Re: What Hath Einstein Writ?

[Hugh Haskell <hhaskell@MINDSPRING.COM>]

Jim Green was overheard saying, in response to some comments of mine:

>  >
>  >But E=mc^2 does permit mass to be converted to energy, and that, of
>  >course, is the whole point of nuclear physics. The energy that binds
>  >the nucleons together comes from the masses of the nucleons
>  >themselves.
>
> I would hold that this is not very good physics -- I don't even think that
> this is a helpful alternative view.
>
> I would say that the equation E=mc^2 can easily (and rightly) be
> interpreted as saying "energy" _is_ "mass" -- if the _negative_ binding
> energy increases, ie if the energy decreases,  the mass also decreases.
> Both "energy" and "mass" are _properties_ of the "system" not "substances"
> to be converted into anything.

I don't think there is anything wrong with the physics, but I'll give
you that I spoke sloppily, and your statement of the situation is
probably more correct, certainly better phrased.

>  >And those nuclei that prove to be radioactive are those
>  >for which there is another possible "daughter" nucleus whose mass is
>  >less than the parent nucleus less the radioactive particles emitted.
>  >Systems always tend to the lowest available energy state.
>
> No, the daughter particles fly out of the nucleus at high energy and hence
> at a high mass.  Later work is done on those particles by people and
> buildings or boiling water (or something)  thus decreasing the energy and
> thus decreasing the mass of the particle.

I disagree. The daughter nuclei are what is left behind, normally
accounting for little if any of the released energy or of the
momentum of the system because they are locked in a lattice (not
necessarily, though. They may be free particles in which case the
laws of conservation of momentum and energy apply to them). The
emitted particles (alpha, beta, or gamma) are what go flying off,
usually carrying with them most of the released energy and available
momentum, which are almost always large enough to require that they
be calculated relativistically. It does not require that I find, or
even consider the "relativistic mass" of the emitted particles, only
their momentum and energy.

> To say that one can't weigh a fast moving particle on a laboratory balance
> is vacuous.  The mass can be determined -- and it is greater than the
> particle when it finally comes to rest -- what ever that may mean.

Perhaps you can elucidate the experimental method of determining the
mass directly when the particle is moving.

>  >One can get
>  >a lot of mileage out of that idea in discussing nuclei. And nowhere
>  >do we ever use "relativistic mass."
>
> Of course one could (and most do).  When one adds up the various masses to
> calculate the binding energy, what mass do you think that the bound
> particles have??????

A total that is less than the individual masses of the particles when
separated. That is not what we have been calling relativistic mass.

>  >So if we don't need it, why use it? I have to come down on the side
>  >of the professional relativists who have not used the term for years,
>  >if not longer.

For example, see David Griffiths "Introduction to Elementary
Particles," Harper & Row, 1987. A pretty good book, but getting a bit
long in the tooth (which makes my point about relativistic mass not
having been a part of the professional high energy folks' vocabulary
for at least some time). On p. 90, following a lengthy derivation of
the relativistic energy-momentum relationship, he adds the following
parenthetical aside:

"Note that I have never mentioned relativistic mass in all this. It
is a superfluous quantity that serves no useful function. In case you
encounter it the definition is m-rel=gamma*m; it has died out because
it differs from E only by a factor of c^2. Whatever could be said
about m-rel could just as well be said about E, for instance, the
'conservation of relativistic mass' is nothing but conservation of
energy, with a factor of c^2 divided out."

He then goes on to point out that classical physics cannot have
particles of 0 mass because they would, by definition have 0 momentum
and energy. But if a particle travels at the speed of light then the
relativistic expressions for p and E take on an indeterminate form
and so it may become possible to define momentum and energy for such
particles.

And of course, we know that such particles exist (in particular,
photons) which have measurable momentum and energy, but zero mass. I
am hard-pressed to justify a "relativistic mass" for such an object.
It certainly doesn't fit the standard definition of relativistic
mass, and while one might talk about an "effective mass," it really
isn't ever necessary to do so.

> To paint a complete picture of nuclear physics I don't see how one can
> escape the concept of "rest mass" v bound (ie relativistic) mass.

So you are saying that the mass of a bound particle at rest (a
deuteron, for example) is "relativistic" whereas the neutron and
proton that gave rise to the deuteron have a "rest mass"??? I think
what people have been talking about here is the mass that one finds
when one multiplies the "rest mass" by the scale factor gamma, which
will always make the number found larger than the original number,
whereas the mass of the deuteron at rest is less than the mass of its
constituent particles when their masses are taken to be "at rest."

> As Joel points out, one can construct a lecture with out the concept of
> relativistic mass, but one should not feel required to do so.  For my part
> (over here in this dark corner) I think that it is silly to try,
>
> If you do, you end up saying things like "mass" converted  to
> "energy"  Ugh!!! Outrageous!!!

Well, I wouldn't go that far, but I will concede that the choice of
phrasing wasn't wise.

> And still no one has answered the iceberg question: compare the weight of
> an iceberg with the weight of the water from the melting berg.  Explain
> this problem to your students without using the concept of relativistic mass!

I think Leigh has pointed out that that question is not a proper one
since whatever difference there might be is so tiny as to be
unmeasurable, even in principle. On the other hand, it is not
unreasonable to talk about, even though we don't at present know how
to measure, the mass of a hydrogen atom in its first excited state.
It should be larger by a few eV/c^2, but the number is still too
small for instruments that we have available to measure. But in this
case, we are talking about atoms that in both cases are "at rest." In
the case of the iceberg/pool of water that condition does not apply,
since the macroscopic state of the water is very different in the two
conditions (ice/liquid).

Hugh


Hugh Haskell
<mailto://hhaskell@mindspring.com>

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