Me:
>>The formula is correct, but your conclusion is not. While n is an
>>integer, f is not.
>If some physical property comes in units, 1c, 2c, 3c, 4c, ... then that
>property (in this case momentum) is digital, the constant c can be pi, e,
>or any real number except zero, it does not matter. The absolute value
>of one unit is not important, the point is that the difference between
>one unit and the next comes in steps, integers. Digital.
That is the whole point, the quantities we are talking about are not
observed in multiples of some invariant "unit". Take muon decay as
an example. To the best of our knowledge, muon is an elementary
particle. It decays into an electron and two neutrinos. If what you
are saying were true, momentum of the electron in the rest frame of
the parent muon would be quantized, meaning it could only have values
from a discrete set, and the spacing between the possible values would
be some constant. This is not observed. Electrons from muon decays
have a continuous momentum spectrum, and the probability distribution
of that spectrum can be calculated with little sweat.
Mass is equivalent to energy, and also doesn't come in discreete
steps. It was fashionable in the middle of this century to split
mass eigenstates by their intrinsic width into "particles" and
"resonances", but in fact there is no logical difference between
the two. What I'm saying is, every particle that decays has a
fuzzy mass. The exact amount of fuzziness depends on how quickly
it decays. For example, a rho is sort of an excited pion, and its
intrinsic width is about ten percent of its mass. It decays into
two pions most of the time. There is nothing discrete about its
mass. You can observe a rho at 770 MeV as easily as at 770 plus
pi MeV. Both particles are equally real.
It is certainly true that some observable quantities, like spin, have
discrete (you prefer "digital") spectra, but energy/momentum/mass do
not.
>If you want to find the momentum of an individual photon, hf, you
>can find it by using the formula hN/X , h is the Planck constant
>again, and N is the number of wave crests the photon makes over a
>distance X. Digital.
>>Another wrong conclusion. One: N is not an integer. A photon that
>>has one "crest" over one meter has one point three crests over one
>>point three meters.
>1.3 crests? Is that anything like 1.3 pregnancies? As I said, as you
>get to the end of distance X there will always be uncertainty over
>whether something is a crest or not, but the uncertainty can be made
>arbitrary small simply by making X larger because that increases the
>number of crests you are certain about and leaves the number you are
>uncertain of unchanged. Now you have better knowledge about what the
>momentum is, but because X is now larger you have less knowledge about
>where the photon is.
No, you cannot count crests of a single photon. By "counting" a single
crest you are destroying the photon. This particular fact is the gist
of the Heisenberg inequality and its most "intuitive" explanation:
"counting" a crest of a photon means measuring its electromagnetic
field, but the photon itself _is_ electromagnetic field, so you are
in fact having it interact with something and therefore destroying
your original photon. You may still be left with a photon, but that
won't be your original photon. Counting crests is just a mental crutch.
A gedankenexperiment if you will.
Now, don't confuse this with interference. Measuring interference is
counting crests of an _ensemble_ of particles. Yes, you can count
one point three crests, for example by fitting a sine function instead
of counting peaks. And yes, adding another crest will improve your
measurement by reducing the error. This is not the same as counting
crests of a single particle. Interference is a beautiful quantum
phenomenon in that it defies "intuition" in a maximal way. Even if
you send particles one by one through a contraption that makes them
interfere, you will still see the interference pattern. The only
"intuitive" (i.e. classical) explanation is that single particles
self-interfere, which is of course horse manure. Quantum mechanics
is simply not intuitive to a classically indoctrinated mind, as you
aptly put it:
> True, but it's not randomness that makes Quantum Mechanics so weird,
> it's spooky action at a distance, retroactive causes, and the question
> of exactly what an observer is and is not that makes it all so bazaar.
> The crazy experiments that proved Bell's inequality is violated should
> not have turned out the way they did, but they did.
Again, don't forget that all these weirdnesses are problems of
interpretation, not of quantum mechanics itself. QM works.
>>Two: the number of crests depends on the frame of reference.
> Everything depends of the frame of reference, except the speed of light.
> All photons will be red shifted or blue shifted, the amount of the shift
> will depends on how fast you're moving relative to the source, and at any
> instant the shift will be a constant for all atoms of any element in the
> source.
True, and the shift is not discrete. It is continuously variable,
so in principle you can make any source emit on any (non-quantized)
wavelength.
>>in the Universe and in my current profession c == h == 1.
>One what?
It doesn't matter. The point is that c and h are more fundamental
than meter or second.
>c and h do not come in the same units, but the units are related,
>sure you could change the size of the units of c and make c whatever
>you want, but then h would have to be fixed.
Exactly, nail on the head. It is the relationship between the two
constants that is fundamental, not the value of the units we choose
to express them in. Or, as Jay Reynolds Freeman noticed:
>>> Would it not be more accurate to say that one can select units
>>>in which c == h == 1, and that the particular values they have in
>>>(say) the MKS system of units are historical accidents based on
>>>the way that system was set up.
>>> It would, of course, be possible to set up units in which c and
>>>h had other values -- say, pi -- in which case the case for what
>>>some posters have referred to as a "digital universe" would be
>>>even more far-fetched.
Regards,
--dv