[Phys-L] (now fixed) RE: Fraction of energy carried off

[caviness at SOUTHERN.EDU (Ken Caviness)]

Ludwik Kowalski wrote:

> On Oct 7, 2005, at 12:20 PM, Ken Caviness wrote:
>
> > A large nucleus at rest decays into 2 particles the ratio of whose
> > masses is  $B&V (J.
>
> What is the <SB&V (J> at the end of the sentence? A transmission error
> of some kind?

Oh oh.  Some of the characters that showed up fine on my composition screen
don't seem to have traveled well.

The symbol 'chi' became "$B&V (J",
the symbol 'equivalent' became "$B"a (J"
the symbol 'gamma' became "$B&C (J".

I'll do a search and replace 'chi' by "r", 'equivalent' by "=", 'gamma' by
"g", that might make the whole thing a little more intelligible:

-------- Now fixed: -----------

From: Forum for Physics Educators [mailto:PHYS-L@list1.ucc.nau.edu] On
Behalf Of Ken Caviness
Sent: Friday, October 07, 2005 3:21 PM
To: PHYS-L@LISTS.NAU.EDU
Subject: Fraction of energy carried off

I have a physics question for you:

(modified from Young & Freedman, 11th ed., 8.88):  A large nucleus at rest
decays into 2 particles the ratio of whose masses is  r.  What fraction of
the total kinetic energy (after the decay) does each particle have?  (This
is a classical, non-relativistic treatment.)

First, feel free to do the problem before scrolling down to see my answer
below.  I tried a couple different methods and show only the one that I like
the most!

Next, notice what this means about how the fraction of total kinetic energy
which each particle carries off depends on the fraction of the total mass
which each particle got.  Interesting, I think!  Seems fundamentally
significant to me.

Finally, and the reason I turn to the Forum for Physics Educators, consider
doing the same problem relativistically.  I don't get the same simple, neat
answer, in fact, I get a complicated mess.  Does this still simplify
somehow?  Or is the peculiarly simple and appealing result only true in the
classical limit?  Maybe a different (and still elegant) result is always
true?  I've tried total energies instead of kinetic energies, still a mess.
Besides, that wouldn't work in the classical limit anyway.  Sigh!  I would
appreciate any thoughts on this.

Thanks,

Ken
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Solution starts below, last chance to solve the problem first.....
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Let the two masses be m and M, where m/M =  r.  Let their speeds be v and V,
respectively.

Conservation of momentum:  0 = p_1(vector) + p_2(vector) ==> p_1 = p_2 = p.

Total Kinetic Energy:  K = K_1 + K_2 = p_1/(2m) + p_2(2M).  We seek the
ratios K_1/K and K_2/K :

K_1/K = K_1/(K_1 + K_2)

      = (p/(2m)) / (p/(2m) + p/(2M))

      = 1/(1 + m/M)

      = 1/(1+r),

K_2/K = 1 - 1/(1+r)

      = (1+r)/(1+r) - 1/(1+r)

      =  r/(1+r).

Multiplying numerators and denominators through by M gives:

K_1/K = 1/(1+r) = M/(M+rM) = M/(M+m),

K_2/K = r/(1+r) = rM/(M+rM) = m/(M+m).

A nifty result.  The fraction of kinetic energy each got is the same as the
fraction of the mass that the _other_ got.  For instance, if an object
explodes into 2 pieces, getting 1/10 and 9/10 of the mass, respectively,
then they carry away 9/10 and 1/10 of the energy of the explosion,
respectively.  A useful corollary lets you treat any collision/rebound of
two objects:  shift into the center of mass coordinate system, in which the
incoming relative momenta are equal in magnitude, and in that coordinate
system the above fraction of energy relationships will hold.  Nice, don't
you agree?

An earlier solution used everything in terms of speeds (p=mv, K=(1/2)mv^2)
to reach the same results, but took longer to get there.  In keeping with
that idea, in the relativistic case I'll use the relativistic relationship
between energy and momentum,

        E^2 = p^2 c^2 + (m c^2)^2,

rather than the formulas relating each to velocity

        p =  g m v, K = (g-1)m c^2,

where  g = 1/sqrt(1-v^2/c^2).

Here we go:

K_1 = E_1 - m c^2 = sqrt( p^2 c^2 + (m c^2)^2 ) - m c^2

K_2 = E_2 - M c^2 = sqrt( p^2 c^2 + (M c^2)^2 ) - M c^2.

Yes, it's the same momentum magnitude p in both cases, since momentum is
conserved.  This should result in some simplification, just as in the
non-relativistic treatment, even though M and m are different.  To proceed,

K_1/K

=          [ sqrt( p^2 c^2 + (m c^2)^2 ) - m c^2 ]
  ------------------------------------------------------------------
  [ sqrt(p^2 c^2+(mc^2)^2) - mc^2 + sqrt(p^2 c^2+(Mc^2)^2) - M c^2 ]

=            [ sqrt( p^2 + m^2 c^2 ) - m c ]
  ------------------------------------------------------------
  [ sqrt( p^2 + m^2 c^2 ) - m c + sqrt( p^2 + M^2 c^2) - M c ]

=            m c [ sqrt((p/(mc))^2 + 1) - 1 ]
  ---------------------------------------------------------------------
  m c [ sqrt((p/(mc))^2 + 1 ) - 1 ] + M c [ sqrt((p/(Mc))^2 + 1 ) - 1 ]

=            m [ sqrt((p/(mc))^2 + 1) - 1 ]
  -----------------------------------------------------------------
  m [ sqrt((p/(mc))^2 + 1 ) - 1 ] + M [ sqrt((p/(Mc))^2 + 1 ) - 1 ]

=            m [ sqrt((p/(mc))^2 + 1) - 1 ]
  -----------------------------------------------------------------
  m [ sqrt((p/(mc))^2 + 1 ) - 1 ] + M [ sqrt((p/(Mc))^2 + 1 ) - 1 ]

Nothing like M/(M+m) showing up here, and I don't see how to proceed without
using classical limit approximations, for example.  Maybe the result is just
a lucky coincidence?  I would be disappointed if that were all there was to
it.