Neutrino mass

I have heard that solar neutrino experiments have shown that neutrinos
oscillate between the three types. This supposedly proves that
neutrinos have mass. This seems to violate the conservation of mass,
or momentum, or both.

On the other hand if neutrinos were massless, they would quite
reasonably be able to oscillate without violating any conservation
laws (they can not change their spin direction, however; so there are
still two fundamental types of neutrinos.)

Why am I wrong?

Andrew Usher

Andrew Usher wrote:
I have heard that solar neutrino experiments have shown that neutrinos
oscillate between the three types. This supposedly proves that
neutrinos have mass.

Not "proves", but provides fairly strong evidence. And even more so if
you combine these results with the ones of the other experiments
(SuperKamiokande, for starters).


This seems to violate the conservation of mass, or momentum, or both.

No. The "mass eigenstates" of the neutrinos are not the same as the
"flavour eigenstates".

To elaborate: according to the theory, there are three neutrinos, let's
call them nu_1, nu_2 and nu_3, which all have a fixed, defined mass.
However, they are *not* identical to the known electron, muon and
tau-neutrinos! In contrast, these "flavour eigenstates" are linear
combinations of the "mass eigenstates", e.g.:
nu_electron = a u_1 + b nu_2 + c nu_3
etc.

Have you ever heard of the Cabbibo-Kobayashi-Maskawa (sp???) matrix?
That's essentially the same, although not for neutrinos, but for quarks.


On the other hand if neutrinos were massless, they would quite
reasonably be able to oscillate without violating any conservation
laws (they can not change their spin direction, however; so there are
still two fundamental types of neutrinos.)

The oscillation occurs *precisely* because the mass eigenstates are
*not* identical to the flavour eigentstates, and the masses are
different. This follows simply from the time evolution of the system
(Schroedinger equation).

Why on earth *should* the neutrinos oscillate if they were massless?


Why am I wrong?

Probably because you don't know enough Quantum Mechanics?



Bye,
Bjoern

Bjoern Feuerbacher wrote in message ...
Andrew Usher wrote:
I have heard that solar neutrino experiments have shown that neutrinos
oscillate between the three types. This supposedly proves that
neutrinos have mass.

Not "proves", but provides fairly strong evidence. And even more so if
you combine these results with the ones of the other experiments
(SuperKamiokande, for starters).


This seems to violate the conservation of mass, or momentum, or both.

No. The "mass eigenstates" of the neutrinos are not the same as the
"flavour eigenstates".

To elaborate: according to the theory, there are three neutrinos, let's
call them nu_1, nu_2 and nu_3, which all have a fixed, defined mass.
However, they are *not* identical to the known electron, muon and
tau-neutrinos! In contrast, these "flavour eigenstates" are linear
combinations of the "mass eigenstates", e.g.:
nu_electron = a u_1 + b nu_2 + c nu_3
etc.


OK, there are three types of neutrinos, each of which is a linear
combination of the three flavors we can observe. So every time we do
observe one, it is randomly one of the three flavors, with
probabilities determined by this matrix.

Now I've heard of an experiment being performed, where they're going
to send electron neutrinos 500 miles through the earth, and see how
many have changed. But if the above is true, the distance shouldn't
matter, and it would be much easier to do it within one facility.

Have you ever heard of the Cabbibo-Kobayashi-Maskawa (sp???) matrix?
That's essentially the same, although not for neutrinos, but for quarks.


Quarks don't oscillate.


On the other hand if neutrinos were massless, they would quite
reasonably be able to oscillate without violating any conservation
laws (they can not change their spin direction, however; so there are
still two fundamental types of neutrinos.)

The oscillation occurs *precisely* because the mass eigenstates are
*not* identical to the flavour eigentstates, and the masses are
different. This follows simply from the time evolution of the system
(Schroedinger equation).


I do know that Schrodinger's equation is time independent (that's why
it confuses people into thinking that electrons are really smeared out
into orbitals - it is giving an average over infinite time.)

Why on earth *should* the neutrinos oscillate if they were massless?


Because they can?


Why am I wrong?

Probably because you don't know enough Quantum Mechanics?



Bye,
Bjoern

Andrew Usher

Dear Andrew Usher:

"Andrew Usher" wrote in message
om...
Bjoern Feuerbacher wrote in message
...
....
The oscillation occurs *precisely* because the mass eigenstates are
*not* identical to the flavour eigentstates, and the masses are
different. This follows simply from the time evolution of the system
(Schroedinger equation).


