| Re: The Uncertainty Principle -
07-08-2008, 11:08 AM
How I understand the uncertainty principle:
Let x and t be position (in meters, or whatever units you choose) and time (in seconds, or whatever units you choose). Let dx and dt be their uncertainties.
Let f and k be frequency (in cycles per second) and wave number (in cycles per meter). Let df and dk be their uncertainties. (k = 1/lambda, where lambda is wavelength)
I'm not sure how it's derived, but there is an uncertainties between their quantities:
dx * dk >= 1/4pi wave cycles (or 1/2 radian)
dt * df >= 1/4pi wave cycles (or 1/2 radian)
So far, all of that was pure math/geometry. Now let's add "matter".
Let E and p be energy and momentum. Let h = Planck's constant, and h-bar = h/2pi.
According to the quantum mechanics:
E = hf
p = hk
multiply both sides of the uncertainty inequalities by h:
dx * d(hk) >= h/4pi
dt * d(hf) >= h/4pi
Substitute terms:
dx * dp >= h-bar/2
dt * dE >= h-bar/2
Side note: We talk of uncertainty between position and speed. Actually, the uncertainty is between position and momentum. Perhaps it's the mass, rather than the speed that is uncertain. If a particle is very short-lived before it decays, there is indeed an uncertainty in it's mass.
A particle at rest still has a frequency of some sort, as though it were spinning in place. It's rest mass is hf/c^2
If an electron's angular momentum is h-bar/2, and it has mass m_e, then it's frequency is m_e * c^2 / h... Can we derive it's radius from this, as well as its linear speed along its radius?
The electron's angular momentum is fixed at h-bar/2, then it's angular position (phase of rotation) is completely uncertain. I wonder if we can measure its phase, causing its angular momentum to become uncertain? |