Question: Does and/or Can Gravity exist indepent of objects?

[K^2]

No, it doesn't work that way. Light doesn't travel in the straight path near black hole. Neither will gravitons, if you want to look at gravity that way. Basically, gravitational lensing will make sure that you are not shielded from anything "on the other side".

 

[cragar]

ok so the G field gets bent around the black hole , as well as light .

 

[K^2]

I wouldn't say it curves the G-field, because curvature IS the G-field. But if you wanted to consider a small linear correction due to an additional mass near black hole, and you treated it as classical gravity, you certainly need to account for overall curvature as well.

 

[Antiphon]

Been reading. What you're saying is consistent with every source I managed to get my hands on, so you're probably right.

Now I'm just trying to figure out what the hell is going on with that.

So, by definition, the energy-stress tensor is given by:

\Delta p^\mu = T^{\mu \nu} n_\nu \Delta V

Where \Delta V is the 3-volume element of hypersurface with normal n_\nu.

Alright. I have choice of coordinate system. I can choose one where p={p₀,0,0,0}. In that case, regardless of choice for n and ΔV, Δp will also have only the 0th component. And that would imply that T^ij=0.

Do you see a flaw in that logic? There probably is one, I just can't spot it.

I think what you've written is a source consisting of a mass density p. The time coordinate is energy density so you must have a "c" multiplying coordinate 0, this I think is usually the case. It looks like you've written T for a "cosmic dust". If your mass were a cube of water then there would be some diagonal terms corresponding to the pressure. *The pressure would originate from surface tension for a small blob but for a planet-sized blob of water the large T00 term would mean there would be a larger pressure originating from the water gravitating itself inward. *A piece of glass on the other hand would have all the off-diagonal shear terms, etc. *For everyday matter the T00 term dominates.*

Your equation looks correct to me. *When the volume element and normal is there you are (I think) essentially integrating the stress tensor over a small volume element. The T00 term should become mass-energy and the diagonal Tij should become net force x distance (units of energy, good), and the off diagonal Tij should become net torque (still units of energy I believe).*

*

 

[Antiphon]

(no idea what asterisks are from in my above post. I use Notes on the iPhone as an editor then paste into the little window. Apologies for the eyesore.)

 

[Antiphon]

so when you say gravity are you talking about the field , is the graviton the excitation of the field , is a gravitational wave a self-sustaining gravitational field and this might be the graviton , is it possible to turn all of the visible universe into gravitons .

The gravitation field when can carry energy ex. gravitational waves. But I can't envision a means of converting all the energy of the universe into just gravity.

There is a facet of the gravitational field I will bring up shortly in a new thread having to do with gravitational potential energy. There is something I don't understand about it. Frankly the forces point the wrong way. I must be making an elementary mistake. Maybe K^2 you will spot it. I will title the thread "Potential energy of gravity".

 

[K^2]

No, this is the actual definition of Stress Energy Tensor. It has to work for general case. For a collection of particles (what your referred to as cosmic dust), it simplifies to this.

T^{\mu\nu} = p^{\mu}u^{\nu}

And then, if the particles are at rest, this is just T⁰⁰ = E and all other terms are zero.

With a solid object under mechanical stress, even when the object is at rest, apparently you get non-zero terms for the stress portion of the tensor. That's the part I don't completely understand.