Two Neutrons in Gravity
Most recent answer: 09/20/2011
Q:
Not too sure how to ask this, but ill try my best...
Suppose i have 2 stable, isolated neutral particles A & B (lets say neutrons). In classical reasoning, if we are to pull A and B further away, gravitation (however insignificant a force on this scale, and im aware that it doesn't apply to this scale of physics nicely) will eventually bring them together. Since it takes energy to move particles apart and energy must be conserved, gravity can be treated as a force of negative magnitude (disregarding the strong force to make it simpler).
I'm aware of the theory that attempts to derive gravitation from entropy alone, and this is similar, but not what im getting at... What eludes me is how would this scenario reflect onto a quantum mechanical view of those neutrons?
in 1D (again, for simplicity) one could imagine the tails of their wavefunctions interacting with eachother. i dont know enough about the specific interference or interaction calculations, but at a glance, the co-interactions of their tails must induce a 'pull' of some sort in response to the total wavefunction of the 2 neutrons having been stretched to form 2 (relatively) distant regions of high magnitute from a total wavefunction describing 2 nearby regions of high magnitude. I could say im looking into how the wave equation of 2 particles, one that is constantly evolving into energetically favorable states if left to its own devices, would be able to tell that it is favorable for the peaks of the wave to be closer than further.
I guess i'm also asking if the above has any validity to it, as it is merely a conclusion im coming to from bits and pieces of quantum theory that i've read into so far, which isn't a lot or to great depth.
thanks.
- Michael (age 21)
Australia
- Michael (age 21)
Australia
A:
This is a very interesting and challenging question. I'll need to deal with it in several stages.
First, the neutrons you refer to are not themselves stable (lifetimes about 15 minutes) , but let's pretend they are for the sake of the discussion. Let's also assume the neutrons are in opposite spin states so we can ignore the special issues connected with the exclusion principle for fermions. I'm also going to neglect the nuclear interactions between the neutrons, which are very important when their wavefunctions overlap, becaue I think those weren't the issues you were worried about.
Second, the picture you have of things settling into the lowest energy state available is incomplete. It sort of assumes the presence of friction, which isn't applicable on the small scale. Any closed system in a state with some particular exact energy will stay in that state forever. When an atom falls from an excited state to the ground state, it emits electromagnetic radiation. That means that once you include electromagnetic effects, the original excited state wasn't quite a state of definite energy, which accounts for why its appearance can change over time.
For neutral particles, one could in principle lose energy by gravitational radiation but that process is so extremely slow that it can be neglected for atomic-scale events for the lifetime of the universe. So let's look at what the gravitationally bound states of the two neutrons would be. They'd look somewhat like ordinary atomic states, but enormously more spread out. A quick and dirty calculation indicates a spread of about a million light-years. In order for the neutrons to be at some more or less definite positions (say spread out over a range of 10-6 m or so) the kinetic energy associated with the narrowness of the wavefunction would be by far, far larger than the gravitational interaction energy between the neutrons. In effect, you just have two neutrons spreading out independently following quantum mechanical laws, not noticing each others' existence.
Mike W.
First, the neutrons you refer to are not themselves stable (lifetimes about 15 minutes) , but let's pretend they are for the sake of the discussion. Let's also assume the neutrons are in opposite spin states so we can ignore the special issues connected with the exclusion principle for fermions. I'm also going to neglect the nuclear interactions between the neutrons, which are very important when their wavefunctions overlap, becaue I think those weren't the issues you were worried about.
Second, the picture you have of things settling into the lowest energy state available is incomplete. It sort of assumes the presence of friction, which isn't applicable on the small scale. Any closed system in a state with some particular exact energy will stay in that state forever. When an atom falls from an excited state to the ground state, it emits electromagnetic radiation. That means that once you include electromagnetic effects, the original excited state wasn't quite a state of definite energy, which accounts for why its appearance can change over time.
For neutral particles, one could in principle lose energy by gravitational radiation but that process is so extremely slow that it can be neglected for atomic-scale events for the lifetime of the universe. So let's look at what the gravitationally bound states of the two neutrons would be. They'd look somewhat like ordinary atomic states, but enormously more spread out. A quick and dirty calculation indicates a spread of about a million light-years. In order for the neutrons to be at some more or less definite positions (say spread out over a range of 10-6 m or so) the kinetic energy associated with the narrowness of the wavefunction would be by far, far larger than the gravitational interaction energy between the neutrons. In effect, you just have two neutrons spreading out independently following quantum mechanical laws, not noticing each others' existence.
Mike W.
(published on 09/20/2011)
Follow-Up #1: excited states and radiation
Q:
This is indeed getting quite interesting... i think i understand. i have one other perspective on this to explore, but first, could you please clarify one point to that answer...
you mentioned that a closed system of definite energy should remain as is so long as it is undisturbed (remains closed). but then you say that an excited atom that emits EM radiation, returning it to a ground state (or at least a state of lower energy), and that it did so as a consequence of it being not quite in a state of definite energy to begin with. it could be how i'm interpreting it, but at face value it implies to me that excited states do not have strictly defined energy levels, which doesnt sound right except when considering a superposition between some possible excited states?
more on topic, i have been reading into loop quantum gravity and its variants. my own explanation of it is basically the fruits of attempting to construct general relativity at the quantum scale, effectively defining a limit to that nasty renormalization issue by defining (or confirming, depending how you look at it) the smallest possible length, area, volume and time (plank units, of course), and by extension, all possible lengths, areas, volumes etc (multiples of the lowest allowed, just like energy states). the result is a fairly accurate model of what i guess you could call quantum relativity and quantum state evolution, all done with a spin-network that evolves in timesteps of plank-time. i believe this was used to infer that some black holes could be older than the universe, ie, 'survived' the big bang if you will.
there is very little information for the layman on the matter... what i haven't seen discussed yet, is how gravity would propagate in such a system, the irony being that it was intended to explain just that. using a spin-network model, would it be more easily understood as graviton propagation as per the standard model (or any particle-based propagator), or does the topology of the spin-network predict a tendency for energy to cluster (ie, can the other fundamental forces be used to describe such tendencies)? the latter idea i find more elegant despite the basis for it disproven - a recent proposal (in the last couple of years) that gravity arises entirely from entropy. i dont remember the name of the author of this, but he used photon world lines to depict degrees of freedom, and showed that there are more degrees of freedom (ie, more entropy) when 2 areas opaque to photons come closer, and at a minimum when they combine to form one area.
sorry about the wall of text, kind of hard to condense it into a simple question.
- Michael (age 21)
Australia
- Michael (age 21)
Australia
A:
"it implies to me that excited states do not have strictly defined energy levels,"
Yep, that's exactly it when you include the quantized electromagnetic field a part of the system, as it is in real life. We usually teach these initially leaving that out, and describe excited states of definite energy in the simpler problem.
Concerning loop quantum gravity, it's too hard for me.,
Mike W.
Yep, that's exactly it when you include the quantized electromagnetic field a part of the system, as it is in real life. We usually teach these initially leaving that out, and describe excited states of definite energy in the simpler problem.
Concerning loop quantum gravity, it's too hard for me.,
Mike W.
(published on 09/22/2011)