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Can someone explain the zero energy universe in a way that makes sense?

The explanation that all positive and negative energies cancel out to zero does not make sense to me, because if that were the case then there should not be anything in existence.


marked as duplicate by knzhou, sammy gerbil, CuriousOne, Community Aug 2 '16 at 2:16

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  • $\begingroup$ It doesn't make sense to me, either, but Krauss, I believe, sold a lot of books with it, so it probably makes a lot of sense to him. :-) $\endgroup$ – CuriousOne Aug 2 '16 at 0:13
  • $\begingroup$ $H\left|\psi\right\rangle = 0$ $\endgroup$ – Count Iblis Aug 2 '16 at 0:19
  • $\begingroup$ @CountIblis: Blessed be the wave function of the Lord. It's just too bad the the resulting quantum Zeno stops the universe, no? $\endgroup$ – CuriousOne Aug 2 '16 at 0:23
  • $\begingroup$ Where have you been reading about this? I don't think your reasoning is compelling : If you start with level ground and dig, you get a hole and a pile of dirt, which is not the same as what you started with. $\endgroup$ – sammy gerbil Aug 2 '16 at 0:30
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Every particle in the universe attracts every other particle, by gravity. There are other forces between them too, but over all, there's lots of mutual attraction going on.

Now, suppose you wanted to rip all the matter apart, so that all the particles were at infinite distances from one another, and there was no more attraction. You would have to add lots of energy, to move the particles away from each other.

It makes sense to think of the gravitational field between two particles as having negative energy, because you have to add energy to overcome it.

The idea of a "zero energy" universe is that the amount of energy you'd have to add to the universe to remove all the attractive forces in this way is equal to the amount of mass that you'd have, once all the particles are separate from each other.


There are a number of motivations for saying the total mass-energy of the universe is zero. One of them is the Hamiltonian constraint for ADM general relativity that has $NH~=~0$, for $N$ the lapse function. The quantum mechanical version of this is $H\Psi[g]~=~0$, which is the same as saying the Schrodinger equation has no $i\partial_t\Psi[g]$ term. Equivalently this also says the Killing vector $K_t~=~\partial_t$ is zero. This is a funny equation, for what it really tells us is that for a general spacetime manifold there is no natural place to put a Gaussian surface in which mass-energy can be measured. Another way to see it is that we can't put the universe on a scale! In effect if the Killing vector does not exist there is no isometry that defines a Noetherian theorem result for the conservation of energy. However, since we do have $H\Psi[g]~=~E\Psi[g]~=~0$ we can interpret this as meaning energy is zero.

The other motivation is supersymmetry. Supersymmetry assigns a boson to every fermion and a fermion to every boson. Bosons have positive zero point energy and fermions have negative zero point energy. If supersymmetry is unbroken then the zero point energy of the quantum field theory of the universe is zero. In a topological field theory perspective this means the total energy is zero, and we are motivated to think this continues to the low energy regime with broken supersymmetry.

The ADM constraint $NH~=~0$ can be seen to work in a Newtonian framework. The FLRW equations are expressed according to a dimensionless parameter. For a radial distance $r$ we set $r~=~ax$, for $x$ a standard distance and a the scale parameter. We then write the first time derivative of the radius as $dr/dt~=~ xda/dt$, and the second derivative $d^2r/dt^2~=~xd^2a/dt^2$. Given galaxies or matter of mass $m$ at a distance $r~=~xa$ the total energy in Newtonian mechanics is $$ E~ =~ \frac{1}{2}mx^2\left(\frac{da}{dt}\right)^2 ~-~ \frac{Gmm'}{xa}, $$ where $m'$ is all the mass-energy in the region to the radial distance $r$. We set the total energy to zero and divide through by $m$. $$ \frac{1}{2}x^2\left(\frac{da}{dt}\right)^2 ~-~ \frac{Gmm'}{xa} ~=~ 0. $$ The mass $m'$ is determined by all the mass in the volume $4\pi r^3/3$ for $r^3 ~=~ x^3a^3$ and as a result the mass is $m' ~=~ 4\pi\rho x^3a^3/3$. Inserting this in we see that $$ \frac{1}{2}x^2\left(\frac{da}{dt}\right)^2 ~-~ \frac{4\pi G\rho x^2a^2}{3}~=~ 0. $$ and the ruler distance $x$ can be removed from consideration. This is a matter of the invariance of ruler measure. Our energy equation is $$ \left(\frac{da}{dt}\right)^2 ~-~ \frac{8\pi G\rho a^2}{3}~=~ 0 $$ This defines the Hubble parameter $H ~=~ (\dot a/a)$ that dependx on the density of mass-energy $$ \left(\frac{\dot a}{a}\right)^2 ~=~ H^2 ~=~ \frac{8\pi G\rho}{3}. $$ This is the energy condition for the FLRW cosmology. This is a standard equation, and beneath it all is the assumption that the total mass-energy of the universe is zero.

