1
$\begingroup$

First,sorry for the clickbate(kinda).

I don't know if this will work or not but It seems worth a shot unfortunately I am stuck on the approach of proving it can't be done.....'cause ofcourse it can't be!!

so my thought process goes as follows:

we take two spheres of sterile neutrinos (best suited as they only interact via gravitational forces) of radius $a$ and mass $m$. now at infinite distance apart they have, as a system ,mass of '2m'. when we bring them closer we can see it's total energy decrease as the gravitation potential energy increases. now we assume at r distance apart they have 0 mass(as a system i.e. their gravitational potential energy overcomes their mass energy) clearly 'r>2a' to be possible and we can have a minimum condition for 'a' as swartzchild radius.

it is elementary to note that it is not possible in this case as they would turn into black hole way before we can have a small enough r(which is greater than 2a).

so what I thought was to have different types of systems I tried out an equilateral triangle all the way to an octagon and also some 3-D shapes although 'the gap'(the gap refers to the diffrence between swarzchild radius and the minimum 'a' require to get 0 mass)seems to be decrasing I can't tell if the gap would be asymptotic or not.

Can anyone please prove that it is impossible for any system to have such an 'a' such that they don't turn into black hole and have so much gravitatinal potential energy to overcome their mass energy i.e. E=mc^2

if it can, I think there can be these reasons(at their decreasing order of probability).

  • sterile can't be captured and then thus can't be morph into spheres.

  • Newtons or Einsteins equations are not so nice for the small scale.

  • sterile neutrinos might also interact with something other than gravitational force....or maybe not interact with gravitational force.

  • negative mass can be formed this way(impossible and I refuse to believe it).

regardless, I just wanted to have fun with the idea and I thought this is an interesting concept.

I would really appreciate any participation. thanks for reading!

$\endgroup$

closed as off-topic by Aaron Stevens, GiorgioP, Ben Crowell, Kyle Kanos, Jon Custer Mar 4 at 19:51

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "We deal with mainstream physics here. Questions about the general correctness of unpublished personal theories are off topic, although specific questions evaluating new theories in the context of established science are usually allowed. For more information, see Is non mainstream physics appropriate for this site?." – Aaron Stevens, GiorgioP, Ben Crowell, Kyle Kanos, Jon Custer
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ It's impossible to find a value for $a$ because the zero of potential energy can be defined at any distance $r$. Absolute potential energy has no practical meaning. Only differences have meaning. BTW, the same is true of kinetic energy. $\endgroup$ – garyp Mar 3 at 16:33
0
$\begingroup$

We could imagine doing some sort of experiment like this:

Binding energy

The two masses (sterile neutrinos or whatever) attract each other due to their mutual gravity. We can tie them to ropes and connect the other end of of the ropes to generators so as the two masses move together we get energy generated to power our light bulbs.

But, as you say, this means we have taken some energy $E$ out of the system, so the mass of our system must have been reduced by the equivalent mass $\Delta M = E/c^2$. This really happens, e.g. in nuclei, where the energy $E$ is the binding energy and $\Delta M$ is the mass deficit. So the question is whether we can take so much energy out that $\Delta M$ becomes larger than $2M$ i.e. the mass minus the mass deficit goes negative?

Sadly this won't work for the simple reason that as we take energy/mass out of the system we reduce the mass of the system and that reduces the gravitational binding energy. After we've taken out an energy equal to the original mass there would be nothing left to generate a gravitational binding energy and we couldn't get any more energy out.

However I think this is an fascinating question and it would be interesting to do an explicit calculation of the change in mass and PE as the objects approach each other.

$\endgroup$
  • $\begingroup$ Wow that's a fantastic explanations,But I don't get it I mean it decreases mass of the system but not the neutrino.I hope I don't sound stupid $\endgroup$ – Avi Mar 3 at 11:37
  • $\begingroup$ @Avi the mass of a bound system is always less than the total masses of its component parts, for exactly the reason I describe above. But the mass of the whole bound system can't be simply split up into various bits i.e. we can't say the mass of one of the neutrinos changes. Don't worry if this is confusing because it's a hard concept. I remember struggling with it when I first learned about it. You find yourself thinking huh, where did the mass go? $\endgroup$ – John Rennie Mar 3 at 11:55
  • $\begingroup$ With all due respect sir,I guess you got what I was asking wrong,I was asking that if the mass of neutrino doesn't change why does their binding energy changes.As, if I am correct, the binding energy depends on the mass of their constituents and not the system(Acc. to Newton atleast) $\endgroup$ – Avi Mar 3 at 12:08
0
$\begingroup$

There is a positive energy theorem that asserts that for a system composed of matter satisfying dominant energy condition the total mass of it (formally, the ADM mass) is strictly positive (with zero only possible for completely empty Minkowski spacetime).

In OP's constructions the problem is that there is no obvious way which would keep the masses from moving once they are in position under the gravitational attraction. While for systems where the gravitational binding energy is only a small part of a rest mass-energy it is easy to build struts or strings that would keep the masses rigidly in place, when the gravitational binding energy becomes comparable to the rest-energy of the system there is no easy way to do it.

If there is a strut strong enough that it would keep the masses from e.g. coalescing into a black hole, yet light enough that its own mass is small compared with other masses of the system, then such a strut would violate the energy condition (and so there is no known material for it). If you do have such material from which such struts could be made, it would be a form of exotic matter and it is indeed conceivable that overall negative mass could be constructed from such material moving in a relativistic way in a system with some more ordinary matter.

Another way to circumvent the positive energy theorem is in spacetimes with positive cosmological constant. For example, recently Mbarek & Paranjape constructed (arXiv:1407.1457) a solution to Einstein equations with a fluid satisfying dominant energy condition (simplified model of a realistic matter) yet having a negative mass in de Sitter spacetime. While the 'mass' of such solution is negative, it is on the background of the positive energy density of the de Sitter spacetime, see this blogpost for a simplified explanation.

The “dark energy” of current cosmological epoch could very well turn out to be a positive cosmological constant, so such solutions could be realizable in our Universe, but the scale for them would be comparable with de Sitter length so it would be billions of light years in size.

$\endgroup$

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