1) From
the quantities given:
dt/
t » GM(1/r1 - 1/r2) » g(dr)/ c2
where G is the Newtonian gravitational constant, M is the Earth's mass, and g is the acceleration of gravity (g = 980 cm/ sec2 in cgs) and c = 3 x 1010 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/s2)(10-4 cm)/ (3 x 1010 cm/sec )2
dt/t » 10-22
and for an interval say t = 0.01 sec, dt =
where G is the Newtonian gravitational constant, M is the Earth's mass, and g is the acceleration of gravity (g = 980 cm/ sec2 in cgs) and c = 3 x 1010 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/s2)(10-4 cm)/ (3 x 1010 cm/sec )2
dt/t » 10-22
and for an interval say t = 0.01 sec, dt =
(10-22
)(0.01 sec) = 10-24 sec
This observation would actually generate a time uncertainty of 10-24 sec- and hence an uncertainty DE in the energy of the photon.,
This 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/ c2
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/ c2 » 1.054
x 10 -10 J/ 3 x 108
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 xo = 0.5 nm, and we are to compute the E-field amplitude: Ez
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:
[
Ez ] 2 = 4 sin2 2p (1 – 0.5) / (1 – 0.5) =
4
sin2 2p
But: sin 2p = 0
so [ Ez ] 2 =
0
Physical
interpretation:
The dimension of Dk relative to Dx = (x
- xo )
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, Dfk = 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 Dfk) 2 = b (Dt)
(d Dfk) = b 1/2 (Dt)1/2
b= (d Dfk) 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 m1/2 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 Dfk ) = ab
= (10 -6 m1/2
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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