Does the cosmic inflation reduce the energy density (inversely) proportional to the volume, or does the inflation "cost" energy? Is space itself "something" created at the expense of energy?
$\begingroup$
$\endgroup$
5
-
1$\begingroup$ possible duplicate of Energy conservation in General Relativity $\endgroup$– John RennieCommented Jul 29, 2014 at 9:00
-
$\begingroup$ I've suggested a duplicate because your question boils down to whether energy is conserved during inflation, and the duplicate question addresses this issue. Generally energy is not conserved in GR because by Noether's theorem energy conservation implies time shift symmetry, and this is violated during inflation and indeed during the current phase of accelerated expansion due to dark energy. If you search for energy conservation relativity you'll find many related questions. $\endgroup$– John RennieCommented Jul 29, 2014 at 9:02
-
$\begingroup$ @JohnRennie looking at two different questions and their answers, i found one that i thought said the energy goes to the "dark energy", while another said it was going into the energy of the gravitational field. Is that related or am i mixing unrelated things here? $\endgroup$– kutschkemCommented Jul 29, 2014 at 9:31
-
1$\begingroup$ There are endless arguments about whether energy is or isn't conserved in the expansion due to dark energy. One view is that it isn't because the amount of dark energy increases with time as the universe expands. An opposing view is that this increase is balanced out by the energy of the gravitational field and overall energy really is conserved. See this article by Luboš Motl and this article by Phil Gibbs for opposing views $\endgroup$– John RennieCommented Jul 29, 2014 at 9:36
-
$\begingroup$ @JohnRennie i think at the heart of my question is whether space itself (the metric, i guess) is "created" at the expense of energy. I guess the answers you pointed me to should make the answer "we don't know, but most explanations don't assume it"? I guess asking whether there are models in which space and energy are interchangeable or the same thing, as mass and energy, should be a seperate question. $\endgroup$– kutschkemCommented Jul 29, 2014 at 10:54
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
4
Ok, this is going to be a math-free-ish answer. Let's get something straight right off the bat. Inflation is over. Inflation refers to the first ~$10^{-34}s$ of the universe after the Big Bang. What we have now is accelerated expansion. Second, asking if this accelerated expansion "costs energy" is not particularly meaningful. What you'll find is that most cosmologists you ask this question to will hem and haw and maybe stutter a bit. Not because they don't know the answer, it's because this question is so non-specific and interpretive that virtually any answer they give will be in some way correct.
However, that first question you ask is nicely posed. Does the expansion reduce the energy density proportional (although perhaps you meant inversely proportional) to the volume? The answer is sometimes. When it comes to matter (dust, dark matter, stars) the answer is a resounding yes. Three atoms in a box represents a specific amount of energy. The same three atoms in a larger box is virtually the same amount of energy, so the energy density decreases with the increase in volume. For radiation and relativistic particles, the answer is no. Because expansion of the universe also red-shifts radiation, which decreases the photon's energy, the radiation energy density decreases not only with an $R^3$-like volume, it decreases like $R^4$ because of the redshift (OK $R$ is not actually what we use, but it's close enough to make a point with). Dark energy energy density also does not decrease. In the standard cosmological model, $\Lambda CDM$, the energy density of dark matter does not change ever. In a way, this can be seen as the expansion costing negative energy. As the universe expands the total amount of dark energy increases.
-
$\begingroup$ You probably wrote the answer before i made the edit, what i was getting at with "is space something" was really more "is space something that is created at the cost of energy", if that makes any sense at all (i was thinking similarly to how matter and energy are exchangeable, or maybe even the same thing, whatever that should mean) $\endgroup$ Commented Jul 29, 2014 at 14:03
-
1$\begingroup$ @kutschkem best not to think of space being "created" so much as being "stretched". This is a particularly useful picture when you get to Jim's bit about radiation energy density going as $R^{-4}$. $\endgroup$ Commented Jul 29, 2014 at 15:23
-
$\begingroup$ @Jim How do we know whether photons lose energy or had less energy to start with? And if you can accept that photons are losing energy, why are you so sure that atoms don't? I'm not saying you're wrong, just trying to gain your opinion on these matters. $\endgroup$– Alan GeeCommented Dec 8, 2018 at 9:43
-
$\begingroup$ @AlanGee The energy of photons is related to their wavelength, which expands with space. The energy of matter at rest (which is the assumed average state of matter) is related to its mass, which doesn't change as the universe expands $\endgroup$– JimCommented Dec 10, 2018 at 12:55
$\begingroup$
$\endgroup$
One of the Friedmann equations is a "conservation" equation: $$\dot \rho_i= -3\frac{\dot a}{a} (\rho_i + p_i) \tag{1}$$ where $\rho_i, p_i$ describes energy density and pression for a particular "fluid" (dust, relativist particle, dark energy/cosmological constant). For each fluid, there is a relation between $p_i$ and $\rho_i$ (respectively $p_i=0, p_i =\frac{\rho_i}{3}$, $p_i = -\rho_i$). One found easily the expressions of the $\rho_i$ as function of $a$, $\rho_i = \rho_i(a)$.
Except for the dark energy/cosmological constant case ($\rho_i= Cte$), one sees that the density is not constant. If $\dot a>0$, then the densities of dust and relativistic particles, are decreasing.
Now, considering energies instead of energy densities (excluding gravitational energy), equation $(1)$ can be recasted in a thermodynamic-like relation: $$ d (\Delta U_i) = d(\rho_i \Delta V) = - p_i d(\Delta V)\tag{2}$$
where $\Delta V = (\frac{4 \pi a^3}{3}) \Delta x \Delta y \Delta z$ represents the physical volume at constants $\Delta x, \Delta y, \Delta z$.
$\Delta U_i$ is the internal energy, due to the i-th fluid, contained in the physical volume $\Delta V$
Considering $\dot a>0$, so that $d(\Delta V)>0$, one sees that the internal energy $\Delta U_i$ is constant for dust, decreases for relativist particles, and increases for dark energy/cosmological constant.
Now, to have a complete behaviour, you must make the sum $d(\Delta U) = \sum\limits_i d(\Delta U_i)$. However, as said before, I am not counting gravitational energy here.