I do know that Schrodinger's equation is time independent (that's why
it confuses people into thinking that electrons are really smeared out
into orbitals - it is giving an average over infinite time.)

Electrons are not billiard balls. Electrons manage to populate the entire
orbital *without* moving (which would create a magnetic field, and allow
energy loss). "Smeared out" is exactly sufficient to describe this. What
else would you propose?

David A. Smith

"N:dlzc D:aol T:com \(dlzc\)" N: dlzc1 D:cox wrote in message news:qY2rc.31771$Md.30744@lakeread05...
Dear Andrew Usher:

"Andrew Usher" wrote in message
om...

I do know that Schrodinger's equation is time independent (that's why
it confuses people into thinking that electrons are really smeared out
into orbitals - it is giving an average over infinite time.)

Electrons are not billiard balls. Electrons manage to populate the entire
orbital *without* moving (which would create a magnetic field, and allow
energy loss). "Smeared out" is exactly sufficient to describe this. What
else would you propose?

David A. Smith

'Energy loss' is forbidden because there are no lower quantum states
to decay to, not because of the presence/absence of a magnetic field.
There is always such a field due to the electron's spin, anyway.

This approach is reasonable for a 1s orbital, but what about a
1,000,000s orbital? Look up 'correspondence principle' (as n,k - inf
the Bohr-Sommerfeld orbits become more and more exact).

Andrew Usher

Dear Andrew Usher:

"Andrew Usher" wrote in message
om...
"N:dlzc D:aol T:com \(dlzc\)" N: dlzc1 D:cox wrote in
message news:qY2rc.31771$Md.30744@lakeread05...
Dear Andrew Usher:

"Andrew Usher" wrote in message
om...

I do know that Schrodinger's equation is time independent (that's why
it confuses people into thinking that electrons are really smeared
out
into orbitals - it is giving an average over infinite time.)

Electrons are not billiard balls. Electrons manage to populate the
entire
orbital *without* moving (which would create a magnetic field, and
allow
energy loss). "Smeared out" is exactly sufficient to describe this.
What
else would you propose?

David A. Smith

'Energy loss' is forbidden because there are no lower quantum states
to decay to, not because of the presence/absence of a magnetic field.
There is always such a field due to the electron's spin, anyway.

The electron's spin is a quantum property. It is conserved, it has two
states, but it has nothing to do with a physical rotation, nor an induced
magentic field intrinsic to the electron.

If the electron moves in its orbital around the nucleus, an magnetic field
would be produced. Such a magnetic field would allow energy to be
withdrawn from the electron, until it spiralled into the nucleus (a
psossibility... electron capture does happen for some isotopes). Since
such does not happen in general, it is consistent with the electron being
"smeared out" across the entire orbital. I know you don't like it, but it
fits the facts.

This approach is reasonable for a 1s orbital, but what about a
1,000,000s orbital? Look up 'correspondence principle' (as n,k - inf
the Bohr-Sommerfeld orbits become more and more exact).

Look up what "spin" is for the electron. Then think about the silliness
you have posted.

David A. Smith

Andrew Usher wrote:
Bjoern Feuerbacher wrote in message ...

Andrew Usher wrote:

I have heard that solar neutrino experiments have shown that neutrinos
oscillate between the three types. This supposedly proves that
neutrinos have mass.

Not "proves", but provides fairly strong evidence. And even more so if
you combine these results with the ones of the other experiments
(SuperKamiokande, for starters).



This seems to violate the conservation of mass, or momentum, or both.

No. The "mass eigenstates" of the neutrinos are not the same as the
"flavour eigenstates".

To elaborate: according to the theory, there are three neutrinos, let's
call them nu_1, nu_2 and nu_3, which all have a fixed, defined mass.
However, they are *not* identical to the known electron, muon and
tau-neutrinos! In contrast, these "flavour eigenstates" are linear
combinations of the "mass eigenstates", e.g.:
nu_electron = a u_1 + b nu_2 + c nu_3
etc.



OK, there are three types of neutrinos, each of which is a linear
combination of the three flavors we can observe. So every time we do
observe one, it is randomly one of the three flavors, with
probabilities determined by this matrix.

