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I was thinking of a universe in which rest mass like spin, and electrical charge is quantized. In this hypothetical universe elementary particles could have an intrinsic/quantized rest mass so long as it is a whole number multiple of $\frac{1}{5}$ of a certain constant. Any rest mass that is not a whole number multiple of this constant has to be attained from interacting with the Higgs Field.

All the fermions of this universe would have some attained rest mass but only some gauge bosons would have attained rest mass. Some gauge bosons would have only intrinsic/quantized rest mass with no attained rest mass while others would be massless.

Would the intrinsic/quantized mass in this hypothetic universe effect how stable an elementary particle would be or would it only be the attained restmass that would effect how stable an elementary particle would be? What else would need to change for particles to have quantized rest mass?

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    $\begingroup$ Somehow, this is going to break the universe, the question is how $\endgroup$ – Hohmannfan Mar 8 '16 at 6:59
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    $\begingroup$ How do we know they aren't quantized in the current universe? The first few Google results seemed to suggest to me that it's not clear if rest mass is quantized in some way or not. For example, see the comment here: physics.stackexchange.com/questions/122/is-rest-mass-quantized "Sure, there's nothing wrong with entertaining the concept, and some theories do predict quantization of mass, although it may be more complicated than just multiples of a certain value. But there's no experimental evidence that mass is actually quantized, at least not for fundamental particles" $\endgroup$ – sumelic Mar 8 '16 at 17:30
  • $\begingroup$ Interesting abstract, I can't wait to read the paper. $\endgroup$ – Scott Downey Mar 16 '16 at 14:11
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The important things to take from the question:

  1. elementary particles could have an intrinsic/quantized rest mass

  2. Any rest mass that is not a whole number multiple of this constant has to be attained from interacting with the Higgs Field

Answer

You have just described exactly how things are (to current physics knowledge) in our own universe, with the quantised mass-constant $= 0$, where any non-zero mass (of the fundamental/elementary particles) is obtained via interaction with the Higgs field/boson. While all fermions attain a mass via the Higgs, not all gauge bosons do (the photon).

If, however, you want a non-zero mass-constant, then this requires breaking charge conservation (electric, weak and strong/colour). Stability is more to do with the total mass, with the decay constant of a particle related to the imaginary part of the mass of that particle.

Explanation

So, where does mass come from (apart from the Higgs)? (here, intrinsic mass will be used to refer to mass that isn't obtained from the Higg's mechanism)

That would be the Standard Model of particle physics (Brace yourselves...): all the known elementary particles and forces (excluding gravity) are described by terms in the following: $$\mathcal{L} = \mathcal{L}_{Higgs} + \mathcal{L}_{gauge} + \mathcal{L}_{lepton} + \mathcal{L}_{quark} + \mathcal{L}_{Yukawa}$$

Each term in the above contains mathematical terms such as $\bar f\cdot \gamma^a g_2 W_a f$ (in this case, interaction of a lepton or quark with part of the electroweak field. Other mathematical terms are not written for brevity), which can be drawn as part of Feynman diagrams, which in turn gives all the maths/statistics required to do particle physics. Including calculating the mass of the elementary particles - a term giving a particle $p$ a mass is written as a term such as $m\bar pp$.

From condition 2, we can ignore $\mathcal{L}_{Higgs}$.

What about the other terms? $\mathcal{L}_{gauge}$ contains the electroweak and strong fields, which have no intrinsic mass. Actually, neither does $\mathcal{L}_{lepton}$ or $\mathcal{L}_{quark}$. So, we're left with $\mathcal{L}_{Yukawa}$

$\mathcal{L}_{Yukawa}$ is, of itself, still quite long, containing terms involving the allowed possible interactions of fermions with the Higgs boson, so can also be ignored by point 2.

Or, the only way that elementary particles can have mass is from the Higgs boson. This includes, to my knowledge, neutrinos.

The reason that this is how things are is that any possible mass terms not arising from some form of interaction with the Higgs in the standard model wouldn't be invariant under what's known as gauge symmetry. Due to what's known as Noether's theorem, this means that making the Standard Model non-invariant under gauge symmetry would break conservation of charge - electric, weak and strong (colour).

Putting it another way, the only way that your top paragraph isn't true (in a universe where the standard model is correct, anyway) is by not having charge conserved. If charge is conserved (and the standard model is correct), then your first paragraph is true, with the constant being $0$.

Conveniently, the Yukawa coupling gives all fermions an 'attained mass'. Also convenient is that the photon, a gauge boson remains massless. However, no bosons have 'only intrinsic/quantized mass with no attained mass' [I've removed the word 'rest' because I'm pedantic about not using the term 'rest mass'] where the intrinsic mass is non-zero (the photon is a gauge boson with zero mass). If you want this to be the case, then feel free to violate charge conservation and the set the mass-constant to be whatever you like!

How stable is an elementary particle?

