Re: question

In article <1137948828.779570.106940@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>, "asdf" <qjohnny2000@xxxxxxxxx> writes:
>I was always under the impression that in any nuclear explosion there
>was always a small amount of mass converted to energy. What am I
>missing.. does the energy itself have a gravitational pull ?
>
You're in the right direction now.

The phrase "mass converted to energy" is a loose and misleading one.
Strictly speaking, mass has no existence independent of energy and
momentum. In relativity mass of a system is defined by the relationship

m^2*c^4 = E^2 - p^2*c^2

where E is the total energy and p is the total momentum of the system.
It is important to realize that this mass is not equal, in general, to
the sum of the masses of the components of the system.

So, what happens whan a nuclear, or chamical or any other process
occurs within this system. Well, since all physical processes obey
conservation of enery and momentum, E and p don't change, thus the
mass of the system doesn't change either. But what may occur is that
some of the energy that was previously "bound" is converted into a
transferrable from, i.e. kinetic energy, and this energy leaves the
system (that's what explosion is). When this happens, then the energy
and mass of what's left behind is less. But that's not because "mass
was converted to energy" but because energy was taken out. If you had
at your disposal a perfectly impervious box, capable of withstanding a
nuclear explosion with nothing leaking out, and put a nuclear bomb
within this box and exploded it, then the the mass of the box with its
contents would've been the same before and after the explosion.

Mati Meron | "When you argue with a fool,
meron@xxxxxxxxxxxxxxxxx | chances are he is doing just the same"
.

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