Not only by this matrix - also by the flavour it was in when it was
created (neutrinos get created in weak reactions in the states nu_e,
u_mu and nu_tau, not in the mass eigenstates nu_1, nu_2 and nu_3) and
the time since the creation.


Now I've heard of an experiment being performed, where they're going
to send electron neutrinos 500 miles through the earth, and see how
many have changed. But if the above is true, the distance shouldn't
matter, and it would be much easier to do it within one facility.

Well, the distance (or, more precisely, the *time*) *does* matter. As I
already mentioned: the oscillation is a simple consequence of the *time
evolution*.


Have you ever heard of the Cabbibo-Kobayashi-Maskawa (sp???) matrix?
That's essentially the same, although not for neutrinos, but for quarks.



Quarks don't oscillate.

Yes, they do. You don't see this directly, since free quarks aren't
observable - but K- and B-mesons, which are *made* from quarks, do
indeed oscillate.


On the other hand if neutrinos were massless, they would quite
reasonably be able to oscillate without violating any conservation
laws (they can not change their spin direction, however; so there are
still two fundamental types of neutrinos.)

The oscillation occurs *precisely* because the mass eigenstates are
*not* identical to the flavour eigentstates, and the masses are
different. This follows simply from the time evolution of the system
(Schroedinger equation).



I do know that Schrodinger's equation is time independent

Err, there is a time-independent from of Schroedinger's equations and a
time-dependent form. The second is more general than the first one.
1) H psi(x,y,z) = E psi(x,y,z)
2) H phi(x,y,z,t) = i hbar d/dt phi(x,y,z,t)

You get the first from the second by making the ansatz
phi(x,y,z,t) = psi(x,y,z) e^(i E t/hbar)
These are the so-called "stationary states". The neutrinos nu_1, nu_2
and nu_3 are stationary states, but nu_e, nu_mu and nu_tau are not.


(that's why
it confuses people into thinking that electrons are really smeared out
into orbitals - it is giving an average over infinite time.)

I don't think that taking the quantum mechanical expectation value has
got anything to do with a time average.


Why on earth *should* the neutrinos oscillate if they were massless?



Because they can?

Please present a bit math on this.

*I* can show you mathematically why neutrinos *have* to oscillate if
there are different masses for them, and if the mass eigenstates are not
equivalent to the flavour eigenstates.

[snip]


Bye,
Bjoern

N:dlzc D:aol T:com (dlzc) wrote:
Dear Andrew Usher:

"Andrew Usher" wrote in message
om...

[snip]


The electron's spin is a quantum property. It is conserved, it has two
states, but it has nothing to do with a physical rotation, nor an induced
magentic field intrinsic to the electron.

Sorry, but you are wrong on this. A magnetic moment is associated with
the spin of the electron (ever heard of the g-factor? what do you think
it describes?), and something with a magnetic moment has a magnetic field.


If the electron moves in its orbital around the nucleus, an magnetic field
would be produced. Such a magnetic field would allow energy to be
withdrawn from the electron,

How does this follow?


until it spiralled into the nucleus

The "spiralling" would happen because the electron would be
*accelerated* (moving in a circular orbit is an accelerated movement).
And an accelerated charge radiates electromagnetic waves. The motion
itself is not important - the *acceleration* is important.


(a psossibility... electron capture does happen for some isotopes).

But has nothing to do with "spiralling" into the nucleus.


Since
such does not happen in general, it is consistent with the electron being
"smeared out" across the entire orbital.

As long as one keeps in mind that this is only a nice little picture,
but not necessarily the "truth" (the "truth" is that the probability
amplitude to find the electron is determined by the wave function, and
that we can't say anything more about it), there is no problem with this
"smearing out", agreed.


I know you don't like it, but it fits the facts.


This approach is reasonable for a 1s orbital, but what about a
1,000,000s orbital? Look up 'correspondence principle' (as n,k - inf
the Bohr-Sommerfeld orbits become more and more exact).


Look up what "spin" is for the electron. Then think about the silliness
you have posted.

What has spin to do with the correspondence principle for high orbitals?


Bye,
Bjoern

Andrew Usher wrote:
"N:dlzc D:aol T:com \(dlzc\)" N: dlzc1 D:cox wrote in message news:qY2rc.31771$Md.30744@lakeread05...

Dear Andrew Usher:

"Andrew Usher" wrote in message
.com...


I do know that Schrodinger's equation is time independent (that's why
it confuses people into thinking that electrons are really smeared out
into orbitals - it is giving an average over infinite time.)

