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Q: Mass in Beta - Decay ( Answered,   1 Comment )
Subject: Mass in Beta - Decay
Category: Miscellaneous
Asked by: puzzledoldman-ga
List Price: $25.00
Posted: 04 Aug 2003 05:52 PDT
Expires: 03 Sep 2003 05:52 PDT
Question ID: 238805
Using the word mass in its invariant form, rather than its
relativistic form, what are the best current estimates of the masses
of the neutron and of the proton, electron, and anti-neutron into
which the neutron is believed sometimes to disintegrate?
Subject: Re: Mass in Beta - Decay
Answered By: mathtalk-ga on 04 Aug 2003 19:51 PDT
Hi, puzzledoldman-ga:

By "invariant form" of mass, I suspect you mean the "rest mass" of a
particle, which is independent of its velocity ("relativistic form"
?).  On the other hand perhaps you are asking for the mass of the
particles to be given in "grams" rather than in millions of electron
volts, which might be another meaning of "relativistic form".  I'll
try to accomodate both points.

The beta-decay of a free neutron produces:

 - a proton

 - an electron (or "beta" particle)

 - an (electron) anti-neutrino

through a "weak interaction".  See here:

[Decay of the Neutron]

The neutron is a baryon, like the proton, and an electron is a lepton
with lepton number 1.  The presence of all three particles in the
"yield" provides for conservation of both baryon number (1 in, 1 out)
and lepton number (because the lepton number of the electron
anti-neutrino is -1).

[Physical Constants and Conversion Factors]

The rest mass of the neutron is experimentally determined to be:

mass of neutron  = 1.6749286(10)e-24 gram

The rest mass of the proton is experimentally determined to be:

mass of proton   = 1.6726231(10)e-24 gram

The rest mass of the electron is experimentally determined to be:

mass of electron = 9.1093897(54)e-28 gram

Experimental determination of the rest mass of the electron
anti-neutrino has proven to be a difficult and perhaps profound issue.
 All known experimental evidence is consistent with a zero rest mass
for the electron neutrino (equiv. for the electron anti-neutrino), but
a tiny mass on the order of 1eV or 1.78e-33 gram has not yet been

For links to papers on measurements of neutrino masses, see here:

[Neutrino Mass - Direct Measurement]

On theoretical grounds it is believed that observations of
"oscillation" between flavors of neutrinos, which are thought to be
three (electron, muon, tau), show that some (presumably all) of the
flavors have mass:

[Neutrinos Have Mass]

Please advise me if further clarification would be useful.

regards, mathtalk-ga

Clarification of Answer by mathtalk-ga on 04 Aug 2003 19:57 PDT
Search Strategy

Keywords: "neutron decay"

Keywords: "mass of electron"

Keywords: "mass of neutrino"

Keywords: "mass equivalent" eV

Keywords: "massless neutrinos" oscillation

Request for Answer Clarification by puzzledoldman-ga on 04 Aug 2003 20:11 PDT
Thanks, very helpful.  But if the masses are invariant how come the
sum of the resulting proton and electron masses isn't about equal to
the mass of the neutron?  Does invariant mean invariant only with
respect to velocity?

Clarification of Answer by mathtalk-ga on 04 Aug 2003 20:32 PDT
Hi, puzzledoldman-ga:

Yes, the mass of the proton plus the mass of the electron is "about"
equal to the mass of the original neutron (note the powers of 10
involved).  See the discussion on the link [Neutron Decay] given

The difference is made more meaningful by converting the mass to the
equivalent energy units (E = mc and all that):

mass of neutron  --> 939.5656 MeV

mass of proton   --> 938.2723 MeV

mass of electron -->   0.5110 MeV

which, after subtractions, leaves 0.7823 MeV unaccounted for by the
masses of the proton + electron.

The same site goes on to explain that if only those two particles were
being produced, then their kinetic energy and momentums would be very
constrained.  Experimental observation, however, showed that the
actual distribution of energy and momentum of the electron was a
smooth, continuous one, and that at the peak the electron's
(relativistic) kinetic energy would account for essentially all of the
"lost energy".  These facts could only be explained by the presence of
a third particle, without charge and (within observable limits)
without mass:

"The mysterious particle was called a neutrino, but it was twenty five
years before unambiguous experimental observation of the neutrino was
made by Cowan and Reines."

Today we call that particle an electron anti-neutrino, meaning that
it's in the electron family of leptons and an anti- particle so that
the lepton numbers 1 for the electron and -1 for the electron
anti-neutrino cancel out to conserve the total lepton number (which
was 0 for the original neutron).

regards, mathtalk-ga

Request for Answer Clarification by puzzledoldman-ga on 05 Aug 2003 11:05 PDT
For mathtalk-ga

Thanks for the references.  Since the missing mass after disintgration
exceeds the mass of the electron I guess I'll have to remain somewhat
puzzled about the meaning of invariant mass.


