Re: Mass and Energy

From: Keith P Walsh (keith.p.walsh_at_btinternet.com)
Date: 09/01/04 

Date: Wed, 1 Sep 2004 09:58:58 +0000 (UTC)

On Mon, 30 Aug 2004 11:42:32 -0700, Uncle Al <UncleAl0@hate.spam.net>
wrote:

>No, you fucking imbecile. Binding energies SUBTRACT from the bound
>masses. No mass (leptons and/or baryons) is converted to energy.
>
>http://t2.lanl.gov/data/astro/molnix96/massd.html
> Find any bound nucleus and compare its mass to the summed mass of its
>individual neutrons and protons. Jackass.

I understand the phenomenon of "binding energy" very well.

(In fact I rather suspect that my understanding of it is more accurate
than yours.)

And the accurate scientific description of this phenomenon does not
contradict any of the assertions that I have put forward in this
discussion.

On the contrary, it supports them perfectly consistently.

The "binding energy" of an atomic nucleus is the amount of energy
given off and dissipated to the surroundings when the constituent
neutrons and protons (which I shall refer to collectively as nucleons)
bind together to form the nucleus.

The amount of energy given off is exactly equal to the difference
between the sum of the rest masses of the constituent "particles"
before binding, and the rest mass of the nucleus formed (this being
the lesser), according to the relation E=mc^2.

The energy released is therefore equal to the mass lost by the system,
and this is the mass (or energy) which is given up by the individual
protons and neutrons in order for them to bind together.

I think that part of your confusion may be due to a common
misconception that if any mass were to be "lost" in a process such as
this then this could only happen in amounts equal to the "loss" of one
or more entire nucleons. And, consequently, no mass can therefore have
been lost because equal numbers of nucleons may be accounted for both
before and after the process has taken place.

But that's not how it works.

I think it helps if, rather than thinking of the nucleons as
indivisible quantities of matter with constant masses regardless of
their bound or unbound state, each nucleon is regarded more as an
"arrangement of energy" which is "doing things", and in doing these
"things" these arrangements of energy create the individual properties
and physical characteristics of what we call protons and neutrons
(mass, charge, etc.).

Then, when the nucleons combine into the bound configuration which
comprises the atomic nucleus, they actually require less of the energy
(mass) which they held as individuals in order to exist in the bound
or "shared" state.

In other words, some of the rest mass of the component particles must
be given up and released as energy in order for the binding of the
nucleons to take place.

And this fully explains why the bound nucleus has less mass than the
sum of the masses of its constituent parts.

The exchange of masses and energies at the molecular level in
"chemical" processes is analogous to what happens in nuclear
reactions, but instead we are then dealing with the amounts of energy
involved in creating or breaking molecular bonds, which are much
smaller.

Nevertheless, the balance *** for calculating the exchanges of
"masses" and "energies" (which, don't forget, are one and the same
thing) is still essentially dependent upon the relationship E=mc^2.

As a general rule I think you should accept that ANY physical process
which yields an amount of useful energy MUST involve a loss of rest
mass from the initial state of the process.

And if you have been reading Mati Meron's contributions to our
discussion you will be aware that he appears to agree with me.

Keith P Walsh

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