Electrons are not billiard balls. Electrons manage to populate the entire
orbital *without* moving (which would create a magnetic field, and allow
energy loss). "Smeared out" is exactly sufficient to describe this. What
else would you propose?

David A. Smith


'Energy loss' is forbidden because there are no lower quantum states
to decay to, not because of the presence/absence of a magnetic field.
There is always such a field due to the electron's spin, anyway.

This approach is reasonable for a 1s orbital, but what about a
1,000,000s orbital? Look up 'correspondence principle' (as n,k - inf
the Bohr-Sommerfeld orbits become more and more exact).

In the limit of high n and l (what do you mean by "k"?), the probability
to find an electron at the distance from the nucleus predicted by Bohr's
model becomes greater and greater. But this has nothing to do with the
electron actually moving on a circular orbit with that radius.


Bye,
Bjoern

Dear Bjoern Feuerbacher:

"Bjoern Feuerbacher" wrote in message
...
N:dlzc D:aol T:com (dlzc) wrote:
Dear Andrew Usher:

"Andrew Usher" wrote in message
om...

[snip]


The electron's spin is a quantum property. It is conserved, it has two
states, but it has nothing to do with a physical rotation, nor an
induced
magentic field intrinsic to the electron.

Sorry, but you are wrong on this. A magnetic moment is associated with
the spin of the electron (ever heard of the g-factor? what do you think
it describes?), and something with a magnetic moment has a magnetic
field.

OK, then it depends on your definition of "magentic field", doesn't it.
The neutron also has a non-zero magnetic moment. The ability to store
energy doesn't require a classically defined magnetic field. Because such
a field stores energy outside the particle. I believe you are incorrect,
Bjoern.

If the electron moves in its orbital around the nucleus, an magnetic
field
would be produced. Such a magnetic field would allow energy to be
withdrawn from the electron,

How does this follow?

Electric motors transfer energy to a moving rotor using magnetic fields. A
moving-electron-in-Bohr-orbital would be a generator, transmitting power to
the space around it, until it dropped into the nucleus. As you note, all
the interesting particles have magnetic moments, and they all would siphon
its energy off.

until it spiralled into the nucleus

The "spiralling" would happen because the electron would be
*accelerated* (moving in a circular orbit is an accelerated movement).
And an accelerated charge radiates electromagnetic waves. The motion
itself is not important - the *acceleration* is important.

And the electron radiates no such energy. It would be accelerated in a
Bohr orbital, and even more so in a Schroedinger orbital. Do a search on
"cyclotron radiation". You get radiated *magnetic* energy from orbiting
charges, even when the D's are switched off.

(a psossibility... electron capture does happen for some isotopes).

But has nothing to do with "spiralling" into the nucleus.

Some would argue that electron capture was an electron ending up in the
nucleus. I wanted to concede that this happened up front, so that I did
not hide anything.

Since
such does not happen in general, it is consistent with the electron
being
"smeared out" across the entire orbital.

As long as one keeps in mind that this is only a nice little picture,
but not necessarily the "truth" (the "truth" is that the probability
amplitude to find the electron is determined by the wave function, and
that we can't say anything more about it), there is no problem with this
"smearing out", agreed.

The electron is not a billiard ball. It is consistent with being
completely distributed into its orbital, once the known, nearly exact
amount of information (namely the binding energy) is transmitted to the
Universe. It is then not moving in a classical Newtonian (or Faraday)
sense, wrt the Universe at large. I don't think we have anything other
than "pretty pictures" to work with, which is what all theories are. I
would ask you to consider what happens with Bose-Einstein condensates when
a roughly equivalent amount of energy is removed from them... their
individual constituents are non-localized also.

I know you don't like it, but it fits the facts.


This approach is reasonable for a 1s orbital, but what about a
1,000,000s orbital? Look up 'correspondence principle' (as n,k - inf
the Bohr-Sommerfeld orbits become more and more exact).


Look up what "spin" is for the electron. Then think about the
silliness
you have posted.

What has spin to do with the correspondence principle for high orbitals?

His argument is based on his misconception of what electron spin is. I
should have cut his last sentence off. I did not respond to it, as I did
not feel it to be germaine to his issues with spin.

Sorry to disagree, Bjoern. You have a level head. If you are still
convinced I am incorrect, let me know and I'll shut up.

David A. Smith