Is a geodesic only 'frictionless' in a flat space? gravity 'steals energy' of objects, right? Two heavenly bodies circling each other loses energy to their interacting gravitationally, so what about one body, moving through a curved space? Will that one also lose energy?

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It's been bugging me for a while, the definition of a 'frictionless gravity'. If I would define gravity as a 'force', it seems to me that I also have to assume it acting on bodies following geodesics through a curved space.

And if that would be true then it seems to me, assuming a straightforward propagation of light, that this also should be relevant for light paths, as we have the equivalence between energy and mass? As I think of it it shouldn't matter for the definition of a geodesic? If space is 'curved' (gravity), or not (flat space). But thinking of binary stars interacting, gravity seem to bleed away 'energy' from them, described as a system?

"Einstein's general theory of relativity explains gravity as a consequence of the curvature of spacetime created by the presence of mass and energy. As two stars orbit each other, gravitational waves are emitted - wrinkles moving out in spacetime. As a result, the binary slowly loses energy, the stars move closer, and the orbital period shortens."

http://www.spacedaily.com/reports/Bizarre_binary_star_system_pushes_study_of_relativity_to_new_limits_999.html=

Ok, the mass quadrupole moment, is that what explains why binary stars is expected to lose energy, making waves, and why a sole body in a curved space, won't? (But if the universe is a 'container of it all', why can't I define that, or a gravitational field, relative a sole body's geodesic in a curved space too?)

"An object's gravitational monopole is just the total amount of its mass.

An object's gravitational dipole is a measure of how much that mass is distributed away from some center in some direction. It's a vector, since it had to convey not only how much the mass is off-center but also which way. Considering some object in the abstract, the natural 'center' to pick is the center of mass, which is the point around which the dipole is zero.

The quadrupole represents how stretched-out along some axis the mass is. A sphere has zero quadrupole. A rod has a quadrupole. A flat disk also has a quadrupole, with the opposite sign of the quadrupole of a rod pointing out from its flat sides. The rod is a sphere stretched along that axis and the disk is a sphere squashed along that axis. In general, objects can have quadrupole moments along three different axes at right angles to each other. (The quadrupole moment is something called a tensor.) ...

... for gravitational radiation you need an oscillating quadrupole moment. The difference is that electric charge comes in two varieties of charge, plus and minus. When you interchange the two charges, as in an oscillating electric dipole, you get a change in the electric field distribution. Gravitational mass, on the other hand, comes in only one sign: plus. There are no minus values. So if you interchange two masses you don't get a change in the gravitational field. Hence, no dipole radiation. "

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On the other tentacle, the universe seem nearly 'flat', as I've seen it defined.