Is mass the source of spacetime?

[PeterDonis]

they write, that a volume of a small ball of testparticles will decrease in time. And that's the basic meaning of "gravity attracts".

More precisely, that the normalized "acceleration" of the volume--the second derivative of the volume with respect to time, divided by the volume--is negative. That means the volume, if it starts from "rest" (zero rate of change with time), will decrease at a rate that increases with time.

If I the mass is increased, will the volume increase or decrease because of the additional mass?

The Ricci tensor does not describe "mass". It describes density of stress-energy. For the simple case of a perfect fluid, the Ricci tensor--more precisely, the piece of it that determines the (negative) "acceleration" of the volume described above--will be $\rho + 3 p$, where $\rho$ is the energy density of the fluid and $p$ is the pressure. So positive #\rho + 3p## means the "acceleration" of the volume is negative--"attractive" gravity. Note that this is for an observer who is inside the fluid, observing the behavior of a small ball of test particles that is also inside the fluid.

can the metric tensor help? diag(B, -A, -r², -r²sin(theta))

Where is this metric tensor coming from?

 

[Angelika10]

The Ricci tensor does not describe "mass". It describes density of stress-energy. For the simple case of a perfect fluid, the Ricci tensor--more precisely, the piece of it that determines the (negative) "acceleration" of the volume described above--will be $\rho + 3 p$, where $\rho$ is the energy density of the fluid and $p$ is the pressure. So positive #\rho + 3p## means the "acceleration" of the volume is negative--"attractive" gravity. Note that this is for an observer who is inside the fluid, observing the behavior of a small ball of test particles that is also inside the fluid.

Hm, ok. I understand this. But the question is: if the energy-momentum-tensor is increased, what happens with the volume?

Where is this metric tensor coming from?

From the standard form of a static, spherically symetric metric.

$ds^2 = B(r)c^2dt^2 -A(r)dr^2 - r^2(d\theta^2+sin^2 \theta d\phi^2)$

so

$(g_{\mu\nu}) = diag (B(r), -A(r), -r^2, -r^2 sin^2\theta)$

 

[PeterDonis]

if the energy-momentum-tensor is increased

What does it even mean to "increase" a tensor? A tensor is not a number.

From the standard form of a static, spherically symetric metric.

A non-vacuum spacetime (i.e., a spacetime with a nonzero Ricci tensor, which requires a nonzero stress-energy tensor) will not necessarily be either static or spherically symmetric.

 

[vanhees71]

More precisely, that the normalized "acceleration" of the volume--the second derivative of the volume with respect to time, divided by the volume--is negative. That means the volume, if it starts from "rest" (zero rate of change with time), will decrease at a rate that increases with time.The Ricci tensor does not describe "mass". It describes density of stress-energy. For the simple case of a perfect fluid, the Ricci tensor--more precisely, the piece of it that determines the (negative) "acceleration" of the volume described above--will be $\rho + 3 p$, where $\rho$ is the energy density of the fluid and $p$ is the pressure. So positive #\rho + 3p## means the "acceleration" of the volume is negative--"attractive" gravity. Note that this is for an observer who is inside the fluid, observing the behavior of a small ball of test particles that is also inside the fluid.

But this is not the Ricci but the energy-momentum tensor, $T_{\mu \nu}=(\epsilon+P) u^{\mu} u^{\nu}-P g^{\mu \nu}$ (for an ideal fluid). The source of the gravitational field (or equivalently space-time curvature) is the energy-momentum-stress of the "matter", as described by the Einstein field equations,
$$R_{\mu \nu}-\frac{1}{2} R g_{\mu \nu} = \kappa T_{\mu \nu}.$$
Here $R_{\mu \nu}$ is the Ricci tensor, $R$ its trace, and $\kappa=8 \pi G/c^4$ with $G$ being Newton's gravitational constant. The sign on the right-hand side depends on the convention, i.e., how you contract the Riemann curvature tensor to the Ricci tensor.

An equivalent form is
$$R_{\mu \nu}=\kappa \left (T_{\mu \nu}-\frac{1}{2} T g_{\mu \nu} \right), \quad T=T_{\mu}^{\mu}.$$

 

[PeterDonis]

this is not the Ricci but the energy-momentum tensor

In the paper by Baez and Bunn that I linked to, they rearrange the field equation so that the Ricci tensor is on the LHS, in the "equivalent form" you give at the end of your post. That is because the Ricci tensor is the tensor that has the direct physical interpretation being discussed (the "acceleration" of the volume, divided by the volume, of a small ball of test particles, caused by "mass" or more precisely density of stress-energy). That is what I was referring to.

 

[catlove]

Tenors (field vectors) are associated with an entity that consitutes matter yet in vacuum there is not matter so tenor theory (guess) is speculative.

 

[PeterDonis]

Tenors (field vectors) are associated with an entity that consitutes matter

Not necessarily. Spacetime itself is described by tensors (metric tensor, Riemann tensor, etc.) even in the absence of matter.

so tenor theory (guess) is speculative.

Your guess is incorrect.

 

[catlove]

Really?

"Maxwell's electrodynamics proceeds in the same unusual way already analyzed in studying his electrostatics. Under the influence of hypotheses which remain vague and undefined in his mind, Maxwell sketches a theory which he never completes, he does not even bother to remove contradictions from it; then he starts changing this theory, he imposes on it essential modifications which he does not notify to his reader; the latter tries in vain to fix the fugitive and intangible thought of the author; just when he thinks he has got it, even the parts of the doctrine dealing with the best studied phenomena are seen to vanish. And yet this strange and disconcerting method led Maxwell to the electromagnetic theory of light!" (Duhem, 1902).

 

[PeterDonis]

Really?

Yes.

Duhem, 1902

A reference from 119 years ago, particularly from a philosopher and not a scientist, more particularly one that makes incorrect claims (Maxwell electrodynamics is perfectly consistent), and even more particularly when the issue it is talking about has nothing whatever to do with tensors, is not a good basis for discussion.