Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

The big bang theory proposes that equal amounts of matter and antimatter were created in the beginning. Shortly afterwards most of it annihilated. Could that have produced enough energy to drive cosmic inflation?

share|improve this question
    
At the beginning of universe, matter/anti-matter are roughly in equilibrium with radiation; matter/anti-matter annihilates constantly into radiation, but radiation is also constantly transforming into matter/anti-matter (pair production). In addition, the result of annihilation may not be radiation; it can be other types of matter/anti-matter. For example, electron-positron annihilation may produce, if energy permits, photons, hadrons/anti-hardons, muon/anti-muon, Z bosons, etc. –  Siyuan Ren Jul 13 '12 at 4:59
    
gave you some points so you can comment. –  anna v Jul 13 '12 at 5:40
    
Forgive my naivety but it appears that there is only ~4% of the original matter left from the big bang thought possible due to cp violation etc the rest of the material presumably was converted to radiation. What sort of time scale are we looking at for this to occur? Thanks for all the answers everybody –  Glenn Reid Jul 15 '12 at 2:15

4 Answers 4

Thermodynamic Aspects

Just to expand more on Karsus Ren's point:

In the early universe we generally like to assume thermal and chemical equilibrium. What this equates to is that reactions are equally likely to happen each way. Matter-Antimatter annihilation is relatively symmetric process. Lets just look at electrons/positron annihilations. This have the form of:

$e^++e^-=>2\gamma$

But we can also have two photons collide to form an electron positron pair.

$2\gamma => e^++e^-$

These two processes have the same amount of energy involved. So if we are in the pre-inflationary universe, with energies up in $10^{27}K = 9\times 10^{16} MeV$, we have photons travelling around with such high energies that the issue resolves somewhat simply. We can actually calculate the number density of fermions (both particles and antiparticles) fairly easily since $m_e=0.5 MeV <<T$.

$n_i=g_i e^{\mu/T} \int \frac{d^3p}{2\pi^3}e^{-E_i/T} = g_i \frac{T^3}{\pi^2}e^{\mu/T}$ (for $m_e<<T$)

Where $g_i$ is a constant determined by the number of spin states of species. We can also get from this calculation that even if there is some small chemical potential (found in the $\mu$ term), it is hugely overshadowed by the $T$ term, so we can easily assume number density equilibrium, so no annihilations.

Cosmological Aspects

In addition, there if there were high energy annihilation events in the hyper-early universe it couldn't cause an inflationary scenario. The expansion rate of the universe is determined by the Friedmann equations in terms of the scale factor which time-evolution is determined by the total energy in radiation, matter, and cosmological constant. Annihilation events would move more fraction of the universe's "inventory" from matter to radiation, making the universe scale factor change from more $a(t)=a_ot^{2/3}$ to $a(t)=a_ot^{1/2}$ (see friedmann equations). Although this means that it would slightly increase expansion speed, it wouldn't create the $a(t)=a_oe^{Ht}$ type behavior seen in the inflationary epoch.

share|improve this answer
    
This is correct, but it doesn't require the Friedmann equation to see it--- it follows from the stress tensor of hot matter alone. –  Ron Maimon Jul 13 '12 at 6:58
    
I suppose it comes down to how pedagogical you want to be. My point was to show that increase in radiation won't cause a "inflationary-like" effect, rather than assuming a priori that only a cosmological constant could cause it. –  Benjamin Horowitz Jul 13 '12 at 7:06
    
Ok, you're right. The only issue I have is that inflation requires a violation of a dominant energy condition, which doesn't happen with any normal matter. This is why it took until the late 1970s to discover it--- Hawking had already essnetially proved no inflation without violation of an energy condition. That a scalar violates this energy condition (or higher order corrections to GR, if you go Starobinsky's way) then became highly notable. –  Ron Maimon Jul 13 '12 at 7:42

Matter antimatter annihilation doesn't produce inflation, because the stress tensor of hot fermions doesn't lead to a cosmological constant. The main property of inflation is that at early times, the universe was dominated by a large positive cosmological constant, and this can be due to a scalar field, it violates the energy condition that the pressure is less than 1/3 the energy density.

For a photon gas, the pressure is exactly 1/3 the energy density. For any massive particle, in the limit of extremely high temperature, the pressure is 1/3 the energy density also, but slightly less because the mass breaks the scale symmetry. When you have scale invariance, the trace of the stress tensor is zero, so the energy component minus 3 times the pressure component must be zero, which explains this universal result.

The only relativistically invariant system that violates this pressure bound is a scalar condensate with a potential. In this case, you can see that a scalar field expectation value can't pick out a rest frame, so the stress tensor must be proportional to the metric tensor, and the pressure is equal to the energy density. This is a gross violation of the dominant energy condition, and it leads to runaway inflation. It is only such a system that can lead to inflation.

Further, the initial conditions of the universe in inflation are not hot at all, they are cold--- the field has an expectation value, which drive the universe to deSitter, and in a causal patch, the deSitter space is otherwise empty. This is a low-entropy initial condition (a cold initial condition), and one way of interpreting this is that the deSitter horizon is too small to allow any large entropy inside the causal patch.

share|improve this answer

No.

In the Big Bang model of the creation of the universe all the energy comes from the initial singularity from which space and time expand. The energies you describe are already created then, at 0,0,0,0 ,though it is an energy momentum tensor that is generated which obeys the equations of General Relativity. A simple minded energy conservation will not work.

It is worth noting that all points in our present space time started at that space time point, back in 0,0,0,0.

share|improve this answer
    
I don't quite follow what argument you are trying to say. It seems like you both argue for and against the idea of energy conservation in cosmological expansion. It should be made clear that there is no reason to believe that conservation of energy holds throughout the universe as, according to GR, there is no clear way to define total energy globally. –  Benjamin Horowitz Jul 13 '12 at 6:45
    
I am trying to say that all manifestations of energy that are used in developing the early universe, come from the initial input. And that the energy momentum tensor in GR says that due to gravitational interactions with the space time continuum part of what we call energy will be balanced there. –  anna v Jul 13 '12 at 7:04

What about interaction graviton/antigraviton to prevent matter-antimatter anhilitaion as a repulsive aspect of gravity?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.