be deflected) to follow the geodesic path defined by warping of space-time in the Sun's vicinity.

In a terrestrial elevator in free fall, by contrast, you are merely experiencing a pseudo-weightlessness because of the lack of any *reaction* force to your weight.

The net force may be described:  F = (ma - mg)

but if a = g

F = 0

Thus, if the elevator's downward acceleration can be made to match the surface gravity (g = 9.8 m/s/s) then there can be no back reaction force and hence you will experience no weight. And, of course, there is no "warping" of the fabric of space-time in the elevator!

From this example, gravity in the general (cosmic) sense is defined by the extent of warping of space-time itself with reference to a particular mass. Gravity in the more general terrestrial (operational) sense, will depend on what the particular values of a, g are for any intertial reference frames.

Lastly, current gravity theories also incorporate gravitational waves, and postulate a hypothetical "carrier" particle to mediate gravity. We believe this carrier is the "graviton".  For more on this, see:

http://en.wikipedia.org/wiki/Graviton