# Yang–Mills and Mass Gap

Discussion in 'Physics & Math' started by Thales, Nov 29, 2017.

1. ### arfa branecall me arfValued Senior Member

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Before putting on the mathematical gloves, perhaps some ordinary English discussion of the likes of isotopic spin (isobaric spin or just isospin).

From Wikipedia:
G. t'Hooft, in the SciAm article I linked earlier, states that isospin symmetry is continuous. This means for instance that proton/neutron isospin can describe a particle in a superposition of proton + neutron. So what about the electric charge of +1 in the case of such a superposition?

Or is it the case that, even though the symmetry is continuous, physically it doesn't exist?

3. ### arfa branecall me arfValued Senior Member

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Another question about the difference between a global and a local symmetry: some authors appear to distinguish between the two by saying only the latter can be a gauge symmetry, however this seems to be contradicted by t'Hooft in his article, and also here at physics.stackexchange by Rod Vance (reply #9).
t'Hooft says something like, by taking a global symmetry and making it local, something needs to be added to a gauge theory, which is a force. Is this otherwise known as symmetry-breaking?

One idea I have about breaking a symmetry is, in the case of a sphere the (global?) symmetry is that all points on the surface are equivalent (for instance, all points have the same fiber over them which is the set of all possible directions a vector at any point can have). If however, the sphere is rotated about an axis, two points become 'special' (the poles) and all other points rotate with the same angular velocity (sort-of "combing" all the direction vectors).

Of course, the sphere can rotate about more than one axis of symmetry, it can precess and so on, but then this additional rotation is relative to another principal axis.

Rotation breaks spherical symmetry because you can then treat the surface as a family of infinitesimal rings rotating about the same axis.

Last edited: Jan 27, 2018

5. ### hansdaValued Senior Member

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Our Earth is spinning with relative to its axis. What do you think is the frequency $f$ for this spin?

7. ### NotEinsteinValued Senior Member

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I believe it's exactly $0.5\:\text{double-sidereal days}^{-1}$, so less than 1.

8. ### hansdaValued Senior Member

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What is the definition of frequency, you are considering here?

9. ### NotEinsteinValued Senior Member

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Just the standard one: the number of occurrences of a repeating event per unit of time.

Edit: Although I think I got the units wrong. It's actually: $0.5\:\text{half-sidereal day}^{-1}$.

10. ### hansdaValued Senior Member

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So, can this repeating be less than one?

11. ### NotEinsteinValued Senior Member

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If your time units are less than a sidereal day, then obviously yes. Likewise, the frequency of the little hand of a clock going around is less than 1 if your time units are less than 1 hour. If something is rotating (at a fixed speed) once an hour, then during half an hour it will rotate halfway. 1/hr = 0.5/half-hr.

12. ### hansdaValued Senior Member

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So, what is the minimum frequency here?

13. ### NotEinsteinValued Senior Member

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Zero, if the object isn't rotating at all. If you want to exclude that, it's the smallest non-zero positive number you can imagine.

14. ### hansdaValued Senior Member

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Consider the equation $w=2\pi f$. What do you think is the minimum frequency here?

15. ### NotEinsteinValued Senior Member

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That's simple: because I based my response in post #70 on just the definition on frequency, it's the same answer: "Zero, if the object isn't rotating at all. If you want to exclude that, it's the smallest non-zero positive number you can imagine."

16. ### hansdaValued Senior Member

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If the frequency is less than one, the cycle is incomplete.

17. ### NotEinsteinValued Senior Member

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Yes, so? I don't see any problem with that?

18. ### arfa branecall me arfValued Senior Member

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A "Feynman" diagram I found on the net. Apparently M is the invariant mass of some exchange particle (hence the difference in energy ΔE is defined by two constants, but corresponds to a "rest energy" (??)). Or maybe it's the "mass gap" (?).

Anyhoo, the last equation gives the range of the interaction. If M = 0, then we have an algebraic problem, namely division by 0, or maybe something unphysical.On the other hand if the exchange particle has zero rest mass, that implies the range is infinite!

The above, btw, is from a public lecture by Kenneth Young on Yang-Mills and the Higgs boson. I'd post a link but the lecture itself isn't all that edifying, he spends about 5 seconds explaining this diagram and that's a bit of a trend.

19. ### arfa branecall me arfValued Senior Member

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But you haven't defined a cycle. You haven't specified what is a "complete" cycle, or whether a cycle is a smooth function, etc.

20. ### NotEinsteinValued Senior Member

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(Note: this is a guess.)
The first formula's LHS is the difference in energy between the incoming and outgoing particles on one side, i.e. the energy available for the red-line particle. It's expressed as a mass here.
The second formula is Heisenberg's uncertainty principle.
The third formula is simply the second one rewritten, with the first formula plugged in.
The fourth formula calculates the distance the particle can travel while complying to Heisenberg's uncertainty principle.

So if the particle is massless, its range is infinite. While I don't dare draw solid conclusions from this because there's all kinds of problems setting M to zero in this derivation, this outcome seems to make sense: photons (massless particles) are stable and will travel forever (if not stopped). They have infinite range, so R being infinite in their case kinda makes sense.

21. ### arfa branecall me arfValued Senior Member

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Isospin symmetry is a global symmetry. Intermediate states between protons and neutrons aren't seen in nature, locally, because of "spontaneous symmetry breaking", via the Higgs mechanism (!).

Since isospin is global, the orthogonal proton/neutron states transform everywhere in the same way (protons become neutrons, neutrons become protons and the strong coupling is invariant under a global transformation).

The Higgs field in effect provides a reference direction for the isospin state of neutrons/protons; the symmetry is local (a gauge symmetry), but hidden--see t'Hooft's article.

22. ### arfa branecall me arfValued Senior Member

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hansda was almost on to something, I have to admit (but what exactly I'm not too sure).

According to Kenneth Krane in Modern Physics 3rd ed., the uncertainty relation between frequency and time is readily derived from the concept of measuring a period T of some waveform. There is an uncertainty in the time measurement: Δt ≈ T. So there is an uncertainty in the period itself: ΔT. Assume this uncertainty is a small fraction of T, i.e. ΔT ~ εT.

Next, take the product: ΔtΔT ~ (T)εT. But f = 1/T and we can now take differentials: $df = -1/T^2 dT$, and "convert" the differentials into absolute differences, so the minus sign can be ignored (we are interested in the magnitudes of the uncertainties). So $Δf = 1/T^2 ΔT$ which yields after substitution: ΔfΔt ~ ε.

Hence, if Δt is large (the duration of a measurement of a period), the uncertainty in frequency is small.

That said, hansda is assuming that $E = mc^2 = hf$ holds. But this means there's a big problem with Einstein's 1905 paper on the photelectric effect, because it says photons have a rest energy and so a finite range. Thus Maxwell's equations are wrong!

Last edited: Jan 30, 2018
23. ### Q-reeusBannedValued Senior Member

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You might be better off trying somewhere like: https://www.quora.com/How-does-the-uncertainty-principle-relate-to-Fourier-transforms
Where on earth did you conclude that from?! Nonsense. Photoelectric effect is all about light absorption and release of electrons from a metal having a particular work function. It doesn't even prove the EM field is quantized as photons - that required more sophisticated approaches much later on.