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So I keep running into this idea on the Internet that only mass has gravity, but isn't mass simply contained energy?

Energy bounces around, or travels at the speed of light, because it is not in the form of an atom, it is in the form of light, photons, quarks, neutrinos, and things we haven't discovered yet.

Wouldn't it make sense to say that if mass can warp space by its existence, then everything that exists must also have that effect as well, but to a much smaller degree, a square of the speed of light smaller, and we could only detect that gravity it had, if it wasn't flying around at the speed of light?

The reason this is important is that light cannot escape a black hole, and I believe that nothing can escape a black hole over time, whether it be light, dark energy, photons, neutrinos, or anything that exists at all, meaning that the fate of the universe changes.

If dark energy gets sucked into black holes because it has gravity by its existence, then the pushing effect it has turns into a pulling effect over time, and then the only logical conclusion is the universe is eternal, moving in an endless cycle of big bang, big crunch, big bang, which also explains why the Big Bang happened, and continues to happen, forever.

If you say that dark energy remains at constant density despite the expansion of the universe, explain why it does this, and why you think that, and why it doesn't have gravity. When was this proven, and what caused the Big Bang?

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"The reason this is important is that light cannot escape a black hole, and I believe that nothing can escape a black hole over time, whether it be light, dark energy, photons, neutrinos, or anything that exists at all, meaning that the fate of the universe changes." : Hawking radiation? –  Gummy bears Jun 30 at 13:02
    
Yeah, Hawking radiation doesn't happen, because it can't, because all energy has gravity. –  rowanman28 Jun 30 at 13:07
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So you're saying that Hawking Radiation is a false concept? –  Gummy bears Jun 30 at 13:08
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OK, when was it ever proven that energy doesn't have gravity, and that only mass does? I don't believe that you could ever prove that, except by looking at light from distant stars, which are affected by gravity. Doesn't that prove that since light is energy, that energy has gravity? If it has gravity because all things that exist bend space to some degree, then everything that exists including dark energy will get sucked into black holes and cause the big crunch. –  rowanman28 Jun 30 at 13:10
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The difference is, energy curves space time a square of the speed of light less than mass, and it's very hard to measure or prove that. –  rowanman28 Jun 30 at 13:14

2 Answers 2

I am going to answer the following questions:

1) Does dark energy have gravity?

2) Why does dark energy have constant density during the expansion of the universe?

First about the term of dark energy - it is more of a label for many attempts of a solution to the problem of an accelerated expansion of the universe. One of the most promising (or say "mainstream") solutions is that of the cosmological constant or vacuum energy. Both of these concepts play an identical role in the equations of standard cosmology but their origin is either from a modified theory of relativity (the cosmological constant $\Lambda$) or particle physics (the vacuum energy). Now to the questions.

1) If we say accelerated expansion of the universe is due to a fundamental constant of the theory of gravity $\Lambda$, Dark energy indeed does not "have" gravity, because it is not really energy, it is just gravitation behaving slightly differently from what we thought on large scales.

If we say accelerated expansion is due to vacuum energy or some other "non-mainstream" energy, we are sure by observation that we do not feel it's "local" gravitational effect, that is lumps of dark energy pulling on objects in space. This means this energy must very homogeneous, ideally constant throughout the whole space. Why? Because if the distribution is completely homogeneous, there isn't any larger lump of dark energy we would be attracted to rather than anywhere else - the gravity just cancels out locally because it is equal in all directions but has an effect on cosmic scales.

2) If we consider vacuum energy, it is an energy inherent per unit volume of space even if it is completely empty (vacuous). So as the volume of a patch of the universe grows (as there is "more space"), there is always a constant energy per volume. (Notice that the overall energy of the universe grows, this is a well known result of standard cosmology.) This can be traced back to a basically postulated property of vacuum in quantum field theory and quantum field theory on curved background.

In the case of the cosmological constant, we don't need any argumentation of this sort, the constant is a property of gravitation.

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Forgive me in advance, this may get overcomplicated. I am going to give you the facts as scaled down as I can but still sufficiently detailed. I think providing you with what we have and allowing you to infer from it is the best way to avoid misrepresenting the answer.

