Wyrm wrote:"Blindingly obvious"? You're appealing to common sense, Simon. The problem is that our common sense is based on prior experience... prior experience that is formed in the complete absence of perfect copy machines. You cannot expect this "blindingly obvious" shit to hold in this arugment about a technology very much outside our prior experience. You must prove it before I will accept it.
Fair enough.
[digs out textbooks for reference purposes]
[cracks knuckles]
To begin:
Axiom: "Fermion" and "boson" are scientific terms, whose meaning is not changed by the existence of new technology.
This needs to be formally stated, because it is implicit in the rest of my argument. I find it difficult to believe that you would object to this principle, though. The alternative is absurd. Redefining "fermion" or "boson" to mean something different in a Star Trek context than what it means
in real life is no more sensible than redefining "heat" or "gravity" to mean something different. The latter would be an abomination against scientific language; so would the former.
I assume for the sake of argument that real life definitions of fermions and bosons can be applied in Star Trek. If so, fermions and bosons can be defined as follows, in terms of their wavefunctions, which I will denote using Dirac bra-ket notation:
P
i j |(N identical bosons)> = |(N identical bosons)>
P
i j |(N identical fermions)> = -|(N identical fermions)>
In somewhat more detail:
The wavefunction of a system of N bosons is unchanged when we apply the permutation operator P
i j, which interchanges the
ith and the
jth particle, where
i and
j are arbitrary. The wavefunction of a system of N bosons is therefore "symmetric;" interchanging any pair of bosons within the system causes no change. It is this property which allows a large number of bosons to occupy a single quantum-mechanical state, allowing for interesting phenomena such as the Bose-Einstein condensate.
The wavefunction of a system of N fermions is "antisymmetric:" interchanging a pair of fermions within the system leaves the waveform unchanged
except that the sign reverses. For many purposes this interchange will not cause a visible difference, since the wavefunction is squared in order to find the probability of locating the particle at a given point in space, and (ψ(x))
2 = (-ψ(x))
2. However, the antisymmetry of a system of fermions forbids any two fermions from occupying the same quantum-mechanical state, leading us to interesting results such as the Pauli exclusion principle.
More detailed explanations are available on request. Non-physicists will probably need one, for which I apologize, but a
really detailed explanation would distract from the main argument.
_________
Now, while it pains me to contradict SapphireFox, the human body can be represented as a system of subatomic particles. It is a very complicated system, but it still is one. And there
is a precedent for treating composites of multiple particles as fermions or bosons: The helium-4 nucleus, for instance, is a boson and can be used to form Bose-Einstein condensates. The helium-3 nucleus, by contrast, is a fermion. Therefore, we can at least imagine claiming that Captain Kirk is a boson or fermion, in the same sense that we can make the claim of helium nuclei.
Consider the physical parameters. At a given time, a Captain Kirk is a collection of... at a rough estimate, something on the order of 1E29 particles.* Moreover, because of the sheer size of the systems involved, the minimum separation distance between Captain Kirks is on the order of a meter, even if we pack them into an enclosed space like sardines.
*I can nail that down more closely if need be. It is not necessary to this argument.
_________
A problem arises when we examine two Kirks in widely separate positions. Consider one Kirk located on the bridge of the
Enterprise and one Kirk located in Iowa. Over any extended period of time (say, a millisecond), each Kirk will undergo random quantum-mechanical processes. Radioactive nuclei within Kirk's body will decay (or not). Each Kirk will absorb ionizing radiation (or not). Electrons in each Kirk's body will absorb photons and rise to excited energy states (or not), and excited electrons in each Kirk's body will emit photons and return to the ground state (or not). All these things happen at random.
Therefore, if we begin with two Kirks at separated positions in spacetime, we see that they are exposed to different patterns of ionizing radiation: even if the
level of radiation is identical, the positions at which individual particles are absorbed will not be. We see that the Kirks will experience different sequences of atomic decays: they may have started with the same number of radioisotope atoms in the same configuration, but there is no assurance that Kirk-1's carbon-14 atoms will decay at the same instants that Kirk-2's will. We see that the question of which atoms in their bodies contain excited electrons depends on which Kirk we examine.
