0
$\begingroup$

First of all, it is important to note that I'm not very savvy in neither general relativity nor any other area of expertise that answering this question may require. Therefore, I mean for the question to be answered in layman's terms.

Okay, now to the question itself: as far as I understand, for a given energy total in a system, its contents can theoretically be put into any other configuration of energies with the same total sum (if that makes sense). Therefore, if that sum is zero, the contents can be configured into, basically, 'nothing', which also has an energy total of zero.

However, with the universe, the only sources of negative energy are the fundamental forces. I think it may be reasonable to put all forces other than gravity aside, as only gravity is generated (sorry for funky wording) by something which itself is another source of energy (mass), and all the other forces eventually cancel out. If we also ignore the dark energy, we are left with a collection of masses that will eventually crunch together. Here is the problem I have with this: where will the mass go? Even if the potential energy of every point in the universe is zero (i.e. all the mass in the universe has now been smushed into a point), there is still mass to deal with, right?

I know I am probably wrong in many, many places in this question, but can you explain the general, underlying flaw behind my reasoning?

Improve this question
$\endgroup$
9
  • 1
    $\begingroup$ Possible duplicate of Total energy of the Universe $\endgroup$
    – StephenG - Help Ukraine
    Commented Jun 18, 2018 at 22:17
  • $\begingroup$ @StephenG certainly not. In fact, I've only posted this question having thoroughly read all the answers in that thread. $\endgroup$
    – Max
    Commented Jun 18, 2018 at 22:20
  • $\begingroup$ Potential energy would be a negative value, but the zero point is arbitrary. So this is one way to set the total energy to zero, for a given configuration. $\endgroup$
    – Peter Diehr
    Commented Jun 18, 2018 at 22:20
  • $\begingroup$ @Peter Diehr that is only a matter of pedantry. In reality, we would have a collection of mass and no potential energy in the sense that it would not be convertible into any other type of energy. In fact, under the standard definition, the point mass would have an infinite negative potential energy, but that doesn't reflect the reality. $\endgroup$
    – Max
    Commented Jun 18, 2018 at 22:28
  • $\begingroup$ $E=mc^2$ ? Really. $\endgroup$
    – JMLCarter
    Commented Jun 18, 2018 at 22:43

2 Answers 2

Reset to default
1
$\begingroup$

The energy density of a gravitational field is negative, so it can serve as a "reservoir" of negative energy. If the total (negative) energy in gravitational fields throughout the universe is precisely equal in magnitude to the (positive) energy throughout the universe that resides in electromagnetic fields, kinetic energy of moving masses, "bare" particle rest masses, etc., then the total energy of the universe can be zero. However, it is not known whether the total energy of the universe is positive, zero, or negative.

Improve this answer
$\endgroup$
10
  • $\begingroup$ But isn't that energy density still proportional to the total rest mass reservoir in the universe? That would mean rest mass would always be greater than GPE. And what about my point mass example? Where would the rest mass energy go? $\endgroup$
    – Max
    Commented Jun 18, 2018 at 22:49
  • $\begingroup$ Total energy of the gravitational field around a mass is the integral of the energy density of that field over all space. The smaller radius a massive object has, the more negative energy is stored in its gravitational field, because a smaller radius provides more volume over which the field energy outside that radius is integrated. Shrink the radius enough, and in principle the negative field energy can be greater in magnitude than the positive rest energy of the mass. $\endgroup$
    – S. McGrew
    Commented Jun 18, 2018 at 23:02
  • $\begingroup$ Again, that is a matter of definition (as far as I understand, at least). In my understanding, a point mass has no potential energy if it's alone in the universe, as it cannot be converted into anything else, e.g. kinetic energy or radiation. $\endgroup$
    – Max
    Commented Jun 18, 2018 at 23:09
  • $\begingroup$ @Max That's not necessarily true. For example, a muon is a point mass that spontaneously converts into two neutrinos and an electron (all of which have kinetic energy even if the muon is initially at rest). $\endgroup$
    – probably_someone
    Commented Jun 18, 2018 at 23:21
  • 1
    $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – probably_someone
    Commented Jun 19, 2018 at 1:06
1
$\begingroup$

It really depends.. Settle for Feynman asserting the energy is conserved in the universe as a whole. Now you can call it $0$ if you wish. :)

According to Noether's Theorem time translation symmetry means conservation of, guess what.

