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

[russ_watters]

But what about with the ship that's 100's of thousands of km from earth? or if an its on its way to mars? The astronauts will still float around right?

If the spaceship's engines are not running, the spaceship is in freefall, so the astronauts will "float". This has *nothing whatsoever* to do with how strong of a gravitational field exists in an area. They could be seconds from crossing into a black hole and still be "floating" inside their spaceship.

 

[Antiphon]

And stress, in this case, just contributes as a form of potential energy. The name stress-energy tensor is derived from its similarity to stress tensor from classical statics. Other than the name, it has nothing actually to do with mechanical stress.

I agree with this except the last point. Mechanical stress does gravitate. In truth since all mechanical stress (including nuclear stress) is a manifestation of stressed fields, these will all contribute to the stress energy tensor and thus gravity.

 

[K^2]

Not at all. The stress in a solid body will only contribute to the $T_{00}$ component. The other 15 components will depend on flux of 4-momentum, which has nothing to do with mechanical stress.

Now, if you are talking about stress in the liquid flow, yes, that is related, but I still wouldn't call it a contribution.

 

[cragar]

You can have gravity with nothing in the space but gravity.

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 .

 

[Antiphon]

Not at all. The stress in a solid body will only contribute to the $T_{00}$ component. The other 15 components will depend on flux of 4-momentum, which has nothing to do with mechanical stress.

Now, if you are talking about stress in the liquid flow, yes, that is related, but I still wouldn't call it a contribution.

Why do you say that? Take the simplest case of a uniform pressure. The stress tensor would have T11=T22=T33 and this definitely contributes as a gravitationl source.

 

[K^2]

How is that? The stress-energy tensor is flux of energy/momentum. You don't have energy or momentum flow if you have stress in solid body.

T11, for example is flux of e1 component of momentum in e1 direction. If nothing is moving in e1 direction, you don't have a flux. Not to mention that momentum is uniformly zero.

 

[Antiphon]

Here's my understanding of it: (0-time, 1,2,3 space)
T00 energy density
Tx0 momentum density
T0x energy flux
Txx (diagonal space terms) pressure
Txy (off diagonal space term) shear

In the above, xy is any of 1,2,3 for the space terms.

I'm pretty sure the pressure and shear terms gravitate. A stellar physicist could confirm it but I believe the pressure I inside a massive star contributes a significant term to stellar gravity under extreme conditions.

 

[K^2]

It's hydrodynamic dynamic pressure and shear. Not static stress.

 

[Antiphon]

Did you mean dynamic twice? In any case you don't need dynamic forces to source gravity. Remember a static stress is a momentum flow viewed from a moving reference frame.

 

[K^2]

Did you mean dynamic twice? In any case you don't need dynamic forces to source gravity. Remember a static stress is a momentum flow viewed from a moving reference frame.

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.

 

[cragar]

Black hole's gravity all comes from mass that's still above event horizon. Things that are still falling in. Of course, to an outside observer, falling in takes an infinite amount of time. So anything that has fallen into black hole is still falling. Still above event horizon. That even includes the star that initially collapsed to produce the black hole. It's still there, just outside event horizon, getting closer and closer but not quite getting there. It's almost infinitely red-shifted by now, of course, so there won't be much in terms of EM radiation escaping, but it can still interact gravitationally and electromagnetically with the stuff outside.

So you are saying that the energy inside the black hole does not contribute to pulling the object in , so what happens when the object is near the event horizon on the safe side , the energy that pulled it there is now on the other side so why wouldn't it pull it back , or does it have enough kinetic energy to keep going ,

 

[K^2]

Where would such an object come from, though? For something to fall into black hole completely, it would have to be formed infinitely long ago. But universe has a finite age. So all of the black holes still have their mass completely outside their event horizons, as far as the rest of the universe is concerned.

 

[cragar]

i guess the mass from the other side of the black hole would pull it in , how do you know the age of the universe is finite .

 

[K^2]

It has to be finite if Big Bang theory has any truth to it, and it's a rather solid one. The estimates on the age could be quite a bit off, but I do not see how it could be infinite.

What "other side" of a black hole? There is nothing on the "other side". There is an event horizon. All the mass is above event horizon. That's your black hole as seen from outside.

 

[cragar]

the mass from the other side would not pull you in because its field would have to travel through the center of the black hole and could not escape the event horizon to reach you , by other side i meant through the black hole and out past the other side of the event horizon . like I am on the tire then the rim then back to the tire . So you don't believe in steady state theory then , so are you saying that time was created at the big bang ,
what came first time or space ,

 

[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.