Consider the magnetic potential energy of the two-Kirk system. Each Kirk will have energy μ
KiB
i, where μ
Ki is the magnetic moment of the
ith Captain Kirk, and B
i is the ambient magnetic field at the
ith Kirk's position.
Imagine that the Kirk standing on the bridge of the
Enterprise experiences no magnetic field (since he's standing in a Faraday cage), while the Kirk in Iowa is exposed to the Earth's magnetic field, B.
Now suppose that we interchange the two Kirks, moving one from Iowa to the
Enterprise and vice versa.
If the two Kirks have identical magnetic moments, the magnetic potential energy is unchanged: the change in energy experienced by the Kirk moved from the
Enterprise to Iowa is exactly cancelled out by the change in energy experienced by the Kirk moving from Iowa to the
Enterprise. The net energy of the system is preserved, and the two Kirks can be interchanged freely, with no observable input or output of energy from the process. The magnetic potential energy of the system remains μ
KB, where B is the ambient magnetic field in Iowa.
But if the two Kirks have been separated in spacetime for periods that are long compared to the time scales of subatomic processes, they need not have identical magnetic moments. Even assuming that the magnetic field in Iowa has not caused some of the Earthside Kirk's particles to flip their spins, some of each Kirk's atoms will have undergone beta decays, which alter their magnetic moment... and it won't be the same atoms, or precisely the same number of atoms, in each case.
Define μ
K1 as the magnetic moment of the first Kirk (currently on the
Enterprise), and μ
K2 as the magnetic moment of the second Kirk (currently in Iowa). If we interchange the two Kirks, a change in the energy of the two-Kirk system equal to ΔE = |(μ
K2 - μ
K1)B| must occur.
________
Now, if we find that ΔE is nonzero for two spatially separated Kirks, we have evidence that the Hamiltonian of the two-Kirk system is not symmetric when we interchange the Kirks. This forces us to conclude that the two Kirks are not identical particles.
And it is extremely probable that ΔE will NOT be zero; see above.
It is at least
possible that the two Kirks will be identical, giving ΔE = 0, at a particular instant in time. It is (for all practical purposes) not possible that this will remain true indefinitely. In the general case, the two Kirks will not be identical particles, even if they were when you put them there.
But if Captain Kirks are not identical particles, then clearly they cannot be bosons or fermions. Therefore:
Captain Kirk is neither a boson nor a fermion, in the general case.
___________
Now, looking this over, I cannot expect all this to be "blindingly obvious" to the average person. Moreover, it is not
strictly impossible for a pair of Kirks to be identical, over extremely short time scales. Short enough that the probability of particle interactions within either Kirk occuring approaches zero, which is
really short when you're multiplying the probability over roughly ten to the 29th particles.
Therefore, I retract my previous claim that it is "blindingly obvious" that Captain Kirk is neither a boson nor a fermion. I replace it with a new claim, one that is functionally similar and better informed, one which I make in light of the fact that the argument above is fairly trivial by the standards of a good physics education:
It is blindingly obvious
to anyone who knows what a fermion or boson is that Captain Kirk is not a fermion or boson
in the general case, with exceptions occuring at extremely low probability and only over extremely short time scales.
This still makes your original argument "Electrons can occupy separate places and not be unique!" disingenuous, assuming that you know what fermions and bosons are, which I believe you do. You are arguing that Captain Kirks
might be identical, when in fact the probability of this happening is extremely small, and is effectively zero over human time scales. Taken over periods of time a human can actually perceive, we can say with extreme confidence that the Captain Kirks will not be identical, and will indeed be
verifiably unique.
Methods of verification include, for instance, measuring the change in energy of the two-Kirk system as we vary the ambient magnetic fields around them.