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I was thinking of a universe in which the W boson has no rest mass. How would this universe be different from our universe?

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    $\begingroup$ I'm voting to close this question as off-topic because it belongs on physics.stackexchange.com $\endgroup$ – Aify Nov 30 '15 at 20:47
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    $\begingroup$ I don't think that this should be closed, in part because Physics would never take this. It's too speculative for there. $\endgroup$ – HDE 226868 Nov 30 '15 at 21:28
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    $\begingroup$ @o.m. I don't think it's too specific, nor do I think that specificity is a bad thing. $\endgroup$ – HDE 226868 Dec 1 '15 at 15:55
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    $\begingroup$ Specificity in a question is good. Being too broad is bad. $\endgroup$ – Euan M Dec 4 '15 at 0:41
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    $\begingroup$ I think it's an excellent question for this site. I've edited the tags a bit, though. My one problem is that "how would this universe be different" seems enormously open-ended. Can the original poster clarify and narrow? $\endgroup$ – CAgrippa Dec 7 '15 at 11:09
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I'm going to make an attempt to explain this from scratch, so here goes (WARNING: this might get complicated):

After symmetry breaking, the mass term for the weak field in the standard model Lagrangian becomes $\mathcal{L}_{W, mass} = -\frac{1}{2\lambda}\left(\mu g_2\right)^2W^{-\nu}W^+_{\nu} = m^2_WW^{-\nu}W^+_{\nu}$. For the Weak bosons to be massless requires $\mu g_2 = 0$. Therefore there are 2 ways to go about this:

  1. $\mu = 0$
  2. $g_2 = 0$

So what are these terms? $\mu = \frac{1}{\sqrt{2}}m_H$, the mass of the Higgs. $\lvert e\rvert = g_2\sin\theta_W$ is the (modulus of the ) charge of an electron. The mass of the Z boson is $m_Z = \frac{m_W}{\cos\theta_W}$

Considering the two cases separately:

1: Setting $\mu = 0$:

Both the Higgs and the Z boson also become massless. The Higgs mechanism no longer exists and electroweak symmetry breaking never occurs. The Higgs potential becomes $V = -\mu^2\phi^*\cdot\phi + \frac{1}{2}\lambda\left(\phi^*\cdot\phi\right)^2 = \frac{1}{2}\lambda\left(\phi^*\cdot\phi\right)^2$ which has a minimum point at $0$ (i.e. $\left\langle\phi_0\right\rangle_{min} = \left\langle\phi_1\right\rangle_{min} = v = 0$). Perturbing about this point with $\phi_0 = \frac{H}{\sqrt{2}}$ gives $V = \frac{\lambda H^4}{8}$

This also causes the mass of all elementary fermions to be $0$, or, the mass of all the elementary particles to be discovered so far is $0$. As a result of this, the mass of the neutron and the proton are equal. Not that it matters, because current understanding has it that the universe would consist of nothing but plasma as a result of constant pair creation and annihilation of, well, all the fundamental particles. That's assuming that the fundamental particles would even be the same.

2: Setting $g_2 = 0$:

The first result is that nothing has electric charge. The concept of an electric charge ceases to exist. The gauge transformation for the gauge bosons becomes is $D_{\nu} = \partial_{\nu} + \frac{i}{2}Ig_1B_{\nu}$ where $I$ is the identity matrix. This gives the single field, $B_{\nu}$. In this case, $v = \frac{\mu}{\sqrt{\lambda}}$ as usual and the mass term becomes $\mathcal{L}_{B, mass} = -\frac{1}{8}v^2g^2_1B_{\nu}B^{\nu} = -\frac{1}{2}m_B^2B_{\nu}B^{\nu}$, giving a mass of $m_B = \frac{vg_1}{2}$. However, the three other gauge fields don't just vanish - they still exist, but are massless. These fields are not the same fields as the photon and W boson fields. Nevertheless, we have 3 massless fields and another with mass. From your comment saying that you want the Z boson to have mass, this is therefore the condition closest to what you want.

This is surprisingly simple: the Higgs term in the Standard Model Lagrangian consists of 3 massless gauge fields and 1 massive one, as opposed to the relative mess that results from the electroweak symmetry breaking of our own universe. The symmetry breaking still occurs, so I'll assume that the fundamental particles are still the same, although the other way round - the SU(2) symmetry is unbroken and the U(1) symmetry is broken.

However, there is a difference: the coupling constant of the 3 massless fields, $g_2 = 0$ and so they just don't interact with anything.

Looking at the leptons: we have 3 generations of $l = \begin{pmatrix} \nu_e \\ e \end{pmatrix}$ and $\bar{e}$. For simplicity, we'll look at the first generation - electrons, as all the others follow the same principle.

The covariant derivatives are $D_{\nu}l = \partial_{\nu}l + \frac{i}{2}g_1B_{\nu}l$ and $D_{\nu}\bar{e} = \partial_{\nu}\bar{e} - ig_1B_{\nu}\bar{e}$, giving the 'charge' of the electron = 'charge' of the neutrino = $-\frac{1}{2}g_1$. The 'charge' of the positron is $g_1$. In the same way, the 'charge' of the quarks is $\frac{1}{6}g_1$ and the 'charge' of the anti-up and anti-down quark are $-\frac{2}{3}g_1$ and $\frac{1}{3}g_1$ respectively. $g_1$ is unknown.