Here is the General Relativity equation that describes how gravity interacts with everything else: $$R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}R=8\pi G_NT_{\mu\nu}$$ On the left side is gravity; it is described by terms that have to do with how space curves, expands, and contracts. On the right side is matter, radiation, etc. Pretty much all forms of energy.

It was determined through observation that the universe as we know it is expanding and that the rate of expansion is accelerating. To account for the acceleration of the expansion, the best-fitting theory includes a constant term: $$R_{\mu\nu}-\frac{1}{2}g_{\mu\nu}(R-2\Lambda)=8\pi G_NT_{\mu\nu}$$ Alternatively, one could choose to express that constant term on the right side of the equation. It makes no physical difference. Accordingly, you can interpret this term as a modification to how gravity affects spacetime (if added to the LHS) or as an additional energy term with a negative pressure (if added to the RHS).

Now, as per original GR, we have equations describing the expansion of the universe: $$\frac{\ddot a}{a}=-\frac{4\pi G_N}{3}\sum_i(\rho_i+3p_i)$$ $a$ represents how much the universe has expanded, $\ddot a$ is the acceleration of the expansion, and $\rho_i$ and $p_i$ are the energy density and the pressure for the $i$th form of energy (the rest are constants). We put in matter, dark matter, radiation, and we even treat the background curvature of the universe as a form of energy here. What we find is that when we treat dark energy as an energy and set its pressure to be equal to negative of its energy density, our equations very closely match the observations. That is sufficient reason to like doing that. However, we also find (via a separate equation) that the energy density of something whose pressure equals the negative of the energy density remains constant for all time. This is unavoidable, it is one of the best fitting models so far.

Furthermore, the constant energy density presents other effects. The energy density of matter or radiation decreases with time. For instance, and this should be intuitive, the energy density of matter decreases like $a^{-3}$. That is, it drops like the cube of the expansion of the universe. And why not? as the universe expands, the volume increases like the expansion cubed; energy density is energy over volume, so it decreases like expansion cubed. Radiation goes like $a^{-4}$, and other energies decrease at varying rates. A constant density means that after a long time, it becomes the dominant term in the above equation. Effectively, after a long time: $$\frac{\ddot a}{a}=\frac{8\pi G_N}{3}\rho_{DE}$$ This barrage of equations might mean nothing at all to you. That is fine. This is an answer to what dark energy is, why it has a constant energy density, and whether it has gravity.

As for whether or not dark energy can fall into a black hole, that is more complicated. From the point of view of modifying gravity, dark energy is not something that can fall into a black hole. However, from the point of view of being an additional energy term, one might think it must be able to fall into a black hole. Truthfully, I don't know that answer but I know it makes an insignificant difference. Dark energy is too weak to have an effect on the scale of black holes. Even the small gravity of the Sun is enough to negate the effect of dark energy throughout the solar system and probably a bit further. As you can see from the last equation, at late times the acceleration of the expansion is positive. Dark energy never brings the universe to a Big Crunch. For matter, radiation, etc, the acceleration terms at late times all are negative. Dark energy is preventing a Big Crunch.

In response to your last questions. This was never "proven". It was postulated decades ago and has been confirmed by many experiments but none that prove it beyond a doubt. As for what caused the Big Bang. The Big Bang was not an event, it was a moment of time. The Big Bang represents the point in time where the equation for $a$ (remember that's the term that represents how much the universe has expanded) goes to zero. That's it. It has no more need for a cause than has any moment of time after it.

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Just a few comments. The "rate of expansion of the universe" is a term strictly used for $H={\dot a}/a$ and it is observed to be roughly constant. However, a given patch of the universe now grows approximately as $\sim a^3\sim e^{3 H({\rm now}) t} $ - so the growth is accelerating, the rate (as given oppositely in your answer) is not. Negative pressure is only a confusing analogy. In any relevant sense, we do not actually presume dark energy to be a perfect fluid. –  Void Jun 30 at 20:04
    
My last comment did not fit in the limit. Your last equation is not correct, it should be $({\dot a}/a)^2 = 8 \pi G \rho_{DM}/3$. Only then it all makes sense. –  Void Jun 30 at 20:05
    
@Void when dark energy dominates, $(\dot a/a)^2=\ddot a/a$ –  Jim Jun 30 at 20:08
    
Okay, that's true, ma' bad. –  Void Jun 30 at 20:35

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