For a non-rigorous thought about how decay of a particle works, let's imagine that we've got a stationary particle* of mass $m$. This gives the energy as $E=mc^2$ and its time evolution is described by the Schrödinger equation $i\hbar\partial_t \Psi = \hat{H}\Psi$. If $\Psi$ is an eigenstate of $\hat{H}$ then this is defined as $i\hbar\partial_t \Psi = E\Psi$ and so, $$\Psi\left( t\right) = e^{-\frac{i}{\hbar}Et}\Psi_0.$$ Now, if $E$ is a complex number, $E = E_R - iE_I$, $$\Psi\left( t\right) = e^{-\frac{i}{\hbar}E_Rt}e^{-\frac{1}{\hbar}E_It}\Psi_0,$$ giving the decay constant as $\frac{1}{\hbar}E_I = \frac{1}{\hbar}m_Ic^2$. Or in other words, the decay constant is related to the imaginary component of the total mass, which means that, so long as the total mass of the particle remains the same, the stability in another universe of that particle is unaffected.

*Stationary relative to you, anyway

Edit: Gauge invariance in the Standard Model

For the purposes of this section, there are 2 types of particle: bosons and fermions.

Bosons are 'force carriers' and so are the result of what can only be described as 'quantising a gauge field'. In simpler terms, if you 'quantise' the electromagnetic (EM) field, you get a photon (or a number of photons). For a photon, the EM field can be written as $A_{\mu}$ or $A^{\mu}$, with the 'mass term' being $-m^2A_{\mu}A^{\mu}$. However, real-life physics dictates that you can perform what's known as a 'gauge transformation', where you can change the value of $A_{\mu}$ without actually changing the value of the electric field, $\mathbf{E}$ (which is the measurable thing, where $A_{\mu}$ isn't directly measurable*). This means that any term that is an intrinsic $A_{\mu}A^{\mu}$ can change value without having any effect on the measured results. However, if you can change such a term around as much as you like, then the mass of the particle (also measurable), can be whatever you like and change between measurements without anything being done to it, which is simply impossible. So, as this is the electric field, charge cannot be conserved. This is due to Maxwell's equation $\nabla\cdot \mathbf{E} = \frac{\rho}{\epsilon_0}$ where $\epsilon_0$ is a constant and $\rho$ is charge. The other gauge bosons have similar terms, only with more complicated maths.

Now taking fermions, say an electron ($e^-$) and positron ($e^+$). The 'mass term' is $-me^+e^- = e^+_Le^-_R + e^+_Re^-_L$ (these are actually 'spinors' - the representation of spin-half particles - and the result is just how the maths says that these spinors are multiplied ('dotted') together - no-one really understands spin). However, the weak isospin (similar to a weak force equivalent to charge) of an electron is $-\frac{1}{2}$ and a positron is $0$. However, just like Kirchoff's current law, if charge is conserved, the sum of the current going in = the sum of the current going out. In the above electron/positron case, the current going in (equivalent to the weak isopin of the positron, $e^+$) doesn't equal the current going out (weak isospin of the electron), so charge isn't conserved. As with the boson case, the same is true for all fermions, so an intrinsic mass term is impossible without violating conservation of charge.

* for some maths, $A^{\mu} = \left(\frac{1}{c}\phi, \mathbf{A} \right)$ and $\mathbf{E} = -\nabla\phi - \partial_t\mathbf{A}$, so $A_{\mu}A^{\mu} = -\frac{1}{c^2}\phi^2 + \vert\mathbf{A}\vert^2$

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  • $\begingroup$ I don't follow your use of left-handed being matter and anti-right-handed being antimatter. The H matrix is only giving one chirality/helicity? $\endgroup$ – JDługosz Feb 24 '17 at 23:16
  • $\begingroup$ How does gauge symmetry prevent intrinsic mass from being present? $\endgroup$ – JDługosz Feb 24 '17 at 23:18
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    $\begingroup$ I think the real point is that rest mass doesn't exist ex nilo; it arises from some mechanism involving the elementary particles such as internal energy of bound systems or interaction with another field. So we might address this with another mechanism for giving mass, such as motion in another dimension. But I think my first sentence is a summary of the answer the OP is interested in. $\endgroup$ – JDługosz Feb 24 '17 at 23:21
  • $\begingroup$ So I'm summing over $f' = l_R$ (right-handed leptons) and $f = l_L$ (left-handed leptons), so $\bar{f'}$ is an anti-right-handed lepton. To be fair, this does mean that 'positrons' should be replaced with 'positrons, anti-muons and anti-taus' - ie. I'm summing over all 3 generations. There are other terms in the Yukawa coupling involving quarks that are similar (involving both chiralities of the quarks) that I've excluded for brevity as they don't add anything to the explanation of what's going on. This gives the matrices H, H' and H'', only 2 can be diagonalised. The other is the CKM matrix $\endgroup$ – Mithrandir24601 Feb 24 '17 at 23:27
  • $\begingroup$ @JDługosz 1st question is answered in the above comment, 2nd question answered by an edit in the answer. I agree with your 3rd comment. Hope that's somewhat helpful! $\endgroup$ – Mithrandir24601 Feb 25 '17 at 0:38
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As far as I understand, this question is unanswerable without more research into the state of our current universe, seeing as we are not sure what fundamental particles could exist, and if we figure out that there are some, surely our own universe would fit into this category

So the answer would be: Would the intrinsic/quantized mass in this hypothetic universe effect how stable an elementary particle would be? Almost certainly not. Stability of elemental particles has to occur for a stable universe, surely.

What else would need to change for particles to have quantized rest mass? Future scientific discovery finds that this is the case in our own universe.

The jury's out.

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