From: ram@zedat.fu-berlin.de
"Paul.B.Andersen" wrote or quoted:
:1n + U-235 → Ba-141 + Kr-92 + 3n
. . .
:Are you claiming that the mass 3.916659E-25 kg
:is not converted to 2.78E-11 J kinetic energy?
When you start in the rest system of "1n + U-235",
its momentum is p=0. As the momentum is conserved,
the momentum of "Ba-141 + Kr-92 + 3n" then also is
p=0, so its kinetic Energy still is zero.
Only when you change the definition of what your
system is, you get kinetic energy. For example,
when you look at only one of the "3n", it has
kinetic energy.
For another example: On my table, a cup is resting.
In my room, it has the momentum of /zero/. The cup
contains gluons. When I look at one of those gluons
in isolation, its momentum is non-zero. This energy
is part of the mass of the cup. So, I have converted
mass into kinetic energy by observing a different
system: What appears as kinetic energy of the gluon
appears as mass of the cup.
>The point is that the fission was not discovered because
>someone thought that since E = mc², it must be possible
>to release energy by splitting atom.
We see kinetic energy because we get the products,
like "3n" is isolation. So, effectively, there is
kinetic energy, because it is apt in this case to
look at only one of the products in isolation.
However, the kinetic energy depends on the definition
of the observed system, as I explained.
>So why did you say that it is a common misconception that:
>"The atomic bomb proved in a very convincing way
> that mass could be converted to energy as predicted by
> Einstein's E = mc²?
(I think I already answered this in my previous post.)
--- SoupGate-DOS v1.05
* Origin: you cannot sedate... all the things you hate (1:229/2)
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