Look at a free particle with momentum $p$ and energy $p^2/2m$.

So it has $0$ energy at rest? No. Einstein relation, and expand once in $p^2$

$$E=c \sqrt{m^2c^2+p^2} \approx mc^2+p^2/2m+...$$

What about the negative energies hiding there? What the hell does that mean anyway? It has an interesting meaning to it, but the important thing is the exclusion principle for electrons for example, and the apparent positive energy they all have, if there is any. They'd rather take a more relaxed lower energy. They cannot because there is a sort of sea of positrons occupying the negative band.

No one likes Smushed "points" of mass/energy. Some variants of the general theory are free of them.

If you do want to think of a crunch of a collection of masses to a small region which looks like a point, the rest of the universe is not "void". Wait a tiny moment and you have the particle but a tad later and not mathematically vanishing velocity. There can be space like interaction from the future point to the present point as felt by the present point, but upon reaching that future nothing is sent back. That's because the the now past particle knows it is coming and counters it with anti-interaction. :). And that's crazy, i.e virtual. that's why it is not infinite, and not 0.

You say you are probably wrong in many ways like it is a bad thing.

If you come to realize you are wrong at some point about something you gain a lot more than reading what you were thinking in an academic paper, for instance.

I'm no expert but I know enough to get numerous absurdities that would counter a $0$ energy pointy, something..?

On the other hand I can just tell you to look at 15 solutions to EFE. I can tell you to go read about the Dirac equation or better to see what it does to GR if taken as matter.

I can say that before and towards your crunch you still have a continuous symmetry and it is lost at the crunch and $Q$, the conserved Noether charge operator now proves a new massless boson crying lost energy due to a new vacuum state (Nambu-Goldstone boson for you). You take a symmetry you get energy, positive, energy. and you also get to avoid the rabbit hole of no more physics :).

Yesterday you turned off dark energy and imagined a "crunch" can occur as a result. Go back a bit because you have to consider dark energy after you switch it on or off... If you don't then you don't get to be wrong! And, you cannot deduce any conclusion about the conditions for the $0$ energy universe you postulate as a question.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Can you elaborate on your last point? If I understand you correctly, the energy is NOT 0? $\endgroup$
    – Max
    Commented Jun 19, 2018 at 22:09
  • $\begingroup$ Added elaborate. $\endgroup$
    – user192234
    Commented Jun 20, 2018 at 0:55
  • $\begingroup$ I know this might be a bit awkward, but I have already figured out an answer myself. The problem was that I considered the definition of potential energy completely arbitrary. Now I know that that definition comes from the assumption that a point infinitely far away from the field is indistinguishable from a point in no field at all. $\endgroup$
    – Max
    Commented Jun 20, 2018 at 1:39
  • $\begingroup$ On the last bit: if I keep dark energy, another question arises: wouldn't that, under the 0-energy hypothesis, then imply a big crunch scenario? As in 'rest energy + kinetic energy = |GPE|', in which case |GPE|>KE? $\endgroup$
    – Max
    Commented Jun 20, 2018 at 1:45
  • $\begingroup$ What field? The ordinary definition is "A field such that if you put a test particle over there (not infinity) in 1 sec from now, say, such and such force would be felt by the test particle". but if you put it infinity far away nothing is felt by no one. there is no concept of potential either way with your answer. $\endgroup$
    – user192234
    Commented Jun 20, 2018 at 2:02

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