  • $\begingroup$ It does not make any sense to me to say that there is no way to apply Noether to the entire universe and then to claim at the same time that the total energy is not only constant but zero. $\endgroup$ – CuriousOne Aug 2 '16 at 1:10
  • $\begingroup$ More problematic, though, is the fact that the same theory predicts that we can only see part of the universe, so how can we know what's happening in the part that we don't see? $\endgroup$ – CuriousOne Aug 2 '16 at 1:23
  • $\begingroup$ Perhaps from a layman standpoint, it might suggested that the Killing vector, $K_t$ should be zero as there is no additional framework or space-time geometry to embed the entire universe into. There naturally is no way to measure the velocity of the entire universe as a whole as opposed to the relative movements of its individual parts. Similarly, there is no interaction of the universe with something 'outside' of it, as the universe, by definition includes all of space-time and energy. $\endgroup$ – dualredlaugh Aug 2 '16 at 2:12
  • $\begingroup$ @dualredlaugh: How does one logically go from "undefined, unmeasurable, most likely meaningless" to "absolutely, exactly zero"? $\endgroup$ – CuriousOne Aug 2 '16 at 2:37
  • $\begingroup$ Mathematically, it is well established that one can construct a manifold with curvature defined at different points in the manifold without having to embed it in a higher dimensional space. In a similar vane, one can mathematically define the concept of a universe having negative or positive or zero curvature without needing additional structure. That doesn't mean that we can take measurements from all of this universe. But, nevertheless, the mathematics can lead to some real predictions. Using this construct for the "universe" naturally leads the time change of the aggregate being zero. $\endgroup$ – dualredlaugh Aug 2 '16 at 2:57

It's nothing very subtle: the positive and negative energy is just in different places. Compare the argument that an electrically neutral universe can't have any electromagnetic phenomena because all of the charge cancels out. It can, if the positive and negative charges are in different places.

  • $\begingroup$ So is it also true that similar to the case of charge, the sign of energy is merely a matter of convention? $\endgroup$ – velut luna Aug 2 '16 at 0:29
  • $\begingroup$ @AlphaGo No, positive and negative energy really are different. There's no symmetry that exchanges them. They are like positive and negative curvature (elliptical and hyperbolic space), whereas the electric charges are like upward and downward curvature, which can be swapped by a vertical reflection. But I think they are similar enough that the analogy works in this answer. $\endgroup$ – benrg Aug 2 '16 at 0:45
  • $\begingroup$ I dont know... to me, when someone says this and that cancels out, to me that means that they annihilate each other and both cease to exist. For example if i tell you i have 1 apple and -1 apple, to me thats the same as saying i have nothing. I dont buy it but hey not my area of expertise at all. $\endgroup$ – Gbailey1988 Aug 2 '16 at 0:45
  • $\begingroup$ @Gbailey1988 Well, when people talk about energy "canceling out" to zero, they mean something like my example with the electric charges. The terminology may not be ideal, but that's what they mean. $\endgroup$ – benrg Aug 2 '16 at 0:49
  • $\begingroup$ How did you measure that? $\endgroup$ – CuriousOne Aug 2 '16 at 1:21

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