>No, this is not true. If it were, the star's mass would be zero.
All it means is that if we were to take all the constituent particles in a star and spread them out at 'infinity' from each other (effectively all across the universe), pushing them apart working against the gravity of the star holding it together, the energy required would be equal to the energy in those particles (from their mass, nuclear forces, etc).
The energy cost of disassembling a star equals the energy in the star, so the net energy of the star is zero. It's just a matter of doing the relevant calculations to see this, which was first done by Pascual Jordan in the 1940s.
I realise it's a ridiculously counter-intuitive result. Apparently when Einstein was told this, he stopped dead in his tracks while crossing a busy road.
> f we were to take all the constituent particles in a star and spread them out at 'infinity' from each other (effectively all across the universe), pushing them apart working against the gravity of the star holding it together, the energy required would be equal to the energy in those particles (from their mass, nuclear forces, etc).
This would require that all of the particles in the end state of the disassembly were massless (so all of the rest mass goes away in the disassembly process); but if they are massless, they can't also have zero kinetic energy (the only kind of energy a massless particle can have in the absence of gravity) or they don't exist at all. So I don't think the scenario you are implicitly relying on here is possible.
> It's just a matter of doing the relevant calculations to see this, which was first done by Pascual Jordan in the 1940s.
Do you have a reference? (And no, I don't mean the pop science references in the "Zero energy universe" Wikipedia article, I mean an actual reference to the published paper by Jordan where he makes the calculations you refer to. Or a more recent paper where someone else makes similar calculations.)
The particles in the end state would be the same mass they were in the star, and at rest in the star’s frame of reference. Why do you think simply moving them would make them massless?
As for references, the Wikipedia article explains the publication history of the idea and has references to the relevant papers. Jordan didn’t publish as it was during the war, as I understand it he just discussed it in his correspondences but there has been work done on it since.
> The particles in the end state would be the same mass they were in the star, and at rest in the star’s frame of reference.
Then the energy remaining at the end of the process would not be zero. Rest mass is energy.
> Why do you think simply moving them would make them massless?
Because you said "the energy required would be equal to the energy in those particles (from their mass, nuclear forces, etc)"; you included "mass" in the energy that would be required to overcome gravity, and would therefore be gone at the end of the process.
> Then the energy remaining at the end of the process would not be zero.
I didn’t say it would be. The final energy would be in the rest mass of the particles, and that’s all the energy there would be, despite just having pumped an entire star masses worth of energy into it to counteract the gravitational energy of the star.
You’re reading what I wrote and then filtering that into something completely different. You’re not actually replying to what I’m saying, but some bizarre distorted misunderstanding of it. To illustrate...
> Because you said "the energy required would be equal to the energy in those particles (from their mass, nuclear forces,
etc)"; you included "mass" in the energy that would be
required to overcome gravity, and would therefore be gone at
the end of the process.
I didn’t say anything about using the mass of the particles to move them about. That is utter nonsense. I just said that’s how much energy it would take. It’s a general point about the amount of energy in the system.
"The energy cost of disassembling a star equals the energy in the star, so the net energy of the star is zero."
> The final energy would be in the rest mass of the particles
Which contradicts the previous statement of yours that I just quoted. Rest mass counts as energy, so the net energy of the star is not zero, it's the star's rest mass.
> despite just having pumped an entire star masses worth of energy into it to counteract the gravitational energy of the star.
No, you would not have to pump an entire star mass's worth of energy into the star to disassemble it. The numbers are easy to run for a typical star, say the Sun, if we assume it to be a spherical distribution of matter with uniform density (an idealization, but it's enough to get the order of magnitude of the energy involved). The gravitational binding energy of such a mass distribution is U = (3/5) G M^2 / R. The numbers for the Sun are:
G = 6.67e-11
M = 1.989e30
R = 6.957e8
This gives U = 2.28e41.
Compare this to the rest energy of the sun, which is M c^2; for c = 299792458 this gives 1.79e47. So the gravitational binding energy of the Sun, which is the amount of energy that would need to be added to the Sun to completely disassemble it, is roughly 1 millionth of the Sun's rest mass.
(Note, btw, that this energy is added to the Sun, so the total energy of the disassembled Sun is larger, by about 1 part per million, than the total energy of the Sun in its current bound state.)
The stars energy is its rest mass plus the energy in its gravitational field. That gravitational energy is negative. Otherwise we have all sorts of problems with conservation laws. The energy model for the early universe with its dispersed array of unbound atoms and photons is a matter of record. There are plenty of referenced articles available and I have already explained how to find them. There’s no point complaining to me about it.
Your calculation of the stars binding energy uses the classical formula, but for a star and the universe generally you need to use relativity. The calculations for this are in the paper linked from reference 5 in the Wikipedia page on the zero energy universe. It’s a free pdf download.
> the Wikipedia article explains the publication history of the idea and has references to the relevant papers
The idea referred to in that article and its references is the speculative hypothesis I mentioned earlier about the total energy of the universe. None of those references do a calculation such as you described, showing that "disassembly" of a gravitating body like a star ends up with zero energy left over.
The calculations are for universes as a whole regardless of the number or nature of any objects within them. If you care, just follow along with the calculations assuming that any universe has just one star in it. There's no difference in the calculations due to actual quantities because everything cancels out, which is rather the point.
It doesn’t actually make much sense to do that for each individual particle, because the strong force will prevent some of them from being separated due to color confinement.
> I realise it's a ridiculously counter-intuitive result. Apparently when Einstein was told this, he stopped dead in his tracks while crossing a busy road.
All it means is that if we were to take all the constituent particles in a star and spread them out at 'infinity' from each other (effectively all across the universe), pushing them apart working against the gravity of the star holding it together, the energy required would be equal to the energy in those particles (from their mass, nuclear forces, etc).
The energy cost of disassembling a star equals the energy in the star, so the net energy of the star is zero. It's just a matter of doing the relevant calculations to see this, which was first done by Pascual Jordan in the 1940s.
I realise it's a ridiculously counter-intuitive result. Apparently when Einstein was told this, he stopped dead in his tracks while crossing a busy road.