Gravitational Waves and Potential

[Naty1]

What's the relationship between gravitational waves and (traditional/stationary) gravitational potential?

Does the (traditional/stationary) potential become a "wave" under acceleration? Or do both exist under acceleration? Do both use energy from the source? I think I saw somewhere that GR predicts gravitational waves...true? Is this related in any way to the Unruh (acceleration) effect? Are these effects (waves, Unruh) evidence of the "absolute" nature of acceleration?

(http://en.wikipedia.org/wiki/Bill_Unruh)

Wiki says:
http://en.wikipedia.org/wiki/Gravitational_Radiation

Distances between objects will increase and decrease rhythmically as the wave passes.

Gravitational waves can penetrate regions that the more familiar waves cannot, providing us with a view of black holes and other mysterious objects in the distant Universe.

In general terms, gravitational waves are radiated by objects whose motion involves acceleration

gravitational waves can pass through any intervening matter without being scattered. Whereas light from distant stars may be blocked out by interstellar dust, for example, gravitational waves will pass through unimpeded

(gravitational) waves are able to rob that source of its energy

 

[tiny-tim]

Hi Naty1!

Here's another wiki link (about the decomposition of the Riemann curvature tensor into the Ricci tensor and the Weyl tensor) which may help: http://en.wikipedia.org/wiki/Ricci_decomposition#Physical_interpretation

The Ricci decomposition can be interpreted physically in Einstein's theory of general relativity, where it is sometimes called the Géhéniau-Debever decomposition. In this theory, the Einstein field equation

G^ab = 8πT^ab​

where T^ab is the stress-energy tensor describing the amount and motion of all matter and all nongravitational field energy and momentum, states that the Ricci tensor-- or equivalently, the Einstein tensor-- represents that part of the gravitational field which is due to the immediate presence of nongravitational energy and momentum.

The Weyl tensor represents the part of the gravitational field which can propagate as a gravitational wave through a region containing no matter or nongravitational fields.
Regions of spacetime in which the Weyl tensor vanishes contain no gravitational radiation and are also conformally flat, which implies for example that light rays passing through such a region exhibit no light bending.

EDIT: and this link from John Baez's website:

As with many things in general relativity, it can help to state the corresponding fact about electricity and magnetism. If you know the charge and current distributions everywhere in space, you might think that that would let you figure out the electric and magnetic fields everywhere. But it doesn't. There's extra information in the fields beyond just what the sources of the fields can tell you. After all, you could have an electromagnetic wave passing by. It needn't have any source, but it still alters the fields.

So in electromagnetism, knowing all about the sources isn't enough to specify the fields. In general relativity, knowing all about the sources (the stress-energy tensor T) isn't enough to tell you all about the curvature. In both cases, you can supplement the source information with some extra initial conditions to get a unique solution. (For example, in electromagnetism you can specify that no electromagnetic waves are zooming in from infinity. That's enough to give you a unique solution to the fields given the sources. For general relativity, you can perform similar feats, although it's technically trickier.)

Anyway, I hope that makes it a little bit clearer why people say that the Weyl part of the curvature has to do with gravitational radiation: the Weyl tensor carries information about the kind of curvature that's independent of the source distribution, sort of like electromagnetic waves are fields that propagate independently of whatever sources are around.

 

[Naty1]

Tiny-tim...great descriptions to get me started...thanks..

 

[Antenna Guy]

from John Baez's website:

...After all, you could have an electromagnetic wave passing by. It needn't have any source, but it still alters the fields.

So in electromagnetism, knowing all about the sources isn't enough to specify the fields...

An electromagnetic wave that needn't have any source?

How does one rationalize something like that?

Regards,

Bill

 

[spidey]

will an accelerated observer also emits gravitational waves?

 

[Naty1]

An electromagnetic wave that needn't have any source?

How does one rationalize something like that?

It's a wave originating from other than the source description reflected in the Ricci tensor. Like an initial condition.

 

[Naty1]

will an accelerated observer also emits gravitational waves?

Sure. Although extremely,extremely small as even massive bodies emit only small amounts of gravitational radiation (power).

And an accelerating observer, in the absence of external gravity, also observes curved space time via the equivalence principle (I think). I think this is why special relativity (no gravity) requires inertial frames to keep space time observations flat.

I still not quite sure whether an accelerating observer measures local light at "c". I believe I read there is a theory that basically says instantaneous readings while accelerating measure "c". These would apparently be inertial observers.
But there are so many references which say "free falling observers" (non accelerating) in general relativity measure light at "c" that I am still suspicious...it may be a matter of context. I know Einstein came close to relying exclusively on free falling (local) frames and there had to be a good reason.

 

[tiny-tim]

Weyl curvature tensor is not related to sources

It's a wave originating from other than the source description reflected in the Ricci tensor. Like an initial condition.

Yup!

The source tensor (the stress-energy tensor) only has ten independent parameters,

but the Riemann curvature tensor has twenty …

Einstein's field equations ("matter tells space how to curve") only relate the source tensor to the Ricci tensor, which of course also has ten parameters (natch ) …

the remaining ten parameters of the Riemann curvature tensor (in other words, the ten parameters of the Weyl tensor) are entirely independent of the source tensor, and therefore independent of all sources (assuming Einstein is right! )

 

[DrGreg]

And an accelerating observer, in the absence of external gravity, also observes curved space time via the equivalence principle (I think). I think this is why special relativity (no gravity) requires inertial frames to keep space time observations flat.

I'd just like to remind you of this reply I gave you in another thread. "Curvature of spacetime" is an intrinsic property of spacetime and does not depend on the observer. In the absence of gravity, spacetime is always "flat" whether you are an inertial observer or not. It's just that (to use the analogy of that other post) non-inertial observers draw a curved grid on flat graph paper.

Non-inertial observers in the absence of gravity do indeed see some (but not all) of the effects that "stationary" observers see near black holes, but you shouldn't attribute those effects to "spacetime curvature" -- attribute them to a non-inertial frame instead.

This is more a question of the established terminology than anything else.

(For the benefit of other readers: as for your last paragraph, we are currently discussing that in another thread.)

 

[Naty1]

I just read your "this reply"...which was # 11 on the other thread ...don't believe I saw that ...excellent description and likely the source of my confusion...

I'll think about it and post tomorrow...