Wolfgang Pauli once said (regarding the neutrino):

I have done a terrible thing. I have postulated a particle that cannot be detected.

Why did he figure it couldn't be detected? Was this because he thought it was massless? According to Wikipedia, "neutrinos were long believed to be massless".

If so, why did they think it was massless? I thought the particle was hypothesised in order to maintain the conservation of momentum in a beta decay. If it was massless, this wouldn't have any effect, right?


6 Answers 6


I thought the particle was hypothesised in order to maintain the conservation of momentum in a beta-decay. If it was massless, this would have no effect, right?

This is where you are confused.

Having no mass does not mean having no momentum.

I think you are probably thinking of momentum as Newtonian mechanics would express it : $P=mv$

However Einstein came up with a relativistic expression linking energy and momentum which is :

$$E^2 = m^2c^4+p^2c^2$$

where $m$ is rest mass.

Now even if rest mass is zero, the particle has energy (like photons do) and you get :

$$p = \frac E c$$

So massless neutrinos would have momentum.

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    $\begingroup$ The point is, all massless particles travel at the speed of light. This is what gives them non-zero momentum. $\endgroup$
    – TonyK
    Commented Jun 9, 2021 at 8:56
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    $\begingroup$ Photons are the best known example of this. They travel at the speed of light and carry both energy and momentum, though having no (rest) mass. $\endgroup$
    – md2perpe
    Commented Jun 9, 2021 at 9:11
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    $\begingroup$ How does this answer "Why was the neutrino thought to be massless?" Inexplicable. $\endgroup$
    – ProfRob
    Commented Jun 11, 2021 at 16:11
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    $\begingroup$ @ProfRob It addresses the OP's misconception that massless particles do not carry momentum, which is the misconception preventing them from accepting (as I read the question) that the neutrino conserved momentum in certain reaction. $\endgroup$ Commented Jun 11, 2021 at 17:21
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    $\begingroup$ What you've said isn't wrong, it just doesn't address the main Q at all. Neutrinos were considered to be massless and it isn't because of a misunderstanding of SR. $\endgroup$
    – ProfRob
    Commented Jun 11, 2021 at 17:26

An interesting aspect not touched on by existing answers at the time of writing.

What was observed about beta decay is that for the same change in the atomic nucleus, the emitted electron (which, being charged, is easily detected) had a wide range of energies. Some electrons come out with high energy, some with low, even though the same initial and final nucleus is involved. So the issue is: how to explain this aspect?

Now it is an interesting feature of energy and momentum processes in general that the decay of one particle into two is a special case, because it has a unique solution. That is, there is only one possible value for the energies of the two decay products. But the decay of one particle into three has a variety of solutions for the energies, depending on how the energy and momentum is divided up between the products. For the first case you have $$ E_0 = E_1 + E_2 \\ {\bf p}_0 = {\bf p}_1 + {\bf p}_2 \tag{1} $$ (where 0 labels the initial particle, 1 and 2 the products) and for the second case you have $$ E_0 = E_1 + E_2 + E_3\\ {\bf p}_0 = {\bf p}_1 + {\bf p}_2 + + {\bf p}_3 \tag{2} $$ In the case of beta decay the initial particle (0) is a neutron, and the decay products (1,2,3) are proton, electron, and anti-neutrino. But of course this was not known at first. It looked as if you had just the initial nucleus, the final nucleus, and an electron (the electron can also be called a beta particle).

Let's solve equations (1). For example, we might have ${\bf p}_0 = 0$ and then ${\bf p}_1 = - {\bf p}_2$ so $|p_1| = |p_2|$. This fact can be put into the energy equation, using the relativistic relation $E^2 - p^2 c^2 = m^2 c^4$ to relate each momentum to the energy and rest mass of that particle. One finds there is enough information to get a unique solution, and one finds $$ E_1 = \frac{M_0^2 + m_1^2 - m_2^2}{2 M_0}c^2 $$

In the case of equations (2), on the other hand, there is a range of possible values for $E_1$ depending on the directions of emission and the way the energy and momentum is divided up between particles 1,2 and 3. This is a well known fact to anyone studying particle physics. So when you see data in which an emitted particle (here, an electron) has a range of energies, i.e. some decays give one energy, some give another, then you may suspect that there are two or more further particles in the decay products, not just one (e.g. a neutron decays to three things not two). But when you see a sharp spike in the energy spectrum, with just one energy for some given decay product, then you think "ah-ha! A two-body decay process is involved!"

At the time of the proposal of the neutrino, the options were, roughly: (i) abandon energy or momentum conservation or both; (ii) propose that there is another particle involved which is hard to detect directly (it is uncharged and has low or perhaps zero mass); (iii) propose that the rest masses of otherwise identical nuclei can be different; (iv) other suggestions. Pauli recommended option (ii), but admitted that there was not much other evidence to go on.

Finally, then, to answer the question: the reason to propose the neutrino was massless is that the observed energy data can be used to constrain the masses. In the case of beta decay, some of the decays give essentially all the energy to the electron, so there is none left over for the neutrino, so its mass must be either zero or small compared to that of the electron.

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    $\begingroup$ Re: “then you may suspect that there are two or more further particles in the decay products,” I believe here you want either “one or more further particles,” or “two or more total particles,” since “two further particles” would indicate a minimum of 3 particles ejected. $\endgroup$
    – KRyan
    Commented Jun 10, 2021 at 22:21
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    $\begingroup$ @KRyan thanks for this. I was counting the total decay products, i.e. (final nucleus) + electron + neutrino = 3. I made an edit to be clearer. $\endgroup$ Commented Jun 10, 2021 at 22:58

Why did he figure it couldn't be detected?

