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Re: the energy



Britton wrote:

It appears to me that

delta E = delta m c^2

might be a safe way to express things to a novice.

Not without a highly restrictive context it's not safe.

The most common, the most important delta E is a change
in kinetic energy, which is *not* included in delta m c^2.

================

As a pedagogical point, IMHO one of the niftiest calculations
in the whole world is to examine what relativity says about
mass and kinetic energy in the low-velocity limit.

Start with the ordinary 3-vector momentum ps = [px, py, pz].
Assume that this is the spatial part of some 4-vector.
Let's call the remaining component E. We shall see that
E is the energy, but we don't necessarily know that at
this stage of the calculation. So the 4-vector momentum
is p = [E, px, py, pz]. This 4-vector has some invariant
norm, which we can call m. Therefore
m^2 = E^2 - ps^2 [in units where c=1]

Solving for E and then expanding to lowest order we find
E = m c^2 + ps^2 / (2m) [including explict c]

This is an utterly straightforward Taylor-series expansion.
No wizardry required.

Therefore, the energy can always be considered the sum of
the rest energy plus the kinetic energy.
-- The rest energy (*not* the total energy, not "the"
energy) is m c^2.
-- For almost all practical purposes (i.e. for v<<c), the
kinetic energy is ps^2 / (2m).

IMHO this calculation is vastly more interesting and
vastly more relevant than pole-vaulters in barns. It
tells us something about the fundamental unity of physics.

===

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. 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. 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.

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.