Clarification of Answer by mathtalk-ga on 05 Aug 2003 20:35 PDT
Hi, puzzledoldman-ga:

It is tempting to say that the "lost" mass of the neutron (after that
of the proton, electron, and anti-neutrino are consider) is converted
into "kinetic energy" of those emitted particles.  This has the germ
of the right idea but is not fully accurate, since there is a total
combination of energy due to a particle's momentum as well as its mass
equivalent.  So let me drop back a bit and try to give a more complete

The old law of conservation of mass was replaced under Einstein's
theory of special relativity by conservation of mass-energy, whose
equivalence is given by the well-known formula E = mc.  For
background definitions, see this:

[Theory of Special Relativity]

You will not find there any reference to "mass in its invariant form",
but you will find mention of the mass of an object at rest ("rest
mass") and how this is related to the "total energy" E of a freely
moving particle.  The formula is:

E = (mc) + (pc)

where m is the "rest mass", c is the speed of light, p is the

p = mv/SQRT(1 - (v/c))

taking into account the particle's relativistic mass when v is the
particle's velocity. With a little algebra one more simply say that
total energy is:

E = Mc

where M = m/SQRT(1 - (v/c)) is the particle's relativistic mass.

Note that the first equation gave a formula for the energy squared,
E.  An intuitive (but incorrect) guess would be that E could be
expressed simply as a sum of the particle's rest mass energy
equivalent plus the particle's "kinetic energy", familiar from
Newtonian mechanics as (1/2)mv.

But although things are not so intuitive, the mathematics of the above
is elegant in unifying the correct relativistic computations with
Newtonian formulas in the limiting case of comparatively slow
velocities v (i.e. compared to the speed of light).

What I'm trying to say is that the missing mass from the beta decay is
fully accounted for by the momentum of the particles yielded,
especially that of the electron and the electron anti-neutrino.  As
mentioned earlier, the experimental measurement of the electron's
momentum shows that it varies smoothly over a range, with a maximum
equal to the entire 0.7823 MeV of missing mass energy equivalent.  The
smoothness of the distribution beneath this shows that the electron
anti-neutrino may (randomly) carry of a certain portion of the energy
release through its momentum.

If the total energies of proton, electron, and electron anti-neutrino
are combined, then the total energy of the original neutron is
entirely accounted for.

Perhaps you have doubts that a portion of the original neutron's rest
mass, which I think you consider "mass in its invariant form", can be
converted into energy.  Yet this is precisely what happens with beta
decay, as also with the fission decay of unstable isotopes.  In a real
sense the strong forces that bind baryons (neutrons and protons)
together in the nucleus of atoms of a stable isotope inhibit the
spontaneous decay of neutrons that would otherwise take place fairly
quickly, were neutrons to be freed from the nucleus.

regards, mathtalk-ga
Subject: Re: Mass in Beta - Decay
From: hedgie-ga on 09 Aug 2003 06:56 PDT
What does that means in  physics?

  Modern physics equations are
formulated so that they are valid in all frames of references.
 We can transform the quantities from one frame of reference into
another, e.g. into one which is moving with respect to the former.

 The laws of transformation are different in Newtonian and
Relativistic physics.  (Both sets of equations are  invariant these
days. It was not so when people were still arguing about
heliocentric vs.geocentric models).

In  Newton's universe the masses before and after an event
add up to the same number.
 In this sense mass is conserved, and is also
truly invariant, i.e. same in all frames. Better way to say it, mass
is a scalar. So is Energy .  Both are conserved.  Event may be
collision, disintegration, etc.

In Einstein's universe it is more complex.  There are quantities which
are conserved, and some of them are scalars, e.g. the lepton and
barion numbers mathtalk is talking about. Some are vectors, and they
have a complex law of transformation.   Since classical physics
approximates relativity  in same cases, we can point
out a quantity which has some properties of classical mass and energy.
This, this correspondence, is source of a lot of semantic confusion
you see on the Internet.   Mass, or more exactly mass-energy is not a
scalar in this aproximation or analogy.      In the context of your 
question it means: there is no  single-number-quantity, which you
can add (for all particles in the event) and get same value, before
and after the event, in any frame. In other words,  there is no exact
equivalent of scalar mass in relativity.
  There are many  quotes about Internet.  One cartoon  says
 'On Internet no one knows  you are a dog'  (or  a hedgehog :-)
I would add to the collection,  the following:
"On Internet, everybody understands relativity. But many
understand it differently than most textbooks." If you are seriously
interested, you may ask for recommended reading, to cover the
area of interest at an appropriate level of complexity. This is just
a free comment. It recommends a critical, skeptical approach to what
you  read.  Particularly when it is on the Internet, and for free.

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