Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

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



I finally got around to looking at the fixed up particle decay
question that Ken Caviness asked about in his post of 09 OCT 05:

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.

OK. The solution is a little nicer if instead of finding the
fraction of total kinetic energy possessed by each of the particles
one finds the ratio of the kinetic energies of the daughter
particles.

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.

It is.

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,

You ought *not* get the same simple answer. The relativistic general
answer depends on *2* parameters. *One* of the parameters the answer
depends on is the mass ratio of the daughter particles, and the
*other* parameter is the total fraction of the initial mass-energy in
the parent particle that is converted to kinetic energy. In the
Newtonian non-relativistic limit this second parameter is precisely
zero and the general expression simplifies because the answer now
depends on only the one parameter of the mass ratio of the daughters.

in fact, I get a complicated mess.

Then you probably got all balled up in the intermediate algebra.

Does this still simplify somehow?

Yes, it simplifies a lot, but the relativistic general case is
intrinsically somewhat more complicated than the special case
limiting Newtonian limit.

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 think the general case is still elegant, but since it depends on
2 parameters instead of 1, it *can't* be as simple as the Newtonian
limit.

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.

Try setting c = 1 and only consider the relevant ratios in your
intermediate calculations. That should keep things a lot simpler.
But even then the intermediate steps *do* involve a lot of somewhat
messy algebra. But when the algebraic smoke clears the final result
is not very messy considering how hairy many of the intermediate
steps were.

Let M be the mass of the parent particle (initially at rest).

Let m_1 and m_2 be the masses of the daughter particles.

Let r == m_1/m_2 be the mass ratio of the daughter particles.

Let q == 1 - (m_1 + m_2)/M be the fraction of initial mass-energy
of the parent particle that gets converted to kinetic energy in the
decay (in the center of momentum frame).

Let K_1/K_2 be the ratio of the kinetic energies of the daughter
particles (taken in the same order as in the definition of r).

In the Newtonian non-relativistic limit q = 0 and the result is:

K_1/K_2 = 1/r .

In the special relativistic general case 0 < q < 1 and we get:

K1/K_2 = (1 + x)/(r - x)

where the dimensionless parameter x is defined as x == q*(r - 1)/2 .
Note that when we set q = 0 in the special non-relativistic case the
value of x zeros out and the full result boils down to the simpler
Newtonian result.

Thanks,

Ken

You are welcome.

David Bowman