I Does the energy in cc also gravitate?

1. Oct 19, 2016

kodama

since there is energy in the vaccuum does it also gravitate?

can energy measured in cc explain mond-dark matter effects?

2. Oct 19, 2016

Staff Emeritus
What do you mean by this? Exactly. This is a topic that has a lot of misconceptions and half-baked definitions that if you don't define it this will become one of those multipage threads that go nowhere.

CC? Carbon Copy?

3. Oct 19, 2016

kodama

the measured cosmological constant

4. Oct 19, 2016

Simon Bridge

... good, and what are you calling "energy in the vaccuum"?
There are lots of things you could mean by that.

Have you tried googling "vaccuum energy and cosmological constant"?

5. Oct 19, 2016

Staff: Mentor

If you mean, does it affect the curvature of spacetime, yes. That's obvious just looking at the Einstein Field Equation.

No. Dark energy or a cosmological constant, the kind of "energy in the vacuum" I assume you are talking about, causes a different kind of spacetime curvature than dark matter (or ordinary matter and radiation). The simplest manifestation of the difference is that dark energy causes the expansion of the universe to accelerate, whereas dark matter, ordinary matter, and radiation all cause it to decelerate.

6. Oct 19, 2016

kodama

is that consistent though? mass-energy is always attractive and additive. shouldn't the total energy contained in a vacuum in a volume the size of a andromeda galaxy cause it to decelerate?

7. Oct 19, 2016

Staff: Mentor

"Mass-energy" is not the source of spacetime curvature. The stress-energy tensor is. The stress-energy tensor includes pressure as well as energy density; and in the case of dark energy, the pressure is negative (tension) and its total effect is larger than the effect of the (positive) energy density, so the net effect is acceleration.

8. Oct 19, 2016

Jorrie

Only if the 'total energy' has positive pressure. The cosmological constant has negative pressure and it can be (loosely) said that the mass-equivalent part of its energy gravitates like normal matter, but that the negative pressure part "pushes", causing cosmic expansion to accelerate. Note that Andromeda galaxy-sized space is far too small for dark energy to be observable.

9. Oct 20, 2016

Staff Emeritus

10. Oct 20, 2016

pervect

Staff Emeritus
If one looks at Einstein's equations, from the wiki https://en.wikipedia.org/w/index.php?title=Cosmological_constant&oldid=743876741, the cosmological constant appears as follows:

$$R_{\mu \nu} -\frac{1}{2}R\,g_{\mu \nu} +\Lambda\,g_{\mu \nu} = {8 \pi G \over c^4} T_{\mu \nu}$$

in which it the cosmological constant $\Lambda$ is just a number. To interpret the cosmological constant as having an energy, one needs to fold it into the stress-energy tensor, by interpreting $\Lambda\,g_{\mu \nu}$ as an adjustment to the right hand side of Einstein's equation rather than a separate term.

If this is done, the "energy" in the resulting modified stress energy tensor gravitates - but the modified stress energy tensor also includes a pressure term as well as an energy term. Both the energy term and the pressure terms gravitate. See Baez's "The Meaning of Einstein's equation" for an explanation of why pressure causes gravity in general relativity. A positive cosmological constant implies a positive equivalent energy density on the right hand side (RHS), and a negative pressure on the RHS. The later term is not intuitive, not mentioned in most popularizations, but yet is very important. The positive energy density would slow down the expansion of the universe, but the negative pressure increases the rate of expansion more than the energy density slows it. Thus the net effect of a positive cosmological constant is to cause the expansion of the universe to accelerate (which is supported by current observations and the $\Lambda$-CDM model. This is the opposite of the effect one would expect if one (incorrectly) did not consider the effect of the pressure terms when the cosmological constant was interpreted as an adjustement to the RHS of Einstein's equations.

11. Oct 20, 2016

kodama

in context of GR cc can only mean cosmological constant.

12. Oct 20, 2016

kodama

regarding other posters, i understand the issue you raise about pressure.

my point is imagine an object, say a hollow ball with mass energy 100 joules, imagine a region of space with 100 joules of photons, now imagine a volume of empty space where the total energy from the vacuum is 100 joules. based on the equivalence principle, do all 3 joules gravitate the same magnitude?

13. Oct 20, 2016

Staff: Mentor

I don't think you do, since you leave it out of the rest of your post, even though it's critical for understanding why the three cases you give do not all behave the same. See below.

The EP is irrelevant here, because the three configurations you mention all have different pressures, and pressure also gravitates. So the cases are not equivalent in any relevant sense.

If we idealize all three cases as homogeneous perfect fluids, then how they "gravitate" can be expressed as follows:

$$\frac{\ddot{V}}{V} = - \left( \rho + 3 p \right)$$

where $V$ is the volume of a small ball of test particles centered on some point in the fluid, $\rho$ is the energy density of the fluid, and $p$ is the pressure of the fluid. For all of the cases you give, the fluid has an equation of state of the form $p = w \rho$, where $w$ is a constant. So we can rewrite the above as:

$$\frac{\ddot{V}}{V} = - \rho \left( 1 + 3 w \right)$$

This tells us the percentage "acceleration" of $V$ as a function of $\rho$ and $w$. Now let's look at your three cases:

(1) The fluid is made of ordinary matter. For ordinary matter, $w = 0$, so we have $\ddot{V} / V = - \rho$. The small ball of test particles accelerates inwards--its volume decreases. This is what we mean when we say that gravity is attractive for ordinary matter.

(2) The fluid is made of photons. For photons, $w = 1/3$, so we have $\ddot{V} / V = - 2 \rho$. So gravity is also attractive for photons, and a given energy density of photons gravitates twice as strongly as the same energy density of ordinary matter. (In practice, the photons we observe have energy densities many orders of magnitude smaller than the energy densities of the ordinary matter we observe, which is why we don't notice this effect.)

(3) The fluid is made of vacuum energy. For vacuum energy, $w = -1$. This means that we have $\ddot{V} / V = 2 \rho$. Notice that the RHS is now positive, not negative; the small ball of test particles accelerates outward, not inward. (In the case of the universe as a whole, this shows up as accelerated expansion instead of decelerated expansion.) So vacuum energy gravitates the opposite way from ordinary matter or photons.