Because the neutrinos interact so weakly that it was for some time believed that it will not be possible to detect. E.g. in the 1934 paper by Bethe and Pierls, where the crossection for neutrino interacting with matter was computed, they concluded that "It is therefore absolutely impossible to observe processes of this kind with the neutrinos created in nuclear transformations."

Was this because he thought it was massless?


I thought the particle was hypothesised in order to maintain the conservation of momentum in a beta-decay. If it was massless, this would have no effect, right?

No, the massless particle carries energy and momentum just as good as the massive one.

To correct you a bit, the neutrinos were hypothesised to explain the violation of the conservation of energy. The observation of the "violation" of the conservation of momentum due to the neutrinos carrying momentum was actually the first evidence that this idea of new "invisible" particle is not that stupid.

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    $\begingroup$ This doesn't address "Why was the neutrino thought to be massless?" (unless you are saying they weren't). $\endgroup$
    – ProfRob
    Commented Jun 11, 2021 at 17:28

Here is a bit more background which might help.

At the time Pauli made that statement, it was known that charged particles (alphas, betas) could be detected and their charges, energies, and masses measured. The techniques needed to detect the existence of neutral particles had not yet been developed, but would appear not too long thereafter, leading to the first detection of the neutron. But that neutron had mass, and it interacted strongly with protons if you knew how to get them to collide. Finding a massless or almost massless particle that did not interact with electromagnetic fields or with protons seemed to be impossible at the time.

The early neutrino experiments suggested that the neutrino had so little mass that it might actually be massless. Later experiments placed upper bounds on the neutrino mass which meant that it in fact had nonzero mass, but that mass was extremely small. Those experiments were enabled by the availability of intense neutrino sources (see below) and by the invention of extremely sensitive and clever detector systems. As those detectors got better, the bounds on the possible mass of the neutrino got tighter.

Bethe and Peierls published their paper at a time when atomic reactors hadn't yet been invented. Once reactors were built, physicists realized that they could be used to produce enough neutrinos per second to be detected despite the extremely low cross-section for neutrino-matter interactions.

Next, huge, high-sensitivity detectors coupled with huge, high-energy particle accelerators allowed even more precise measurements to be made with extremely intense beams of neutrinos, and enabled the researchers to devise experiments specifically designed to reveal the neutrino's deepest secrets.

  • $\begingroup$ This answer is well and good. Is there a a lower bound on the mass? An upper bound along still admits the possibility that neutrinos are massless. $\endgroup$ Commented Jun 27 at 2:58

The following properties suggested masslessness:

  • Neutrinos appear in neutron decay to conserve lepton number l.
  • Neutrons appear in neutron decay to preserve 1/2 h angular momentum. If it were only missing linear momentum they could have postulated a photon and taken quite a while to discover they were wrong. The photon has angular momentum of h so there was already not a rule that said you needed mass to carry angular momentum.
  • Neutrinos have a definite chirality. Assigning them a mass means they don't travel at c which was at the time incompatible with the math for particles with a definite chirality.

In short, massless neutrinos fit the physics theories from the 1950s through the 1990s well and assigning them a mass requires developing new math, so they were assumed to be massless until proven otherwise. This is not lazyness; this is predicting results from the math.

The first hint there was a problem was the solar neutrino flux (when we could finally measure it) was found to be between 1/3 and 1/4 the expected value.

The current situation is the holding that neutrinos have mass based on it being the best explanation of the missing neutrino flux; the mechanism having now been reproduced in the world lab (that is; using disjoint instruments measuring the same source beaming through the earth). Thus, l is not conserved.

Attempts to measure the speed of the neutrino travel as distinct from c have proven so far beyond definitive reach. This would prove it for certain; but it's really hard to measure the existence of a rest mass of less than 1ev that way.


Why did he figure it couldn't be detected? Was this because he thought it was massless?

No. "Detection" in the context of particle physics is close to being synonymous with "interaction"; particles are detected through their interaction with other particles. These interactions are through the four forces: gravity, weak interaction, electromagnetism, and strong interaction.

Gravity is extremely weak, and only in large scale structures is it significant. In particle physics, it's generally too small to be noticeable. Detecting a neutrino through its gravitational attraction of other particles is well beyond current capabilities.

Electromagnetic force is, depending on the scale, the first or second strongest force. It is generally the easiest interaction to detect. A particle being electrically neutral makes it much harder to work with.

Neutron are electrically neutral, but they interact with the strong interaction, so they are more difficult to detect than proton, but advances in particle physics has allowed them physicists to work with them.

Neutrinos don't experience the strong interaction, they're electrically neutral, and like other particles, their mass is negligible when it comes to gravitational attraction. That leaves only the weak interaction, which is a millionth the strength of the strong interaction. Strictly speaking, the very fact that neutrino can be emitted means that that interaction can be time reversed, and so it is theoretically possible to detect them, the issue is how likely that interaction is.

If so, why did they think it was massless?

Their mass is extremely small, both in an absolute sense (an electron is half a million times as massive) and in terms of comparison to their energy. As a result, their velocity is more than 99.9999999999% of the speed of light. Velocity is a major distinguishing property between massive and massless particles: particles with mass travel at less than $c$, and massless particles travel at exactly $c$. With such a high velocity, it is difficult to use this property to distinguish neutrinos from massless particles.

If it was massless, this wouldn't have any effect, right?

The term "mass" can be ambiguous. It can refer to rest mass or relativistic mass. In this context, it means rest mass, and particles with zero rest mass can still carry momentum. In the formula "momentum = mass x velocity", mass is the relativistic mass.


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