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Re: the energy



Two very nice examples. Thank you, John Denker!

One question about your second example:

Quoting John Denker <jsd@AV8N.COM>:

Another nifty calculation is the bevatron-design-energy
calculation. The objective is to manufacture antiprotons.
Consider the reaction where an accelerator smashes a
fast-moving proton #1 into a stationary proton #2 to
produce a proton/antiproton pair:
proton + proton --> proton + proton + proton + antiproton

The question is, what is the minimum voltage the accelerator
must produce in order to make this reaction go?

(Note that starting with the given reactants on the LHS,
the RHS is the simplest thing we can have that includes
an antiproton while upholding the conservation laws.)

Write down the [energy,momentum] 4-vector for the RHS
in the center-of-mass frame. Hint: the lowest energy
at which the reaction can take place is the case where
the four products on the RHS have zero velocity relative
to each other.

Yes, got it.

Then analyze the LHS in the lab frame.
Hint: both sides have the same invariant norm. Also
proton #2 has px=py=pz=0 in the lab frame.

Ok.

At this
point you have one equation in one unknown, namely the
KE of proton #1. The whole calculation takes only a
few lines and a few minutes. The result is simple and
IMHO interesting.

Hmmm, I'm evidently not plugging in the formulas as you are.

Using p = gamma m v, E = gamma m c^2, K = (gamma-1) m c^2, I get:

(4mc^2) = sqrt( E^2 - (pc)^2 )
= sqrt( (mc^2 + gamma mc^2)^2 - (gamma m v c)^2 )

[Here the LHS is the invariant you mention, the RHS is formed from the total
energy and momentum of the two protons before the collision, in the lab frame,
recognizing that one is stationary the other moving with some v resulting in
some gamma = 1/sqrt(1-v^2/c^2).]

Solving gives: gamma = 7. Then v = (1-gamma^-2)c ~~ .98 c

And K = (gamma-1)mc^2 = 6 mc^2 = 6 rest energy of a proton = 6(938 MeV)

It's very slick, but this way has the disadvantage of not being very clear to
people unused to gammas in their equations, and it still involves some non-
transparent algebraic manipulation.

I'm sure you have some other steps in mind that make it clearer. But I still
haven't found anything better than just applying conservation of energy and
momentum in the Lab Frame, and occasionally applying (eq1) E^2 = E0^2 + (pc)^2:

E: E + E0 = E'
p: p + 0 = p'

[E and p are the energy and momentum of the moving proton before the collision,
E' and p' that of the all 4 protons after the collision.]

Squaring the energy equation:

(E + E0)^2 = E'^2

Expanding the LHS, and applying (eq1) to the RHS (there are 4 particles after
the collision, so use 4E0):

==> E^2 + 2 E E0 + E0^2 = (4E0)^2 + (p'c)^2

Applying (eq1) to the LHS:

==> (E0^2 + (pc)^2) + 2 E E0 + E0^2 = 16 E0^2 + (p'c)^2

Now plug in p = p' and simplify:

==> E0^2 + 2 E E0 + E0^2 = 16 E0^2
==> 2 E E0 = 14 E0^2
==> E = 7 E0

And, since K = E - E0,

K = 6 E0

I recognize that my repeated application of (eq1) is equivalent to explicitly
examining the invariant norm of the 4-vector, but I wonder, would you be
willing to show us your steps? I can't get it to come out simpler than the
above, in which the algebra is easy but the 4-vector-ness of the approach is
not obvious. Also, I'm using the total energy all the way through. It looks a
lot messier when I put in the kinetic energy.

Note that each of these examples is a pretty strong
advertisement for the 4-vector approach, demonstrating
that you can get interesting results from relativity
without having to write out any Lorentz transformations.

Also, to get back to the point of this thread: these
exercises, individually and (especially) collectively
serve to clarify the ideas of total energy, rest energy,
and kinetic energy.

Yes, very nice! Thanks again,

Ken Caviness
Physics @ Southern