1) From
the quantities given:

dt/
t » GM(1/r1 - 1/r2) » g(dr)/ c

where G is the Newtonian gravitational constant, M is the Earth's mass, and g is the acceleration of gravity (g = 980 cm/ sec

From the hypothetical data given in the post, the box deflection (r2 - r1)was 0.001 mm = 0.0001cm, then:

dt/t ~ (980 cm/s

dt/t » 10

and for an interval say t = 0.01 sec, dt =

^{2}where G is the Newtonian gravitational constant, M is the Earth's mass, and g is the acceleration of gravity (g = 980 cm/ sec

^{2}in cgs) and c = 3 x 10^{10}cm/sec.From the hypothetical data given in the post, the box deflection (r2 - r1)was 0.001 mm = 0.0001cm, then:

dt/t ~ (980 cm/s

^{2})(10^{-4}cm)/ (3 x 10^{10}cm/sec )^{2}dt/t » 10

^{-22}and for an interval say t = 0.01 sec, dt =

(10

This observation would actually generate a time uncertainty of 10

^{-22})(0.01 sec) = 10^{-24}secThis observation would actually generate a time uncertainty of 10

^{-24}sec- and hence an uncertainty DE in the energy of the photon.,
The
mass uncertainty is D
m = DE/ c

^{2}
DE Dt

__>__h/ 2p so DE » 1.054 x 10^{-34}J-s/ 10^{-24}s
DE » 1.054
x 10

^{-10}J
Therefore:
D m = DE/ c

^{2 }» 1.054 x 10^{-10}J/ 3 x 10^{8}ms^{-1}
D m » 3.5 x
x 10

^{-17}kg
And
the uncertainty in weight is:

D W » D m
g » 3.5 x
x 10

^{-17}kg (9.81 N/ kg)
Or: D W »
1.2 x 10

^{-17}N
2) The experimental result:

S = (A1,A2)I + (A1,A2)II + (A1,A2,)III + (A1,A2)IV =

2.65

__+__0.10
is at the cusp of necessary threshold (2.70) for basic confirmation. However, the uncertainty (

__+__0.10) suggests that at least some of the results might be confirmed but this would have to be subject to careful re-test.
3) We have a theoretical wavelength, l = 2p / Dk, where Dk is the expected width
of the wave packet. If Dk = 0.5 nm, and x = 1 nm
with x

_{o}= 0.5 nm, and we are to compute the E-field amplitude: E_{z }**Solution**: We know

Dk = p / x = p / (0.5 nm) = 2p
nm, but 2p
(nm) > 2 nm.

The maximum of the wave packet is
approximated closely by the square of the amplitude:

[
E

_{z}]^{2}= 4 sin^{2}2p (1 – 0.5) / (1 – 0.5) =
4
sin

^{2}2p
But: sin

^{ }2p = 0 so [ E_{z}]^{2}= 0
Physical
interpretation:

The dimension of Dk relative to Dx = (x
- x

_{o})
Implies

*no wave packet can exist*.
4) In a particular experiment to test Bohmian
quantum mechanics on a computer, the uncertainty in one input turns out to be: Dt = 10

^{-39}s and in the other, Df_{k}= 10^{-51}m. From this data, find the quantity b. Then compose a form of the Uncertainty principle and obtain the product ab. (Where ab plays the same role as ħ in the conventional form of the Uncertainty principle.) Comment on how your product ab compares to ħ.

**Solution:**

__(__

__d__

__D__

__f__

_{k}__)__

^{2}= b (Dt)

__(__

__d__

__D__

__f__

_{k}__)__= b

^{1/2}(Dt)

^{1/2}

b=

__(____d____D____f___{k}__)__^{ 2}/ (Dt) = (10^{-51}m)^{2}/ (10^{-39}s ) =
(10

^{-63}m)^{2}/ s
(Note:
a is a constant of proportionality so let a = 1)

Bohm
notes that

__p___{k}also fluctuates at random over the given range so:
d p

_{k}= a b^{1/2}/ (Dt)^{1/2}=
(10

^{-51}m)^{1/2}/ (10^{-39}s )^{ 1/2}= 10^{-6}m^{1/2}s -^{1/2}
Combining
all the preceding results one finally gets a relation reflective of the
Heisenberg principle, but time independent:

d p

_{k}(d Df_{k }) = ab = (10^{-6}m^{1/2}s^{-1/2})( 10^{-51}m)
= (10

^{-57}m^{3/2}s^{-1/2})
This
is analogous to Heisenberg’s principle, cf.

dp dq

__<__ħ except that the units of ħ are J-s.
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