The act of measuring requires one to impart some energy (photons or other)
or matter upon the particle. Upon the object being measured, the object
may instantly increase in mass or change velocity. Over time this energy
will be transferred back to its environment as it evaporates...
On
The measuring system can either transfer energy from itself to the
system being measured or do the reverse and transfer energy from the
system being measured to itself.
harry
On Sun, Aug 19, 2012 at 6:58 AM, ChemE Stewart cheme...@gmail.com wrote:
The act of measuring requires one to impart
I agree with that. Either way you have changed the measured.
On Sunday, August 19, 2012, Harry Veeder wrote:
The measuring system can either transfer energy from itself to the
system being measured or do the reverse and transfer energy from the
system being measured to itself.
harry
On
Actually, I tend agree with Robin that measuring cannot increase the
energy of the particle. My question reflects my own attempt to
understand why it is so. Now that I have thought about it, it is
because one doesn't measure energy per se. Most measurements are
really the result of calculations
Hello Harry,
To be really precise, though, an energy measurement of a particle in a
superposition of energy eigenstates might find it in one of the states
higher than the weighted average energy of its wavefunction. So, you
might say that the measurement increased its energy, but over many such
Hi LP,
I haven't read the paper, but I don't disagree with claim. In fact it
should not be unexpected.
Even in a macroscopic system a concentration energy can come about as
a result of energy being transferred from the measuring system to the
system being measured. Of course, such a measuring
BTW, I appear to contradict myself when I said measuring cannot
increase the energy of the particle
vs I agree with the claim that measuring can concentrate energy in a
system. In the former, I mean I don't accept the idea that measuring
can somehow increase the energy the particle without the
Pardon for this very late postscript, time is hard to find.
I believe you assume a wave function totally confined in all 3-dimensions.
This is probably not what was intended. It is easy to find papers
describing crystal/lattice channel conduction of much higher energy
particles (electrons,
In reply to pagnu...@htdconnect.com's message of Fri, 17 Aug 2012 13:11:31
-0400 (EDT):
Hi,
[snip]
Pardon for this very late postscript, time is hard to find.
I believe you assume a wave function totally confined in all 3-dimensions.
This is probably not what was intended. It is easy to find
On Fri, Aug 17, 2012 at 8:57 PM, mix...@bigpond.com wrote:
In reply to pagnu...@htdconnect.com's message of Fri, 17 Aug 2012 13:11:31
-0400 (EDT):
Hi,
[snip]
Pardon for this very late postscript, time is hard to find.
I believe you assume a wave function totally confined in all
Hello Harry,
You asked --
So, the measuring instrument itself will produce energy, if it is used
to precisely measure the energy of a particle?
Probably not.
But maybe there are subtleties that obey the 2nd Law of Thermodynamics,
but allow for some counterintuitive effects. For example, refer
Good questions, Robin
I wish I remembered solid state physics better, but I am not sure that
sure that your estimates are correct in a crystal lattice where the energy
eigenfunctions are nonlocal and span the entire crystal, but acquire more
nodes when they gain energy.
If my memory is correct,
Brillouin's ICCF-17 paper [1] states:
Brillouin's lattice stimulation reverses the natural decay of neutrons to
protons and Beta particles, catalyzing this endothermic step. Constraining
a proton spatially in a lattice causes the lattice energy to be highly
uncertain. With the Hamiltonian of the
I agree that you do have particles constrained thermodynamically and
spatially within the void(s) cracks of the lattice. Why a proton causes
the lattice energy to be uncertain escapes me but might be true. I can
understand how some collapsed matter would keep things uncertain since
it is
I also am not sure why a hot neutron might not be created as opposed to a
Cold Neutron from this. If there were some type of collapsed matter
triggering the event I can understand loss of momentum to all radiation
escaping due to the extra quantum gravitational pull to be overcome.
On Wed, Aug
CE,
Localizing the wave function of a proton (or an electron), i.e., making it
more narrow --- for instance,
+--+
| |
__ | |
/ \ | |
Once the energy spread of the active particles reaches the threshold needed
for electron-capture, presumably the probability of their reaching much
higher energies is minimal, i.e., they will be converted into neutrons
before that. So the resulting neutron is born cold with little extra
energy.
Lou suggests:
If so, the effectiveness of the stimulus could be quite sensitive to
waveform shape and frequency.
Absolutely it would... It wouldn't surprise me for Celani's and Rossi's
cells, that increasing the resistance heater temp will increase the rate of
reaction, but at some point it will
Here's an older article I found:
http://www.lenr-canr.org/acrobat/GodesRquantumfus.pdf
Jeff
On Wed, Aug 15, 2012 at 9:09 PM, MarkI-ZeroPoint zeropo...@charter.netwrote:
Lou suggests:
If so, the effectiveness of the stimulus could be quite sensitive to
waveform shape and frequency.
In reply to pagnu...@htdconnect.com's message of Wed, 15 Aug 2012 14:54:29
-0400 (EDT):
Hi,
[snip]
Brillouin's lattice stimulation reverses the natural decay of neutrons to
protons and Beta particles, catalyzing this endothermic step. Constraining
a proton spatially in a lattice causes the
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