In turn, this means that the charge of a proton = charge of a neutron = - charge of an electron. Neutrally charged particles are entirely possible (theoretically). However, there is now a major difference between anti-particles and particles... [Aside: I'm absolutely stunned] An anti-proton has charge $-g_1$ and an anti-neutron has charge 0. Anti-matter works in a completely different way to normal matter.

In terms of the electric force, we have an equivalent force, only it has a finite range. Very, very surprisingly, the anti-matter in this universe works similar to the anti-matter in our universe, only with a short range EM force and each anti-proton needs to be coupled to 2 electrons, or to a neutron and an electron, or to 2 neutrons (if such a thing is possible). The matter on the other hand, is very different - each proton and each neutrino has to be coupled to an electron to create a neutrally charged particle.

Apart from that, the universe should have expanded from the Big Bang in a way reasonably similar to our own - things equivalent to stars should still be able to form, although Neutron stars probably wouldn't exist. Supporting life on planets would be the tricky bit due to the range of radiation from these 'stars', although this range isn't defined in the model (i.e. in an alternate universe, the value could be obtained by experiment), so could be made different in an alternate universe (tweak things so that $m_B$ is small enough). One other difference is that we wouldn't be able to see other galaxies for the same reason.

Also, it's not inconceivable that something could be made to travel faster than this force.

To sum up:

You want the coupling constant $g_2 = 0$. This creates a (relatively) short-range equivalent to the EM force and changes the nature of composite particles (e.g. protons), although they can still exist. Neutron stars are impossible and other galaxies could not be observed.

Source: Basic ideas from Mark Srednicki - Quantum Field Theory

Edit: see also What would our knowledge of physics look like without astronomical observations?

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  • $\begingroup$ How can anti particles be other than T symmetry = CP reversal? How could pair production work in your “stunning” case? I have to suspect a mistake or something deeper is not being explained. $\endgroup$ – JDługosz Jul 28 '16 at 17:48
  • $\begingroup$ I'm confused re “no electric charge” yet 'charge' and electric force are discussed. $\endgroup$ – JDługosz Jul 28 '16 at 17:56
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    $\begingroup$ “CP reversal is electric charge and parity reversal.” all kinds of charge…a quark pair production will be a red quark and an antired antiquark, involving the strong force. Anything that’s concerved must sum to zero in the pair, and there is no way to choose a preferred distribution so the antiparticle must be symmetricly the equal and opposite. $\endgroup$ – JDługosz Jul 29 '16 at 1:32
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    $\begingroup$ @Mithrandir24601 +1 you forgot a 3rd option though: leaving all the parameters constant and switching off the SU(2) coupling of the Higgs. that should also cause the least severe change since most particle properties would be unaffected $\endgroup$ – Wolpertinger Jul 29 '16 at 8:11
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    $\begingroup$ @Numrok As for turning of the SU(2) coupling of the Higgs - if you do that, then you're turning off all SU(2) couplings of the Higgs, which includes Yukawa coupling and so electrons/positrons and quarks/anitquarks don't have mass and we're left with case 1 again. This is as a result of switching off the SU(2) coupling of the Higgs changing the underlying representation of the Higgs to $\left(1, 1, -\frac{1}{2}\right)$ and so creating a gauge invariant Yukawa-like term in the Lagrangian isn't possible any more (we no longer have a gauge-group singlet) $\endgroup$ – Mithrandir24601 Jul 29 '16 at 9:34
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If W was massless because of lack of a Higgs mechanism, then everything would be different.

But maybe W just happens to be massless, without affecting all particles. Well, there is still a relationship between W, Z, and photon and the electroweak unification. Turns out that the photon we know isn't the "original" rock-bottom thing, but a remixture of the 4 forms of the electroweak bosons. If W wasn't picked out due to its mass, that might not happen.

We'd have a different breakdown, or none at all, of electromagnetic and weak interactions. We'd have degrees of freedom in the original fields end up in different places.

You might read up on electro-weak spontaneous symmetry breaking. Relate that to the readers, and pick up the story where something happens differently in that narrative. That would be a good way to introduce it to the readers, too.

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  • $\begingroup$ Would cp violation happen if the W Boson was massless but it was just the W Boson? $\endgroup$ – Anders Gustafson Dec 2 '15 at 0:12
  • $\begingroup$ Quallify "just the W". You mean no Z and no photon? But in general CP violation does not require mass-giving: the real puzzle is, given that nature can distinguish left and right, why the strong force doesn't. $\endgroup$ – JDługosz Dec 2 '15 at 0:55
  • $\begingroup$ What I mean by "just the W" is that the W Boson would be massless but other particles would still have mass. $\endgroup$ – Anders Gustafson Dec 2 '15 at 1:08
  • $\begingroup$ I think my 4th paragraph is the way to proceed. $\endgroup$ – JDługosz Dec 2 '15 at 1:25

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