Phew!
Ok, finished the video series. Narrator has a good voice, and doesn't go too fast or to slow.
Everything sounds legit to me (and very cool!), but the devil is (as they say) in the details.
So he starts with this equation: V
t = ƒ
ƛ
V
t = The speed of the quantum transmission
ƒ = The frequency of the emitted photon
ƛ = The wavelength of the photon at the moment of transmission
Now, Frank picks a value of 1.094 x 10^6 m/s for V
t and in the video it is explained where this constant comes from. I believe it was
empirically determined.
The point of this equation is that with V
t known, and the frequency of an emitted photon measurable, we can arrive at the wavelength
of the photon at the moment of transmission.
Next, he takes a formula describing capacitance between two square plates: C = e
0A / D
C = capacitance
e
0 = permittivity of free space
A = area
D = distance
He then substitutes ƛ^2 for A, and ƛ/2 for D, resulting in an equation that looks like C = e
0ƛ^2 / (ƛ/2). This step seems fishy to
me. He's using wavelength as a substitution for distance and area? Maybe this was explained in the video, but if it was, I didn't get it, and I still
don't.
This simplifies further to C = 2e
0ƛ and since ƛ = V
t / ƒ (as per the first equation, we end up with
a formula that looks like this:
C = 2e
0V
t / ƒ
If there is funny business going on in his math, it is with regards to this above equation! This just looks like unit manipulation to me. How
can capacitance be expressed as a function of frequency at the moment of quantum transmission? Capacitance has nothing to do with the wavelength or
frequency of photons. But I'm not a physicist, so what do I know?
Anyways, next we take a formula for energy stored in a capacitor: E = Q^2 / 2C
E = energy
Q = charge (in coulombs)
C = capacitance
He then subs in his 2e
0V
t / ƒ in place of C in the denominator, and we arrive at:
E = [Q^2 / 4e
0V
t] ƒ
And since Einstein's equation for the photo-electric effect is E = h ƒ
We can isolate [Q^2 / 4e
0V
t] = h
Tada!
We've derived Plank's constant as a function of charge squared over speed of quantum transmission.
And just to make sure they're actually equivalent....
Q = 1.60217646 × 10-19 coulombs
V
t = 1.094 x 10^6 m/s
e
0 = 8.85418782 × 10-12 m-3 kg-1 s4 A2
(2.56696942 × 10-38) / 4(8.85418782 × 10-12)(1.094 x 10^6) =
6.62513376 × 10-34 m3 kg / s2
Which equals Plank's constant to within significant digits.
So, he's not lying - he does derive plank's constant. But as to whether or not that means anything, or has any relevance, I really have no idea.
I'm not a physicist, I just know how to do algebra. I'll leave it up to the self-described "experts" to explain all this. At least now you can all
argue about the actual math, rather than nothing at all.
I mean, seriously, none of the skeptics in this thread so far have even bothered to watch the flippin' video. Now you don't have to. Here's the math.
Debunk away.
edit on 1-11-2010 by RedBird because: (no reason given)